The behaviour of gases 2016 1 CHAPTER ONE MATTER 1.0 Introduction Long before the science of chemistry was established, materials were described as existing in one of three physical states. There are either rigid, solid objects, having a definite volume and a fixed shape, nonrigid liquids, having no fixed shape other than that of their containers but having definite volumes or gases, which have neither fixed shape nor fixed volume. The techniques used for handling various materials depend on their physical states as well as their chemical properties. While it is comparatively easy to handle liquids and solids, it is not as convenient to measure out a quantity of a gas. Fortunately, except under rather extreme conditions, all gases have similar physical properties, and the chemical identity of the substance does not influence those properties. For example, all gases expand when they are heated in a nonrigid container and contract when they are cooled or subjected to increased pressure. They readily diffuse through other gases. Any quantity of gas will occupy the entire volume of its container, regardless of the size of the container. 1.1. States of Matter Matter is anything that has mass and occupies space. All the material things in the universe are composed of matter, including anything we can touch as well as the planets in the solar system and all the stars in the sky. It is composed of tiny particles such as atoms, molecules, or ions and can exist in three physical states- solid, liquid and gas. Solid State In the solid state, the individual particles of a substance are in fixed positions with respect to each other because there is not enough thermal energy to overcome the intermolecular interactions between
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The behaviour of gases 2016
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CHAPTER ONE
MATTER
1.0 Introduction Long before the science of chemistry was established, materials
were described as existing in one of three physical states. There are either rigid, solid objects, having a definite volume and a fixed shape, nonrigid liquids, having no fixed shape other than that of their containers but having definite volumes or gases, which have neither fixed shape nor fixed volume.
The techniques used for handling various materials depend on their physical states as well as their chemical properties. While it is comparatively easy to handle liquids and solids, it is not as convenient to measure out a quantity of a gas. Fortunately, except under rather extreme conditions, all gases have similar physical properties, and the chemical identity of the substance does not influence those properties. For example, all gases expand when they are heated in a nonrigid container and contract when they are cooled or subjected to increased pressure. They readily diffuse through other gases. Any quantity of gas will occupy the entire volume of its container, regardless of the size of the container.
1.1. States of Matter
Matter is anything that has mass and occupies space. All the material things in the universe are composed of matter, including anything we can touch as well as the planets in the solar system and all the stars in the sky. It is composed of tiny particles such as atoms, molecules, or ions and can exist in three physical states- solid, liquid and gas. Solid State
In the solid state, the individual particles of a substance are in fixed positions with respect to each other because there is not enough thermal energy to overcome the intermolecular interactions between
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the particles. As a result, solids have a definite shape, volume and are incompressible. Most solids are hard, but some (like waxes) are relatively soft. Some large crystals look the way they do because of the regular arrangement of atoms (ions) in their crystal structure. Solids usually have their constituent particles arranged in a regular, three-dimensional array of alternating positive and negative ions called a crystal. Some solids, especially those composed of large molecules, cannot easily organize their particles in such regular crystals and exist as amorphous (literally, βwithout formβ) solids. Glass is one example of an amorphous solid.
Liquid State A liquid is a nearly incompressible fluid that conforms to the
shape of its container but retains a (nearly) constant volume independent of pressure. The volume is definite if the temperature and pressure are constant. The molecules have enough energy to move relative to each other and the structure is mobile. Gaseous State
Gases consist of tiny particles widely spaced (Figure 1.1). Under typical conditions, the average distance between gas particles is about ten times their diameter. Because of these large distances, the volume occupied by the particles themselves is very small compared to the volume of the empty space around them. For a gas at room temperature and pressure, the gas particles themselves occupy about 0.1% of the total volume. The other 99.9% of the total volume is empty space (whereas in liquids and solids, about 70% of the volume is occupied by particles). Because of the large distances between gas particles, the attractions or repulsions among them are weak.
The particles in a gas are in rapid and continuous motion. For example, the average velocity of nitrogen molecules, N2, at 20 Β°C is about 500 m/s. As the temperature of a gas increases, the particlesβ velocity increases. The average velocity of nitrogen molecules at 100 Β°C is about 575 m/s.
The particles in a gas are constantly colliding with the walls of the container and with each other. Because of these collisions, the gas particles are constantly changing their direction of motion and their velocity. In a typical situation, a gas particle moves a very short distance between collisions. For example, oxygen, O2, molecules at normal temperatures and pressures move an average of 10-7 m between collisions.
Fig.1.1. A Representation of the Solid, Liquid, and Gas States The various characteristics or properties of the states of matter discussed above are summarized in table 1.1 below.
Table 1.1. Characteristics of the Three States of Matter Characteristic Solid Liquid Gas
Shape Definite conforms to the shape of its container
1.2. Phase Transition Phase transition is a term used to describe a state of change of
matter from one state to another. The state or phase of a given set of matter can change depending on pressure and temperature conditions, transitioning to other phases as these conditions change to favour their existence; for example, solid transitions to liquid with an increase in temperature. Near absolute zero, a substance exists as a solid. As heat is added to this substance it melts into a liquid at its melting point, boils into a gas at its boiling point, and if heated high enough would enter a plasma state in which the electrons are so energized that they leave their parent atoms. 1.2.1. Melting point
This is the temperature at which the solid and liquid forms of a pure substance can exist at equilibrium. As heat is applied to a solid, its temperature will increase until the melting point is reached. More heat then will convert the solid into a liquid with no temperature change. When the entire solid has melted, additional heat will raise the temperature of the liquid. The melting temperature of crystalline solids is a characteristic figure and is used to identify pure compounds and elements. Most mixtures and amorphous solids melt over a range of temperatures.
The melting temperature of a solid is generally considered to be the same as the freezing point of the corresponding liquid; because a liquid may freeze in different crystal systems and because impurities lower the freezing point, however, the actual freezing point may not be the same as the melting point. Thus, for characterizing a substance, the melting point is preferred. A typical example is the change of solid ice to liquid water as shown below. H2O(s) β H2O(l) (melting, fusion) Ice, snow liquid water
1.2.2. Freezing point This is the temperature at which a liquid becomes a solid. As
with the melting point, increased pressure usually raises the freezing point. The freezing point is lower than the melting point in the case of mixtures and for certain organic compounds such as fats. As a mixture freezes, the solid that forms first usually has a composition different from that of the liquid, and formation of the solid changes the composition of the remaining liquid, usually in a way that steadily lowers the freezing point. This principle is used in purifying mixtures, successive melting and freezing gradually separating the components. The heat of fusion (heat that must be applied to melt a solid), must be removed from the liquid to freeze it. Some liquids can be supercooled i.e., cooled below the freezing point without solid crystals forming. Putting a seed crystal into a supercooled liquid triggers freezing, whereupon the release of the heat of fusion raises the temperature rapidly to the freezing point. Freezing of liquid water to ice is a common example. H2O(l) β H2O(s) (freezing) liquid water Ice 1.2.3. Condensation
This is change of a gas to either liquid or solid state, generally upon a surface that is cooler than the adjacent gas. The change of vapour to solid is sometimes called deposition. A substance condenses when the pressure exerted by its vapour exceeds the vapour pressure of the liquid or solid phase of the substance at the temperature of the surface where condensation occurs. Heat is released when a vapour condenses. Unless this heat is removed, the surface temperature will increase until it is equal to that of the surrounding vapour. In the atmosphere, however, there is an abundant supply of aerosols, which serve as nuclei, called condensation nuclei, on which water vapour may condense. Some are hygroscopic (moisture-attracting), and condensation begins on them when the relative humidity is less than
100 percent, but other nuclei require some supersaturation before condensation begins. Condensation accounts for the formation of dew (liquid water formed by condensation of water vapour from the atmosphere), and Frost (solid water formed by direct condensation of water vapour from the atmosphere without first forming liquid water). H2O(g) β H2O(l) (condensation) Water vapour dew
This refers to the conversion of a substance from the liquid or solid phase into the gaseous (vapour) phase. Heat must be supplied to a solid or liquid to effect vaporization. If the surroundings do not supply enough heat, it may come from the system itself as a reduction in temperature. The atoms or molecules of a liquid or solid are held together by cohesive forces, and these forces must be overcome in separating the atoms or molecules to form the vapour; the heat of vaporization is a direct measure of these cohesive forces. H2O(l) β H2O(g) (vaporization) Liquid water water vapour 1.2.5. Sublimation
The change of a solid directly to the vapour without its becoming liquid is specifically referred to as sublimation. Although the vapor pressure of many solids is quite low, some (usually molecular solids) have appreciable vapor pressure. Ice, for instance, has a vapour pressure of 4.7 mmHg at 0oC. For this reason, a pile of snow slowly disappears in winter even though the temperature is too low for it to melt. The snow is being changed directly to water vapour.
H2O(s) β H2O(g) (sublimation) Ice, snow Water vapour
Sublimation can be used to purify solids such as impure iodine that readily vaporize. Impure iodine is heated in a beaker so that it vaporizes, leaving nonvolatile impurities behind. The vapour crystallizes on the bottom surface of a dish containing ice that rests on top of the beaker. Freeze-drying of foods is a commercial application of sublimation. Brewed coffee, for example, is frozen and placed in a vacuum to remove water vapour. The ice continues to sublime until it is all gone, leaving freeze-dried coffee. Most freeze-dried foods are easily reconstituted by adding water. The following diagram summarizes these phase transitions.
Fig.1.2. Diagram showing the nomenclature for the different phase transitions.
1.3. Heat of Phase Transition Any change of state involves the addition or removal of energy
as heat to or from the substance. A simple experiment shows that this is the case. Suppose you add heat at a constant rate to a beaker containing ice at -20oC. In Figure 1.3 below, we have plotted the temperature of the different phases of water as heat is added. The
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temperature of the ice begins to rise from -20oC, as you would expect; the addition of heat normally raises the temperature of a substance. At 0oC, the ice begins to melt, so that you get a beaker of ice in water. Note the flat region in the curve, labeled ice and water. Why is this region flat? It means that heat is being added to the system without a change in temperature; the temperature remains at 0oC. This temperature, of course, is the melting point of ice. The heat being added is energy required to melt ice to water at the same temperature. The intermolecular forces binding water molecules to specific sites in the solid phase must be partially broken to allow water molecules the ability to slide over one another easily, as happens in the liquid state. Note the flat regions for each of the phase transitions. Because heat is being added at a constant rate, the length of each flat region is proportional to the heat of phase transition.
Fig. 1.3. Heating curve for water: Heat is being added at a constant rate to a system containing water. Note the flat regions of the curve. When heat is added during a phase transition, the temperature does not change.
The heat needed for the melting of a solid is called the heat of
fusion (or enthalpy of fusion) and is denoted βHfus. For ice, the heat of fusion is 6.01 kJ per mole. H2O(s) β H2O(l); βHfus = 6.01 kJ/mol
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The heat needed for the vaporization of a liquid is called the heat of vaporization (or enthalpy of vaporization) and is denoted βHvap. At 100oC, the heat of vaporization of water is 40.7 kJ per mole. H2O(l) β H2O(g); βHvap = 40.7 kJ/mol Note that much more heat is required for vaporization than for melting. Melting needs only enough energy for the molecules to escape from their sites in the solid. For vaporization, enough energy must be supplied to break most of the intermolecular attractions. A refrigerator relies on the cooling effect accompanying vaporization. The mechanism contains an enclosed gas that can be liquefied under pressure, such as ammonia or 1,1,1,2-tetrafluoroethane, CH2FCF3. As the liquid is allowed to evaporate, it absorbs heat and thus cools its surroundings (the interior space of the refrigerator). Gas from the evaporation is recycled to a compressor and then to a condenser, where it is liquefied again. Heat leaves the condenser, going into the surrounding air. 1.4. Pressure of Gases
The molecules of a gas, being in continuous motion, frequently strike the inner walls of their container. As they do so, they immediately bounce off without loss of kinetic energy, but the reversal of direction (acceleration) imparts a force to the container walls. This force, divided by the total surface area on which it acts, is the pressure of the gas.
The pressure of a gas is observed by measuring the pressure that must be applied externally in order to keep the gas from expanding or contracting. To visualize this, imagine some gas trapped in a cylinder having one end enclosed by a freely moving piston. In order to keep the gas in the container, a certain amount of weight (more precisely, a force, f) must be placed on the piston so as to exactly balance the force exerted by the gas on the bottom of the piston, and tending to push it up. The pressure of the gas (P) is simply the quotient f/A, where A is the cross-section area of the piston.
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Example 1.1. If a force of 16N is pressed against an area of 2.44 m2, what is the pressure in pascals?
Solution Given force, F = 16N, area, A = 2.44 m2
Apply the relationship,
π = πΉ
π΄
π = 16π
2.44m2
= 6.57ππβ2
1.4.1. Pressure Units The unit of pressure in the SI system is the pascal (Pa), defined
as a force of one newton per square metre (1 Nmβ2 = 1 kg mβ1 sβ2 ). In chemistry, it is more common to express pressures in units of atmospheres or torr:
1 atm = 101325 Pa = 760 torr.
The older unit millimetre of mercury (mm Hg) is almost the same as the torr; it is defined as one mm of level difference in a mercury barometer at 0Β°C. In meteorology, the pressure unit most commonly used is the bar:
1 bar = 106 N mβ2 = 0.987 atm. For conversion purposes,
1 atm = 760 torr =760 mmHg = 1.01325 Γ 105 Nm-2
Example 1.2. How many atmospheres are in 1547mmHg
Solution Use the conversion factor; 1 ππ‘π = 760 πππ»π β΄ π₯ ππ‘π = 1547 πππ»π Cross multiplying and making π₯ the subject gives
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π₯ = 1 ππ‘π Γ1547 πππ»π
760 πππ»π
π₯ = 2.04 ππ‘π
Example 1.3. Write the conversion factor to determine how many mmHg are in 9.65 atm.
Solution Use the same conversion factor as in example 1.2 above 1 ππ‘π = 760 πππ»π β΄ 9.65 ππ‘π = π₯ πππ»π Cross multiplying and making π₯ the subject give
π₯ = 9.65 ππ‘π Γ760 πππ»π
1 ππ‘π
π₯ = 7334 πππ»π
Example 1.4. How many torr are in 1.56 atm
Solution
Use the conversion factor;
1 ππ‘π = 760 π‘πππ
β΄ 1.56 ππ‘π = π₯ π‘πππ
Cross multiplying and making π₯ the subject give
π₯ = 1.56ππ‘π Γ760 π‘πππ
1 ππ‘π
π₯ = 1190 π‘πππ
Example 1.5. Blood pressures are expressed in mmHg. What would be the blood pressure in atm if a patientβs systolic and diastolic blood pressures are 120 mmHg and 82 mmHg respectively? (In medicine,
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such a blood pressure would be reported as β120/82β, spoken as βone hundred twenty over eighty-twoβ ).
Solution Use the same conversion factor as in example one above 1 ππ‘π = 760 πππ»π β΄ 9.65 ππ‘π = π₯ πππ»π
Cross multiplying and making π₯ the subject give
Use the same conversion factor as in example one above
1.4.2. Atmospheric Pressure This is defined as the force per unit area exerted against a surface by the weight of the air above that surface. In most circumstances atmospheric pressure is closely approximated by the hydrostatic pressure caused by the weight of air above the measurement point. On a given plane, low-pressure areas have less atmospheric mass above their location, whereas high-pressure areas have more atmospheric mass above their location. Likewise, as elevation (altitude) increases, there is less overlying atmospheric mass, so that atmospheric pressure decreases with increasing elevation.
1.4.3. Measurement of Gas Pressure A barometer is piece of lab equipment specifically designed to
measure the atmospheric pressure. Invented in the early 17th century by the Italian EVANGELISTA TORRICELLI. The barometer consists of a vertical glass tube closed at the top and evacuated, and open at the bottom, where it is immersed in a dish of a liquid. The atmospheric pressure acting on this liquid will force it up into the evacuated tube until the weight of the liquid column exactly balances the atmospheric pressure. If the liquid is mercury, the height supported will be about 760 cm; this height corresponds to standard atmospheric pressure.
Fig. 1.4. A simple barometer
The formula for this pressure in the atmosphere is derived as shown below: πππππ = ππππ Γ ππππππππππππ
or π = ππ or mg
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Earth's acceleration of objects is based on its gravitational field and equals approximately 9.80665 m s-2. Additionally, since pressure is the force per the unit area being measured, then
π· = π Γ π Γ π Where d = density, g = gravity and h = height of the liquid or gas. Example 1.6. Mercury has a density of 13.6 g/cm3 and water has a density of 1.00 g/cm3. If a column of mercury has a height of 755 mm, how high would a corresponding column of water be in feet? Solution: Let us begin by setting the pressures equal:
Pmercury = Pwater Since π· = π Γ π Γ π We can write:
= 10268 ππ = 33.7 ππ‘ 1.4.4. The Manometer
A modification of the barometer, the U-tube manometer, provides a simple device for measuring the pressure of any gas in a container. There are a variety of manometer designs. A simple,
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common design is to seal a length of glass tubing and bend the glass tube into a U-shape. The glass tube is then filled with a liquid, typically mercury, so that all trapped air is removed from the sealed end of the tube. The glass tube is then positioned with the curved region at the bottom. The mercury settles to the bottom.
After the mercury settles to the bottom of the manometer, a vacuum is produced in the sealed tube. The open tube is connected to the system whose pressure is being measured. In the sealed tube, there is no gas to exert a force on the mercury (except for some mercury vapor). In the tube connected to the system, the gas in the system exerts a force on the mercury. The net result is that the column of mercury in the sealed tube is higher than that in the unsealed tube. The difference in the heights of the columns of mercury is a measure of the pressure of gas in the system.
In the open-tube manometer, the pressure of the gas is given by h (the difference in mercury levels) in units of torr or mmHg. Atmospheric pressure pushes on the mercury from one direction, and the gas in the container pushes from the other direction. In a manometer, since the gas in the bulb is pushing more than the atmospheric pressure, you add the atmospheric pressure to the height difference:
Pgas > Patm
Gas pressure = atmospheric pressure + h (height of the mercury) Pgas < Patm Gas pressure = atmospheric pressure - h (height of the mercury) The closed-tube manometer look similar to regular manometers except that the end that is open to the atmospheric pressure in a regular manometer is sealed and contains a vacuum. In these systems, the difference in mercury levels (in mmHg) is equal to the pressure in torr.
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Fig.1.5. The Manometer
Example 1.7. Find the pressures using the manometer set up below.
Solution
since Pgas > Patm
Pgas= Patm + h
Pgas= (755 + 24 )mmHg
=779mmHg
since Pgas < Patm Pgas= Patm β h
Pgas= (763 β35)g
Pgas= 728 mmHg
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Example 1.8. Suppose you want to construct a closed-end manometer to measure gas pressures in the range 0.000β0.200 atm. Because of the toxicity of mercury, you decide to use water rather than mercury. How tall a column of water do you need? (The density of water is 1.00 g/cm3; the density of mercury is 13.6 g/cm3). Solution Given: pressure range and densities of water and mercury, column height unknown. Strategy: Step 1. Calculate the height of a column of mercury corresponding to 0.200 atm in millimeters of mercury. This is the height needed for a mercury-filled column. Step 2. From the given densities, use a proportion to compute the height needed for a water-filled column. In millimeters of mercury, a gas pressure of 0.200 atm 1atm = 760mmHg
β΄ 0.200 atm will be 0.200 ππ‘π Γ760πππ»π
1 ππ‘π
= 152 πππ»π Using a mercury manometer, you would need a mercury column of at least 152 mm high.
Because water is less dense than mercury, you need a taller column of water to achieve the same pressure as a given column of mercury. The height needed for a water-filled column corresponding to a pressure of 0.200 atm is proportional to the ratio of the density of mercury to the density of water; Using π· = π Γ π Γ π Where d = density, g = gravity and h = height of the liquid or gas. Let us begin by setting the pressures equal:
Pmercury = Pwater We can then write: π Γ ππ»π Γ ππ»π = π Γ ππ€ππ‘ππ Γ ππ€ππ‘ππ
= 2070 ππ Comment: it takes a taller column of a less dense liquid to achieve the same pressure. 1.4.5. Effect of Pressure on the volume of gases
For a gas whose volume is not fixed, increasing the pressure will cause the gas to contract (reducing the volume), and decreasing the pressure will cause the gas to expand (increasing the volume). If the volume is fixed, then increasing the pressure will increase the temperature, and decreasing the pressure will decrease the temperature. 1.4.6. Simple Pressure Related Applications
β’ Drinking straw: A drinking straw is used by creating a
suction with your mouth. Actually this causes a decrease in air pressure on the inside of the straw. Since the atmospheric pressure is greater on the outside of the straw, liquid is forced into and up the straw.
β’ Siphon: With a siphon water can be made to flow "uphill". A siphon can be started by filling the tube with water (perhaps by suction). Once started, atmospheric pressure upon the surface of the upper container forces water up the short tube to replace water flowing out of the long tube. 1.5. Density of a Gas
This is defined as mass divided by the volume of a gas
= 1.428π/πΏ Example 1.10. A 0.0125g sample of a gas with an empirical formula of CHF2 is placed in a 165-mL flask. It has a pressure of 13.7 mm Hg at 22.5 Β°C. What is the molecular formula of the compound? [R= 0.8206L atm mol-1K-1]
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Solution Collect the available data and convert as necessary to agree with the given unit of R then find the value of density from which the molecular weight of the
gas can be determined using the relation: π = π Γ π
π π
Mass of gas sample = 0.0125g Volume = 165 mL = 0.156 L Temperature, T = 22.5Β°C = 295.7K
Pressure, P = 13.7 mm Hg = 1 atm Γ 13.7 πππ»π
M = 102ππππβ1 The molecular formula is (CHF2)2 or C2H2F4. Example 1.11. If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC, what is the pressure of O2 & H2O?[0.8206 πΏ ππ‘π πππβ1 πΎβ1 , H =1, 0 = 16, ] Solution Step 1: Write the balanced chemical reaction. Step 2: Calculate the moles of each product. Step 3: Find the pressure of each via PV = nRT Equation of reaction : 2H2O2(l) β 2H2O (g) + O2 (g)
From the equation of reaction, 2 mol of 2H2O2 produce 2 mol of H2O and a mol of O2.
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Therefore mol of H2O2 = πππ π
πππππ πππ π
= 0.11π
34π/πππ
=0.0032 mol
mol of O2 = 1
2 Γ 0.0032 mol of H2O2
=0.0016 mol mol of H2O = 1 Γ 0.0032 mol of H2O2
=0.0032 mol Using PV = nRT to calculate the pressure of the gases
= 0.032 atm Example 1.12. A chemist has synthesized a greenish-yellow gaseous compound of chlorine and oxygen and finds that its density is 8.14 g/L at 47Β°C and 3.15 atm. Calculate the molar mass of the compound and determine its molecular formula. Solution We can calculate the molar mass of a gas if we know its density, temperature, and pressure. The molecular formula of the compound must be consistent with its molar mass. What temperature unit should we use? Data provided density = 8.14 g/L T = 47Β°C = 320 K P = 3.15 atm
= 67.9 ππππβ1 We can determine the molecular formula of the compound by trial and error, using only the knowledge of the molar masses of chlorine (35.45 g) and oxygen (16.00 g). We know that a compound containing one Cl atom and one O atom would have a molar mass of 51.45 g, which is too low, while the molar mass of a compound made up of two Cl atoms and one O atom is 86.90 g, which is too high. Thus, the compound must contain one Cl atom and two O atoms and have the formula ClO2 , which has a molar mass of 67.45 g. Example 1.13. The density of a gaseous organic compound is 3.38 g/L at 40Β°C and 1.97 atm. What is its molar mass? Solution Data provided d = 3.38 g/L T = 40Β°C = 313 K P = 1.97 atm
= 44.0 ππππβ1 1.5.1. The effects of temperature on density
The density of a gas depends quite strongly on its temperature, so hot air has a smaller density than does cold air; colder air is more dense than hot air. From everyday experience, we know that something is dense if it tries to drop, which is why a stone drops to the bottom of a pond and a coin sinks to the bottom of a pan of water. This relative motion occurs because both the stone and the coin have higher densities than does water, so they drop. Similarly, we are more dense than air and will drop if we fall off a roof. Just like the coin in water, cold air sinks because it is denser than warmer air. We sometimes see
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this situation stated as warm air βdisplacesβ the cold air, which subsequently takes its place. Alternatively, we say βwarm air risesβ, which explains why we place our clothes above a radiator to dry them, rather than below it.
Light entering the room above the radiator passes through these pockets of warm air as they rise through colder air, and therefore passes through regions of different density. The rays of light bend in transit as they pass from region to region, much in the same way as light twists when it passes through a glass of water. We say the light is refracted. The eye responds to light, and interprets these refractions and twists as different intensities.
So we see swirling eddy (or βconvectiveβ) patterns above a radiator because the density of air is a function of temperature. If all the air had the same temperature, then no such difference in density would exist, and hence we would see no refraction and no eddy currents β which is the case in the summer when the radiator is switched off. Then again, we can sometimes see a βheat hazeβ above a hot road, which is caused by exactly the same phenomenon. 1.6. Temperature ` This is the numerical measure of the degree of hotness or coldness of a body. It is an important property of any gas. If two bodies are at different temperatures, heat will flow from the warmer to the cooler one until their temperatures are the same. This is the principle on which thermometry is based; the temperature of an object is measured indirectly by placing a calibrated device known as a thermometer in contact with it. When thermal equilibrium is obtained, the temperature of the thermometer is the same as the temperature of the object. 1.6.1. Temperature Scale
A thermometer makes use of some temperature-dependent quantity, such as the density of a liquid, to allow the temperature to be found indirectly through some easily measured quantity such as the
length of a mercury column. The resulting scale of temperature is entirely arbitrary; it is defined by locating its zero point, and the size of the degree unit.
Celsius temperature scale locates the zero point at the freezing temperature of water; the Celsius degree (C Β°) is defined as 1/100 of the difference between the freezing and boiling temperatures of water at 1 atm pressure. The older Fahrenheit scale placed the zero point at the coldest temperature it was possible to obtain at the time (by mixing salt and ice.) The 100Β° point was set with body temperature (later found to be 98.6Β°F.) On this scale, water freezes at 32Β°F and boils at 212Β°F. The Fahrenheit scale is a finer one than the Celsius scale; there are 180 Fahrenheit degrees in the same temperature interval that contains 100 Celsius degrees, so 1FΒ° = 9/5 C . Since the zero points are also different by 32F, conversion between temperatures expressed on the two scales requires the addition or subtraction of this offset, as well as multiplication by the ratio of the degree size. These selections allow us to write the following relations.
t(oF) = 9
5 t(oC) + 32
t(oC) = 9
5 t(oF) β 32
Where
t(oF) is the temperature in degree Fahrenheit and
t(oC) is the temperature in degree Celsius.
1.6.2. Absolute temperature In 1787 the French mathematician and physicist JACQUES
CHARLES discovered that for each Celsius degree that the temperature of a gas is lowered, the volume of the gas will diminish by 1/273 of its volume at 0Β°C. The obvious implication of this is that if the temperature could be reduced to β273Β°C, the volume of the gas would contract to zero. Of course, all real gases condense to liquids before this
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happens, but at sufficiently low pressures their volumes are linear functions of the temperature (Charles' Law), and extrapolation of a plot of volume as a function of temperature predicts zero volume at -273Β°C. This temperature, known as absolute zero, corresponds to the total absence of thermal energy.
Because the Kelvin scale is based on an absolute, rather than on an arbitrary zero of temperature, it plays a special significance in scientific calculations; most fundamental physical relations involving temperature are expressed mathematically in terms of absolute temperature. The diagram below compares the different temperature scales with respect to boiling and freezing point of water.
Fig. 1.6. Comparison of Temperature Scales (Schematic) 1.6.3. Conversion between Celsius and Kelvin Scale
In order to covert temperature in degree Celsius to temperature in Kelvin, the expression below is used. toC = (273 + t )K = T (K)
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Where t is the temperature on the Celsius scale, T is the temperature on the Kelvin scale. Example 1.14. Covert the following temperatures to Kelvin scale:
a. 27oC; b. -10oC. Solutions
a. Using the relationship
toC = (273 + t )K = T (K)
27oC = (273 + 27)K = 300K
b. toC = (273 + t )K = T (K)
-10oC = (273- 10)K = 263K
In order to convert absolute temperature T K to degree Celsius, 273 is simply subtracted from the value. Example 1.15. Covert the following temperatures to degree Celsius:
a. 298K b. 25K
Solutions a. Using the relationship
toC = (273 + t )K = T (K)
toC = (298 β 273) oC = 25oC
b. Using the relationship
toC = (273 + t )K = T (K)
toC = (25 β 273) oC = β 248 oC
1.7. The Volume of Gas The volume of a gas is simply the space in which the molecules
of the gas are free to move. If we have a mixture of gases, such as air, the various gases will coexist within the same volume. In these
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respects, gases are very different from liquids and solids, the two condensed states of matter. The volume of a gas can be measured by trapping it above mercury in a calibrated tube known as a gas burette (fig. 1.7). The SI unit of volume is the cubic meter, but in chemistry the liter and the milliliter (mL) are commonly used.
Fig. 1.7. Gas burette
It is important to bear in mind, however, that the volume of a gas varies with both the temperature and the pressure, so reporting the volume alone is not very useful. A common practice is to measure the volume of the gas under the ambient temperature and atmospheric pressure, and then to correct the observed volume to what it would be at standard atmospheric pressure and some fixed temperature, usually 0Β° C or 25Β°C. The table below shows some commonly used volume measurement units and their conversion factor.
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1.8. Effect of Temperature on the volume of gases If the volume of the container is not fixed, increasing the
temperature will cause a gas to expand (increase the volume), and contract when cooled (decreasing the volume). This would be the case for a gas inside a piston, or inside a rubber balloon. If the volume is fixed, then increasing the temperature will increase the pressure, and decreasing the temperature will decrease the pressure. This would be the case for a gas in a closed solid container, like a canister or sealed metal box. Why does thunder accompany lightning?
Lightning is one of the most impressive and yet frightening manifestations of nature. It reminds us just how powerful nature can be. Lightning is quite a simple phenomenon. Just before a storm breaks, perhaps following a period of hot, fine weather, we often note how the air feels βtenseβ. In fact, we are expressing an experiential truth: the air contains a great number of ions β charged particles. The existence of a large charge on the Earth is mirrored by a large charge in the upper atmosphere. The only difference between these two charges is that the Earth bears a positive charge and the atmosphere bears a negative charge.
Accumulation of a charge difference between the Earth and the upper atmosphere cannot proceed indefinitely. The charges must eventually equalize somehow: in practice, negative charge in the upper atmosphere passes through the air to neutralize the positive charge on the Earth. The way we see this charge conducted between the Earth and the sky is lightning: in effect, air is ionized to make it a conductor, allowing electrons in the clouds and upper atmosphere to conduct through the air to the Earthβs surface. This movement of electrical charge is a current, which we see as lightning. Incidentally, ionized air emits light, which explains why we see lightning. Lightning comprises a massive amount of energy, so the local air through which it conducts tends to heat up to as much as a few thousand degrees centigrade. And we have already seen how air expands when warmed, e.g. as described
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mathematically by Charlesβs law. In fact, the air through which the lightning passes increases in volume to an almost unbelievable extent because of its rise in temperature. And the expansion is very rapid. 1.9. Standard Temperature and Pressure, s.t.p.
Suppose two scientists work on the same research project, but one resides in the far north of the Arctic Circle and the other lives near the equator. Even if everything else is the same β such as the air pressure, the source of the chemicals and the manufacturers of the equipment β the difference between the temperatures in the two laboratories will cause their results to differ widely. For example, the βroom energyβ RT will differ. One scientist will not be able to repeat the experiments of the other, which is always bad science.
An experiment should always be performed at known temperature. Furthermore, the temperature should be constant throughout the course of the experiment, and should be noted in the laboratory notebook. But to enable complete consistency, sets of universally accepted arbitrary standards were devised and are called a set of standard conditions. βStandard pressureβ was set as 1 atm and βStandard temperatureβ has the value of 0oC (273 K). If both the pressure and the temperature are maintained at these standard conditions, then we say the measurement was performed at βstandard temperature and pressureβ, which is universally abbreviated to βs.t.p.β If the scientists at the equator and the Arctic Circle perform their work in thermostatically controlled rooms, both at s.t.p., then the results of their experiments will be identical. If we know the volume of a sample of a gas at any condition, we can easily calculate the volume it would have as an ideal gas at STP by employing the combined gas law. 1.10. Molar volume of a gas
The volume occupied by one mole of a gas under any conditions of temperature and pressure is called the molar volume, Vm. The molar volume of an ideal gas depends on the conditions of temperature and pressure; at s.t.p. it is 22.4 L (or 22400 cm3).
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How did we arrive at this value? It is simply the volume of 1.00 mol of gas at STP At s.t.p, pressure (P) = 1atm, temperature (T) = 27K, for one mole of gas, n = 1, R= 0.0821 L atm mol-1K-1 Using ideal gas equation to calculate the volume
The molar volumes of all gases are the same when measured at the same temperature and pressure. But the molar masses of different gases will vary. This means that different gases will have different densities (different masses per unit volume). If we know the molecular weight of a gas, we can calculate its density.
More importantly, if we can measure the density of an unknown gas, we have a convenient means of estimating its molecular weight. This is one of many important examples of how a macroscopic measurement (one made on bulk matter) can yield microscopic information (that is, about molecular-scale objects).
Determination of the molecular weight of a gas from its density is known as the Dumas method, after the French chemist JEAN DUMAS (1800-1840) who developed it. One simply measures the weight of a known volume of gas and converts this volume to its STP equivalent, using Boyle's and Charles' laws. The weight of the gas divided by its STP volume yields the density of the gas, and the density multiplied by 22.4Lmolβ1 gives the molecular weight. Pay careful attention to the examples of gas density calculations shown below.
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Example 1.16. Calculate the approximate molar mass of a gas whose measured density is 3.33 g/L at 30oC and 780 torr. Solution. Data provided
Molar mass?
Density = 3.33 g/L
Volume = 1L
Temperature,T = 30oC = (30 +273)K Pressure, P = 780 torr = (780/760) atm From the ideal gas equation, the number of moles contained in one litre of the gas is
Therefore, π = ππ£ But mass (m) = number of mole (n) Γ molar mass (M)
Therefore ππ£ = ππ
M= π Γπ£
π
Substituting gives
M = 33π πΏβ1 Γ1.0 πΏ
0.0413 πππ
= 80.6gmol-1
Example 1.17. The density of air at 15OC and 1.00 atm is 1.23g/L. What is the molar mass of the air?
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Solution First calculate the mole of air from which the molar mass can be gotten. Data provided Density = 1.23 g/L Volume = 1L Temperature,T = 15oC = (15 +273)K = 288K Pressure, P = 1 atm Molar mass? From the ideal gas equation, the number of moles contained in one litre of the air is
2.1. Introduction Experience has shown that several properties of a gas can be
related to each other under certain conditions. The properties are pressure (P), volume (V), temperature (T, in kelvins), and amount of material expressed in moles (n). What we find is that a sample of gas cannot have any random values for these properties. Instead, only certain values, dictated by some simple mathematical relationships, will occur. These properties and other variables such as rate of diffusion of any gaseous substance bear a simple mathematical relationship to each other. These are collectively called gas laws. 2.2. Pressure β Volume Relationship
Robert Boyle (1627β1691), an Irish physical scientist, discovered that the volume of a given sample of a gas at a constant temperature is inversely proportional to its pressure. This generalization, known as Boyleβs law, applies approximately to any gas, no matter what its composition. (It does not apply to liquids or solids.)
Inverse proportionality occurs when one variable gets larger by the same factor as another gets smaller. For example, average speed and the time required to travel a certain distance are inversely proportional. If we double our speed, the time it takes us to complete the trip is halved. Similarly, if the pressure on a given sample of gas at a given temperature is doubled (increased by a factor of 2), its volume is halved (decreased by a factor of 2).
Boyle might have observed the following data on volume and pressure for a given sample of gas at a given temperature, under four different sets of conditions:
4 0.500 8.00 Note that tabulating data is very helpful when two or more variables are being considered. The units are usually included in the column headings in such a table. The data in the table show that the product of the volume (V) and the pressure (P) is a constant. The table may be expanded to show this relationship:
PV = K (Where K = constant of proportionality). A more useful form of the law can be written as: P1V1= P2V2
Where V1 and P1 refer to the original volume and pressure, V2 and P2 refer to the volume and pressure under the new or changed conditions.
If we place the values of P on the horizontal axis and the values of V on the vertical axis, plot the preceding tabulated values for P and V, and smoothly connect the points, we get a curve that can tell us what the volume will be at any intermediate pressure (Figure 2.1a). We can also plot 1/V versus P and get a straight line through the origin (Figure 2.1b).
Fig. 2.1. Graphical illustration of Boyleβs law: (a) Plot of V versus P. (b) Plot of 1/V versus P. 2.3. Kinetic Theory and Boyleβs Law
The pressure of gas is due to continuous collision of the gaseous molecules with the walls of the container. At constant temperature, the average kinetic energy of the gas molecules is constant. If the size of the container is reduced to a half (volume reduces), the frequency of collision of the gas molecules with the walls of the container will be doubled. This is due to the fact that the distance to the walls has been reduced to a half. Therefore, the gas pressures will double the initial value.
On the other hand if the volume of the container (size) is doubled, the frequency of collision of the gas molecules with the walls of the container will become reduced by a half, since the distance between the molecules before colliding with the walls has been doubled. Hence the pressure will be half of the initial value. Example 2.1. A certain mass of a gas occupies 400cm3 at 1.0 Γ 105 Nm-2. Calculate its volume when the pressure is 4.0 Γ 105 Nm-2 at constant temperature.
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Solution In trying to solve this kind of problem, it is always good to collect the given information together so as to easily detect the variable you are asked to find. Data provided; P1 = 1.0 Γ 105 Nm-2, V1 =400cm3, P2 = 2.0 Γ 105 Nm-2, V2 = ? According to boyleβs law, P1V1= P2V2
Making V2 the subject,
V2 = P1V1
P2
On substituting, V2 = 1 Γ 105 Γ400
2 Γ 105 = 200cm3
Example 2.2. If 4 Liters of methane gas has a pressure of 1.0 atm, what will be the pressure of the gas if we squish it down so it has a volume of 2.5 L? Solution Data provided; P1 = 1.0 atm V1 = 4.0L P2 = ? V2 = 2.5 L According to boyleβs law, P1V1= P2V2
Making P2 the subject,
P2 = p1v1
v2
P2 = 1.0 Γ 4
2.5 = 1.6 ππ‘π
Example 2.3. A 3.50-L sample of gas has a pressure of 0.750 atm. Calculate the volume after its pressure is increased to 1.50 atm at constant temperature. Solution Alternatively, data collection can be in the form of table as shown below
Pressure Volume 1 0.750 atm 3.50 L
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2 1.50 atm ? Using P1V1= P2V2
Making V2 the subject,
V2 = p1v1
P2
Substitution of the values into the equation yields
V2 = 0.750 Γ3.50
1.50= 1.75 πΏ
Note that multiplying the pressure by 2 causes the volume to be reduced to half. Example 2.4. A sample of gas initially occupies 35.0 mL at 1.50 atm. Calculate the pressure required to reduce its volume to 20.5 mL at constant temperature. Solution Data collection
Pressure Volume 1 1.50 atm 35.0 mL 2 ? 20.5 mL Using P1V1= P2V2
Making P2 the subject,
P2 = p1v1
V2
Substitution of the values into the equation yields
P2 = 1.50 Γ35.0
20.5= 2.56 ππ‘π
Note that the units of pressure and volume must be the same on each side of the equation P1V1= P2V2 . If the units given in a problem are not the same, one or more of the units must be converted. Example 2.5. A 1.45-L sample of gas has a pressure of 0.950 atm. Calculate the volume after its pressure is increased to 787 torr at constant temperature. Solution Because the pressures are given in two different units, one of them must be changed.
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Pressure Volume 1 0.950 atm 1.45 L
2 787 π‘πππ 1 ππ‘π
760 π‘πππ = 1.036 ππ‘π ?
Using P1V1= P2V2
Making V2 the subject,
V2 = p1v1
P2
Substitution of the values into the equation yields
V2 = 0.950 Γ1.45
1.036
= 1.33 πΏ Alternatively, we can change 0.950 atm to torr and still arrive at the same answer. (722 torr) (1.45 L) = (787 torr)V2
V2 = 1.33 πΏ Note: 1 atm = 760 torr Example 2.6. Calculate the initial volume of a sample of gas at 1.20 atm if its volume is changed to 70.4 mL as its pressure is changed to 744 torr at constant temperature Solution Data collection
Pressure Volume 1 1.20 atm ?
2 744 π‘πππ 1 ππ‘π
760 π‘πππ = 0.979 ππ‘π 70.4 L
Using P1V1= P2V2
Making V1 the subject,
V1 = p2v2
P1
Substitution of the values into the equation yields
V1 = 0.979 Γ70.4
1.2
= 57.4 πΏ
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Example 2.7. Calculate the pressure required to change a 3.38-L sample of gas initially at 1.15 atm to 925 mL, at constant temperature. Solution Collect the data and convert 925 mL to L (mL β‘ cm3, 1000mL = 1L)
Pressure Volume 1 1.15 atm 3.38 L 2 ? 925 mL = 0.925 L Using P1V1= P2V2
Making P2 the subject,
P2 = p1v1
V2
Substitution of the values into the equation yields
P2 = 1.15 Γ3.38
0.925
= 4.20 ππ‘π The pressure must be raised to 4.20 atm. Practice questions
1. State Boyleβs law (i) in words (ii) mathematically 2. Explain Boyleβs law in terms of kinetic theory. 3. Fill the following gaps: (Measurements at constant
temperatures).
Initial pressure Initial volume Final pressure Final volume
1.0 Γ 105 Nm-2 300cm3 1.5 Γ 105 Nm-2 -
1.0 Γ 105 Nm-2 225cm3 - 900cm3
- 3.50dm3 760 mmHg 700 cm3
800 mm Hg 300cm3 650 mmHg -
4. 30dm3 of oxygen at 10 atmospheres is placed in a 20dm3
container. Calculate the new pressure if temperature is kept constant.
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5. Calculate the initial pressure of a 485-mL sample of gas that has been changed at constant temperature to 1.16 L and 1.18 atm.
2.4. Temperature β Volume Relationship
In 1787, 125 years after Boyle published the law that bears his name, J. A. C. Charles (1746β1823) discovered a law relating the volume of a given sample of gas to its absolute temperature. It took more than a century to discover this law because of the requirement that the temperature be absolute.
The volume of a sample of gas varies with the temperature, as shown in Table 2.1 and plotted in Figure 2.2(a) for a particular sample. Although the volume changes with the Celsius temperature, the relationship is not a direct proportionality. That is, when the Celsius temperature doubles, the volume does not double, all other factors being held constant. On the graph, the plotted points form a straight line, but the line does not pass through the origin. For a direct proportionality to exist, the straight line must pass through the origin. If the straight line corresponding to the points in Table 12.1 is extended until the volume reaches 0 L, the Celsius temperature is -273K (Figure 2.2b). Charles defined a new temperature scale in which the lowest possible temperature is absolute, corresponding to -273K. This temperature is called absolute zero. Table 2.1 Temperature and Volume Data for a Particular Sample of Gas at a Given Pressure
Temperature(Β°C) Volume(L)
1 0 0.400
2 100 0.548
3 200 0.692
4 300 0.840
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(a) (b)
Fig. 2.2. Dependence of Volume on Temperature at Constant Pressure (a) Plot of the data given in Table 2.1. (b) Extension of the line in part (a) to absolute zero, with the Kelvin scale added to the horizontal axis. We can state Charlesβ findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature. This means an increase in the temperature of a fixed mass of a gas leads to a corresponding increase in the volume of the gas by the same proportion, and vice versa, with the proviso that pressure remains the same. Mathematically expression of the law; [ Vβ T ]P
[ V= KT ] P
[ V/π = K ] P
(Where K = constant of proportionality).
A more useful form of the law can be express as:
π1
π1 =
π2
π2
Where V1 and T1 refer to the original volume and pressure, V2 and T2 refer to the volume and pressure under the new or changed conditions.
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2.5. Kinetic Theory and Charlesβ Law As the temperature of the gas molecules increase, the average
kinetic energy is equally raised, and hence, the average velocity of gas molecules. The gas molecules move more rapidly colliding with one another and more frequently with the walls of the container. For gas pressure to remain constant, the volume of the container must be increased with an increase in temperature. Example 2.8. Assume that the volume of a balloon filled with H2 is 1.00 L at 25Β°C. Calculate the volume of the balloon when it is cooled to -78Β°C in a low-temperature bath made by adding dry ice to acetone. Solution Collect the given information and convert as necessary Data provided; V1 = 1.00L,
T1 = 250C = (25 + 273)K = 298K
T2 = -780C = (273 - 78)K =195K
V2 = ?
Applying Charlesβ law,
π1
π1 =
π2
π2
V2 = 195 Γ1.00
293
= 0.65L
Example 2.9. The volume of a fixed mass of gas measured at
atmospheric pressure and 260C is 3.0 dm3. Calculate the volume at
1270C and at the same pressure.
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Solution
Data provided;
V1 = 3.0 dm3,
T1 = 260C = (25 + 273)K = 299K
T2 = 1270C = (273 + )K =400 K
V2 = ?
Applying Charlesβ law,
π1
π1 =
π2
π2
V2 = 400 Γ3.00
299
= 4.0 dm3
Example 2.10. If 250cm3 of a gas at s.t.p. is heated to 270C at constant
pressure, calculate its new volume.
Solution
Data provided;
V1 = 250 cm3,
T1 = s.t = 273K
T2 = 270C = (273 +27 )K =300 K
V2 = ? Applying Charlesβ law,
V1/ T1 = V2/ T2
V2 = V1 Γ T2/ T1
V2 = 250 Γ 300
273
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= 274.7 cm3
Example 2.11. Show that the data in Table below prove (a) that the
Celsius temperature is not directly proportional to volume and (b) that
the Kelvin temperature is directly proportional to volume.
Temperature and Volume data for a particular Sample of gas at a given pressure
Solution
As the absolute temperature 273 K is increased to 373 K or 473 K, the
volume increases to 373/273 = 1.37 or 473/273 = 1.37 times the original
volume. The ratio of V to T is constant (see Table above). The volume is
directly proportional to absolute temperature.
Example 2.12. Calculate the Celsius temperature to which a 678-mL
sample of gas at 0oC must be heated at constant pressure for the
volume to change to 0.896 L.
Solution
Data provided
V1 = 678 mL = 0.678 L
T1 = 0oC = 273K
V2 = 0.896 L
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T2 = ?
Using the relationship
π1
π1 =
π2
π2
T2 = 273 Γ0.896
0.678
= 361 πΎπΎ
= (361 β 273)oC
= 88oC
Note: 1000 mL = 1L
Example 2.13. Calculate the original temperature of a 0.456-mL gas
sample if it is expanded at constant pressure to 1.75 L at 55Β°C.
Solution
Data provided
V1 = 0.456 mL = 0.000456 L
T1 = ?
V2 = 1.75 L
T2 = 55 OC = (273 + 55)K = 238K
Using the relationship below and making T1 the subject;
π1
π1 =
π2
π2
T1 = 238 Γ0.000456
1.75
= 0.1 πΎK
= (0.1 β 273) oC
= β272.9 oC
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Example 2.14. A plastic bag of peanuts is laid on a windowsill in the
sun, where its temperature increases from 20OC to 30OC. If the original
volume is 100.0 cm3, what is the final volume after warming?
Solution
Data collection
V1 = 100 cm3 T1 = 20OC = 293 K
V2 = ? T2 = 30 oC = 303 K Using the relationship below and making V2 the subject and substituting;
π1
π1 =
π2
π2
V2= 303 Γ100
293
= 103.4 cm3
Example 2.15. The temperature of a 4.00 L sample of gas is changed
from 10.0 Β°C to 20.0 Β°C. What will the volume of this gas be at the new
temperature if the pressure is held constant?
Solution
Data collection
V1 = 4.00L
T1 = 10OC = 283 K
V2 = ?
T2 = 20 OC = 293 K
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Using the relationship below and making V2 the subject and
substituting;
π1
π1 =
π2
π2
π2 = π2π1
π1
= 293 Γ4.00
283
= 4.1 πΏ
Example 2.16. Carbon dioxide is usually formed when gasoline is
burned. If 30.0 L of CO2 is produced at a temperature of 1.00 x103 Β°C
and allowed to reach room temperature (25.0 Β°C) without any pressure
changes, what is the new volume of the carbon dioxide?
V2 = ? T2 = 25 OC = 298 K Using the relationship below and making V2 the subject and
substituting;
π1
π1 =
π2
π2
π2 = π2π1
π1
= 298 Γ 30.00
1273
= 7.0 πΏ
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Example 2.17. The volume of a gas syringe which contains 56.05
milliliters was raised to 67.7 milliliters at 107.5 oC. Determine the initial
temperature of the gas?
Solution
Data collection
V1 = 56.05 mm = 0.05605L
T1 =
V2 = 67.7 mm = 0.068L
T2 = 107.5 OC = 380.5 K
Using the relationship below and making T1 the subject and
substituting;
π1
π1 =
π2
π2
π1 = π2π1
π2
= 380.5 Γ 0.05605
0.068
= 313.6 πΎ
= (313.6 β 273) = 40.6 oC
Example 2.18. If 15.0 liters of neon at 25.0 Β°C is allowed to expand to
45.0 liters, what is the new temperature?
Solution
Data provided
V1 = 15.0L
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T1 = 25 Β°C = (273 + 25)K = 298 K
V2 = 45.0 L
T2 = ?
Using the relationship below and making T2 the subject and
substituting;
π1
π1 =
π2
π2
π2 = π1π2
π1
= 298 Γ 45.00
15
= 294 πΎ
Example 2.19. A balloon has a volume of 2500.0 mL on a day when the
temperature is 30.0 Β°C. If the temperature at night falls to 10.0 Β°C, what
will be the volume of the balloon if the pressure remains constant?
Solution
Data collection
V1 = 2500 mL
T1 = 30OC = 303 K
V2 = ?
T2 = 10 OC = 283 K
Using the relationship below and making V2 the subject and
substituting;
π1
π1 =
π2
π2
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π2 = π2π1
π1
= 283 Γ2500.00
303
= 2335 ππΏ
2.6. Temperature-Pressure Relationship
Boyleβs Law is the relationship between Pressure and Volume but does not address temperature. How does temperature change affect the properties of a sample of gas? Recall that temperature is a measure of the average kinetic energy of particles. As the particles of a substance move faster, the substanceβs temperature increases. The particles bump into each other and the sides of the container more often.
How would this affect a system where the volume is closed and constant? This observation was first made by Gay-Lussac. He observed that pressure has a direct proportional link with temperature of a sample of gas in a closed container (volume constant). Properly put, this law states that at constant volume, the pressure of a fixed mass of a gas is directly proportional to its absolute temperature. The law is expressed mathematically as follows:
π β π (Constant volume)
π
π= π
A more useful form of the law can be express as:
π1
π1 =
π2
π2
Where P1 and T1 refer to the original pressure and temperature, P2 and T2 refer to the pressure and temperature under the new or changed conditions. Note: in solving or addressing mathematical problems with this law, the temperature must be expressed in Kelvin and the pressure in a standard uint.
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Example 2.20. 10.0 L of a gas is found to exert 97.0 kPa at 25.0Β°C. What
would be the required temperature (in Celsius) to change the pressure
Example 2.21. If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0 Β°C, what would the final temperature of the gas be? Solution Data provided P1 = 15 atm
Example 2.22. A 30.0 L sample of nitrogen inside a metal container at 20.0 Β°C was placed inside an oven whose temperature is 50.0 Β°C. The pressure inside the container at 20.0 Β°C was 3.00 atm. What is the pressure of the nitrogen after its temperature was increased?
Solution
Collect data and convert temperatures to Kevin
P1 = 3.00 atm
T1 = 25.0Β°C = ( 20 + 273)K = 293.0 K
P2 = ?
T2 = 50.0Β°C = ( 50 + 273)K = 323.0 K
Applying π1
π1 =
π2
π2 and making P2 the subject
P2 = π2π1
π1
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= 323.0 πΎ Γ3.00 ππ‘π
293 ππ‘π
= 3.3 atm
Example 2.23. The temperature of a sample of gas in a steel container at 30.0 kPa is increased from β100.0 Β°C to 1.00 x 103 Β°C. What is the final pressure inside the tank?
Solution
Collect data and convert temperatures to Kevin
P1 = 30 kPa
T1 = β100.0 Β°C = (β100.0 + 273)K = 173.0 K
P2 = ?
T2 = 1.00 x 103 Β°C = (1.00 x 103 + 273)K = 1273.0 K
Applying π1
π1 =
π2
π2 and making P2 the subject
P2 = π2π1
π1
= 1273 πΎ Γ30 πππ
173 πΎ
= 220 kPa
2.7. The Combined Gas Law Boyleβs and Charlesβ laws may be merged into one law, called the
combined gas law, expressed in equation form as derived below: From Boyleβ law: Vβ 1/π (T constant)
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From Charlesβ law: V β π (P constant) V β 1/π β T π = ππ/π ππ
π = k
That is, for a given sample of a gas, PV/T remains constant, and therefore
π1 π1
π1=
π2 π2
π2 (a given sample of a gas)
This expression is a mathematical statement of the combined (or general) gas law. In words, the volume of given sample of a gas is inversely proportional to its pressure and directly proportional to its absolute temperature.
Note that if the temperature is constant, T1 = T2, then the expression reduces to the equation for Boyleβs law, P1V1 = P2V2. Alternatively, if the pressure is constant, P1 = P2, the expression is equivalent to Charlesβ law, V1/T1 = V2/T2. When the initial volume V1 of a gas at temperature T1 and pressure P1 is subjected to changes in temperature to T2 and pressure to P2, its new volume V2 is obtained from the equation.
To apply this gas law, the amount of gas should remain constant. As with the other gas laws, the temperature must be expressed in kelvins, and the units on the similar quantities should be the same. Because of the dependence on three quantities at the same time, it is difficult to tell in advance what will happen to one property of a gas sample as two other properties change. The best way to know is to work it out mathematically.
Example 2.24. A certain mass of a gas occupies 330 cm3 at 27oC and 9.0
Γ 104 Nm-2 pressure. Calculate its volume at s.t.p. (s.p = 1.0 Γ 105 Nm-2).
Solution
Write the given data down, convert as variable to appropriate units
and substitute into the form to find the unknown.
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Data provided:
V1 = 330 cm3
P1 = 9.0 Γ 104 Nm-2
T1 = 27oC = (27 + 273)K = 300K
T2 = s.t. = 273K
P2 = s.p. = 1.0 Γ 105Nm-2
V2 = ?
Using the gas equation:
π1 π1
π1 =
π2 π2
π2
Making V2 the subject of the formula:
V2 = π1π1π2
π2π1
= 9.0 Γ 104 Γ 330 Γ 273
1.0 Γ 105 Γ 300
= 270 cm3
Example 2.25. Calculate the volume of a sample of gas originally
occupying 908 mL at 717 torr and 20OC after its temperature and
pressure are changed to 72OC and 1.07 atm.
Solution
In attempting this problem, the volume can be stated in millilitres in
both states. The pressure can be stated in atmospheres in both but the
temperature must be in kelvins in both states.
Data provided
V1 = 908 mL
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P1 = 717
760 ππ‘π = 0.94 ππ‘π
T1 = 20oC = (20 + 273)K = 293K
T2 = 72 oC = 345K
P2 = 1.07 atm
V2 = ?
Using the gas equation:
π1 π1
π1 =
π2 π2
π2
Making V2 the subject of the formula:
V2 = π1π1π2
π2π1
= 0.94 Γ 908 Γ 345
1.07 Γ 293
= 943 mL
Example 2.27. Calculate the original volume of a sample of gas that is
at 700 torr and 22 oC before its volume, temperature, and pressure are
changed to 998 mL, 82Β°C, and 2.07 atm
Solution
Data provided
V1 = ?
P1 = 700
760 ππ‘π = 0.92 ππ‘π
T1 = 22oC = (22 + 273)K = 295K
T2 = 82 oC = 355K
P2 = 2.07 atm
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V2 = 998 mL
Using the gas equation:
π1 π1
π1 =
π2 π2
π2
Making V1 the subject of the formula:
V1 = π2π2π1
π1π2
= 2.07 Γ 998 Γ 298
0.92 Γ 355
= 1884 mL
Example 2.28. 17.3-mL sample of gas originally at standard
temperature and pressure is changed to 10.9 mL at 678 torr. Calculate
its final temperature in degrees
Celsius.
Solution
Data provided
V1 = 17.3 mL
P1 = s.p. = 760 torr
T1 = s.t. = 273 K
T2 = ?
P2 = 678 torr
V2 =10.9 mL
Using the gas equation:
π1 π1
π1 =
π2 π2
π2
Making T2 the subject of the formula:
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T2 = π2π2π1
π1π1
= 678 Γ 10.9 Γ 273
760 Γ 17.3
= 153 K
Converting to degree Celsius
= (153 β 273) oC
= 120 oC
Example 2.29. Calculate the volume at standard temperature and
pressure of a sample of gas that has a volume of 49.7 mL at 52Β°C and
811 torr.
Solution
Data provided
V1 = 49.7 mL
P1 = 811
760 ππ‘π = 1.07 ππ‘π
T1 = 52Β°C = 325 K
T2 = s.t. = 273 K
P2 = s.p. = 1 atm
V2 =?
Using the gas equation:
π1 π1
π1 =
π2 π2
π2
Making T2 the subject of the formula:
V2 = π1π1π2
π2π1
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= 1.07 Γ 49.7 Γ 273
1.0 Γ 325
= 45 mL
Example 2.30. Calculate the new volume after a 2.00-L sample of gas at
1.50 atm and 127oC is changed to 27oC at 3.50 atm.
Solution
Data provided
V1 = 2.00 L
P1 =1.50 ππ‘π
T1 = 127Β°C = 400 K
T2 = 27oC =300 K
P2 = 3.50 atm
V2 =?
Using the gas equation:
π1 π1
π1 =
π2 π2
π2
Making T2 the subject of the formula:
V2 = π1π1π2
π2π1
= 1.5 Γ 2.00 Γ 300
3.50 Γ 400
= 0.64 L
Example 2.31. 500.0 liters of a gas are prepared at 700.0 mmHg and
200.0 Β°C. The gas is placed into a tank under high pressure. When the
tank cools to 20.0 Β°C, the pressure of the gas is 30.0 atm. What is the
2.8. Relationship between Amount and Volume 2.8.1. Gay-Lussac's Law of Combining Volumes
In the same 1808 article in which Gay-Lussac published his observations on the thermal expansion of gases, he pointed out that when two gases react, they do so in volume ratios that can always be expressed as small whole numbers. This came to be known as the Law of combining volumes. Example 2.39. Ammonium carbonate decomposes when heated to yield carbon dioxide, ammonia, and water vapour. Calculate the ratio
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of the (separate) volume of ammonia to that of water vapour, each at 450Β°C and 1.00 atm. Solution
The mole ratio of the gases, given in the balanced equation, is 2 mol NH3: 1 mol CO2: 1 molH2O
The ammonia and water vapour are separated and measured at the given temperature and pressure. The ratio of their volumes can be calculated as follows:
πππ» 3
ππ»2π=
πππ» 3 π π /π
ππ»2π π π /π
Because R is a constant and both T and P are the same for the two gases, this equation reduces to
πππ» 3
ππ»2π
= πππ» 3
ππ»2π
The right side of this equation is the ratio of the numbers of molesβthe ratio given by the balanced chemical equation. The left side of the equation is the ratio of the volumes, so the ratio given by the balanced chemical equation is equal to the volume ratio under these conditions. The ratio is 2: 1. Example 2.40. If 2.00 L H2 of and 1.00 L of both at standard temperature and pressure, are allowed to react, will the water vapor they form at 250Β°C and 1.00 atm occupy 2.00 L? Solution 2H2 (g) + O2 (g) β 2H2O (g)
The volumes of H2 and O2 that react are in the ratio given in the balanced equation because the two gases have the same temperature and pressure. The volume of water vapour formed is not in that ratio, however, because its temperature is different. Its volume will be much greater than 2 L. 2.8.2. Avogadro's Law
The work of the Italian scientist Amedeo Avogadro complemented the studies of Boyle, Charles, and Gay-Lussac. In 1811
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he published a hypothesis stating that at the same temperature and pressure, equal volumes of different gases contain the same number of molecules (or atoms if the gas is monatomic). This law states that equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of molecule. Mathematically: V β π (at constant T and P) π = ππ
π
π= π
Where V is the volume of gas, n is the number of molecules and π is the proportionality constant.
This law relates the volume of a fixed mass of a gas to the number of molecules it contains. It shows that the volume occupied by a gas depends on the number of molecules it contains, at a given temperature and pressure. An increase in the number of gas molecules leads to an increase in gas volume, and vice versa.
According to Avogadroβs law we see that when two gases react with each other, their reacting volumes have a simple ratio to each other. If the product is a gas, its volume is related to the volume of the reactants by a simple ratio (a fact demonstrated earlier by Gay-Lussac). For example, consider the synthesis of ammonia from molecular hydrogen and molecular nitrogen:
3H2(g) + N2(g) β 2NH3(g) 3 mol 1 mol 2 mol
Because, at the same temperature and pressure, the volumes of gases are directly proportional to the number of moles of the gases present, we can now write
The volume ratio of molecular hydrogen to molecular nitrogen is 3:1, and that of ammonia (the product) to molecular hydrogen and molecular nitrogen combined (the reactants) is 2:4, or 1:2.
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Example 2.41. 50 cm3 of sulphur (IV) oxide were produced at s.t.p. when some quantity of powdered sulphur was burnt in excess oxygen. (a) Write a balanced chemical equation for the reaction. (b) Calculate the volume of oxygen used up during the reaction. (c) Which of the laws is applicable? State the law. Solution (a). S(g) + O2(g) β SO2(g)
(b). From the balanced chemical equation in (a) above; At s.t.p: 22400 cm3 of SO2 required 22400 cm3 of O2 Hence 1 cm3 of SO2 will require 1 cm3 of O2
β΄ 50 cm3 of SO2 will use 50 cm3 of O2
(c). Avogadroβs law is applicable in (b) above and it state that at the same temperature and pressure equal volume of gases contain the same number of molecules. 2.9. The Ideal Gas Law
So far, the gas laws we have used have focused on changing one or more properties of the gas, such as its volume, pressure, or temperature. There is one gas law that relates all the independent properties of a gas under any particular condition, rather than a change in conditions. This gas law is called the ideal gas law. The general ideal gas equation is a combination of Boyleβs, Charlesβ and Avogadroβs laws involving the four gas variables: pressure (P), volume (V), number of mole of gas (n), and temperature (T).
From Boyleβ law: Vβ 1/π (T constant) From Charlesβ law: V β π (P constant) From Avogadroβs law: V β π (P,T constant) V β 1/π β T β π
V = R Γ 1/π Γ T Γ π
PV = nRT
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In this equation, P is pressure, V is volume, n is amount of moles, and T is temperature. R is called the ideal gas law constant and is a proportionality constant that relates the values of pressure, volume, amount, and temperature of a gas sample. The variables in this equation do not have the subscripts i and f to indicate an initial condition and a final condition. The ideal gas law relates the four independent properties of a gas under any conditions.
2.10. Evaluation of the Gas Constant, R The gas constant can be expressed in various units, all having
the dimension of energy per degree per mol. From the general equation PV = nRT we get:
π = ππ
ππ
Where P is pressure, V is volume, n is amount, and T is temperature. R is most easily calculated from the fact that the hypothetical volume of an ideal gas is 22.4L at STP (273.K and 1 atm).
i. If volume is expressed in liters and pressure in atmospheres, then the proper value of R is as follows:
π = ππ
ππ
= 1.0 ππ‘π Γ22.414 πΏ
1.0 πππ Γ273.15 πΎ
R= 0.08206 atm L mol-1K-1
ii. if pressure is in Nm-2 and volume in m3 then the proper value of R is thus:
Thus far we have concentrated on the behaviour of pure
gaseous substances, but experimental studies very often involve
mixtures of gases. For example, for a study of air pollution, we may be
interested in the pressure-volume-temperature relationship of a sample
of air, which contains several gases. In this case, and all cases involving
mixtures of gases, the total gas pressure is related to partial pressures,
that is, the pressures of individual gas components in the mixture. In 1801
Dalton formulated a law, now known as Daltonβs law of partial
pressures, which states that the total pressure of a mixture of gases is just
the sum of the pressures that each gas would exert if it were present alone.
Figure 2.3 illustrates Daltonβs law.
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Figure 2.3. Schematic illustration of Daltonβs law of partial pressures. Consider a case in which two gases, A and B, are in a container of volume V. The pressure exerted by gas A, according to the ideal gas equation, is
ππ΄ =ππ΄π π
π
Where nA is the number of moles of A present. Similarly, the pressure exerted by gas B is
ππ΅ =ππ΅π π
π
In a mixture of gases A and B, the total pressure PT is the result of the collisions of both types of molecules, A and B, with the walls of the container. Thus, according to Daltonβs law, ππ = ππ΄ + ππ΅
= ππ΄π π
π +
ππ΅π π
π
= π π
π (ππ΄ + ππ΅)
= ππ π
π
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Where n, the total number of moles of gases present, is given by n = nA + nB, and PA and PB are the partial pressures of gases A and B, respectively. For a mixture of gases, then, PT depends only on the total number of moles of gas present, not on the nature of the gas molecules. In general, the total pressure of a mixture of gases is given by ππ = π1 + π2 + π3 + β β β β Where P1, P2, P3, . . . are the partial pressures of components 1, 2, 3, . . . . To see how each partial pressure is related to the total pressure, consider again the case of a mixture of two gases A and B. Dividing PA by PT, we obtain
ππ΄
ππ =
ππ΄π ππ
(ππ΄ +ππ΅) π ππ
= ππ΄
ππ΄ +ππ΅
= ππ΄
Where XA is called the mole fraction of A. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of one component to the number of moles of all components present. In general, the mole fraction of component i in a mixture is given by
ππ = π π
ππ
Where ni and nT are the number of moles of component i and the total number of moles present, respectively. The mole fraction is always smaller than 1. We can now express the partial pressure of A as ππ΄ = ππ΄ππ Similarly, ππ΅ = ππ΅ππ Note that the sum of the mole fractions for a mixture of gases must be unity. If only two components are present, then
ππ΄ + ππ΅ = ππ΄
ππ΄ + ππ΅ +
ππ΅
ππ΄ + ππ΅= 1
If a system contains more than two gases, then the partial pressure of the ith component is related to the total pressure by ππ = ππππ
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How are partial pressures determined? A manometer can measure only the total pressure of a gaseous mixture. To obtain the partial pressures, we need to know the mole fractions of the components, which would involve elaborate chemical analyses. The most direct method of measuring partial pressures is using a mass spectrometer. The relative intensities of the peaks in a mass spectrum are directly proportional to the amounts, and hence to the mole fractions, of the gases present. Example 2.65. A mixture of gases contains 3.85 moles of neon (Ne), 0.92 mole of argon (Ar), and 2.59 moles of xenon (Xe). Calculate the partial pressures of the gases if the total pressure is 2.50 atm at a certain temperature. Solution The partial pressure of a gas is equal to the product of its mole fraction and the total pressure (PT) Given data Mole of Ne = 3.85 moles Mole of Ar = 0.92 moles Mole of Xe = 2.59 moles Total pressure, PT = 2.5
Mole fraction of neon (πππ ) = πππ
= 0.523 Therefore πππ = πππππ = 0.523 Γ 2.50 = 1.31 ππ‘π Similarly, we can calculate the mole fraction of argon and its partial pressure:
Mole fraction of Argon (ππ΄π ) = ππ΄π
= 0.125 Therefore ππ΄π = ππ΄πππ = 0.125 Γ 2.50 = 0.313 ππ‘π Finally, we calculate the mole fraction of xenon and its partial pressure:
Mole fraction of Xenon (πππ ) = πππ
= 0.352 Therefore πππ = πππππ = 0.352 Γ 2.50 = 0.88 ππ‘π Check: The individual partial pressures must be less than the total pressure and make sure that the sum of the partial pressures is equal to the total pressure; that is, (1.31 + 0.313 + 0.880) atm = 2.50 atm. Example 2.66. A sample of natural gas contains 8.24 moles of methane (CH4), 0.421 mole of ethane (C2H6), and 0.116 mole of propane (C3H8). If the total pressure of the gases is 1.37 atm, what are the partial pressures of the gases? Solution Data provided
Mole of methane = 8.24 moles
Moles of ethane = 0.421 moles
Mole of propane = 0.116 moles
Total pressure = 1.37 atm
Mole fraction of neon (ππππ‘ ππππ ) = ππππ‘ ππππ
Whenever a gas is collected over water, it becomes wet by water vapour. Since water vapour is a gas, it exerts its own pressure, and a mixture of gases is obtained. The pressure exerted is, therefore, the sum of the partial pressures of the gas and that of the water vapour at that temperature, i.e. PTotal = Pgas +Pwater vapour Pgas = PTotal β Pwater vapour Daltonβs law of partial pressures is useful for calculating volumes of gases collected over water. For example, when potassium chlorate (KClO3) is heated, it decomposes to KCl and O2:
2KClO3(s) β 2KCl(s) + 3O2(g) The oxygen gas can be collected over water, as shown in Figure 2.4. Initially, the inverted bottle is completely filled with water. As oxygen gas is generated, the gas bubbles rise to the top and displace water from the bottle. This method of collecting a gas is based on the assumptions that the gas does not react with water and that it is not appreciably soluble in it. These assumptions are valid for oxygen gas, but not for gases such as NH3, which dissolves readily in water. The oxygen gas collected in this way is not pure, however, because water vapour is also present in the bottle. The total gas pressure is equal to the sum of the pressures exerted by the oxygen gas and the water vapour: PT = P π2 + Pπ»2O Consequently, we must allow for the pressure caused by the presence of water vapour when we calculate the amount of O2 generated. Table 2.2 shows the pressure of water vapour at various temperatures.
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Table 2.2. pressure of water vapour at various temperatures Temperature Water vapour pressure
Figure 2.4: An apparatus for collecting gas over water. The oxygen generated by heating potassium chlorate (KClO3) in the presence of a small amount of manganese dioxide (MnO2), which speeds up the reaction, is bubbled through water and collected in a bottle as shown. Water originally present in the bottle is pushed into the trough by the oxygen gas. Example 2.67. Oxygen gas generated by the decomposition of potassium chlorate is collected as shown in Figure 2.4. The volume of oxygen collected at 26Β°C and atmospheric pressure of 771 mmHg is 141 mL. Calculate the mass (in grams) of oxygen gas obtained. The pressure of the water vapour at 26Β°C is 25.2 mmHg. [O = 16] Solution To solve for the mass of O2 generated, we must first calculate the partial pressure of O2 in the mixture. What gas law do we need? How do we convert pressure of O2 gas to mass of O2 in grams? Data provided PT = 771 mmHg Pπ»2O = 25.2 mmHg P π2 = ? V = 141mL = (141/1000)L = 0.141 L Molar mass (M) = 32.0 g/mol T = 26Β°C = (26 + 273)K = 299 K R = 0.082 atm L K-1 mol-1 From Daltonβs law of partial pressures we know that PT = P π2 + Pπ»2O P π2 = PT β Pπ»2O = 771 mmHg β 25.2 mmHg
= 746 mmHg = (746/760) atm = 0.98 atm
From the ideal gas equation we write ππ = ππ π
= 0.180 g Example 2.68. Hydrogen gas generated when calcium metal reacts with water is collected as shown in Figure 5.14. The volume of gas collected at 30Β°C and pressure of 988 mmHg is 641 mL. What is the mass (in grams) of the hydrogen gas obtained? The pressure of water vapor at 30Β°C is 31.82 mmHg. [H =1]. Solution Data provided PT = 988 mmHg Pπ»2O = 31.82 mmHg P π»2 = ? V = 641mL = (641/1000)L = 0.641 L Molar mass (M) = 2.0 g/mol T = 30Β°C = (30 + 273)K = 303 K R = 0.082 atm L K-1 mol-1 From Daltonβs law of partial pressures we know that PT = P π2 + Pπ»2O P π»2 = PT β Pπ»2O = 988 mmHg β 31.82 mmHg
= 956.18 mmHg = (956.18/760) atm = 1.26 atm
From the ideal gas equation we write ππ = ππ π
= 0.065 g Example 2.69. 20 dm3 of hydrogen were collected over water at 17oC and 79.7KNm-2 pressure. Calculate (i) pressure of dry hydrogen at this temperature. (ii) volume of dry hydrogen gas at s.t.p. (vapour pressure of water is 1.9 KNm-2 at 17oC; 1atm = 101.3 KNm-2 ). Solution Data given:
PTotal = 79.7KNm-2
Pwater vapour = 1.9 KNm-2 at 17oC
Pgas = ?
(i) According to Daltonβs law:
Pgas = PTotal β Pwater vapour
p(H2) = (79.7 β 1.9 ) KNm-2
= 77.8 KNm-2
(ii) To find the volume of dry hydrogen gas at s.t.p.
Data provided.
P1= 77.8 KNm-2
V1= 20 dm3
T1 = 17oC = (17 + 273)K = 290K
P2 = s.p. = 101.3 KNm-2
T2 = s.t. = 273K
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V2 = ? Using the general gas equation and substituting, P1V1/T1 = P2V2/T2
V2 = 77.8 Γ20 Γ273
101.3 Γ290 = 14.5 dm3
Example 2.70. (i). If I try to put a 1.00-L sample of O2 at 300 K and 1.00 atm plus a 1.00-L sample of N2 at 300 K and 1.00 atm into a rigid 1.00-L container at 300 K, will they fit? (ii) If so, what will be their total volume and total pressure? Solution (i) The gases will fit; gases expand or contract to fill their containers. (b) The total volume is the volume of the containerβ1.00 L. The temperature is 300 K, given in the problem. The total pressure is the sum of the two partial pressures. Partial pressure is the pressure of each gas (as if the other were not present). The oxygen pressure is 1.00 atm. The oxygen has been moved from a 1.00-L container at 300 K to another 1.00-L container at 300 K, and so its pressure does not change. The nitrogen pressure is 1.00 atm for the same reason. The total pressure is 1.00 atm + 1.00 atm = 2.00 atm. Example 2.71. A 1.00-L sample of O2 at 300 K and 1.00 atm plus a 0.500-L sample of N2 at 300 K and 1.00 atm are put into a rigid 1.00-L container at 300 K. What will be their total volume, temperature, and total pressure? Solution The total volume is the volume of the containerβ1.00 L. The temperature is 300 K, given in the problem. The total pressure is the sum of the two partial pressures. The oxygen pressure is 1.00 atm. The nitrogen pressure is 0.500 atm, since it was moved from 0.500 L at 1.00 atm to 1.00 L at the same temperature (Boyleβs law). The total pressure is 1.00 atm + 0.500 atm = 1.50 atm
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Example 272. A 1.00-L sample of O2 at 300 K and 1.00 atm plus a 0.500-L sample of N2 at 300 K and 1.00 atm are put into a rigid 1.00-L container at 300 K. What will be their total volume, temperature, and total pressure? Calculate the number of moles of O2 both before and after mixing. Solution Data provided P = 1.0 atm T = 300 K V = 1.0 L R = 0.082 atm L K-1 mol-1 n = ? Applying the ideal gas equation to find n before mixing ππ = ππ π
Making n the subject, and substituting:
n = ππ
π π
= 1 atm Γ 1.0 L
0.082 atm L mol β1 Kβ1 Γ 300 K
= 0.0406 mol of O2 gas
After mixing
nπ2 = ππ2π
π π
= 1 atm Γ 1.0 L
0.082 atm L mol β1 Kβ1 Γ 300 K
= 0.0406 mol of O2 gas
There is no change in the number of moles of oxygen gas before and
after mixing.
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Example 2.73. Calculate the volume of 1.00 mol of H2O at 1.00-atm pressure and a temperature of 25oC. Solution Water (H2O) is not a gas under these conditions, and so the equation PV = nRT does not apply. (The ideal gas law can be used for water vapour, e.g., water over 100 oC at 1 atm or water at lower temperatures mixed with air). At 1-atm pressure and 25 oC, water is a liquid with a density of about 1.00 g/mL. 1 mol of H2O contain 18g
= 18 mL Example 2.74. A container holds three gases: oxygen, carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm, respectively. What is the total pressure inside the container? Solution
Data provided
Poxygen = 2.00 atm
Pcarbon dioxide = 3.00 atm
Phelium = 4.00 atm
PT = ?
From Daltonβs law of partial pressure,
PT = sum of the individual partial pressures in the reaction vessel
= Poxygen + Pcarbon dioxide + Phelium
= (2.0 + 3.0 +4.0) atm = 9.0 atm
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Example 2.75. A tank contains 480.0 grams of oxygen and 80.00 grams of helium at a total pressure of 7.00 atmospheres. Calculate the following. a) How many moles of O2 are in the tank? b) How many moles of He are in the tank? c) Total moles of gas in tank. d) Mole fraction of O2.
e) Mole fraction of He.
f) Partial pressure of O2.
g) Partial pressure of He.
Solution
Let us first find the mole of the gases with the simple relation (n = m/M)
Data given:
Mass of O2 = 480.0 g
Mass of helium = 80g
Total pressure (PT) = 7.00 atm
a). Mole of O2 in the tank (n) = πππ π (π)
πππππ πππ π (π)
= 480 π
32 π/πππ
= 15 mol of O2
b). Mole of He in the tank (n) = πππ π (π)
πππππ πππ π (π)
= 80 π
4.0 π/πππ
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= 20 mol of He
c). Total mole of gas in the tank = mole of O2 + mole of He
= (15 + 20)mole
= 35 moles
d). Mole fraction of O2 (ππ2) =
ππ2
ππ2 + ππ»π
= 15 πππ
15 πππ +20 πππ
= 0.4286
e). Mole fraction of He (ππ»π ) = ππ»π
ππ2+ ππ»π
= 20 πππ
15 πππ +20 πππ
= 0.5714 f). Partial pressure of O2 (ππ2
) = ππ2ππ
= 0.4286 Γ 7.00 = 3.0 atm g). Partial pressure of He (ππ»π) = ππ»πππ = 0.5714Γ 7.00 = 3.99 atm Example 2.76. A tank contains 5.00 moles of O2, 3.00 moles of neon, 6.00 moles of H2S, and 4.00 moles of argon at a total pressure of 1620.0 mmHg. Calculate the following. a) Total moles of gas in tank b) Mole fraction of gases c) Partial pressure of gases d) Pressure fraction of gases
= 0.22 atm Example 2.77. A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0Β°C and the total pressure in the cylinder. [H = 1, C = 12]. Solution Let us first find the mole of the gases with the simple relation (n = m/M) then apply ideal gas equation to find the partial pressure of the gases.
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Data given:
Mass of methane = 1813 g
Mass of ethane = 336g
T = 22.0Β°C = 295 K
V = 20.0 L
Mole of methane in the cylinder (n) = πππ π (π)
πππππ πππ π (π)
But molar mass of methane (CH4) = [12 + (1Γ 4)] = 16 g/mol
n = 1813 π
16 π/πππ
= 113 mol of methane
Mole of ethane in the cylinder (n) = πππ π (π)
πππππ πππ π (π)
But molar mass of methane (C2H6) = [(12 Γ2) + (1Γ 6)] = 30 g/mol
n = 338 π
30 π/πππ
= 11.3 mol of ethane
Partial pressure of methane (π) =ππππ‘ ππππ Γ π π
Total pressure in the cylinder (PT) = Pmethane + Pethane
= (136 + 13.6)atm
= 149.4 ππ‘π 2.13. Gas Diffusion and Effusion 2.13.1. Gas Diffusion
A direct demonstration of random motion is provided by diffusion, the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. Despite the fact that molecular speeds are very great, the diffusion process takes a relatively long time to complete. For example, when a bottle of concentrated ammonia solution is opened at one end of a laboratory bench, it takes some time before a person at the other end of the bench can smell it. The reason is that a molecule experiences numerous collisions while moving from one end of the bench to the other. Thus, diffusion of gases always happens gradually, and not instantly as molecular speeds seem to suggest. Furthermore, because the root-mean square speed of a light gas is greater than that of a heavier gas, a lighter gas will diffuse through a certain space more quickly than will a heavier gas. Figure 2.5 illustrates gaseous diffusion.
In 1832 the Scottish chemist Thomas Graham found that under the same conditions of temperature and pressure, rates of diffusion for gases are inversely proportional to the square roots of their molar masses. This statement, now known as Grahamβs law of diffusion, is expressed mathematically as
π1
π2=
π2
π1
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Where r1 and r2 are the diffusion rates of gases 1 and 2, and M1 and M2 are their molar masses, respectively.
Figure 2.5: A demonstration of gas diffusion. NH3 gas (from a bottle containing aqueous ammonia) combines with HCl gas (from a bottle containing hydrochloric acid) to form solid NH4Cl. Because NH3 is lighter and therefore diffuses faster, solid NH4Cl first appears nearer the HCl bottle (on the right). 2.13.2. Gas Effusion
Whereas diffusion is a process by which one gas gradually mixes with another, effusion is the process by which a gas under pressure escapes from one compartment of a container to another by passing through a small opening. Figure 2.6 shows the effusion of a gas into a vacuum. Although effusion differs from diffusion in nature, the rate of effusion of a gas has the same form as Grahamβs law of diffusion [see Equation for diffusion of a gas]. A helium-filled rubber balloon deflates faster than an air-filled one because the rate of effusion through the pores of the rubber is faster for the lighter helium atoms than for the air molecules. Industrially, gas effusion is used to separate uranium isotopes in the forms of gaseous 235UF6 and 238UF6. By subjecting the gases to many stages of effusion, scientists were able to obtain highly enriched 235U isotope, which was used in the construction of atomic bombs during World War II.
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Figure 2.6: Gas effusion. Gas molecules move from a high-pressure region (left) to a low-pressure one through a pinhole. Example 2.78. A flammable gas made up only of carbon and hydrogen is found to effuse through a porous barrier in 3.50 min. Under the same conditions of temperature and pressure, it takes an equal volume of chlorine gas 7.34 min to effuse through the same barrier. Calculate the molar mass of the unknown gas, and suggest what this gas might be. Solution Data given Time for effusion of chlorine = 7.34 min Time of effusion of unknown gas X = 3.5 min Molar mass of chlorine gas = 70.90 g/mol Molar mass of X = ? Strategy We find the rate of effusion of both gases and the use grahamβs law to find the molar mass of gas X
Rate of effusion of Cl2 = π ππππππ’πππ ππ ππππππππ
7.34 πππ
Rate of effusion of X = π ππππππ’πππ ππ π
ππ = 16.1 π/πππ Because the molar mass of carbon is 12.01 g and that of hydrogen is 1.008 g, the gas is methane (CH4). Note: Because lighter gases effuse faster than heavier gases, the molar mass of the unknown gas must be smaller than that of chlorine gas. Indeed, the molar mass of methane (16.04 g) is less than the molar mass of chlorine gas (70.90 g). Example 2.79. Methane effuses through a small opening in the side of a container at rate of 1.3 mols -1. An unknown gas X effuses through the same opening at the rate of 5.42 mols-1 when maintained at the same temperature and pressure as methane. Determine the molar mass of the unknown gas. [H= 1, C= 12]
ππ = 16 Γ 5.7528 = 92. 045 Therefore molar mass of gas X is 92.045g/mol Example 2.80. The time required for a volume of O2 to diffuse through opening is 40seconds. Calculate the molar mass of gas which requires 50seconds for the same volume to diffuse through the same opening under the same conditions [O = 16]. Solution Data provided Rate of diffusion of oxygen = 40s
Rate of diffusion of gas X = 50s
Molar mass of oxygen = 32.0g/mol
Molar mass of gas X = ?
Rate of diffusion of O2 = π ππππππ’πππ ππ ππ₯π¦πππ
35π
Rate of diffusion of X = π ππππππ’πππ ππ π
Example 2.81. If equal amounts of hydrogen and argon are placed in a porous container and allowed to escape, which gas will escape faster and why? Solution If equal amounts of hydrogen and argon are placed in a porous container and allowed to escape, hydrogen gas will effuse faster than argon because hydrogen is a lighter gas (with molecular mass of 2.0g/mol) than argon (with molecular mass of 39.95g/mol) and the rate of effusion varies inversely with molecular weight of a gas. Example 2.82. The time required for a volume of gas y to effuse through a small hole was 112.2seconds. The time required for the same volume of oxygen was 84.7seconds. Calculate the molecular weight of gas y. Solution Data provided Rate of effusion of oxygen = 48.7s
Rate of effusion of gas y = 112.2s
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Molar mass of oxygen = 32.0g/mol
Molar mass of gas y = ?
Rate of effusion of O2 = π ππππππ’πππ ππ ππ₯π¦πππ
48.5π
Rate of effusion of y = π ππππππ’πππ ππ π
Example 2.83. Consider the reaction represented by the following equation:
State what would happen to the vapour density of N2O4 as the temperature of the system is increased. If the system is cooled, would
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the gases become lighter or darker in colour? Explain your answer in each case. Solution The density becomes lighter as the temperature of the system increases. If the system is cooled, the product becomes lighter in colour. At low temperature, dinitrogen (IV) oxide (N2O4) predominates. Hence, the gases become lighter while at high temperature, N2O4 dissolves to nitrogen (IV) oxide (NO2) molecules and the gases becomes darker in colour and lighter in density.
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CHAPTER THREE
THE KINETIC MOLECULAR THEORY OF GASES
3.1 Introduction
The gas laws help us to predict the behaviour of gases, but they do not explain what happens at the molecular level to cause the changes we observe in the macroscopic world. For example, why does a gas expand on heating?
This section introduces the kinetic molecular theory of gases, which explains the gas laws and when extended, also explains some properties of liquids and solids.
Five postulates explain why gases behave as they do: 1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. 2. The volume of the molecules is insignificant compared with the volume occupied by the gas. 3. Forces between the molecules are negligible, except when the molecules collide with one another. 4. Molecular collisions are perfectly elastic; that is, no energy is lost when the molecules collide. 5. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy. The average kinetic energy of a molecule is given by
πΎπΈ = 12 ππ’2
Where m is the mass of the molecule and u is its speed. The horizontal
bar denotes an average value. The quantity π’2 is called mean square speed; it is the average of the square of the speeds of all the molecules:
π’2 = π’12 + π’2
2 + β β β π’π2
Where N is the number of molecules. Assumption 5 enables us to write πΎπΈ πΌ π
12 ππ’2 πΌ π
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πΎπΈ = 12 ππ’2 = πΆπ
Where C is the proportionality constant and T is the absolute temperature.
According to the kinetic molecular theory, gas pressure is the result of collisions between molecules and the walls of their container. It depends on the frequency of collision per unit area and on how βhardβ the molecules strike the wall. The theory also provides a molecular interpretation of temperature. According to Equation above, the absolute temperature of a gas is a measure of the average kinetic energy of the molecules. In other words, the absolute temperature is a measure of the random motion of the moleculesβthe higher the temperature, the more energetic the molecules. Because it is related to the temperature of the gas sample, random molecular motion is sometimes referred to as thermal motion. 3.2. Application to the Gas Laws
Although the kinetic theory of gases is based on a rather simple model, the mathematical details involved are very complex. However, on a qualitative basis, it is possible to use the theory to account for the general properties of substances in the gaseous state. The following examples illustrate the range of its utility: β’ Compressibility of Gases. Because molecules in the gas phase are separated by large distances (assumption 1), gases can be compressed easily to occupy less volume. β’ Boyleβs Law. The pressure exerted by a gas results from the impact of its molecules on the walls of the container. The collision rate, or the number of molecular collisions with the walls per second, is proportional to the number density (that is, number of molecules per unit volume) of the gas. Decreasing the volume of a given amount of gas increases its number density and hence its collision rate. For this reason, the pressure of a gas is inversely proportional to the volume it occupies; as volume decreases, pressure increases and vice versa. β’ Charlesβs Law. Because the average kinetic energy of gas molecules is proportional to the sampleβs absolute temperature (assumption 5),
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raising the temperature increases the average kinetic energy. Consequently, molecules will collide with the walls of the container more frequently and with greater impact if the gas is heated, and thus the pressure increases. The volume of gas will expand until the gas pressure is balanced by the constant external pressure. β’ Avogadroβs Law. We have shown that the pressure of a gas is directly proportional to both the density and the temperature of the gas. Because the mass of the gas is directly proportional to the number of moles (n) of the gas, we can represent density by n/V. Therefore,
π β π
π π
For two gases, 1 and 2, we write
π1 β π1π1
π1 = πΆ
π1π1
π1
π2 β π2π2
π2 = πΆ
π2π2
π2
Where C is the proportionality constant. Thus, for two gases under the same conditions of pressure, volume, and temperature (that is, when P1 = P2, T1 = T2, and V1 = V2), it follows that n1 = n2, which is a mathematical expression of Avogadroβs law. β’ Daltonβs Law of Partial Pressures. If molecules do not attract or repel one another (assumption 3), then the pressure exerted by one type of molecule is unaffected by the presence of another gas. Consequently, the total pressure is given by the sum of individual gas pressures. 3.3. Distribution of Molecular Speeds
The kinetic theory of gases enables us to investigate molecular motion in more detail. Suppose we have a large number of gas molecules, say, 1 mole, in a container. As long as we hold the temperature constant, the average kinetic energy and the mean square speed will remain unchanged as time passes. As you might expect, the motion of the molecules is totally random and unpredictable. At a given instant, how many molecules are moving at a particular speed? To answer this question Maxwell analyzed the behaviour of gas molecules at different temperatures.
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Figure 3.1(a) shows typical Maxwell speed distribution curves for nitrogen gas at three different temperatures. At a given temperature, the distribution curve tells us the number of molecules moving at a certain speed. The peak of each curve represents the most probable speed, that is, the speed of the largest number of molecules. Note that the most probable speed increases as temperature increases (the peak shifts toward the right). Furthermore, the curve also begins to flatten out with increasing temperature, indicating that larger numbers of molecules are moving at greater speed. Figure 3.1(b) shows the speed distributions of three gases at the same temperature. The difference in the curves can be explained by noting that lighter molecules move faster, on average, than heavier ones.
(a)
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(b) Figure 3.1 (a) The distribution of speeds for nitrogen gas at three different temperatures. At the higher temperatures, more molecules are moving at faster speeds. (b) The distribution of speeds for three gases at 300 K. At a given temperature, the lighter molecules are moving faster, on the average. 3.4. Root-Mean-Square Speed
How fast does a molecule move, on the average, at any temperature T? One way to estimate molecular speed is to calculate the root-mean-square (rms) speed (urms), which is an average molecular speed. One of the results of the kinetic theory of gases is that the total kinetic
energy of a mole of any gas equals 3
2π π. Earlier we saw that the
average kinetic energy of one molecule is 1 2 ππ’2 and so we can write
ππ΄ 12 ππ’2 =
3
2π π
Where NA is Avogadroβs number and m is the mass of a single molecule. Because NAm = M, where M is the molar mass, this equation can be rearranged to give
π’2 = 3π π
π
Taking the square root of both sides gives
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π’2 = π’πππ = 3π π
π (3.1)
Equation (3.1) shows that the root-mean-square speed of a gas increases with the square root of its temperature (in kelvins). Because M appears in the denominator, it follows that the heavier the gas, the more slowly its molecules move. If we substitute 8.314 J/K-1 mol-1 for R and convert the molar mass to kg/mol, then urms will be calculated in meters per second (m/s). Example 3.1. Calculate the root-mean-square speeds of helium atoms and nitrogen molecules in m/s at 25Β°C. Strategy
To calculate the root-mean-square speed we use π’πππ = 3π π
π
expressed in m/s and convert temperature to kelvin Solution To calculate urms, the units of R should be 8.314 J/K-1 mol-1 and, because 1 J= 1 kg m2s-2, the molar mass must be in kg/mol. The molar mass of He is 4.003 g/mol, or 4.003 Γ 10-23 kg/mol. 25Β°C = (25 + 273)K = 298 K
The calculation in Example 3.1 has an interesting relationship to the composition of Earthβs atmosphere. Unlike Jupiter, Earth does not have appreciable amounts of hydrogen or helium in its atmosphere. Why is this the case? A smaller planet than Jupiter, Earth has a weaker gravitational attraction for these lighter molecules. A fairly straightforward calculation shows that to escape Earthβs gravitational field, a molecule must possess an escape velocity equal to or greater than 1.1 Γ 104 m/s. Because the average speed of helium is considerably greater than that of molecular nitrogen or molecular oxygen, more helium atoms escape from Earthβs atmosphere into outer space. Consequently, only a trace amount of helium is present in our atmosphere. On the other hand, Jupiter, with a mass about 320 times greater than that of Earth, retains both heavy and light gases in its atmosphere.
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3.5. Ideal and Real Gases Any gas that obeys the gas laws at all temperatures and pressures is called an ideal or perfect gas. Hence, the gas the equation: PV = nRT is applicable to ideal gases only. Real gases only obey gas laws under normal conditions of temperature and pressure. 3.6. Differences between ideal gas and real gases
1. An ideal gas obeys the gas laws at all temperatures and pressure, while a real gas obeys the gas laws under normal conditions of temperature and pressure.
2. The actual volume of the molecules of an ideal gas is negligible compared with the volume of the container, while the actual volume of the molecules of a real gas is not negligible i.e. molecules of a real gas occupy space.
3. In an ideal gas there are no intermolecular attractions at all temperatures and pressures, while intermolecular attraction is strong and appreciable in a real gas at high pressure and low temperature.
3.7. Deviation from Ideal Behaviour The gas laws and the kinetic molecular theory assume that
molecules in the gaseous state do not exert any force, either attractive or repulsive, on one another. The other assumption is that the volume of the molecules is negligibly small compared with that of the container. A gas that satisfies these two conditions is said to exhibit ideal behaviour.
Although we can assume that real gases behave like an ideal gas, we cannot expect them to do so under all conditions. For example, without intermolecular forces, gases could not condense to form liquids. The important question is: Under what conditions will gases most likely exhibit nonideal behaviour?
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Figure 3.2 shows PV/RT plotted against P for three real gases and an ideal gas at a given temperature. This graph provides a test of ideal gas behaviour. According to the ideal gas equation (for 1 mole of gas), PV/RT equals 1, regardless of the actual gas pressure. (When n = 1, PV = nRT becomes PV = RT, or PV/RT = 1.) For real gases, this is true only at moderately low pressures (β€ 5 atm); significant deviations occur as pressure increases. Attractive forces operate among molecules at relatively short distances. At atmospheric pressure, the molecules in a gas are far apart and the attractive forces are negligible. At high pressures, the density of the gas increases; the molecules are much closer to one another. Intermolecular forces can then be significant enough to affect the motion of the molecules, and the gas will not behave ideally.
Another way to observe the nonideal behaviour of gases is to lower the temperature. Cooling a gas decreases the moleculesβ average kinetic energy, which in a sense deprives molecules of the drive they need to break from their mutual attraction.
To study real gases accurately, then, we need to modify the ideal gas equation, taking into account intermolecular forces and finite molecular volumes. Such an analysis was first made by the Dutch physicist J. D. van der Waals in 1873. Besides being mathematically simple, van der Waalsβs treatment provides us with an interpretation of real gas behaviour at the molecular level.
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Figure 3.2: Plot of PV/RT versus P of 1 mole of a gas at 0Β°C. For 1 mole of an ideal gas, PV/RT is equal to 1, no matter what the pressure of the gas is. For real gases, we observe various deviations from ideality at high pressures. At very low pressures, all gases exhibit ideal behavior; that is, their PV/RT values all converge to 1 as P approaches zero
Consider the approach of a particular molecule toward the wall of a container. The intermolecular attractions exerted by its neighbours tend to soften the impact made by this molecule against the wall. The overall effect is a lower gas pressure exerted by an ideal gas, Pideal, is related to the experimentally measured; that is, observed pressure, Pobs, by the equation
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Where a is a constant and n and V are the number of moles and volume of the gas, respectively. The correction term for pressure (an2/V2) can be understood as follows. The intermolecular interaction that gives rise to nonideal behavior depends on how frequently any two molecules approach each other closely. The number of such βencountersβ increases with the square of the number of molecules per unit volume, (n/V)2, because the presence of each of the two molecules in a particular region is proportional to n/V and so a is just a proportionality constant. The quantity Pideal is the pressure we would measure if there were no intermolecular attractions.
Another correction concerns the volume occupied by the gas molecules. In the ideal gas equation, V represents the volume of the container. However, each molecule does occupy a finite, although small, intrinsic volume, so the effective volume of the gas becomes (V β nb), where n is the number of moles of the gas and b is a constant. The term nb represents the volume occupied by n moles of the gas.
Having taken into account the corrections for pressure and volume, we can rewrite the ideal gas equation as follows:
Equation (3.2), relating P, V, T, and n for a nonideal gas, is known as the van der Waals equation. The van der Waals constants a and b are selected to give the best possible agreement between Equation (3.2) and observed behaviour of a particular gas. Table 3.1 lists the values of a and b for a number of gases. The value of a indicates how strongly molecules of a given type of gas attract one another. We see that helium atoms have the weakest attraction for one another, because
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helium has the smallest a value. There is also a rough correlation between molecular size and b. Generally, the larger the molecule (or atom), the greater b is, but the relationship between b and molecular (or atomic) size is not a simple one.
Example 3.3. Given that 2.75 moles of CO2 occupy 4.70 L at 53Β°C, calculate the pressure of the gas (in atm) using (a) the ideal gas equation and (b) the van der Waals equation. Solution Strategy To calculate the pressure of CO2 using the ideal gas equation and van der Waals equation, we proceed by applying the ideal gas equation and then find the correction terms in van der Waals equation and substitute in the equation to get the pressure. (a).
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Data provided: V = 4.70 L T = 53Β°C = (53 + 273) K = 326 K n = 2.75 mol R = 0.0821 L atm/K -1 mol-1 Substituting these values in the ideal gas equation, we write ππ = ππ π
P = 14.5 atm Example 3.4. Using the data shown in Table 3.1, calculate the pressure exerted by 4.37 moles of nitrogen gas confined in a volume of 2.45 L at 38Β°C using van der Waals equation. Solution Data provided V = 2.45 L T =38Β°C = (38 + 273) K = 311 K n = 4.37 mol R = 0.0821 L atm/K -1 mol-1 Now let us apply equation 3.2, to find the pressure. From Table 3.1, we have
a = 1.39 atm L2/mol2 b = 0.0913 L/mol
so that the correction terms for pressure and volume are
= 0.399 L Finally, substituting these values in the van der Waals equation, we have
(P + 4.4 atm) (2.45 L β 0.399 L)= (4.37 mol) (0.0821 L atm/K -1 mol-1) (311 K) P = 50. 0 atm
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Example 3.5. Using (a) the ideal gas law equation and (b) Van der Waalβs eqauation, cal culate the pressure exerted by 50.0g of carbon (IV) oxide in 1.00 L vessel at 25oC. [Find values of a and b in table 3.1] Solution Convert the mass of CO2 to mole and temperature to Kelvin. Data provided a = 3.592 atm L2/mol2 b = 0.043 L/mol V = 1.00 L T =25Β°C = (25 + 273) K = 298 K R = 0.0821 L atm/K -1 mol-1
Mass of CO2 = 50.0g
Number of mole of CO2 (n) = πππ π
πππππ πππ π
But molar mass of CO2 = 12 + (16 Γ 2) = 44 g/mol
β΄ (n) = 50 π
44 π/πππ= 1.136 πππ
(a). Let us substitute these values in the ideal gas equation, we write ππ = ππ π
= 0.0488 L Finally, substituting these values in the van der Waals equation, we have
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(P + 4.635 atm) (1.00 L β 0.0488 L) = (1.136 mol) (0.0821 L atm/K -1 mol-1) (298 K) P = 24.61 atm Example 3.6. Oxygen is supplied to hospital and chemical laboratories in large steel cylinders. Typically, such a cylinder has an internal volume of 28.0 litres and contains 6.80kg of oxygen. Use Van der Waalβs equation to estimate the pressure inside such cylinder at 20oC. Solution Data provided a = 1.36 atm L2/mol2 b = 0.032 L/mol V = 28.0 L T =20Β°C = (20 + 273) K = 293 K R = 0.0821 L atm/K -1 mol-1
Mass of O2 = 6.8kg = 6800g
Number of mole of O2 (n) = πππ π
πππππ πππ π
But molar mass of O2 = (16 Γ 2) = 32 g/mol
β΄ (n) = 6800 π
32 π/πππ= 212.5 πππ
Let us apply Van der Wallβs equation to find the pressure.
so that the correction terms for pressure and volume are
Finally, substituting these values in the van der Waals equation, we have
(P + 78.3 atm) (28 L β 6.8 L)= (212.5 mol) (0.0821 L atm/K -1 mol-1) (293 K) P = 162.82 atm Example 3.7. Use Van der Waalβs equation, calculate the pressure exerted by 1 mol of ammonia at 0OC in a volume of
(a) 1 .0 litre and (b) 0.05 litre
Solution Use Van der Waalβs equation. Convert temperatures to kelvins. Data provided a = 4.17 atm L2/mol2 b = 0.037 L/mol R = 0.0821 L atm/K -1 mol-1
(a) V = 1.0 L T =0Β°C = (0 + 273) K = 273 K n = 1 mol Applying Van der Waalβs equation to find the pressure;
0.0821 L atmKβ1 molβ1 Γ 273 K π + 1668 ππ‘ππππβ1 0.013 πΏ/πππ = 22.4 L atm 0.013π πΏ/πππ + 21.68 πΏ ππ‘π = 22.4 L atm 0.013π πΏ/πππ = (22.4 β 21.68 )πΏππ‘π 0.013π πΏ/πππ = 0.716 πΏππ‘π
π = 0.716 πΏππ‘π
0.013π πΏ/ππππ
= 55.0 ππ‘π πππβ1 3.8. Intermolecular Forces Intermolecular forces are attractive forces between molecules. Intermolecular forces are responsible for the nonideal behaviour of gases. They exert even more influence in the condensed phases of matterβliquids and solids. As the temperature of a gas drops, the average kinetic energy of its molecules decreases. Eventually, at a sufficiently low temperature, the molecules no longer have enough energy to break away from the attraction of neighbouring molecules.
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At this point, the molecules aggregate to form small drops of liquid. This transition from the gaseous to the liquid phase is known as condensation.
In contrast to intermolecular forces, intramolecular forces hold atoms together in a molecule. Intramolecular forces stabilize individual molecules, whereas intermolecular forces are primarily responsible for the bulk properties of matter (for example, melting point and boiling point).
Generally, intermolecular forces are much weaker than intramolecular forces. Much less energy is usually required to evaporate a liquid than to break the bonds in the molecules of the liquid. For example, it takes about 41 kJ of energy to vaporize 1 mole of water at its boiling point; but about 930 kJ of energy are necessary to break the two OβH bonds in 1 mole of water molecules. The boiling points of substances often reflect the strength of the intermolecular forces operating among the molecules.
At the boiling point, enough energy must be supplied to overcome the attractive forces among molecules before they can enter the vapour phase. If it takes more energy to separate molecules of substance A than of substance B because A molecules are held together by stronger intermolecular forces, then the boiling point of A is higher than that of B. The same principle applies also to the melting points of the substances. In general, the melting points of substances increase with the strength of the intermolecular forces.
To discuss the properties of condensed matter, we must understand the different types of intermolecular forces. Dipole-dipole, dipole-induced dipole, and dispersion forces make up what chemists commonly refer to as van der Waals forces, after the Dutch physicist Johannes van der Waals. Ions and dipoles are attracted to one another by electrostatic forces called ion-dipole forces, which are not van der Waals forces. Hydrogen bonding is a particularly strong type of dipole-dipole interaction. Because only a few elements can participate in hydrogen bond formation, it is treated as a separate category. Depending on the phase of a substance, the nature of chemical bonds,
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and the types of elements present, more than one type of interaction may contribute to the total attraction between molecules, as we will see below. 3.8.1. Dipole-Dipole Forces
Dipole-dipole forces are attractive forces between polar molecules, that is, between molecules that possess dipole moments. Their origin is electrostatic, and they can be understood in terms of Coulombβs law. The larger the dipole moment, the greater the force. 3.8.2. Ion-Dipole Forces
Coulombβs law also explains ion-dipole forces, which attract an ion (either a cation or an anion) and a polar molecule to each other. The strength of this interaction depends on the charge and size of the ion and on the magnitude of the dipole moment and size of the molecule. The charges on cations are generally more concentrated, because cations are usually smaller than anions. Therefore, a cation interacts more strongly with dipoles than does an anion having a charge of the same magnitude. 3.8.3. Dispersion Forces What attractive interaction occurs in nonpolar substances? To answer this question, let us consider the arrangement shown in Figure 3.2 below.
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Figure 3.2: (a) Spherical charge distribution in a helium atom. (b) Distortion caused by the approach of a cation. (c) Distortion caused by the approach of a dipole. If we place an ion or a polar molecule near an atom (or a nonpolar molecule), the electron distribution of the atom (or molecule) is distorted by the force exerted by the ion or the polar molecule, resulting in a kind of dipole. The dipole in the atom (or nonpolar molecule) is said to be an induced dipole because the separation of positive and negative charges in the atom (or nonpolar molecule) is due to the proximity of an ion or a polar molecule. The attractive interaction between an ion and the induced dipole is called ion-induced dipole interaction, and the attractive interaction between a polar molecule and the induced dipole is called dipole-induced dipole interaction. The likelihood of a dipole moment being induced depends not only on the charge on the ion or the strength of the dipole but also on the polarizability of the atom or moleculeβthat is, the ease with which the electron distribution in the atom (or molecule) can be distorted. Generally, the larger the number of electrons and the more diffuse the electron cloud in the atom or molecule, the greater its polarizability. By diffuse cloud we mean an electron cloud that is spread over an appreciable volume, so that the electrons are not held tightly by the nucleus.
Figure 3.3: Induced dipoles interacting with each other. Such patterns exist only momentarily; new arrangements are formed in the next instant. This type of interaction is responsible for the condensation of nonpolar gases.
Polarizability allows gases containing atoms or nonpolar molecules (for example, He and N2) to condense. In a helium atom, the electrons are moving at some distance from the nucleus. At any instant
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it is likely that the atom has a dipole moment created by the specific positions of the electrons. This dipole moment is called an instantaneous dipole because it lasts for just a tiny fraction of a second. In the next instant, the electrons are in different locations and the atom has a new instantaneous dipole, and so on. Averaged over time (that is, the time it takes to make a dipole moment measurement), however, the atom has no dipole moment because the instantaneous dipoles all cancel one another. In a collection of He atoms, an instantaneous dipole of one He atom can induce a dipole in each of its nearest neighbors (Figure 3.3). At the next moment, a different instantaneous dipole can create temporary dipoles in the surrounding He atoms. The important point is that this kind of interaction produces dispersion forces, attractive forces that arise as a result of temporary dipoles induced in atoms or molecules. At very low temperatures (and reduced atomic speeds), dispersion forces are strong enough to hold He atoms together, causing the gas to condense. The attraction between nonpolar molecules can be explained similarly.
A quantum mechanical interpretation of temporary dipoles was provided by the German physicist Fritz London in 1930. London showed that the magnitude of this attractive interaction is directly proportional to the polarizability of the atom or molecule. As we might expect, dispersion forces may be quite weak. This is certainly true for helium, which has a boiling point of only 4.2 K, or 2269Β°C. (Note that helium has only two electrons, which are tightly held in the 1s orbital. Therefore, the helium atom has a low polarizability.)
Dispersion forces, which are also called London forces, usually increase with molar mass because molecules with larger molar mass tend to have more electrons, and dispersion forces increase in strength with the number of electrons. Furthermore, larger molar mass often means a bigger atom whose electron distribution is more easily disturbed because the outer electrons are less tightly held by the nuclei. Table 3.2 compares the melting points of similar substances that consist of nonpolar molecules. As expected, the melting point increases as the number of electrons in the molecule increases. Because these are all
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nonpolar molecules, the only attractive intermolecular forces present are the dispersion forces.
In many cases, dispersion forces are comparable to or even greater than the dipole-dipole forces between polar molecules. For a dramatic illustration, let us compare the boiling points of CH3F (278.4Β°C) and CCl4 (76.5Β°C). Although CH3F has a dipole moment of 1.8 D, it boils at a much lower temperature than CCl4, a nonpolar molecule. CCl4 boils at a higher temperature simply because it contains more electrons. As a result, the dispersion forces between CCl4 molecules are stronger than the dispersion forces plus the dipole-dipole forces between CH3F molecules. (Keep in mind that dispersion forces exist among species of all types, whether they are neutral or bear a net charge and whether they are polar or nonpolar.)
Table 3.2 melting points of similar Nonpolar compounds Compound Melting point (OC) CH4 β182.5 CF4 β150.0 CCl4 β23.0 CBr4 90.0 Cl4 171.0 Example 3.8. What type(s) of intermolecular forces exist between the following pairs: (a) HBr and H2S, (b) Cl2 and CBr4, (c) I2 and NO3
β, (d) NH3 and C6H6? Strategy Classify the species into three categories: ionic, polar (possessing a dipole moment), and nonpolar. Keep in mind that dispersion forces exist between all species.
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Solution (a) Both HBr and H2S are polar molecules.
Therefore, the intermolecular forces present are dipole-dipole forces, as well as dispersion forces.
(b) Both Cl2 and CBr4 are nonpolar, so there are only dispersion forces between these molecules.
(c) I2 is a homonuclear diatomic molecule and therefore
nonpolar, so the forces between it and the ion NO3β are ion-
induced dipole forces and dispersion forces.
(d) NH3 is polar, and C6H6 is nonpolar. The forces are dipole-induced dipole forces and dispersion forces.
3.9. The Hydrogen Bond
Normally, the boiling points of a series of similar compounds containing elements in the same periodic group increase with increasing molar mass. This increase in boiling point is due to the
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increase in dispersion forces for molecules with more electrons. Hydrogen compounds of Group 4A follow this trend, as Figure 3.4 shows. The lightest compound, CH4, has the lowest boiling point, and the heaviest compound, SnH4, has the highest boiling point. However, hydrogen compounds of the elements in Groups 5A, 6A, and 7A do not follow this trend. In each of these series, the lightest compound (NH3, H2O, and HF) has the highest boiling point, contrary to our expectations based on molar mass. This observation must mean that there are stronger intermolecular attractions in NH3, H2O, and HF, compared to other molecules in the same groups. In fact, this particularly strong type of intermolecular attraction is called the hydrogen bond, which is a special type of dipole-dipole interaction between the hydrogen atom in a polar bond, such as Nβ H, Oβ H, or Fβ H, and an electronegative O, N, or F atom. The interaction is written
AβH β’β’β’β’ B or AβH β’β’β’ A A and B represent O, N, or F; AβH is one molecule or part of a molecule and B is a part of another molecule; and the dotted line represents the hydrogen bond. The three atoms usually lie in a straight line, but the angle AHB (or AHA) can deviate as much as 30Β° from linearity. Note that the O, N, and F atoms all possess at least one lone pair that can interact with the hydrogen atom in hydrogen bonding.
The average energy of a hydrogen bond is quite large for a dipole-dipole interaction (up to 40 kJ/mol). Thus, hydrogen bonds have a powerful effect on the structures and properties of many compounds. Figure 3.5 shows several examples of hydrogen bonding.
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Figure 3.4: Boiling points of the hydrogen compounds of Groups 4A, 5A, 6A, and 7A elements. Although normally we expect the boiling point to increase as we move down a group, we see that three compounds (NH3, H2O, and HF) behave differently. The anomaly can be explained in terms of intermolecular hydrogen bonding.
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Figure 3.5: Hydrogen bonding in water, ammonia, and hydrogen fluoride. Solid lines represent covalent bonds, and dotted lines represent hydrogen bonds.
The strength of a hydrogen bond is determined by the coulombic interaction between the lone-pair electrons of the electronegative atom and the hydrogen nucleus. For example, fluorine is more electronegative than oxygen, and so we would expect a stronger hydrogen bond to exist in liquid HF than in H2O. In the liquid phase, the HF molecules form zigzag chains:
The boiling point of HF is lower than that of water because each H2O takes part in four intermolecular hydrogen bonds. Therefore, the forces holding the molecules together are stronger in H2O than in HF. Example 3.9. Which of the following can form hydrogen bonds with water? CH3OCH3, CH4, F
β, HCOOH, Na+.
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Strategy A species can form hydrogen bonds with water if it contains one of the three electronegative elements (F, O, or N) or it has an H atom bonded to one of these three elements. Solution There are no electronegative elements (F, O, or N) in either CH4 or Na+. Therefore, only CH3OCH3, Fβ and, HCOOH can form hydrogen bonds with water.
The intermolecular forces discussed so far are all attractive in
nature. Keep in mind, though, that molecules also exert repulsive forces on one another. Thus, when two molecules approach each other, the repulsion between the electrons and between the nuclei in the molecules comes into play. The magnitude of the repulsive force rises very steeply as the distance separating the molecules in a condensed phase decreases. This is the reason that liquids and solids are so hard to compress. In these phases, the molecules are already in close contact with one another, and so they greatly resist being compressed further. Example 3.10. What intermolecular forces besides dispersion forces, if any, exist in each substance? Are any of these substances solids at room temperature? 1. potassium chloride (KCl) 2. ethanol (C2H5OH) 3. bromine (Br2)
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Solution 1. Potassium chloride is composed of ions, so the intermolecular interaction in potassium chloride is ionic forces. Because ionic interactions are strong, it might be expected that potassium chloride is a solid at room temperature. 2. Ethanol has a hydrogen atom attached to an oxygen atom, so it would experience hydrogen bonding. If the hydrogen bonding is strong enough, ethanol might be a solid at room temperature, but it is difficult to know for certain. (Ethanol is actually a liquid at room temperature.) 3. Elemental bromine has two bromine atoms covalently bonded to each other. Because the atoms on either side of the covalent bond are the same, the electrons in the covalent bond are shared equally, and the bond is a nonpolar covalent bond. Thus, diatomic bromine does not have any intermolecular forces other than dispersion forces. It is unlikely to be a solid at room temperature unless the dispersion forces are strong enough. Bromine is a liquid at room temperature. 3.10. Intermolecular Forces at Low Temperature, High Molecular Weight, and High Pressure
At low temperatures where the gas molecules have lower kinetic energies, the contribution for attractive forces increases, which the ideal gas law does not account for.
The universal attractive force, or London dispersion force, also generally increases with molecular weight. The London dispersion force is caused by correlated movements of the electrons in interacting molecules. Electrons that belong to different molecules start "fleeing" and avoiding each other at the short intermolecular distances, which is frequently described as formation of "instantaneous dipoles" that attract each other.
Finally, as a gas is compressed and pressure increases, repulsive forces from the gas molecules oppose the decrease in
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volume. The frequency of collisions also increases at higher pressure, thereby increasing the contribution of these intermolecular forces.
3.11. The Mean Free Path of Gas Molecules
The motion of a molecule in a gas is complicated. Besides colliding with the walls of the confinement vessel, the molecules collide with each other. A useful parameter to describe this motion is the mean free path. The mean free path is the average distance traversed by a molecule between collisions. The mean free path of a molecule is related to its size; the larger its size the shorter its mean free path.
Suppose the gas molecules are spherical and have a diameter d. Two gas molecules will collide if their centers are separated by less than 2d. Suppose the average time between collisions is βt. During this time, the molecule travels a distance v . βt, and sweeps a volume equal to see diagrams below
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If on average it experiences one collision, the number of molecules in the volume V must be 1. If N is the number of molecules per unit volume, this means that
or
The time interval βt defined in this manner is the mean time between collisions, and the mean free path is given by
Here we have assumed that only one molecule is moving while all others are stationary. If we carry out the calculation correctly (all molecules moving), the following relation is obtained for the mean free path:
The Number Density
For N1 stationary particles, the number of molecules per unit volume
V
N
V
NNd
1
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Collision Frequency
Z1 = ππ2ππ
The Mean Collision Time
The mean collision time is average time elapsed between successive collisions.
coll 1
π1 =
1
ππ2ππ
Factors affecting mean free path
1. Density: As gas density increases, the molecules become closer to each other. Therefore, they are more likely to run into each other, so the mean free path decreases.
2. Radius of molecule: increasing the radius of the molecules will decrease the space between them, causing them to run into each other thereby decreasing the mean free path.
3. Pressure, temperature, and other factors that affect density can indirectly affect mean free path.
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CHAPTER FOUR
THE MOLE CONCEPT
4.1 Introduction Most chemical experiments involve enormous numbers of
atoms or molecules. In order to estimate the quantities of various chemical substances, Chemists adopted a convenient concept involving collection of elementary units such as atoms, molecules, ions, or electrons of a chemical substance. This is concept is βthe moleβ.
This quantity is sometimes referred to as the chemical amount. In Latin mole means a "massive heap" of material. It is convenient to think of a chemical mole as such. Visualizing a mole as a pile of particles, however, is just one way to understand this concept. A sample of a substance has a mass, volume (generally used with gases), and number of particles that is proportional to the chemical amount (measured in moles) of the sample. For example, one mole of oxygen gas (O2 ) occupies a volume of 22.4 L at standard temperature and pressure (STP; 0Β°C and 1 atm), has a mass of 31.998 grams, and contains about 6.022 Γ 10 23 molecules of oxygen. Measuring one of these quantities allows the calculation of the others and this is frequently done in stoichiometry.
The mole is the amount of a chemical substance which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12; its symbol is "mol." When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.
The term "mole" commonly is used to represent the number of molecules (or atoms) in a quantity of material; that is, one mole of molecules = Avogadro's number of molecules. In this sense, a mole is a dimensionless number, just as a dozen means 12. For example, you could talk about a mole of caterpillars, meaning 6.023 x 1023
caterpillars. However, the number is seldom useful except in talking about molecules or atoms. The metric prefixes commonly are used to give such units as millimoles, nanomoles, or picomoles. 4.2. Avogadro's Number
We have noted that one mole of a substance always contains a certain number of molecules (or atoms), regardless of the substance involved. This number is called Avogadroβs number (in honor of the scientist who first suggested the concept, long before the value of the number could be determined). The number represented as NA is 6.023 x 1023. We know that the number of particles (atoms or molecules) in 1 mole of a substance is 6.022Γ1023 atoms or molecules. We can therefore say: 1 mole of carbon atoms weighs 12.0 g and contains 6.023 x 1023 atoms 1 mole of sodium atoms weighs 23.0 g and contains 6.023 x 1023 atoms 4.3. Calculations Using Mole Concept
Example 4.1. How many molecules are there in 20.0 g of benzene, C6H6? Solution:
First find how many moles of C6H6 there are in 20.0 g, then use
Avogadro's number to find the number of molecules.
Molar mass of C6H6 = (6 x 12.0) + (6 x 1.0)
= 78.0 g/mol
Moles of C6H6 = πππ π ππ πΆ6π»6
Number of molecules = moles Γ Avogadroβs number
=(0.256 πππππ )(6.023 x 1023 molecules/mole)
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= 1.54 Γ 1023 molecules
Example 4.2. How many atoms are represented by 3.00 moles of calcium (Ca) ? Solution:
3.00 mol Ca Γ 6.023 x 1023
1 .00 mol Ca
= 1.81 x 1024 atoms Ca
Example 4.3. How many atoms are represented by 3 g of calcium (Ca) ?
[Ca = 40, Avogadro constant = 6.023 x 1023]
Solution
40 g (1 mol) of calcium contains 6.023 x 1023 atoms
1 g of calcium will contain 6.023 x 1023
40 atoms
β΄ 3 g will contain 6.023 x 1023 Γ 3
40 atoms
= 4.5 Γ 1022 atoms
Example 4.4. What is the mass of 6.02 Γ 1024 atoms of magnesium many atoms are represented by 3 g of calcium (Ca) ? [Mg = 24, Avogadro constant = 6.023 x 1023] Solution 6.023 x 1023 atoms of magnesium weigh 24 g (1 mol)
β΄ 6.02 Γ 1024 atoms weigh 6.02 x 1024 Γ24
6.023 x 1023
= (24 Γ 10)g
= 240 g
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Example 4.5. How many moles are in 9.03 Γ 1023 atoms of sodium? [Na = 23, Avogadro constant = 6.023 x 1023] Solution Let y represent the amount of Na 6.023 x 1023 atoms β‘ 1 mole 9.03 Γ 1023 atoms β‘ y mole
y = 9.03 x 1023 Γ 1.0
6.023 x 1023
= 1.50 mol Example 4.6. Calculate the mass of sodium, which would contain the same number of atoms as 9.0 g of carbon? [Na = 23, C = 12] Solution Since equal amount of two or more elements contain the same number of atoms, then: 1 mole (12.0 g) of carbon contains the same number of atoms as 1 mole (23.0 g) of sodium. i.e. 12.0 g of carbon β‘ 23.0 g of sodium
β΄ 9.0 g of carbon β‘ 9.0 Γ 23.0
12
= 30.0 g of sodium
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CHAPTER FIVE
GASES IN CHEMICAL REACTION 5.1. Introduction
Gases that are involved in chemical reactions obey the same laws of stoichiometry that apply to substances in any other state therefore, the ideal gas law can be used to calculate the quantities of gaseous substances involved in a reaction and then those results used to find the quantities of other substances. Figure 5.1 presents the conversions allowed by the ideal gas law to determine the number of moles of a gaseous reactant or product.
Figure 5.1: Mole Conversions, Including Application of the Ideal Gas Law to Determine the Number of Moles of a Gaseous Reactant or Product. Example 5.1. How many liters of oxygen gas at 21oC and 1.13 atm can be prepared by thermal decomposition of 0.950 g of KClO3? [K = 39, Cl = 35.5, O = 16]
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Solution We first find the number of mole of KClO3 from which the number of mole of oxygen can be determine through the balanced equation and finally apply ideal gas equation to find the volume of oxygen that can be prepared.
Solution
Data provided and necessary conversion
T = 21oC = (21 +273)K = 294K
P = 1.13 atm
Mass of KClO3 = 0.950 g
V = ?
Molar mass of KClO3 = [39 + 35.5 + (16Γ 3)] = 122.5 g/mol
Example 5.2. A chemist decomposes 1.06 g of Hg2O in a sealed system. The oxygen produced has a pressure of 0.514 atm and a volume of 62.5 mL at 35Β°C Calculate the value of R from these data. Solution Data provided and necessary conversion T = 35oC = (35 +273)K = 308K P = 0.514 atm Mass of Hg2O = 1.06 g V = 62.5 mL = (62.5/1000) L = 0.0625 L Molar mass of Hg2O = [ (200.5 Γ 2 )+ 16] = 417 g/mol
Mole of Hg2O (n) = πππ π
πππππ πππ π
= 1.06 g
417 π/πππ
= 0.00254 mol of Hg2O
The number of moles of O2 produced is
0.00254 mol of Hg2O (1 πππ π2
2 πππ π»π2π) = 0.00127 mol O2
π = ππ
ππ =
0.514 atm (0.0625 L)
0.00127 πππ (308 πΎ)
= 0.0821 L atm K-1mol-1
Example 5.3. As N2 and H2 react to form NH3 in a large cylinder at 500Β°C, what happens to (a) the total number of atoms? (b) the total number of molecules? (c) the total pressure? Solution (a) The number of atoms stays the same, as is true for all reactions. That is the basis for the balanced chemical equation.
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(b) N2 (g) + 3H2 (g) β 2NH3 (g) The total number of moles of gas decreases as this reaction proceeds, so the number of molecules also decreases. (c) The total pressure decreases as the total number of moles of gas decreases. Example 5.4. How many litres of CO2 at STP can be prepared by the complete thermal decomposition of 0.150 mol of Ca(HCO3)2. The products are CaO, CO2 and H2O. Solution Data provided T = 273K P = 1.0 atm R = = 0.0821 L atm K-1mol-1 Mole of Ca(HCO3)2 = 0.150 mol V = ?
The number of moles of CO2 produced is
0.150 mol of Ca(HCO3)2 (2 πππ π2
1 πππ πΆπ(π»πΆπ3)2) = 0.30 mol CO2
6.1. Introduction Few chemical reactions proceed in only one direction. Most are,
at least to some extent, reversible. At the start of a reversible process, the reaction proceeds toward the formation of products. As soon as some product molecules are formed, the reverse processβthat is, the formation of reactant molecules from product moleculesβbegins to take place. When the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products no longer change with time, chemical equilibrium is reached. Chemical equilibrium is a dynamic process. As such, the rate of product formation and conversion back to reactant molecules is constant. Note that a chemical equilibrium reaction involves different substances as reactants and products. Equilibrium between two phases of the same substance is called physical equilibrium because the changes that occur are physical processes. The vaporization of water in a closed container at a given temperature is an example of physical equilibrium. In this instance, the number of H2O molecules leaving and the number returning to the liquid phase are equal: H2O(l) β H2O(g)
(The double arrow means that the reaction is reversible.) The study of physical equilibrium yields useful information,
such as the equilibrium vapour pressure. However, chemists are particularly interested in chemical equilibrium processes, such as the reversible reaction involving nitrogen dioxide (NO2) and dinitrogen tetroxide (N2O4). The progress of the reaction
N2O4(g) β 2NO2(g) can be monitored easily because N2O4 is a colorless gas, whereas NO2 has a darkbrown colour that makes it sometimes visible in polluted air. Suppose that a known amount of N2O4 is injected into an evacuated flask. Some brown colour appears immediately, indicating the
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formation of NO2 molecules. The colour intensifies as the dissociation of N2O4 continues until eventually equilibrium is reached. Beyond that point, no further change in colour is observed. By experiment we find that we can also reach the equilibrium state by starting with pure NO2 or with a mixture of NO2 and N2O4. In each case, we observe an initial change in colour, caused either by the formation of NO2 (if the colour intensifies) or by the depletion of NO2 (if the colour fades), and then the final state in which the colour of NO2 no longer changes. Depending on the temperature of the reacting system and on the initial amounts of NO2 and N2O4, the concentrations of NO2 and N2O4 at equilibrium differ from system to system (Figure 6.1).
Figure 6.1: Change in the concentrations of NO2 and N2O4 with time, in three situations. (a) Initially only NO2 is present. (b) Initially only N2O4 is present. (c) Initially a mixture of NO2 and N2O4 is present. In each case, equilibrium is established to the right of the vertical line. The NO2βN2O4 System at 6.2. The Equilibrium Constant
Let us consider the following reversible reaction: aA + bB βcC + d D
in which a, b, c, and d are the stoichiometric coefficients for the reacting species A, B, C, and D. The equilibrium constant for the reaction at a particular temperature is
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πΎ = [πΆ]π [π·]π
[π΄]π [π΅]π ( 6.1)
Equation (6.1) is the mathematical form of the law of mass action. It relates the concentrations of reactants and products at equilibrium in terms of a quantity called the equilibrium constant. The equilibrium constant is defined by a quotient. The numerator is obtained by multiplying together the equilibrium concentrations of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation. The same procedure is applied to the equilibrium concentrations of reactants to obtain the denominator. This formulation is based on purely empirical evidence, such as the study of reactions like NO2βN2O4.
The equilibrium constant has its origin in thermodynamics, however, we can gain some insight into K by considering the kinetics of chemical reactions. Let us suppose that this reversible reaction occurs via a mechanism of a single elementary step in both the forward and reverse directions:
in which kf and kr are the rate constants for the forward and reverse directions, respectively. At equilibrium, when no net changes occur, the two rates must be equal:
ratef = rater Or
kf[A][B]2 = kr[AB2]
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πΎπ
πΎπ=
[π΄π΅2]
π΄ [π΅]2
Because both kf and kr are constants at a given temperature, their ratio is also a constant, which is equal to the equilibrium constant Kc.
πΎπ
πΎπ= πΎπ =
[π΄π΅2]
π΄ [π΅]2
So Kc is a constant regardless of the equilibrium concentrations of the reacting species because it is always equal to kf/kr, the quotient of two quantities that are themselves constant at a given temperature. Because rate constants are temperature-dependent, it follows that the equilibrium constant must also change with temperature.
Finally, we note that if the equilibrium constant is much greater than 1 (that is, K > 1), the equilibrium will lie to the right of the reaction arrows and favour the products. Conversely, if the equilibrium constant is much smaller than 1 (that is, K Λ 1), the equilibrium will lie to the left and favour the reactants (Figure 6.2).
Figure 6.2: (a) At equilibrium, there are more products than reactants, and the equilibrium is said to lie to the right. (b) In the opposite situation, when there are more reactants than products, the equilibrium is said to lie to the left.
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6.3. Ways of Expressing Equilibrium Constants To use equilibrium constants, we must express them in terms of
the reactant and product concentrations. Our only guidance is the law of mass action [Equation (6.1)]. However, because the concentrations of the reactants and products can be expressed in different units and because the reacting species are not always in the same phase, there may be more than one way to express the equilibrium constant for the same reaction. To begin with, we will consider reactions in which the reactants and products are in the same phase. 6.4. Homogeneous Equilibria
The term homogeneous equilibrium applies to reactions in which all reacting species are in the same phase. An example of homogeneous gas-phase equilibrium is the dissociation of N2O4. The equilibrium constant is
πΎπ = [ππ2]2
[π2π4]
Note that the subscript in Kc denotes that the concentrations of the reacting species are expressed in moles per liter. The concentrations of reactants and products in gaseous reactions can also be expressed in terms of their partial pressures. At constant temperature the pressure P of a gas is directly related to the concentration in moles per liter of the gas; that is, P = (n/V)RT. Thus, for the equilibrium process
N2O4(g) β 2NO2(g)
We can write
πΎπ = π2ππ2
ππ2π4
in which Pππ2 and ππ2π4 are the equilibrium partial pressures (in atmospheres) of NO2 and N2O4, respectively. The subscript in KP tells us that equilibrium concentrations are expressed in terms of pressure.
In general, Kc is not equal to Kp, because the partial pressures of reactants and products are not equal to their concentrations expressed
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in moles per liter. A simple relationship between KP and Kc can be derived as follows. Let us consider this equilibrium in the gas phase:
aA(g) β bB(g)
in which a and b are stoichiometric coefficients. The equilibrium constant Kc is
πΎπ = [π΅]π
[π΄]π
and the expression for KP is
πΎπ = πππ΅
πππ΄
in which PA and PB are the partial pressures of A and B. Assuming ideal gas behaviour, ππ΄π = ππ΄π π
ππ΄ = ππ΄π π
π
in which V is the volume of the container in liters. Also, ππ΅π = ππ΅π π
ππ΅ = ππ΅π π
π
Substituting these relations into the expression for KP, we obtain
Now both nA/V and nB/V have the units of moles per liter and can be replaced by [A] and [B], so that
in which
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βπ = π β π = moles of gaseous products β moles of gaseous reactants
Because pressure is usually expressed in atmospheres, the gas constant R is given by 0.0821 L atm/K mol, and we can write the relationship between KP and Kc as
πΎπ = πΎπΆ(0.0821T)βπ (6.2) In general, KP β Kc except in the special case when βn = 0. In that case, Equation (5.2) can be written as πΎπ = πΎπΆ(0.0821T)0 πΎπ = πΎπΆ 6.5. Equilibrium Constant and Units
Note that it is general practice not to include units for the equilibrium constant. In thermodynamics, the equilibrium constant is defined in terms of activities rather than concentrations. For an ideal system, the activity of a substance is the ratio of its concentration (or partial pressure) to a standard value, which is 1 M (or 1 atm). This procedure eliminates all units but does not alter the numerical parts of the concentration or pressure. Consequently, K has no units. Example 6.1. Write expressions for KP if applicable, for the following reversible reactions at equilibrium: (a). HF (aq) + H2O (l) β H3O+ (aq) + F- (aq)
(b). 2NO (g) + O2(g) β 2NO2 (g)
Strategy Keep in mind the following facts: (1) the KP expression applies only to gaseous reactions and (2) the concentration of solvent (usually water) does not appear in the equilibrium constant expression. Solution
(a) Because there are no gases present, KP does not apply.
(b) πΎπ = π2ππ2
π2ππ ππ2
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Example 6.2 The equilibrium constant KP for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorine PCl5(g) β PCl3(g) + Cl2(g)
is found to be 1.05 at 250Β°C. If the equilibrium partial pressures of PCl5 and PCl3 are 0.973 atm and 0.548 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250Β°C. Strategy The concentrations of the reacting gases are given in atm, so we can express the equilibrium constant in KP. From the known KP value and the equilibrium pressures of PCl3 and PCl5, we can solve for PπΆπ2. Solution Data given Partial pressures of PCl5 = 0.973 atm Partial pressures of PCl3 = 0.548 atm πΎπ = 1.05 Partial pressures of Cl2 = ? First, we write KP in terms of the partial pressures of the reacting species
πΎπ = πππΆπ 3 ππΆπ2
ππΆπ5
Knowing the partial pressures, we write
1.05 = 0.548 ( ππΆπ2 )
(0.973)
ππΆπ2=
1.05 Γ0.973
0.548
= 1.86 atm Example 6.3 The equilibrium constant KP for the reaction 2NO2 β 2NO + O2 (g)
is 158 at 1000 K. Calculate PO2 if PNO2 = 0.400 atm and PNO = 0.270 atm. Solution Data given
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Partial pressures of NO2 = 0.400 atm Partial pressures of NO = 0.270 atm πΎπ = 158 Partial pressures of O2 =?
πΎπ = π2
ππ ππ2
π2ππ 2
158 = (0.270)2 Γ ππ2
(0.400)2
ππ2 =
0.16 (158)
0.073
= 346 atm Example 6.4 Methanol (CH3OH) is manufactured industrially by the reaction CO(g) + 2H2(g) β CH3OH(g)
The equilibrium constant (Kc) for the reaction is 10.5 at 220Β°C. What is the value of KP at this temperature? Strategy Apply the relationship between Kc and KP is given by Equation (6.2). What is the change in the number of moles of gases from reactants to product? Recall that βn = moles of gaseous products β moles of gaseous reactants. Convert temperature to kelvins Solution Data given Kc = 10.5
T = 220oC = (220 + 273)K = 493 K
βn = (1β3) = β2
Applying πΎπ = πΎπΆ(0.0821T)βπ
πΎπ = (10.5)(0.0821 Γ 493)-2
= 6.41Γ 10-3
Example 6.5 For the reaction N2(g) + 3H2(g) β 2NH3(g)
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KP is 4.3 Γ 10-4 at 375Β°C. Calculate Kc for the reaction. Solution
Data given
Kp = 4.3 Γ 10-4
T = 375oC = (375 + 273)K = 648 K
βn = (2β4) = β2
Applying πΎπ = πΎπΆ(0.0821T)βπ
= πΎπΆ (0.0821 Γ 648)-2
4.3 Γ 10-4 = πΎπΆ (3.5 Γ 10-4)
πΎπΆ = 4.3 Γ 10β4
3.5 Γ 10β4
= 1.2
6.6. Heterogeneous Equilibria A reversible reaction involving reactants and products that are in
different phases leads to a heterogeneous equilibrium. For example, when calcium carbonate is heated in a closed vessel, this equilibrium is attained:
CaCO3(s) β CaO(s) + CO2(g)
The two solids and one gas constitute three separate phases. At equilibrium, we might write the equilibrium constant in terms of partial pressure for gases as πΎπ = ππΆπ2
The equilibrium constant in this case is numerically equal to the pressure of CO2 gas, an easily measurable quantity. Example 6.6 Consider the following heterogeneous equilibrium:
CaCO3(s) β CaO(s) + CO2(g)
At 800Β°C, the pressure of CO2 is 0.236 atm. Calculate (a) KP and (b) Kc for the reaction at this temperature.
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Strategy Remember that pure solids do not appear in the equilibrium constant expression. The relationship between KP and Kc is given by Equation (6.2). Solution (a) Using πΎπ = ππΆπ2
= 2.68 Γ 10-3 Example 6.7 Consider the following equilibrium at 395 K:
NH4HS(s) β NH3(g) + H2S(g) The partial pressure of each gas is 0.265 atm. Calculate KP and Kc for the reaction. Solution
Data given
T = 395oC = (395 + 273)K = 670 K
βn = 2
πππ»3= 0.265
ππ»2π = 0.265
Kp = ?
Kc = ?
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πΎπ = πππ»3 ππ»2π
= 0.265 (0.265)
= 0.07
Applying πΎπ = πΎπΆ(0.0821T)βπ
= πΎπΆ (0.0821 Γ 670)2
0.07 = πΎπΆ (3025)
πΎπΆ = 0.07
3025
= 2.3 Γ 10-5
Example 6.8 Starting with a 3: 1 mixture of H2 and N2 at 450.0Β°C, the equilibrium mixture is found to be 9.6% NH3, 22.6% N2, and 67.8% H2 by volume. The total pressure is 50.0 atm. Calculate KP and Kc.
The reaction is N2 + 3H2 β 2NH3. Solution: According to Dalton's law of partial pressures, the partial pressure of a gas in a mixture is given by the product of its volume fraction and the total pressure. Therefore the equilibrium pressure of each gas is πππ»3
= (0.096)(50.0 atm) = 4.8 atm
ππ2 = (0.226)(50.0 atm) =11.3 atm
ππ»2 = (0.678)(50.0 atm) = 33.9 atm
Total pressure = 50.0 atm T = 450.0Β°C = (450 + 273)K = 723 K βn = (2β4) = β2 The equilibrium expression for the reaction is
πΎπ = π2
ππ» 3
π3π»2 ππ2
Substituting the data gives
= (4.80)2
(33.9)3 Γ(11.3)
= 23.04
(440227 )
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= 5.2 Γ 10-5
Applying πΎπ = πΎπΆ(0.0821T)βπ to find KC
= πΎπΆ (0.0821 Γ 723)-2
5.2 Γ 10-5 = πΎπΆ (2.8 Γ 10-4)
πΎπΆ = 5.2 Γ 10β5
2.8 Γ 10β4
= 0.184 6.7. The Effect of Change in Partial Pressure of One Gas on Kp and Equilibrium Position
It often is important to know the yield of a chemical reactionβthat is, the percentage of reactants converted to products. The following example shows how this yield may be calculated, and how conditions may be altered to increase the yield. Example 6.9 Kp = 54.4 at 355.0Β°C for the reaction H2 + I2 β 2HI. What percentage of I2 will be converted to HI if 0.20 mole each of H2 and I2 are mixed and allowed to come to equilibrium at 355.0Β°C and a total pressure of 0.50 atm? Solution: Assume that X moles each of H2 and I2 are used up in reaching equilibrium to give 2X moles of HI, in accordance with the chemical equation, leaving 0.20 - X moles each of H2 and I2. The partial pressure of each gas is given by the product of its mole fraction and the total pressure.
ππ»πΌ =
2π
0.4 (0.50 ππ‘π)
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ππ»2= ππΌ2
= 0.2 βπ
0.4 (0.50 ππ‘π)
πΎπ = π2
π»πΌ
ππ»2 ππΌ2
= [
2π
0.4 (0.50 ππ‘π )]2
[ 0.2 βπ
0.4 (0.50 ππ‘π )]2
54.4 = (2π
0.2 βπ)2
Taking the square root of each side, we obtain
7.4 = 2π
0.2 βπ
2X = 7.4 (0.2 β X) 2X = 1.48 β 7.4X 9.4X = 1.48 X = 0.157 = moles of H2 and I2 used up
Percentage conversion (yield) = 0.157
0.200 Γ 100 = 78.5%
Example 6.10 What percentage of I2 will be converted to HI at equilibrium at 355.0Β°C, if 0.200 mole of I2 is mixed with 2.00 moles of H2 at total pressure of 0.50 atm? Solution: In this problem, it is advantageous first to assume that the large excess of H2 will use almost the entire amount of I2, leaving only X moles of it unused. In general, it is always advantageous to let X represent the smallest unknown entity because it often simplifies the mathematical solution. If X moles of I2 are not used, then 0.20 β X moles are used. For every mole of I2 used up, one of H2 is used up, and two of HI are formed. Proceeding as in the last problem, the number of moles of each component at equilibrium is
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The partial pressure of each component will be the mole fraction of each times the total pressure, as follows.
ππ»2=
1.80+ π
2.20 (0.50 ππ‘π)
ππΌ2 =
π
2.20 (0.50 ππ‘π)
ππ»πΌ = 0.40 β 2π
2.20 (0.50 ππ‘π)
When we substitute these partial pressures into the expression for Kp, we get an expression that will be tedious to solve unless we make a reasonable approximation: we assume that X is negligible in comparison with 0.20 and 1.80.
πΎπ = 54.4 = [
0.40 β 2π
2.20 (0.50 )]2
1.80+ π
2.20 (0.50 ) [
π
2.20 (0.50 )]
β (0.40)2
πΌ.80 (π)
X = (0.40)2
πΌ.80 (54.4)
= 0.0016 moles of I2 not used
0.200 β 0.0016 = 0.1984 moles of I2 used
Percentage of I2 used = 0.1984
0.200 Γ 100 = 99.2%
Note that the wise decision to let X = the amount of I2 not used instead of the amount of I2 that was used really did simplify the solution by making it possible to neglect X when added to or subtracted from larger numbers. If we had solved the quadratic equation instead, we
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would have found that 99.197% of the I2 had been used up. This is a common method of simplifying a mathematical problem, and at the end you can always check to see whether your answer really is negligible compared to what you said it was. Many chemists say that if X is less than 10.0% of what it is added to or subtracted from, it is okay to neglect it. The preceding problem illustrates the fact that, although the value of Kp does not change with changes in concentration, the equilibrium position will change to use up part of the excess of any one reagent that has been added. In this problem, the large excess of H2 shifts the equilibrium position to the right, causing more of the I2 to be used up (99.2% compared to 78.5%) than when H2 and I2 are mixed in equal proportions. Advantage may be taken of this principle by using a large excess of a cheap chemical to convert the maximum amount of an expensive chemical to a desired product. In this case I2, the more expensive chemical, is made to yield more HI by using more of the cheaper H2. 6.8. The Percentage Decomposition of Gases
Many gases decompose into simpler ones at elevated temperatures, and it often is important to know the extent to which decomposition takes place. Example 6.11 Kp = 1.78 atm at 250.0Β°C for the decomposition reaction PCl5(g) β PCl3(g) + Cl2(g). Calculate the percentage of PC15 that dissociates if 0.0500 mole of PC15 is placed in a closed vessel at 250.0Β°C and 2.00 atm pressure. Solution: Although you are told that you are starting with 0.0500 mole PCl5, this piece of information is not needed to find the percentage dissociation at the given pressure and temperature. If you were asked for the volume of the reaction vessel, then you would need to know the actual number of moles; otherwise not. To answer the question that is asked, it is simpler to just start with one mole (don't worry about the volume)
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and assume that X moles of PCl5 dissociate to give X moles each of PCl3 and Cl2 and 1 - X moles of PCl5 at equilibrium.
The partial pressures are given by the mole fractions times the total pressure, and are substituted into the Kp expression, to give
πΎπ = 1.78 =
π
1+π 2.0 ππ‘π
π
1+π 2.0 ππ‘π
1 βπ
1+π 0.50 ππ‘π
= 2π2
1βπ (1+π)
1.78 = 2π2
1β π2
1.78 β 1.78X2 = 2X2
π2 = 1.78
3.78= 0.478
π = 0.478
= 0.686 moles PCl5 dissociate
Percentage of PCl5 dissociated = 0.686
1.00 Γ 100 = 68.6%
This was not a difficult quadratic equation to solve but, even if it had been, it would not be possible to neglect X compared to 1.00; it is too large. If we had neglected X, we would have obtained the extremely erroneous answer of 94.3% dissociated. If Kp is very large (or very small), it means that the equilibrium position lies far to the right (or to
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the left). In either of these cases it is possible to choose X so that it will be very small and amenable to a simplified math solution. The value of Kp for the PCl5 equilibrium is neither very large nor very small, and hence it never will be possible to neglect X.
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CHAPTER SEVEN
WORK OF EXPANSION AND COMPRESSION OF GAS
7.1. Introduction
We have seen that work can be defined as force F multiplied by distance d: w = Fd
In thermodynamics, work has a broader meaning that includes mechanical work (for example, a crane lifting a steel beam), electrical work (a battery supplying electrons to light the bulb of a flashlight), and surface work (blowing up a soap bubble). In this section we will concentrate on mechanical work.
One way to illustrate mechanical work is to study the expansion or compression of a gas. Many chemical and biological processes involve gas volume changes. Breathing and exhaling air involves the expansion and contraction of the tiny sacs called alveoli in the lungs. Another example is the internal combustion engine of the automobile. The successive expansion and compression of the cylinders due to the combustion of the gasoline-air mixture provide power to the vehicle. Figure 6.1 shows a gas in a cylinder fitted with a weightless, frictionless movable piston at a certain temperature, pressure, and volume. As it expands, the gas pushes the piston upward against a constant opposing external atmospheric pressure P. The work done by the gas on the surroundings is
Work done = π Γ π΄ Γ π
But π΄ Γ π = π
Therefore π€ = πβπ
w = βPβV (7.1)
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Where βV, the change in volume, is given by Vf β Vi. The minus sign in Equation (7.1) takes care of the sign convention for w. For gas expansion (work done by the system), βV > 0, so βPβV is a negative quantity. For gas compression (work done on the system), βV Λ 0, and βPβV is a positive quantity. Note that ββPβVβ is often referred to as βP-Vβ work.
Figure 7.1: The expansion of a gas against a constant external pressure (such as atmospheric pressure). The gas is in a cylinder fitted with a weightless movable piston. The work done is given by βPβV. According to Equation (6.1), the units for work done by or on a gas are liters atmospheres. To express the work done in the more familiar unit of joules, we use the conversion factor 1 L . atm = 101.3 J Example 7.1. A certain gas expands in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 1.2 atm. Strategy A simple sketch of the situation is helpful here:
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The work done in gas expansion is equal to the product of the external, opposing pressure and the change in volume. Solution (a) Because the external pressure is zero, no work is done in the expansion:
w = βPβV = β (0)(6.0 β 2.0) L = 0 (b) The external, opposing pressure is 1.2 atm, so
w = βPβV = β (1.2 atm) (6.0β 2.0) L
= β 4.8 L . atm To convert the answer to joules, we write
w = β 4.8 L . atm Γ 101.3π½
1 πΏ.ππ‘π
= β 4.9 Γ 102 J Because this is gas expansion (work is done by the system on the surroundings), the work done has a negative sign. Example 7.2. A gas expands from 264 mL to 971 mL at constant temperature. Calculate the work done (in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 4.00 atm. Solution This is similar to above example. Convert volume to litre.
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(a) Because the external pressure is zero, no work is done in the expansion:
w = βPβV
= β (0)(971 β 264) mL
= β (0)(527/1000) L
= 0
(b) The external, opposing pressure is 4.0 atm, so
w = βPβV
= β (4.0 atm) (971 β 264) Ml
=β (4.0 atm) (971 β 264)
=β (4.0 atm)(527/1000) L
= β (4.0 atm)(0.527) L
= β 2.108 L. atm
To convert the answer to joules, we write
w = β 2.108 L . atm Γ 101.3π½
1 πΏ.ππ‘π
= β 2.13 Γ 102 J Example 7.3. The work done when a gas is compressed in a cylinder is 387 J. During this process, there is a heat transfer of 152 J from the gas to the surroundings. Calculate the energy change for this process. Strategy Compression is work done on the gas, so what is the sign for w? Heat is released by the gas to the surroundings. Is this an endothermic or exothermic process? What is the sign for q? Solution To calculate the energy change of the gas, we need βU = q + w. Work of compression is positive and because heat is released by the gas, q is negative. Therefore, we have
βU = q + w
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= β152 J + 387 J = 235 J
As a result, the energy of the gas increases by 235 J. Example 7.4. A gas expands and does P-V work on the surroundings equal to 279 J. At the same time, it absorbs 216 J of heat from the surroundings. What is the change in energy of the system? Expansion is work done by system on surrounding, so what is the sign for w? Heat is absorbed from the surroundings. Is this an endothermic or exothermic process? What is the sign for q? Solution To calculate the energy change of the gas, we use βU = q + w. Work of expansion is negative and because heat is absorbed from the surroundings, q is positive. Therefore, we have βU = q + w
= 216 J β 279 J = β 63 J
7.2. Enthalpy
Internal energy of a system is a state function, and this property is used in the discussion of any change in the heat content of the chemical reaction. Consider the reaction:
2CO(g) +O2(g) β 2CO2(g)
The reactants CO and oxygen are the initial states of the atoms and the product CO2 gives the final state. Since internal energy is a state function, the energy change βE associated with the reaction depends only on the initial and final states but not on the path taken by the reaction.
The energy change, βE of each chemical reaction run at constant temperature is a measure of the relative bond strength of reactants and products. The expression for pressure β volume (P-V) work shows than βE can be measured using the equation
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βπΈ = π β π€
= q β πππ₯ππ£π£2
π£1
For reaction at constant volume
V1=V2,
βV = 0
Therefore βE = q β 0 (constant volume)
So βE = ππ
(Change in internal energy is equal to the heat absorbed by the system when the process occurs at constants volume).
Change in internal energy can easily be measured using closed vessel such as bomb calorimeter but most often than not, chemical reactions are run at constant pressure like using open vessels, instead of constant volume, such that heat absorbed by the system is neither equal to ππ£ or βE.
We shall therefore develop another state function called the enthalpy which takes care of both the internal energy and work done by the system.
π, π πππ π are all state functions therefore π + ππ (internal energy plus work of expansion) will therefore give us another state function represented by π» (π + ππ) - enthalpy.
Addition of heat at constant pressure result to increase in enthalpy. The enthalpy change is equal to the heat absorbed only when the process is carried out at constant pressure. Because of equality between βH and ππ , the enthalpy is often called heat content of a system.
For a process at constant pressure in which heat is evolved by the system to the surroundings, βH and ππ are both negative. This is
because enthalpy of the final state is lower than the enthalpy of the initial state.
βH = π»π β π»π , negative
This process is exothermic. There is rise in temperature, on the other hand when the system absorbs heat from the surroundings both βH and ππ are the positive, the process is endothermic. There is fall in
temperature.
Our results so far show that
βE = ππ
βH = ππ
βπ» = βπΈ + β(ππ)
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= βπΈ + π2π2 β π1π1
βH and βE differ only by the difference in the PV products of the final and initial states.
If gases are produced or consumed in a chemical reaction, β(ππ) may be quite appreciable and βH and βE differ significantly.
For ideal gases reacting at constant temperature producing gases as the only products,
aA (g) + bB (g) β cC (g) + dD (g)
ππ (products) = (π + π)π π
ππ (reactants) = (π + π)π π
Therefore β(ππ) is given by the expression
Example 7.9. Calculate the work done on the surrounding when one mole of water is vaporized at 100oC and 1atm. Giving molar volume of liquid water as 18cm3 mol-1 and molar volume of steam as 24dm3 mol-1
at 1atm
Solution: Work done on the surrounding is = πβπ Molar volume of liquid water is 18cm3 mol-1 Molar volume of steam = 24dm3 mol-1
Therefore the volume of 1 mole of liquid water is negligible
1 atm = 1.01325Γ 105ππβ2
24dm3 = 24Γ 10β3m3
Work done = 1.01325Γ 105ππβ2 Γ 24Γ 10β3m3 mol-1
= 24.319 Γ 102 Nm mol-1
= 2.43 kJ mol-1
Example 7.10. If the enthalpy change for the process in question 3 above is 40.70 kJ mol-1, what is the change in internal energy βE?
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Solution
From βπ» = βπΈ + β(ππ)
βπΈ = βπ» β β(ππ)
Where βH = 40.70 kJ mol-1
βE = 40.70 kJ mol-1 β 2. 43 kJ mol-1
= 38.27 kJ mol-1
Example 7.11. For the combustion of benzene according to the equation
C6H6 (l ) + 71
2 O2 (g) β 6CO2 (g) + 3H2O (l)
If the heat of reaction at constant pressure is
βH25oC = β3267.62 kJ mol-1, find βE
Solution:
There is contraction of gaseous volume from 7.5 to 6 moles of gas.
Hence βn = 6 β 7.5 = β1.5
βπ» = βπΈ + β(ππ)
= βπΈ + βππ π
βπΈ = βπ» β βππ π
= βH β (β 1.5Γ 8.314 Γ 298.2)J
= (β 3267.62 + 3.72) kJ
= β 3263.90 kJ
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Example 7.12. To vaporize 100.0g of CCl4 at its normal boiling point, 349.0K and 1atm; 19.5kJ of heat is required. Calculate βHvap for CCl4 and compare it with βE for the same process.
Solution:
Molar mass of CCl4 = 154.0g mol-1
Number of moles of CCl4 in 100g = 100π
154π πππ β1
= 0.65 mol
For 0.65mol, enthalpy change = 19.5 kJ
For 1mol, enthalpy change will be 1 πππ πΆπΆπ4
0.65 πππΓ 19.5 kJ
= 30.0 kJ
β΄ βπ»vap = 30.0K
βπ» = βπΈ + β ππ
βπΈ = βπ» β β(ππ)
= βπ»vap β βππ π
βn = 1 because CCl4 (l) β CCl4 (g)
Volume of CCl4 (l) is negligible compared to CCl4 (g)
For βn = 1 at 349.0K,
βE = 30.0KJ β 1mol Γ 8.314JK-1mol-1) (349K)
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= 30.0 kJ β 2.90 kJ
= 27.1 kJ
Thus of the 30.0KJ of energy transferred from the surroundings in the form of heat, 27.1KJ is used to increase the internal energy of the molecules βE and 2.9KJ is used to expand the resulting vapour, β(ππ).
7.3. Heat Capacities of Gases The amount of heat expressed in joules, necessary to produce a standard change of 1oC in one gram of material is called specific heat. The product of the specific heat and the molar mass of a substance is the heat required to raise the temperature of one mole of that substance by 1oC. This is called the molar heat capacity and is a positive number which has the unit of joules per mole β degree (J.mol-1 K-1). Since the heat capacity C is the amount of heat needed to produce a temperature change of 1oC, it would appear that the quantity of heat required to produce a total temperature change, βT, is
q = CβT = C(T2 β T1).
Heat is not a state function therefore it does not depend on the initial and final states only but on how the process is carried out. The equation above says nothing about the process of producing the temperature change. We can remove this vagueness by defining two molar heat capacities for gases, one Cp for processes at constant pressure the other Cv for processes at constant volume.
Thus Cp = ππ
βπ and Cv = ππ£
βπ
The equivalent definitions in terms of infinitesimal changes are:
Cp = πππ
ππ =
ππ»
ππ and
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Cv = πππ£
ππ =
ππΈ
ππ
From the above, we can find that the heat required to change the temperature of n moles of material from T1 to T2 is:
qp = ππΆππππ2
π1 for process at constant pressure and
qv = ππΆπ£πππ2
π1 for process at constant volume
Near room temperature, πΆπ and πΆπ£ are constants independent of
temperature therefore:
qp = nCp (T2 β T1)
qv = nCv (T2 β T1)
For an ideal gas, there is a relationship between Cp and Cv. This relationship is shown easily by combing the definition of enthalpy with definitions of Cp and Cv. Thus for one mole of an ideal gas,
π» = πΈ + ππ
Differentiating the above equation gives
ππ» = ππΈ + π(ππ)
Dividing through by ππ
ππ»
ππ =
ππΈ
ππ +
π(ππ)
ππ
Cp = Cv + π(ππ)
ππ (Cp =
ππ»
ππ , Cv =
ππΈ
ππ )
Cp = Cv + π(π π)
ππ (PV = RT)
Cp = Cv + π ππ
ππ (R is gas constant)
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Cp = Cv + R
The heat capacity at constant pressure Cp is always larger than the heat capacity at constant volume Cv because PV β work is done when gas is heated at constant pressure. The relationship is visualized quite easily for an ideal gas. When one mole of ideal gas is heated at constant pressure, the work done in pushing back the piston is PβV = RβT. For a 1oC change in temperature, the amount of work done is equal to R, and this is just the extra energy required to heat a mole of ideal gas at constant pressure over that required to heat it through 1oC at constant volume.
Example 6.13. How much heat is required to raise the temperature of 10g of argon through 10OC at (a) constant volume (b) constant pressure?
[ Cv = 12.468J, Cp = 20.794J, Ar = 40]
Solution:
Atomic mass of Ar, a mono atomic gas = 40gmol-1
No. of moles in 10g = 10π
40ππππ β1 = 0.25mol
(a). qv = nCvβT
= 0.25 molΓ 12.468π½πΎβ1 molβ1 Γ 10πΎ
= 31.17J = 31J
(b). qp = nCpβT
= 0.25molΓ 20.794JK-1 mol-1Γ 10πΎ
= 51.985J = 51.99J = 52J
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Example 6.14. Suppose that 1.00kJ of heat is transferred to 2.00mol of argon at 298.0K, 1atm. What will the final temperature Tf be if the heat is transferred at (a) Constant volume (b) Constant pressure? Calculate the energy change βE in each case. [ Cv = 12.468 JK-1 mol-1, Cp = 20.794 JK-1 mol-1]
Note: the expression for βE involves Cv even though the process is carried out at constant pressure. The difference of 400J between the input qp and βE is the work done by the gas as it expands.
7.4. Reversible Isothermal and Adiabatic Processes
Before discussing adiabatic processes, let us look at reversible isothermal expansion of gases. Isothermal expansion of gas is the expansion carried out at constant temperature. The maximum work that is obtainable from the isothermal expansion of an ideal gas is easily calculated using the ideal equation of state,
ππ = ππ π
P = ππ ππ
In reversible isothermal expansion,
W = β ππππ2
π1
Substituting for p,
Wrev = β ππ π
π
π2
π1ππ
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= β nRT ππ
π
π£2
π£1
= β ππ ππΌππ2
π1
= β 2.303ππ π ππππ2
π1
Since V2 > V1, and the logarithm is positive then Wrev < 0. A negative W indicates that work is done by the system on the surroundings.
In compression, the final volume V2 is less than V1 so Wrev is positive. The positive value means that work is done on the gas.
Example 7.15. A mole of CH4 expands reversibly and isothermally from 1dm3 to 50 dm3 at 250C. Calculate the work by the gas in joules assuming the gas is ideal.
7.5. Expression of Wrev in terms of Pressure Let us consider a situation in which a change in pressure causes an infinitesimal change in volume of the gas. This means there are
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changes in both pressure and volume, so work done by the system will be W = β π(ππ). Since it is assumed that the change in pressure is more pronounced, above equation can be expressed as W = β πππ. But V is not constant, rather a function of pressure, so using general gas equation of state
PV = nRT
V = ππ π
π
Substituting for V in the above equation W =β ππ π
When P2 is greater than P1, work is done ON the system by the surroundings.
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Example 7.16. One mole of nitrogen at 250C and 1.01325Γ 105Nm-2 is expanded reversibly and isothermally to a pressure of 1.32Γ 104 Nm-2 (a) what is the value of Wrev? (b) What is the value of Wrev if the gas is expanded against a constant pressure of 1.32Γ 104 Nm-2 ?
Example 7.17. An ideal gas expanded reversibly and isothermally from 10 bar to 1 bar at 250C. What are the values of
(a) W per mole?
(b) q per mole
(c) βE and
(d) βH
Solution:
(a). For one mole,
W = RTlnπ2
π1
= (8.314JK-1mol-1) (298.15K)ln1
10
= - 5707.67J
= - 5.71kJ
(b). For isothermal expansion,
q = β w = β (β 5.71kJ) = 5.71kJ
(c). βE = 0, since this is an ideal gas
(d) ππ» = ππΈ + π(ππ) = 0 (Expansion is not at constant pressure).
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7.6. Adiabatic Expansion An adiabatic process is one in which there is neither gain nor loss of heat, that is one in which the system under investigation is thermally isolated from its environment such that q = 0. In an adiabatic expansion, work is done at the expense of the internal energy of the gas, resulting in drop of temperature. When a gas expands adiabatically to a larger volume and a lower pressure, the volume is usually smaller than it would be for an isothermal expansion to the same pressure. The work done by isothermal reversible expansion of a gas is always larger than the work done by the adiabatic expansion. The energy for doing the additional work in isothermal expansion is provided by heat absorbed from constant temperature reservoir. The energy for doing work in the adiabatic expansion comes only from the cooling of the gas itself.
Now let us consider the reversible adiabatic expansion of one mole of ideal gas. Since for an adiabatic process, dq = 0, then by the first law of thermodynamics,
Since work, PV, is done at the expense of internal energy
dE = β PdV
CvdT = β PdV
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For one mole of gas,
PV = RT
P = π π
π
CvdT = β π π
π dV
πΆπ£ππ
π = β R .
ππ
π
If V1 is the volume of the gas at T1 and V2 at T2 and Cv is independent of temperature,
πΆπ£ππ
π
π2
π1 = β π
ππ
π
π2
π1
Within these limits, on integration,
Cvlnπ2
π1 = β Rln
π2
π1
For ideal gas during expansion V2>V1 so T2<T1, there is temperature drop in adiabatic expansion. In other words the gas cools. Conversely, adiabatic compression of gas produces an increase in temperature.
Example 7.18. Calculate the temperature increase of helium if a mole of it is compressed adiabatically and reversibly from 44.8dm3 at 10C to 22.4dm3
[Cv for He = 12.55JK-1 mol-1, R = 8.314JK-1 mol-1]
Solution:
V1 = 44.8dm3, V2 = 22.4dm3
T1 = 273.15K , T2 = ?
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Cv = 12.55K-1 mol-1 R = 8.314JK-1 mol-1
Use equation Cvlnπ2
π1 = β Rln
π2
π1
=12.55JK-1 mol-1 ln(π2
273.15) = β 8.314JK-1mol-1ln
22.4
44.8
= ln π2
273.15πΎ = β
8.314
12.55 ln
22.4
44.8
lnT2 β ln273.15 = β 0.6625ln(1
2)
lnT2 = β 0.6625 Γ β 0.6931 + ππ273.15
lnT2 = 0.4592 + 5.6100
= 6.0692
T2 = 432.34K
Temperature increase = 432.34 β 273.15
= 159.19oC
β 159oC
Above calculation shows that the compression was carried out so
rapidly that there was no heat transfer to the container but sufficiently
slow to make it reversible.
To find the change in pressure, we use the equation,
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Cv lnπ2
π1 = β Rln
π2
π1
lnπ2
π1 = β
π
πΆπ£ In
π2
π1
= π
πΆπ£ In
π1
π2
Taking the antilog of both sides reduces the equation to
π2
π1 = (
π1
π2) R/Cv
From Cp = Cv + R,
R = Cp β Cv
β΄ π2
π1 =
π1
π2
πΆπβ πΆπ
πΆπ
= π1
π2 Ξ³ β1
Where Ξ³ = πΆπ
πΆπ£
Example 7.19. A mole of argon is allowed to expand adiabatically and
reversibly from a pressure of 10 bar (106 Pa) and 298.15K to 1 bar (105
Pa). What is the final temperature and how much work is done by the
gas? [Cp = 5
2 R, Cv =
3
2 R ]
Solution:
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From Cp ln π2
π1 = π ππ
π2
π1
5
2 R ln
π2
π1= π ππ
π2
π1
2.5 Γ logπ2
π1 = log
π2
π1
logπ2
π1 =
1
2.5 πππ (0.1)
log π2 β log π1 = 1
2.5 log 0.1
log π2 = log(298.15 ) β 0.4
= 2.4744 β 0.4
= 2.0744
T2 = 102.0744
β΄ π2 = 118.70πΎ
Alternatively,
5
2πΌπ
π2
π1 = πΌπ
π2
π1
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2. 5 πΌπ π2
298.15 = πΌπ (
105ππ
106ππ)
= πΌππ2
298.15 =
1
2.5 πΌπ
1
10
ln T2 β In 298.15 = β0.9210
ππ T2 = πΌπ 298.15 β 0.9210
= 5.6979 β 0.9210
= 4.7766
= 118.70K
(b). W = πΆππππ2
π1= πΆπ π2 β π1
= 3
2R π2βπ1
= 3
2 ( 8.31455JK-1mol-1)(118.7β298.15πΎ)
=β2.238kJmol-1
The maximum work that can be done by the gas is 2.24kJmol-1
Example 7.20. An ideal monatomic gas at 298.15K and 105Nm-2 is expanded in reversible adiabatic process to a final pressure of 5Γ
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104ππβ2. Calculate: (a) final temperature (b) q per mole (c) W per
Note: for infinitesimal increase in volume dV at the pressure P, work done by the gas is PdV. Since this work is accomplished at the expense of the internal energy of the gas, the internal energy must decrease by an amount dE.
Example 7.21. A tank contains 20dm3 of compressed N2 at 10 bar (106 Pa) and 250C. Calculate w when the gas is allowed to expand reversibly to 1 bar (105Pa) pressure (a) isothermally (b) adiabatically.
π = πrev per mole Γ ππ. ππ πππππ
= β 5.71Γ 8.07ππ½
= β 46.08 ππ½
= β 46.1 ππ½
(b). For adiabatic expansion, Cp = 29.125JKmol-1, Cv = 20.811JK-1mol-1, R = 8.314JK-1mol-1.
Cp ln π2
π1 = π ππ
π2
π1
29.125 ln T2
298.15K = 8.314ln
8.314
29.125 ln (
105 Nm β2
106 Nm β2)
ln T2 β ln 298.15 = 8.314
29.125 ln (
1
10)
= β 0.6573
ln T2 = ln 298.15 β 0.6573
= 5.6976 β 0.6573
= 5.0403
β΄ π2 = 154.5πΎ
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W = πΆππππ2
π1 = Cv (T2-T1)
= (20.811JK-1mol-1)(154.5 β 298.15K) = (20.811)(β143.65)Jmol-1 = β2.99kJmol-1 For 8.07moles W = β2.99kJmol-1Γ 8.07πππππ
= β 24.12kJ 7.7. Degrees of freedom and equipartition of energy
For each atom in a solid or gas phase, three coordinates have to be specified to describe the atomβs position β a single atom has 3 degrees of freedom for its motion. A solid or a molecule composed of N atoms has 3N degrees of freedom. We can also think about the number of degrees of freedom as the number of ways to absorb energy. The theorem of equipartition of energy (classical mechanics) states that in thermal equilibrium the same average energy is associated with each independent degree of freedom and that the energy is Β½kBT. For the interacting atoms, e.g. liquid or solid, for each atom we have
Β½ kBT for kinetic energy and Β½ kBT for potential energy - equality of kinetic and potential energy in harmonic approximation is addressed by the virial theorem of classical mechanics. Based on equipartition principle, we can calculate heat capacity of the ideal gas of atoms - each atom has 3 degrees of freedom and internal energy of 3/2kBT. The molar internal energy U=3/2NAkBT=3/2RT and the molar heat capacity under conditions of constant volume is Cv =[dU/dT]V=3/2R In an ideal gas of molecules only internal vibrational degrees of freedom have potential energy associated with them. For example, a diatomic molecule has 3 translational + 2 rotational + 1 vibrational = 6 total degrees of freedom. Potential energy contributes Β½ kBT only to the energy of the vibrational degree of freedom, and Umolecule = 7/2kBT if all degrees of freedom are βfullyβ excited.
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CHAPTER EIGHT
CHEMICAL KINETICS
We have learnt on several occasions that a balanced chemical equation
is a chemical statement that gives the mole ratios of reactants and
products as well as the ratios of formula units. A balanced chemical
equation as ordinarily written provides valuable chemical information
as to the masses, or volumes (if gases are involved) and is therefore an
essential quantitative tool for calculating product yields from amounts
of reacting substances. However, a balanced chemical equation tells us
nothing about how fast or quickly chemical changes occur, or what
energy changes are associated with the molecular interaction in a given
chemical reaction. Knowing how quickly a chemical reaction occurs is
a crucial factor in how the reaction affects its surroundings. Therefore,
knowing the rate of a chemical reaction and the energy changes
associated with the molecular interaction during the reaction are
integral to understanding the reaction.
The questions of βhow fast does the reaction goβ? and βwhat
conditions or factors bring about variations in speedβ in a given
chemical reaction are the subject of this chapter.
The concept of rate applies to a number of phenomenon in our
daily life. For example the change in distance by an athlete over time is
the running rate of the athlete. The number of soap bars that are
produced in a given time is the rate of production of soap etc. We
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apply the same principle in chemical reaction. This time as products
are formed reactants are used up and rate (speed) of a chemical
reaction can be expressed as the ratio of the change in the
concentration of a reactant (or product) to a change in time. The study
that deals with the movement/motion-the speeds, or rates of chemical
reactions is known as chemical kinetics.
We know that any reaction can be represented by the general equation
Reactants β products
This equation tells us that, during the course of a reaction, reactant
molecules are consumed while product molecules are formed. As a
result, we can follow the progress
of a reaction by monitoring either the decrease in concentration of the
reactants or the increase in concentration of the products.
Let us consider a simple reaction in which A molecules are converted
to B molecules (for example, the conversion of cis-1,2-dichloroethylene
to trans-1,2-dichloroethylene):
A β B
The decrease in the number of A molecules leads to increase in the
number of B molecules with time. In general, it is more convenient to
express the rate in terms of change in concentration with time. Thus,
for the preceding reaction we can express the rate as
π ππ‘π = β β[π΄]
βπ‘ ππ π ππ‘π =
β[π΅]
βπ‘
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in which Ξ[A] and Ξ[B] are the changes in concentration (in molarity)
over a period Ξt. Because the concentration of A decreases during the
time interval, Ξ[A] is a negative quantity. The rate of a reaction is a
positive quantity, so a minus sign is needed in the rate expression to
make the rate positive. On the other hand, the rate of product
formation does not require a minus sign because Ξ[B] is a positive
quantity (the concentration of B increases with time).
For more complex reactions, we must be careful in writing the rate
expression.
Consider, for example, the reaction
2A β B
Two moles of A disappear for each mole of B that formsβthat is, the
rate at which B forms is one half the rate at which A disappears. We
write the rate as either
πππ‘π = β 1
2 β[π΄]
βπ‘ ππ πππ‘π =
β[π΅]
βπ‘
Consider the following hypothetical reaction between reactants
A and B to form products C and D
aA + bB β cC + dD
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( a, b, c and d are the stoichiometric coefficients of A, B, C, and D
respectively). The rate of this reaction is the speed at which A or B is
consumed or, alternatively, the speed at which C or D is formed.
Mathematically this is given by
πππ‘π = β1
π β[π΄]
βπ‘=
β1
π β[π΅]
βπ‘=
1
π β[πΆ]
βπ‘=
1
π β[π·]
βπ‘
Figure 8.1: The rate of reaction A βB, represented as the decrease of A molecules with time and as the increase of B molecules with time. Example 8.1
Write the rate expressions for the following reactions in terms of the
disappearance of the reactants and the appearance of the products:
(a). I-(aq) + OCl-(aq) β Cl-(aq) + OI- (aq)
(b). 3O2 (g) β 2O3 (g)
(c). 4NH3(g) + 5O2(g) β 4NO(g) + 6H2O (g)
Solution
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(a). Because each of the stoichiometric coefficients equals 1
πππ‘π = β β[πΌβ]
βπ‘=
β[ππΆπΌβ]
βπ‘=
β[πΆπβ]
βπ‘=
β[ππΌβ]
βπ‘
(b). Here the coefficients are 3 and 2, so
πππ‘π = β 1
3 β[π2]
βπ‘=
1
2 β[π3]
βπ‘
(c) In this reaction
πππ‘π = β 1
4 β[ππ»3]
βπ‘= β
1
5 β[πΆπ2]
βπ‘=
1
4 β[ππ]
βπ‘=
1
6 β[π»2π]
βπ‘
Practice Exercise
Write the rate expression for the following reaction:
Example 8.2
Consider the reaction
4NO2 (g) + O2 (g) β 2N2O5 (g)
Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.037 M/s. (a) At what rate is N2O5 being formed? (b) At what rate is NO2 reacting?
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Strategy To calculate the rate of formation of N2O5 and disappearance of NO2, we need to express the rate of the reaction in terms of the stoichiometric coefficients as in Example 8.1:
πππ‘π = β 1
4 β[ππ2]
βπ‘ = β
β[π2]
βπ‘ =
1
2 β[π2π5]
βπ‘
We are given
β[π2]
βπ‘ = 0.037 M/s
Where the minus sign shows that the concentration of O2 is decreasing
where m and n are the rate law exponents and indicate the order of the
reaction with respect to the corresponding reactants. The values of m
and n for a given reaction must be determined experimentally and do
not change with temperature.
This relationship given by equation 8.1 can be expressed in a general
equation given below, called the rate law equation.
Rate = k [A]m [B]n β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦β¦β¦β¦.8.2
The rate law equation expresses the relationship between the
concentrations of the reactants and the rate of the reaction. The letter k
represents a proportionality constant called the rate constant and it
indicates how fast or slow a reaction is proceeding. A small rate
constant indicates a slow reaction and a large rate constant indicates a
fast reaction. The value of k for a given reaction is temperature
dependent and is constant under constant temperature and pressure
conditions.
The exponents m and n do not necessarily correspond to the
stoichiometric coefficients of their reactants. Usually the value of a rate
law exponent is 1 or 2. But, seldom values of 0, 3; and even fractions
can occur. If the exponent for a given reactant is 1, then the reaction is
said to be first order with respect to that reactant. Similarly, if the
exponent of a reactant is 2, the reaction is said to be second order in
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this reactant. For example, the rate law equation below represents a
reaction that is first order in A, second order in B, and third order (1 +
2) overall.
Rate = k[A]1[B]2
For example, the reaction between nitric oxide and ozone
NO(g) + O3(g) β NO2(g) + O2(g)
is first order in nitric oxide and first order in ozone. The rate law
equation for this reaction is:
Rate = k{NO]1[O3]1
The overall order of the reaction is 1 + 1 = 2.
Determining reaction Orders
The order of a reaction with respect to its reactants can be determined
by running a series of experiments each of which starts with a different
set of reactant concentrations and the initial rate is obtained. The
experiments are designed to change one reactant concentration while
keeping the other constant. This method of determining order of a
reaction is known as the initial rate method.
Example 8.7
For the reaction A + B β products, the following rate data were
obtained in three separate experiments:
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a) What is the order of the reaction with respect to A and B?
b) What is the rate constant, k?
Solution
a) The general rate expression for this reaction is Rate = k[A]x [B]y and
the values of x and y must be determined from the above data.
The data obtained shows that the rate doubles in experiment number 2
than it was in experiment number 1 when the concentration of B is
doubled keeping A constant. In such a condition, when the rate of a
reaction doubles by doubling the concentration of a given reactant,
order of the reaction with respect to that reactant is 1.
In experiment number 3 the rate is found not to change, than it was in
experiment number 2, when the concentration of A is doubled. This
indicates that the rate does not depend on the concentration of A. This
means, order of the reaction with respect to A is zero.
The experimentally determined rate equation will then be Rate =
k[A]o[B] = k[B] and the overall order = 1
The order with respect to each reactant can also be determined by
calculation as follows:
Rate1 = 2 x 10-5 M min-1 = k[2]x [1]y
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Rate2 = 4 x 10-5 M min-1 = k[2]x [2]y
Taking the ratio of the two rates we have
The order of the reaction with respect to the reactant A can be
calculated by taking the ratio of the rate expressions of experiment
number 2 and 3.
Thus, the experimentally determined rate equation is given by
Rate = k[A]x[B]y = K[A]o[B] = k[B]
b) Once we get the order of the reaction the rate constant can be
calculated by taking the data obtained in any of the three experiments.
Consider experiment number 1
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Example 8.8
For the reaction X + 2Y + 2Z β products, the following rate data were
obtained:
a) What is the order of each reactant in the system?
b) What is the value of the rate constant?
c) What is the rate of disappearance of X in experiment 4?
Solution
a) In the first experiment the rate of the reaction was determined to be
1.0 x 10-6 M/min when the concentrations of all reactants were kept at
0.1 M.
In the second experiment the rate was found to be tripled when the
concentrations of X and Y were tripled and that of Z was kept constant.
This indicates that the rate of the reaction depends on the
concentration on either X or Y or on both of them.
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In the third experiment, the rate remained the same as in experiment
number 1 when the concentration of X was quadrupled and those of Y
and Z were kept constant. This fact indicates that, the rate of the
reaction does not depend on the concentration of X.
Therefore, Rate Ξ± [X]0
Going back to experiment number 2, the rate was tripled when the
concentrations of X and Y were each tripled. But we have decided that
the rate does not depend on the concentration of X. Therefore, the rate
was tripled when the concentration of Y was tripled.
Thus, Rate Ξ± [Y]1
When we compare experiments 2 and 4, neglecting X; the
concentration of Y is kept constant while that of Z is tripled. As a result
the rate was found to increase by a factor of 9.
Thus, Rate Ξ± [Z]2 since 9 = 32
b) k can be evaluated from any one of the 4 data sets, once orders are
known; e.g., expt. 1.
Rate = k[X]0[Y]1[Z]2
c) When [Z] = 0.15, [Y] = 0.15, and [X] = 0.125, from balanced equation,
Rate = k[Y] [Z]2
= 10-3 M-2 / min (0.15) (0.15)2 = 3.4 x 10-6 M/min
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Relation between Reactant Concentrations and Time Rate laws enable us to calculate the rate of a reaction from the rate
constant and reactant concentrations. They can also be converted into
equations that enable us to determine the concentrations of reactants at
any time during the course of a reaction. We will illustrate this
application by considering first one of the simplest kind of rate
lawsβthat applying to reactions that are first order overall.
First-Order Reactions
A first-order reaction is a reaction whose rate depends on the reactant
concentration raised to the first power. In a first-order reaction of the type
Aβ product
the rate is
From the rate law, we also know that
Thus,
We can determine the units of the first-order rate constant k by
transposing:
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Because the unit for β[A] and [A] is M and that for βt is s, the unit for k
is
Derivation of first order reaction
The preceding first order rate in differential form becomes
Half-Life
The half-life of a reaction, t1/2, is the time required for the concentration of
a reactant to decrease to half of its initial concentration.
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By the definition of half-life,
The equation above tells us that the half-life of a first-order reaction is
independent of the initial concentration of the reactant. Measuring the
half-life of a reaction is one way to determine the rate constant of a
first-order reaction.
Example 8.9
The decomposition of ethane (C2H6) to methyl radicals is a first-order
reaction with a rate constant of 5.36 Γ 10-4 s-1 at 700Β°C:
Calculate the half-life of the reaction in minutes.
Strategy
To calculate the half-life of a first-order reaction, we use half life
equation above. A
conversion is needed to express the half-life in minutes.
Solution
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For a first-order reaction, we only need the rate constant to calculate
the half-life of the reaction.
Using
Second-Order Reactions
A second-order reaction is a reaction whose rate depends on the
concentration of one reactant raised to the second power or on the
concentrations of two different reactants, each raised to the first power. The
simpler type involves only one kind of reactant molecule:
A β product
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Another type of second-order reaction is
A + B β product
and the rate law is given by
rate = k[A][B]
The reaction is first order in A and first order in B, so it has an overall
reaction order of 2.
Using calculus, we can obtain the following expressions for βAβ
productβ second-order reactions:
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The equation above is a result of
Zero-Order Reactions
First- and second-order reactions are the most common reaction types.
Reactions whose order is zero are rare. For a zero-order reaction
A β product
the rate law is given by rate = k[A]0
= k Thus, the rate of a zero-order reaction is a constant, independent of
reactant concentration.
Using calculus, we can show that
The Equation above has the form of a linear equation. A plot of [A]t
versus t gives a straight line with slope = -k and y intercept = [A]0. To
calculate the half-life of a zero-order reaction, we set [A]t = [A]0/2 in
the equation above equation and obtain
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Many of the known zero-order reactions take place on a metal surface.
An example is the decomposition of nitrous oxide (N2O) to nitrogen
and oxygen in the presence of platinum (Pt):
2N2O(g) β 2N2(g) + O2(g)
When all the binding sites on Pt are occupied, the rate becomes
constant regardless of the amount of N2O present in the gas phase.
Third-order and higher order reactions are quite complex; they are not
presented in this book. Table 8.1 summarizes the kinetics of zero-order,
first-order, and second order reactions.
Table 8.1: Summary of the Kinetics of zero-order, first order, and second-order
reactions
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Activation Energy and Temperature Dependence of Rate Constants
With very few exceptions, reaction rates increase with increasing
temperature. For example, much less time is required to hard-boil an
egg at 100Β°C (about 10 min) than at 80Β°C (about 30 min). Conversely,
an effective way to preserve foods is to store them at subzero
temperatures, thereby slowing the rate of bacterial decay. Figure 8.2
shows a typical example of the relationship between the rate constant
of a reaction and temperature. To explain this behavior, we must ask
how reactions get started in the first place.
Figure 8.2: Dependence of rate constant on temperature. The rate constants of most reactions increase with increasing temperature.
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The Collision Theory of Chemical Kinetics
The kinetic molecular theory of gases states that gas molecules
frequently collide with one another. Therefore it seems logical to
assumeβand it is generally trueβthat chemical reactions occur as a
result of collisions between reacting molecules.
In terms of the collision theory of chemical kinetics, then, we
expect the rate of a reaction to be directly proportional to the number
of molecular collisions per second, or to the frequency of molecular
collisions:
This simple relationship explains the dependence of reaction rate on
concentration.
Consider the reaction of A molecules with B molecules to form
some product. Suppose that each product molecule is formed by the
direct combination of an A molecule and a B molecule. If we doubled
the concentration of A, say, then the number of A-B collisions would
also double, because, in any given volume, there would be twice as
many A molecules that could collide with B molecules. Consequently,
the rate would increase by a factor of 2. Similarly, doubling the
concentration of B molecules would increase the rate twofold. Thus, we
can express the rate law as
Rate = k[A][B]
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The reaction is first order in both A and B and obeys second-order
kinetics.
The collision theory is intuitively appealing, but the
relationship between rate and molecular collisions is more complicated
than you might expect. The implication of the collision theory is that a
reaction always occurs when an A and a B molecule collide. However,
not all collisions lead to reactions. Calculations based on the kinetic
molecular theory show that, at ordinary pressures (say, 1 atm) and
temperatures (say, 298 K), there are about 1 Γ 1027 binary collisions
(collisions between two molecules) in 1 mL of volume every second, in
the gas phase. Even more collisions per second occur in liquids. If
every binary collision led to a product, then most reactions would be
complete almost instantaneously. In practice, we find that the rates of
reactions differ greatly. This means that, in many cases, collisions alone
do not guarantee that a reaction will take place.
Any molecule in motion possesses kinetic energy; the faster it moves,
the greater the kinetic energy. When molecules collide, part of their
kinetic energy is converted to vibrational energy. If the initial kinetic
energies are large, then the colliding molecules
will vibrate so strongly as to break some of the chemical bonds. This
bond fracture is the first step toward product formation. If the initial
kinetic energies are small, the molecules will merely bounce off each
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other intact. Energetically speaking, there is some minimum collision
energy below which no reaction occurs.
We postulate that, to react, the colliding molecules must have a
total kinetic energy equal to or greater than the activation energy (Ea),
which is the minimum amount of energy required to initiate a chemical
reaction. Lacking this energy, the molecules remain intact, and no
change results from the collision. The species temporarily formed by the
reactant molecules as a result of the collision before they form the product is
called the activated complex (also called the transition state).
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Figure 8.3: Potential energy profiles for (a) exothermic and (b) endothermic reactions. These plots show the change in potential energy as reactants A and B are converted to products C and D. The transition state is a highly unstable species with a high potential energy. The activation energy is defined for the forward reaction in both (a) and (b). Note that the products C and D are more stable than the reactants in (a) and less stable than those in (b). Figure 8.3 shows two different potential energy profiles for the
reaction
A + B β C + D
If the products are more stable than the reactants, then the reaction will
be accompanied
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by a release of heat; that is, the reaction is exothermic [Figure 8.3(a)].
On the other hand, if the products are less stable than the reactants,
then heat will be absorbed by the reacting mixture from the
surroundings and we have an endothermic reaction [Figure 8.3(b)]. In
both cases, we plot the potential energy of the reacting system versus
the progress of the reaction. Qualitatively, these plots show the
potential energy changes as reactants are converted to products.
We can think of activation energy as a barrier that prevents less
energetic molecules from reacting. Because the number of reactant
molecules in an ordinary reaction is very large, the speeds, and hence
also the kinetic energies of the molecules, vary greatly. Normally, only
a small fraction of the colliding moleculesβthe fastest-moving onesβ
have enough kinetic energy to exceed the activation energy. These
molecules can therefore take part in the reaction. The increase in the
rate (or the rate constant) with temperature can now be explained: The
speeds of the molecules obey the Maxwell distributions shown in
earlier chapters. Compare the speed distributions at two different
temperatures. Because more high-energy molecules are present at the
higher temperature, the rate of product formation is also greater at the
higher temperature.
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The Arrhenius Equation
The dependence of the rate constant of a reaction on temperature can
be expressed by this equation, now known as the Arrhenius equation:
in which Ea is the activation energy of the reaction (in kilojoules per
mole), R is the gas constant (8.314 J/K . mol), T is the absolute
temperature, and e is the base of the natural logarithm scale. The
quantity A represents the collision frequency and is called the frequency
factor. It can be treated as a constant for a given reacting system over a
fairly wide temperature range. The equation above shows that the rate
constant is directly proportional to A and, therefore, to the collision
frequency. Further, because of the minus sign associated with the
exponent Ea/RT, the rate constant decreases with increasing activation
energy and increases with increasing temperature. This equation can
be expressed in a more useful form by taking the natural logarithm of
both sides:
Equation above can take the form of a linear equation:
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Thus, a plot of ln k versus 1/T gives a straight line whose slope m is
equal to βEa/R and whose intercept b with the ordinate (the y-axis) is
ln A.
Example 8.10
The rate constants for the decomposition of acetaldehyde
were measured at five different temperatures. The data are shown in
the table. Plot ln k versus 1/T, and determine the activation energy (in
kJ/mol) for the reaction. This reaction has been experimentally shown
to be β3/2β order in CH3CHO, so k has the units of 1/M1 /2 . s.
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Strategy
Consider the Arrhenius equation written as a linear equation
A plot of ln k versus 1/T (y versus x) will produce a straight line with a
slope equal to βEa/R. Thus, the activation energy can be determined
from the slope of the plot.
Solution
First, we convert the data to the following table:
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Figure 8.4: Plot of ln k versus 1/T. The slope of the line is equal to βEa/R
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A plot of these data yields the graph in Figure 8.4. The slope of the line
is calculated from two pairs of coordinates:
From the linear form of Equation
An equation relating the rate constants k1 and k2 at temperatures T1 and
T2 can be used to calculate the activation energy or to find the rate
constant at another temperature if the activation energy is known. To
derive such an equation we start with equation below:
Subtracting ln k2 from ln k1 gives
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The rate constant of a first-order reaction is 4.68 Γ10-2 s-1 at 298 K. What
is the rate constant at 375 K if the activation energy for the reaction is
33.1 kJ/mol?
Strategy
A modified form of the Arrhenius equation relates two rate constants
at two different temperatures [see above]. Make sure the units of R and
Ea are consistent.
Solution
The data are
Substituting in Equation
gives
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We convert Ea to units of J/mol to match the units of R. Solving the
equation gives
Reaction Mechanisms
As we mentioned earlier, an overall balanced chemical equation does
not tell us much about how a reaction actually takes place. In many
cases, it merely represents the sum of several elementary steps, or
elementary reactions, a series of simple reactions that represent the progress of
the overall reaction at the molecular level. The term for the sequence of
elementary steps that leads to product formation is reaction mechanism.
The reaction mechanism is comparable to the route of travel followed
during a trip; the overall chemical equation specifies only the origin
and destination.
As an example of a reaction mechanism, let us consider the
reaction between nitric oxide and oxygen:
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2NO(g) + O2(g) β 2NO2(g)
We know that the products are not formed directly from the collision
of two NO molecules with an O2 molecule because N2O2 is detected
during the course of the reaction. Let us assume that the reaction
actually takes place via two elementary steps as follows:
In the first elementary step, two NO molecules collide to form a N2O2
molecule. This event is followed by the reaction between N2O2 and O2
to give two molecules of NO2. The net chemical equation, which
represents the overall change, is given by the sum of the elementary
steps:
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Species such as N2O2 are called intermediates because they appear in the
mechanism of the reaction (that is, the elementary steps) but not in the overall
balanced equation. Keep in mind that an intermediate is always formed
in an early elementary step and consumed in a later elementary step.
The molecularity of a reaction is the number of molecules reacting
in an elementary step. These molecules may be of the same or different
types. Each of the elementary steps just discussed is called a
bimolecular reaction, an elementary step that involves two molecules. An
example of a unimolecular reaction, an elementary step in which only one
reacting molecule participates, is the conversion of cyclopropane to
propene is an example. Very few termolecular reactions, reactions that
involve the participation of three molecules in one elementary step, are
known, because the simultaneous encounter of three molecules is a far
less likely event than a bimolecular collision.
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REFERENCES
Atu Ausaji A (2006). Fundamental of physical chemistry. Owerri, Nigeria, Supreme Publishers.
Atkins, P.W. & Friedman, R. (2008). Quanta, Matter and Change: A Molecular Approach to Physical Change. Oxford: Oxford University Press.
Atkins, P. W. (1986). Physical Chemistry. (3rd ed.). Oxford: Oxford University Press
David E. Goldberg (2007). Fundermental of Chemistry. 5th Ed. New York, The McGraw-Hills Companies.