1 Thermochemistry The study of heat change in chemical reactions.

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Thermochemistry

The study of heat change in chemical reactions

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Thermochemistry is part of a broader subject called thermodynamics, which is the scientific study of the interconversion of heat and other kinds of energy.

The first law of thermodynamics, which is based on the law of conservation of energy, states that energy can be converted from one form to another, but cannot be created or destroyed.

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Types of Energy

• Thermal energy - The energy associated with the random motion of atoms and molecules.– Temperature - the average kinetic

energy of the particles in a sample of matter

– Heat (q)- the total kinetic energy of the particles in a sample of matter

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Kinetic energy - the energy produced by a moving object (various types of molecular motion and the movement of electrons within molecules).

Potential energy - energy available by virtue of an object’s position (determined by the attractive and repulsive forces at the atomic level) .

It is impossible to measure all these contributions accurately, so we cannot calculate the total energy of a system with any certainty. Changes in energy, on the other hand, can be determined experimentally.

The internal energy of a system has two components: kinetic and potential energy.

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• Chemical energy - The energy stored within the structural units of chemical substances; its quantity is determined by the type and arrangement of constituent atoms. – When substances participate in chemical

reactions, chemical energy is released, stored, or converted to other forms of energy.

– Chemical energy can be considered a form of potential energy because it is associated with the relative positions and arrangements of atoms within a given substance.

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Energy can assume many different forms that are interconvertible. When one form of energy disappears, some other form of energy (of equal magnitude) must appear, and vice versa.

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Energy Changes in Chemical Reactions

Almost all chemical reactions absorb or produce (release) energy, generally in the form of heat.

In order to analyze energy changes associated with chemical reactions we must first define the terms system and surroundings.

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For example, in an acid-base neutralization experiment, the system would be the contents of the beaker, the 50.0 mL of HCl to which 50.0 mL of NaOH are added.

• System - the specific part of the universe that is of interest to us. For chemists, systems usually include substances involved in chemical and physical changes.

• Surroundings - the rest of the universe outside the system.

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2H2(g) + O2(g) 2H2O(l) + energy

The combustion of hydrogen gas is one of many chemical reactions that release considerable quantities of energy.

System - ?Surroundings - ?

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Since energy cannot be created or destroyed, any energy lost by the system must be gained by the surroundings. Thus the heat generated by the combustion process is transferred from the system to its surroundings. This is an example of an exothermic process, which is any process which gives off heat - that is, transfers thermal energy to the surroundings.

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heat + 2HgO(s) 2Hg(l) + O2(g)

The decomposition of mercury (II) oxide is an endothermic process, in which heat has to be supplied to the system by the surroundings.

System - ?

Surroundings - ?

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Breaking chemical bonds is an endothermic process, bond formation is exothermic. In an exothermic reaction, more energy is given off by bond formation than is absorbed breaking the bonds (therefore the overall process releases energy); in an endothermic reaction the opposite is true, more energy is absorbed during bond breaking than is released during bond formation.

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Calorimetry

Our discussion of calorimetry, the measurement of heat changes, will depend on an understanding of specific heat and heat capacity.

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Specific Heat and Heat Capacity

Specific heat capacity (c) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius.

• Specific heat capacity is an intensive property, its magnitude doesn’t depend on the amount of the substance present

• Specific heat capacity has the units J/g.oC or kJ/g.oC

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• If we know the specific heat capacity and the mass of a substance in grams, (m), then the change in the sample’s temperature (T = Tfinal - Tinitial) will tell us the amount of heat (q) that has been absorbed or released in a particular process.

q = mcT

The sign convention for q is positive for endothermic processes and negative for exothermic processes.

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Example: A 466g sample of water is heated from 8.50 oC to 74.60 oC. Calculate the amount of heat absorbed by the water in kJ.

q = mcT

q = (466g)(4.184 J/g.oC)(74.60 oC - 8.50 oC)

q = 128878 J

q = 129 kJ

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The heat capacity (C) of a substance is the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius.

• Heat capacity is an extensive property (it varies dependent on the amount of substance present).

• Heat capacity has the units J/oC or kJ/oC • The relationship between the heat capacity

and specific heat of a substance isC = mc

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Determine the heat capacity for 60.0g of water.

C = mc251 J/oC = (60.0g)(4.184 J/g.oC)

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Calorimeters

In a laboratory, heat changes in physical and chemical processes are measured with a calorimeter, a closed container designed specifically for this purpose.

• The special design of a calorimeter allows us to assume that no heat is lost to the surroundings during the time it takes to make measurements.

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• If we know the heat capacity of a substance, C, then the change in the sample’s temperature (T = Tfinal - Tinitial) will tell us the amount of heat (q) that has been absorbed or released in a particular process.

q = CT

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A certain calorimeter has a heat capacity of 1.55 kJ/oC. Calculate the heat absorbed by a calorimeter in kJ that increases in temperature from 23.60 oC to 25.87 oC.

q = CT

q = (1.55 kJ/oC)(25.87 oC - 23.60 oC)

q = 3.52 kJ

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Heat of combustion, the amount of energy given off when a certain amount of a substance is “burned” in the presence of oxygen, is usually measured by placing a known mass of a compound in a steel container called a constant-volume bomb calorimeter. (With this being a steel container, its volume obviously doesn’t change, but the pressure certainly would.)

This calorimeter is filled with oxygen before the sample is placed in the cup. The sample is ignited electrically, and the heat produced by the reaction can be accurately determined by measuring the temperature increase in the known amount of surrounding water.

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Because no heat enters or leaves the system throughout the process, heat lost by the reaction must be equal to the heat gained by the calorimeter and water, therefore, we can write…

qrxn = -(qwater + qcalorimeter)

where qwater is obtained by q = mcT

and qcalorimeter is determined byq = CT

Heat lost = exothermic, so the kJ would be negative

Heat gained = endothermic, so the kJ would be positive, which

is why the formula is –(qwater + qcalorimeter)

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A quantity of 1.435g of naphthalene (C10H8), a pungent-smelling substance used in moth repellents, was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.17 oC to 25.84 oC. If the mass of water surrounding the calorimeter was exactly 2000g and the heat capacity of the bomb calorimeter was 1.80 kJ/ oC, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.

q = -(qwater + qcalorimeter)

qwater = mcT

= (2000.g)(4.184J/g.oC)(25.84 oC - 20.17 oC)

= 47400 J or 47.4 kJ

qcalorimeter = CT

= (1.80 kJ/oC)(25.84 oC - 20.17 oC)

= 10.2 kJ

q = -(47.4 kJ + 10.2 kJ) = -57.6 kJ

-57.6 kJ = -5140 kJ__

1.435 g C10H8 128.173 g C10H8

The molar mass of naphthalene is 128.173 g/mol, so the heat of combustion of 1 mol of naphthalene can be calculated…

heat lost by the naphthalene = heat gained by water and the calorimeter

Notice, the calorimeter and the water have the same final and initial temps

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A simpler device to the constant-volume calorimeter is the constant-pressure calorimeter used to determine the heat changes for noncombustion reactions. An operable constant-pressure calorimeter can be constructed from two Styrofoam coffee cups.

This device measures the heat effects of a variety of reactions, such as acid-base neutralization, as well as the heat of solution and heat of dilution.

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As the following example shows, we can study the heat of neutralization of acid-base reactions using a constant-pressure calorimeter.

1.00 x 102 mL of 0.500 M HCl is mixed with 1.00 x 102 mL of 0.500 M NaOH in a constant-pressure calorimeter that has a heat capacity of 335 J/oC. The initial temperature of the HCl and NaOH solution is the same, 22.50 oC, and the final temperature of the mixed solution is 24.90 oC. Calculate the heat of neutralization for the following reaction

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)(Assume the densities and specific heats of the solutions are the same as for water.)

Assuming that no heat is lost to the surroundings, we write

qrxn = -(qsoln + qcalorimeter)

qsoln = mcT= (1.00 x 102g + 1.00 x 102g)(4.184 J/g.oC)(24.90 oC - 22.50 oC)= 2010 J or 2.01 kJ

qcalorimeter= CT= (335 J/oC )(24.90 oC - 22.50 oC)= 804 J or .804 kJ

qrxn = -(2.01 kJ + .804 kJ) = -2.81 kJ

As you will see on the next slide, this is for the formation of .0500 mol H2O

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0.500 M HCl = 0.500 mol HCl = 0.0500 mol HCl 1 L .1 L

yields .0500 mol0.500 M NaOH = 0.500 mol NaOH = 0.0500 mol NaOH H2O

1 L .1 L

To calculate the heat of neutralization for this reaction, you need to know the number of kJ released when 1 mol of HCl reacts with 1 mol of NaOH. (Remember, the coefficients in a thermochemical equation represent moles.)

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

-2.81 kJ = -56.2 kJ

0.0500 mol 1 mol

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Enthalpy

To quantify the heat flow into or out of a system, chemists use a property called enthalpy, represented by the symbol H. Enthalpy is an extensive property; its magnitude depends on the amount of the substance present.

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It is the change in enthalpy, H, that we measure in a chemical reaction. The change of enthalpy of a reaction, Hrxn commonly represented as just H, is the difference between the enthalpies of the products and the enthalpies of the reactants:

H = H(products) - H(reactants)

Enthalpy of a Reaction

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In other words, H represents the heat given off or absorbed during a reaction.

• The enthalpy of a reaction can be positive or negative, depending on the process.

• For an endothermic process H is positive. For an exothermic process H is negative.

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Thermochemical equations show the enthalpy changes as well as the mass relationships in a thermochemical reaction.

Thermochemical equations

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In a thermochemical equation...• the stoichiometric coefficients always refer to the

number of moles of a substance.• When we reverse an equation, we change the roles

of reactants and products. Consequently, the magnitude of H for the equation remains the same, but its sign changes.

• If we multiply or divide the coefficients by a factor, then H must also change by the same factor.

• When writing thermochemical equations, we must always specify the physical states of all reactants and products, because they help determine the actual enthalpy changes.

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Examples of thermochemical equations

H2O(s) H2O(l) H = 6.01 kJ

H2O(l) H2O(s) H = -6.01 kJ

2H2O(l) 2H2O(s) H = -12.02 kJ

Thermochemical equations show the enthalpy changes that occur in a physical or chemical process.

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How much energy is absorbed or released when 25.0 g of water is frozen?

25.0 g H2O x1 mol H2O

18.02 g H2O= 1.39 mol H2O

H2O(l) H2O(s) H = -6.01 kJ

1 mol H2O

-6.01 kJ

1.39 mol H2O=-8.34 kJ

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It is impossible to determine the absolute enthalpy of a substance. This problem is similar to the one geographers face in expressing the elevations of specific mountains or valleys. Rather than trying to devise some type of “absolute” elevation scale (perhaps based on distance from the center of the Earth?), by common agreement all geographic heights and depths are expressed relative to sea level, an arbitrary reference with a defined elevation of “zero” meters or feet. Similarly, chemists have agreed on an arbitrary reference point for enthalpy.

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The “sea level” reference point for all enthalpy expressions is called the standard enthalpy of formation, (H0

f), which is defined as the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. Elements are said to be in the standard state at 1 atm, hence the term “standard enthalpy”. The superscript “0” represents standard-state conditions (1 atm), and the subscript “f” stands for formation. Although the standard state does not specify a temperature, we will always use H0

f values measured at 25 oC.

Enthalpy of Formation

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By convention, the standard enthalpy of formation of any element in its most stable form is zero.

Using your reference packet, state the most stable (allotrope) form for molecular oxygen

O2 - because the H0f for O2 = 0, and

the H0f for O3 = 0

State the most stable allotrope for carbonC(graphite) is the most stable

allotrope

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You can usually get Hf values for compounds from your thermochemistry reference packet.

Hf that aren’t listed in the reference packet can be determined in one of two ways, the direct method or the indirect method.

• Both of these methods require an understanding of thermochemical equations.

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The Direct Method for Determining Hf

• Suppose we want to know the enthalpy of formation of solid calcium oxide.

CaO(s) + CO2(g) CaCO3(s) Horxn = -177.8

kJ/mol

Horxn = nHo

f(products) - nHof(reactants)

-177.8 kJ/mol = 1 mol(-1206.9) - [1 mol(x) + 1 mol(-393.5)]

Hfo for CaO(s) = -635.6 kJ

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The Indirect Method for Determining Hf

In some cases, a reaction proceeds too slowly, or side reactions produce substances other than the desired compound for Hf to be determined directly. In these cases, Ho

f can be determined by an indirect approach, which is based on the law of heat summation (or simply Hess’s law).

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• Hess’s law can be stated as follows: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. In other words, if we break down the reaction of interest into a series of reactions for which Ho

rxn can be measured, we can calculate Ho

rxn for the overall reaction.

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What is H0f for methane?

We would represent the synthesis of CH4 from its elements as

C(graphite) + 2H2(g) CH4(g)

However, this reaction does not take place as written, so we cannot measure the enthalpy change directly. We must use Hess’s law.

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The general rule in applying Hess’s law is that we should arrange a series of chemical equations (corresponding to a series of steps) in such a way that, when added together, all species will cancel except for the reactants and products that appear in the overall reaction.

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The following reactions involving C(graphite), H2(g) and CH4(g) with O2(g) have all been studied and the H0

rxn values are accurately known

(a) C(graphite) + O2(g) CO2(g) H0rxn = -393.5 kJ

(b) 2H2(g) + O2(g) 2H2O(l) H0rxn = -571.6 kJ

(c) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H0rxn = -890.4 kJ

Remember, we are trying to determine the H0f for

C(graphite) + 2H2(g) CH4(g)

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Since we want to obtain one equation containing only C and H2 as reactants and CH4 as a product, we must reverse (c) to get

(d) CO2(g) + 2H2O(l) CH4(g) + 2O2(g) H0rxn = 890.4 kJ

The next step is to add equations (a), (b), and (d)

(a) C(graphite) + O2(g) CO2(g) H0rxn = -393.5 kJ

(b) 2H2(g) + O2(g) 2H2O(l) H0rxn = -571.6 kJ

(d) CO2(g) + 2H2O(l) CH4(g) + 2O2(g) H0rxn = 890.4 kJ

C(graphite) + 2H2(g) CH4(g) H0rxn = -

74.7 kJAs you can see, all the nonessential species (O2, CO2, and H2O) cancel in this operation. Because the above equation represents the synthesis of 1 mol of CH4 from its elements, we have H0

f(CH4) = -74.7 kJ/mol.

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Once we know the standard enthalpies of formation, we can calculate the standard enthalpy of reaction, H0

rxn, defined as the enthalpy of a reaction carried out at 1 atm.

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Consider this hypothetical reactionaA + bB cC + dD

where a, b, c, and d are stoichiometric coefficients.

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For this reaction H0rxn is given by

H0rxn = [cH0

f(C) + dH0f(D)] - [aH0

f(A) + bH0f(B)]

We can generalize this equation as:

H0rxn = nH0

f(products) – nH0f(reactants)

where n denotes the stoichiometric coefficients for the reactants and products, and (sigma) means “the sum of”.

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Calculate the kJ of heat released when 1 mol of liquid pentaborane, B5H9, is exposed to gaseous oxygen producing solid diboron trioxide and liquid water.

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(*Note - Hf values vary slightly depending on which reference book you choose, these values are from Raymond Chang Chemistry, 6th Edition)

Substance Hf

(kJ/mol)

O2(g) 0

B2O3(s) -1263.6

H2O(l) -285.8

B5O9(l) 73.2

Remember, these are kJ released or absorbed per one mole

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H = H(products) - H(reactants)

H0rxn = [cH0

f(C) + dH0f(D)] - [aH0

f(A) + bH0f(B)]

H0rxn = -4518.3 kJ

H0rxn = [2.5(-1263.6) + 4.5(-285.8)] - [1(73.2)

+6(0)]

B5H9(l) + 6O2(g) 2.5 B2O3(s) + 4.5 H2O(l)

Notice, this was balanced in such a way that you only started with 1 mol of B5H9

If it had been written with 2B5H9, you would have to divide the final kJ by 2 to get the amount of heat released when 1 mol of pentaborane reacted with oxygen.

aA + bB cC + dD

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Hess’s Law can also be used to calculate Hrxn. The following slide shows the steps involved in forming sodium chloride from elemental sodium and molecular chlorine.

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Although we have focused so far on the thermal energy effects resulting from chemical reactions, many physical processes, such as the melting of ice or the condensation of a vapor, also involve the absorption or release of heat.

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Change of State Processes for Water

Thermal energy increases the kinetic energy of molecules, allowing them to break free of intermolecular forces. With enough added thermal energy, molecules in a solid enter the liquid phase and molecules in a liquid enter the gas phase.

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Enthalpy changes dramatically during a change of state.

The added thermal energy not only causes a change of state but also increases the molar enthalpy of water.

Water’s freezing point

_____ oC

_____ K

Water’s boiling point

_____ oC

_____ K

0

273

373

100

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• The molar enthalpy change that occurs during melting is called the molar enthalpy of fusion, or heat of fusion (Hfus). It is the difference between the enthalpy of 1 mol of a substance in its liquid and solid states. Water’s Hfus equals 6.01 kJ/mol

Since H is a positive value, this is an endothermic process, as expected for an energy-absorbing change of melting ice.

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25.0 g H2O x 1 mol H2O = 1.39 mol H2O

18.02 g8.35 kJ6.01 kJ =

1 mol H2O 1.39 mol H2O

H2O(s) H2O(l) H = 6.01 kJ

Then determine the mol of H2O you are melting

Use a ratio to determine the

kJ of heat absorbed

Calculate the heat absorbed when 25.0 g of H2O(s) is converted to H2O(l).

to solve this, you need to use water’s heat of fusion…

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• The molar enthalpy change that occurs during evaporation is called the molar enthalpy of vaporization, or heat of vaporization (Hvap). It is the difference between the enthalpy of 1 mol of a substance in its liquid and gaseous states.

Water’s Hvap equals 40.79 kJ/mol

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Heat of Solution and Dilution

Enthalpy changes occur when a solute dissolves in a solvent or when a solution is diluted.

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Heat of SolutionIn the vast majority of cases, dissolving a solute

in a solvent produces measurable heat change. The heat change of solution, or change in enthalpy of solution, Hsoln, is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. The quantity Hsoln represents the difference between the enthalpy of the final solution and the enthalpies of its original components (solute and solvent) before they are mixed.

• Like other enthalpy changes, Hsoln is positive for endothermic processes and negative for exothermic processes.

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Consider the heat of solution of a process in which an ionic compound is the solute and water is the solvent. For example, what happens when solid NaCl dissolves in water? In solid NaCl, the Na+ and the Cl- ions are held together by strong positive-negative (electrostatic) forces, but when a small crystal of NaCl dissolves in water, the three

dimensional network of ions breaks into its individual units.

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The energy required to completely separate one mole of a solid ionic compound into gaseous ions is called lattice energy (U). The lattice energy of NaCl is 788 kJ/mol. In other words, we would need to supply 788 kJ of energy to break 1 mole of solid NaCl into 1 mole of Na+ and Cl- ions.

NaCl(s) Na+(g) + Cl-(g) U = 788 kJ

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Next the separated Na+ and Cl- ions are stabilized in solution by their interaction with water molecules. These ions are said to be hydrated.

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The enthalpy change associated with the hydration process is called the heat of hydration, Hhydr

• Heat of hydration is a negative quantity for cations and anions.

Na+(g) + Cl-(g) Na+(aq) + Cl-(aq) Hhydr = -784 kJ

H2O

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Applying Hess’s law, it is possible to consider Hsoln as the sum of two related quantities, lattice energy (U) and heat of hydration (Hhydr):

NaCl(s) Na+(aq) + Cl-(aq)Hsoln = ?

H2O

NaCl(s) Na+(g) + Cl-(g) U = 788 kJ

Na+(g) + Cl-(g) Na+(aq) + Cl-(aq) Hhydr = -784 kJ

NaCl(s) Na+(aq) + Cl-(aq) Hsoln = 4 kJ

H2O

H2O

Therefore, when 1 mol of NaCl dissolves in water, 4 kJ of heat will be absorbed from the surroundings

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The solution process for NaCl.

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Heat of Dilution

When a previously prepared solution is diluted, that is, when more solvent is added to lower the overall concentration of the solute, additional heat is usually given off or absorbed. The heat of dilution is the heat change associated with the dilution process.

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Therefore, always be cautious when working on a dilution procedure in the laboratory. Because of its highly exothermic heat of dilution, concentrated sulfuric acid (H2SO4) poses a particularly hazardous problem if its concentration must be reduced by mixing it with additional water. The process is so exothermic that you must NEVER attempt to dilute the concentrated acid by adding water to it. The heat generated could cause the acid solution to boil and splatter.

• The recommended procedure is to add the concentrated acid slowly to the water (while constantly stirring).

“Do as you oughter, add acid to water.”

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THE END