Chemistry Chapter 01

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Chapter 1Thermochemistry

CHEMISTRYThe Central Science

David P. White

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Kinetic Energy and Potential Energy Kinetic energy is the energy of motion:

Potential energy is the energy an object possesses by

virtue of its position.

Potential energy can be converted into kinetic energy.Example: a bicyclist at the top of a hill.

The Nature of Energy

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Kinetic Energy and Potential Energy

Electrostatic potential energy,Ed, is the attraction

between two oppositely charged particles, Q1 and Q2, a

distance dapart:

The constant = 8.99 109 J-m/C2. (a constant ofproportionality)

If the two particles are of opposite charge, thenEd is the

electrostatic repulsion between them.


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Kinetic Energy and Potential Energy

When Q1 and Q2 have the same sign (Q1=+ve, Q2=+ve),

its repels each other, pushing them apart, then Ed is

positive.

For opposite sign (Q1=+ve, Q2=-ve), they are attract

each other pulling them each other, then Ed is negative.


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Units of Energy

SI Unit for energy is the joule, J:

We sometimes use the calorie instead of the joule:

1 cal = 4.184 J (exactly)

A nutritional Calorie:

1 Cal = 1000 cal = 1 kcal


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Systems and Surroundings System: part of the universe we are interested in.

Surroundings: the rest of the universe.


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Definitions:

System and Surroundings

The system includes themolecules we want to

study (here, the hydrogenand oxygen molecules).

The surroundings areeverything else (here, the

cylinder and piston).


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Transferring Energy: Work and Heat

Force is a push or pull on an object. Work is the product of force applied to an object over a

distance:

Energy is the work done to move an object against aforce.

Heat is the transfer of energy between two objects.

Therefore, Energy is the capacity to do work or transferheat.


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Heat

Energy can also betransferred as heat.

Heat flows fromwarmer objects tocooler objects.


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Internal Energy

Internal Energy: total energy of a system.

Cannot measure absolute internal energy.

Change in internal energy,

The First Law of

Thermodynamics

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Relating E to Heat and Work Energy cannot be created or destroyed.

Energy of (system + surroundings) is constant.

Any energy transferred from a system must be transferred

to the surroundings (and vice versa).

From the first law of thermodynamics:

when a system undergoes a physical or chemical change, thechange in internal energy is given by the heat added to or

absorbed by the system plus the work done on or by the

system:

The First Law of

Thermodynamics

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The First Law of

Thermodynamics

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Exothermic and EndothermicProcesses

Endothermic: absorbs heat from the surroundings.

Exothermic: transfers heat to the surroundings. An endothermic reaction feels cold.

An exothermic reaction feels hot.

The First Law of

Thermodynamics

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Exchange of Heat between


When heat is absorbed by the system fromthe surroundings, the process is endothermic.


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Exchange of Heat between


When heat is absorbed by the system fromthe surroundings, the process is endothermic.

When heat is released by the system into thesurroundings, the process is exothermic.


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State Functions

State function: a property of a system that depends only

on the initial and final states of system, not on how the

internal energy is used. (depends only on the physical state (P,T,etc) of the system and not on the route used to achieve the current state).

The First Law of

Thermodynamics

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State Functions

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However, q and w are not state functions.

Whether the battery is shorted out or is discharged by

running the fan, its Eis the same.


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Chemical reactions can absorb or release heat. However, they also have the ability to do work.

For example, when a gas is produced, then the gas

produced can be used to push a piston, thus doing work.

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

The work performed by the above reaction is called

pressure-volume work.

When the pressure is constant,

(when work is done on the surrounding,

w0 and positive.)

Enthalpy

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Enthalpy,H: Heat transferred between the system andsurroundings carried out under constant pressure.

(internal energy + the product of the pressure and

volume of the system)

Enthalpy is a state function becauseE, P & Vare all state

functions.

If the process occurs at constant pressure,

Enthalpy

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Since we know that

We can write

When H, is positive, the system gains heat from the

surroundings. (endothermic)

When H, is negative, the surroundings gain heat from

the system. (exothermic)

Enthalpy

VPw

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Enthalpy

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For a reaction:

Enthalpy is an extensive property (magnitude His

directly proportional to amount):

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H= -802 kJ2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) H= 1604 kJ

Enthalpies of Reaction

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When we reverse a reaction, we change the sign ofH:

CO2(g) + 2H2O(g) CH4(g) + 2O2(g) H= +802 kJ

Change in enthalpy depends on state:

2H2O(g) 2H2O(l) H= -88 kJ

H2O(l) H2O(g) H= +44 kJ

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

Enthalpies of Reaction

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Heat Capacity and Specific Heat

Calorimetry = measurement of heat flow by expt

Calorimeter = apparatus that measures heat flow.

Heat capacity = the amount of energy required to raisethe temperature of an object (by one degree).

Molar heat capacity = heat capacity of 1 mol of a

substance. Specific heat = specific heat capacity = heat capacity of 1

g of a substance.

Calorimetry

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Calorimetry

Constant Pressure Calorimetry

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Calorimetry

Reaction carried out under

constant volume.

Use a bomb calorimeter.

Usually study combustion.

Ccal is the heat capacity of

the calorimeter.

Bomb Calorimetry (Constant Volume

Calorimetry)

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Hesss law: if a reaction is carried out in a number of

steps, Hfor the overall reaction is the sum ofHfor

each individual step.

For example:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H= -802 kJ

2H2O(g) 2H2O(l) H= -88 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H= -890 kJ

Hesss Law

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Note that:

H1 = H2 + H3

Hesss Law

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SAMPLE EXERCISE 5.8 Using Hesss Law to Calculate H

The enthalpy of reaction for the combustion of C to CO2 is 393.5 kJ/mol C, and the enthalpy for thecombustion of CO to CO

2is 283.0 kJ/mol CO:

Using these data, calculate the enthalpy for the combustion of C to CO:

Solution

In order to use equations (1) and (2), we arrange them so that C(s) is on the reactant side and CO(g)is on the product side of the arrow, as in the target reaction, equation (3). Because equation (1) hasC(s) as a reactant, we can use that equation just as it is. We need to turn equation (2) around,however, so that CO(g) is a product. Remember that when reactions are turned around, the sign of His reversed. We arrange the two equations so that they can be added to give the desired equation:


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SAMPLE EXERCISE 5.8 continued

When we add the two equations, CO2(g) appears on both sides of the arrow and therefore cancels out.Likewise, is eliminated from each side.


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SAMPLE EXERCISE 5.9 Using Three Equations with Hesss Law to Calculate H

Calculate Hfor the reaction

given the following chemical equations and their respective enthalpy changes:

Solution

Analyze: We are given a chemical equation and asked to calculate its Husing three chemicalequations and their associated enthalpy changes.

Plan: We will use Hesss law, summing the three equations or their reverses and multiplying each byan appropriate coefficient so that they add to give the net equation for the reaction of interest. At thesame time, we keep track of the Hvalues, reversing their signs if the reactions are reversed andmultiplying them by whatever coefficient is employed in the equation.

Solve:Because the target equation has C

2H

2as a product, we turn the first equation around; the

sign of H is therefore changed. The desired equation has 2 C(s) as a reactant, so we multiply thesecond equation and its Hby 2. Because the target equation has as a reactant, we keep the thirdequation as it is. We then add the three equations and their enthalpy changes in accordance withHesss law:


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When the equations are added, there are on both sides of the arrow. Theseare canceled in writing the net equation.


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If 1 mol of compound is formed from its constituent

elements, then the enthalpy change for the reaction is

called the enthalpy of formation, Hof .

Standard conditions (standard state): 1 atm and 25oC

(298 K).

Standard enthalpy,

H

o

, is the enthalpy measured wheneverything is in its standard state.

Standard enthalpy of formation: 1 mol of compound is

formed from substances in their standard states.

Enthalpies of Formation

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If there is more than one state for a substance under

standard conditions, the more stable one is used.

Standard enthalpy of formation of the most stable form ofan element is zero.


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Using Enthalpies of Formation tocalculate Enthalpies of Reaction

We use Hess Law to calculate enthalpies of a reaction

from enthalpies of formation.


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Using Enthalpies of Formation tocalculate Enthalpies of Reaction

For a reaction


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SAMPLE EXERCISE 5.10 Identifying Equations associated with Enthalpies of Formation

For which of the following reactions at 25C would the enthalpy change represent a standard enthalpy

of formation? For those where it does not, what changes would need to be made in the reactionconditions?

Solution

Analyze:The standard enthalpy of formation is represented by a reaction in which each reactant is

an element in its standard state and the product is one mole of the compound.

Plan: To solve these problems, we need to examine each equation to determine, first of all, whetherthe reaction is one in which a substance is formed from the elements. Next, we need to determinewhether the reactant elements are in their standard states.

Solve: In (a) Na2O is formed from the elements sodium and oxygen in their proper states, a solidand O2 gas, respectively. Therefore, the enthalpy change for reaction (a) corresponds to a standard

enthalpy of formation.In (b) potassium is given as a liquid. It must be changed to the solid form, its standard state atroom temperature. Furthermore, two moles of product are formed, so the enthalpy change for thereaction as written is twice the standard enthalpy of formation of KCl(s). The proper equation for theformation reaction is


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Reaction (c) does not form a substance from its elements. Instead, a substance decomposes to itselements, so this reaction must be reversed. Next, the element carbon is given as diamond, whereasgraphite is the lowest-energy solid form of carbon at room temperature and 1 atm pressure. Theequation that correctly represents the enthalpy of formation of glucose from its elements is


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SAMPLE EXERCISE 5.11 Calculating an Enthalpy of Reaction fromEnthalpies of Formation

(a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to CO2(g)

and H2O(l). (b) Compare the quantity of heat produced by combustion of 1.00 g propane to thatproduced by 1.00 g benzene.

SolutionAnalyze: (a) We are given a reaction [combustion of C6H6(l) to form CO2(g) and H2O(l)] and askedto calculate its standard enthalpy change, H(b) We then need to compare the quantity of heatproduced by combustion of 1.00 g C6H6 with that produced by 1.00 g of C3H8, whose combustion was

treated above in the text.Plan: (a) We need to write the balanced equation for the combustion of C6H6. We then look upvalues in Appendix C or inTable 5.3 and apply Equation 5.31 to calculate the enthalpy change for thereaction. (b) We use the molar mass of C6H6 to change the enthalpy change per mole to that pergram. We similarly use the molar mass of C3H8 and the enthalpy change per mole calculated in thetext above to calculate the enthalpy change per gram of that substance.

Solve: (a) We know that a combustion reaction involves O2(g) as a reactant. Thus, the balancedequation for the combustion reaction of 1 mol C6H6(l) is

SAMPLE EXERCISE 5 11 continued


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We can calculate Hfor this reaction by using Equation 5.31 and data in Table 5.3. Remember tomultiply the

value for each substance in the reaction by that substances stoichiometric coefficient. Recall

also that

for any element in its most stable form under standard conditions, so

Comment: Both propane and benzene are hydrocarbons. As a rule, the energy obtained from thecombustion of a gram of hydrocarbon is between 40 and 50 kJ.


(b) From the example worked in the text, H =2220 kJ for the combustion of 1 mol of propane. In

part (a) of this exercise we determined that H=3267 kJ for the combustion of 1 mol benzene. Todetermine the heat of combustion per gram of each substance, we use the molar masses to convertmoles to grams:


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Foods

Fuel value = energy released when 1 g of substance is

burned.

1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal. Energy in our bodies comes from carbohydrates and fats

(mostly).

Intestines: carbohydrates converted into glucose which

react with O2 to produce energy.C6H12O6 + 6O2 6CO2 + 6H2O, H = -2816 kJ

Fats break down as follows:2C57H110O6 + 163O2 114CO2 + 110H2O, H

o = -75,520 kJ

Foods and Fuels

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Foods Fats: store excess energy; are insoluble in water, so are

good for energy storage.

Foods and Fuels

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Fuels

In 2000 the United States consumed 1.03 1017 kJ of

fuel.

Most fuel derived from petroleum and natural gas. Remainder from coal, nuclear, and hydroelectric.

Fossil fuels are not renewable.

Foods and Fuels

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Foods and

Fuels

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Fuels

Fuel value = energy released when 1 g of substance is

burned.

Hydrogen has great potential as a fuel with a fuel value of142 kJ/g.

Foods and Fuels

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Each Cl- ion is surrounded by six Na+ ions in NaCl

molecules

There is a regular arrangement of Na

+

and Cl

-

in 3D. Note that the ions are packed as closely as possible.

Note that it is not easy to find

a molecular formula to describe

the ionic lattice.

Lattice Energy in Ions


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Energetics of Ionic Bond

Formation

The formation of Na+(g) and Cl-(g) from Na(g) and

Cl(g) is endothermic.

Why is the formation of Na(s) exothermic? The reaction NaCl(s) Na+(g) + Cl-(g) is endothermic

(H= +788 kJ/mol).

The formation of a crystal lattice from the ions in the

gas phase is exothermic:

Na+(g) + Cl-(g) NaCl(s) H= -788 kJ/mol


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Lattice Energy

Lattice energy: the energy required to completely

separate a mole of an ionic solid into its gaseous ions.

Lattice energy depends on the charges on the ions and

the sizes of the ions:

is a constant (8.99 x 109 Jm/C2), Q1 and Q2 are the

charges on the ions, and dis the distance between ions.

d

QQEl

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L tti i


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Lattice energy increases as

- the charges on the ions increase

- the distance between the ions decreases.


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Chemistry Chapter 01

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