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## Chemistry: Atoms First Julia Burdge & Jason Overby

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Chemistry: Atoms First

Julia Burdge & Jason Overby

Energy Changes in Chemical Reactions

Kent L. McCorkle

Cosumnes River College

Sacramento, CA

Thermochemistry

10.1 Energy and Energy Changes

10.2 Introduction to Thermodynamics

States and State Functions

The First Law of Thermodynamics

Work and Heat

10.3 Enthalpy

Reactions Carried Out at Constant Volume or at Constant Pressure

Enthalpy and Enthalpy Changes

Thermochemical Equations

10.4 Calorimetry

Specific Heat and Heat Capacity

Constant-Pressure Calorimetry

Constant-Volume Calorimetry

10.5 Hess’s Law

10.6 Standard Enthalpies of Formation

10.7 Bond Enthalpy and the Stability of Covalent Molecules

10.8 Lattice Energy and the Stability of Ionic Compounds

The Born-Haber Cycle

Comparison of Ionic and Covalent Compounds

10.1

System

The system is a part of the universe that is of specific interest.

The surroundings constitute the rest of the universe outside the system.

The system is usually defined as the substances involved in chemical and physical changes.

Surroundings

Universe = System + Surroundings

Thermochemistry is the study of heat (the transfer of thermal energy) in chemical reactions.

Heat is the transfer of thermal energy.

Heat is either absorbed or released during a process.

heat

Surroundings

2H2(g) + O2(g)

2H2O(l) + energy

Energy and Energy Changes

An exothermic processoccurs when heat is transferred from the system to the surroundings.

“Feels hot!”

heat

Surroundings

Universe = System + Surroundings

energy + 2HgO(s)

2Hg(l) + O2(g)

Energy and Energy Changes

An endothermic processoccurs when heat is transferred from the surroundings to the system.

“Feels cold”

heat

Surroundings

Universe = System + Surroundings

Introduction to Thermodynamics

10.2

Thermodynamics is the study of the interconversion of heat and other kinds of energy.

In thermodynamics, there are three types

of systems:

An open systemcan exchange mass and

energy with the surroundings.

A closed system allows the transfer

of energy but not mass.

An isolated systemdoes not exchange

either mass or energy with its

surroundings.

Introduction to Thermodynamics

State functionsare properties that are determined by the state of the system, regardless of how that condition was achieved.

The magnitude of change depends only on the initial and final states of the system.

- Energy
- Pressure
- Volume
- Temperature

ΔUsys = –ΔUsurr

The First Law of Thermodynamics

The first law of thermodynamicsstates that energy can be converted from one form to another, but cannot be created or destroyed.

ΔU is the change in the internal energy.

“sys” and “surr” denote system and surroundings, respectively.

ΔU = Uf – Ui; the difference in the energies of the initial and final states.

Work and Heat

The overall change in the system’s internal energy is given by:

q is heat

- q is positive for an endothermic process (heat absorbed by the system)
- q is negative for an exothermic process (heat released by the system)

w is work

- w is positive for work done on the system
- w is negative for work done by the system

Work and Heat

Calculate the overall change in internal energy, ΔU, (in joules) for a system that absorbs 188 J of heat and does 141 J of work on its surroundings.

StrategyCombine the two contributions to internal energy using ΔU = q + w and the sign conventions for q and w.

SolutionThe system absorbs heat, so q is positive. The system does work on the surroundings, so w is negative.

ΔU = q + w = 188 J + (-141 J) = 47 J

Think About ItConsult Table 10.1 to make sure you have used the proper sign conventions for q and w.

10.3

2NaN3(s)

2Na(s) + 3N2(g)

Sodium azide detonates to give a large quantity of nitrogen gas.

Under constant volume conditions, pressure increases:

2Na(s) + 3N2(g)

Enthalpy

Sodium azide detonates to give a large quantity of nitrogen gas.

Under constant volume conditions, pressure increases:

Enthalpy

Pressure-volume, or PV work, is done when there is a volume change under constant pressure.

P is the external opposing pressure.

ΔV is the change in the volume of the container.

1 L∙atm

1 L

1000 mL

101.3 J

1 L∙atm

1 L

1000 mL

Worked Example 10.2

Determine the work done (in joules) when a sample of gas extends from 552 mL to 891 mL at constant temperature (a) against a constant pressure of 1.25 atm, (b) against a constant pressure of 1.00 atm, and (c) against a vacuum (1 L∙atm = 101.3 J).

StrategyDetermine change in volume (ΔV), identify the external pressure (P), and use w = −PΔV to calculate w. The result will be in L∙atm; use the equality 1 L∙atm = 101.3 J to convert to joules.

SolutionΔV = (891 – 552)mL = 339 mL. (a) P = 1.25 atm, (b) P = 1.00 atm, (c) P = 0 atm.

(a) w = -(1.25 atm)(339 mL)

(b) w = -(1.00 atm)(339 mL)

= -42.9 J

= -34.3 J

1 L∙atm

1 L

1000 mL

Worked Example 10.2 (cont.)

Solution

(c) w = -(0 atm)(339 mL)

= 0 J

Think About ItRemember that the negative sign in the answers to part (a) and (b) indicate that the system does work on the surroundings. When an expansion happens against a vacuum, no work is done. This example illustrates that work is not a state function. For an equivalent change in volume, the work varies depending on external pressure against which the expansion must occur.

Enthalpy

Pressure-volume, or PV work, is done when there is a volume change under constant pressure.

w = −PΔV

substitute

ΔU = q + w

ΔU = q − PΔV

When a change occurs at constant volume, ΔV = 0 and no work is done.

Enthalpy

The thermodynamic function of a system called enthalpy (H) is defined by the equation:

A note about SI units:

Pressure: pascal; 1Pa = 1 kg/(m . s2)

Volume: cubic meters; m3

PV: 1kg/(m . s2) x m3 = 1(kg . m2)/s2 = 1 J

Enthalpy: joules

U, P, V, and H are all state functions.

ΔH = ΔU + PΔV

(2)

Rearrange to solve for ΔU:

ΔU = ΔH + PΔV

(3)

Remember, qp:

qp = ΔU + ΔV

(4)

Substitute equation (3) into equation (4) and solve:

qp = (ΔH − PΔV) + PΔV

(5)

for a constant-pressure process

qp = ΔH

Enthalpy and Enthalpy Changes

For any process, the change in enthalpy is:

ΔH = ΔU + Δ(PV)

(1)

ΔH = H(products) – H(reactants)

Enthalpy and Enthalpy Changes

The enthalpy of reaction (ΔH)is the difference between the enthalpies of the products and the enthalpies of the reactants:

Assumes reactions in the lab occur at constant pressure

ΔH > 0 (positive) endothermic process

ΔH < 0 (negative) exothermic process

H2O(l) ΔH = +6.01 kJ/mol

Thermochemical Equations

Concepts to consider:

- Is this a constant pressure process?
- What is the system?
- What are the surroundings?
- ΔH > 0 endothermic

CO2(g) + 2H2O(l) ΔH = −890.4 kJ/mol

Thermochemical Equations

Concepts to consider:

- Is this a constant pressure process?
- What is the system?
- What are the surroundings?
- ΔH < 0 exothermic

mole of reaction as written

Thermochemical Equations

Enthalpy is an extensive property.

Extensive properties are dependent on the amount of matter involved.

H2O(l) → H2O(g) ΔH = +44 kJ/mol

2H2O(l) → 2H2O(g) ΔH = +88 kJ/mol

Double the amount of matter

Double the enthalpy

CO2(g) + 2H2O(g)

ΔH = −802.4 kJ/mol

different states

different enthalpies

CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)

ΔH = +890.4 kJ/mol

Thermochemical Equations

The following guidelines are useful when considering thermochemical equations:

1) Always specify the physical states of reactants and products becausethey help determine the actual enthapy changes.

CO2(g) + 2H2O(g)

ΔH = −802.4 kJ/mol

2CH4(g) + 4O2(g)

2CO2(g) + 4H2O(g)

ΔH = −1604.8 kJ/mol

CH4(g) + 2O2(g)

CO2(g) + 2H2O(g)

ΔH = −802.4 kJ/mol

CO2(g) + 2H2O(g)

CH4(g) + 2O2(g)

ΔH = +802.4 kJ/mol

Thermochemical Equations

The following guidelines are useful when considering thermochemical equations:

2)When multiplying an equation by a factor (n), multiply the ΔH value by same factor.

3)Reversing an equation changes the sign but not the magnitude of ΔH.

180.2 g C6H12O6

Worked Example 10.3

Given the thermochemical equation for photosynthesis,

6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g) ΔH = +2803 kJ/mol

calculate the solar energy required to produce 75.0 g of C6H12O6.

StrategyThe thermochemical equation shows that for every mole of C6H12O6 produced, 2803 kJ is absorbed. We need to find out how much energy is absorbed for the production of 75.0 g of C6H12O6. We must first find out how many moles there are in 75.0 g of C6H12O6.

The molar mass of C6H12O6 is 180.2 g/mol, so 75.0 g of C6H12O6 is

75.0 g C6H12O6 ×

We will multiply the thermochemical equation, including the enthalpy change, by 0.416, in order to write the equation in terms of the appropriate amount of C6H12O6.

= 0.416 mol C6H12O6

Solution

(0.416 mol)[6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g)]

and (0.416 mol)(ΔH) = (0.416 mol)(2803 kJ/mol) gives

2.50H2O(l) + 2.50CO2(g) → 0.416C6H12O6(s) + 2.50O2(g) ΔH = +1.17×103 kJ

Therefore, 1.17×103 kJ of energy in the form of sunlight is consumed in the production of 75.0 g of C6H12O6. Note that the “per mole” units in ΔH are canceled when we multiply the thermochemical equation by the number of moles of C6H12O6.

Think About ItThe specified amount of C6H12O6 is less than half a mole. Therefore, we should expect the associated enthalpy change to be less than half that specified in the thermochemical equation for the production of 1 mole of C6H12O6.

10.4

Calorimetry is the measurement of heat changes.

Heat changes are measured in a device called a calorimeter.

The specific heat(s) of a substance is the amount of heat required to raise the temperature of 1 g of the substance by 1°C.

1 kg water

Specific heat capacity of water

Specific Heat and Heat Capacity

The heat capacity (C) is the amount of heat required to raise the temperature of an object by 1°C.

The “object” may be a given quantity of a particular substance.

Specific heat capacity has units of J/(g• °C)

Heat capacity has units of J/°C

q = CΔT

Specific Heat and Heat Capacity

The heat associated with a temperature change may be calculated:

m is the mass.

s is the specific heat.

ΔT is the change in temperature (ΔT = Tfinal – Tinitial).

C is the heat capacity.

g∙°C

Worked Example 10.4

Calculate the amount of heat (in kJ) required to heat 255 g of water from 25.2°C to 90.5°C.

StrategyUse q = msΔT to calculate q. s = 4.184 J/g∙°C, m = 255 g, ΔT = 90.5°C – 25.2°C = 65.3°C.

Solution

q =

× 255 g × 65.3°C = 6.97×104 J or 69.7 kJ

Think About ItLook carefully at the cancellation of units and make sure that the number of kilojoules is smaller than the number of joules. It is a common error to multiply by 1000 instead of dividing in conversions of this kind.

Calculate the amount of heat required to heat 1.01 kg of water from 0.05°C to 35.81°C.

Solution:

Step 1: Use the equation q = msΔT to calculate q.

A coffee-cup calorimeter may be used to measure the heat exchange for a variety of reactions at constant pressure:

Heat of neutralization:

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

Heat of ionization:

H2O(l) → H+(aq) + OH‒(aq)

Heat of fusion:

H2O(s) → H2O(l)

Heat of vaporization:

H2O(l) → H2O(g)

qsurr = msΔT

qsys = −qsurr

Calorimetry

Concepts to consider for coffee-cup calorimetry:

qP = ΔH

System: reactants and products (the reaction)

Surroundings: water in the calorimeter

For an exothermic reaction:

- the system loses heat
- the surroundings gain (absorb) heat

The minus sign is used to keep sign conventions consistent.

A metal pellet with a mass of 100.0 g, originally at 88.4°C, is dropped into 125 g of water originally at 25.1°C. The final temperature of both pellet and the water is 31.3°C. Calculate the heat capacity C (in J/°C) of the pellet.

StrategyWater constitutes the surroundings; the pellet is the system. Use qsurr = msΔT to determine the heat absorbed by the water; then use q = CΔT to determine the heat capacity of the metal pellet.

mwater = 125 g, swater = 4.184 J/g∙°C, and ΔTwater = 31.3°C – 25.1°C = 6.2°C. The heat absorbed by the water must be released by the pellet: qwater = -qpellet, mpellet = 100.0 g, and ΔTpellet = 31.3°C – 88.4°C = -57.1°C.

g∙°C

Worked Example 10.5 (cont.)

Solution

qwater =

Thus,

qpellet = -3242.6 J

From q = CΔT we have

-3242.6 J = Cpellet × (-57.1°C)

Thus,

Cpellet = 57 J/°C

× 125 g × 6.2°C = 3242.6 J

Think About ItThe units cancel properly to give appropriate units for heat capacity. Moreover, ΔTpellet is a negative number because the temperature of the pellet decreases.

Constant-Volume Calorimetry

Constant volume calorimetry is carried out in a device known as a constant-volume bomb.

A constant-volume calorimeter is an isolated system.

Bomb calorimeters are typically used to determine heats of combustion.

Constant-Volume Calorimetry

To calculate qcal, the heat capacity of the calorimeter must be known.

qcal = CcalΔT

qrxn = −qcal

Think About ItAccording to the label on the cookie package, a service size is four cookies, or 29 g, and each serving contains 150 Cal. Convert the energy per gram to Calories per serving to verify the result.

21.5 kJ

g

1 Cal

4.184 kJ

29 g

serving

×

= 1.5×102 Cal/serving

×

1.559×102 kJ

7.25 g

Worked Example 10.6

A Famous Amos bite-sized chocolate chip cookie weighing 7.25 g is burned in a bomb calorimeter to determine its energy content. The heat capacity of the calorimeter is 39.97 kJ/°C. During the combustion, the temperature of the water in the calorimeter increases by 3.90°C. Calculate the energy content (in kJ/g) of the cookie.

StrategyUse qrxn = -CcalΔT to calculate the heat released by the combustion of the cookie. Divide the heat released by the mass of the cookie to determine its energy content per gram Ccal = 39.97 kJ/°C and ΔT = 3.90°C.

Solution

qrxn = -CcalΔT = -(39.97 kJ/°C)(3.90°C) = -1.559×102 kJ

Because energy content is a positive quantity, we write

energy content per gram =

= 21.5 kJ/g

10.5

CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)

ΔH = −890.4 kJ/mol

2H2O(l)

2H2O(g)

ΔH = +88.0 kJ/mol

CH4(g) + 2O2(g)

CO2(g) + 2H2O(g)

ΔH = −802.4 kJ/mol

Hess’s lawstates that the change in enthalpy for a stepwise process is the sum of the enthalpy changes for each of the steps.

CH4(g) + 2O2(g)

ΔH = −802.4 kJ

ΔH = −890.4 kJ

CO2(g) + 2H2O(g)

ΔH = +88.0 kJ

CO2(g) + 2H2O(l)

When applying Hess’s Law:

1) Manipulate thermochemical equations in a manner that gives the overall desired equation.

2)Remember the rules for manipulating thermochemical equations:

3)Add theΔH for each step after proper manipulation.

4) Process is useful for calculating enthalpies that cannot be found directly.

Always specify the physical states of reactants and products because they help determine the actual enthalpy changes.

When multiplying an equation by a factor (n), multiply the ΔH value by same factor.

Reversing an equation changes the sign but not the magnitude of ΔH.

2

Worked Example 10.7

Given the following thermochemical equations,

determine the enthalpy change for the reaction

NO(g) + O(g) → NO2(g)

NO(g) + O3(g) → NO2(g) + O2(g)

ΔH = –198.9 kJ/mol

ΔH = –142.3 kJ/mol

O3(g) → O2(g)

ΔH = +495 kJ/mol

O2(g) → 2O(g)

StrategyArrange the given thermochemical equations so that they sum to the desired equation. Make the corresponding changes to the enthalpy changes, and add them to get the desired enthalpy change.

2

1

2

3

2

1

2

Worked Example 10.7 (cont.)

Solution The first equation has NO as a reactant with the correct coefficients, so we will use it as is.

The second equation must be reversed so that the O3 introduced by the first equation will cancel (O3 is not part of the overall chemical equation). We also must change the sign on the corresponding ΔH value.

These two steps sum to give:

NO(g) + O3(g) → NO2(g) + O2(g)

ΔH = –198.9 kJ/mol

ΔH = +142.3 kJ/mol

O2(g) → O3(g)

NO(g) + O3(g) → NO2(g) + O2(g)

ΔH = –198.9 kJ/mol

ΔH = +142.3 kJ/mol

+

O2(g)

O2(g) → O3(g)

NO(g) + O2(g) → NO2(g)

ΔH = –56.6 kJ/mol

2

1

2

3

2

1

2

Worked Example 10.7 (cont.)

Solution We then replace O2 on the left with O by incorporating the last equation. To do so, we divide the third equation by 2 and reverse its direction. As a result, we must also divide ΔH value by 2 and change its sign.

Finally, we sum all the steps and add their enthalpy changes.

O(g) → O2(g)

ΔH = –247.5 kJ/mol

NO(g) + O3(g) → NO2(g) + O2(g)

ΔH = –198.9 kJ/mol

ΔH = +142.3 kJ/mol

O2(g) → O3(g)

O(g) → O2(g)

ΔH = –247.5 kJ/mol

+

ΔH = –304 kJ/mol

NO(g) + O(g) → NO2(g)

Think About ItDouble-check the cancellation of identical items–especially where fractions are involved.

Standard Enthalpies of Formation

10.6

C(graphite) + O2(g)

CO2(g)

ΔHf° = −393.5 kJ/mol

The standard enthalpy of formation (ΔHf°) is defined as the heat change that results when 1 mole of a compound is formed from its constituent elements in their standard states.

Elements in standard states

1 mole of product

Standard Enthalpies of Formation

The standard enthalpy of formation (ΔHf°) is defined as the heat change that results when 1 mole of a compound is formed from its constituent elements in their standard states.

The superscripted degree sign denotes standard conditions.

- 1 atm pressure for gases
- 1 M concentration for solutions

“f” stands for formation.

ΔHf° for an element in its most stable form is zero.

ΔHf° for many substances are tabulated in Appendix 2 of the textbook.

ΔH°rxn = [cΔHf°(C) + dΔHf°(D) ] – [aΔHf°(A) + bΔHf°(B)]

ΔH°rxn = ΣnΔHf°(products) – ΣmΔHf°(reactants)

Standard Enthalpies of Formation

The standard enthalpy of reaction (ΔH°rxn) is defined as the enthalpy of a reaction carried out under standard conditions.

aA + bB → cC + dD

n and m are the stoichiometric coefficients for the reactants and products.

Using data from Appendix 2, calculate ΔH°rxn for Ag+(aq) + Cl-(aq) → AgCl(s).

StrategyUse ΔH°rxn = ΣnΔHf°(products) – ΣmΔHf°(reactants) and ΔHf° values from Appendix 2 to calculate ΔH°rxn. The ΔHf° values for Ag+(aq), Cl-(aq), and AgCl(s) are +105.9, –167.2, and –127.0 kJ/mol, respectively.

Solution

ΔH°rxn = ΔHf°(AgCl) – [ΔHf°(Ag+) + ΔHf°(Cl-)]

= –127.0 kJ/mol – [(+105.9 kJ/mol) + (–167.2 kJ/mol)]

= –127.0 kJ/mol – (–61.3 kJ/mol) = –65.7 kJ/mol

Think About ItWatch out for misplaced or missing minus signs. This is an easy place to lose track of them.

2

Worked Example 10.9

Given the following information, calculate the standard enthalpy of formation of acetylene (C2H2) from its constituent elements:

C(graphite) + O2(g) → CO2(g)

ΔH°rxn= –393.5 kJ/mol

(1)

ΔH°rxn= –285.5 kJ/mol

H2(g) + O2(g) → H2O(l)

(2)

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l)

ΔH°rxn= –2598.8 kJ/mol

(3)

StrategyArrange the equations that are provided so that they will sum to the desired equation. This may require reversing or multiplying one or more of the equations. For any such change, the corresponding change must also be made to the ΔH°rxn value. The desired equation, corresponding to the standard enthalpy of formation of acetylene, is

2C(graphite) + H2(g) → C2H2(g)

2

Think About ItRemember that a ΔH°rxn is only a ΔH°f when there is just one product, just one mole produced, and all the reactants are elements in their standard states.

5

2

1

2

5

2

Worked Example 10.9 (cont.)

Solution We multiply Equation (1) and its ΔH°rxn value by 2:

We include Equation (2) and its ΔH°rxn value as is:

We reverse Equation (3) and divide it by 2 (i.e., multiply through by 1/2):

Summing the resulting equations and the corresponding ΔH°rxn values:

2C(graphite) + 2O2(g) → 2CO2(g)

ΔH°rxn= –787.0 kJ/mol

ΔH°rxn= –285.5 kJ/mol

H2(g) + O2(g) → H2O(l)

ΔH°rxn= +1299.4 kJ/mol

2CO2(g) + H2O(l) → C2H2(g) + O2(g)

2C(graphite) + 2O2(g) → 2CO2(g)

ΔH°rxn= –787.0 kJ/mol

ΔH°rxn= –285.5 kJ/mol

H2(g) + O2(g) → H2O(l)

ΔH°rxn= +1299.4 kJ/mol

2CO2(g) + H2O(l) → C2H2(g) + O2(g)

ΔH°f= +226.6 kJ/mol

2C(graphite) + H2(g) → C2H2(g)

Bond Enthalpy and the Stability of Covalent Molecules

10.7

ΔH° = ΣBE(reactants) – ΣBE(products)

The bond enthalpyis the enthalpy change associated with breaking a bond in 1 mole of gaseous molecule.

The enthalpy for a gas phase reaction is given by:

H2(g) → H(g) + H(g) ΔH° = 436.4 kJ/mol

ΔH° = total energy input – total energy released

bonds formed

bonds broken

Bond Enthalpy and the Stability of Covalent Molecules

Bond enthalpy change in an exothermic reaction.:

Bond Enthalpy and the Stability of Covalent Molecules

Bond enthalpy change in an endothermic reaction:

Think About ItUse equation ΔH°rxn = ΣnΔHf°(products) – ΣmΔHf°(reactants) and data from Appendix 2 to calculate this enthalpy of reaction again; then compare your results using the two approaches. The difference in this case is due to two things: most tabulated bond enthalpies are averages and, by convention, we show the product of combustion as liquid water–but average bond enthalpies apply to species in the gas phase, where there is little or no influence exerted by neighboring molecules.

Worked Example 10.10

Use bond enthalpies from Table 10.4 to estimate the enthalpy of reaction for the combustion of methane:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

StrategyDraw Lewis structures to determine what bonds are to be broken and what bonds are to be formed.

Bonds to break: 4 C–H and 2 O=O

Bonds to form: 2 C=O and 4 H–O

Bond enthalpies from Table 10.4: 414 kJ/mol (C–H), 498.7 kJ/mol (O=O), 799 kJ/mol (C=O in CO2), and 460 kJ/mol (H–O).

Solution

[4(414 kJ/mol) + 2(498.7 kJ/mol)] – [2(799 kJ/mol) + 4(460 kJ/mol)] = –785 kJ/mol

Lattice Energy and the Stability of Ionic Compounds

10.8

1

2

A Born-Haber cycleis a cycle that relates the lattice energy of an ionic compound to quantities that can be measured.

Na(s) + Cl2(g) → Na+(g) + Cl-(g)

Lattice Energy and the Stability of Ionic Compounds

1

2

A Born-Haber cycleis a cycle that relates the lattice energy of an ionic compound to quantities that can be measured.

Na(s) + Cl2(g) → Na+(g) + Cl-(g)

Think About ItCompare this value to that for NaCl in Figure 10.14 (787 kJ/mol). Both compounds contain the same anion (Cl-) and both have cations with the same charge (+1), so the relative sizes of the cations will determine the relative strengths of their lattic energies. Because Cs+ is larger than Na+, the lattice energy of CsCl is smaller than the lattice energy of NaCl.

Worked Example 10.11

Using data from Figure 4.8 and 4.10 and Appendix 2, calculate the lattice energy of cesium chloride (CsCl).

StrategyUsing the Born-Haber Cycle Figure as a guide, combine pertinent thermodynamic data and Hess’s law to calculate the lattice energy.

From Figure 4.8, IE1(Cs) = 376 kJ/mol. From Figure 4.10, EA1(Cl) = 349.0 kJ/mol. From Appendix 2, ΔH°f [Cs(g)] = 76.50 kJ/mol, ΔH°f [Cl(g)] = 121.7 kJ/mol, ΔH°f [CsCl(s)] = – 422.8 kJ/mol. Because we are interested in magnitudes only, we can use the absolute values of the thermodynamic data. And, because only the standard heat of formation of CsCl(s) is a negative number, it is only one for which the sign changes.

Solution

{ΔH °f [Cs(g)] + ΔH °f [Cl(g)] + IE1(Cs) + ΔH °f [CsCl(s)]} – EA1(Cl) = lattice energy

= (76.50 kJ/mol + 121.7 kJ/mol + 376 kJ/mol + 422.8 kJ/mol) – 349.0 kJ/mol

= 648 kJ/mol

- A hot piece of copper (at 98.7oC, specific heat = 0.385 J/g•oC) weighs 34.6486 g. When placed in room temperature water, it is calculated that 915.1 J of heat are released by the metal.
- What gains heat?
- What loses heat?
- What is the final temperature of the metal?
- Watch signs!!!!

- Given the following equations:
- 2CO2 (g) O2 (g) + 2CO (g) H = 566.0 kJ
- ½ N2 (g) + ½ O2 (g) NO (g) H = 90.3 kJ
- Calculate the enthalpy change for:
- 2CO (g) + 2NO (g) 2CO2 (g) + N2 (g) H = ?

63

- Use Standard Heat of Formation values to calculate the enthalpy of reaction for:
- C6H12O6(s) C2H5OH(l) + CO2(g)
- Hint: Is the equation balanced?
- Hof (C6H12O6(s)) = -1260.0 kJ/mol
- Hof (C2H5OH(l)) = -277.7 kJ/mol
- Hof (CO2(g)) = -393.5 kJ/mol

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