1. Example 18.1 Predicting the Sign of Entropy Change
Predict the sign of ΔS for each process.
a. H2O(g) H2O(l)
b. Solid carbon dioxide sublimes.
c. 2 N2O(g) N2(g) + O2(g)
Solution
a. Since a gas has a greater entropy than a liquid, the entropy decreases and ΔS is negative.
b. Since a solid has a lower entropy than a gas, the entropy increases and ΔS is positive.
c. Since the number of moles of gas increases, the entropy increases and ΔS is positive.
For Practice 18.1
Predict the sign of ΔS for each process.
a. the boiling of water
b. I2(g) I2(s)
c. CaCO3(s) CaO(s) + CO2(g)

2. Example 18.2 Calculating 𝚫S for a Change of State
Calculate the change in entropy that occurs in the system when 25.0 g of water condenses from a gas to a liquid at the normal boiling point of water (100.0 °C).
Solution
Because the condensation is occurring at the boiling point of water and the temperature is constant, you can use Equation 18.1 to calculate the change in entropy of the system.
Gather the necessary quantities. Look up the value of the enthalpy of vaporization for water at its boiling point in Table The enthalpy of condensation is the same in value but opposite in sign.
Calculate the temperature in K.
ΔHvap = 40.7 kJ/mol
ΔHcondensation = –40.7 kJ/mol
T(K) = T(°C) =
= K

3. Example 18.2 Calculating 𝚫S for a Change of State
Continued
Substitute into Equation 18.1 to calculate the change in entropy for the system.
Check
The answer has the correct units for entropy (J/K). The answer is negative as you would expect for the condensation of a gas to a liquid.
For Practice 18.2
Calculate the change in entropy that occurs in the system when 10.0 g of acetone (C3H6O) vaporizes from a liquid to a gas at its normal boiling point (56.1 °C).

4. C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) ΔHrxn = –2044 kJ
Example 18.3 Calculating Entropy Changes in the Surroundings
Consider the combustion of propane gas:
C3H8(g) + 5 O2(g) CO2(g) + 4 H2O(g) ΔHrxn = –2044 kJ
a. Calculate the entropy change in the surroundings when this reaction occurs at 25 °C.
b. Determine the sign of the entropy change for the system.
c. Determine the sign of the entropy change for the universe. Is the reaction spontaneous?
Solution
a. The entropy change of the surroundings is given by Equation Substitute the value of ΔHrxn and the temperature in kelvins and calculate ΔSsurr.
b. Determine the number of moles of gas on each side of the reaction. An increase in the number of moles of gas implies a positive ΔSsys.
C3H8(g) + 5 O2(g) CO2(g) + 4 H2O(g)
6 mol gas mol gas
ΔSsys is positive.

5. Example 18.3 Calculating Entropy Changes in the Surroundings
Continued
c. The change in entropy of the universe is the sum of the entropy changes of the system and the surroundings. If the entropy changes of the system and surroundings are both the same sign, the entropy change for the universe also has the same sign.
Therefore, ΔSuniv is positive and the reaction is spontaneous.
For Practice 18.3
Consider the reaction between nitrogen and oxygen gas to form dinitrogen monoxide:
2 N2(g) + O2(g) N2O(g) ΔHrxn = kJ
a. Calculate the entropy change in the surroundings when this reaction occurs at 25 °C.
b. Determine the sign of the entropy change for the system.
c. Determine the sign of the entropy change for the universe. Is the reaction spontaneous?
For More Practice 18.3
For a reaction, ΔHrxn = –107 kJ and ΔSrxn = 285 J/K. At what temperature is the change in entropy for this reaction equal to the change in entropy for the surroundings?

6. CCl4(g) C(s, graphite) + 2 Cl2(g) ΔH = +95.7 kJ; ΔS = +142.2 J/K
Example 18.4 Calculating Gibbs Free Energy Changes and Predicting Spontaneity from 𝚫H and 𝚫S
Consider the reaction for the decomposition of carbon tetrachloride gas:
CCl4(g) C(s, graphite) + 2 Cl2(g) ΔH = kJ; ΔS = J/K
a. Calculate ΔG at 25 °C and determine whether the reaction is spontaneous.
b. If the reaction is not spontaneous at 25 °C, determine at what temperature (if any) the reaction becomes spontaneous.
Solution
a. Use Equation 18.8 to calculate ΔG from the given values of ΔH and ΔS. The temperature must be in kelvins. Be sure to express both ΔH and ΔS in the same units (usually joules).
The reaction is not spontaneous.

7. C2H4(g) + H2(g) C2H6(g) ΔH = –137.5 kJ; ΔS = –120.5 J/K
Example 18.4 Calculating Gibbs Free Energy Changes and Predicting Spontaneity from 𝚫H and 𝚫S
Continued
b. Since ΔS is positive, ΔG becomes more negative with increasing temperature. To determine the temperature at which the reaction becomes spontaneous, use Equation 18.8 to find the temperature at which ΔG changes from positive to negative (set ΔG = 0 and solve for T). The reaction is spontaneous above this temperature.
For Practice 18.4
Consider the reaction:
C2H4(g) + H2(g) C2H6(g) ΔH = –137.5 kJ; ΔS = –120.5 J/K
Calculate ΔG at 25 °C and determine whether the reaction is spontaneous. Does ΔG become more negative or more
Positive as the temperature increases?

8. 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
Example 18.5 Calculating Standard Entropy Changes (𝚫S°rxn)
Calculate ΔS°rxn for the balanced chemical equation:
4 NH3(g) + 5 O2(g) NO(g) + 6 H2O(g)
Solution
Begin by looking up the standard entropy for each reactant and product in Appendix IIB. Always note the correct state—(g), (l), (aq), or (s)—for each reactant and product.
Calculate ΔS°rxn by substituting the appropriate values into Equation Remember to include the stoichiometric coefficients in your calculation.

9. 2 H2S(g) + 3 O2(g) 2 H2O(g) + 2 SO2(g)
Example 18.5 Calculating Standard Entropy Changes (𝚫S°rxn)
Continued
Check
Notice that ΔS°rxn is positive, as you would expect for a reaction in which the number of moles of gas increases.
For Practice 18.5
Calculate ΔS°rxn for the balanced chemical equation:
2 H2S(g) + 3 O2(g) H2O(g) + 2 SO2(g)

10. Example 18.6 Calculating the Standard Change in Free Energy for a Reaction Using 𝚫G°rxn = 𝚫H°rxn − T𝚫S°rxn
One of the possible initial steps in the formation of acid rain is the oxidation of the pollutant SO2 to SO3 by the reaction:
SO2(g) O2(g) SO3(g)
Calculate ΔG°rxn at 25 °C and determine whether the reaction is spontaneous.
Solution
Begin by looking up (in Appendix IIB) the standard enthalpy of formation and the standard entropy for each reactant and product.
Calculate ΔH°rxn using Equation 6.15.

11. Example 18.6 Calculating the Standard Change in Free Energy for a Reaction Using 𝚫G°rxn = 𝚫H°rxn − T𝚫S°rxn
Continued
Calculate ΔS°rxn using Equation
Calculate ΔG°rxn using the calculated values of ΔH°rxn and ΔS°rxn and Equation Convert the temperature to
kelvins.
The reaction is spontaneous at this temperature because ΔG°rxn is negative.

12. Example 18.6 Calculating the Standard Change in Free Energy for a Reaction Using 𝚫G°rxn = 𝚫H°rxn − T𝚫S°rxn
Continued
For Practice 18.6
Consider the oxidation of NO to NO2:
NO(g) + 12 O2(g) NO2(g)
Calculate ΔG°rxn at 25 °C and determine whether the reaction is spontaneous at standard conditions.

13. Example 18.7 Estimating the Standard Change in Free Energy for a Reaction at a Temperature Other Than 25 °C Using 𝚫G°rxn = 𝚫H°rxn − T𝚫S°rxn
For the reaction in Example 18.6, estimate the value of ΔG°rxn at 125 °C. Is the reaction more or less spontaneous at this elevated temperature; that is, is the value of ΔG°rxn more negative (more spontaneous) or more positive (less
spontaneous)?
Solution
Estimate ΔG°rxn at the new temperature using the calculated values of ΔH°rxn and ΔS°rxn from Example For T, convert the given temperature to kelvins. Make sure to use the same units for ΔH°rxn and ΔS°rxn (usually joules).
Since the value of ΔG°rxn at this elevated temperature is less negative (or more positive) than the value of ΔG°rxn at 25 °C (which is –70.9 kJ), the reaction is less spontaneous.
For Practice 18.7
For the reaction in For Practice 18.6, calculate the value of ΔG°rxn at –55 °C. Is the reaction more spontaneous (more negative ΔG°rxn) or less spontaneous (more positive ΔG°rxn) at the lower temperature?

14. CH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4 O3(g)
Example 18.8 Calculating 𝚫G°rxn from Standard Free Energies of Formation
Ozone in the lower atmosphere is a pollutant that can form by the following reaction involving the oxidation of
unburned hydrocarbons:
CH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4 O3(g)
Use the standard free energies of formation to determine ΔG°rxn for this reaction at 25 °C.
Solution
Begin by looking up (in Appendix IIB) the standard free energies of formation for each reactant and product. Remember that the standard free energy of formation of a pure element in its standard state is zero.

15. 2 CO(g) + 2 NO(g) 2 CO2(g) + N2(g)
Example 18.8 Calculating 𝚫G°rxn from Standard Free Energies of Formation
Continued
Calculate ΔG°rxn by substituting into Equation
For Practice 18.8
One of the reactions that occurs within a catalytic converter in the exhaust pipe of a car is the simultaneous oxidation of carbon monoxide and reduction of NO (both of which are harmful pollutants):
2 CO(g) + 2 NO(g) CO2(g) + N2(g)
Use standard free energies of formation to determine ΔG°rxn for this reaction at 25 °C. Is the reaction spontaneous at
Standard conditions?
For More Practice 18.8
In For Practice 18.8, you calculated ΔG°rxn for the simultaneous oxidation of carbon monoxide and reduction of NO using standard free energies of formation. Calculate ΔG°rxn for that reaction again at 25 °C, only this time use ΔG°rxn = ΔH°rxn – TΔS°rxn. How do the two values compare? Use your results to calculate ΔG°rxn at K and explain why you could not calculate ΔG°rxn at K using tabulated standard free energies of formation.

16. Example 18.9 Calculating 𝚫G°rxn for a Stepwise Reaction
Find ΔG°rxn for the reaction:
3 C(s) + 4 H2(g) C3H8(g)
Use the following reactions with known ΔG°rxn values:
C3H8(g) + 5 O2(g) CO2(g) + 4 H2O(g) ΔG°rxn = –2074 kJ
C(s) + O2(g) CO2(g) ΔG°rxn = –394.4 kJ
2 H2(g) + O2(g) H2O(g) ΔG°rxn = –457.1 k
Solution
To work this problem, you need to manipulate the given reactions with known values of ΔG°rxn in such a way as to get the reactants of interest on the left, the products of interest on the right, and other species to cancel.
Since the first reaction has C3H8 as a reactant, and the reaction of interest has C3H8 as a product, reverse the first reaction and change the sign of ΔG°rxn.
3 CO2(g) + 4 H2O(g) C3H8(g) + 5 O2(g) ΔG°rxn = kJ
The second reaction has C as a reactant and CO2 as a product, as required in the reaction of interest. However, the coefficient for C is 1, and in the reaction of interest, the coefficient for C is 3. Therefore, multiply this equation and its
ΔG°rxn by 3.
3 × [C(s) + O2(g) CO2(g)] ΔG°rxn = 3 × (–394.4 kJ)
= –1183 kJ

17. Example 18.9 Calculating 𝚫G°rxn for a Stepwise Reaction
Continued
In the third reaction H2(g) is a reactant, as required. However, the coefficient for H2 is 2, and in the reaction of interest, the coefficient for H2 is 4. Multiply this reaction and its ΔG°rxn by 2.
2 × [2 H2(g) + O2(g) H2O(g)] ΔG°rxn = 2 × (–457.1 kJ)
= –914.2
Lastly, rewrite the three reactions after multiplying through by the indicated factors and show how they sum to the reaction of interest. ΔG°rxn for the reaction of interest is then the sum of the ΔG’s for the steps.

18. Example 18.9 Calculating 𝚫G°rxn for a Stepwise Reaction
Continued
For Practice 18.9
Find ΔG°rxn for the reaction:
N2O(g) + NO2(g) NO(g)
Use the following reactions with known ΔG°rxn values:
2 NO(g) + O2(g) NO2(g) ΔG°rxn = –71.2 kJ
N2(g) + O2(g) NO(g) ΔG°rxn = kJ
2 N2O(g) N2(g) + O2(g) ΔG°rxn = –207.4 kJ

19. Example 18.10 Calculating 𝚫G°rxn under Nonstandard Conditions
Consider the reaction at 298 K:
2 NO(g) + O2(g) NO2(g) ΔG°rxn = –71.2 kJ
Calculate ΔGrxn under these conditions:
PNO = atm; PO2 = atm; PNO2 = 2.00 atm
Is the reaction more or less spontaneous under these conditions than under standard conditions?
Solution
Use the law of mass action to calculate Q.
Substitute Q, T, and ΔG°rxn into Equation to calculate ΔGrxn. (Since the units of R include joules, write ΔG°rxn in joules.)
The reaction is spontaneous under these conditions, but less spontaneous than it would be under standard conditions (because ΔGrxn is less negative than ΔG°rxn).

20. Example 18.10 Calculating 𝚫G°rxn under Nonstandard Conditions
Continued
Check
The calculated result is consistent with what you would expect based on Le Châtelier’s principle; increasing the
concentration of the products and decreasing the concentration of the reactants relative to standard conditions should
make the reaction less spontaneous than it was under standard conditions.
For Practice 18.10
Consider the reaction at 298 K:
2 H2S(g) + SO2(g) S(s, rhombic) + 2 H2O(g) ΔG°rxn = –102 kJ
Calculate ΔGrxn under these conditions:
PH2S = 2.00 atm; PSO2 = 1.50 atm; PH2O = atm
Is the reaction more or less spontaneous under these conditions than under standard conditions?

21. Example 18.11 The Equilibrium Constant and 𝚫G°rxn
Use tabulated free energies of formation to calculate the equilibrium constant for the reaction at 298 K:
N2O4(g) NO2(g)
Solution
Look up (in Appendix IIB) the standard free energies of formation for each reactant and product.
Calculate ΔG°rxn by substituting into Equation

22. Example 18.11 The Equilibrium Constant and 𝚫G°rxn
Continued
Calculate K from ΔG°rxn by solving Equation for K and substituting the values of ΔG°rxn and temperature.
For Practice 18.11
Calculate ΔG°rxn at 298 K for the reaction:
I2(g) + Cl2(g) ICl(g) Kp = 81.9