4.4 Practice Problems
Attempt these problems as if they were real exam questions in an exam environment.
Only look up information if you get severely stuck. Never look at the solution until you have exhausted all efforts to solve the problem.
Which of following has the largest entropy?
- Na(s)
- Na(g)
- Ne(g)
- He(g)
Solution
Which of the following has the largest entropy?
- H2(g)
- H2O(s)
- H2O(l)
- H2O(g)
Solution
Answer: D
In this series, the largest molecule (by mass) in the gas phase has the largest entropy. The standard entropy values (in J mol–1 K–1) for 1 mole of each sample is given below.
- H2(g): 130.7
- H2(s): 41
- H2(l): 70
- H2(g): 188.8
The entropy of a substance increases when undergoing which process?
- freezing
- vaporization
- condensation
Solution
Answer: B
Substances transitioning into the gas phase will experience the largest change in entropy relative to liquids or solids.
Predict the entropy change for the following reaction:
\[\mathrm{C_3H_8}(g) + \mathrm{5O_2}(g) \longrightarrow \mathrm{3CO_2}(g) + \mathrm{4H_2O}(g)\]
- ΔS > 0
- ΔS < 0
- ΔS = 0
Solution
Answer: A
The reaction begins with six moles of gas and results in 7 moles of gas. More gas means higher entropy. Therefore, the entropy change of the reaction is positive.
Predict the entropy change for the following reaction.
\[3\mathrm{CO_2}(g) + 4\mathrm{H_2O}(g) \longrightarrow \mathrm{C_3H_8}(g) + 5\mathrm{O_2}(g)\]
- ΔS > 0
- ΔS = 0
- ΔS < 0
- cannot be predicted
Solution
Answer: C
The reaction begins with 7 moles of gas and results in 6 moles of gas. Less gas means lower entropy. Therefore, the entropy change of the reaction is negative.
What is the temperature (in °C) at which the following reaction is at equilibrium?
\[\mathrm{CCl_4}(g) \longrightarrow \mathrm{C}(s,\text{graphite}) + 2\mathrm{Cl_2}(g) \\[1.5ex] \Delta H^{\circ} = 95.7~\mathrm{kJ~mol^{-1}} \quad \Delta S^{\circ} = 142.2~\mathrm{J~mol^{-1}~K^{-1}}\]
- 53
- 239
- 400
- 673
Solution
Answer: C
We use the Gibbs free energy equation for this question.
\[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]
At equilibrium, ΔG = 0. Rearrange the equation and solve for temperature.
\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta S^{\circ} \\[1.5ex] 0 &= \Delta H^{\circ} - T\Delta S^{\circ} \\[1.5ex] T\Delta S^{\circ} &= \Delta H^{\circ} \\[2.5ex] T &= \dfrac{\Delta H^{\circ}}{\Delta S^{\circ}} \\[1.5ex] &= \dfrac{95.7~\mathrm{kJ~mol^{-1}}}{142.2~\mathrm{J~mol^{-1}~K^{-1}} \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right )} \\[1.5ex] &= 672.9~\mathrm{K} \end{align*}\]
Convert temperature to °C.
\[672.9~\mathrm{K} - 273.15 = 399.8~^{\circ}\mathrm{C}\]
What is the temperature (in °C) at which the following reaction switches between being spontaneous and non-spontaneous according to the following thermodynamic values?
\[\mathrm{H_2O}(l) \longrightarrow \mathrm{H_2O}(g) \\[1.5ex] \Delta H^{\circ} = 44.01~\mathrm{kJ~mol^{-1}} \quad \Delta S^{\circ} = 118.8~\mathrm{J~mol^{-1}~K^{-1}}\]
- 40.2
- 97.3
- 148.6
- 370.5
Solution
Answer: B
We use the Gibbs free energy equation for this question.
\[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]
At equilibrium, ΔG = 0. Rearrange the equation and solve for temperature.
\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta S^{\circ} \\[1.5ex] 0 &= \Delta H^{\circ} - T\Delta S^{\circ} \\[1.5ex] T\Delta S^{\circ} &= \Delta H^{\circ} \\[2.5ex] T &= \dfrac{\Delta H^{\circ}}{\Delta S^{\circ}} \\[1.5ex] &= \dfrac{44.01~\mathrm{kJ~mol^{-1}}}{118.8~\mathrm{J~mol^{-1}~K^{-1}} \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right )} \\[1.5ex] &= 370.45~\mathrm{K} \end{align*}\]
Convert temperature to °C.
\[370.45~\mathrm{K} - 273.15 = 97.3~^{\circ}\mathrm{C}\]
Liquid water is in equilibrium with water vapor with a vapor pressure of 0.0313 atm (at 25 °C). What is ΔGrxn (in kJ) for this process at equilibrium given the following information?
\[\mathrm{H_2O}(l) \longrightarrow \mathrm{H_2O}(g) \quad \Delta G^{\circ} = 8.59~\mathrm{kJ~mol^{-1}}\]
- –12
- 0
- 54
- 134
Solution
Answer: B
Processes at equilibrium have a ΔG = 0.
What is ΔS°rxn (in J K–1) for the following reaction at 25 °C?
\[\mathrm{4NH_3}(g) + 5\mathrm{O_2}(g) \longrightarrow \mathrm{NO}(g) + \mathrm{6H_2O}(g)\]
Substance S° (J mol–1 K–1) H2O(g) 188.8 NH3(g) 192.8 O2(g) 205.2 - 1.6
- 88
- 104
- 178.8
Solution
Answer: D
What is the sign for ΔH and ΔS, respectively, for a reaction that is spontaneous at any temperature?
- –,+
- +,+
- +,–
- –,–
Solution
Answer: A
What is ΔGrxn (in kJ mol–1) for the following reaction at 25 °C?
\[\mathrm{SO_2}(g) + \frac{1}{2}\mathrm{O_2}(g) \longrightarrow \mathrm{SO_3}(g)\]
Substance ΔH°f (kJ mol–1) S° (J mol–1 K–1) O2(g) 0 205.2 SO2(g) -296.83 248.2 SO3(g) -395.72 256.76 - –234.8
- –70.9
- –40.3
- 66.4
Solution
Answer: B
What is ΔS°rxn (in J) for the following reaction at 25 °C?
\[\mathrm{CH_3OH}(l) + 3\mathrm{O_2}(g) \longrightarrow \mathrm{2CO_2}(g) + \mathrm{4H_2O}(l)\]
Substance S° (J mol–1 K–1) CH3OH(l) 126.8 O2(g) 205.2 CO2(g) 213.8 H2O(l) 70 - –208.4
- –161.6
- –125.5
- 88.9
Solution
Answer: B
A reaction is found to have a ΔG° = –123.45 kJ mol–1. Predict K.
- K > 0
- K ≈ 0
- K < 0
Solution
Answer: A
What is ΔG (in kJ) for the following reaction when PN2 = 0.870 atm, PH2 = 0.250 atm and PNH3 = 12.9 atm (at 25 °C)?
- 6.3
- 9.68
- 34.1
- 54.4
Solution
Answer: B