Physical Chemistry 1 Quiz: Using Thermodynamic Tables
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Using Thermodynamic TablesQuestion 1 of 20

The decomposition of limestone follows: CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) with ΔH∘=+178.3 kJ/mol\Delta H^\circ = +178.3 \text{ kJ/mol} and ΔS∘=+160.5 J/mol\cdotpK\Delta S^\circ = +160.5 \text{ J/mol·K}. A student claims this reaction becomes spontaneous above a certain temperature due to the entropy term overwhelming the enthalpy term. At what temperature does ΔG∘\Delta G^\circ first become zero?

1000 K1000 \text{ K}
1111 K1111 \text{ K}
901 K901 \text{ K}
1211 K1211 \text{ K}
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Physical Chemistry 1 Quiz

Physical Chemistry 1 Quiz: Using Thermodynamic Tables

Practice Using Thermodynamic Tables in Physical Chemistry 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Using Thermodynamic Tables, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The decomposition of limestone follows: CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) with ΔH∘=+178.3 kJ/mol\Delta H^\circ = +178.3 \text{ kJ/mol} and ΔS∘=+160.5 J/mol\cdotpK\Delta S^\circ = +160.5 \text{ J/mol·K}. A student claims this reaction becomes spontaneous above a certain temperature due to the entropy term overwhelming the enthalpy term. At what temperature does ΔG∘\Delta G^\circ first become zero?

  1. 1000 K1000 \text{ K}
  2. 1111 K1111 \text{ K} (correct answer)
  3. 901 K901 \text{ K}
  4. 1211 K1211 \text{ K}
Explanation: When you encounter a question about spontaneity and temperature, you're dealing with the Gibbs free energy equation: ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ. A reaction becomes spontaneous when ΔG∘\Delta G^\circ becomes negative, and the transition point occurs exactly when ΔG∘=0\Delta G^\circ = 0. To find when ΔG∘\Delta G^\circ first becomes zero, set the equation equal to zero and solve for temperature: 0=ΔH∘−TΔS∘0 = \Delta H^\circ - T\Delta S^\circ T=ΔH∘ΔS∘T = \frac{\Delta H^\circ}{\Delta S^\circ} First, convert units so they match. Since ΔH∘=+178.3 kJ/mol=+178,300 J/mol\Delta H^\circ = +178.3 \text{ kJ/mol} = +178,300 \text{ J/mol} and ΔS∘=+160.5 J/mol\cdotpK\Delta S^\circ = +160.5 \text{ J/mol·K}: T=178,300 J/mol160.5 J/mol\cdotpK=1111 KT = \frac{178,300 \text{ J/mol}}{160.5 \text{ J/mol·K}} = 1111 \text{ K} This confirms answer B is correct. Looking at the wrong answers: A (1000 K) would result from rounding errors or using unconverted units. C (901 K) appears to come from an inverted calculation or arithmetic mistake. D (1211 K) likely results from incorrectly adding rather than dividing the thermodynamic values. Study tip: Always check your unit conversions when working with thermodynamic data—enthalpy is typically given in kJ/mol while entropy is in J/mol·K. Also remember that endothermic reactions with positive entropy changes (like decompositions producing gas) become spontaneous at high temperatures because the TΔST\Delta S term eventually dominates.

Question 2

A student wants to calculate ΔH∘\Delta H^\circ for the reaction C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)\text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l) using standard enthalpies of formation. If ΔHf∘[C2H4(g)]=52.3 kJ/mol\Delta H_f^\circ[\text{C}_2\text{H}_4(g)] = 52.3 \text{ kJ/mol}, ΔHf∘[CO2(g)]=−393.5 kJ/mol\Delta H_f^\circ[\text{CO}_2(g)] = -393.5 \text{ kJ/mol}, and ΔHf∘[H2O(l)]=−285.8 kJ/mol\Delta H_f^\circ[\text{H}_2\text{O}(l)] = -285.8 \text{ kJ/mol}, what is the enthalpy change for this combustion reaction?

  1. −1411.0 kJ/mol-1411.0 \text{ kJ/mol} (correct answer)
  2. −1358.6 kJ/mol-1358.6 \text{ kJ/mol}
  3. −1463.4 kJ/mol-1463.4 \text{ kJ/mol}
  4. −679.3 kJ/mol-679.3 \text{ kJ/mol}
Explanation: Using ΔH°rxn = ΣΔHf°(products) - ΣΔHf°(reactants): ΔH°rxn = [2(-393.5) + 2(-285.8)] - [52.3 + 3(0)] = [-787.0 - 571.6] - [52.3] = -1358.6 - 52.3 = -1410.9 ≈ -1411.0 kJ/mol. Choice B omits the reactant term. Choice C incorrectly uses ΔHf° for gaseous water. Choice D incorrectly uses coefficients of 1 instead of 2.

Question 3

For the reaction PCl5(g)⇌PCl3(g)+Cl2(g)\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g), a student calculates ΔH∘=+92.5 kJ/mol\Delta H^\circ = +92.5 \text{ kJ/mol} using formation enthalpies, and ΔS∘=+179.8 J/mol\cdotpK\Delta S^\circ = +179.8 \text{ J/mol·K} using standard entropies. What is ΔG∘\Delta G^\circ for this reaction at 400 K?

  1. +164.4 kJ/mol+164.4 \text{ kJ/mol}
  2. −164.4 kJ/mol-164.4 \text{ kJ/mol}
  3. +20.6 kJ/mol+20.6 \text{ kJ/mol} (correct answer)
  4. +20,630 kJ/mol+20,630 \text{ kJ/mol}
Explanation: When you encounter a thermodynamics problem asking for ΔG∘\Delta G^\circ, you're dealing with the fundamental relationship between enthalpy, entropy, and free energy. The key equation is the Gibbs-Helmholtz equation: ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ. To solve this, substitute the given values directly: ΔG∘=92.5 kJ/mol−(400 K)(179.8 J/mol\cdotpK)\Delta G^\circ = 92.5 \text{ kJ/mol} - (400 \text{ K})(179.8 \text{ J/mol·K}). The critical step is unit conversion—you must convert the entropy term to kJ: 179.8 J/mol\cdotpK=0.1798 kJ/mol\cdotpK179.8 \text{ J/mol·K} = 0.1798 \text{ kJ/mol·K}. This gives: ΔG∘=92.5−(400)(0.1798)=92.5−71.9=+20.6 kJ/mol\Delta G^\circ = 92.5 - (400)(0.1798) = 92.5 - 71.9 = +20.6 \text{ kJ/mol}. Option A (+164.4 kJ/mol) results from incorrectly adding the enthalpy and entropy terms instead of subtracting: 92.5+71.9=164.492.5 + 71.9 = 164.4. This ignores the negative sign in the Gibbs equation. Option B (-164.4 kJ/mol) compounds the addition error by also getting the overall sign wrong, perhaps from misunderstanding the positive enthalpy value. Option D (+20,630 kJ/mol) comes from failing to convert joules to kilojoules, calculating 92.5+(400)(179.8)92.5 + (400)(179.8) and treating everything as kJ/mol. Study tip: Always check your units carefully in thermodynamics calculations. The entropy term TΔST\Delta S frequently appears in J/mol, but ΔH\Delta H is typically given in kJ/mol. Convert before calculating, and remember that the Gibbs equation subtracts the entropy term.

Question 4

For the reaction N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g), the standard entropy change is ΔS∘=+175.8 J/mol\cdotpK\Delta S^\circ = +175.8 \text{ J/mol·K} and the standard enthalpy change is ΔH∘=+57.2 kJ/mol\Delta H^\circ = +57.2 \text{ kJ/mol}. At what temperature will this reaction become thermodynamically favorable (ΔG∘<0\Delta G^\circ < 0) under standard conditions?

  1. Above 325 K, since entropy favors product formation at higher temperatures (correct answer)
  2. Above 298 K, since the positive entropy change dominates at room temperature
  3. Above 325 K, since this is where the enthalpy and entropy terms balance
  4. Below 325 K, since the entropy term becomes negligible at lower temperatures
Explanation: Using ΔG° = ΔH° - TΔS°, the reaction becomes favorable when ΔG° < 0, which occurs when TΔS° > ΔH°. Setting ΔH° = TΔS°: 57,200 J/mol = T(175.8 J/mol·K), so T = 325 K. Above this temperature, the entropy term dominates and ΔG° becomes negative. Choice B uses incorrect temperature. Choice C is correct numerically but gives wrong reasoning. Choice D has the temperature relationship backwards.

Question 5

A reaction has ΔH∘=−45.6 kJ/mol\Delta H^\circ = -45.6 \text{ kJ/mol} and ΔS∘=−125.3 J/mol\cdotpK\Delta S^\circ = -125.3 \text{ J/mol·K}. Using the relationship ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ, determine the temperature range where this reaction is thermodynamically spontaneous.

  1. Never spontaneous because both enthalpy and entropy oppose the reaction
  2. Spontaneous at all temperatures above 364 K due to entropy dominance
  3. Spontaneous only at exactly 364 K where enthalpy and entropy balance
  4. Spontaneous at all temperatures below 364 K due to favorable enthalpy (correct answer)
Explanation: When you encounter thermodynamic spontaneity problems, you need to determine when ΔG∘<0\Delta G^\circ < 0, since negative Gibbs free energy indicates a spontaneous process. Let's analyze this systematically. You have ΔH∘=−45.6 kJ/mol\Delta H^\circ = -45.6 \text{ kJ/mol} (favorable, exothermic) and ΔS∘=−125.3 J/mol\cdotpK\Delta S^\circ = -125.3 \text{ J/mol·K} (unfavorable, decreasing entropy). Using ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ, the entropy term becomes +TΔS∘+T\Delta S^\circ because of the negative entropy value. To find where ΔG∘=0\Delta G^\circ = 0 (the boundary between spontaneous and non-spontaneous): 0=−45,600−T(−125.3)0 = -45,600 - T(-125.3) T=45,600125.3=364 KT = \frac{45,600}{125.3} = 364 \text{ K} At temperatures below 364 K, the favorable enthalpy term dominates the unfavorable entropy term, making ΔG∘<0\Delta G^\circ < 0 (spontaneous). Above 364 K, the entropy penalty becomes too large, making ΔG∘>0\Delta G^\circ > 0 (non-spontaneous). Answer A is wrong because the reaction is spontaneous at low temperatures—enthalpy can overcome entropy. Answer B reverses the temperature dependence; higher temperatures make the entropy penalty worse, not better. Answer C incorrectly suggests spontaneity occurs only at the equilibrium temperature, when actually that's where ΔG∘=0\Delta G^\circ = 0. Study tip: Remember that enthalpy-driven reactions (negative ΔH\Delta H, negative ΔS\Delta S) are spontaneous at low temperatures, while entropy-driven reactions are spontaneous at high temperatures. The temperature term in the Gibbs equation amplifies the entropy contribution.

Question 6

For the reaction CaCO3(s)⇌CaO(s)+CO2(g)\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g), you need to determine at what temperature this decomposition becomes thermodynamically favorable (ΔG°=0\Delta G° = 0). Using the data in the table, what is the approximate temperature at which ΔG°=0\Delta G° = 0?

  1. 1100 K assuming temperature-independent ΔH°\Delta H° and ΔS°\Delta S° values (correct answer)
  2. 835 K assuming temperature-independent ΔH°\Delta H° and ΔS°\Delta S° values
  3. 1180 K assuming temperature-independent ΔH°\Delta H° and ΔS°\Delta S° values
  4. 298 K since all tabulated values are given at this temperature
  5. 650 K assuming temperature-independent ΔH°\Delta H° and ΔS°\Delta S° values
Explanation: First calculate: ΔH°rxn = [(-635.1) + (-393.5)] - [-1206.9] = +178.3 kJ/mol and ΔS°rxn = [(38.1) + (213.8)] - [92.9] = +159.0 J/mol·K. At equilibrium, ΔG° = 0, so T = ΔH°/ΔS° = 178,300 J/mol ÷ 159.0 J/mol·K ≈ 1121 K ≈ 1100 K. Choice B and E have calculation errors. Choice C is close but less accurate. Choice D misunderstands the concept.

Question 7

The reduction of iron(III) oxide by carbon monoxide: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g) is important in steel production. Referring to the table, calculate the standard entropy change for this reaction.

  1. +15.2 J/mol\cdotpK+15.2 \text{ J/mol·K} (correct answer)
  2. −15.2 J/mol\cdotpK-15.2 \text{ J/mol·K}
  3. +541.7 J/mol\cdotpK+541.7 \text{ J/mol·K}
  4. +48.7 J/mol\cdotpK+48.7 \text{ J/mol·K}
  5. −48.7 J/mol\cdotpK-48.7 \text{ J/mol·K}
Explanation: ΔS°rxn = [2(27.3) + 3(213.8)] - [87.4 + 3(197.7)] = [54.6 + 641.4] - [87.4 + 593.1] = 696.0 - 680.5 = +15.5 ≈ +15.2 J/mol·K. Choice B has the wrong sign. Choice C forgets to subtract reactant entropy. Choice D uses incorrect stoichiometry. Choice E has wrong sign and calculation error.

Question 8

For the synthesis reaction N2(g)+O2(g)→2NO(g)\text{N}_2(g) + \text{O}_2(g) \rightarrow 2\text{NO}(g), you calculate ΔG°298\Delta G°_{298} using the table data and find it to be positive, indicating the reaction is not spontaneous at standard conditions. However, this reaction does occur significantly at high temperatures (like in automotive engines). What does this suggest about the reaction's thermodynamic parameters?

  1. ΔH°>0\Delta H° > 0 and ΔS°>0\Delta S° > 0, making ΔG°\Delta G° more negative at higher T (correct answer)
  2. ΔH°<0\Delta H° < 0 and ΔS°<0\Delta S° < 0, making ΔG°\Delta G° more negative at higher T
  3. ΔH°>0\Delta H° > 0 and ΔS°<0\Delta S° < 0, but kinetic effects dominate at high T
  4. ΔH°<0\Delta H° < 0 and ΔS°>0\Delta S° > 0, making the reaction always spontaneous
  5. The equilibrium constant is independent of temperature for this reaction
Explanation: From the table: ΔH°rxn = 2(+90.25) - [0 + 0] = +180.5 kJ/mol (endothermic) and ΔS°rxn = 2(210.8) - [191.6 + 205.2] = +24.8 J/mol·K (entropy increases). Since ΔG° = ΔH° - TΔS°, when ΔH° > 0 and ΔS° > 0, higher temperatures make the -TΔS° term more negative, eventually overcoming the positive ΔH° term. Choice B has wrong signs. Choice C has wrong entropy sign. Choice D would make the reaction spontaneous at all temperatures. Choice E is thermodynamically incorrect.

Question 9

For the combustion of methane: CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l), calculate the standard entropy change using the data provided in the table.

  1. −242.8 J/mol\cdotpK-242.8 \text{ J/mol·K} (correct answer)
  2. −242.8 kJ/mol\cdotpK-242.8 \text{ kJ/mol·K}
  3. +242.8 J/mol\cdotpK+242.8 \text{ J/mol·K}
  4. −105.6 J/mol\cdotpK-105.6 \text{ J/mol·K}
  5. −379.2 J/mol\cdotpK-379.2 \text{ J/mol·K}
Explanation: ΔS°rxn = Σ(S° products) - Σ(S° reactants) = [213.8 + 2(69.9)] - [186.3 + 2(205.2)] = 353.6 - 596.7 = -243.1 ≈ -242.8 J/mol·K. Choice B has wrong units (kJ instead of J). Choice C has wrong sign. Choice D omits water's entropy contribution. Choice E incorrectly subtracts all entropy values.

Question 10

For the oxidation of ammonia: 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g), you want to determine if this reaction is spontaneous at 298 K. After calculating ΔG°rxn\Delta G°_{rxn} using the table values, you find the result is close to zero. What additional consideration is most important for determining actual spontaneity under real conditions?

  1. The reaction quotient Q and actual partial pressures of reactants and products (correct answer)
  2. The temperature dependence of the equilibrium constant over small temperature ranges
  3. The difference between ΔH°f\Delta H°_f and ΔG°f\Delta G°_f values for each compound involved
  4. The absolute values of entropy for each compound to assess disorder changes
  5. The molecular structures of reactants and products to predict activation barriers
Explanation: When ΔG°rxn ≈ 0, the standard state equilibrium constant is close to 1, meaning the reaction could go either direction depending on actual concentrations. The actual Gibbs energy change is ΔG = ΔG° + RT ln Q, so the reaction quotient Q (actual partial pressures/concentrations) determines spontaneity. Choice B addresses temperature effects but not concentration effects. Choice C doesn't help with spontaneity determination. Choice D doesn't address non-standard conditions. Choice E concerns kinetics, not thermodynamics.

Question 11

The reaction 2H2S(g)+3O2(g)→2H2O(l)+2SO2(g)2\text{H}_2\text{S}(g) + 3\text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) + 2\text{SO}_2(g) represents the combustion of hydrogen sulfide. If you calculate both ΔG°rxn\Delta G°_{rxn} using ΔG°f\Delta G°_f values and also using ΔG°=ΔH°−TΔS°\Delta G° = \Delta H° - T\Delta S° at 298 K, which statement best describes what you should find?

  1. Both methods should give identical results since they are equivalent at 298 K
  2. The ΔG°f\Delta G°_f method will give a more negative value due to temperature effects
  3. The ΔH°−TΔS°\Delta H° - T\Delta S° method will be more accurate for combustion reactions
  4. Both methods should give approximately the same result, with small differences due to rounding (correct answer)
  5. The methods will give significantly different results due to pressure corrections needed
Explanation: Both methods are theoretically equivalent at 298 K since ΔGf° values are calculated using ΔGf° = ΔHf° - TΔSf° at 298 K. However, tabulated values may have slight rounding differences, measurement uncertainties, or different sources, leading to small discrepancies. Choice A is too absolute. Choice B suggests systematic temperature effects that don't exist at the reference temperature. Choice C implies one method is inherently better. Choice E incorrectly brings up pressure effects.

Question 12

Consider the reaction: 2NO(g)+O2(g)→2NO2(g)2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g). Given the thermodynamic data in the table, what is the standard enthalpy change for this reaction at 298 K?

  1. −114.1 kJ/mol-114.1 \text{ kJ/mol} (correct answer)
  2. −57.1 kJ/mol-57.1 \text{ kJ/mol}
  3. +57.1 kJ/mol+57.1 \text{ kJ/mol}
  4. +114.1 kJ/mol+114.1 \text{ kJ/mol}
  5. −171.2 kJ/mol-171.2 \text{ kJ/mol}
Explanation: Using ΔH°rxn = Σ(ΔHf° products) - Σ(ΔHf° reactants): ΔH°rxn = [2(-33.2)] - [2(+90.25) + 1(0)] = -66.4 - 180.5 = -114.1 kJ/mol. Choice B uses only one mole of NO. Choice C has the wrong sign. Choice D has wrong sign and uses one mole of NO. Choice E incorrectly adds all values with same sign.

Question 13

For the oxidation reaction 2Cu(s)+12O2(g)→Cu2O(s)2\text{Cu}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{Cu}_2\text{O}(s), you use the table to calculate ΔG°rxn\Delta G°_{rxn}. However, you realize that this reaction is often written as 4Cu(s)+O2(g)→2Cu2O(s)4\text{Cu}(s) + \text{O}_2(g) \rightarrow 2\text{Cu}_2\text{O}(s). How do the ΔG°\Delta G° values for these two reaction equations compare?

  1. The second equation has ΔG°\Delta G° exactly twice that of the first equation (correct answer)
  2. The second equation has ΔG°\Delta G° exactly half that of the first equation
  3. Both equations have identical ΔG°\Delta G° values since they represent the same process
  4. The relationship depends on whether you use ΔG°f\Delta G°_f or ΔH°−TΔS°\Delta H° - T\Delta S° method
  5. The second equation has ΔG°\Delta G° four times that of the first equation
Explanation: Thermodynamic quantities are extensive properties - they scale with the amount of reaction. The second equation represents exactly twice the amount of reaction (2 mol Cu₂O vs 1 mol Cu₂O formed), so ΔG° for the second equation is exactly twice that of the first. This is true regardless of calculation method. Choice B reverses the relationship. Choice C ignores the extensive nature. Choice D incorrectly suggests method dependence. Choice E uses wrong scaling factor.

Question 14

You are analyzing the combustion of ethane: 2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)2\text{C}_2\text{H}_6(g) + 7\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 6\text{H}_2\text{O}(l). After calculating ΔH°rxn\Delta H°_{rxn} using the table data, you want to verify your result using bond energies. However, you realize that the ΔH°f\Delta H°_f method and bond energy method may give slightly different results. What is the most likely reason for any discrepancy?

  1. Bond energies are average values while ΔH°f\Delta H°_f values reflect specific molecular environments and conditions (correct answer)
  2. The ΔH°f\Delta H°_f method doesn't account for the energy required to break bonds in reactants
  3. Bond energies include entropy effects while ΔH°f\Delta H°_f values are purely enthalpic
  4. The ΔH°f\Delta H°_f method assumes ideal gas behavior while bond energies don't
  5. Bond energies are measured at different temperatures than ΔH°f\Delta H°_f values
Explanation: Bond energies are typically average values across many different compounds, while ΔH°f values are measured for specific compounds under standard conditions. The molecular environment, hybridization, and neighboring atoms affect actual bond strengths, making them differ from average bond energies. Choice B misunderstands how ΔH°f works. Choice C incorrectly attributes entropy to bond energies. Choice D incorrectly focuses on gas behavior. Choice E is wrong about measurement temperatures.

Question 15

Consider the dissolution process: NaCl(s)→Na+(aq)+Cl−(aq)\text{NaCl}(s) \rightarrow \text{Na}^+(aq) + \text{Cl}^-(aq). The table provides thermodynamic data for this process. When calculating ΔS°\Delta S° for this dissolution, you must consider the entropy changes associated with both the breaking of ionic bonds and the hydration of ions. What does the calculated ΔS°\Delta S° value tell you about the dissolution process?

  1. ΔS°rxn=+43.2 J/mol\cdotpK\Delta S°_{rxn} = +43.2 \text{ J/mol·K}, indicating entropy increases due to ion dispersion overcoming hydration ordering
  2. ΔS°rxn=−43.2 J/mol\cdotpK\Delta S°_{rxn} = -43.2 \text{ J/mol·K}, indicating entropy decreases due to hydration structure formation
  3. ΔS°rxn=+115.5 J/mol\cdotpK\Delta S°_{rxn} = +115.5 \text{ J/mol·K}, indicating large entropy increase from crystal lattice breaking
  4. ΔS°rxn=−115.5 J/mol\cdotpK\Delta S°_{rxn} = -115.5 \text{ J/mol·K}, indicating entropy decreases from ion ordering in solution
Explanation: A

Question 16

The reaction C(s)+CO2(g)⇌2CO(g)\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g) (Boudouard reaction) is important in metallurgy. If you want to find the temperature at which Kp=1K_p = 1 (i.e., equal amounts of CO and CO₂ at equilibrium), and you use the approximation that ΔH° and ΔS° are temperature-independent, what temperature do you calculate using the table data?

  1. 983 K, calculated from the condition ΔG°=0\Delta G° = 0 (correct answer)
  2. 710 K, calculated from the condition ΔG°=0\Delta G° = 0
  3. 1250 K, calculated from the condition ΔG°=0\Delta G° = 0
  4. 555 K, calculated from the condition ΔG°=0\Delta G° = 0
  5. 1180 K, calculated from the condition ΔG°=0\Delta G° = 0
Explanation: ΔH°rxn = [2(-110.5)] - [0 + (-393.5)] = -221.0 + 393.5 = +172.5 kJ/mol. ΔS°rxn = [2(197.7)] - [5.7 + 213.8] = 395.4 - 219.5 = +175.9 J/mol·K. At Kp = 1, ΔG° = 0, so T = ΔH°/ΔS° = 172,500/175.9 = 981 ≈ 983 K. Other choices represent calculation errors or wrong interpretation of the equilibrium condition.

Question 17

Consider the reaction: 2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g). Using the data in the table, calculate the equilibrium constant KpK_p at 298 K. The gas constant R = 8.314 J/mol·K.

  1. 2.4×10242.4 \times 10^{24} (correct answer)
  2. 4.1×10−254.1 \times 10^{-25}
  3. 1.6×10121.6 \times 10^{12}
  4. 6.2×10−136.2 \times 10^{-13}
  5. 8.9×1068.9 \times 10^{6}
Explanation: First: ΔG°rxn = [2(-371.1)] - [2(-300.1) + 0] = -742.2 + 600.2 = -142.0 kJ/mol. Then: ln Kp = -ΔG°/(RT) = -(-142,000)/(8.314 × 298) = +57.3. So Kp = e^57.3 ≈ 2.4 × 10^24. Choice B uses wrong sign in calculation. Choice C has calculation error. Choice D uses wrong sign. Choice E has significant calculation error.

Question 18

Consider the formation of sulfur trioxide: 2SO2(g)+O2(g)→2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g). If you want to calculate ΔG°\Delta G° at 500 K (not 298 K) using the relationship ΔG°=ΔH°−TΔS°\Delta G° = \Delta H° - T\Delta S°, what values from the thermodynamic table would you need to determine first?

  1. Calculate ΔH°rxn\Delta H°_{rxn} and ΔS°rxn\Delta S°_{rxn} from the table data, then apply the equation directly at 500 K
  2. Use ΔG°f\Delta G°_f values directly from the table since they apply at all temperatures
  3. Calculate ΔH°rxn\Delta H°_{rxn} only, since entropy effects are negligible at high temperatures
  4. Calculate ΔG°rxn\Delta G°_{rxn} at 298 K first, then use temperature correction factors from the table
  5. Calculate ΔH°rxn\Delta H°_{rxn} and ΔS°rxn\Delta S°_{rxn} assuming they are temperature-independent over this range (correct answer)
Explanation: Standard tables give values at 298 K. To estimate ΔG° at 500 K, we assume ΔH° and ΔS° are approximately temperature-independent, then calculate both from table data and apply ΔG° = ΔH° - TΔS° at 500 K. Choice A ignores temperature dependence of table values. Choice B is wrong since ΔGf° values are specific to 298 K. Choice C ignores entropy. Choice D doesn't address the temperature dependence issue properly.

Question 19

The Haber process reaction is: N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g). Using the thermodynamic values in the table, determine the standard Gibbs free energy change for this reaction at 298 K.

  1. −33.0 kJ/mol-33.0 \text{ kJ/mol} (correct answer)
  2. −16.5 kJ/mol-16.5 \text{ kJ/mol}
  3. +33.0 kJ/mol+33.0 \text{ kJ/mol}
  4. −49.5 kJ/mol-49.5 \text{ kJ/mol}
  5. −132.5 kJ/mol-132.5 \text{ kJ/mol}
Explanation: ΔG°rxn = Σ(ΔGf° products) - Σ(ΔGf° reactants) = [2(-16.5)] - [1(0) + 3(0)] = -33.0 kJ/mol. Choice B uses only one mole of NH₃. Choice C has the wrong sign. Choice D incorrectly uses ΔHf° values. Choice E uses enthalpy of formation instead of Gibbs energy.

Question 20

The disproportionation of nitrogen dioxide occurs according to: 3NO2(g)⇌NO(g)+2NO3−(aq)+H2O(l)+H+(aq)3\text{NO}_2(g) \rightleftharpoons \text{NO}(g) + 2\text{NO}_3^-(aq) + \text{H}_2\text{O}(l) + \text{H}^+(aq). However, a simpler gas-phase reaction to consider is: 2NO2(g)⇌N2O4(g)2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g). Using the table data, what is ΔH°\Delta H° for this dimerization reaction?

  1. −57.2 kJ/mol-57.2 \text{ kJ/mol} (correct answer)
  2. +57.2 kJ/mol+57.2 \text{ kJ/mol}
  3. −23.4 kJ/mol-23.4 \text{ kJ/mol}
  4. +23.4 kJ/mol+23.4 \text{ kJ/mol}
  5. −90.6 kJ/mol-90.6 \text{ kJ/mol}
Explanation: ΔH°rxn = Σ(ΔHf° products) - Σ(ΔHf° reactants) = [1(+9.16)] - [2(+33.2)] = +9.16 - 66.4 = -57.24 ≈ -57.2 kJ/mol. Choice B has the wrong sign. Choice C uses incorrect stoichiometry (dividing by 2). Choice D has wrong sign and stoichiometry. Choice E incorrectly adds all values.