Using Thermodynamic Tables - Physical Chemistry 1

1. You are analyzing the combustion of ethane: $2\text{C}_2\text{H}_6(g) + 7\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 6\text{H}_2\text{O}(l)$. After calculating $\Delta H°_{rxn}$ using the table data, you want to verify your result using bond energies. However, you realize that the $\Delta H°_f$ method and bond energy method may give slightly different results. What is the most likely reason for any discrepancy?
   Answer: Bond energies are average values while $\Delta H°_f$ values reflect specific molecular environments and conditions 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.
2. Consider the reaction: $2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g)$. Given the thermodynamic data in the table, what is the standard enthalpy change for this reaction at 298 K?
   Answer: $-114.1 \text{ kJ/mol}$ 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.
3. Consider the reaction: $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$. Using the data in the table, calculate the equilibrium constant $K_p$ at 298 K. The gas constant R = 8.314 J/mol·K.
   Answer: $2.4 \times 10^{24}$ 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.
4. For the synthesis reaction $\text{N}_2(g) + \text{O}_2(g) \rightarrow 2\text{NO}(g)$, you calculate $\Delta G°_{298}$ 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?
   Answer: $\Delta H° > 0$ and $\Delta S° > 0$, making $\Delta G°$ more negative at higher T 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.
5. The decomposition of limestone follows: $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$ with $\Delta H^\circ = +178.3 \text{ kJ/mol}$ and $\Delta S^\circ = +160.5 \text{ J/mol}\cdot\text{K}$. A student claims this reaction becomes spontaneous above a certain temperature due to the entropy term overwhelming the enthalpy term. At what temperature does $\Delta G^\circ$ first become zero?
   Answer: $1111 \text{ K}$ When you encounter a question about spontaneity and temperature, you're dealing with the Gibbs free energy equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. A reaction becomes spontaneous when $\Delta G^\circ$ becomes negative, and the transition point occurs exactly when $\Delta G^\circ = 0$. To find when $\Delta G^\circ$ first becomes zero, set the equation equal to zero and solve for temperature: $0 = \Delta H^\circ - T\Delta S^\circ$ $T = \frac{\Delta H^\circ}{\Delta S^\circ}$ First, convert units so they match. Since $\Delta H^\circ = +178.3 \text{ kJ/mol} = +178,300 \text{ J/mol}$ and $\Delta S^\circ = +160.5 \text{ J/mol}\cdot\text{K}$: $T = \frac{178,300 \text{ J/mol}}{160.5 \text{ J/mol}\cdot\text{K}} = 1111 \text{ K}$ 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 $T\Delta S$ term eventually dominates.
6. For the oxidation reaction $2\text{Cu}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{Cu}_2\text{O}(s)$, you use the table to calculate $\Delta G°_{rxn}$. However, you realize that this reaction is often written as $4\text{Cu}(s) + \text{O}_2(g) \rightarrow 2\text{Cu}_2\text{O}(s)$. How do the $\Delta G°$ values for these two reaction equations compare?
   Answer: The second equation has $\Delta G°$ exactly twice that of the first equation 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.
7. Consider the formation of sulfur trioxide: $2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)$. If you want to calculate $\Delta G°$ at 500 K (not 298 K) using the relationship $\Delta G° = \Delta H° - T\Delta S°$, what values from the thermodynamic table would you need to determine first?
   Answer: Calculate $\Delta H°_{rxn}$ and $\Delta S°_{rxn}$ assuming they are temperature-independent over this range 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.
8. For the reaction $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$, you need to determine at what temperature this decomposition becomes thermodynamically favorable ($\Delta G° = 0$). Using the data in the table, what is the approximate temperature at which $\Delta G° = 0$?
   Answer: 1100 K assuming temperature-independent $\Delta H°$ and $\Delta S°$ values 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.
9. For the reaction $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$, a student calculates $\Delta H^\circ = +92.5 \text{ kJ/mol}$ using formation enthalpies, and $\Delta S^\circ = +179.8 \text{ J/mol}\cdot\text{K}$ using standard entropies. What is $\Delta G^\circ$ for this reaction at 400 K?
   Answer: $+20.6 \text{ kJ/mol}$ When you encounter a thermodynamics problem asking for $\Delta G^\circ$, you're dealing with the fundamental relationship between enthalpy, entropy, and free energy. The key equation is the Gibbs-Helmholtz equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. To solve this, substitute the given values directly: $\Delta G^\circ = 92.5 \text{ kJ/mol} - (400 \text{ K})(179.8 \text{ J/mol}\cdot\text{K})$. The critical step is unit conversion—you must convert the entropy term to kJ: $179.8 \text{ J/mol}\cdot\text{K} = 0.1798 \text{ kJ/mol}\cdot\text{K}$. This gives: $\Delta G^\circ = 92.5 - (400)(0.1798) = 92.5 - 71.9 = +20.6 \text{ kJ/mol}$. Option A (+164.4 kJ/mol) results from incorrectly adding the enthalpy and entropy terms instead of subtracting: $92.5 + 71.9 = 164.4$. 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 $92.5 + (400)(179.8)$ and treating everything as kJ/mol. Study tip: Always check your units carefully in thermodynamics calculations. The entropy term $T\Delta S$ frequently appears in J/mol, but $\Delta H$ is typically given in kJ/mol. Convert before calculating, and remember that the Gibbs equation subtracts the entropy term.
10. For the reaction $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, the standard entropy change is $\Delta S^\circ = +175.8 \text{ J/mol}\cdot\text{K}$ and the standard enthalpy change is $\Delta H^\circ = +57.2 \text{ kJ/mol}$. At what temperature will this reaction become thermodynamically favorable ($\Delta G^\circ < 0$) under standard conditions?
    Answer: Above 325 K, since entropy favors product formation at higher temperatures 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.
11. A student wants to calculate $\Delta H^\circ$ for the reaction $\text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l)$ using standard enthalpies of formation. If $\Delta H_f^\circ[\text{C}_2\text{H}_4(g)] = 52.3 \text{ kJ/mol}$, $\Delta H_f^\circ[\text{CO}_2(g)] = -393.5 \text{ kJ/mol}$, and $\Delta H_f^\circ[\text{H}_2\text{O}(l)] = -285.8 \text{ kJ/mol}$, what is the enthalpy change for this combustion reaction?
    Answer: $-1411.0 \text{ kJ/mol}$ 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.
12. The reaction $2\text{H}_2\text{S}(g) + 3\text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) + 2\text{SO}_2(g)$ represents the combustion of hydrogen sulfide. If you calculate both $\Delta G°_{rxn}$ using $\Delta G°_f$ values and also using $\Delta G° = \Delta H° - T\Delta S°$ at 298 K, which statement best describes what you should find?
    Answer: Both methods should give approximately the same result, with small differences due to rounding 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.
13. The disproportionation of nitrogen dioxide occurs according to: $3\text{NO}_2(g) \rightleftharpoons \text{NO}(g) + 2\text{NO}_3^-(aq) + \text{H}_2\text{O}(l) + \text{H}^+(aq)$. However, a simpler gas-phase reaction to consider is: $2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$. Using the table data, what is $\Delta H°$ for this dimerization reaction?
    Answer: $-57.2 \text{ kJ/mol}$ 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.
14. The Haber process reaction is: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$. Using the thermodynamic values in the table, determine the standard Gibbs free energy change for this reaction at 298 K.
    Answer: $-33.0 \text{ kJ/mol}$ 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.
15. For the combustion of methane: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$, calculate the standard entropy change using the data provided in the table.
    Answer: $-242.8 \text{ J/mol}\cdot\text{K}$ 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.
16. A reaction has $\Delta H^\circ = -45.6 \text{ kJ/mol}$ and $\Delta S^\circ = -125.3 \text{ J/mol}\cdot\text{K}$. Using the relationship $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, determine the temperature range where this reaction is thermodynamically spontaneous.
    Answer: Spontaneous at all temperatures below 364 K due to favorable enthalpy When you encounter thermodynamic spontaneity problems, you need to determine when $\Delta G^\circ < 0$, since negative Gibbs free energy indicates a spontaneous process. Let's analyze this systematically. You have $\Delta H^\circ = -45.6 \text{ kJ/mol}$ (favorable, exothermic) and $\Delta S^\circ = -125.3 \text{ J/mol}\cdot\text{K}$ (unfavorable, decreasing entropy). Using $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, the entropy term becomes $+T\Delta S^\circ$ because of the negative entropy value. To find where $\Delta G^\circ = 0$ (the boundary between spontaneous and non-spontaneous): $0 = -45,600 - T(-125.3)$ $T = \frac{45,600}{125.3} = 364 \text{ K}$ At temperatures below 364 K, the favorable enthalpy term dominates the unfavorable entropy term, making $\Delta G^\circ < 0$ (spontaneous). Above 364 K, the entropy penalty becomes too large, making $\Delta G^\circ > 0$ (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 $\Delta G^\circ = 0$. Study tip: Remember that enthalpy-driven reactions (negative $\Delta H$, negative $\Delta S$) 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.
17. For the oxidation of ammonia: $4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g)$, you want to determine if this reaction is spontaneous at 298 K. After calculating $\Delta G°_{rxn}$ 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?
    Answer: The reaction quotient Q and actual partial pressures of reactants and products 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.
18. The reaction $\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)$ (Boudouard reaction) is important in metallurgy. If you want to find the temperature at which $K_p = 1$ (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?
    Answer: 983 K, calculated from the condition $\Delta G° = 0$ 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.
19. The reduction of iron(III) oxide by carbon monoxide: $\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)$ is important in steel production. Referring to the table, calculate the standard entropy change for this reaction.
    Answer: $+15.2 \text{ J/mol}\cdot\text{K}$ 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.

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