IB Chemistry Quiz: Apply Entropy And Spontaneity
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Apply Entropy And SpontaneityQuestion 1 of 20

For the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the standard enthalpy change, \Delta H^⦵, is -92 kJ mol⁻¹ and the standard entropy change, \Delta S^⦵, is -199 J K⁻¹ mol⁻¹. What is the standard Gibbs free energy change, \Delta G^⦵, in kJ mol⁻¹ for this reaction at 298 K?

-33
+59200
-151
-59300
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IB Chemistry Quiz

IB Chemistry Quiz: Apply Entropy And Spontaneity

Practice Apply Entropy And Spontaneity in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Apply Entropy And Spontaneity, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

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Question 1

For the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the standard enthalpy change, \Delta H^⦵, is -92 kJ mol⁻¹ and the standard entropy change, \Delta S^⦵, is -199 J K⁻¹ mol⁻¹. What is the standard Gibbs free energy change, \Delta G^⦵, in kJ mol⁻¹ for this reaction at 298 K?

  1. -33 (correct answer)
  2. +59200
  3. -151
  4. -59300
Explanation: Use the equation \Delta G^⦵ = \Delta H^⦵ - T\Delta S^⦵. It is crucial to ensure the units are consistent. Convert \Delta S^⦵ to kJ K⁻¹ mol⁻¹ by dividing by 1000: \Delta S^⦵ = -0.199 kJ K⁻¹ mol⁻¹. \Delta G^⦵ = -92\text{ kJ mol⁻¹} - (298\text{ K} \times -0.199\text{ kJ K⁻¹ mol⁻¹}) = -92 - (-59.3) = -92 + 59.3 = -32.7 kJ mol⁻¹. This is approximately -33 kJ mol⁻¹.

Question 2

For which of the following processes is the entropy change, ΔS\Delta S, negative?

  1. 2SO₃(g) → 2SO₂(g) + O₂(g)
  2. H₂O(l) → H₂O(g)
  3. NaCl(s) → Na⁺(aq) + Cl⁻(aq)
  4. Ag⁺(aq) + Cl⁻(aq) → AgCl(s) (correct answer)
Explanation: Entropy is a measure of disorder. A negative entropy change (ΔS<0\Delta S < 0) means the system becomes more ordered. In option D, two moles of mobile aqueous ions combine to form one mole of a highly ordered solid precipitate, representing a significant decrease in disorder. Option A shows an increase in the moles of gas (2 mol → 3 mol), so ΔS>0\Delta S > 0. Option B is vaporization (liquid to gas), a large increase in disorder, so ΔS>0\Delta S > 0. Option C is dissolving a solid into mobile ions, an increase in disorder, so ΔS>0\Delta S > 0.

Question 3

The conversion of diamond to graphite is spontaneous under standard conditions: C(s, diamond) → C(s, graphite), \Delta G^⦵ = -2.9 kJ mol⁻¹. Why do diamonds persist indefinitely at room temperature?

  1. The reaction has a very large positive enthalpy change.
  2. The reaction is readily reversible under atmospheric pressure.
  3. The reaction has a very high activation energy. (correct answer)
  4. The entropy change for the conversion is highly negative.
Explanation: Thermodynamic spontaneity (indicated by a negative ΔG\Delta G) does not provide information about the rate of a reaction. The conversion of diamond to graphite is kinetically very slow because it has an extremely high activation energy. This kinetic barrier prevents the reaction from occurring at a noticeable rate under normal conditions, even though it is thermodynamically favourable.

Question 4

A spontaneous electrochemical cell (voltaic cell) operates under standard conditions. Which combination of signs for \Delta G^⦵ and standard cell potential, E^⦵_{cell}, is correct?

  1. \Delta G^⦵ is positive, E^⦵_{cell} is positive.
  2. \Delta G^⦵ is negative, E^⦵_{cell} is negative.
  3. \Delta G^⦵ is positive, E^⦵_{cell} is negative.
  4. \Delta G^⦵ is negative, E^⦵_{cell} is positive. (correct answer)
Explanation: A spontaneous process has a negative Gibbs free energy change (\Delta G^⦵ < 0). The relationship between Gibbs free energy and cell potential is \Delta G^⦵ = -nFE^⦵_{cell}, where n (moles of electrons) and F (Faraday's constant) are positive values. For \Delta G^⦵ to be negative, the standard cell potential, E^⦵_{cell}, must be positive. Therefore, a spontaneous voltaic cell has \Delta G^⦵ < 0 and E^⦵_{cell} > 0.

Question 5

A reaction has a standard equilibrium constant, K, of 3.5 x 10⁻⁴ at 298 K. What is the value of \Delta G^⦵ in kJ mol⁻¹? (R = 8.31 J K⁻¹ mol⁻¹)

  1. -20.0
  2. +20.0 (correct answer)
  3. -8.6
  4. +8.6
Explanation: The governing equation is \Delta G^⦵ = -RT \ln K. Given K = 3.5 x 10⁻⁴, R = 8.31 J K⁻¹ mol⁻¹, and T = 298 K. First, calculate ln⁡K\ln K: ln⁡(3.5×10−4)≈−7.96\ln(3.5 \times 10^{-4}) \approx -7.96. Now, calculate \Delta G^⦵: \Delta G^⦵ = -(8.31 \text{ J K⁻¹ mol⁻¹})(298 \text{ K})(-7.96) \approx +19700 J mol⁻¹. Finally, convert to kJ mol⁻¹ by dividing by 1000: \Delta G^⦵ \approx +19.7 kJ mol⁻¹. This is closest to +20.0 kJ mol⁻¹.

Question 6

Which change is expected to cause the largest increase in the entropy of a system?

  1. Heating 1 mole of nitrogen gas from 300 K to 350 K at constant volume.
  2. Allowing 1 mole of nitrogen gas to expand from a volume of 1 dm³ to 2 dm³ at constant temperature.
  3. Condensing 1 mole of nitrogen gas to liquid nitrogen at its boiling point.
  4. Subliming 1 mole of solid carbon dioxide at its sublimation point. (correct answer)
Explanation: Entropy changes are largest for processes involving a phase change, particularly to the gaseous state. Sublimation (solid to gas) involves a massive increase in volume and molecular freedom, resulting in a very large increase in entropy. Condensing a gas (option C) causes a large decrease in entropy. Heating a gas (option A) and expanding a gas (option B) both increase entropy, but these changes are typically much smaller than the entropy change associated with a phase transition from a condensed phase (solid or liquid) to the gas phase.

Question 7

A certain endothermic reaction has a positive standard entropy change. Which statement correctly describes the spontaneity of this reaction?

  1. The reaction is spontaneous at all temperatures.
  2. The reaction is non-spontaneous at all temperatures.
  3. The reaction is spontaneous only at low temperatures.
  4. The reaction is spontaneous only at high temperatures. (correct answer)
Explanation: The spontaneity of a reaction is determined by the sign of the Gibbs free energy change, ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S. The reaction is endothermic, so ΔH>0\Delta H > 0. The entropy change is positive, so ΔS>0\Delta S > 0. In this case, the ΔH\Delta H term is positive, which disfavors spontaneity, while the −TΔS-T\Delta S term is negative, which favors spontaneity. The reaction will be spontaneous (ΔG<0\Delta G < 0) when the magnitude of the −TΔS-T\Delta S term is greater than the magnitude of the ΔH\Delta H term. This occurs at sufficiently high temperatures.

Question 8

The thermal decomposition of calcium carbonate is shown below: CaCO₃(s) → CaO(s) + CO₂(g). This reaction is endothermic. Assuming standard pressure, above what approximate temperature will this reaction become spontaneous? (Given \Delta H^⦵ = +178 kJ mol⁻¹ and \Delta S^⦵ = +161 J K⁻¹ mol⁻¹)

  1. 1.1 K
  2. 298 K
  3. 1106 K (correct answer)
  4. The reaction is never spontaneous.
Explanation: A reaction becomes spontaneous when ΔG\Delta G changes from positive to negative. The crossover point occurs when ΔG=0\Delta G = 0. At this point, ΔH=TΔS\Delta H = T\Delta S, so T=ΔH/ΔST = \Delta H / \Delta S. We must use consistent units. Convert ΔH\Delta H to J mol⁻¹: 178 kJ mol⁻¹=178000 J mol⁻¹178 \text{ kJ mol⁻¹} = 178000 \text{ J mol⁻¹}. Now calculate T: T=(178000 J mol⁻¹)/(161 J K⁻¹ mol⁻¹)≈1105.6T = (178000 \text{ J mol⁻¹}) / (161 \text{ J K⁻¹ mol⁻¹}) \approx 1105.6 K. Since both ΔH\Delta H and ΔS\Delta S are positive, the reaction becomes spontaneous at temperatures above this value. So, T > 1106 K.

Question 9

A reaction is non-spontaneous at 298 K with \Delta H^⦵ = +90 kJ mol⁻¹ and \Delta G^⦵ = +30 kJ mol⁻¹. Assuming \Delta H^⦵ and \Delta S^⦵ are constant with temperature, above what temperature will this reaction become spontaneous?

  1. 100 K
  2. 298 K
  3. 333 K
  4. 450 K (correct answer)
Explanation: This is a two-step problem. First, we must find \Delta S^⦵ at 298 K using the given data. From \Delta G^⦵ = \Delta H^⦵ - T\Delta S^⦵, we rearrange to find \Delta S^⦵ = (\Delta H^⦵ - \Delta G^⦵) / T. \Delta S^⦵ = (+90 \text{ kJ mol⁻¹} - (+30 \text{ kJ mol⁻¹})) / 298 \text{ K} = 60 \text{ kJ mol⁻¹} / 298 \text{ K} \approx 0.201 kJ K⁻¹ mol⁻¹. The reaction becomes spontaneous when ΔG<0\Delta G < 0. The crossover temperature is where ΔG=0\Delta G = 0, so T = \Delta H^⦵ / \Delta S^⦵. Using the calculated \Delta S^⦵: T=90 kJ mol⁻¹/0.201 kJ K⁻¹ mol⁻¹≈447T = 90 \text{ kJ mol⁻¹} / 0.201 \text{ kJ K⁻¹ mol⁻¹} \approx 447 K. The reaction becomes spontaneous above this temperature, so approximately 450 K.

Question 10

For the reaction A(g) + B(g) → C(g), \Delta G^⦵ is positive at a certain temperature T. The reaction quotient is Q. Under which condition must the reaction be spontaneous (ΔG<0\Delta G < 0) in the forward direction?

  1. When Q = K
  2. When Q > 1
  3. When Q is sufficiently larger than K
  4. When Q is sufficiently smaller than K (correct answer)
Explanation: The relationship between standard and non-standard Gibbs free energy is \Delta G = \Delta G^⦵ + RT \ln Q. We are also given that \Delta G^⦵ is positive. For the reaction to be spontaneous, ΔG\Delta G must be negative. This requires the RTln⁡QRT \ln Q term to be negative and have a greater magnitude than \Delta G^⦵. For RTln⁡QRT \ln Q to be negative, ln⁡Q\ln Q must be negative, which means Q<1Q < 1. Furthermore, we know that at equilibrium ΔG=0\Delta G = 0 and Q=KQ = K, so \Delta G^⦵ = -RT \ln K. Since \Delta G^⦵ is positive, K must be less than 1. For the forward reaction to proceed spontaneously, the system must be far from equilibrium in a state where there are many more reactants than products compared to the equilibrium state, meaning Q must be significantly smaller than K.

Question 11

At 298 K, Reaction X has \Delta G^⦵ = -25 kJ mol⁻¹ and Reaction Y has \Delta G^⦵ = +10 kJ mol⁻¹. What can be concluded about their respective equilibrium constants, Kx and Ky?

  1. Kx > 1 and Ky < 1 (correct answer)
  2. Kx < 1 and Ky > 1
  3. Both Kx and Ky are greater than 1.
  4. Both Kx and Ky are less than 1.
Explanation: The relationship \Delta G^⦵ = -RT \ln K dictates the connection between \Delta G^⦵ and K. For Reaction X, \Delta G^⦵ is negative, which means ln⁡Kx\ln Kx must be positive, and therefore Kx > 1. This indicates that products are favoured at equilibrium. For Reaction Y, \Delta G^⦵ is positive, which means ln⁡Ky\ln Ky must be negative, and therefore Ky < 1. This indicates that reactants are favoured at equilibrium.

Question 12

A chemical engineer needs to determine the minimum temperature at which the reaction 2ZnS(s)+3O2(g)→2ZnO(s)+2SO2(g)2\text{ZnS}(s) + 3\text{O}_2(g) \rightarrow 2\text{ZnO}(s) + 2\text{SO}_2(g) becomes non-spontaneous. Given: ΔH∘=−878 kJ mol−1\Delta H^\circ = -878 \text{ kJ mol}^{-1} and ΔS∘=−32 J mol−1K−1\Delta S^\circ = -32 \text{ J mol}^{-1}\text{K}^{-1}. What is this temperature and what does it represent?

  1. 27,438 K; the temperature above which the reaction becomes thermodynamically unfavorable due to entropy effects (correct answer)
  2. 2,744 K; the temperature above which the reaction becomes thermodynamically unfavorable due to entropy effects
  3. 27,438 K; the temperature above which the reaction rate becomes too slow to be practical
  4. 2,744 K; the temperature above which the equilibrium shifts completely toward reactants
Explanation: The reaction becomes non-spontaneous when ΔG > 0. At the boundary, ΔG = 0: ΔH - TΔS = 0, so T = ΔH/ΔS = (-878,000 J/mol)/(-32 J/mol·K) = 27,438 K. Above this temperature, the negative ΔS makes -TΔS positive and large enough to overcome the negative ΔH. Choice B uses incorrect unit conversion. Choice C confuses thermodynamics with kinetics. Choice D misrepresents what happens at this temperature boundary.

Question 13

A reaction has ΔH=+85 kJ mol−1\Delta H = +85 \text{ kJ mol}^{-1} and ΔS=+180 J mol−1K−1\Delta S = +180 \text{ J mol}^{-1} \text{K}^{-1}. At what temperature does this reaction become thermodynamically favorable, and what is the primary driving force at temperatures well above this point?

  1. 472 K; the reaction is driven primarily by the favorable enthalpy change
  2. 472 K; the reaction is driven primarily by the favorable entropy change (correct answer)
  3. 189 K; the reaction is driven primarily by the favorable entropy change
  4. 189 K; the reaction is driven primarily by the favorable enthalpy change
Explanation: The reaction becomes spontaneous when ΔG < 0. Using ΔG = ΔH - TΔS: 0 = 85,000 J/mol - T(180 J/mol·K), so T = 472 K. At temperatures above 472 K, the -TΔS term dominates because ΔS is positive and large, making entropy the primary driving force. Choice A incorrectly identifies enthalpy as the driving force despite ΔH being positive (unfavorable). Choices C and D use incorrect temperature calculation (likely dividing by 1000 instead of converting ΔH to J).

Question 14

For the equilibrium CaCO3(s)⇌CaO(s)+CO2(g)\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g), ΔH∘=+178 kJ mol−1\Delta H^\circ = +178 \text{ kJ mol}^{-1} and ΔS∘=+161 J mol−1K−1\Delta S^\circ = +161 \text{ J mol}^{-1} \text{K}^{-1}. Which statement best explains why this decomposition becomes more favorable at higher temperatures?

  1. The positive ΔS increases the equilibrium constant, and higher temperatures provide activation energy to overcome the endothermic barrier
  2. The entropy term -TΔS becomes more negative at higher temperatures, eventually overcoming the positive enthalpy contribution to make ΔG negative
  3. The entropy term TΔS becomes more positive at higher temperatures, eventually overcoming the positive enthalpy contribution to make ΔG negative (correct answer)
  4. Higher temperatures increase the kinetic energy of reactant molecules, making bond breaking more favorable and reducing the effective ΔH
Explanation: Since ΔG = ΔH - TΔS and both ΔH and ΔS are positive, at high temperatures the TΔS term (positive) will eventually exceed ΔH, making ΔG negative and the reaction spontaneous. Choice B incorrectly states -TΔS becomes more negative (it becomes more positive since ΔS > 0). Choice A confuses thermodynamics with kinetics. Choice D incorrectly suggests temperature affects the thermodynamic ΔH value.

Question 15

A biochemical reaction has Keq=1.8×10−3K_{eq} = 1.8 \times 10^{-3} at body temperature (310 K). If this reaction is coupled to ATP hydrolysis (ΔG∘=−30.5 kJ mol−1\Delta G^\circ = -30.5 \text{ kJ mol}^{-1} at 310 K), what is the equilibrium constant for the overall coupled process?

  1. 2.1×1022.1 \times 10^{2}, because the coupling adds the free energy changes algebraically
  2. 1.1×1081.1 \times 10^{8}, because the coupling multiplies the equilibrium constants directly
  3. 1.1×1081.1 \times 10^{8}, because the coupling adds the free energy changes and affects K exponentially (correct answer)
  4. 2.1×1022.1 \times 10^{2}, because the coupling multiplies the individual free energy changes
Explanation: First find ΔG° for the original reaction: ΔG° = -RT ln K = -(8.314)(310) ln(1.8×10⁻³) = +16.0 kJ/mol. For the coupled process: ΔG°(total) = 16.0 + (-30.5) = -14.5 kJ/mol. Then: K(coupled) = exp(-ΔG°/RT) = exp(14,500/2577) = 1.1×10⁸. Choice A incorrectly just adds the K values. Choice B gets the right answer but with wrong reasoning (you don't multiply K values). Choice D combines wrong reasoning with wrong answer.

Question 16

Two students measure the equilibrium constant for the same reaction at different temperatures and obtain K₁ = 2.5 × 10⁴ at T₁ = 298 K and K₂ = 8.1 × 10² at T₂ = 398 K. What can be concluded about the thermodynamic parameters of this reaction?

  1. ΔH° < 0 and ΔS° < 0, because K decreases with temperature and the reaction is spontaneous (correct answer)
  2. ΔH° > 0 and ΔS° > 0, because higher temperature favors the reverse reaction direction
  3. ΔH° < 0 and ΔS° > 0, because the reaction is highly spontaneous at both temperatures
  4. ΔH° > 0 and ΔS° < 0, because K decreases significantly as temperature increases
Explanation: Since K decreases with increasing temperature, the reaction is exothermic (ΔH° < 0) per Le Châtelier's principle. Both K values are >> 1, indicating ΔG° < 0 (spontaneous). At 298K: ΔG° = -RT ln K = -24.8 kJ/mol. At 398K: ΔG° = -22.9 kJ/mol. Since ΔG becomes less negative as T increases despite negative ΔH, the entropy term -TΔS must be becoming less favorable, meaning ΔS° < 0. Choice B incorrectly assigns signs. Choice C assumes ΔS° > 0. Choice D incorrectly makes ΔH° > 0.

Question 17

The dissolution of ammonium nitrate in water is endothermic (ΔH>0\Delta H > 0) yet occurs spontaneously at room temperature. If the temperature of this solution is increased, what happens to the solubility and why?

  1. Solubility increases because higher temperature provides more energy to overcome the endothermic dissolution process
  2. Solubility remains constant because the equilibrium constant is independent of temperature for dissolution processes
  3. Solubility decreases because the positive ΔH contribution to ΔG becomes larger relative to the TΔS term
  4. Solubility increases because the entropy change becomes more favorable at higher temperatures, making ΔG more negative (correct answer)
Explanation: When you encounter a question about spontaneous endothermic processes, you need to consider both enthalpy and entropy contributions to Gibbs free energy using ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S. For ammonium nitrate dissolution to be spontaneous despite being endothermic (ΔH>0\Delta H > 0), the entropy term TΔST\Delta S must be large enough to make ΔG<0\Delta G < 0. This makes sense because dissolving creates disorder as the solid crystal breaks apart into mobile ions in solution. When temperature increases, the TΔST\Delta S term becomes more negative (larger in magnitude), making ΔG\Delta G more negative and increasing solubility. Answer D correctly identifies this relationship - higher temperatures amplify the favorable entropy contribution. Answer A contains a common misconception. While higher temperature does provide more energy, this explanation focuses only on kinetics rather than thermodynamics. The question asks about equilibrium solubility, not reaction rate. Answer B is fundamentally wrong because equilibrium constants are temperature-dependent for all processes. The relationship follows ln⁡K=−ΔHRT+ΔSR\ln K = -\frac{\Delta H}{RT} + \frac{\Delta S}{R}, clearly showing temperature dependence. Answer C reverses the correct logic. While the ΔH\Delta H term does become relatively larger at higher temperatures, the TΔST\Delta S term grows even faster since it's directly proportional to temperature. The entropy term dominates, making dissolution more favorable. Study tip: For IB Chemistry thermodynamics questions, always consider both ΔH\Delta H and ΔS\Delta S contributions to ΔG\Delta G. Remember that temperature amplifies the entropy term's effect on spontaneity.

Question 18

Consider two reactions at 298 K: Reaction 1: ΔG∘=−15 kJ mol−1\Delta G^\circ = -15 \text{ kJ mol}^{-1}, ΔH∘=−8 kJ mol−1\Delta H^\circ = -8 \text{ kJ mol}^{-1} and Reaction 2: ΔG∘=−15 kJ mol−1\Delta G^\circ = -15 \text{ kJ mol}^{-1}, ΔH∘=−25 kJ mol−1\Delta H^\circ = -25 \text{ kJ mol}^{-1}. How will the spontaneity of these reactions compare at significantly higher temperatures?

  1. Both reactions will remain equally spontaneous since they have the same ΔG° at 298 K
  2. Reaction 1 will become more spontaneous because it has a more positive ΔS° value (correct answer)
  3. Reaction 2 will become more spontaneous because it releases more energy through bond formation
  4. Reaction 2 will become less spontaneous because its negative ΔS° value opposes spontaneity at high temperature
Explanation: First, calculate ΔS° for each: Reaction 1: ΔS° = (ΔH° - ΔG°)/T = (-8000 + 15000)/298 = +23.5 J/mol·K; Reaction 2: ΔS° = (-25000 + 15000)/298 = -33.6 J/mol·K. At higher temperatures, ΔG = ΔH - TΔS, so Reaction 1 (positive ΔS°) becomes more favorable while Reaction 2 (negative ΔS°) becomes less favorable. Choice A ignores temperature dependence. Choices C and D focus on enthalpy rather than the entropy effect.

Question 19

A reaction has a standard Gibbs free energy change, \Delta G^⦵, of +20.5 kJ mol⁻¹ at 298 K. What can be deduced about the equilibrium constant, K, for this reaction? (R = 8.31 J K⁻¹ mol⁻¹)

  1. K is much greater than 1.
  2. K is much less than 1. (correct answer)
  3. K is approximately equal to 1.
  4. K is approximately equal to 0.
Explanation: The relationship between \Delta G^⦵ and K is given by \Delta G^⦵ = -RT \ln K. Since \Delta G^⦵ is positive (+20.5 kJ mol⁻¹), the term −RTln⁡K-RT \ln K must be positive. As R and T are positive constants, ln⁡K\ln K must be negative. For ln⁡K\ln K to be negative, K must be less than 1. A positive \Delta G^⦵ indicates a non-spontaneous reaction under standard conditions, meaning reactants are favoured at equilibrium, which corresponds to K < 1.

Question 20

A reaction is spontaneous at 280 K but non-spontaneous at 300 K under standard conditions. What must be the signs of \Delta H^⦵ and \Delta S^⦵?

  1. \Delta H^⦵ > 0 and \Delta S^⦵ > 0
  2. \Delta H^⦵ < 0 and \Delta S^⦵ < 0 (correct answer)
  3. \Delta H^⦵ < 0 and \Delta S^⦵ > 0
  4. \Delta H^⦵ > 0 and \Delta S^⦵ < 0
Explanation: The spontaneity of the reaction is temperature-dependent and it is spontaneous at lower temperatures but not at higher temperatures. This behaviour occurs when both \Delta H^⦵ and \Delta S^⦵ are negative. The Gibbs free energy equation is \Delta G^⦵ = \Delta H^⦵ - T\Delta S^⦵. If \Delta H^⦵ is negative and \Delta S^⦵ is negative, the equation becomes \Delta G^⦵ = (\text{negative}) - T(\text{negative}) = (\text{negative}) + (\text{positive}). At low T, the negative \Delta H^⦵ term dominates, making \Delta G^⦵ negative (spontaneous). At high T, the positive -T\Delta S^⦵ term dominates, making \Delta G^⦵ positive (non-spontaneous).