Apply Entropy & Spontaneity - IB Chemistry

Consider two reactions at 298 K: Reaction 1: $\Delta G^\circ = -15 \text{ kJ mol}^{-1}$ , $\Delta H^\circ = -8 \text{ kJ mol}^{-1}$ and Reaction 2: $\Delta G^\circ = -15 \text{ kJ mol}^{-1}$ , $\Delta H^\circ = -25 \text{ kJ mol}^{-1}$ . How will the spontaneity of these reactions compare at significantly higher temperatures?
Answer: Reaction 1 will become more spontaneous because it has a more positive ΔS° value 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.
The dissolution of ammonium nitrate in water is a spontaneous endothermic process. Which statement correctly describes the thermodynamic changes for this dissolution?
Answer: $\Delta H > 0$ , $\Delta S > 0$ , $\Delta G < 0$ The process is described as spontaneous, which means the Gibbs free energy change must be negative ( $\Delta G < 0$ ). It is also described as endothermic, which means the enthalpy change is positive ( $\Delta H > 0$ ). For the process to be spontaneous despite being endothermic, the entropy term must drive the reaction. From the equation $\Delta G = \Delta H - T\Delta S$ , if $\Delta G$ is negative and $\Delta H$ is positive, the $-T\Delta S$ term must be negative and large in magnitude. This requires the entropy change, $\Delta S$ , to be positive. This is expected, as a solid crystal dissolving into aqueous ions represents an increase in disorder.
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 ( $\Delta G < 0$ ) in the forward direction?
Answer: When Q is sufficiently smaller than K 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, $\Delta G$ must be negative. This requires the $RT \ln Q$ term to be negative and have a greater magnitude than $\Delta G^⦵$ . For $RT \ln Q$ to be negative, $\ln Q$ must be negative, which means $Q < 1$ . Furthermore, we know that at equilibrium $\Delta G = 0$ and $Q = K$ , 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.
Which of the following substances, all at 298 K and 100 kPa, has the highest standard molar entropy (S⦵)?
Answer: C₃H₈(g) Standard molar entropy generally increases with: 1) phase changes from solid to liquid to gas, and 2) increasing molecular complexity (more atoms, greater molar mass). Comparing the options: H₂O(s) has the lowest entropy as a solid. C₂H₅OH(l) is a liquid, so its entropy is higher than solid water. Both Ne(g) and C₃H₈(g) are gases, which have much higher entropy than liquids or solids. Between the two gases, propane (C₃H₈) is a much more complex molecule than neon (Ne), with more atoms and more possible vibrational and rotational modes. Therefore, C₃H₈(g) has the highest standard molar entropy.
What is the best explanation for why the entropy of a sample of water increases when it is heated from 20 °C to 50 °C?
Answer: There is a wider range of accessible molecular speeds and vibrational energies. Heating a substance increases the thermal energy available to it. This energy is distributed among the various motional states (translational, rotational, vibrational) of the molecules. At a higher temperature, the distribution of molecular speeds (kinetic energies) becomes broader, and more vibrational and rotational energy levels become populated. This increase in the number of accessible microstates corresponds to a greater disorder and therefore a higher entropy.
For a hypothetical reaction, $\Delta H^⦵ = +40$ kJ mol⁻¹ and $\Delta S^⦵ = +200$ J K⁻¹ mol⁻¹. At what temperature does the reaction mixture exist at equilibrium under standard pressure?
Answer: 200 K Equilibrium under standard conditions corresponds to the state where $\Delta G^⦵ = 0$ . Using the Gibbs free energy equation, $0 = \Delta H^⦵ - T_{eq}\Delta S^⦵$ . Rearranging gives $T_{eq} = \Delta H^⦵ / \Delta S^⦵$ . It is essential to use consistent units. Convert $\Delta S^⦵$ to kJ K⁻¹ mol⁻¹: $200 \text{ J K⁻¹ mol⁻¹} = 0.200 \text{ kJ K⁻¹ mol⁻¹}$ . Now, calculate the equilibrium temperature: $T_{eq} = (40 \text{ kJ mol⁻¹}) / (0.200 \text{ kJ K⁻¹ mol⁻¹}) = 200$ K.
For which of the following processes is the entropy change, $\Delta S$ , negative?
Answer: Ag⁺(aq) + Cl⁻(aq) → AgCl(s) Entropy is a measure of disorder. A negative entropy change ( $\Delta S < 0$ ) 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 $\Delta S > 0$ . Option B is vaporization (liquid to gas), a large increase in disorder, so $\Delta S > 0$ . Option C is dissolving a solid into mobile ions, an increase in disorder, so $\Delta S > 0$ .
A certain endothermic reaction has a positive standard entropy change. Which statement correctly describes the spontaneity of this reaction?
Answer: The reaction is spontaneous only at high temperatures. The spontaneity of a reaction is determined by the sign of the Gibbs free energy change, $\Delta G = \Delta H - T\Delta S$ . The reaction is endothermic, so $\Delta H > 0$ . The entropy change is positive, so $\Delta S > 0$ . In this case, the $\Delta H$ term is positive, which disfavors spontaneity, while the $-T\Delta S$ term is negative, which favors spontaneity. The reaction will be spontaneous ( $\Delta G < 0$ ) when the magnitude of the $-T\Delta S$ term is greater than the magnitude of the $\Delta H$ term. This occurs at sufficiently high temperatures.
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)
Answer: 1106 K A reaction becomes spontaneous when $\Delta G$ changes from positive to negative. The crossover point occurs when $\Delta G = 0$ . At this point, $\Delta H = T\Delta S$ , so $T = \Delta H / \Delta S$ . We must use consistent units. Convert $\Delta H$ to J mol⁻¹: $178 \text{ kJ mol⁻¹} = 178000 \text{ J mol⁻¹}$ . Now calculate T: $T = (178000 \text{ J mol⁻¹}) / (161 \text{ J K⁻¹ mol⁻¹}) \approx 1105.6$ K. Since both $\Delta H$ and $\Delta S$ are positive, the reaction becomes spontaneous at temperatures above this value. So, T > 1106 K.
A spontaneous electrochemical cell (voltaic cell) operates under standard conditions. Which combination of signs for $\Delta G^{\circ}$ and standard cell potential, $E^{\circ}_{cell}$ , is correct?
Answer: $\Delta G^{\circ}$ is negative, $E^{\circ}_{cell}$ is positive. 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 $E^⦵_{cell} > 0$ .
The dissolution of ammonium nitrate in water is endothermic ( $\Delta H > 0$ ) yet occurs spontaneously at room temperature. If the temperature of this solution is increased, what happens to the solubility and why?
Answer: Solubility increases because the entropy change becomes more favorable at higher temperatures, making ΔG more negative When you encounter a question about spontaneous endothermic processes, you need to consider both enthalpy and entropy contributions to Gibbs free energy using $\Delta G = \Delta H - T\Delta S$ . For ammonium nitrate dissolution to be spontaneous despite being endothermic ( $\Delta H > 0$ ), the entropy term $T\Delta S$ must be large enough to make $\Delta G < 0$ . This makes sense because dissolving creates disorder as the solid crystal breaks apart into mobile ions in solution. When temperature increases, the $T\Delta S$ term becomes more negative (larger in magnitude), making $\Delta G$ 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 $\ln K = -\frac{\Delta H}{RT} + \frac{\Delta S}{R}$ , clearly showing temperature dependence. Answer C reverses the correct logic. While the $\Delta H$ term does become relatively larger at higher temperatures, the $T\Delta S$ 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 $\Delta H$ and $\Delta S$ contributions to $\Delta G$ . Remember that temperature amplifies the entropy term's effect on spontaneity.
A reaction is spontaneous at 280 K but non-spontaneous at 300 K under standard conditions. What must be the signs of $\Delta H^{\circ}$ and $\Delta S^{\circ}$ ?
Answer: $\Delta H^{\circ} < 0$ and $\Delta S^{\circ} < 0$ 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).
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?
Answer: ΔH° < 0 and ΔS° < 0, because K decreases with temperature and the reaction is spontaneous 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.
A biochemical reaction has $K_{eq} = 1.8 \times 10^{-3}$ at body temperature (310 K). If this reaction is coupled to ATP hydrolysis ( $\Delta G^\circ = -30.5 \text{ kJ mol}^{-1}$ at 310 K), what is the equilibrium constant for the overall coupled process?
Answer: $1.1 \times 10^{8}$ , because the coupling adds the free energy changes and affects K exponentially 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.
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?
Answer: Kx > 1 and Ky < 1 The relationship $\Delta G^⦵ = -RT \ln K$ dictates the connection between $\Delta G^⦵$ and K. For Reaction X, $\Delta G^⦵$ is negative, which means $\ln Kx$ must be positive, and therefore Kx > 1. This indicates that products are favoured at equilibrium. For Reaction Y, $\Delta G^⦵$ is positive, which means $\ln Ky$ must be negative, and therefore Ky < 1. This indicates that reactants are favoured at equilibrium.
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)
Answer: +20.0 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 $\ln K$ : $\ln(3.5 \times 10^{-4}) \approx -7.96$ . 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.
A reaction has $\Delta H = +85 \text{ kJ mol}^{-1}$ and $\Delta S = +180 \text{ J mol}^{-1} \text{K}^{-1}$ . At what temperature does this reaction become thermodynamically favorable, and what is the primary driving force at temperatures well above this point?
Answer: 472 K; the reaction is driven primarily by the favorable entropy change 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).
Consider the following reaction at 298 K: 2NO₂(g) ⇌ N₂O₄(g). Given $\Delta H^⦵ = -57.2$ kJ mol⁻¹ and $\Delta S^⦵ = -175.8$ J K⁻¹ mol⁻¹, which statement is correct?
Answer: The reaction is spontaneous at 298 K but becomes non-spontaneous at high temperatures. First, calculate $\Delta G^⦵$ at 298 K to determine spontaneity. $\Delta G^⦵ = \Delta H^⦵ - T\Delta S^⦵ = -57.2 \text{ kJ mol⁻¹} - (298 \text{ K})(-0.1758 \text{ kJ K⁻¹ mol⁻¹}) = -57.2 - (-52.4) = -4.8$ kJ mol⁻¹. Since $\Delta G^⦵$ is negative, the reaction is spontaneous at 298 K. Now, analyze the temperature dependence. Both $\Delta H^⦵$ and $\Delta S^⦵$ are negative. The $-T\Delta S^⦵$ term is positive and its magnitude increases with temperature. At a sufficiently high temperature, this positive term will overcome the negative $\Delta H^⦵$ term, making $\Delta G^⦵$ positive and the reaction non-spontaneous.
Which change is expected to cause the largest increase in the entropy of a system?
Answer: Subliming 1 mole of solid carbon dioxide at its sublimation point. 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.
A chemical reaction has $\Delta H^⦵ = -150.0$ kJ mol⁻¹ and $\Delta G^⦵ = -105.0$ kJ mol⁻¹ at 25 °C. What is the standard entropy change, $\Delta S^⦵$ , in J K⁻¹ mol⁻¹?
Answer: -151 First, convert the temperature to Kelvin: T = 25 + 273 = 298 K. Rearrange the Gibbs free energy equation $\Delta G = \Delta H - T\Delta S$ to solve for $\Delta S$ : $T\Delta S = \Delta H - \Delta G$ , so $\Delta S = (\Delta H - \Delta G) / T$ . Substitute the given values (ensuring they are in the same energy unit, kJ): $\Delta S = (-150.0 \text{ kJ mol⁻¹} - (-105.0 \text{ kJ mol⁻¹})) / 298 \text{ K} = (-45.0 \text{ kJ mol⁻¹}) / 298 \text{ K} \approx -0.151$ kJ K⁻¹ mol⁻¹. To convert to J K⁻¹ mol⁻¹, multiply by 1000: $-0.151 \times 1000 = -151$ J K⁻¹ mol⁻¹.

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