Understand Entropy & Spontaneity - IB Chemistry

1. Consider the decomposition of dinitrogen tetroxide: N₂O₄(g) → 2NO₂(g). This reaction is endothermic. Which statement correctly describes the thermodynamic driving forces for this reaction?
   Answer: The reaction is entropy-driven because the number of moles of gas increases. The reaction involves one mole of gaseous reactant forming two moles of gaseous products. This represents an increase in disorder, so the entropy change (ΔS) is positive. A positive ΔS provides a driving force for the reaction, making it entropy-driven. Since the reaction is endothermic (ΔH > 0), it is not enthalpy-driven. A reaction with ΔH > 0 and ΔS > 0 becomes spontaneous at high temperatures when the TΔS term overcomes the positive ΔH, making ΔG negative. Therefore, stating it is entropy-driven is the correct description of the favourable thermodynamic factor.
2. A reaction has a standard Gibbs free energy change, ΔG⦵, of +35 kJ mol⁻¹. What can be deduced about this reaction under standard conditions?
   Answer: The forward reaction is non-spontaneous and the value of the equilibrium constant, K, is less than 1. A positive ΔG⦵ indicates that the forward reaction is non-spontaneous under standard conditions. The relationship between ΔG⦵ and K is given by ΔG⦵ = −RT lnK. If ΔG⦵ is positive, then lnK must be negative, which means K must be less than 1. This indicates that at equilibrium, the reactants are favoured over the products. The reverse reaction would have a ΔG⦵ of -35 kJ mol⁻¹, making it spontaneous.
3. For a given reaction, plotting Gibbs free energy (G) against the extent of reaction shows that the products have a lower Gibbs free energy than the reactants at a certain temperature and pressure. What can be concluded?
   Answer: The forward reaction is spontaneous. The change in Gibbs free energy (ΔG) for a reaction is the difference between the Gibbs free energy of the products and the reactants (G_products - G_reactants). If the products have a lower Gibbs free energy than the reactants, this difference is negative (ΔG < 0). A negative ΔG is the criterion for a spontaneous reaction under those conditions. The sign of ΔG does not provide direct information about the enthalpy change (exothermic/endothermic) without knowing the entropy change, nor does it provide any information about the reaction rate.
4. Consider the following reaction at 298 K: 2H₂(g) + O₂(g) → 2H₂O(l). Given ΔG⦵ = -474 kJ mol⁻¹. If the same reaction produces gaseous water, 2H₂(g) + O₂(g) → 2H₂O(g), how would the new ΔG⦵ compare to -474 kJ mol⁻¹?
   Answer: It would be less negative, because H₂O(g) has higher enthalpy than H₂O(l). We can think of the reaction to form H₂O(g) as a two-step process via Hess's Law: 1) 2H₂(g) + O₂(g) → 2H₂O(l), ΔG₁ = -474 kJ. 2) 2H₂O(l) → 2H₂O(g), ΔG₂. The overall reaction is the sum. The vaporization of water at 298 K is non-spontaneous, so ΔG₂ is positive. Adding a positive value to -474 kJ will result in a less negative value. Alternatively, ΔH for forming H₂O(g) is less negative (less exothermic) than for H₂O(l), and while ΔS is less negative (more favorable), the enthalpy change is the dominant factor. Thus, the overall reaction to form gaseous water is less spontaneous (less negative ΔG⦵) than the reaction to form liquid water.
5. The condensation of steam, H₂O(g) → H₂O(l), is a spontaneous process at 90 °C and 100 kPa. Which statement correctly describes the entropy changes for this process?
   Answer: ΔS_system < 0 and ΔS_universe > 0 The system is the water undergoing condensation. The process goes from a gas to a liquid, which is a decrease in disorder, so ΔS_system is negative. Condensation is an exothermic process (releases heat), so the surroundings gain heat, and ΔS_surroundings is positive. For a process to be spontaneous, the total entropy change of the universe (ΔS_universe = ΔS_system + ΔS_surroundings) must be positive. Therefore, ΔS_system < 0 and ΔS_universe > 0 correctly describes the process.
6. Which of the following processes is expected to have the largest increase in entropy?
   Answer: CaCO₃(s) → CaO(s) + CO₂(g) Entropy change (ΔS) is largely influenced by the change in the number of moles of gas (Δn_gas). A large increase in moles of gas corresponds to a large increase in entropy. A: Δn_gas = 1 - 1 = 0. B: Δn_gas = 1 - 0 = +1. A solid produces a gas, a significant increase in disorder. C: Δn_gas = 2 - 3 = -1. D: Δn_gas = 2 - 4 = -2. Process B involves the creation of a mole of gas from solid reactants, representing the largest increase in disorder and thus the largest positive ΔS.
7. The standard enthalpy of formation of liquid water is -286 kJ mol⁻¹ and the standard enthalpy of formation of water vapour is -242 kJ mol⁻¹. The standard entropy change for the vaporization of water, H₂O(l) → H₂O(g), is +119 J K⁻¹ mol⁻¹. What is the standard Gibbs free energy change, ΔG⦵, for the vaporization of water at 298 K?
   Answer: +8.6 kJ mol⁻¹ First, calculate the enthalpy change for the vaporization: ΔH⦵ = Δ$H_f$⦵(H₂O, g) - Δ$H_f$⦵(H₂O, l) = (-242) - (-286) = +44 kJ mol⁻¹. Then, use the Gibbs free energy equation, ΔG⦵ = ΔH⦵ - TΔS⦵. Ensure consistent units: ΔS⦵ = +0.119 kJ K⁻¹ mol⁻¹. ΔG⦵ = (+44 kJ mol⁻¹) - (298 K * +0.119 kJ K⁻¹ mol⁻¹) = 44 - 35.462 = +8.538 kJ mol⁻¹. The positive value indicates that vaporization is non-spontaneous at standard conditions (298 K).
8. Consider the oxidation of iron: 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s). This reaction is highly exothermic (ΔH < 0) and spontaneous. What can be deduced about the sign of the entropy change, ΔS, for the system?
   Answer: ΔS must be negative because gaseous reactants are converted into a solid product. The entropy change of the system (ΔS_system) is determined by the change in disorder. In this reaction, 3 moles of a highly disordered gas (O₂) are consumed to form a highly ordered solid product (Fe₂O₃). This represents a significant decrease in the disorder of the system. Therefore, ΔS for the system must be negative. The reaction is spontaneous because the large negative ΔH (exothermic) outweighs the unfavorable negative TΔS term, making ΔG negative (ΔG = ΔH - TΔS).
9. The Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), has ΔH⦵ = −92 kJ mol⁻¹ and ΔS⦵ = −199 J K⁻¹ mol⁻¹. A chemist suggests that increasing the temperature will make the reaction more thermodynamically favourable. Evaluate this suggestion.
   Answer: The suggestion is incorrect because increasing T will make the TΔS term more negative, thus making ΔG less negative. Thermodynamic favourability is assessed by ΔG. The equation is ΔG = ΔH - TΔS. Here, ΔH is negative and ΔS is negative. The term -TΔS is therefore -T(-199) = +199T. So, ΔG = -92000 + 199T (in Joules). As temperature (T) increases, the positive term (+199T) becomes larger, making ΔG less negative (more positive). Thus, increasing the temperature makes the reaction less thermodynamically favourable. The suggestion is incorrect. Note that while increasing temperature increases the reaction rate, it shifts the equilibrium to the left and makes the reaction less spontaneous.
10. At 298 K, the standard Gibbs free energy of formation for $\text{NO}(g)$ is $+86.6 \text{ kJ mol}^{-1}$. Despite this positive value, NO is observed in car exhaust and lightning strikes. Which explanation best accounts for this apparent contradiction?
    Answer: The positive $\Delta G°_f$ indicates thermodynamic instability, but kinetic factors allow NO formation under high-energy conditions despite thermodynamic unfavorability $\Delta G°_f > 0$ indicates NO is thermodynamically unstable relative to its elements under standard conditions. However, kinetics allows its formation under high-energy conditions (high temperature, electrical discharge) even though it's thermodynamically unfavorable. Once formed, NO persists due to kinetic barriers to decomposition. Choice A incorrectly suggests high temperature makes $\Delta G < 0$ for formation. Choice C wrongly dismisses the relevance of thermodynamic data. Choice D incorrectly assumes observation implies thermodynamic favorability.
11. The thermal decomposition of silver carbonate is shown: Ag₂CO₃(s) → Ag₂O(s) + CO₂(g). Given that this reaction is non-spontaneous at 298 K but becomes spontaneous at temperatures above 483 K, what must be the signs of ΔH and ΔS for the reaction?
    Answer: ΔH is positive, ΔS is positive. The reaction becomes spontaneous at high temperatures. This temperature-dependent spontaneity occurs when the signs of ΔH and ΔS are the same. Let's analyze ΔG = ΔH - TΔS. If the reaction becomes spontaneous (ΔG < 0) as T increases, the -TΔS term must be overcoming a positive ΔH. This requires ΔS to be positive. Therefore, both ΔH and ΔS must be positive. This is consistent with the reaction: breaking bonds in Ag₂CO₃ requires energy (endothermic, ΔH > 0), and producing a mole of gas from a solid increases disorder (ΔS > 0).
12. A certain chemical reaction has a standard enthalpy change, ΔH⦵, of +120 kJ mol⁻¹ and a standard entropy change, ΔS⦵, of +300 J K⁻¹ mol⁻¹. Under what condition will this reaction be spontaneous?
    Answer: The reaction is spontaneous only at temperatures above 400 K. For a reaction to be spontaneous, the Gibbs free energy change, ΔG, must be negative. The relationship is given by ΔG = ΔH − TΔS. In this case, ΔH is positive (unfavourable) and ΔS is positive (favourable). For ΔG to be negative, the term TΔS must be larger than ΔH. This occurs at high temperatures. The crossover temperature where ΔG = 0 is T = ΔH / ΔS. It is crucial to use consistent units: T = (120000 J mol⁻¹) / (300 J K⁻¹ mol⁻¹) = 400 K. Therefore, the reaction becomes spontaneous at temperatures above 400 K.
13. For a reaction system at equilibrium, $\Delta G = 0$. If the temperature is suddenly increased while keeping pressure constant, and both $\Delta H > 0$ and $\Delta S > 0$ for the forward reaction, what happens to the system?
    Answer: $\Delta G$ becomes negative, causing the equilibrium to shift forward because the $T\Delta S$ term increases more than $\Delta H$ When you encounter equilibrium problems involving temperature changes, focus on the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$. At equilibrium, $\Delta G = 0$, but this changes when external conditions shift. When temperature increases suddenly, the system is temporarily knocked out of equilibrium. Since both $\Delta H > 0$ (endothermic) and $\Delta S > 0$ (entropy increases), the $-T\Delta S$ term becomes more negative as temperature rises. This makes $\Delta G$ negative for the forward reaction, driving the equilibrium forward until a new equilibrium establishes where $\Delta G = 0$ again. Option A incorrectly assumes $\Delta G$ stays zero instantly. While equilibrium systems do re-establish $\Delta G = 0$, there's a period where the system shifts to reach this new state. Option B misunderstands Le Chatelier's principle. Yes, increasing temperature shifts endothermic equilibria forward, not backward. The reasoning about $\Delta G$ becoming positive is also wrong given our thermodynamic analysis. Option D suggests we need specific values, but we can determine the direction qualitatively. Since both $\Delta H$ and $\Delta S$ are positive, higher temperature always makes $\Delta G$ more negative for the forward reaction. Remember this pattern: for reactions where $\Delta H > 0$ and $\Delta S > 0$, increasing temperature favors the forward direction because the entropy term ($T\Delta S$) grows faster than the constant enthalpy term. This is why such reactions become more spontaneous at higher temperatures.
14. For a reaction, ΔH⦵ = -200 kJ mol⁻¹ and ΔS⦵ = -150 J K⁻¹ mol⁻¹. Above which approximate temperature will this reaction become non-spontaneous?
    Answer: 1330 K The reaction becomes non-spontaneous when ΔG > 0. The crossover point is at ΔG = 0, where T = ΔH / ΔS. It is essential to use consistent units. ΔH⦵ = -200000 J mol⁻¹ and ΔS⦵ = -150 J K⁻¹ mol⁻¹. So, T = (-200000 J mol⁻¹) / (-150 J K⁻¹ mol⁻¹) ≈ 1333 K. Since ΔH is negative and ΔS is negative, the reaction is enthalpy-driven and spontaneous at temperatures below this value. It will become non-spontaneous at temperatures above approximately 1330 K.
15. The dissolution of ammonium nitrate in water is endothermic ($\Delta H > 0$) yet occurs spontaneously at room temperature. Based on the thermodynamic principles of entropy and spontaneity, what must be true about this process?
    Answer: $\Delta S_{system} > 0$ and $|T\Delta S_{system}| > |\Delta H|$, making $\Delta G < 0$ despite the positive enthalpy change When you encounter questions about spontaneous endothermic processes, you need to think beyond just enthalpy and consider the complete thermodynamic picture using Gibbs free energy. For any process to be spontaneous, $\Delta G$ must be negative. The Gibbs free energy equation is $\Delta G = \Delta H - T\Delta S$. Since ammonium nitrate dissolution is both endothermic ($\Delta H > 0$) and spontaneous ($\Delta G < 0$), the entropy term $T\Delta S$ must be large enough and positive enough to overcome the positive enthalpy change. When ionic solids dissolve, they break apart into individual ions that move freely in solution, dramatically increasing the system's entropy ($\Delta S_{system} > 0$). For this process to be spontaneous despite absorbing heat, the magnitude of the favorable entropy term $|T\Delta S_{system}|$ must exceed the unfavorable enthalpy term $|\Delta H|$, making $\Delta G < 0$. This is exactly what option D describes. Option A is wrong because endothermic spontaneous processes don't violate thermodynamics—they're actually common examples of entropy-driven processes. Option B incorrectly states that $\Delta S_{system} < 0$, when dissolving an ionic solid actually increases system entropy significantly. Option C is wrong because $\Delta H$ is genuinely positive for this dissolution—the process does absorb heat from surroundings. Study tip: Remember that spontaneity depends on $\Delta G$, not just $\Delta H$. When you see "endothermic but spontaneous," immediately think about entropy increases overcoming unfavorable enthalpy through the Gibbs equation.
16. A chemical reaction has $\Delta H = -85 \text{ kJ mol}^{-1}$ and $\Delta S = -120 \text{ J K}^{-1} \text{mol}^{-1}$. At what temperature does this reaction transition from being spontaneous to non-spontaneous?
    Answer: 708 K, because this is where $\Delta G$ changes from negative to positive as temperature increases At equilibrium, $\Delta G = 0$, so $\Delta H = T\Delta S$. Therefore $T = \frac{\Delta H}{\Delta S} = \frac{-85000 \text{ J mol}^{-1}}{-120 \text{ J K}^{-1}\text{mol}^{-1}} = 708 \text{ K}$. Since both $\Delta H$ and $\Delta S$ are negative, the reaction is spontaneous at low temperatures ($\Delta G < 0$) but becomes non-spontaneous at high temperatures ($\Delta G > 0$). Choice B uses incorrect calculation (85/0.12 = 708, but student might divide 85/0.195 ≈ 435). Choice C has correct temperature but wrong direction of spontaneity change. Choice D has both wrong temperature and wrong direction.
17. The value of ΔG for a reaction is determined to be -50 kJ mol⁻¹ at a particular temperature and pressure. Which statement is a valid conclusion?
    Answer: The position of equilibrium lies strongly towards the products. A negative value for ΔG indicates that a reaction is spontaneous in the forward direction. The magnitude of ΔG is related to the position of equilibrium through the equation ΔG⦵ = -RTlnK. A large negative ΔG corresponds to a large value for the equilibrium constant K (K > 1), meaning the concentration of products at equilibrium is much greater than the concentration of reactants. ΔG provides no information about the rate of reaction. A spontaneous reaction is not necessarily fast. The sign of ΔG does not directly indicate whether a reaction is endothermic or exothermic without knowing ΔS. The reverse reaction would have a ΔG of +50 kJ mol⁻¹ and would be non-spontaneous.
18. The dissolution of ammonium nitrate in water is an endothermic process that is spontaneous. Which statement provides the best explanation for its spontaneity?
    Answer: The large positive entropy change of the system outweighs the positive enthalpy change. The process is endothermic, meaning ΔH is positive. For a process with a positive ΔH to be spontaneous (ΔG < 0), the entropy change (ΔS) must be positive and the TΔS term must be larger in magnitude than ΔH. When solid ammonium nitrate dissolves, its ions dissociate and become hydrated, a large increase in disorder. This large positive ΔS is the driving force that makes the TΔS term overcome the positive ΔH, resulting in a negative ΔG.
19. Calculate the standard Gibbs free energy change, ΔG⦵, in kJ mol⁻¹ for the synthesis of methanol at 298 K.
    Answer: -25.1 kJ mol⁻¹ First, calculate the standard enthalpy change for the reaction: ΔH⦵ = ΣΔ$H_f$⦵(products) − ΣΔ$H_f$⦵(reactants) = [(-201.2)] - [(-110.5) + 2(0)] = -90.7 kJ mol⁻¹. Next, calculate the standard entropy change: ΔS⦵ = ΣS⦵(products) − ΣS⦵(reactants) = [(239.8)] - [(197.7) + 2(130.7)] = 239.8 - 459.1 = -219.3 J K⁻¹ mol⁻¹. Convert ΔS⦵ to kJ K⁻¹ mol⁻¹: -0.2193 kJ K⁻¹ mol⁻¹. Finally, calculate ΔG⦵ using ΔG⦵ = ΔH⦵ − TΔS⦵: ΔG⦵ = -90.7 kJ mol⁻¹ - (298 K)(-0.2193 kJ K⁻¹ mol⁻¹) = -90.7 + 65.35 = -25.35 kJ mol⁻¹. The closest value is -25.1 kJ mol⁻¹.
20. Consider two processes: Process X has $\Delta S_{sys} = +25 \text{ J K}^{-1}$ and $\Delta S_{surr} = -15 \text{ J K}^{-1}$. Process Y has $\Delta S_{sys} = -20 \text{ J K}^{-1}$ and $\Delta S_{surr} = +35 \text{ J K}^{-1}$. Which statement correctly describes the spontaneity of these processes?
    Answer: Both processes are spontaneous because $\Delta S_{universe} > 0$ for both, with Process X having $\Delta S_{univ} = +10 \text{ J K}^{-1}$ and Process Y having $\Delta S_{univ} = +15 \text{ J K}^{-1}$ Spontaneity is determined by $\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}$. For Process X: $\Delta S_{univ} = +25 + (-15) = +10 \text{ J K}^{-1} > 0$ (spontaneous). For Process Y: $\Delta S_{univ} = -20 + (+35) = +15 \text{ J K}^{-1} > 0$ (spontaneous). Choice B incorrectly suggests only negative $\Delta S_{sys}$ can lead to spontaneity. Choice C incorrectly requires positive $\Delta S_{sys}$. Choice D misunderstands that individual negative values don't prevent spontaneity if the sum is positive.

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