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

A reaction is spontaneous at all temperatures. Which combination of signs for ΔH and ΔS describes such a reaction?

ΔH is positive, ΔS is positive.
ΔH is positive, ΔS is negative.
ΔH is negative, ΔS is positive.
ΔH is negative, ΔS is negative.
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IB Chemistry Quiz

IB Chemistry Quiz: Understand Entropy And Spontaneity

Practice Understand 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.

What this quiz covers

This quiz focuses on Understand 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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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

A reaction is spontaneous at all temperatures. Which combination of signs for ΔH and ΔS describes such a reaction?

  1. ΔH is positive, ΔS is positive.
  2. ΔH is positive, ΔS is negative.
  3. ΔH is negative, ΔS is positive. (correct answer)
  4. ΔH is negative, ΔS is negative.
Explanation: Spontaneity is determined by ΔG = ΔH - TΔS. For a reaction to be spontaneous at all temperatures, ΔG must be negative regardless of the value of T (which is always positive in Kelvin). If ΔH is negative (favourable) and ΔS is positive (favourable), the term -TΔS will always be negative. Adding two negative numbers (ΔH and -TΔS) will always result in a negative ΔG. Therefore, a negative ΔH and a positive ΔS ensure spontaneity at all temperatures.

Question 2

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?

  1. The reaction is endothermic.
  2. The reaction is exothermic.
  3. The reaction rate is high.
  4. The forward reaction is spontaneous. (correct answer)
Explanation: 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.

Question 3

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?

  1. It would be more negative, because H₂O(g) has higher entropy than H₂O(l).
  2. It would be less negative, because the vaporization of water is a non-spontaneous process.
  3. It would be the same, because the reactants are identical.
  4. It would be less negative, because H₂O(g) has higher enthalpy than H₂O(l). (correct answer)
Explanation: 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.

Question 4

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?

  1. ΔS must be positive because the reaction is spontaneous.
  2. ΔS must be negative because gaseous reactants are converted into a solid product. (correct answer)
  3. ΔS is approximately zero because the number of solid moles changes.
  4. The sign of ΔS cannot be determined without more information.
Explanation: 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).

Question 5

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?

  1. ΔH is positive, ΔS is positive. (correct answer)
  2. ΔH is positive, ΔS is negative.
  3. ΔH is negative, ΔS is positive.
  4. ΔH is negative, ΔS is negative.
Explanation: 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).

Question 6

Consider the reaction for the synthesis of methanol: CO(g) + 2H₂(g) ⇌ CH₃OH(g). The standard thermodynamic data at 298 K are provided: ΔH_f⦵ (CO, g) = -110.5 kJ mol⁻¹ ΔH_f⦵ (CH₃OH, g) = -201.2 kJ mol⁻¹ S⦵ (CO, g) = 197.7 J K⁻¹ mol⁻¹ S⦵ (H₂, g) = 130.7 J K⁻¹ mol⁻¹ S⦵ (CH₃OH, g) = 239.8 J K⁻¹ mol⁻¹

Calculate the standard Gibbs free energy change, ΔG⦵, in kJ mol⁻¹ for the synthesis of methanol at 298 K.

  1. +29.1 kJ mol⁻¹
  2. -29.1 kJ mol⁻¹
  3. -152.0 kJ mol⁻¹
  4. -25.1 kJ mol⁻¹ (correct answer)
Explanation: First, calculate the standard enthalpy change for the reaction: ΔH⦵ = ΣΔH_f⦵(products) − ΣΔH_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.

Question 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?

  1. +8.6 kJ mol⁻¹ (correct answer)
  2. -35.5 kJ mol⁻¹
  3. +79.4 kJ mol⁻¹
  4. -528 kJ mol⁻¹
Explanation: First, calculate the enthalpy change for the vaporization: ΔH⦵ = ΔH_f⦵(H₂O, g) - ΔH_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).

Question 8

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.

  1. The suggestion is correct because increasing T increases the kinetic energy of molecules.
  2. The suggestion is incorrect because increasing T will make the TΔS term more negative, thus making ΔG less negative. (correct answer)
  3. The suggestion is correct because the reaction is exothermic, and adding heat always favours the products.
  4. The suggestion is incorrect because ΔH is negative, meaning the reaction is already favourable at all temperatures.
Explanation: 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.

Question 9

Which change will result in a decrease in the entropy of the system?

  1. Dissolving NaCl(s) in water.
  2. Expanding a sample of neon gas into a larger volume.
  3. Heating a sample of solid iron from 300 K to 400 K.
  4. Deposition of iodine gas to form solid iodine, I₂(g) → I₂(s). (correct answer)
Explanation: Entropy is a measure of disorder or randomness. A decrease in entropy means the system becomes more ordered. A: Dissolving a solid increases entropy as ions become mobile. B: Gas expansion increases the volume available, increasing disorder. C: Heating increases the kinetic energy and vibrational motion of atoms, increasing entropy. D: Deposition is the phase change from gas to solid. This is a significant increase in order as particles go from random, high-speed motion to a fixed lattice, resulting in a decrease in the entropy of the system.

Question 10

The dissolution of ammonium nitrate in water is endothermic (ΔH>0\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?

  1. The process violates the second law of thermodynamics because spontaneous endothermic processes are thermodynamically impossible
  2. ΔSsystem<0\Delta S_{system} < 0 but ΔSsurroundings>0\Delta S_{surroundings} > 0 due to heat absorption, resulting in overall entropy increase
  3. ΔH\Delta H must actually be negative because endothermic processes cannot be spontaneous according to thermodynamic laws
  4. ΔSsystem>0\Delta S_{system} > 0 and ∣TΔSsystem∣>∣ΔH∣|T\Delta S_{system}| > |\Delta H|, making ΔG<0\Delta G < 0 despite the positive enthalpy change (correct answer)
Explanation: 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, ΔG\Delta G must be negative. The Gibbs free energy equation is ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S. Since ammonium nitrate dissolution is both endothermic (ΔH>0\Delta H > 0) and spontaneous (ΔG<0\Delta G < 0), the entropy term TΔST\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 (ΔSsystem>0\Delta S_{system} > 0). For this process to be spontaneous despite absorbing heat, the magnitude of the favorable entropy term ∣TΔSsystem∣|T\Delta S_{system}| must exceed the unfavorable enthalpy term ∣ΔH∣|\Delta H|, making ΔG<0\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 ΔSsystem<0\Delta S_{system} < 0, when dissolving an ionic solid actually increases system entropy significantly. Option C is wrong because ΔH\Delta H is genuinely positive for this dissolution—the process does absorb heat from surroundings. Study tip: Remember that spontaneity depends on ΔG\Delta G, not just ΔH\Delta H. When you see "endothermic but spontaneous," immediately think about entropy increases overcoming unfavorable enthalpy through the Gibbs equation.

Question 11

At 298 K, the standard Gibbs free energy of formation for NO(g)\text{NO}(g) is +86.6 kJ mol−1+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?

  1. High temperatures in engines and lightning provide sufficient thermal energy to make ΔG<0\Delta G < 0 for NO formation, overcoming the unfavorable standard conditions
  2. The positive ΔG°f\Delta G°_f indicates thermodynamic instability, but kinetic factors allow NO formation under high-energy conditions despite thermodynamic unfavorability (correct answer)
  3. Standard conditions don't apply to these systems, so the ΔG°f\Delta G°_f value is irrelevant for predicting spontaneity in real-world scenarios
  4. The measurement of ΔG°f\Delta G°_f must be incorrect because spontaneous formation of NO proves the value should be negative
Explanation: ΔG°f>0\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 ΔG<0\Delta G < 0 for formation. Choice C wrongly dismisses the relevance of thermodynamic data. Choice D incorrectly assumes observation implies thermodynamic favorability.

Question 12

The vaporization of water at 298 K has ΔHvap=+44.0 kJ mol−1\Delta H_{vap} = +44.0 \text{ kJ mol}^{-1} and ΔSvap=+118 J K−1mol−1\Delta S_{vap} = +118 \text{ J K}^{-1} \text{mol}^{-1}. Why is liquid water stable at this temperature despite the positive entropy change favoring vaporization?

  1. Liquid water violates thermodynamic predictions, demonstrating that kinetic factors always override thermodynamic spontaneity in phase transitions
  2. The entropy change is not large enough to overcome the enthalpy penalty, requiring temperatures above 373 K where TΔS>ΔHT\Delta S > \Delta H
  3. The large positive ΔH\Delta H creates ΔG=+44.0−(298)(0.118)=+8.8 kJ mol−1>0\Delta G = +44.0 - (298)(0.118) = +8.8 \text{ kJ mol}^{-1} > 0, making vaporization thermodynamically unfavorable (correct answer)
  4. The positive ΔS\Delta S value is incorrect because liquids always have higher entropy than gases due to stronger intermolecular interactions
Explanation: When you encounter thermodynamics problems involving phase transitions, the key is using the Gibbs free energy equation ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S to determine spontaneity. A process is thermodynamically favorable when ΔG<0\Delta G < 0. Let's calculate ΔG\Delta G for water vaporization at 298 K. Converting the entropy change to consistent units: ΔS=+118 J K−1mol−1=+0.118 kJ K−1mol−1\Delta S = +118 \text{ J K}^{-1}\text{mol}^{-1} = +0.118 \text{ kJ K}^{-1}\text{mol}^{-1}. Now: ΔG=ΔH−TΔS=+44.0−(298)(0.118)=+44.0−35.2=+8.8 kJ mol−1\Delta G = \Delta H - T\Delta S = +44.0 - (298)(0.118) = +44.0 - 35.2 = +8.8 \text{ kJ mol}^{-1}. Since ΔG>0\Delta G > 0, vaporization is thermodynamically unfavorable at 298 K, explaining why liquid water is stable despite the positive entropy change. Answer A incorrectly suggests thermodynamics fails to predict behavior—thermodynamics works perfectly here, showing vaporization is unfavorable. Answer B contains a calculation error, stating you need temperatures above 373 K, but setting ΔG=0\Delta G = 0 gives T=ΔH/ΔS=44.0/0.118=373 KT = \Delta H/\Delta S = 44.0/0.118 = 373 \text{ K}, meaning vaporization becomes favorable exactly at 373 K (water's boiling point). Answer D shows a fundamental misunderstanding—gases always have higher entropy than liquids because gas molecules have greater freedom of movement. Remember: positive ΔS\Delta S doesn't guarantee spontaneity. You must consider both enthalpy and entropy through the Gibbs equation. At low temperatures, the ΔH\Delta H term dominates; at high temperatures, the TΔST\Delta S term takes over.

Question 13

For a reaction system at equilibrium, ΔG=0\Delta G = 0. If the temperature is suddenly increased while keeping pressure constant, and both ΔH>0\Delta H > 0 and ΔS>0\Delta S > 0 for the forward reaction, what happens to the system?

  1. ΔG\Delta G remains zero because equilibrium systems automatically adjust to maintain ΔG=0\Delta G = 0 regardless of temperature changes
  2. ΔG\Delta G becomes positive, causing the equilibrium to shift backward because increased temperature always opposes endothermic reactions
  3. ΔG\Delta G becomes negative, causing the equilibrium to shift forward because the TΔST\Delta S term increases more than ΔH\Delta H (correct answer)
  4. The direction of shift cannot be determined without knowing the specific values of ΔH\Delta H and ΔS\Delta S at the new temperature
Explanation: When you encounter equilibrium problems involving temperature changes, focus on the Gibbs free energy equation: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S. At equilibrium, ΔG=0\Delta G = 0, but this changes when external conditions shift. When temperature increases suddenly, the system is temporarily knocked out of equilibrium. Since both ΔH>0\Delta H > 0 (endothermic) and ΔS>0\Delta S > 0 (entropy increases), the −TΔS-T\Delta S term becomes more negative as temperature rises. This makes ΔG\Delta G negative for the forward reaction, driving the equilibrium forward until a new equilibrium establishes where ΔG=0\Delta G = 0 again. Option A incorrectly assumes ΔG\Delta G stays zero instantly. While equilibrium systems do re-establish ΔG=0\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 ΔG\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 ΔH\Delta H and ΔS\Delta S are positive, higher temperature always makes ΔG\Delta G more negative for the forward reaction. Remember this pattern: for reactions where ΔH>0\Delta H > 0 and ΔS>0\Delta S > 0, increasing temperature favors the forward direction because the entropy term (TΔST\Delta S) grows faster than the constant enthalpy term. This is why such reactions become more spontaneous at higher temperatures.

Question 14

A chemical reaction has ΔH=−85 kJ mol−1\Delta H = -85 \text{ kJ mol}^{-1} and ΔS=−120 J K−1mol−1\Delta S = -120 \text{ J K}^{-1} \text{mol}^{-1}. At what temperature does this reaction transition from being spontaneous to non-spontaneous?

  1. 708 K, because this is where ΔG\Delta G changes from negative to positive as temperature increases (correct answer)
  2. 435 K, because this is where ΔG\Delta G changes from positive to negative as temperature increases
  3. 708 K, because this is where ΔG\Delta G changes from positive to negative as temperature increases
  4. 435 K, because this is where ΔG\Delta G changes from negative to positive as temperature increases
Explanation: At equilibrium, ΔG=0\Delta G = 0, so ΔH=TΔS\Delta H = T\Delta S. Therefore T=ΔHΔS=−85000 J mol−1−120 J K−1mol−1=708 KT = \frac{\Delta H}{\Delta S} = \frac{-85000 \text{ J mol}^{-1}}{-120 \text{ J K}^{-1}\text{mol}^{-1}} = 708 \text{ K}. Since both ΔH\Delta H and ΔS\Delta S are negative, the reaction is spontaneous at low temperatures (ΔG<0\Delta G < 0) but becomes non-spontaneous at high temperatures (ΔG>0\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.

Question 15

For the reaction 2SO2(g)+O2(g)→2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g), ΔH°=−198 kJ\Delta H° = -198 \text{ kJ} and ΔS°=−188 J K−1\Delta S° = -188 \text{ J K}^{-1}. At what temperature range will this reaction be spontaneous, and what drives the spontaneity?

  1. Non-spontaneous at all temperatures because the negative entropy change violates the second law of thermodynamics
  2. Spontaneous above 1053 K, driven primarily by the entropy change which becomes favorable at higher temperatures
  3. Spontaneous at all temperatures because the large negative enthalpy change always overcomes the entropy penalty
  4. Spontaneous below 1053 K, driven primarily by the favorable enthalpy change which dominates at lower temperatures (correct answer)
Explanation: When you encounter a question about reaction spontaneity with given enthalpy and entropy values, you need to apply the Gibbs free energy equation: ΔG°=ΔH°−TΔS°\Delta G° = \Delta H° - T\Delta S°. A reaction is spontaneous when ΔG°<0\Delta G° < 0. For this reaction, both ΔH°\Delta H° and ΔS°\Delta S° are negative. Let's find the temperature where ΔG°=0\Delta G° = 0: 0=−198,000 J−T(−188 J K−1)0 = -198,000 \text{ J} - T(-188 \text{ J K}^{-1}). Solving gives T=1053 KT = 1053 \text{ K}. At temperatures below 1053 K, the TΔS°T\Delta S° term is smaller in magnitude than ΔH°\Delta H°, so ΔG°\Delta G° remains negative and the reaction is spontaneous. The favorable enthalpy change (bond formation releases energy) drives spontaneity at lower temperatures. Option A is wrong because negative entropy doesn't violate thermodynamics—it simply means the system becomes more ordered, which is common when gases combine to form fewer gas molecules. Option B incorrectly states the reaction is spontaneous above 1053 K, when actually ΔG°\Delta G° becomes positive (non-spontaneous) as temperature increases beyond this point. Option C is wrong because even large negative enthalpy changes can be overcome by unfavorable entropy at sufficiently high temperatures. Remember that for reactions with negative ΔH°\Delta H° and negative ΔS°\Delta S°, there's always a critical temperature above which the entropy penalty outweighs the enthalpy benefit. Calculate this crossover point to determine the spontaneity range.

Question 16

Consider two processes: Process X has ΔSsys=+25 J K−1\Delta S_{sys} = +25 \text{ J K}^{-1} and ΔSsurr=−15 J K−1\Delta S_{surr} = -15 \text{ J K}^{-1}. Process Y has ΔSsys=−20 J K−1\Delta S_{sys} = -20 \text{ J K}^{-1} and ΔSsurr=+35 J K−1\Delta S_{surr} = +35 \text{ J K}^{-1}. Which statement correctly describes the spontaneity of these processes?

  1. Both processes are spontaneous because ΔSuniverse>0\Delta S_{universe} > 0 for both, with Process X having ΔSuniv=+10 J K−1\Delta S_{univ} = +10 \text{ J K}^{-1} and Process Y having ΔSuniv=+15 J K−1\Delta S_{univ} = +15 \text{ J K}^{-1} (correct answer)
  2. Process X is non-spontaneous and Process Y is spontaneous because only systems with negative ΔSsys\Delta S_{sys} can be spontaneous when ΔSsurr\Delta S_{surr} is sufficiently positive
  3. Process X is spontaneous and Process Y is non-spontaneous because ΔSsys\Delta S_{sys} must be positive for spontaneity regardless of ΔSsurr\Delta S_{surr} values
  4. Neither process is spontaneous because one has negative ΔSsys\Delta S_{sys} and the other has negative ΔSsurr\Delta S_{surr}, violating the entropy requirement
Explanation: Spontaneity is determined by ΔSuniverse=ΔSsystem+ΔSsurroundings\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}. For Process X: ΔSuniv=+25+(−15)=+10 J K−1>0\Delta S_{univ} = +25 + (-15) = +10 \text{ J K}^{-1} > 0 (spontaneous). For Process Y: ΔSuniv=−20+(+35)=+15 J K−1>0\Delta S_{univ} = -20 + (+35) = +15 \text{ J K}^{-1} > 0 (spontaneous). Choice B incorrectly suggests only negative ΔSsys\Delta S_{sys} can lead to spontaneity. Choice C incorrectly requires positive ΔSsys\Delta S_{sys}. Choice D misunderstands that individual negative values don't prevent spontaneity if the sum is positive.

Question 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?

  1. The reaction will proceed rapidly to completion.
  2. The position of equilibrium lies strongly towards the products. (correct answer)
  3. The reaction is strongly endothermic.
  4. The reverse reaction also has a negative ΔG.
Explanation: 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.

Question 18

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?

  1. ΔS_system > 0 and ΔS_surroundings > 0
  2. ΔS_system < 0 and ΔS_universe > 0 (correct answer)
  3. ΔS_system > 0 and ΔS_universe < 0
  4. ΔS_system < 0 and ΔS_surroundings < 0
Explanation: 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.

Question 19

Which substance, under standard conditions, has the highest standard molar entropy, S⦵?

  1. Na(s)
  2. H₂O(l)
  3. CO₂(g) (correct answer)
  4. C(s, diamond)
Explanation: Standard molar entropy depends on the state of matter, molar mass, and molecular complexity. Gases have significantly higher entropy than liquids or solids due to greater freedom of movement. Among the choices, CO₂(g) is the only substance that is a gas at standard conditions. Na(s), H₂O(l), and C(s, diamond) are solid or liquid and will have much lower entropy values. Therefore, CO₂(g) will have the highest standard molar entropy.

Question 20

The dissolution of ammonium nitrate in water is an endothermic process that is spontaneous. Which statement provides the best explanation for its spontaneity?

  1. The enthalpy of the system decreases significantly upon dissolution.
  2. The temperature of the surroundings increases, leading to a favourable entropy change.
  3. The large positive entropy change of the system outweighs the positive enthalpy change. (correct answer)
  4. The process is spontaneous because all dissolution processes increase the entropy of the universe.
Explanation: 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.