Gibbs Free Energy and Thermodynamic Favorability - AP Chemistry

1. The reaction $2\text{NO}_2(g)\rightarrow \text{N}_2\text{O}_4(g)$ is exothermic ($\Delta H<0$) and results in fewer moles of gas, so $\Delta S<0$. Based on these signs, which statement best describes the thermodynamic favorability of the reaction?
   Answer: The reaction is spontaneous only at low temperature. This question tests understanding of Gibbs free energy and thermodynamic favorability. For the dimerization reaction 2NO₂(g) → N₂O₄(g), we have ΔH < 0 (exothermic) and ΔS < 0 (fewer moles of gas). Using ΔG = ΔH - TΔS, at low temperatures, the negative ΔH term dominates over the small positive -TΔS term, making ΔG < 0 and the reaction spontaneous. At high temperatures, the -TΔS term becomes large and positive, potentially making ΔG > 0 and the reaction nonspontaneous. Choice A incorrectly assumes that all exothermic reactions are spontaneous at all temperatures, ignoring the unfavorable entropy change. The principle is that reactions with ΔH < 0 and ΔS < 0 are spontaneous only at temperatures below T = ΔH/ΔS.
2. For the reaction $\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)$, $\Delta H < 0$ and $\Delta S < 0$. Under which conditions is the reaction thermodynamically favorable (spontaneous)?
   Answer: Spontaneous only at low temperature. This question assesses the skill of Gibbs free energy and thermodynamic favorability. The Gibbs free energy change, ΔG, determines spontaneity, where ΔG < 0 indicates a spontaneous reaction. For this reaction with ΔH < 0 and ΔS < 0, ΔG = ΔH - TΔS means negative ΔH favors, positive -TΔS opposes. At low temperatures, ΔG is negative; at high temperatures, positive. A tempting distractor is choice C, which assumes exothermic reactions are always spontaneous, ignoring entropy's role. Assess spontaneity via ΔG, not stirring which affects mixing rate.
3. A reaction has $\Delta H<0$ and $\Delta S<0$. Which statement best describes its thermodynamic favorability at high temperature?
   Answer: Nonspontaneous at high temperature This question assesses the skill of Gibbs free energy and thermodynamic favorability. With ΔH < 0 and ΔS < 0, ΔG = ΔH - TΔS becomes positive at high temperatures as -TΔS grows large positive. At low T, ΔG is negative. This shows nonspontaneity at high T. A tempting distractor is choice A, 'Spontaneous at all temperatures,' based on the misconception that exothermic reactions are invariably spontaneous. Evaluate ΔG at specific temperatures to ascertain thermodynamic behavior.
4. A reaction has $\Delta H>0$ and $\Delta S<0$. Which statement best describes its thermodynamic favorability at any temperature?
   Answer: Nonspontaneous at all temperatures This question assesses the skill of Gibbs free energy and thermodynamic favorability. For ΔH > 0 and ΔS < 0, ΔG = ΔH - TΔS is always positive since both terms are positive. Positive ΔH and positive -TΔS ensure nonspontaneity at any temperature. No temperature can make ΔG negative. A tempting distractor is choice A, 'Spontaneous only at high temperature,' which wrongly assumes negative ΔS could be overcome by high T. Spontaneity is determined solely by ΔG's sign, integrating all factors.
5. For the decomposition $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$, $\Delta H > 0$ and $\Delta S > 0$. Under which conditions is the reaction thermodynamically favorable (spontaneous)?
   Answer: Spontaneous only at high temperature. This question assesses the skill of Gibbs free energy and thermodynamic favorability. The Gibbs free energy change, ΔG, determines spontaneity, where ΔG < 0 indicates a spontaneous reaction. For this decomposition with ΔH > 0 and ΔS > 0, ΔG = ΔH - TΔS shows the positive ΔH opposes, but positive ΔS favors via -TΔS. At high temperatures, -TΔS dominates, making ΔG negative; at low temperatures, ΔG is positive. A tempting distractor is choice B, which assumes endothermic reactions are always nonspontaneous, ignoring entropy's role at high temperatures. Evaluate spontaneity using ΔG's temperature dependence, not just ΔH's sign.
6. A sample of solid ammonium nitrate dissolves in water: $\text{NH}_4\text{NO}_3(s) \rightarrow \text{NH}_4^+(aq) + \text{NO}_3^-(aq)$. For this process, $\Delta H > 0$ and $\Delta S > 0$. Under which conditions is the dissolution thermodynamically favorable (spontaneous)?
   Answer: Spontaneous only at high temperature. This question assesses the skill of Gibbs free energy and thermodynamic favorability. The Gibbs free energy change, ΔG, determines spontaneity, where ΔG < 0 indicates a spontaneous process. For this dissolution with ΔH > 0 and ΔS > 0, ΔG = ΔH - TΔS shows that the positive ΔH opposes spontaneity, but the positive ΔS makes -TΔS negative, favoring it. At high temperatures, the TΔS term dominates, making ΔG negative and the process spontaneous; at low temperatures, ΔH dominates, making ΔG positive. A tempting distractor is choice A, which incorrectly assumes endothermic processes are nonspontaneous at all temperatures, disregarding the entropy increase. Always evaluate spontaneity based on ΔG, not just the sign of ΔH.
7. For the reaction $\mathrm{2NO_2(g) \rightarrow 2NO(g) + O_2(g)}$, $\Delta H>0$ and $\Delta S>0$. Which statement best describes the thermodynamic favorability (spontaneity) of the reaction as written?
   Answer: Spontaneous only at high temperature This question tests understanding of Gibbs free energy and thermodynamic favorability. For NO₂ decomposition, we have ΔH > 0 (endothermic) and ΔS > 0 (increase in entropy as 2 moles of gas form 3 moles). Using ΔG = ΔH - TΔS, when both terms are positive, spontaneity depends on their relative magnitudes. At low temperatures, the positive ΔH dominates over the small TΔS term, making ΔG positive and the reaction nonspontaneous. At high temperatures, the TΔS term becomes large enough to overcome the positive ΔH, making ΔG negative and the reaction spontaneous. Choice E incorrectly states that negative ΔS would make the reaction spontaneous, but with positive ΔH, this would make it nonspontaneous at all temperatures. Remember: when ΔH > 0 and ΔS > 0, reactions become spontaneous only at high temperatures.
8. For the reaction $\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)}$, $\Delta H<0$ and $\Delta S<0$. Which statement best describes when the reaction is thermodynamically favorable (spontaneous) as written?
   Answer: Spontaneous only at low temperature This question tests understanding of Gibbs free energy and thermodynamic favorability. For water formation from gases, we have ΔH < 0 (highly exothermic) and ΔS < 0 (decrease in entropy as gases form liquid). Using ΔG = ΔH - TΔS, when both terms are negative, the -TΔS becomes positive, opposing the negative ΔH. At low temperatures, the TΔS term is small, so the negative ΔH dominates, making ΔG negative and the reaction spontaneous. At high temperatures, the positive TΔS term becomes large enough to overcome the negative ΔH, making ΔG positive and the reaction nonspontaneous. Choice E incorrectly relates reaction speed to spontaneity—thermodynamic favorability depends on ΔG, not kinetics. Remember: when ΔH < 0 and ΔS < 0, reactions are spontaneous only at low temperatures.
9. A reaction has $\Delta G=-8\ \text{kJ mol}^{-1}$ at 298 K for the reaction as written. Which statement describes the thermodynamic favorability at 298 K?
   Answer: Spontaneous at 298 K This question assesses the skill of Gibbs free energy and thermodynamic favorability. A negative ΔG (-8 kJ/mol) at 298 K means the reaction is spontaneous at that temperature. ΔG = ΔH - TΔS predicts spontaneity when negative, regardless of individual ΔH or ΔS signs. This indicates the forward reaction is thermodynamically favored at 298 K. A tempting distractor is choice A, 'Nonspontaneous at 298 K,' which might arise from misreading the sign of ΔG. Always verify the sign of ΔG for spontaneity, as it integrates enthalpy, entropy, and temperature effects.
10. Hydrogen and chlorine form hydrogen chloride: $\text{H}_2(g)+\text{Cl}_2(g)\rightarrow 2\text{HCl}(g)$. For this reaction, $\Delta H<0$ and $\Delta S<0$. At room temperature (moderate temperature), which statement best describes the thermodynamic favorability?
    Answer: Spontaneous only at low temperature This question assesses the skill of Gibbs free energy and thermodynamic favorability. ΔG = ΔH - TΔS determines if a reaction is spontaneous when negative. With ΔH < 0 and ΔS < 0, at moderate room temperature, ΔG is typically negative as |ΔH| > |TΔS|, but it becomes positive at higher temperatures. Thus, the reaction is spontaneous only at low to moderate temperatures. A tempting distractor is choice D, 'Spontaneous at all temperatures,' which assumes exothermic reactions are always spontaneous, ignoring the entropy penalty at high temperatures. Evaluate spontaneity using ΔG, not just the sign of ΔH or reaction rate.
11. A student studies the dissolution process $\text{NH}_4\text{NO}_3(s) \rightarrow \text{NH}_4^+(aq) + \text{NO}_3^-(aq)$ For this process, $\Delta H > 0$ and $\Delta S > 0$. Under which temperature condition, if any, is the process thermodynamically favorable (spontaneous) based on $\Delta G = \Delta H - T\Delta S$?
    Answer: The process is thermodynamically favorable (spontaneous) only at high temperature. This question assesses the skill of Gibbs free energy and thermodynamic favorability. ΔG = ΔH - TΔS determines spontaneity if negative. For this dissolution with ΔH > 0 and ΔS > 0, at high temperatures, TΔS exceeds ΔH, making ΔG negative. At low T, it's positive and nonspontaneous. A tempting distractor is choice C, 'Spontaneous at all temperatures,' which ignores the need for high T to overcome positive ΔH. Apply the ΔG equation to find temperature thresholds for phase or dissolution processes.
12. Consider the reaction $\text{CaCO}_3(s)\rightarrow \text{CaO}(s)+\text{CO}_2(g)$ For this reaction, $\Delta H>0$ and $\Delta S>0$. Which statement best describes when the reaction is thermodynamically favorable (spontaneous) as written?
    Answer: Spontaneous only at high temperature This question assesses the skill of Gibbs free energy and thermodynamic favorability. With $\Delta H > 0$ and $\Delta S > 0$ for this decomposition, $\Delta G = \Delta H - T \Delta S$ is negative when $T \Delta S > \Delta H$, which requires high temperatures. At low temperatures, the entropy contribution is too small, resulting in positive $\Delta G$ and nonspontaneity. Therefore, the reaction is thermodynamically favorable only at high temperatures. A tempting distractor is 'Spontaneous at all temperatures,' which is incorrect because it assumes gas production always ensures spontaneity, ignoring the endothermic enthalpy's temperature dependence. Focus on the signs of $\Delta H$ and $\Delta S$ to predict temperature ranges for spontaneity via $\Delta G$.
13. A phase change is studied: $\mathrm{H_2O(l) \rightarrow H_2O(s)}$ at $1\ \mathrm{atm}$. For freezing, $\Delta H<0$ and $\Delta S<0$. Which statement best describes when freezing is thermodynamically favorable (spontaneous)?
    Answer: Spontaneous only at low temperature This question tests understanding of Gibbs free energy and thermodynamic favorability. For water freezing, we have ΔH < 0 (exothermic, heat released) and ΔS < 0 (decrease in entropy as liquid becomes ordered solid). Using ΔG = ΔH - TΔS, when both terms are negative, the -TΔS becomes positive, opposing the negative ΔH. At low temperatures (below 0°C at 1 atm), the TΔS term is small, so the negative ΔH dominates, making ΔG negative and freezing spontaneous. At high temperatures (above 0°C), the positive TΔS term overcomes the negative ΔH, making ΔG positive and freezing nonspontaneous. Choice E incorrectly suggests that catalysts affect thermodynamic favorability, but they only change reaction rates, not whether freezing is spontaneous. Remember: when ΔH < 0 and ΔS < 0, processes are spontaneous only at low temperatures.
14. A process has $\Delta H>0$ and $\Delta S<0$ under the stated conditions. Which statement best describes the thermodynamic favorability (spontaneity) of this process?
    Answer: Nonspontaneous at all temperatures This question tests understanding of Gibbs free energy and thermodynamic favorability. With ΔH > 0 (endothermic) and ΔS < 0 (decrease in entropy), we can analyze using ΔG = ΔH - TΔS. The positive ΔH contributes positively to ΔG, making the process less favorable. The negative ΔS means -TΔS is positive (since -T × negative = positive), also contributing positively to ΔG. Since both terms in the ΔG equation are positive regardless of temperature, ΔG will always be positive, making the process nonspontaneous at all temperatures. Choice E incorrectly suggests that positive ΔH could make a process spontaneous, but positive ΔH actually opposes spontaneity. Remember: when ΔH > 0 and ΔS < 0, both terms work against spontaneity, making the process nonspontaneous at all temperatures.
15. For the reaction $2\text{NO}(g) \rightarrow \text{N}_2(g) + \text{O}_2(g)$, $\Delta H < 0$ and $\Delta S > 0$. Which statement best describes the thermodynamic favorability of the reaction as written?
    Answer: Spontaneous at all temperatures. This question assesses the skill of Gibbs free energy and thermodynamic favorability. The Gibbs free energy change, ΔG, determines spontaneity, where ΔG < 0 indicates a spontaneous reaction. For this reaction with ΔH < 0 and ΔS > 0, both terms favor spontaneity: negative ΔH and negative -TΔS. Thus, ΔG is always negative regardless of temperature. A tempting distractor is choice B, which assumes a temperature dependence when both signs favor ΔG < 0. Spontaneity is governed by ΔG's sign, not by how quickly the reaction occurs.
16. Consider the decomposition reaction $2\text{HgO}(s)\rightarrow 2\text{Hg}(l)+\text{O}_2(g)$ For this reaction, $\Delta H>0$ and $\Delta S>0$. At which temperature regime is the reaction thermodynamically favorable (spontaneous)?
    Answer: Spontaneous only at high temperature This question assesses the skill of Gibbs free energy and thermodynamic favorability. For this decomposition with $\Delta H > 0$ and $\Delta S > 0$, $\Delta G = \Delta H - T\Delta S$ becomes negative at high temperatures where $T\Delta S$ outweighs $\Delta H$. At low temperatures, $\Delta G$ is positive because the entropy term is insufficient to overcome the endothermic enthalpy. Hence, the reaction is spontaneous only in the high-temperature regime. A tempting distractor is 'Spontaneous at all temperatures,' which is incorrect because it assumes positive entropy always drives spontaneity, ignoring the endothermic nature requiring high temperature. To predict favorability, always consider how temperature modulates the contributions of $\Delta H$ and $\Delta S$ to $\Delta G$.
17. A student studies the reaction $2\text{NO}(g)+\text{O}_2(g)\rightarrow 2\text{NO}_2(g)$ Thermodynamic signs for the reaction are $\Delta H<0$ and $\Delta S<0$. Under which conditions is the reaction thermodynamically favorable (spontaneous)?
    Answer: Spontaneous only at low temperature This question assesses the skill of Gibbs free energy and thermodynamic favorability. Spontaneity is governed by $\Delta G = \Delta H - T \Delta S$, and for $\Delta H < 0$ and $\Delta S < 0$, $\Delta G$ is negative when $|\Delta H| > T|\Delta S|$, which happens at low temperatures. At high temperatures, the $-T \Delta S$ term becomes more positive, making $\Delta G$ positive and the reaction nonspontaneous. Therefore, the reaction is thermodynamically favorable only at low temperatures. A tempting distractor is 'Spontaneous at all temperatures,' which is incorrect because it assumes exothermic reactions are always spontaneous, disregarding the negative entropy effect at high temperatures. Remember, spontaneity depends on the sign of $\Delta G$, not solely on the sign of $\Delta H$.
18. A reaction is measured at $350\ \mathrm{K}$ and found to have $\Delta G=-6.0\ \mathrm{kJ\,mol^{-1}}$. Which statement best describes the thermodynamic favorability of the reaction at $350\ \mathrm{K}$?
    Answer: Spontaneous at $350\ \mathrm{K}$ This question tests understanding of Gibbs free energy and thermodynamic favorability. A negative ΔG value (-6.0 kJ/mol) at 350 K directly indicates that the reaction is spontaneous under these specific conditions. When ΔG < 0, the reaction proceeds spontaneously in the forward direction as written. The negative value means the products are thermodynamically favored over the reactants at this temperature. This tells us nothing definitive about other temperatures—the reaction might be spontaneous or nonspontaneous at different temperatures depending on the signs and magnitudes of ΔH and ΔS. Choice E incorrectly focuses on ΔH alone, but spontaneity is determined by ΔG, which includes both enthalpy and entropy contributions. Remember: a negative ΔG at a given temperature means the reaction is spontaneous at that temperature, regardless of individual ΔH or ΔS values.
19. The reaction $\mathrm{2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)}$ has $\Delta H<0$ and $\Delta S<0$. Under which conditions is the reaction thermodynamically favorable (spontaneous) as written?
    Answer: Spontaneous only at low temperature This question tests understanding of Gibbs free energy and thermodynamic favorability. For SO₃ formation, we have ΔH < 0 (exothermic) and ΔS < 0 (decrease in entropy as 3 moles of gas form 2 moles). Using ΔG = ΔH - TΔS, when both terms are negative, the -TΔS becomes positive, opposing the negative ΔH. At low temperatures, the TΔS term is small, so the negative ΔH dominates, making ΔG negative and the reaction spontaneous. At high temperatures, the positive TΔS term becomes large enough to overcome the negative ΔH, making ΔG positive and the reaction nonspontaneous. Choice E incorrectly suggests that gas production determines spontaneity, but the key factor is the overall entropy change, not just the presence of gaseous products. Remember: when ΔH < 0 and ΔS < 0, reactions are spontaneous only at low temperatures.
20. A reaction has $\Delta H > 0$ and $\Delta S < 0$. At $298\ \text{K}$, which statement best describes the thermodynamic favorability (spontaneity) of the reaction as written?
    Answer: Nonspontaneous at all temperatures. This question assesses the skill of Gibbs free energy and thermodynamic favorability. The Gibbs free energy change, ΔG, determines spontaneity, where ΔG < 0 indicates a spontaneous reaction. For this reaction with ΔH > 0 and ΔS < 0, ΔG = ΔH - TΔS is always positive at any temperature. Therefore, the reaction is nonspontaneous at all temperatures, including 298 K. A tempting distractor is choice C, which assumes positive ΔH could lead to spontaneity, ignoring combined effects. Evaluate spontaneity using ΔG, not reaction pace.

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