Free Energy of Dissolution - AP Chemistry

A solute dissolves with $\Delta H_{\text{soln}} < 0$ and $\Delta S_{\text{soln}} < 0$ . Which change would make the dissolution more thermodynamically favored?
Answer: Decrease the temperature, because $-T\Delta S$ becomes less positive when $\Delta S<0$ . This question tests understanding of how to optimize thermodynamic favorability when ΔH < 0 and ΔS < 0. With an exothermic dissolution (ΔH < 0) and decreased disorder (ΔS < 0), the process has a favorable enthalpy term but an unfavorable entropy term in ΔG = ΔH - TΔS. Since -TΔS is positive when ΔS < 0, lowering the temperature reduces the magnitude of this unfavorable term, making ΔG more negative and dissolution more favored. At lower T, the favorable ΔH term dominates over the smaller unfavorable -TΔS term. Choice A incorrectly suggests increasing temperature, which would make the positive -TΔS term larger and ΔG less negative. The strategy for optimizing processes with opposing enthalpy and entropy is to adjust temperature to minimize the unfavorable contribution.
For a particular dissolution at $25^\circ\text{C}$ , the solution warms ( $\Delta H_{\text{soln}} < 0$ ) and the dissolved particles become more dispersed ( $\Delta S_{\text{soln}} > 0$ ). Which statement about $\Delta G_{\text{soln}}$ is most consistent with these observations?
Answer: $\Delta G_{\text{soln}}$ is negative because both terms in $\Delta G=\Delta H-T\Delta S$ favor spontaneity. This question tests understanding of free energy of dissolution when both thermodynamic factors favor the process. With the solution warming (ΔH < 0, exothermic) and particles dispersing (ΔS > 0), both terms in ΔG = ΔH - TΔS contribute negatively to the free energy change. The negative ΔH directly makes ΔG more negative, while the positive ΔS makes -TΔS negative, also contributing to a negative ΔG. When both enthalpy and entropy favor a process, ΔG must be negative, indicating the dissolution is thermodynamically favorable at 25°C. Choice B incorrectly interprets warming as requiring energy input, confusing the direction of heat flow in exothermic processes. The fundamental principle is that processes releasing heat (ΔH < 0) and increasing disorder (ΔS > 0) are always thermodynamically favorable.
For dissolving a solid in water, a student determines $\Delta H_{\text{soln}}>0$ and $\Delta S_{\text{soln}}<0$ . Which statement best describes the thermodynamic favorability?
Answer: Not favored at any temperature because both terms make $\Delta G$ positive This question evaluates determining overall thermodynamic favorability from ΔH_soln and ΔS_soln signs. With ΔH > 0 and ΔS < 0, both terms are positive in ΔG = ΔH - TΔS (since -TΔS > 0), so ΔG > 0 always, meaning not favored at any temperature, per choice D. Temperature changes only worsen it, as -TΔS increases with T. No conditions make it spontaneous. Choice C is tempting but false, claiming solutions always more disordered, overlooking solvent effects that can decrease entropy, a common overgeneralization. Always check if both terms oppose spontaneity, as that precludes favorability at any T.
A salt dissolves in water: $\text{QX}(s) \rightarrow \text{Q}^+(aq)+\text{X}^-(aq)$ . Breaking the crystal lattice requires energy, so $\Delta H_{\text{soln}}>0$ , but the number of dispersed particles increases, so $\Delta S_{\text{soln}}>0$ . Under which conditions is dissolution thermodynamically favored?
Answer: Favored at high temperature because $T\Delta S$ can outweigh $\Delta H$ This question evaluates the skill of predicting temperature-dependent spontaneity for dissolution based on ΔH_soln and ΔS_soln signs. Here, ΔH_soln > 0 from breaking the lattice, but ΔS_soln > 0 from particle dispersion, so ΔG = ΔH - TΔS can be negative when TΔS exceeds ΔH. Therefore, dissolution is favored at high temperatures where the entropy term dominates, as in choice B. At low temperatures, ΔH prevails, making ΔG positive. Choice D is incorrect because it wrongly claims endothermic dissolutions are always nonspontaneous, missing that positive ΔS can drive them at high T, a misconception confusing enthalpy with overall free energy. Always use the ΔG equation to assess how temperature modulates the balance between enthalpy and entropy terms.
A salt $GH(s)$ dissolves in water: $GH(s) \rightarrow G^+(aq)+H^-(aq)$ . The process has $\Delta H_{\text{soln}} < 0$ and $\Delta S_{\text{soln}} > 0$ . A student argues the dissolution is thermodynamically favored because it happens quickly when the solution is stirred. Which statement best evaluates the student's claim?
Answer: Incorrect, because rate does not determine spontaneity; $\Delta H<0$ and $\Delta S>0$ indicate it is favored This question tests the distinction between kinetics (rate) and thermodynamics (spontaneity) in dissolution. The salt has ΔH < 0 and ΔS > 0, which means ΔG = ΔH - TΔS is negative at all temperatures (both terms favor spontaneity). The student incorrectly links stirring rate to thermodynamic favorability, but stirring only affects how quickly equilibrium is reached, not whether the process is spontaneous. Choice D incorrectly claims ionic dissolutions must have negative entropy due to hydration shells, but the given positive ΔS shows that ion dispersion can dominate over ordering effects. The key principle is that thermodynamic spontaneity (ΔG < 0) is independent of kinetic factors like stirring or dissolution rate.
A solute dissolves with $\Delta H_{\text{soln}}>0$ and $\Delta S_{\text{soln}}<0$ . Which statement about changing temperature is accurate?
Answer: No temperature can make dissolution favored because $\Delta G=\Delta H-T\Delta S$ stays positive This question tests understanding temperature's impact on favorability for ΔH > 0 and ΔS < 0 in dissolution. Both terms make ΔG positive: ΔH > 0 and -TΔS > 0, so ΔG > 0 always, meaning no temperature favors it, as in choice C. Increasing T makes -TΔS more positive, worsening it. No change can help. Choice A is incorrect, claiming high T favors it by absorbing heat, confusing Le Chatelier with free energy signs, a principle misapplication. For cases where both terms oppose, conclude nonspontaneity at all T without exceptions.
A student measures thermodynamic signs for dissolving a solid in water and finds $\Delta H_{\text{soln}}<0$ and $\Delta S_{\text{soln}}>0$ . Which conclusion about $\Delta G$ for dissolution is most appropriate at 298 K?
Answer: $\Delta G$ must be negative because both terms favor spontaneity This question evaluates inferring the sign of ΔG for dissolution from given ΔH_soln and ΔS_soln at standard temperature. With ΔH_soln < 0 and ΔS_soln > 0, both terms make ΔG negative via ΔG = ΔH - TΔS, as -TΔS is negative. At 298 K, this ensures ΔG < 0, indicating spontaneity, as in choice B. The conclusion holds because no temperature can make ΔG positive in this case. Choice A is incorrect, claiming ΔG positive due to solids' low entropy, confusing system entropy change with initial state entropy, a common mix-up. When signs align for spontaneity, confirm by noting both contribute negatively to ΔG regardless of T.
A salt $QX(s)$ dissolves in water: $QX(s) \rightarrow Q^+(aq)+X^-(aq)$ . Calorimetry shows $\Delta H_{\text{soln}} < 0$ , and analysis of the system indicates $\Delta S_{\text{soln}} > 0$ . Which statement best describes whether dissolution is thermodynamically favored?
Answer: Favored at all temperatures, because $\Delta H<0$ and $\Delta S>0$ make $\Delta G<0$ This question tests recognition of thermodynamically favorable conditions when both driving forces favor dissolution. The salt dissolution has ΔH < 0 (exothermic, favorable) and ΔS > 0 (increased disorder, favorable). Applying ΔG = ΔH - TΔS, we see that ΔH contributes a negative value and -TΔS also contributes a negative value (since ΔS > 0). Therefore, ΔG is negative at all temperatures, making the dissolution spontaneous regardless of temperature. Choice E incorrectly confuses kinetics (stirring affects rate) with thermodynamics (spontaneity depends on ΔG, not stirring). The key insight is that when both enthalpy and entropy changes favor the process (ΔH < 0 and ΔS > 0), the dissolution is thermodynamically favored at all temperatures.
When solid $\text{Ca(OH)}_2$ is added to water at $25^\circ\text{C}$ , the container warms (so $\Delta H_{\text{soln}} < 0$ ), but the solution becomes more structured due to strong ion–dipole interactions (so $\Delta S_{\text{soln}} < 0$ ). Based on $\Delta G=\Delta H-T\Delta S$ , is dissolution thermodynamically favored at $25^\circ\text{C}$ ?
Answer: It depends; with $\Delta H<0$ and $\Delta S<0$ , dissolution is favored at lower temperatures and may not be favored at higher temperatures. This question tests understanding of free energy of dissolution when both ΔH and ΔS are negative. For Ca(OH)₂ dissolution, the solution warms (ΔH < 0, exothermic) and becomes more ordered (ΔS < 0, decreased disorder). Using ΔG = ΔH - TΔS, with ΔH < 0 and ΔS < 0, we have a favorable enthalpy term and an unfavorable entropy term (-TΔS becomes positive). At low temperatures, the ΔH term dominates and ΔG < 0 (favored); at high temperatures, the -TΔS term may dominate and ΔG > 0 (not favored). Choice C incorrectly states that both negative terms guarantee negative ΔG, failing to recognize that -TΔS is positive when ΔS < 0. The strategy is to identify temperature-dependent favorability: when ΔH and ΔS have the same sign, temperature determines which term dominates.
A molecular solid dissolves in water with $\Delta H_{\text{soln}} > 0$ (the solution cools) and $\Delta S_{\text{soln}} < 0$ (water becomes more ordered around the solute). Under these conditions, is dissolution thermodynamically favored at any temperature?
Answer: No, because with $\Delta H>0$ and $\Delta S<0$ , $\Delta G=\Delta H-T\Delta S$ is positive at all temperatures. This question tests understanding of free energy of dissolution when ΔH > 0 and ΔS < 0. With the solution cooling (ΔH > 0, endothermic) and water becoming more ordered (ΔS < 0), both terms in ΔG = ΔH - TΔS are unfavorable for dissolution. The positive ΔH contributes positively to ΔG, and the negative ΔS makes -TΔS positive, also contributing positively to ΔG. Since both terms are positive, ΔG > 0 at all temperatures, meaning dissolution is never thermodynamically favored. Choice C incorrectly suggests high temperatures could favor dissolution, misunderstanding that increasing T makes the already positive -TΔS term even more positive. The key insight is recognizing when both enthalpy and entropy oppose a process, making it unfavorable at all temperatures.
A nonelectrolyte dissolves in water: $\text{B}(l) \rightarrow \text{B}(aq)$ . The dissolution releases heat ( $\Delta H_{\text{soln}}<0$ ), but strong solvent ordering around B decreases entropy ( $\Delta S_{\text{soln}}<0$ ). When is dissolution thermodynamically favored?
Answer: Favored only at low temperature because the $-T\Delta S$ term is smaller This question tests determining the temperature range for favored dissolution given ΔH_soln and ΔS_soln signs. The process is exothermic (ΔH_soln < 0) but decreases entropy (ΔS_soln < 0) due to solvent ordering, so ΔG = ΔH - TΔS where -TΔS is positive. At low temperatures, the small magnitude of -TΔS allows the negative ΔH to make ΔG negative, favoring dissolution, per choice A. At high temperatures, -TΔS becomes large positive, potentially making ΔG positive. Choice C is misleading as it assumes ΔH < 0 ensures spontaneity, ignoring that negative ΔS can outweigh it at high T, a common error in overlooking temperature's role. For such problems, calculate the temperature where ΔG = 0 to define low versus high T boundaries.
A student observes that dissolving a solid in water is exothermic ( $\Delta H_{\text{soln}} < 0$ ). Additional evidence suggests that the solvent becomes more ordered around the solute ( $\Delta S_{\text{soln}} < 0$ ). At $25^\circ\text{C}$ , which statement best describes whether dissolution is thermodynamically favored?
Answer: It may be favored at $25^\circ\text{C}$ , but lower temperatures favor it more than higher temperatures. This question tests understanding of temperature-dependent thermodynamic favorability when ΔH < 0 and ΔS < 0. With an exothermic dissolution (ΔH < 0, favorable) and increased ordering (ΔS < 0, unfavorable), the sign of ΔG = ΔH - TΔS depends on which term dominates. At low temperatures, the favorable ΔH term dominates over the smaller unfavorable -TΔS term, making ΔG < 0 and dissolution favored. As temperature increases, the positive -TΔS term grows larger, eventually making ΔG > 0 and dissolution unfavored. At 25°C, dissolution may be favored if |ΔH| > |TΔS|, and lower temperatures would favor it even more. Choice A incorrectly assumes negative ΔH always guarantees favorable dissolution, ignoring the entropy contribution. The strategy is to recognize that exothermic processes with negative entropy changes are favored at low temperatures where enthalpy dominates.
Dissolving solute E in water has $\Delta H_{\text{soln}}<0$ and $\Delta S_{\text{soln}}<0$ . Which temperature change would make dissolution less thermodynamically favored?
Answer: Increase temperature, because the unfavorable entropy term is amplified This question tests identifying temperature changes that reduce favorability for dissolution with ΔH < 0 and ΔS < 0. Increasing temperature amplifies the unfavorable -TΔS (positive), potentially making ΔG positive when it outweighs negative ΔH, per choice A. This makes dissolution less favored. Lower T would enhance favorability. Choice E is incorrect, stating temperature only affects rate, confusing kinetics with thermodynamics, as ΔG explicitly depends on T. To assess temperature effects, consider how T scales the entropy term in ΔG.
Dissolving a solid in water is observed to be endothermic ( $\Delta H_{\text{soln}} > 0$ ) and to increase disorder overall ( $\Delta S_{\text{soln}} > 0$ ). Compared with $25^\circ\text{C}$ , at which condition is dissolution more thermodynamically favored?
Answer: At higher temperature, because the $-T\Delta S$ term becomes more negative when $\Delta S>0$ . This question tests understanding of temperature effects on free energy of dissolution when ΔH > 0 and ΔS > 0. For an endothermic dissolution (ΔH > 0) with increased disorder (ΔS > 0), the ΔG = ΔH - TΔS equation shows competing terms: positive ΔH opposes dissolution while positive ΔS favors it through the -TΔS term. At higher temperatures, the magnitude of -TΔS increases (becomes more negative), making ΔG more likely to be negative and dissolution more favored. At lower temperatures, the positive ΔH term dominates, making dissolution less favored. Choice A incorrectly suggests endothermic processes are favored at low T, confusing thermodynamic favorability with kinetic effects. The key principle is that entropy-driven processes become more favorable as temperature increases.
A student dissolves solid $\text{NH}_4\text{NO}_3$ in water at $25^\circ\text{C}$ . The solution becomes noticeably colder (so $\Delta H_{\text{soln}} > 0$ ), and the ions disperse throughout the solvent (so $\Delta S_{\text{soln}} > 0$ ). Under these conditions, is the dissolution thermodynamically favored?
Answer: Yes, because $\Delta G=\Delta H-T\Delta S$ is likely negative when both $\Delta H$ and $\Delta S$ are positive at $25^\circ\text{C}$ . This question tests understanding of free energy of dissolution and how to apply the Gibbs equation to determine thermodynamic favorability. When NH₄NO₃ dissolves, the solution cools (ΔH > 0, endothermic) and ions disperse (ΔS > 0, increased disorder). Using ΔG = ΔH - TΔS, with both positive ΔH and positive ΔS, the sign of ΔG depends on the relative magnitudes of ΔH and TΔS. At 25°C (298 K), if TΔS > ΔH, then ΔG < 0 and dissolution is thermodynamically favored, which is typically the case for NH₄NO₃. Choice C incorrectly assumes endothermic processes are never spontaneous, ignoring the entropy contribution to free energy. The key strategy is to evaluate both enthalpy and entropy contributions to ΔG, remembering that positive entropy changes favor spontaneity at higher temperatures.
A salt dissolves in water and releases heat ( $\Delta H_{\text{soln}} < 0$ ). The ions disperse without significant ordering of water, so $\Delta S_{\text{soln}} > 0$ . Under which condition is the dissolution thermodynamically favored?
Answer: Favored at all temperatures because $\Delta H<0$ and $\Delta S>0$ make $\Delta G<0$ This question tests understanding of thermodynamically favorable dissolution conditions. For a dissolution with ΔH < 0 (exothermic) and ΔS > 0 (increased disorder), we apply ΔG = ΔH - TΔS. The negative ΔH contributes a negative value to ΔG (favorable), and since ΔS is positive, -TΔS is also negative (favorable). Both terms make negative contributions to ΔG, ensuring ΔG < 0 at all temperatures, so the dissolution is always thermodynamically favored. Students often incorrectly choose option C, thinking that exothermic processes somehow stop at high temperatures, confusing thermodynamics with kinetics or equilibrium position. The strategy is to recognize that when ΔH < 0 and ΔS > 0, both terms in the Gibbs equation favor spontaneity regardless of temperature.
A solute dissolves in water with $\Delta H_{\text{soln}}<0$ and $\Delta S_{\text{soln}}<0$ . At sufficiently high temperature, what is the sign of $\Delta G$ most likely to be for dissolution?
Answer: Positive, because the $-T\Delta S$ term becomes more positive as $T$ increases This question assesses predicting the sign of ΔG at high temperatures given ΔH_soln and ΔS_soln signs. With ΔH < 0 and ΔS < 0, at high T, the -TΔS term (positive since ΔS < 0) becomes large, likely making ΔG = ΔH + (-TΔS, large positive) positive, as in choice B. This occurs because entropy's unfavorable effect grows with T. At low T, ΔG is negative. Choice A is incorrect, assuming exothermic always spontaneous, ignoring negative ΔS's temperature-dependent impact, a misconception about enthalpy dominance. For processes with opposing signs, evaluate ΔG at extreme temperatures using the equation's behavior.
A student claims: "If a solute dissolves, then $\Delta S_{\text{soln}}$ must be positive." Which scenario provides a counterexample while still allowing dissolution to be thermodynamically favored under some conditions?
Answer: $\Delta H_{\text{soln}}<0$ and $\Delta S_{\text{soln}}<0$ , favored at low temperature This question assesses providing a counterexample to a claim about ΔS_soln in dissolution using thermodynamic principles. The claim assumes ΔS > 0 for dissolution, but ΔH < 0 and ΔS < 0 allows favorability at low T, where ΔH dominates in ΔG, countering it as in choice A. This shows entropy can decrease yet spontaneity occur if enthalpy compensates. High T would make it nonspontaneous. Choice B is wrong as that combination never favors dissolution, not providing a counterexample, misconstruing conditions for spontaneity. Seek scenarios where negative ΔS is offset by negative ΔH at appropriate temperatures to challenge entropy assumptions.
A dissolution in water at $25^\circ\text{C}$ is observed to cool the solution ( $\Delta H_{\text{soln}} > 0$ ). The student also concludes that the overall entropy change is positive ( $\Delta S_{\text{soln}} > 0$ ). Which statement is correct about thermodynamic favorability at $25^\circ\text{C}$ ?
Answer: It may be favored at $25^\circ\text{C}$ , because the positive $\Delta S$ can make $\Delta G$ negative even when $\Delta H$ is positive. This question tests understanding of free energy of dissolution when ΔH > 0 and ΔS > 0. With the solution cooling (ΔH > 0, endothermic) and overall entropy increasing (ΔS > 0), the ΔG = ΔH - TΔS equation contains competing terms: positive ΔH opposes dissolution while positive ΔS favors it through the negative -TΔS term. At 25°C (298 K), if the magnitude of TΔS exceeds ΔH, then ΔG < 0 and dissolution is thermodynamically favored, which is often the case for salts like NH₄NO₃. Choice A incorrectly claims endothermic processes cannot be spontaneous, ignoring that entropy can drive spontaneity. The key principle is that endothermic processes can be thermodynamically favorable when the entropy increase is sufficient to overcome the unfavorable enthalpy change.
A solute dissolves in water with $\Delta H_{\text{soln}}>0$ and $\Delta S_{\text{soln}}>0$ . At very low temperature, which is most likely true about thermodynamic favorability?
Answer: Not favored, because the $T\Delta S$ term is too small to offset $\Delta H$ This question evaluates predicting favorability at low temperatures for given ΔH_soln and ΔS_soln. With ΔH > 0 and ΔS > 0, at very low T, TΔS is small, so ΔG ≈ ΔH > 0, making it not favored, as in choice B. The entropy term needs higher T to offset enthalpy. Low T prevents this. Choice A is wrong, claiming positive entropy always makes ΔG negative, ignoring enthalpy's role at low T, a common entropy overemphasis. For entropy-driven processes, recognize low T limits TΔS, potentially keeping ΔG positive.

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