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AP Chemistry · Learn by Concept

AP Chemistry Help: Free Energy Of Dissolution

Review real example questions for Free Energy Of Dissolution in AP Chemistry.

Question 1 / 10

0 of 10 answered

A salt dissolves in water according to: $\text{MX}(s) \rightarrow \text{M}^+(aq)+\text{X}^-(aq)$ . For this dissolution, $\Delta H_{\text{soln}}<0$ and $\Delta S_{\text{soln}}>0$ . Under which conditions is the dissolution thermodynamically favored (i.e., $\Delta G<0$ )?

All questions

Question 1

Favored only at high temperature because entropy is positive
Not favored at any temperature because dissolving requires breaking ionic bonds
Favored only if the solid dissolves quickly because fast processes are spontaneous
Favored only at low temperature because the process is exothermic
Favored at all temperatures because $\Delta H<0$ and $\Delta S>0$ (correct answer)

Explanation: This question tests the ability to determine the temperature conditions for thermodynamic favorability of dissolution using the signs of ΔH_soln and ΔS_soln. The dissolution is exothermic (ΔH_soln < 0) and increases entropy (ΔS_soln > 0), so both terms contribute to a negative ΔG via ΔG = ΔH - TΔS. Since ΔH is negative and -TΔS is also negative (because ΔS > 0), ΔG remains negative regardless of temperature. Thus, the process is favored at all temperatures, as stated in choice C. Choice D is incorrect because it assumes breaking ionic bonds always prevents dissolution, overlooking that hydration energy can compensate and make ΔH negative overall, a common misconception about lattice energy dominating. To analyze dissolution spontaneity, evaluate how the signs of ΔH and ΔS influence ΔG across temperature ranges.

Question 2

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?

No, because dissolving always decreases entropy due to hydration ordering water molecules.
Yes, because the temperature drops, so the process must be spontaneous.
No, because an endothermic dissolution ( $\Delta H>0$ ) is never thermodynamically favored.
Yes, because a faster dissolving solid is always more thermodynamically favored.
Yes, because $\Delta G=\Delta H-T\Delta S$ is likely negative when both $\Delta H$ and $\Delta S$ are positive at $25^\circ\text{C}$ . (correct answer)

Explanation: 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 B 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.

Question 3

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?

Favored, because positive entropy always makes $\Delta G$ negative
Not favored, because the $T\Delta S$ term is too small to offset $\Delta H$ (correct answer)
Favored, because endothermic dissolutions require cold conditions
Always at equilibrium, so $\Delta G=0$ regardless of temperature
Favored only if the solute is ground up to increase surface area

Explanation: 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.

Question 4

Two salts dissolve in separate beakers of water at the same pressure. Salt 1 has $\Delta H_{\text{soln}}<0$ and $\Delta S_{\text{soln}}<0$ . Salt 2 has $\Delta H_{\text{soln}}>0$ and $\Delta S_{\text{soln}}>0$ . Which statement correctly compares when each dissolution is thermodynamically favored?

Salt 1 is favored at high $T$ ; Salt 2 is favored at low $T$
Salt 1 is favored at low $T$ ; Salt 2 is favored at high $T$ (correct answer)
Both are favored at all temperatures because dissolution always increases entropy
Neither is favored at any temperature because one term is unfavorable in each case
Both are favored only if stirred because stirring makes $\Delta H$ negative

Explanation: This question assesses comparing thermodynamic favorability conditions for two dissolutions with different ΔH_soln and ΔS_soln signs. Salt 1 has ΔH < 0 and ΔS < 0, favored at low T where -TΔS is small, allowing ΔH to dominate in ΔG. Salt 2 has ΔH > 0 and ΔS > 0, favored at high T where TΔS overcomes ΔH. Thus, choice B correctly states Salt 1 at low T and Salt 2 at high T. Choice C errs by claiming both favored at all T, assuming entropy always increases, which ignores specific signs and temperature effects, a misconception about universal dissolution behavior. To compare processes, classify them by ΔH and ΔS signs and recall standard temperature dependencies for each combination.

Question 5

Dissolving solute H has $\Delta H_{\text{soln}}<0$ and $\Delta S_{\text{soln}}>0$ . A student argues the dissolution might still be nonspontaneous at some temperatures. Which evaluation is correct?

Correct, because $\Delta G$ becomes positive at high temperature when $T\Delta S$ grows
Correct, because dissolution always reaches $\Delta G=0$ immediately
Correct, because exothermic processes are spontaneous only at low temperature
Incorrect, because spontaneity depends on stirring and particle size, not $\Delta H$ and $\Delta S$
Incorrect, because $\Delta G$ is negative at all temperatures when $\Delta H<0$ and $\Delta S>0$ (correct answer)

Explanation: This question evaluates critiquing a claim about temperature effects on spontaneity for ΔH < 0 and ΔS > 0. The student's argument is incorrect because this combination makes ΔG < 0 at all T, as both terms are negative, per choice C. No temperature renders it nonspontaneous. The claim overlooks perpetual favorability. Choice A is misleading, suggesting ΔG positive at high T, but -TΔS becomes more negative, enhancing favorability, a calculation error misconception. To evaluate such claims, plug signs into ΔG and check if positivity is possible across T.

Question 6

A student compares dissolving two different solids in water. Solid C has $\Delta H_{\text{soln}}<0$ , $\Delta S_{\text{soln}}<0$ . Solid D has $\Delta H_{\text{soln}}>0$ , $\Delta S_{\text{soln}}<0$ . Which statement is correct?

C can be favored at low $T$ , whereas D is not favored at any $T$ (correct answer)
C can be favored at high $T$ , whereas D is favored at low $T$
Both are favored at high $T$ because $T\Delta S$ dominates
Both are favored at all $T$ because dissolution increases entropy
Neither can be favored unless the solids dissolve rapidly

Explanation: This question assesses comparing favorability for two solids with different ΔH_soln and ΔS_soln signs. Solid C (ΔH < 0, ΔS < 0) can be favored at low T where -TΔS is small, making ΔG negative. Solid D (ΔH > 0, ΔS < 0) has ΔG always positive, not favored at any T. Thus, choice A correctly distinguishes them. Choice B reverses the conditions, mistakenly swapping temperature dependencies, a misconception from confusing sign impacts on ΔG. Classify each case by ΔH and ΔS, then apply standard rules for when ΔG < 0.

Question 7

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?

$\Delta G_{\text{soln}}$ is negative because both terms in $\Delta G=\Delta H-T\Delta S$ favor spontaneity. (correct answer)
$\Delta G_{\text{soln}}$ is positive because warming implies energy is required to dissolve.
$\Delta G_{\text{soln}}$ must be zero because dissolution is an equilibrium process.
$\Delta G_{\text{soln}}$ cannot be predicted without the dissolution rate.
$\Delta G_{\text{soln}}$ is positive because dissolving always decreases entropy of the system.

Explanation: 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.

Question 8

A solute dissolves in water and releases heat ( $\Delta H_{\text{soln}}<0$ ). The dissolution also decreases entropy ( $\Delta S_{\text{soln}}<0$ ). Which best describes the sign of $\Delta G$ at low temperature?

Negative, because the favorable enthalpy term can dominate when $T$ is small (correct answer)
Positive, because any negative entropy makes dissolution nonspontaneous
Zero, because exothermic dissolutions must reach equilibrium instantly
Negative, because dissolution always increases entropy overall
Cannot be predicted without knowing how fast the solid dissolves

Explanation: This question tests predicting ΔG sign at low temperature for exothermic dissolution with negative ΔS_soln. ΔH < 0 and ΔS < 0 mean at low T, -TΔS (positive) is small, so favorable ΔH dominates, making ΔG negative, as in choice A. This favors spontaneity. High T could reverse it. Choice B errs by saying negative entropy always prevents spontaneity, overlooking enthalpy's role at low T, a common overstatement of entropy's importance. For opposing signs, focus on low T favoring enthalpy-driven processes in ΔG calculations.

Question 9

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?

Favored only at low temperature because the $-T\Delta S$ term is smaller (correct answer)
Favored only at high temperature because exothermic processes proceed faster
Favored at all temperatures because $\Delta H<0$ guarantees spontaneity
Not favored at any temperature because $\Delta S<0$ prevents dissolution
Favored only if the solute is finely powdered because that increases $\Delta G$

Explanation: 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.

Question 10

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?

It is always favored because $\Delta H<0$ guarantees $\Delta G<0$ .
It is never favored because $\Delta S<0$ guarantees $\Delta G>0$ .
It may be favored at $25^\circ\text{C}$ , but lower temperatures favor it more than higher temperatures. (correct answer)
It is favored only at high temperatures because $T\Delta S$ becomes large.
It is favored if the solid dissolves quickly, and not favored if it dissolves slowly.

Explanation: 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.