AP Chemistry Quiz: Free Energy Of Dissolution

What this quiz covers

This quiz focuses on Free Energy Of Dissolution, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

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.

Question 1

1. Favored only at high temperature because entropy is positive
2. Not favored at any temperature because dissolving requires breaking ionic bonds
3. Favored only if the solid dissolves quickly because fast processes are spontaneous
4. Favored only at low temperature because the process is exothermic
5. 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

1. No, because dissolving always decreases entropy due to hydration ordering water molecules.
2. Yes, because the temperature drops, so the process must be spontaneous.
3. No, because an endothermic dissolution ($\Delta H>0$) is never thermodynamically favored.
4. Yes, because a faster dissolving solid is always more thermodynamically favored.
5. 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

1. Favored, because positive entropy always makes $\Delta G$ negative
2. Not favored, because the $T\Delta S$ term is too small to offset $\Delta H$ (correct answer)
3. Favored, because endothermic dissolutions require cold conditions
4. Always at equilibrium, so $\Delta G=0$ regardless of temperature
5. 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

1. Salt 1 is favored at high $T$; Salt 2 is favored at low $T$
2. Salt 1 is favored at low $T$; Salt 2 is favored at high $T$ (correct answer)
3. Both are favored at all temperatures because dissolution always increases entropy
4. Neither is favored at any temperature because one term is unfavorable in each case
5. 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

1. Correct, because $\Delta G$ becomes positive at high temperature when $T\Delta S$ grows
2. Correct, because dissolution always reaches $\Delta G=0$ immediately
3. Correct, because exothermic processes are spontaneous only at low temperature
4. Incorrect, because spontaneity depends on stirring and particle size, not $\Delta H$ and $\Delta S$
5. 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

1. C can be favored at low $T$, whereas D is not favored at any $T$ (correct answer)
2. C can be favored at high $T$, whereas D is favored at low $T$
3. Both are favored at high $T$ because $T\Delta S$ dominates
4. Both are favored at all $T$ because dissolution increases entropy
5. 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

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

1. Negative, because the favorable enthalpy term can dominate when $T$ is small (correct answer)
2. Positive, because any negative entropy makes dissolution nonspontaneous
3. Zero, because exothermic dissolutions must reach equilibrium instantly
4. Negative, because dissolution always increases entropy overall
5. 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

1. Favored only at low temperature because the $-T\Delta S$ term is smaller (correct answer)
2. Favored only at high temperature because exothermic processes proceed faster
3. Favored at all temperatures because $\Delta H<0$ guarantees spontaneity
4. Not favored at any temperature because $\Delta S<0$ prevents dissolution
5. 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

1. It is always favored because $\Delta H<0$ guarantees $\Delta G<0$.
2. It is never favored because $\Delta S<0$ guarantees $\Delta G>0$.
3. It may be favored at $25^\circ\text{C}$, but lower temperatures favor it more than higher temperatures. (correct answer)
4. It is favored only at high temperatures because $T\Delta S$ becomes large.
5. 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.

Question 11

1. At lower temperature, because endothermic processes are favored when $T$ is small.
2. At higher temperature, because the $-T\Delta S$ term becomes more negative when $\Delta S>0$. (correct answer)
3. At lower temperature, because $\Delta S>0$ matters more when $T$ is small.
4. At the same temperature, because thermodynamic favorability depends only on $\Delta H$.
5. At higher temperature, because faster dissolving at higher $T$ means $\Delta G$ must be more negative.

Explanation: 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.

Question 12

1. Not favored at any temperature because solids are more stable than aqueous ions
2. Favored only at high temperatures because $T\Delta S$ must be large
3. Favored only at low temperatures because exothermic processes stop at high $T$
4. Favored at all temperatures because $\Delta H<0$ and $\Delta S>0$ make $\Delta G<0$ (correct answer)
5. Favored only if the solution is stirred, since stirring changes $\Delta G$

Explanation: 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.

Question 13

1. Favored at all temperatures because dissolution always increases entropy
2. Not favored at any temperature because $\Delta H>0$ and $\Delta S<0$ (correct answer)
3. Favored only at low temperature because entropy is negative
4. Favored only at high temperature because endothermic processes need heat
5. Favored if stirred vigorously because that increases $\Delta S$ of the system

Explanation: This question assesses understanding of how ΔH_soln and ΔS_soln signs determine if dissolution is thermodynamically favored at any temperature. The process is endothermic (ΔH_soln > 0) and decreases entropy (ΔS_soln < 0) due to water ordering, making both ΔH and -TΔS positive in ΔG = ΔH - TΔS. Consequently, ΔG is always positive, so dissolution is not favored at any temperature, matching choice D. This occurs because neither term supports spontaneity, and increasing temperature worsens it by making -TΔS more positive. Choice C is tempting but wrong as it assumes dissolution always increases entropy, ignoring cases where solvent structuring reduces overall entropy. When predicting favorability, systematically check if ΔG can be negative by considering the interplay of ΔH, ΔS, and T.

Question 14

1. Spontaneous at all temperatures because entropy is positive
2. Spontaneous at no temperature because enthalpy is positive
3. Spontaneous at sufficiently high temperature because $\Delta G=\Delta H-T\Delta S$ (correct answer)
4. Spontaneous only at low temperature because $T\Delta S$ is minimized
5. Spontaneous only if the dissolving occurs quickly after mixing

Explanation: This question evaluates determining spontaneity conditions for positive ΔH_soln and ΔS_soln in dissolution. With ΔH > 0 and ΔS > 0, ΔG = ΔH - TΔS becomes negative at high T when TΔS > ΔH, making it spontaneous then, per choice C. At low T, ΔG > 0. Temperature modulates the entropy drive. Choice B is incorrect, claiming no spontaneity due to positive enthalpy, ignoring entropy's potential to overcome it, a misconception undervaluing TΔS. Remember for endothermic processes with positive ΔS, high temperatures enable spontaneity by enhancing the entropy contribution.

Question 15

1. Negative, because exothermic dissolutions are always spontaneous
2. Zero, because increasing temperature forces equilibrium
3. Cannot be determined, because thermodynamics depends on how fast it dissolves
4. Positive, because the $-T\Delta S$ term becomes more positive as $T$ increases (correct answer)
5. Negative, because entropy always increases when a solute dissolves

Explanation: 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.

Question 16

1. Both processes are favored because any dissolution increases entropy overall.
2. Neither process is favored because $\Delta H>0$ makes $\Delta G$ positive.
3. Process 1 may be favored, but Process 2 is not favored at any temperature. (correct answer)
4. Process 2 may be favored, but Process 1 is not favored at any temperature.
5. Both processes are favored at high temperature because dissolving is faster.

Explanation: This question tests understanding of free energy of dissolution for two different entropy scenarios with positive enthalpy. Process 1 has ΔH > 0 and ΔS > 0, giving competing terms in ΔG = ΔH - TΔS; at sufficiently high temperatures, -TΔS can overcome positive ΔH, making dissolution favorable. Process 2 has ΔH > 0 and ΔS < 0, meaning both terms contribute positively to ΔG (positive ΔH and positive -TΔS), making ΔG > 0 at all temperatures and dissolution never favorable. At 25°C, Process 1 may or may not be favored depending on the magnitudes of ΔH and TΔS, while Process 2 is definitely not favored. Choice D incorrectly reverses which process can be favorable, misunderstanding the signs in the Gibbs equation. The key is recognizing that processes with both terms opposing spontaneity (Process 2) are never favorable at any temperature.

Question 17

1. Favored at high temperature because the process absorbs heat
2. Favored at low temperature because $\Delta S$ is negative
3. Favored at all temperatures because solutions are more disordered
4. Not favored at any temperature because both terms make $\Delta G$ positive (correct answer)
5. Favored only if shaken because that increases the entropy enough to change the sign

Explanation: 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.

Question 18

1. Temperature affects only the rate of dissolving, so favorability cannot change
2. No temperature can make dissolution favored because $\Delta G=\Delta H-T\Delta S$ stays positive (correct answer)
3. Lowering temperature can make dissolution favored because it reduces the enthalpy cost
4. Either high or low temperature can make dissolution favored because $\Delta H$ changes sign
5. Raising temperature can make dissolution favored because it increases molecular motion

Explanation: 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.

Question 19

1. $\Delta G$ must be positive because solids have low entropy
2. $\Delta G$ must be negative because both terms favor spontaneity (correct answer)
3. $\Delta G$ must be zero because dissolution is an equilibrium process
4. $\Delta G$ cannot be predicted without knowing the rate of dissolving
5. $\Delta G$ must be positive because exothermic processes require activation energy

Explanation: 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.

Question 20

1. $\Delta H_{\text{soln}}<0$, because then $\Delta G$ must be positive at 298 K
2. $\Delta H_{\text{soln}}>0$, because $\Delta H$ can outweigh $T\Delta S$ at 298 K (correct answer)
3. $\Delta H_{\text{soln}}=0$, because only entropy determines spontaneity
4. Either sign, because spontaneity depends only on dissolution rate
5. $\Delta H_{\text{soln}}<0$, because exothermic dissolutions are never spontaneous

Explanation: This question assesses inferring ΔH_soln sign from ΔS_soln > 0 and non-favorability at 298 K. For ΔS > 0 but ΔG > 0 at 298 K, ΔH must be > 0 and large enough that ΔH > TΔS, consistent with choice B. This makes enthalpy the barrier at that temperature. If ΔH < 0, ΔG would be negative. Choice A is wrong, as negative ΔH with positive ΔS ensures ΔG < 0, contradicting the information, a sign mismatch misconception. When given ΔG outcome and one parameter, deduce the other using ΔG = ΔH - TΔS at specified T.