Free Energy of Dissolution
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AP Chemistry › Free Energy of Dissolution
A molecular solid dissolves in water: $\text{A}(s) \rightarrow \text{A}(aq)$. The dissolution is observed to cool the solution, so $\Delta H_{\text{soln}}>0$, and the ordering of water around A causes $\Delta S_{\text{soln}}<0$. At which temperatures, if any, is dissolution thermodynamically favored?
Favored only at low temperature because entropy is negative
Not favored at any temperature because $\Delta H>0$ and $\Delta S<0$
Favored only at high temperature because endothermic processes need heat
Favored if stirred vigorously because that increases $\Delta S$ of the system
Favored at all temperatures because dissolution always increases entropy
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.
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., $\Delta G<0$)?
Favored only if the solid dissolves quickly because fast processes are spontaneous
Favored only at low temperature because the process is exothermic
Not favored at any temperature because dissolving requires breaking ionic bonds
Favored only at high temperature because entropy is positive
Favored at all temperatures because $\Delta H<0$ and $\Delta S>0$
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.
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 a faster dissolving solid is always more thermodynamically favored.
No, because an endothermic dissolution ($\Delta H>0$) is never thermodynamically favored.
Yes, because the temperature drops, so the process must be spontaneous.
Yes, because $\Delta G=\Delta H-T\Delta S$ is likely negative when both $\Delta H$ and $\Delta S$ are positive at $25^\circ\text{C}$.
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.
For dissolving solute F in water, $\Delta H_{\text{soln}}$ is positive and $\Delta S_{\text{soln}}$ is positive. Which statement about spontaneity is correct?
Spontaneous only if the dissolving occurs quickly after mixing
Spontaneous only at low temperature because $T\Delta S$ is minimized
Spontaneous at sufficiently high temperature because $\Delta G=\Delta H-T\Delta S$
Spontaneous at all temperatures because entropy is positive
Spontaneous at no temperature because enthalpy is positive
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.
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 endothermic dissolutions require cold conditions
Always at equilibrium, so $\Delta G=0$ regardless of temperature
Not favored, because the $T\Delta S$ term is too small to offset $\Delta H$
Favored only if the solute is ground up to increase surface area
Favored, because positive entropy always makes $\Delta G$ negative
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.
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?
Negative, because exothermic dissolutions are always spontaneous
Zero, because increasing temperature forces equilibrium
Cannot be determined, because thermodynamics depends on how fast it dissolves
Positive, because the $-T\Delta S$ term becomes more positive as $T$ increases
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.
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 $T$
Both are favored only if stirred because stirring makes $\Delta H$ negative
Neither is favored at any temperature because one term is unfavorable in each case
Salt 1 is favored at low $T$; Salt 2 is favored at high $T$
Both are favored at all temperatures because dissolution always increases entropy
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.
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?
Favored at high temperature because the process absorbs heat
Favored at all temperatures because solutions are more disordered
Favored at low temperature because $\Delta S$ is negative
Favored only if shaken because that increases the entropy enough to change the sign
Not favored at any temperature because both terms make $\Delta G$ positive
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.
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?
Incorrect, because spontaneity depends on stirring and particle size, not $\Delta H$ and $\Delta S$
Correct, because $\Delta G$ becomes positive at high temperature when $T\Delta S$ grows
Correct, because exothermic processes are spontaneous only at low temperature
Correct, because dissolution always reaches $\Delta G=0$ immediately
Incorrect, because $\Delta G$ is negative at all temperatures when $\Delta H<0$ and $\Delta S>0$
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.
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 $T$
Both are favored at all $T$ because dissolution increases entropy
C can be favored at high $T$, whereas D is favored at low $T$
Neither can be favored unless the solids dissolve rapidly
Both are favored at high $T$ because $T\Delta S$ dominates
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.