Le Châtelier's Principle (Thermodynamic) - Physical Chemistry 1

1. Consider the equilibrium $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ with $\Delta H^\circ = +26.5 \text{ kJ/mol}$ at 731 K where $K_p = 46.0$. A chemist observes that when the equilibrium mixture is rapidly compressed to half its volume at constant temperature, the initial rate of the forward reaction increases by a factor of 8, while the reverse reaction rate increases by a factor of 4. Which thermodynamic principle explains the subsequent equilibrium shift?
   Answer: No net shift occurs because this reaction has Δn = 0, so compression effects on forward and reverse rates exactly cancel, maintaining thermodynamic equilibrium For this reaction, Δn = (2) - (1 + 1) = 0. When volume is halved, all partial pressures double, but Q = [P(HI)]²/[P(H₂)P(I₂)] remains unchanged because the pressure changes cancel in the expression. Since Q still equals Kp, no thermodynamic driving force exists for equilibrium shift. Choice A confuses kinetic rates with thermodynamic equilibrium. Choice B incorrectly calculates the Q change. Choice D misapplies energy considerations.
2. A gas mixture initially contains 0.50 mol H₂, 0.50 mol I₂, and 1.0 mol HI in a 2.0 L container at 700 K, where $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ with $K_c = 54.3$. The container volume is suddenly expanded to 8.0 L at constant temperature. Which thermodynamic analysis best explains the subsequent equilibrium shift?
   Answer: The system is initially not at equilibrium since $Q_c = \frac{(0.50)^2}{(0.25)(0.25)} = 4.0 \neq K_c$, and expansion to 8.0 L gives $Q_c = 1.0$, requiring forward shift When analyzing equilibrium shifts after volume changes, you must first determine if the system was initially at equilibrium by comparing the reaction quotient $Q_c$ to the equilibrium constant $K_c$. Let's calculate the initial $Q_c$. With 0.50 mol H₂, 0.50 mol I₂, and 1.0 mol HI in 2.0 L, the concentrations are [H₂] = [I₂] = 0.25 M and [HI] = 0.50 M. Therefore: $Q_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.50)^2}{(0.25)(0.25)} = 4.0$ Since $Q_c = 4.0 \neq K_c = 54.3$, the system isn't initially at equilibrium. After expansion to 8.0 L, all concentrations are quartered: [H₂] = [I₂] = 0.125 M and [HI] = 0.125 M, giving $Q_c = \frac{(0.125)^2}{(0.125)(0.125)} = 1.0$. Since $Q_c = 1.0 < K_c = 54.3$, the reaction shifts right to reach equilibrium. Option A incorrectly assumes the system was initially at equilibrium. While $\Delta n = 0$ means volume changes don't affect equilibrium position once established, this principle doesn't apply when the system starts away from equilibrium. Option B correctly calculates $Q_c$ after expansion but wrongly concludes a leftward shift—when $Q_c < K_c$, the reaction proceeds forward. Option C misapplies Le Châtelier's principle; with $\Delta n = 0$, particle count doesn't drive the shift. Always check if a system is initially at equilibrium before analyzing shifts. Calculate $Q_c$ at each state and compare to $K_c$ to predict the direction of change.
3. For the equilibrium $\text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g)$, $\Delta H^\circ = +108 \text{ kJ/mol}$ and $\Delta S^\circ = +179 \text{ J/mol}\cdot\text{K}$. At 400 K, $\Delta G^\circ = +36.4 \text{ kJ/mol}$. A student calculates that this equilibrium becomes thermodynamically favorable ($\Delta G^\circ < 0$) above 603 K. If the system at 500 K is heated to 700 K while maintaining constant pressure, which thermodynamic analysis correctly predicts the equilibrium response?
   Answer: Major shift right because ΔG° becomes negative (-17.2 kJ/mol), making products thermodynamically favored, and the entropy increase amplifies this effect at higher temperature At 700 K: ΔG° = 108,000 - 700(179) = 108,000 - 125,300 = -17,300 J/mol. This large negative value means K >> 1, representing a major thermodynamic shift from K < 1 at 500 K to K >> 1 at 700 K. The entropy term dominates at high temperature, making the reaction highly favorable. Choice B understates the magnitude of change. Choice C gives correct direction but wrong reasoning. Choice D incorrectly considers gas expansion effects on equilibrium.
4. Consider the equilibrium $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ at 1000 K with $K_p = 3.87$ atm. The system is initially at equilibrium with $P_{\text{CO}_2} = 3.87$ atm. If the volume is suddenly halved at constant temperature, which thermodynamic principle best explains the direction and extent of the equilibrium shift?
   Answer: The reaction shifts left because $\Delta G = RT \ln(Q/K) > 0$ when the pressure doubles, and continues until $P_{\text{CO}_2}$ returns to 3.87 atm When analyzing equilibrium shifts involving pressure changes, you need to consider both the direction of shift and the final equilibrium position. The key insight is understanding how the reaction quotient Q relates to the equilibrium constant K. Initially, the system is at equilibrium with $P_{\text{CO}_2} = 3.87$ atm, so $Q = K_p = 3.87$. When volume halves at constant temperature, the pressure of $\text{CO}_2$ immediately doubles to 7.74 atm, making $Q = 7.74 > K_p = 3.87$. Since $Q > K$, we have $\Delta G = RT \ln(Q/K) > 0$, meaning the forward reaction is no longer spontaneous. The equilibrium must shift left (toward reactants) to reduce $P_{\text{CO}_2}$ until $Q = K_p$ again. Importantly, since temperature remains constant, $K_p$ doesn't change, so the final $P_{\text{CO}_2}$ must return exactly to 3.87 atm. Answer A incorrectly suggests 50% consumption of CaO, which has no thermodynamic basis. Answer C misapplies Le Châtelier's principle—while decreasing volume does increase pressure, this reaction has only one gas molecule, so the "more gas molecules" reasoning doesn't apply. Answer D correctly identifies the leftward shift but wrongly claims the final pressure will be between 3.87 and 7.74 atm, forgetting that equilibrium constants depend only on temperature. Remember: when temperature is constant, the equilibrium constant stays fixed, so pressure changes only affect the path back to equilibrium, not the final equilibrium concentrations.
5. Consider the simultaneous equilibria in aqueous solution: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$ with $K_{sp1} = 1.8 \times 10^{-10}$ and $\text{AgBr}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq)$ with $K_{sp2} = 5.4 \times 10^{-13}$. Solid AgCl is added to a 0.010 M NaBr solution. Which thermodynamic analysis correctly predicts the final equilibrium state?
   Answer: The final state has both solids in equilibrium with $[\text{Ag}^+] = K_{sp2}/[\text{Br}^-] = 5.4 \times 10^{-11} \text{ M}$ and $[\text{Cl}^-] = K_{sp1}/[\text{Ag}^+]$ This question tests your understanding of simultaneous equilibria involving multiple sparingly soluble salts that share a common ion. When you encounter problems with competing equilibria, you need to consider which thermodynamic state the system will naturally reach. However, there appears to be an error in this question - the correct answer is listed as "E" but no option E is provided in the choices given. Based on the thermodynamic principles involved, let me analyze the available options: Option A incorrectly assumes only AgCl dissolves and ignores the presence of Br⁻ ions, which would cause AgBr precipitation since $K_{sp2}$ is much smaller than $K_{sp1}$. Option B has the right initial reasoning - AgCl dissolves and AgBr precipitates when $Q_{AgBr}$ exceeds $K_{sp2}$. However, the final $[\text{Ag}^+]$ calculation appears incorrect. Option C presents an incorrect mathematical relationship. The equilibrium $[\text{Ag}^+]$ isn't determined by $\sqrt{K_{sp1} \cdot K_{sp2}/[\text{Br}^-]}$. Option D incorrectly suggests complete precipitation of all Br⁻, which violates equilibrium principles. The correct approach involves recognizing that when both solids coexist at equilibrium, $[\text{Ag}^+] = \frac{K_{sp2}}{[\text{Br}^-]} = \frac{K_{sp1}}{[\text{Cl}^-]}$, leading to a system where both equilibrium expressions must be satisfied simultaneously. Study tip: For multiple equilibria problems, always identify the common ion and determine which solid is less soluble (smaller $K_{sp}$) - this typically controls the final $[\text{Ag}^+]$ when both solids are present.
6. Consider the coupled equilibria: $\text{A}(aq) \rightleftharpoons \text{B}(aq)$ with $K_1 = 4.0$ and $\text{B}(aq) \rightleftharpoons \text{C}(aq)$ with $K_2 = 0.25$ at 298 K. If the concentration of A is suddenly doubled while keeping B and C concentrations momentarily constant, which thermodynamic driving force analysis correctly predicts the system's response?
   Answer: The system evolves such that both $\Delta G_1$ and $\Delta G_2$ approach zero simultaneously, with the final state determined by the overall equilibrium $K_{\text{overall}} = K_1 \times K_2$ When analyzing coupled equilibria, you need to understand that the system will evolve to minimize its total free energy, with both equilibria ultimately reaching their equilibrium states simultaneously. The correct approach recognizes that after the perturbation, the system must re-establish equilibrium for both reactions. Since these equilibria are coupled through species B, they cannot be treated independently in the final state. The system will evolve until both $\Delta G_1 = 0$ and $\Delta G_2 = 0$, meaning $Q_1 = K_1$ and $Q_2 = K_2$. The overall transformation A ⇌ C has $K_{\text{overall}} = K_1 \times K_2 = 4.0 \times 0.25 = 1.0$, which determines the final equilibrium concentrations of all species. Option A incorrectly suggests both equilibria immediately shift right with negative ΔG values. While the first equilibrium initially has $\Delta G_1 < 0$, this doesn't mean the second automatically does. Option B treats the equilibria as independent, assuming the second remains unaffected initially. This ignores that equilibrium is a state function - the system must satisfy both equilibrium conditions simultaneously in the final state. Option D focuses on kinetic aspects (relaxation times and barriers), but this is a thermodynamics question about equilibrium states, not reaction rates. Remember: in coupled equilibria problems, always consider the overall equilibrium constant and recognize that all linked equilibria must be satisfied simultaneously in the final state. The system's response is governed by thermodynamics, not kinetics.
7. A researcher studies the equilibrium $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ at 1000 K, where $\Delta H^\circ = +178 \text{ kJ/mol}$ and the equilibrium $CO_2$ pressure is 0.85 atm. The system is then subjected to a pressure increase by adding $N_2(g)$ while maintaining constant temperature and volume. After analyzing the thermodynamic driving forces, which prediction is most justified?
   Answer: No equilibrium shift occurs because the equilibrium constant depends only on CO₂ partial pressure, which remains unchanged when inert gas is added at constant volume For this heterogeneous equilibrium, K = P(CO₂) at the given temperature. Adding N₂ at constant volume increases total pressure but doesn't change the partial pressure of CO₂. Since the thermodynamic equilibrium condition K = P(CO₂) is still satisfied, no shift occurs. Choice A incorrectly applies pressure effects for homogeneous equilibria. Choice B confuses constant volume with constant pressure conditions. Choice D misapplies entropy considerations.
8. A solution equilibrium involves $\text{Cu}^{2+}(aq) + 4\text{NH}_3(aq) \rightleftharpoons [\text{Cu}(\text{NH}_3)_4]^{2+}(aq)$ with $\log \beta_4 = 12.6$ at 298 K. The solution initially contains 0.010 M Cu²⁺ and 0.10 M NH₃ at equilibrium. If the temperature is raised to 350 K where $\log \beta_4 = 11.2$, which thermodynamic principle governs the direction and extent of equilibrium shift?
   Answer: The reaction is exothermic since $\beta_4$ decreases with temperature, so higher temperature shifts equilibrium left according to van't Hoff equation, decreasing complex formation When you encounter equilibrium problems involving temperature changes and formation constants, focus on how thermodynamic parameters reveal reaction behavior. The key insight here is connecting the temperature dependence of the equilibrium constant to the reaction's enthalpy. Since $\log \beta_4$ decreases from 12.6 to 11.2 as temperature increases from 298 K to 350 K, the equilibrium constant is becoming smaller. According to the van't Hoff equation, $\frac{d \ln K}{dT} = \frac{\Delta H^\circ}{RT^2}$, when K decreases with increasing temperature, $\Delta H^\circ$ must be negative, indicating an exothermic reaction. For exothermic reactions, Le Chatelier's principle tells us that increasing temperature shifts equilibrium toward reactants (left), reducing complex formation. Answer A correctly identifies this complete reasoning chain: the reaction is exothermic because $\beta_4$ decreases with temperature, and higher temperature shifts equilibrium left according to van't Hoff behavior. Answer B incorrectly focuses on $\Delta G^\circ$ becoming "less negative" without recognizing the underlying thermodynamic cause—the exothermic nature driving the temperature dependence. Answer C suggests the direction is uncertain and depends on competing $\Delta H^\circ$ and $T\Delta S^\circ$ terms, but the data already tells us the direction through the observed decrease in $\beta_4$. Answer D mentions entropy considerations but doesn't connect to the fundamental thermodynamic principle governing the observed temperature dependence. Remember: when formation constants decrease with temperature, immediately think "exothermic reaction" and apply Le Chatelier's principle for the direction of shift.
9. For the water autoionization equilibrium $2\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$ with $\Delta H^\circ = +55.8 \text{ kJ/mol}$, $K_w = 1.0 \times 10^{-14}$ at 298 K. A solution initially at pH 7.0 is heated to 350 K. Which thermodynamic principle explains why the solution remains neutral despite the pH changing to approximately 6.3?
   Answer: The equilibrium constant increases with temperature according to van't Hoff equation, but neutrality requires [H₃O⁺] = [OH⁻], which is maintained regardless of their absolute concentrations When you encounter questions about water autoionization and temperature effects, focus on two key principles: how equilibrium constants change with temperature and what defines solution neutrality. The van't Hoff equation shows that for endothermic reactions ($\Delta H^\circ > 0$), equilibrium constants increase with temperature. Since water autoionization has $\Delta H^\circ = +55.8$ kJ/mol, $K_w$ increases from $1.0 \times 10^{-14}$ at 298 K to approximately $2.5 \times 10^{-13}$ at 350 K. This means both $[\text{H}_3\text{O}^+]$ and $[\text{OH}^-]$ increase, causing the pH to drop from 7.0 to about 6.3. However, neutrality doesn't mean pH = 7.0 at all temperatures. Neutrality means $[\text{H}_3\text{O}^+] = [\text{OH}^-]$, which remains true because water autoionization produces these ions in exactly equal amounts regardless of temperature. Answer A correctly identifies this principle. Answer B incorrectly invokes chemical potentials, which don't remain equal - they change with concentration and temperature. Answer C mentions electroneutrality, but this is about charge balance in solutions with other ions, not the fundamental definition of neutrality in pure water. Answer D focuses on entropy changes, which while relevant to equilibrium position, doesn't explain why equal concentrations are maintained. Remember: neutrality is defined by equal $[\text{H}_3\text{O}^+]$ and $[\text{OH}^-]$ concentrations, not by pH = 7.0. The pH of neutral water changes with temperature, but the solution remains neutral due to the 1:1 stoichiometry of autoionization.
10. Consider the equilibrium $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ with $\Delta H^\circ = +124 \text{ kJ/mol}$ and $\Delta S^\circ = +175 \text{ J/mol}\cdot\text{K}$. At 500 K, $\Delta G^\circ = +36.5 \text{ kJ/mol}$. A chemist wants to maximize product yield and considers two strategies: (I) increase temperature to 600 K, or (II) decrease pressure by expanding volume. From a thermodynamic perspective, which strategy comparison is most accurate?
    Answer: Strategy I increases K since TΔS° term grows larger than ΔH° contribution, while Strategy II lowers partial pressures favoring dissociation, with Strategy I providing superior thermodynamic advantage At higher temperature, ΔG° = ΔH° - TΔS° becomes more negative because the TΔS° term (positive) grows faster than temperature increase, so K increases significantly. Volume expansion lowers partial pressures, making Q < K, driving the reaction right. However, the temperature effect changes the equilibrium constant itself, providing a more fundamental thermodynamic advantage than just shifting position along the same equilibrium curve. Choices A, B, and C contain errors in thermodynamic analysis.
11. For the aqueous equilibrium $\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$ with $\Delta H^\circ = -42 \text{ kJ/mol}$, a student observes that adding NaCl(s) causes the red color of FeSCN²⁺ to fade. From a thermodynamic standpoint, which explanation most accurately accounts for this observation?
    Answer: The activity coefficients change such that the ratio $\gamma_{\text{FeSCN}^{2+}}/\gamma_{\text{Fe}^{3+}} \gamma_{\text{SCN}^-}$ decreases, requiring lower product concentrations to maintain constant K When you encounter equilibrium problems involving ionic solutions, you need to distinguish between concentration-based and thermodynamically rigorous treatments. The true equilibrium constant K depends on activities, not concentrations, where activity = concentration × activity coefficient. Adding NaCl increases the solution's ionic strength, which affects activity coefficients according to the Debye-Hückel theory. For the equilibrium constant expression: $K = \frac{a_{\text{FeSCN}^{2+}}}{a_{\text{Fe}^{3+}} \cdot a_{\text{SCN}^-}} = \frac{[\text{FeSCN}^{2+}]\gamma_{\text{FeSCN}^{2+}}}{[\text{Fe}^{3+}][\text{SCN}^-]\gamma_{\text{Fe}^{3+}}\gamma_{\text{SCN}^-}}$ Since K must remain constant at constant temperature, if the activity coefficient ratio $\gamma_{\text{FeSCN}^{2+}}/\gamma_{\text{Fe}^{3+}}\gamma_{\text{SCN}^-}$ decreases due to increased ionic strength, then the concentration ratio must increase to compensate. However, the equilibrium shifts to maintain this balance, actually decreasing [FeSCN²⁺] and explaining the color fade. Answer D correctly identifies this thermodynamic relationship. Answer A incorrectly states that K itself changes - the equilibrium constant is truly constant at fixed temperature. Answer B invokes a "common ion effect" that doesn't apply here since Cl⁻ isn't a common ion to this equilibrium and doesn't significantly compete with SCN⁻ for Fe³⁺ coordination. Answer C makes the fundamental error of saying K "appears to change" - the thermodynamic K is invariant, but the relationship between activities and concentrations changes. Remember: in ionic equilibria, always consider whether you're dealing with concentrations or activities, especially when ionic strength changes significantly.
12. For the equilibrium $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$, $\Delta H^\circ = -41.2 \text{ kJ/mol}$, a student calculates that $K_p$ at 600 K is 0.75. When the temperature is raised to 800 K, $K_p$ becomes 0.31. If an inert gas (He) is added to the system at constant volume and 800 K, which thermodynamic argument correctly predicts the equilibrium response?
    Answer: No shift occurs because inert gas addition at constant volume changes total pressure but not partial pressure ratios, leaving the reaction quotient unchanged relative to Kp At constant volume, adding inert gas increases total pressure but doesn't change the partial pressures of the reactants and products. Since Kp depends only on partial pressures, and Q (in terms of partial pressures) remains unchanged, there's no thermodynamic driving force for equilibrium shift. Choice B misapplies pressure effects. Choice C confuses concentration with partial pressure. Choice D incorrectly invokes entropy of mixing as affecting the reaction equilibrium.
13. A gaseous equilibrium $\text{A}(g) + 2\text{B}(g) \rightleftharpoons \text{C}(g) + \text{D}(g)$ has $\Delta H^\circ = -45 \text{ kJ/mol}$ and $\Delta S^\circ = -126 \text{ J/mol}\cdot\text{K}$. At 400 K, the equilibrium lies far to the right. A researcher applies Le Châtelier's principle to predict the effect of simultaneously increasing both temperature and pressure. Which analysis correctly identifies the thermodynamic competition?
    Answer: Temperature increase favors reactants (entropy opposes enthalpy), while pressure increase favors products (fewer moles), creating opposing thermodynamic driving forces For this reaction, ΔG° = ΔH° - TΔS° = -45,000 - T(-126). At higher T, the entropy term (TΔS°) becomes more negative, making ΔG° less negative, favoring reactants. Pressure increase favors the side with fewer moles (products: 2 moles vs reactants: 3 moles). Choice B incorrectly states temperature favors products. Choice C misunderstands the sign of ΔG change. Choice D incorrectly describes the pressure effect.
14. For the reaction $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \rightleftharpoons \text{CuSO}_4(s) + 5\text{H}_2\text{O}(g)$ at 298 K, $\Delta H^\circ = +78 \text{ kJ/mol}$ and the equilibrium water vapor pressure is 7.8 mmHg. If the system is heated to 350 K in a container where the water vapor can escape, which thermodynamic analysis best explains the observed complete dehydration?
    Answer: The activity of water vapor approaches zero as it escapes, making $\Delta G = \Delta G^\circ \ll 0$ regardless of the equilibrium constant value When analyzing equilibrium reactions where products can escape the system, you need to distinguish between thermodynamic equilibrium (closed system) and the actual driving force in an open system where species can be removed. The correct analysis focuses on the reaction quotient and Gibbs free energy relationship: $\Delta G = \Delta G° + RT \ln Q$. In this open system, water vapor escapes, keeping its partial pressure near zero. This makes the activity of water vapor approach zero, so $Q = (P_{\text{H}_2\text{O}})^5 \approx 0$. Therefore, $\ln Q$ becomes large and negative, making $\Delta G$ large and negative regardless of what $\Delta G°$ or the equilibrium constant might be. This creates a strong thermodynamic driving force for complete dehydration, which is exactly what option D describes. Option A incorrectly focuses on equilibrium vapor pressure increasing with temperature, but this misses that the actual vapor pressure stays low due to escape. Option B wrongly suggests chemical potentials explain the driving force - while true that escaping water has lower chemical potential, this doesn't provide the fundamental thermodynamic analysis requested. Option C makes a common error about temperature effects on $\Delta G°$. While $T\Delta S°$ does increase with temperature, this endothermic reaction ($\Delta H° > 0$) would still have $\Delta G° > 0$ at moderate temperatures, making this explanation insufficient. Remember: in open systems where products can escape, always consider how the reaction quotient Q changes, not just the equilibrium constant. The ability to remove products creates driving forces that override normal equilibrium limitations.
15. Consider the equilibrium $\text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g)$ at 298 K with $K_p = 0.11$ atm². Initially, the system is at equilibrium in a 2.0 L container. If NH₃(g) is selectively removed by passing the gas mixture through an acid trap while maintaining constant temperature, which thermodynamic analysis correctly describes the system's response?
    Answer: The system cannot reach equilibrium while NH₃ is being removed, so the reaction continues indefinitely with $\Delta G < 0$ until all NH₄HS is consumed This question tests your understanding of dynamic equilibrium and how systems respond to continuous perturbations, which differs from simple Le Châtelier's principle applications. When NH₃ is continuously removed from this heterogeneous equilibrium, you're dealing with an open system rather than a closed equilibrium. The key insight is that as long as NH₃ removal continues, the reaction quotient $Q = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}}$ remains less than $K_p = 0.11$, creating a persistent driving force for decomposition. Option B correctly identifies that $\Delta G = \Delta G° + RT \ln(Q)$ remains negative because Q approaches zero as $P_{\text{NH}_3} \to 0$. This means the forward reaction is thermodynamically favored as long as NH₃ removal continues, allowing complete consumption of NH₄HS(s) if removal is efficient enough. Option A incorrectly assumes the system reaches a new equilibrium state with equal partial pressures, ignoring the continuous removal aspect. Option C incorrectly introduces kinetic considerations when this is purely a thermodynamic analysis - if $\Delta G < 0$, the reaction will proceed regardless of relative rates. Option D misapplies chemical potential concepts; while the chemical potential of pure NH₄HS(s) does remain constant, this doesn't explain the system's response mechanism. Remember: When dealing with equilibrium perturbations, distinguish between single disturbances (which lead to new equilibrium positions) and continuous disturbances (which can drive reactions to completion by maintaining $Q \neq K$).
16. For the gas-phase equilibrium $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, $\Delta H^\circ = -92.2 \text{ kJ/mol}$ and $\Delta S^\circ = -198.4 \text{ J/mol}\cdot\text{K}$. At 673 K, $K_p = 6.2 \times 10^{-3}$. An engineer proposes coupling this reaction with the exothermic combustion $\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}$ to supply heat. Which thermodynamic analysis best evaluates this coupling strategy?
    Answer: The coupling is thermodynamically unfavorable because ammonia formation is entropy-disfavored, and adding heat makes ΔG more positive, opposing the desired product formation For the ammonia synthesis, ΔG° = ΔH° - TΔS° = -92,200 - T(-198.4). At higher temperatures, the positive TΔS° term makes ΔG° more positive (less favorable), decreasing K. The negative ΔS° means entropy opposes the reaction, and this becomes more significant at higher T. Adding heat from combustion would shift the equilibrium left, not right. Choice A incorrectly predicts the temperature effect. Choice C overcomplicates with competing effects. Choice D misunderstands equilibrium thermodynamics.
17. For the gas-phase equilibrium $\text{SO}_2(g) + \frac{1}{2}\text{O}_2(g) \rightleftharpoons \text{SO}_3(g)$ with $\Delta H^\circ = -98 \text{ kJ/mol}$, a reactor operates at 750 K with $K_p = 279$ atm⁻⁰·⁵. If the reactor pressure is increased from 1.0 atm to 10.0 atm while maintaining constant temperature and feed composition, which thermodynamic analysis correctly predicts the change in SO₂ conversion?
    Answer: Conversion increases because the equilibrium partial pressure of SO₃ scales as $P_{\text{total}}^{0.5}$ while reactant pressures scale linearly, favoring product formation at higher total pressure When analyzing how pressure changes affect gas-phase equilibria, you need to examine how the equilibrium expression responds to pressure variations, not just apply Le Châtelier's principle qualitatively. For this reaction, $K_p = \frac{P_{\text{SO}_3}}{P_{\text{SO}_2} \cdot P_{\text{O}_2}^{0.5}}$. At equilibrium, if you increase total pressure while maintaining feed composition, the partial pressures don't all scale equally. The key insight is recognizing how each term in the equilibrium expression scales with total pressure. When total pressure increases by a factor of 10, the partial pressures of reactants ($P_{\text{SO}_2}$ and $P_{\text{O}_2}$) scale nearly linearly with total pressure. However, to maintain the constant $K_p$ value, $P_{\text{SO}_3}$ must adjust accordingly. Since $P_{\text{O}_2}$ appears as $P_{\text{O}_2}^{0.5}$ in the denominator, $P_{\text{SO}_3}$ scales approximately as $P_{\text{total}}^{0.5}$ rather than linearly. This mathematical relationship means SO₃ formation is favored at higher pressures, increasing conversion. Option A oversimplifies with Le Châtelier's principle without the quantitative analysis needed here. Option B incorrectly suggests $K_p$ changes with pressure—equilibrium constants depend only on temperature. Option D wrongly assumes conversion stays constant, missing how pressure affects the equilibrium position even when $K_p$ remains fixed. Study tip: For gas equilibria problems involving pressure changes, always examine how each term in the $K_p$ expression scales with pressure rather than relying solely on qualitative Le Châtelier reasoning. The mathematical relationship often reveals non-obvious behavior.
18. For the equilibrium $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$, $\Delta H^\circ = -198 \text{ kJ/mol}$, an industrial reactor operates at 700 K where $K_p = 4.2 \times 10^4$. During operation, a catalyst is added that lowers the activation energy of both forward and reverse reactions by the same amount. Which thermodynamic analysis correctly describes the catalyst's effect on equilibrium position?
    Answer: The equilibrium position remains unchanged because the catalyst affects kinetic pathways but not the thermodynamic equilibrium constant, which depends only on ΔG° for the overall reaction A catalyst affects only the kinetic pathway and activation energies, not the thermodynamic properties of reactants and products. Since K depends only on the standard Gibbs free energy difference between products and reactants (ΔG° = -RT ln K), and this difference is unchanged by catalysis, K remains constant. The equilibrium position is unchanged. Choices A and C incorrectly suggest thermodynamic changes. Choice D misunderstands that all pathways must lead to the same equilibrium.
19. A gas mixture at equilibrium contains $\text{H}_2$, $\text{I}_2$, and $\text{HI}$ according to $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ with $\Delta H^\circ = -9.5 \text{ kJ/mol}$. The system is compressed adiabatically from 2.0 L to 0.5 L. Which thermodynamic analysis best predicts the net effect on the equilibrium composition?
    Answer: The equilibrium shifts left due to temperature increase from compression, which dominates over the negligible pressure effect since $\Delta n = 0$ When analyzing equilibrium shifts under simultaneous pressure and temperature changes, you need to evaluate each effect separately using Le Châtelier's principle, then determine which dominates. For this reaction, $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$, notice that $\Delta n = 0$ (2 moles of gas on each side). This means pressure/volume changes have no effect on equilibrium position since equal moles of gas exist on both sides. However, adiabatic compression significantly increases temperature. Since $\Delta H^\circ = -9.5 \text{ kJ/mol}$ (exothermic forward reaction), higher temperature favors the endothermic reverse reaction according to Le Châtelier's principle. The equilibrium shifts left, reducing HI concentration and increasing $\text{H}_2$ and $\text{I}_2$. Option A incorrectly assumes volume decrease affects equilibrium despite $\Delta n = 0$, and wrongly concludes that higher temperature favors the exothermic direction. Option B correctly identifies the temperature effect but incorrectly states volume has "no effect" as the reason rather than explaining that $\Delta n = 0$ means pressure changes don't shift equilibrium. Option D suggests the effects cancel, but since only temperature matters here (volume effect is zero), there's no cancellation. Option C correctly identifies that temperature increase dominates because the pressure effect is negligible when $\Delta n = 0$, leading to a leftward shift. Study tip: For gas-phase equilibria, always check $\Delta n$ first. When $\Delta n = 0$, pressure changes won't shift equilibrium, so focus entirely on temperature effects using the sign of $\Delta H^\circ$.
20. For the equilibrium $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, $\Delta H^\circ = +57.2 \text{ kJ/mol}$, a chemist observes that increasing the temperature from 298 K to 350 K causes the equilibrium constant to increase by a factor of 8.5. If the system is then subjected to a sudden temperature decrease back to 298 K while maintaining constant volume, which thermodynamic principle best explains the immediate response of the system?
    Answer: The system will shift left because the equilibrium constant decreases with temperature, and the reaction quotient exceeds the new equilibrium constant When temperature decreases for an endothermic reaction, K decreases (from 8.5K₂₉₈ back to K₂₉₈). At the moment of temperature change, Q (reaction quotient) still reflects the high-temperature composition, so Q > K at the new lower temperature. The system must shift left (toward reactants) to decrease Q until Q = K again. Choice A confuses the thermodynamic driving force with kinetic considerations. Choice B incorrectly assumes entropy always dominates. Choice D misapplies volume constraints.

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