Physical Chemistry 1 Quiz: Le Chateliers Principle Thermodynamic

What this quiz covers

This quiz focuses on Le Chateliers Principle Thermodynamic, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 1.

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. The entropy change favors the reverse reaction at higher temperature because there are equal moles of gas on both sides
2. The equilibrium constant decreases because $\ln(K_2/K_1) = -\Delta H^\circ/R(1/T_2 - 1/T_1)$ yields a negative value when $\Delta H^\circ < 0$ (correct answer)
3. The Gibbs free energy change becomes more positive because the $T\Delta S^\circ$ term dominates over the $\Delta H^\circ$ term at higher temperature
4. The reaction quotient Q increases relative to K because the partial pressures of products increase more rapidly than reactants with temperature
5. The activation energy for the reverse reaction becomes lower than the forward reaction when temperature increases for exothermic processes

Question 2

1. The chemical potential of each species decreases proportionally, favoring the side with fewer particles to minimize total Gibbs free energy
2. The reaction quotient Q becomes less than K because partial pressures of all species decrease, requiring net forward reaction to restore equilibrium (correct answer)
3. The entropy of mixing increases more significantly for the products side, making $\Delta G_{rxn}$ more negative for the forward direction
4. The activity coefficients of the gaseous species change non-ideally due to increased intermolecular interactions at higher total pressure
5. The standard state chemical potentials shift because the reference pressure effectively changes when inert gas is added to the system

Question 3

1. The reaction shifts left because $\Delta G = RT \ln(Q/K) > 0$ when the pressure doubles, and continues until 50% of CaO is consumed
2. 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 (correct answer)
3. The reaction shifts right because Le Châtelier's principle favors the side with more gas molecules when volume decreases
4. The reaction shifts left because increasing pressure favors fewer gas molecules, but the final $P_{\text{CO}_2}$ will be between 3.87 and 7.74 atm
5. No shift occurs because the equilibrium constant depends only on temperature, so $P_{\text{CO}_2}$ must remain at 3.87 atm regardless of volume changes

Question 4

1. The equilibrium shifts right due to volume decrease, and the temperature increase further favors products because the reaction is slightly exothermic
2. The equilibrium shifts left due to temperature increase from compression, and the volume decrease has no effect because $\Delta n = 0$
3. The equilibrium shifts left due to temperature increase from compression, which dominates over the negligible pressure effect since $\Delta n = 0$ (correct answer)
4. No net shift occurs because the volume effect exactly cancels the temperature effect when $\Delta H^\circ$ is small and $\Delta n = 0$
5. The direction depends on the initial composition, but the magnitude of shift is determined by the relative magnitudes of $\partial \ln K/\partial T$ and $\partial \ln K/\partial P$

Question 5

1. Both equilibria shift right because $\Delta G_1 < 0$ and $\Delta G_2 < 0$ when A concentration increases, creating a cascade effect toward C formation
2. The first equilibrium shifts right with $\Delta G_1 = RT \ln(Q_1/K_1) < 0$, while the second remains unaffected until B concentration changes significantly
3. 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 $K_{\text{overall}} = K_1 \times K_2$ (correct answer)
4. The first equilibrium shifts right immediately, then the second equilibrium responds to the B concentration increase, with relaxation times determined by the relative magnitudes of the kinetic barriers
5. The direction of the second equilibrium depends on whether the rate of B formation from the first equilibrium exceeds the rate of B consumption in the second equilibrium

Question 6

1. The activity coefficient of H⁺ decreases more than that of A⁻ due to higher charge density, making the effective $K_a = K_a^\circ \gamma_{\text{H}^+} \gamma_{\text{A}^-}/\gamma_{\text{HA}}$ smaller
2. The ionic strength effect increases the activity coefficients of all ionic species equally, but HA is unaffected, shifting the equilibrium toward the molecular form
3. The Debye-Hückel effect decreases the activity coefficients of charged species, with $\gamma_{\text{H}^+}$ decreasing more than $\gamma_{\text{A}^-}$, effectively increasing the thermodynamic $K_a$
4. The activity coefficient ratio $\gamma_{\text{H}^+} \gamma_{\text{A}^-}/\gamma_{\text{HA}}$ decreases with increasing ionic strength, requiring lower [H⁺] to maintain constant thermodynamic $K_a$ (correct answer)
5. The standard chemical potential of H⁺ becomes more positive due to increased electrostatic interactions, shifting the equilibrium position toward the conjugate base form

Question 7

1. The catalyst provides an alternative reaction pathway with lower activation energy, but the difference in chemical potentials between products and reactants ($\Delta G_{rxn}$) remains constant
2. The equilibrium constant depends only on standard thermodynamic properties ($\Delta G^\circ = -RT \ln K$), which are independent of reaction mechanism and kinetic factors (correct answer)
3. The principle of microscopic reversibility ensures that any catalytic enhancement of the forward rate constant is exactly matched by enhancement of the reverse rate constant
4. The catalyst changes the activation energies equally for both directions, maintaining the same ratio $k_f/k_r = K$ according to transition state theory
5. The entropy of activation changes equally for forward and reverse reactions, leaving the equilibrium constant $K = \exp(-\Delta G^\circ/RT)$ unaffected by catalytic mechanisms

Question 8

1. The equilibrium constant increases exponentially with temperature according to van't Hoff equation, making $P_{\text{H}_2\text{O}}^{\text{eq}}$ much larger than the actual vapor pressure
2. The chemical potential of gaseous water becomes much lower than that of bound water at higher temperature, creating a large negative $\Delta G_{rxn}$
3. The entropy term $T\Delta S^\circ$ becomes dominant over $\Delta H^\circ$ at higher temperature, making $\Delta G^\circ$ large and negative
4. The activity of water vapor approaches zero as it escapes, making $\Delta G = \Delta G^\circ \ll 0$ regardless of the equilibrium constant value (correct answer)
5. The vapor pressure of water increases exponentially with temperature while the equilibrium vapor pressure increases more slowly, ensuring $P_{\text{actual}} < P_{\text{equilibrium}}$

Question 9

1. 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 (correct answer)
2. The chemical potentials of H₃O⁺ and OH⁻ remain equal at all temperatures because they are produced in equimolar amounts from autoionization
3. The electroneutrality condition requires charge balance, so [H₃O⁺] must equal [OH⁻] even as their product increases with temperature
4. The entropy increase from autoionization becomes more favorable at higher temperature, but equal production stoichiometry ensures [H₃O⁺] = [OH⁻]
5. The activity coefficients of H₃O⁺ and OH⁻ change equally with temperature, maintaining equal thermodynamic activities despite changing concentrations

Question 10

1. Conversion increases because higher pressure favors the side with fewer gas molecules ($\Delta n = -0.5$), shifting equilibrium toward SO₃ formation according to Le Châtelier's principle
2. Conversion increases because $K_p$ has units of atm⁻⁰·⁵, making the equilibrium constant effectively larger at higher pressure when expressed in terms of partial pressures
3. 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 (correct answer)
4. Conversion remains constant because $K_p$ depends only on temperature, and the equilibrium composition is determined solely by the equilibrium constant value
5. Conversion increases because the reaction quotient $Q_p$ decreases relative to $K_p$ when all partial pressures increase proportionally, driving the reaction forward

Question 11

1. 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 (correct answer)
2. The equilibrium shifts left because $\Delta G^\circ = -RT \ln \beta_4$ becomes less negative at higher temperature, making complex formation less thermodynamically favorable
3. The direction depends on the relative magnitudes of $\Delta H^\circ$ and $T\Delta S^\circ$ terms in $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ at the new temperature
4. The equilibrium shifts left because the entropy of the aqueous system increases when the highly organized complex dissociates into smaller, more mobile ions
5. The chemical potential of NH₃ changes more significantly than Cu²⁺ with temperature due to differences in solvation, driving the equilibrium toward the less solvated state

Question 12

1. Only AgCl dissolves because its $K_{sp}$ is larger, with final $[\text{Ag}^+] = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5} \text{ M}$
2. AgCl dissolves until $Q_{AgBr} = [\text{Ag}^+][\text{Br}^-] = K_{sp2}$, then AgBr precipitates while AgCl continues dissolving until both solids coexist with $[\text{Ag}^+] = 5.4 \times 10^{-11} \text{ M}$
3. The system reaches equilibrium with both solids present when the chemical potential of Ag⁺ is simultaneously consistent with both equilibrium expressions, giving $[\text{Ag}^+] = \sqrt{K_{sp1} \cdot K_{sp2}/[\text{Br}^-]}$
4. AgBr precipitates completely, converting all Br⁻ to solid, followed by AgCl dissolution until $[\text{Ag}^+][\text{Cl}^-] = K_{sp1}$ with $[\text{Cl}^-]$ determined by mass balance
5. The final state has both solids in equilibrium with $[\text{Ag}^+] = K_{sp2}/[\text{Br}^-] = 5.4 \times 10^{-11} \text{ M}$ and $[\text{Cl}^-] = K_{sp1}/[\text{Ag}^+]$ (correct answer)

Question 13

1. The ionic strength increase causes activity coefficients to change according to the Debye-Hückel theory, effectively altering the equilibrium constant
2. The common ion effect from Cl⁻ shifts the equilibrium left because chloride competes with SCN⁻ for coordination sites on Fe³⁺
3. The increased ionic strength decreases the activity coefficients of the charged species, making the thermodynamic equilibrium constant appear to change
4. 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 (correct answer)
5. The enthalpy change becomes less negative due to increased ion-ion interactions, shifting the equilibrium toward reactants via temperature-dependent effects

Question 14

1. No shift occurs because $\Delta n = 0$ for this reaction, making the equilibrium constant independent of pressure or volume changes at constant temperature
2. The equilibrium shifts left because $Q_c = \frac{(0.125)^2}{(0.125)(0.125)} = 1.0 < K_c = 54.3$, requiring net reverse reaction to reach equilibrium
3. The equilibrium shifts right because expansion decreases all concentrations equally, but the equilibrium constant expression favors the side with more particles when concentrations are low
4. 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 (correct answer)
5. The equilibrium position remains unchanged because both $Q_c$ and $K_c$ scale identically with concentration changes when $\Delta n = 0$

Question 15

1. The equilibrium shifts right with $\Delta G < 0$ because $Q = P_{\text{NH}_3} P_{\text{H}_2\text{S}} < K_p$ after NH₃ removal, continuing until $P_{\text{NH}_3} = P_{\text{H}_2\text{S}} = \sqrt{0.11} = 0.33$ atm
2. The reaction proceeds right as long as NH₃ is removed, with the driving force $\Delta G = \Delta G^\circ + RT \ln(Q)$ remaining negative because $Q \to 0$ as $P_{\text{NH}_3} \to 0$
3. The equilibrium shifts right, but the final state depends on the relative rates of NH₃ removal versus NH₄HS decomposition, not just thermodynamic considerations
4. The chemical potential of NH₄HS(s) remains constant, but the chemical potentials of the gaseous products decrease as they are removed, creating a persistent thermodynamic driving force
5. The system cannot reach equilibrium while NH₃ is being removed, so the reaction continues indefinitely with $\Delta G < 0$ until all NH₄HS is consumed (correct answer)

Question 16

1. Temperature increase favors reactants (entropy opposes enthalpy), while pressure increase favors products (fewer moles), creating opposing thermodynamic driving forces (correct answer)
2. Temperature increase favors products (exothermic forward reaction), while pressure increase favors reactants (fewer gaseous particles), creating opposing thermodynamic driving forces
3. Temperature increase favors reactants (positive ΔG becomes more positive), while pressure increase favors products (Le Châtelier volume effect), creating opposing thermodynamic driving forces
4. Temperature increase favors products (negative ΔH dominates), while pressure increase favors reactants (∆n < 0 means volume decrease opposed), creating opposing thermodynamic driving forces

Question 17

1. 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 (correct answer)
2. Forward shift occurs because inert gas increases total pressure, and the system responds by minimizing pressure through Le Châtelier's principle despite equal mole numbers
3. Reverse shift occurs because inert gas dilution decreases the effective concentration driving force, reducing the thermodynamic favorability of product formation
4. Forward shift occurs because inert gas addition increases entropy of mixing, which supplements the negative enthalpy change to make ΔG more negative

Question 18

1. Strategy I increases K due to positive ΔH, while Strategy II shifts equilibrium right due to volume increase, but Strategy I provides greater yield enhancement through exponential K dependence
2. Strategy I decreases ΔG° making it less positive due to entropy dominance, while Strategy II decreases Q below K triggering rightward shift, with comparable effectiveness for yield improvement
3. Strategy I makes ΔG° more negative due to temperature-entropy coupling, while Strategy II exploits mole number increase favoring products, with Strategy II being more effective thermodynamically
4. 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 (correct answer)

Question 19

1. 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 (correct answer)
2. Minor shift right because although ΔG° becomes negative, the equilibrium constant only increases moderately due to the logarithmic relationship between K and ΔG°
3. Major shift right because the endothermic reaction is favored at higher temperature, and the positive entropy change provides additional thermodynamic driving force
4. Moderate shift right because ΔG° becomes more negative due to entropy dominance, but gas expansion at constant pressure partially opposes the concentration increase

Question 20

1. All three changes synergistically increase CO yield: higher temperature increases K, CO₂ removal shifts equilibrium right, and excess carbon provides unlimited reactant supply
2. Temperature increase and CO₂ removal both shift equilibrium right through different thermodynamic mechanisms, while carbon addition has no effect since it's a pure solid phase (correct answer)
3. Temperature increase opposes CO₂ removal effects because higher K reduces the thermodynamic driving force created by CO₂ depletion, while carbon addition enhances surface area effects
4. CO₂ removal creates the strongest driving force by continuously maintaining Q < K, while temperature increase supplements this effect, and carbon addition provides kinetic enhancement only