Equilibrium MCQs with Answers – Part 3 (NEET Chemistry)

Explanation: The Assertion compares three experimentally distinct disturbances. Change I adds a gaseous product at unchanged temperature and volume. This raises the reaction quotient because the \(\mathrm{SO_3}\) partial pressure appears squared in the numerator. The equilibrium constant remains unchanged. A reverse reaction tendency is therefore produced. Change II compresses the ideal gas mixture isothermally. The reacting-gas partial pressures increase, but the equilibrium constant remains unchanged because temperature is fixed. For the balanced reaction, \[ \Delta n_g=2-(2+1)=-1 \] Uniform compression lowers the reaction quotient relative to its original value. The forward reaction is consequently favoured. Change III increases temperature. Because the forward reaction is exothermic, the equilibrium constant decreases when temperature rises. The original equilibrium composition need not satisfy the new-temperature equilibrium constant. Therefore, a net reverse reaction tendency can develop during subsequent re-equilibration. Unlike the first two disturbances, the temperature change alters the equilibrium constant itself. The Reason correctly distinguishes changes in the reaction quotient from changes in the equilibrium constant. It also recognises that the reaction quotient may depend on temperature through the activities or partial pressures appropriate to the actual state. The conclusion does not require Q to change in every possible temperature-changing arrangement; the key point is that temperature can change K, whereas concentration and volume disturbances at fixed temperature do not. Both the Assertion and Reason are true, and the Reason explains the Assertion.

Explanation: \(\textbf{Identify the original equilibrium condition:}\) The system is initially at equilibrium. Therefore, its original reaction quotient equals the equilibrium constant at \(T_1\). \[ Q_1=K_1 \] \(\textbf{Determine the pressure contribution:}\) For the balanced reaction, \[ Q=\frac{a_B}{a_A^2} \] The sudden compression doubles both gaseous partial pressures. Using the same standard-pressure convention, their activities also double. Therefore, \[ Q_2=\frac{2a_B}{(2a_A)^2} \] \[ Q_2=\frac12Q_1 \] Consequently, \[ Q_2=\frac{K_1}{2} \] \(\textbf{Determine the temperature contribution:}\) The forward reaction is exothermic. Increasing temperature therefore decreases its equilibrium constant. Thus, \[ K_2\lt K_1 \] However, the problem does not specify the numerical value of \(K_2\). \(\textbf{Test a possible forward reaction tendency:}\) Suppose the new equilibrium constant were \[ K_2=\frac34K_1 \] This is consistent with a decrease upon heating. Then, \[ Q_2=\frac12K_1\lt K_2 \] The forward reaction would be favoured. \(\textbf{Test a possible reverse reaction tendency:}\) Alternatively, suppose \[ K_2=\frac14K_1 \] This also represents a decrease upon heating. Now, \[ Q_2=\frac12K_1\gt K_2 \] The reverse reaction would be favoured. \(\textbf{Identify the limiting case:}\) If \[ K_2=\frac12K_1 \] then the disturbed quotient would equal the new equilibrium constant. No net reaction tendency would exist. \(\textbf{Interpret the competing effects:}\) Compression favours the side containing fewer gaseous molecules, which is B. Heating favours the endothermic reverse direction. Knowing that these effects oppose one another does not establish which is stronger. The final prediction requires a comparison of the actual post-disturbance quotient with the new-temperature equilibrium constant. The question illustrates an important limitation of qualitative Le Chatelier reasoning when two disturbances act simultaneously. A unique direction must not be invented when the supplied information permits different outcomes.

Explanation: \(\textbf{Identify the pressure equilibrium expression:}\) For the balanced reaction, \[ Q_p=\frac{p_C}{p_Ap_B^2} \] The equilibrium constant remains unchanged in both experiments because temperature remains constant. \(\textbf{Analyse experiment P at constant volume:}\) At fixed temperature and volume, the partial pressure of each gas is proportional to its mole amount. Adding A increases its amount from \(10.0\) to \(12.0\,\mathrm{mol}\). The amounts and partial pressures of B and C remain unchanged immediately after addition. Therefore, \[ \frac{Q_{p,P}}{K_p}=\frac{10.0}{12.0} \] \[ \frac{Q_{p,P}}{K_p}=\frac56 \] Since this ratio is smaller than one, experiment P favours the forward reaction. \(\textbf{Determine the original mole-number total:}\) Before either operation, \[ n_{\mathrm{total},1}=10.0+1.00+1.00 \] \[ n_{\mathrm{total},1}=12.0\,\mathrm{mol} \] \(\textbf{Determine the new mole-number total in Q:}\) After adding A, \[ n_{\mathrm{total},2}=12.0+1.00+1.00 \] \[ n_{\mathrm{total},2}=14.0\,\mathrm{mol} \] \(\textbf{Account for the freely moving piston:}\) At fixed total pressure, \[ p_i=\frac{n_i}{n_{\mathrm{total}}}P \] For the stated reaction, \[ Q_p=\frac{n_Cn_{\mathrm{total}}^2}{n_An_B^2P^2} \] \(\textbf{Compare the quotients before and after addition:}\) Because total pressure is unchanged, \[ \frac{Q_{p,Q}}{K_p} = \frac{(1.00)(14.0)^2}{(12.0)(1.00)^2} \div \frac{(1.00)(12.0)^2}{(10.0)(1.00)^2} \] \[ \frac{Q_{p,Q}}{K_p} = \frac{14^2\times10}{12^3} \] \[ \frac{Q_{p,Q}}{K_p}=\frac{245}{216} \] \[ \frac{Q_{p,Q}}{K_p}\approx1.134 \] \(\textbf{Determine the second reaction tendency:}\) Since \[ Q_{p,Q}\gt K_p \] experiment Q favours the reverse reaction. \(\textbf{Interpret the counterintuitive result:}\) Adding a reactant does not necessarily favour the forward direction when total pressure is held constant by expansion. In experiment Q, introducing A increases the total number of gaseous moles and expands the vessel. The accompanying dilution of B is especially important because its partial pressure is squared in the denominator. For the stated initial composition, this expansion effect outweighs the direct addition of A. Thus, identical chemical additions can produce opposite initial reaction tendencies under different mechanical constraints. The reaction direction must be established from the complete reaction quotient rather than from the identity of the added substance alone.

Explanation: \(\textbf{Identify the initial dynamic equilibrium:}\) At equilibrium, the opposing rates are equal. \[ r_{f,1}=r_{r,1}=0.30\,\mathrm{mol\,L^{-1}\,s^{-1}} \] These rates are nonzero because equilibrium is dynamic. \(\textbf{Determine the concentration changes:}\) Reducing volume to one-third increases every gaseous concentration by a factor of three. Immediately after compression, \[ [A]_2=3[A]_1 \] \[ [B]_2=3[B]_1 \] \(\textbf{Calculate the forward rate:}\) The forward elementary rate law is second order in A. \[ r_{f,2}=k_f[A]_2^2 \] \[ r_{f,2}=k_f(3[A]_1)^2 \] \[ r_{f,2}=9r_{f,1} \] \[ r_{f,2}=9(0.30) \] \[ r_{f,2}=2.70\,\mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Calculate the reverse rate:}\) The reverse rate law has the same overall concentration power. \[ r_{r,2}=k_r[B]_2^2 \] \[ r_{r,2}=k_r(3[B]_1)^2 \] \[ r_{r,2}=9r_{r,1} \] \[ r_{r,2}=2.70\,\mathrm{mol\,L^{-1}\,s^{-1}} \] \(\textbf{Examine the reaction quotient:}\) For the balanced reaction, \[ Q_c=\frac{[B]^2}{[A]^2} \] After compression, \[ Q_{c,2}=\frac{(3[B]_1)^2}{(3[A]_1)^2} \] \[ Q_{c,2}=Q_{c,1} \] Since the original mixture was at equilibrium, \[ Q_{c,2}=K_c \] \(\textbf{Interpret the distinction between rate and shift:}\) Both forward and reverse rates increase substantially. Nevertheless, they remain equal. The equilibrium composition ratio is unchanged because the reaction has the same gaseous stoichiometric mole number on both sides. Compression therefore causes no net equilibrium shift under the ideal conditions. An increase in the absolute rate of microscopic reactions must not be confused with a change in the net reaction direction. \(\textbf{Final answer:}\) \[ \boxed{r_{f,2}=r_{r,2}=2.70\,\mathrm{mol\,L^{-1}\,s^{-1}}} \]

Explanation: \(\textbf{Identify the effect of constant-pressure inert-gas addition:}\) Adding inert gas at constant total pressure increases the vessel volume. The reacting-gas mole amounts remain unchanged immediately during the operation. Since the volume doubles, their concentrations are halved. \[ [\mathrm{H_2}]_2=\frac12[\mathrm{H_2}]_1 \] \[ [\mathrm{I_2}]_2=\frac12[\mathrm{I_2}]_1 \] \[ [\mathrm{HI}]_2=\frac12[\mathrm{HI}]_1 \] \(\textbf{Determine the forward-rate change:}\) \[ r_{f,2}=k_f[\mathrm{H_2}]_2[\mathrm{I_2}]_2 \] Substituting, \[ r_{f,2} =k_f\left(\frac12[\mathrm{H_2}]_1\right) \left(\frac12[\mathrm{I_2}]_1\right) \] \[ r_{f,2}=\frac14r_{f,1} \] \(\textbf{Determine the reverse-rate change:}\) \[ r_{r,2}=k_r[\mathrm{HI}]_2^2 \] \[ r_{r,2}=k_r\left(\frac12[\mathrm{HI}]_1\right)^2 \] \[ r_{r,2}=\frac14r_{r,1} \] \(\textbf{Examine the equilibrium quotient:}\) For the reaction as written, \[ Q_c=\frac{[\mathrm{HI}]^2} {[\mathrm{H_2}][\mathrm{I_2}]} \] Every concentration is halved, so \[ Q_{c,2} = \frac{(\frac12[\mathrm{HI}]_1)^2} {(\frac12[\mathrm{H_2}]_1)(\frac12[\mathrm{I_2}]_1)} \] \[ Q_{c,2}=Q_{c,1} \] The quotient remains equal to the equilibrium constant. \(\textbf{Interpret the unusual kinetic result:}\) Both opposing reaction rates decrease by the same factor. Their equality is preserved, so no net chemical reaction is required. The reacting-species mole amounts remain unchanged even though their concentrations and partial pressures decrease. The equilibrium constant also remains unchanged because temperature is fixed. This illustrates that reducing the absolute rates of two opposing elementary processes does not necessarily disturb their kinetic equilibrium. The inert gas changes the volume and concentrations but not the equilibrium mole composition for this zero-\(\Delta n_g\) reaction. \(\textbf{Final answer:}\) \[ \frac{r_{f,2}}{r_{f,1}}= \frac{r_{r,2}}{r_{r,1}}=\frac14 \]

Explanation: \(\textbf{Identify the mathematical transformation:}\) The rewritten equation is the reverse of the original reaction with every coefficient multiplied by one-half. Reversing an equation reciprocates its equilibrium constant. Multiplying all coefficients by one-half then raises that reciprocal constant to the power one-half. \(\textbf{Construct the transformed constant:}\) \[ K'=\left(\frac1K\right)^{1/2} \] \(\textbf{Evaluate the lower-temperature constant:}\) \[ K'_1=\left(\frac1{0.25}\right)^{1/2} \] \[ K'_1=\sqrt4 \] \[ K'_1=2.00 \] \(\textbf{Evaluate the higher-temperature constant:}\) \[ K'_2=\left(\frac1{1.00}\right)^{1/2} \] \[ K'_2=1.00 \] \(\textbf{Infer the reaction enthalpy sign:}\) The original reaction's equilibrium constant increases from \(0.25\) to \(1.00\) as temperature rises. This behaviour corresponds to an endothermic original forward reaction. The rewritten reaction proceeds in the reverse chemical direction. Reversing a reaction reverses the sign of its enthalpy change. Therefore, the rewritten forward reaction is exothermic. \(\textbf{Verify using the temperature trend:}\) For an exothermic forward reaction, increasing temperature decreases the equilibrium constant. The transformed constant decreases from two to one as temperature rises. This agrees with the exothermic classification. \(\textbf{Interpret the coefficient change:}\) Changing the written stoichiometric coefficients changes the numerical equilibrium constant. It does not change the underlying physical equilibrium. The effect of reversing the reaction must be combined with the effect of halving its coefficients. Taking only the reciprocal, without the square root, would produce an incorrect constant for the rewritten equation. Similarly, retaining the original enthalpy sign after reversing the chemical reaction would be inconsistent. \(\textbf{Final answer:}\) \[ \boxed{K'_1=2.00,\quad K'_2=1.00} \] The rewritten forward reaction is exothermic.

Explanation: \(\textbf{Separate the two initial pressure contributions:}\) The saturated water-vapour partial pressure is \[ p_{\mathrm{H_2O},1}=0.040\,\mathrm{bar} \] The remaining pressure is contributed by the non-reactive gas. \[ p_{\mathrm{gas},1}=P_1-p_{\mathrm{H_2O},1} \] \[ p_{\mathrm{gas},1}=1.000-0.040 \] \[ p_{\mathrm{gas},1}=0.960\,\mathrm{bar} \] \(\textbf{Determine the instantaneous expansion response:}\) The volume is doubled at constant temperature. Before appreciable evaporation occurs, the number of gas-phase molecules of each species remains unchanged. Each partial pressure is therefore halved. \[ p_{\mathrm{H_2O},\mathrm{immediate}}=\frac{0.040}{2} \] \[ p_{\mathrm{H_2O},\mathrm{immediate}}=0.020\,\mathrm{bar} \] For the non-reactive gas, \[ p_{\mathrm{gas},\mathrm{immediate}}=\frac{0.960}{2} \] \[ p_{\mathrm{gas},\mathrm{immediate}}=0.480\,\mathrm{bar} \] \(\textbf{Calculate the immediate total pressure:}\) \[ P_{\mathrm{immediate}}=0.020+0.480 \] \[ P_{\mathrm{immediate}}=0.500\,\mathrm{bar} \] \(\textbf{Identify the liquid-vapour response:}\) The water-vapour partial pressure immediately after expansion is below its saturated value. Because liquid water remains, additional water evaporates. Evaporation continues until the equilibrium water-vapour pressure is restored. Therefore, \[ p_{\mathrm{H_2O},\mathrm{final}}=0.040\,\mathrm{bar} \] \(\textbf{Determine the final pressure of the non-reactive gas:}\) The non-reactive gas does not dissolve or react under the stated conditions. Its mole amount and the final vessel volume remain unchanged after the instantaneous expansion. Therefore, \[ p_{\mathrm{gas},\mathrm{final}}=0.480\,\mathrm{bar} \] \(\textbf{Calculate the final total pressure:}\) \[ P_{\mathrm{final}}=0.480+0.040 \] \[ P_{\mathrm{final}}=0.520\,\mathrm{bar} \] \(\textbf{Interpret the two stages:}\) The instantaneous pressure drop is determined by mechanical expansion. The later pressure increase from \(0.500\) to \(0.520\,\mathrm{bar}\) results from net evaporation of water. The saturated vapour pressure is restored without restoring the original total pressure. The non-reactive gas remains diluted because its mole amount does not change. This comparison shows why the partial pressure of a saturated vapour must be distinguished from the total pressure of a gas mixture. \(\textbf{Final answer:}\) \[ \boxed{P_{\mathrm{immediate}}=0.500\,\mathrm{bar},\quad P_{\mathrm{final}}=0.520\,\mathrm{bar}} \]

Explanation: \(\textbf{Determine the original equilibrium constant:}\) For the reaction, \[ K_c=\frac{[C]}{[A][B]} \] At the original equilibrium, \[ K_c=\frac{2.00}{1.00\times2.00} \] \[ K_c=1.00\,\mathrm{L\,mol^{-1}} \] \(\textbf{Apply the selective product removal:}\) Half of C is removed. Therefore, \[ [C]_{\mathrm{after\ I}}=1.00\,\mathrm{M} \] The concentrations of A and B remain at their original values before any appreciable reaction occurs. \[ [A]_{\mathrm{after\ I}}=1.00\,\mathrm{M} \] \[ [B]_{\mathrm{after\ I}}=2.00\,\mathrm{M} \] \(\textbf{Apply the reactant addition:}\) Operation II increases the concentration of B by \(1.00\,\mathrm{M}\). Thus, \[ [B]_{\mathrm{after\ II}}=2.00+1.00 \] \[ [B]_{\mathrm{after\ II}}=3.00\,\mathrm{M} \] The other instantaneous concentrations remain unchanged. \(\textbf{Evaluate the final reaction quotient:}\) \[ Q_c=\frac{1.00}{1.00\times3.00} \] \[ Q_c=\frac13\,\mathrm{L\,mol^{-1}} \] \(\textbf{Compare with the equilibrium constant:}\) \[ Q_c\lt K_c \] The forward reaction is favoured. \(\textbf{Interpret the combined disturbances:}\) Removing product decreases the numerator of the reaction quotient. Adding B increases the denominator. Both operations therefore favour the forward reaction in this particular sequence. Their effects reinforce one another rather than compensate. The equilibrium constant remains unchanged because temperature remains constant. The final instantaneous direction is determined by the composition after both operations, not by assigning a separate permanent shift to each operation. \(\textbf{Final answer:}\) \[ \boxed{Q_c=\frac13\,\mathrm{L\,mol^{-1}}} \] The forward reaction is initially favoured.

Explanation: \(\textbf{Construct the heterogeneous pressure equilibrium expression:}\) The activity of pure solid carbon remains constant and does not appear in the conventional pressure quotient. For the reaction as written, \[ K_p=\frac{p_{\mathrm{CO}}^2}{p_{\mathrm{CO_2}}} \] \(\textbf{Calculate the original equilibrium constant:}\) \[ K_p=\frac{(1.00)^2}{0.50} \] \[ K_p=2.00\,\mathrm{bar} \] \(\textbf{Apply the instantaneous expansion:}\) At constant temperature, doubling the volume halves the partial pressures of the gases before any appreciable reaction occurs. \[ p_{\mathrm{CO_2},2}=\frac{0.50}{2} \] \[ p_{\mathrm{CO_2},2}=0.25\,\mathrm{bar} \] \[ p_{\mathrm{CO},2}=\frac{1.00}{2} \] \[ p_{\mathrm{CO},2}=0.50\,\mathrm{bar} \] \(\textbf{Determine the new reaction quotient:}\) \[ Q_{p,2}=\frac{(0.50)^2}{0.25} \] \[ Q_{p,2}=\frac{0.25}{0.25} \] \[ Q_{p,2}=1.00\,\mathrm{bar} \] \(\textbf{Compare with the unchanged equilibrium constant:}\) \[ Q_{p,2}\lt K_p \] The forward reaction is therefore favoured. \(\textbf{Interpret the gaseous mole-number change:}\) The forward reaction consumes one gaseous molecule of carbon dioxide and produces two gaseous molecules of carbon monoxide. Pure solid carbon is excluded from the gaseous mole count. Thus, \[ \Delta n_g=2-1=+1 \] Expansion favours the side containing more gaseous molecules. The pressure-quotient calculation is consistent with this qualitative prediction. \(\textbf{Recognise the role of the solid phase:}\) The solid carbon must remain present for the forward reaction to proceed. Its quantity does not appear explicitly in the pressure equilibrium expression while it remains a pure solid phase. The equilibrium constant itself is unchanged because the experiment is isothermal. The system subsequently produces additional CO until the pressure quotient returns to the equilibrium value. \(\textbf{Final answer:}\) \[ \boxed{Q_{p,\mathrm{immediate}}=1.00\,\mathrm{bar}} \] The forward reaction is favoured.

Explanation: \(\textbf{Infer the thermal nature from equilibrium-constant trends:}\) For a reversible reaction, increasing temperature favours the endothermic direction. When the equilibrium constant for the forward reaction increases upon heating, the forward direction is endothermic. The problem states that the equilibrium constants of both P and Q increase with temperature. Therefore, both forward reactions are endothermic. \(\textbf{Evaluate the gas-mole change for P:}\) For reaction P, \[ A(g)\rightleftharpoons2B(g) \] the gaseous stoichiometric mole-number change is \[ \Delta n_{g,P}=2-1 \] \[ \Delta n_{g,P}=+1 \] The forward reaction produces more gaseous molecules. An isothermal compression therefore favours the reverse direction. \(\textbf{Evaluate the gas-mole change for Q:}\) For reaction Q, \[ 2C(g)\rightleftharpoons D(g) \] the gaseous mole-number change is \[ \Delta n_{g,Q}=1-2 \] \[ \Delta n_{g,Q}=-1 \] The forward reaction produces fewer gaseous molecules. Compression therefore favours the forward direction. \(\textbf{Distinguish pressure effects from temperature effects:}\) The thermal character of the forward reaction is inferred from the experimental temperature dependence of K. The pressure response is inferred independently from the balanced gaseous stoichiometry. Both reactions can be endothermic while responding oppositely to compression. This is because enthalpy change and gaseous mole-number change describe different properties of the reaction. An isothermal compression changes the instantaneous quotient but not K. An increase in temperature changes K and favours the endothermic direction. Therefore, the correct comparison is that both forward reactions are endothermic, compression opposes P and compression favours Q.

Explanation: For the balanced reaction, \[ Q_p=\frac{p_{\mathrm{SO_3}}^2}{p_{\mathrm{SO_2}}^2p_{\mathrm{O_2}}} \] Operation P increases the partial pressure of oxygen, a reactant. This increases the denominator of the reaction quotient. Therefore, the quotient decreases and the forward reaction is favoured. P matches effect 2. Operation Q removes sulphur trioxide, a product. The numerator of the quotient decreases. Therefore, the quotient again falls below the equilibrium constant and the forward reaction is favoured. Q also matches effect 2. A matching question can legitimately assign more than one entry to the same effect if the listed relationships are not one-to-one. Operation R decreases the vessel volume at constant temperature. For the balanced reaction, \[ \Delta n_g=2-(2+1)=-1 \] Uniform compression decreases Q relative to K because the reactant side contains more gaseous stoichiometric moles. The forward reaction is favoured. R matches effect 4. Operation S introduces a catalyst into a mixture already at equilibrium. A catalyst increases the rates of both opposing processes. It does not change the equilibrium constant or equilibrium composition at fixed temperature. S matches effect 3. However, List II contains effect 1, which is not required by any operation. The correct relationships are therefore P–2, Q–2, R–4 and S–3. The important conceptual distinction is that different operations may produce the same direction of equilibrium response through different changes in the reaction quotient. The matching must be based on chemistry rather than an assumed requirement that every effect be used exactly once.

Explanation: The graph contains two physically distinct stages. The discontinuity at \(t_1\) represents an imposed change that occurs before appreciable chemical reaction. The later gradual variations represent re-equilibration. Only the concentration curve of A jumps instantaneously. Therefore, the operation must directly affect A rather than B or C. An isothermal compression would change the concentrations of all gaseous species instantaneously. It is inconsistent with the graph because only A has a discontinuity. Selective removal of C would cause an instantaneous downward jump in the C curve. That is also inconsistent with the observation. Selective addition of B would produce an instantaneous change in B rather than A. Thus, selective addition of A is the operation consistent with the graph. For the reaction, \[ A(g)+B(g)\rightleftharpoons C(g) \] adding A decreases the reaction quotient at fixed temperature. The forward reaction is consequently favoured. During re-equilibration, A and B are consumed while C is formed. This agrees with the subsequent directions of all three curves. Because A is consumed during this adjustment, its final concentration must be smaller than its value immediately after the addition. However, its final concentration need not be smaller than its original equilibrium concentration before A was added. The graph therefore distinguishes the imposed instantaneous concentration change from the compensating chemical response. The final plateau represents a new dynamic equilibrium rather than a return of every species to its original concentration. Only option A simultaneously explains the discontinuity, the subsequent reaction direction and the final comparison requested.

Explanation: \(\textbf{Construct the original equilibrium relationship:}\) For the reaction, \[ K_{c,1}=\frac{[C]_1}{[A]_1[B]_1} \] The mixture is initially at equilibrium, so its original reaction quotient equals \(K_{c,1}\). \(\textbf{Determine the quotient after the concentration change:}\) Every gaseous concentration is doubled instantaneously. Thus, \[ Q_{c,2} = \frac{2[C]_1}{(2[A]_1)(2[B]_1)} \] \[ Q_{c,2}=\frac12 \frac{[C]_1}{[A]_1[B]_1} \] Therefore, \[ Q_{c,2}=\frac12K_{c,1} \] \(\textbf{Apply the new-temperature constant:}\) The problem states \[ K_{c,2}=\frac14K_{c,1} \] \(\textbf{Compare the disturbed quotient with equilibrium:}\) \[ \frac{Q_{c,2}}{K_{c,2}} = \frac{\frac12K_{c,1}}{\frac14K_{c,1}} \] \[ \frac{Q_{c,2}}{K_{c,2}}=2 \] Because this ratio is greater than one, the reverse reaction is favoured. \(\textbf{Interpret the combined changes:}\) Doubling all concentrations decreases the quotient for this reaction because the denominator contains two concentration factors, while the numerator contains one. If temperature had remained unchanged, this concentration disturbance would have favoured the forward reaction. However, the new temperature reduces the equilibrium constant by an even larger factor. Therefore, the disturbed quotient now lies above the new equilibrium constant. The reverse tendency is the result of the combined operation. Knowing only that concentrations increase is insufficient to predict the outcome when temperature also changes. Similarly, knowing only that the equilibrium constant decreases would not establish the direction without considering the quotient. The supplied mole amounts are not individually needed because the original equilibrium condition allows the concentration factors to cancel. \(\textbf{Final answer:}\) \[ \boxed{\frac{Q_{c,2}}{K_{c,2}}=2} \] The reverse reaction is immediately favoured.

Explanation: Statement I is correct because the equilibrium constant for a defined reaction is determined by temperature under the stated conditions. A catalyst modifies the reaction pathway but does not alter the equilibrium constant. Statement II is incorrect because an increase in both opposing rates does not require a shift in equilibrium composition. At dynamic equilibrium, the opposing rates are equal. If catalysis increases both rates consistently with the unchanged equilibrium constant, their equality is preserved at the equilibrium composition. Thus, faster microscopic transformations can coexist with unchanged macroscopic composition. Statement III is correct. A catalyst introduced into an already equilibrated system can increase the rates of the forward and reverse reactions. The common equilibrium rate can therefore become greater than before catalyst addition. Nevertheless, the net reaction rate remains zero because the opposing rates remain equal. Statement IV is also correct under the stated conditions. When a catalyst is added to a nonequilibrium mixture, it can shorten the time required to reach equilibrium. The final equilibrium composition is still governed by the same equilibrium constant and material-conservation constraints. Therefore, the catalyst does not create a new thermodynamic equilibrium state merely by accelerating the reaction. This distinction explains why catalysts are useful in industrial equilibria even when they do not increase equilibrium yields. Temperature, pressure and composition changes must be distinguished from catalytic changes. A catalyst acts primarily through reaction kinetics rather than a change in equilibrium thermodynamics. Thus, Statements I, III and IV are correct.

Explanation: \(\textbf{Identify the original equilibrium quotient:}\) The original mixture is at equilibrium. Therefore, \[ Q_1=K_1=4.0 \] \(\textbf{Determine the pressure scaling from compression:}\) The gaseous reaction converts one molecule of A into two molecules of B. Therefore, \[ \Delta n_g=2-1=+1 \] A uniform increase in reacting-gas partial pressures by a factor f multiplies the quotient by \(f^{\Delta n_g}\). \(\textbf{Apply the specified volume change:}\) The vessel volume decreases to one-third. Immediately after compression, the mole amounts are unchanged, so the gas concentrations increase by a factor of three. For an ideal gas at the new temperature, the pressure activities also contain the temperature-dependent factor. However, because the final temperature differs from the initial temperature, the partial pressures are multiplied by the factor \[ f=\frac{T_2}{T_1}\times3 \] rather than necessarily by three. \(\textbf{Recognise the missing temperature ratio:}\) The problem supplies the equilibrium constants \(K_1=4.0\) and \(K_2=12.0\), but it does not provide the actual values or ratio of \(T_2\) and \(T_1\). For the pressure-activity quotient, \[ Q_2=Q_1f \] Therefore, \[ Q_2=4.0\left(3\frac{T_2}{T_1}\right) \] \[ Q_2=12.0\frac{T_2}{T_1} \] Because \(T_2\gt T_1\), \[ Q_2\gt 12.0=K_2 \] \(\textbf{Determine the actual reaction direction:}\) The reverse reaction is favoured. Heating increases the equilibrium constant, but the combined heating and compression increase the pressure quotient by an even greater factor under the specified volume change. The temperature ratio need not be numerically supplied because its inequality is sufficient. The correct direction is therefore reverse. \(\textbf{Physical interpretation:}\) When temperature changes in a rigid or compressed ideal-gas system, pressures change through both the volume ratio and the absolute-temperature ratio. Using only the volume factor while ignoring thermal pressure changes is incorrect when working with pressure activities. The analysis illustrates why simultaneous temperature and volume changes require careful definition of the instantaneous quotient. \(\textbf{Final answer:}\) \[ \boxed{\frac{Q_2}{K_2}=\frac{T_2}{T_1}\gt 1} \] The reverse reaction is immediately favoured.

Explanation: The forward conversion of sulphur dioxide and oxygen into sulphur trioxide is exothermic. It also decreases the number of gaseous stoichiometric moles. These properties govern the response to pressure and temperature. Statement I is correct because increasing pressure at constant temperature favours the side containing fewer gaseous molecules. The reactant side contains three gaseous stoichiometric moles, while the product side contains two. Therefore, compression favours sulphur trioxide formation. Statement II is correct when the stated kinetic qualification is retained. Lower temperature favours the exothermic forward direction and increases the corresponding equilibrium constant. However, lowering temperature can also slow the rate of reaction. A suitable catalyst can improve the reaction rate without changing the equilibrium constant at that temperature. Thus, temperature selection and catalysis can be combined to obtain a more favourable equilibrium with a practical rate of approach. Statement III is correct because removal of sulphur trioxide reduces the product contribution to the reaction quotient. The system responds by favouring further forward reaction when reactants remain available. Continuous product removal can consequently maintain a forward driving force. Statement IV is incorrect because heating an exothermic equilibrium decreases the equilibrium constant for the forward reaction. Higher temperature may improve reaction speed, but it does not improve the equilibrium product favourability for this exothermic direction. The proposed changes must be distinguished according to whether they modify thermodynamic equilibrium, instantaneous composition or reaction kinetics. Pressure and product removal change the reacting-system quotient at fixed temperature. Temperature changes the equilibrium constant. Catalysis accelerates the opposing processes without changing equilibrium favourability. Therefore, changes I, II and III are consistent with greater net sulphur trioxide formation.

Explanation: \(\textbf{Establish the original pressure equilibrium relationship:}\) For the reaction as written, \[ K_p=\frac{p_C}{p_Ap_B} \] The original mixture is at equilibrium. Therefore, \[ K_p=\frac{2.00}{2.00\times1.00} \] \[ K_p=1.00\,\mathrm{bar^{-1}} \] \(\textbf{Analyse the selective removal of A:}\) Half of A is removed at constant temperature and volume. Its partial pressure is therefore halved. The other gaseous partial pressures remain unchanged immediately. \[ p_{A,1}=1.00\,\mathrm{bar} \] \[ p_{B,1}=1.00\,\mathrm{bar} \] \[ p_{C,1}=2.00\,\mathrm{bar} \] The reaction quotient becomes \[ Q_{p,1}=\frac{2.00}{1.00\times1.00} \] \[ Q_{p,1}=2.00\,\mathrm{bar^{-1}} \] At this intermediate stage, the reverse reaction is favoured. \(\textbf{Apply the subsequent compression:}\) The volume is reduced to half before appreciable reaction occurs. At unchanged temperature, each existing partial pressure doubles. Thus, \[ p_{A,2}=2.00\,\mathrm{bar} \] \[ p_{B,2}=2.00\,\mathrm{bar} \] \[ p_{C,2}=4.00\,\mathrm{bar} \] \(\textbf{Evaluate the final instantaneous quotient:}\) \[ Q_{p,2}=\frac{4.00}{2.00\times2.00} \] \[ Q_{p,2}=1.00\,\mathrm{bar^{-1}} \] Therefore, \[ Q_{p,2}=K_p \] \(\textbf{Interpret the combined operations:}\) Removing A initially creates a reverse reaction tendency. Compression subsequently decreases the quotient because the reaction converts two gaseous stoichiometric moles into one. In this experiment, compression exactly compensates for the earlier removal. The mixture therefore satisfies the equilibrium condition immediately after both operations. Its individual partial pressures differ from those at the original equilibrium, but equilibrium does not require the same absolute partial pressures in every possible equilibrium state. The temperature remains constant, so the equilibrium constant is unchanged. The result illustrates why the final reaction tendency must be determined from the complete sequence rather than from either disturbance considered separately. \(\textbf{Final answer:}\) \[ \boxed{Q_{p,2}=K_p=1.00\,\mathrm{bar^{-1}}} \]

Explanation: \(\textbf{Write the original equilibrium condition:}\) For the reaction, \[ K_p=\frac{p_B^2}{p_A} \] The original equilibrium mixture satisfies \[ Q_{p,0}=K_p \] \(\textbf{Determine the effect of adding B:}\) The partial pressure of B increases by a factor of \(1.50\). The partial pressure of A is initially unchanged. Therefore, \[ Q_{p,1}=\frac{(1.50p_{B,0})^2}{p_{A,0}} \] \[ Q_{p,1}=(1.50)^2Q_{p,0} \] \[ Q_{p,1}=2.25K_p \] The mixture is initially reverse-favoured after the addition because its quotient exceeds K. \(\textbf{Express the effect of expansion:}\) Let the required volume multiplication factor be f. At constant temperature and unchanged gaseous mole amounts during expansion, every partial pressure is divided by f. Thus, \[ p_{A,2}=\frac{p_{A,1}}f \] \[ p_{B,2}=\frac{p_{B,1}}f \] The new quotient becomes \[ Q_{p,2}=\frac{(p_{B,1}/f)^2}{p_{A,1}/f} \] Simplifying, \[ Q_{p,2}=\frac{Q_{p,1}}f \] \(\textbf{Impose the required equilibrium condition:}\) For the mixture to return exactly to equilibrium, \[ Q_{p,2}=K_p \] Therefore, \[ \frac{2.25K_p}{f}=K_p \] Cancelling the positive equilibrium constant, \[ f=2.25 \] \(\textbf{Check the physical direction:}\) The forward reaction produces more gaseous molecules than it consumes. Expansion therefore favours the forward direction. This opposes the reverse tendency created by adding B. The volume increase calculated here produces exactly the required change in the pressure quotient. A smaller expansion would leave the quotient above K, whereas a larger expansion would lower it below K. The calculation does not require the numerical equilibrium constant or original partial pressures because their common factors cancel. \(\textbf{Final answer:}\) \[ \boxed{\frac{V_{\mathrm{final}}}{V_{\mathrm{after\ addition}}}=2.25} \]

Explanation: \(\textbf{Establish the initial equilibrium quotient:}\) The original mixture is at equilibrium. Therefore, \[ Q_1=K_1=9.0 \] \(\textbf{Determine the instantaneous pressure multiplication:}\) For an ideal gas, \[ p_i=\frac{n_iRT}{V} \] The temperature doubles. The vessel volume is halved. Because the gaseous mole amounts do not change during the rapid operation, every partial pressure increases by \[ f=\frac{T_2}{T_1}\frac{V_1}{V_2} \] \[ f=2\times2 \] \[ f=4 \] \(\textbf{Construct the disturbed pressure quotient:}\) For the reaction as written, \[ Q=\frac{a_B^2}{a_A} \] Every gaseous activity is multiplied by four under the unchanged standard-pressure convention. Therefore, \[ Q_2=\frac{(4a_{B,1})^2}{4a_{A,1}} \] \[ Q_2=4Q_1 \] \[ Q_2=4(9.0) \] \[ Q_2=36.0 \] \(\textbf{Compare with the new equilibrium constant:}\) The equilibrium constant at the higher temperature is \[ K_2=12.0 \] Thus, \[ \frac{Q_2}{K_2}=\frac{36.0}{12.0} \] \[ \frac{Q_2}{K_2}=3.0 \] Because \(Q_2\gt K_2\), the reverse reaction is favoured. \(\textbf{Interpret the competing disturbances:}\) Heating favours the endothermic forward direction and increases K. Compression favours the reverse direction because the reverse reaction has fewer gaseous molecules. The effects oppose one another, but their magnitudes are not equal. The actual pressure increase includes both the volume reduction and the temperature rise. Using only the volume factor would underestimate the new quotient. The calculated quotient exceeds the new equilibrium constant, so the reverse pressure-related tendency dominates immediately after the combined operation. \(\textbf{Final answer:}\) \[ \boxed{\frac{Q_2}{K_2}=3.0} \] The reverse reaction is immediately favoured.

Explanation: \(\textbf{Identify the elementary kinetic relationships:}\) For an elementary reversible first-order reaction, \[ r_f=k_f[A] \] \[ r_r=k_r[B] \] At the original equilibrium, \[ r_{f,0}=r_{r,0}=3.0\,\mathrm{mmol\,L^{-1}\,s^{-1}} \] \(\textbf{Determine the catalyst-induced rate changes:}\) The catalyst doubles both rate constants. The composition is initially unchanged. Therefore, \[ r_{f,1}=2r_{f,0} \] \[ r_{f,1}=6.0\,\mathrm{mmol\,L^{-1}\,s^{-1}} \] Similarly, \[ r_{r,1}=2r_{r,0} \] \[ r_{r,1}=6.0\,\mathrm{mmol\,L^{-1}\,s^{-1}} \] The forward and reverse rates remain equal after catalyst addition. \(\textbf{Analyse the subsequent selective addition:}\) The concentration of A is suddenly tripled. The catalyst remains present, so the catalysed rate constants continue to apply. Because the forward elementary rate is directly proportional to [A], \[ r_{f,2}=3r_{f,1} \] \[ r_{f,2}=3(6.0) \] \[ r_{f,2}=18.0\,\mathrm{mmol\,L^{-1}\,s^{-1}} \] B is not changed instantaneously. Therefore, \[ r_{r,2}=6.0\,\mathrm{mmol\,L^{-1}\,s^{-1}} \] \(\textbf{Determine the net kinetic tendency:}\) \[ r_{\mathrm{net},2}=r_{f,2}-r_{r,2} \] \[ r_{\mathrm{net},2}=18.0-6.0 \] \[ r_{\mathrm{net},2}=+12.0\,\mathrm{mmol\,L^{-1}\,s^{-1}} \] The positive net rate indicates forward reaction. \(\textbf{Distinguish the two operations:}\) Adding the catalyst accelerates both opposing processes while retaining their equality at equilibrium. Adding A changes the reaction quotient and produces an imbalance between the opposing rates. The catalyst does not cause the forward shift by itself. The system subsequently consumes some A and forms B until the opposing rates again become equal. \(\textbf{Final answer:}\) \[ (r_f,r_r)_{\mathrm{II}}=(6,6),\quad(r_f,r_r)_{\mathrm{III}}=(18,6)\,\mathrm{mmol\,L^{-1}\,s^{-1}} \] The immediate net reaction in Stage III is forward.