Equilibrium MCQs with Answers – Part 6 (NEET Chemistry)

Explanation: \(\textbf{Convert the target pH into an equilibrium composition condition:}\) The acidic-buffer relation is \[ pH=pK_a+\log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} \] The target is \[ pH=pK_a \] Therefore, \[ \log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}=0 \] Hence, \[ \frac{[\mathrm{A^-}]}{[\mathrm{HA}]}=1 \] The final solution must contain equal amounts of HA and A⁻. \(\textbf{Express the contribution from P:}\) If P contributes volume \(V_P\) in litres, \[ n_{\mathrm{HA},P}=0.200V_P \] \[ n_{\mathrm{A^-},P}=0.100V_P \] Thus, each litre of P contributes an excess of \[ 0.200-0.100 \] \[ =0.100\,\mathrm{mol} \] of HA over A⁻. \(\textbf{Express the contribution from Q:}\) If Q contributes volume \(V_Q\) in litres, \[ n_{\mathrm{HA},Q}=0.100V_Q \] \[ n_{\mathrm{A^-},Q}=0.300V_Q \] Each litre of Q contributes an excess of \[ 0.300-0.100 \] \[ =0.200\,\mathrm{mol} \] of A⁻ over HA. \(\textbf{Apply equality of the final conjugate-component amounts:}\) The total HA amount is \[ n_{\mathrm{HA,total}}=0.200V_P+0.100V_Q \] The total conjugate-base amount is \[ n_{\mathrm{A^-,total}}=0.100V_P+0.300V_Q \] At the target pH, \[ 0.200V_P+0.100V_Q = 0.100V_P+0.300V_Q \] Rearranging, \[ 0.100V_P=0.200V_Q \] Therefore, \[ \frac{V_P}{V_Q}=2 \] \[ V_P:V_Q=2:1 \] \(\textbf{Verify the target composition:}\) Choose, for example, \[ V_P=2.00\,\mathrm{L} \] \[ V_Q=1.00\,\mathrm{L} \] The total HA amount is \[ n_{\mathrm{HA,total}}=0.200(2.00)+0.100(1.00) \] \[ n_{\mathrm{HA,total}}=0.500\,\mathrm{mol} \] The total A⁻ amount is \[ n_{\mathrm{A^-,total}}=0.100(2.00)+0.300(1.00) \] \[ n_{\mathrm{A^-,total}}=0.500\,\mathrm{mol} \] Their ratio is one, as required. \(\textbf{Interpret the reverse mixture design:}\) The target pH fixes the desired final conjugate-component ratio. The two available buffers have different initial ratios and concentrations. Therefore, the required mixing volumes must be determined from the amounts of each chemical component, not from the arithmetic average of the starting pH values. The acid excess contributed by P must compensate for the conjugate-base excess contributed by Q. This provides a direct design method for obtaining a specified pH from two buffers containing the same conjugate pair. \(\textbf{Final answer:}\) \[ \boxed{V_P:V_Q=2:1} \]

Explanation: Statement I is correct. The target pH satisfies \[ pH=pK_a+\log_{10}3 \] The buffer equation therefore requires \[ \frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}}=3 \] Let x moles of HA be neutralized by NaOH: \[ \mathrm{HA+OH^-\rightarrow A^-+H_2O} \] The resulting conjugate-base amount is x. The remaining HA amount is \[ 0.200-x \] Thus, \[ \frac{x}{0.200-x}=3 \] Cross-multiplying, \[ x=0.600-3x \] \[ 4x=0.600 \] \[ x=0.150\,\mathrm{mol} \] The remaining HA amount is \[ 0.200-0.150=0.0500\,\mathrm{mol} \] Statement II is correct under the stipulated buffer approximation. Dilution lowers both conjugate-component concentrations by the same volume factor. Consequently, the concentration ratio remains approximately \[ \frac{[\mathrm{A^-}]}{[\mathrm{HA}]}=3 \] The pH therefore remains close to \[ pK_a+\log_{10}3 \] Statement III is incorrect. Dilution decreases the amounts of HA and A⁻ contained in a fixed final volume of solution. Although the buffer pH may remain approximately unchanged, the amount of strong acid that a fixed aliquot can absorb before exhausting A⁻ decreases. The total amount of A⁻ in the entire diluted batch remains approximately unchanged, but its amount per unit volume decreases. These two comparisons must not be confused. Statement IV is correct. Moderate dilution does not remove HA or A⁻ from the solution. Both conjugate components remain present and can continue to undergo the relevant acid-base reactions. The solution is therefore not equivalent to pure water merely because its buffer-component concentrations are lower. The example distinguishes pH, component concentration, total chemical inventory and buffering capacity per unit volume. Statements I, II and IV are correct.

Explanation: \(\textbf{Identify the relevant successive deprotonations:}\) The triprotic acid undergoes \[ \mathrm{H_3A\rightleftharpoons H^++H_2A^-} \] \[ \mathrm{H_2A^-\rightleftharpoons H^++HA^{2-}} \] \[ \mathrm{HA^{2-}\rightleftharpoons H^++A^{3-}} \] The starting salts supply H₂A⁻ and HA²⁻. Added hydroxide ions can remove the remaining transferable protons stepwise. \(\textbf{Determine the starting ionic inventories:}\) \[ n_{\mathrm{H_2A^-},0}=0.120\,\mathrm{mol} \] \[ n_{\mathrm{HA^{2-}},0}=0.0200\,\mathrm{mol} \] The added hydroxide amount is \[ n_{\mathrm{OH^-}}=0.150\,\mathrm{mol} \] \(\textbf{Complete the first relevant neutralization stage:}\) Hydroxide converts H₂A⁻ into HA²⁻: \[ \mathrm{H_2A^-+OH^-\rightarrow HA^{2-}+H_2O} \] Consuming the entire initial H₂A⁻ inventory requires \[ n_{\mathrm{OH^-,stage\,1}}=0.120\,\mathrm{mol} \] The hydroxide amount remaining is \[ n_{\mathrm{OH^-,remaining}}=0.150-0.120 \] \[ n_{\mathrm{OH^-,remaining}}=0.0300\,\mathrm{mol} \] \(\textbf{Determine the intermediate HA²⁻ inventory:}\) After complete conversion of the initial H₂A⁻, \[ n_{\mathrm{HA^{2-}}}=0.0200+0.120 \] \[ n_{\mathrm{HA^{2-}}}=0.140\,\mathrm{mol} \] \(\textbf{Apply the next deprotonation stage:}\) The remaining hydroxide reacts through \[ \mathrm{HA^{2-}+OH^-\rightarrow A^{3-}+H_2O} \] Therefore, \[ n_{\mathrm{A^{3-}},f}=0.0300\,\mathrm{mol} \] The remaining HA²⁻ amount is \[ n_{\mathrm{HA^{2-}},f}=0.140-0.0300 \] \[ n_{\mathrm{HA^{2-}},f}=0.110\,\mathrm{mol} \] \(\textbf{Identify the final conjugate pair:}\) The principal final components are HA²⁻ and A³⁻. Their equilibrium is governed by the third acid-ionization constant: \[ \mathrm{HA^{2-}\rightleftharpoons H^++A^{3-}} \] Thus, the relevant pKa is \[ pK_{a3}=10.000 \] \(\textbf{Calculate the final pH:}\) \[ pH=pK_{a3}+\log_{10} \frac{n_{\mathrm{A^{3-}}}}{n_{\mathrm{HA^{2-}}}} \] Substituting, \[ pH=10.000+\log_{10}\frac{0.0300}{0.110} \] \[ pH=10.000+\log_{10}\frac3{11} \] \[ pH=10.000-\log_{10}\frac{11}{3} \] \[ pH=10.000-0.5643 \] \[ pH=9.4357 \] \[ pH\approx9.436 \] \(\textbf{Check the material inventory:}\) The original acid-derived amount is \[ 0.120+0.0200=0.140\,\mathrm{mol} \] The final amount is \[ 0.110+0.0300=0.140\,\mathrm{mol} \] Thus, all acid-derived formula units are accounted for. \(\textbf{Interpret the buffer transition:}\) The initial solution is based on the H₂A⁻/HA²⁻ pair. However, the amount of added strong base exceeds that required to convert all H₂A⁻ into HA²⁻. The remaining hydroxide begins converting HA²⁻ into A³⁻. Consequently, the dominant final buffer pair is governed by Ka₃ rather than Ka₂. A calculation that continues using the original conjugate pair after its acidic component has been exhausted would be invalid. The problem requires identifying the correct final ionization stage before applying the buffer equation. \(\textbf{Final answer:}\) \[ \boxed{\frac{n_{\mathrm{A^{3-}}}}{n_{\mathrm{HA^{2-}}}}=\frac3{11},\qquad pH\approx9.436} \]

Explanation: \(\textbf{Determine the initial conjugate-component amounts in P:}\) Because the initial pH equals pKa, \[ \frac{[\mathrm{A^-}]}{[\mathrm{HA}]}=1 \] The total analytical conjugate-pair concentration in P is \(0.400\,\mathrm{M}\). Therefore, each component has concentration \[ [\mathrm{HA}]_P=[\mathrm{A^-}]_P=0.200\,\mathrm{M} \] In \(0.1000\,\mathrm{L}\), \[ n_{\mathrm{HA},P}=n_{\mathrm{A^-},P} =(0.200)(0.1000) \] \[ n_{\mathrm{HA},P}=n_{\mathrm{A^-},P}=0.0200\,\mathrm{mol} \] \(\textbf{Determine the initial component amounts in Q:}\) The total analytical concentration in Q is \(0.100\,\mathrm{M}\). Thus, \[ [\mathrm{HA}]_Q=[\mathrm{A^-}]_Q=0.0500\,\mathrm{M} \] The amount of each component is \[ n_{\mathrm{HA},Q}=n_{\mathrm{A^-},Q} =(0.0500)(0.1000) \] \[ n_{\mathrm{HA},Q}=n_{\mathrm{A^-},Q}=0.00500\,\mathrm{mol} \] \(\textbf{Apply the HCl addition in P:}\) The reaction is \[ \mathrm{H^++A^-\rightarrow HA} \] After \(0.00500\,\mathrm{mol}\) HCl is added, \[ n_{\mathrm{A^-},P,f}=0.0200-0.00500 \] \[ n_{\mathrm{A^-},P,f}=0.0150\,\mathrm{mol} \] \[ n_{\mathrm{HA},P,f}=0.0200+0.00500 \] \[ n_{\mathrm{HA},P,f}=0.0250\,\mathrm{mol} \] \(\textbf{Calculate the final pH in P:}\) \[ pH_P=5.000+\log_{10}\frac{0.0150}{0.0250} \] \[ pH_P=5.000+\log_{10}(0.600) \] \[ pH_P\approx5.000-0.22185 \] \[ pH_P\approx4.778 \] \(\textbf{Apply the same HCl addition in Q:}\) Q initially contains only \(0.00500\,\mathrm{mol}\) conjugate base. The HCl amount is also \(0.00500\,\mathrm{mol}\). Therefore, the original analytical A⁻ inventory is stoichiometrically consumed: \[ n_{\mathrm{A^-},Q,f}\approx0 \] The weak-acid amount becomes \[ n_{\mathrm{HA},Q,f}=0.00500+0.00500 \] \[ n_{\mathrm{HA},Q,f}=0.0100\,\mathrm{mol} \] \(\textbf{Recognize the failure of the buffer approximation in Q:}\) The ordinary buffer equation cannot be applied using a zero conjugate-base amount. However, HA remains a weak acid and can ionize in water. The final analytical HA concentration is \[ C_{\mathrm{HA},Q}=\frac{0.0100}{0.1000} \] \[ C_{\mathrm{HA},Q}=0.100\,\mathrm{M} \] \(\textbf{Determine Q's actual weak-acid equilibrium pH:}\) Let x represent the concentration of HA ionized. Then, \[ K_a=\frac{x^2}{0.100-x} \] With small ionization, \[ x\approx\sqrt{K_aC} \] \[ x\approx\sqrt{(1.00\times10^{-5})(0.100)} \] \[ x\approx1.00\times10^{-3}\,\mathrm{M} \] Therefore, \[ pH_Q\approx3.000 \] \(\textbf{Interpret the competing calculations:}\) P retains appreciable amounts of both HA and A⁻ and remains an acidic buffer. Q exhausts the added-salt conjugate-base inventory. It no longer satisfies the usual buffer approximation and must instead be treated as an aqueous weak-acid solution. The equality between the added HCl amount and the initial A⁻ amount does not imply that the final solution is neutral. Weak-acid ionization remains possible after the stoichiometric conversion. The correct final pH is therefore close to 3.00, not 4.00. Option A reports both the buffered final pH of P and the weak-acid equilibrium pH of Q.

Explanation: \(\textbf{Recover the equilibrium constant from the original mixture:}\) For the reaction \[ \mathrm{A+B\rightleftharpoons2C} \] the concentration equilibrium constant is \[ K_c=\frac{[\mathrm C]^2}{[\mathrm A][\mathrm B]} \] Substituting the original equilibrium concentrations, \[ K_c=\frac{(1.20)^2}{(0.400)(0.400)} \] \[ K_c=\frac{1.44}{0.160} \] \[ K_c=9.00 \] \(\textbf{Identify the instantaneous composition after product removal:}\) The removal of \(0.600\,\mathrm{mol}\) C from \(1.00\,\mathrm{L}\) decreases its concentration by \(0.600\,\mathrm{M}\). Therefore, immediately after removal, \[ [\mathrm A]_0=0.400\,\mathrm{M} \] \[ [\mathrm B]_0=0.400\,\mathrm{M} \] \[ [\mathrm C]_0=1.20-0.600 \] \[ [\mathrm C]_0=0.600\,\mathrm{M} \] \(\textbf{Determine the direction of re-equilibration:}\) The instantaneous reaction quotient is \[ Q_c=\frac{(0.600)^2}{(0.400)(0.400)} \] \[ Q_c=\frac{0.360}{0.160} \] \[ Q_c=2.25 \] Because \[ Q_c\lt K_c \] the forward reaction is favoured. Additional C forms at the expense of A and B. \(\textbf{Express the stoichiometric concentration changes:}\) Let x molarity of A react during re-equilibration. The reaction consumes A and B equally and forms twice as much C. Therefore, \[ [\mathrm A]_{\mathrm{eq}}=0.400-x \] \[ [\mathrm B]_{\mathrm{eq}}=0.400-x \] \[ [\mathrm C]_{\mathrm{eq}}=0.600+2x \] \(\textbf{Apply the unchanged equilibrium constant:}\) \[ 9.00=\frac{(0.600+2x)^2}{(0.400-x)^2} \] Taking the positive square root, \[ 3.00=\frac{0.600+2x}{0.400-x} \] Therefore, \[ 0.600+2x=1.20-3x \] \[ 5x=0.600 \] \[ x=0.120\,\mathrm{M} \] \(\textbf{Recover the final concentrations:}\) \[ [\mathrm A]_{\mathrm{eq}}=0.400-0.120 \] \[ [\mathrm A]_{\mathrm{eq}}=0.280\,\mathrm{M} \] \[ [\mathrm C]_{\mathrm{eq}}=0.600+2(0.120) \] \[ [\mathrm C]_{\mathrm{eq}}=0.840\,\mathrm{M} \] \(\textbf{Verify the result:}\) The final B concentration also equals \(0.280\,\mathrm{M}\). Substituting, \[ Q_{c,\mathrm{final}} =\frac{(0.840)^2}{(0.280)(0.280)} \] \[ Q_{c,\mathrm{final}}=9.00 \] This equals the original Kc. \(\textbf{Interpret the incomplete replacement of removed product:}\) Product removal initially favours the forward reaction. However, equilibrium is restored before all the removed C has necessarily been replaced. The final C concentration is \(0.840\,\mathrm{M}\), which remains below its original value of \(1.20\,\mathrm{M}\). The equilibrium constant remains unchanged because the temperature is unchanged. The final composition is determined by the new material inventory and the same equilibrium condition. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm A]_{\mathrm{eq}}=0.280\,\mathrm{M},\qquad [\mathrm C]_{\mathrm{eq}}=0.840\,\mathrm{M}} \]

Explanation: \(\textbf{Determine the equilibrium constant:}\) The balanced equation is \[ \mathrm{2A\rightleftharpoons B+C} \] Therefore, \[ K_c=\frac{[\mathrm B][\mathrm C]}{[\mathrm A]^2} \] The verified equilibrium mixture gives \[ K_c=\frac{(0.400)(0.400)}{(0.200)^2} \] \[ K_c=\frac{0.160}{0.0400} \] \[ K_c=4.00 \] \(\textbf{Evaluate the reaction quotient for P:}\) \[ Q_{c,P}=\frac{(0.200)(0.200)}{(0.200)^2} \] \[ Q_{c,P}=1.00 \] Since \[ Q_{c,P}\lt K_c \] the forward reaction is favoured. A is consumed while B and C are formed. \(\textbf{Evaluate the quotient for Q:}\) \[ Q_{c,Q}=\frac{(0.200)(0.200)}{(0.100)^2} \] \[ Q_{c,Q}=4.00 \] Therefore, \[ Q_{c,Q}=K_c \] Q is already at equilibrium for the specified reaction. No net reaction advancement is favoured. \(\textbf{Evaluate the quotient for R:}\) \[ Q_{c,R}=\frac{(0.600)(0.600)}{(0.200)^2} \] \[ Q_{c,R}=9.00 \] Since \[ Q_{c,R}\gt K_c \] the reverse reaction is favoured. B and C are consumed while A is produced. \(\textbf{Interpret the composition comparison:}\) The same reaction has one equilibrium constant at the specified temperature. Nevertheless, different mixtures can lie on opposite sides of equilibrium. A mixture with smaller absolute concentrations is not automatically farther from equilibrium. Likewise, increasing product concentrations does not necessarily mean that the forward reaction is favoured. The decisive quantity is the complete reaction quotient with the correct stoichiometric powers. Mixture Q satisfies the equilibrium condition despite having concentrations different from those of the independently verified equilibrium mixture. This demonstrates that equilibrium composition need not be unique across systems with different conserved material inventories. The direction sequence is forward, equilibrium and reverse.

Explanation: \(\textbf{Construct the equilibrium expression:}\) For ammonia formation, \[ K_c=\frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3} \] The equilibrium expression contains two powers of product concentration and four powers of reactant concentrations. \(\textbf{Determine how uniform concentration scaling affects the quotient:}\) Suppose every concentration is multiplied by a common factor f. The new quotient is \[ Q_c'= \frac{(f[\mathrm{NH_3}])^2} {(f[\mathrm{N_2}])(f[\mathrm{H_2}])^3} \] Collecting powers of f, \[ Q_c'=\frac{f^2}{f^4}Q_c \] \[ Q_c'=f^{-2}Q_c \] The original mixture satisfies \[ Q_c=K_c \] \(\textbf{Evaluate mixture P:}\) For P, \[ f=2 \] Therefore, \[ Q_{c,P}=2^{-2}K_c \] \[ Q_{c,P}=\frac{K_c}{4} \] Hence, \[ Q_{c,P}\lt K_c \] The forward ammonia-forming reaction is favoured. \(\textbf{Evaluate mixture Q:}\) For Q, \[ f=\frac12 \] Therefore, \[ Q_{c,Q}=\left(\frac12\right)^{-2}K_c \] \[ Q_{c,Q}=4K_c \] Thus, \[ Q_{c,Q}\gt K_c \] The reverse reaction is favoured. \(\textbf{Check the physical interpretation:}\) Ammonia formation converts four moles of gaseous reactants into two moles of gaseous product according to the balanced equation. Higher concentrations favour the direction with fewer gaseous particles in this comparison. Lower concentrations favour the direction with more gaseous particles. Uniform scaling preserves simple concentration ratios such as \([\mathrm{NH_3}]/[\mathrm{N_2}]\), but it does not preserve the equilibrium quotient when the total powers in its numerator and denominator differ. The equilibrium constant remains unchanged because the temperature is unchanged. The different tendencies arise entirely from the compositions of the two new mixtures. The conclusion follows from reaction stoichiometry and the transformation of the complete equilibrium expression.

Explanation: \(\textbf{Construct the heterogeneous equilibrium expression:}\) The reaction is \[ \mathrm{NH_4HS(s)\rightleftharpoons NH_3(g)+H_2S(g)} \] The activity of the pure solid is unity. Therefore, under the stated partial-pressure convention, \[ K_p=p_{\mathrm{NH_3}}p_{\mathrm{H_2S}} \] \(\textbf{Calculate the original equilibrium constant:}\) \[ K_p=(0.400)(0.400) \] \[ K_p=0.160\,\mathrm{bar^2} \] \(\textbf{Identify the gas pressures immediately after NH₃ addition:}\) The original ammonia partial pressure is \(0.400\,\mathrm{bar}\). The added ammonia increases this by \(0.600\,\mathrm{bar}\). Thus, \[ p_{\mathrm{NH_3},0}=0.400+0.600 \] \[ p_{\mathrm{NH_3},0}=1.00\,\mathrm{bar} \] The H₂S partial pressure has not yet changed: \[ p_{\mathrm{H_2S},0}=0.400\,\mathrm{bar} \] \(\textbf{Evaluate the immediate reaction quotient:}\) \[ Q_p=(1.00)(0.400) \] \[ Q_p=0.400\,\mathrm{bar^2} \] Since \[ Q_p\gt K_p \] the reverse reaction is favoured. NH₃ and H₂S combine to form additional NH₄HS solid. \(\textbf{Represent the pressure changes:}\) At constant temperature and volume, gas partial pressures are proportional to their respective mole numbers. The reverse reaction consumes NH₃ and H₂S in a \(1:1\) ratio. Let their common pressure decrease be x. Then, \[ p_{\mathrm{NH_3,eq}}=1.00-x \] \[ p_{\mathrm{H_2S,eq}}=0.400-x \] \(\textbf{Apply the unchanged equilibrium condition:}\) \[ (1.00-x)(0.400-x)=0.160 \] Expanding, \[ 0.400-1.40x+x^2=0.160 \] \[ x^2-1.40x+0.240=0 \] Factorising, \[ (x-0.200)(x-1.20)=0 \] \(\textbf{Reject the physically impossible root:}\) The value \(x=1.20\,\mathrm{bar}\) would exceed the available H₂S partial pressure. It would give a negative final gas pressure. Therefore, \[ x=0.200\,\mathrm{bar} \] \(\textbf{Recover the final pressures:}\) \[ p_{\mathrm{NH_3,eq}}=1.00-0.200 \] \[ p_{\mathrm{NH_3,eq}}=0.800\,\mathrm{bar} \] \[ p_{\mathrm{H_2S,eq}}=0.400-0.200 \] \[ p_{\mathrm{H_2S,eq}}=0.200\,\mathrm{bar} \] \(\textbf{Verify the equilibrium product:}\) \[ (0.800)(0.200)=0.160\,\mathrm{bar^2} \] This agrees with Kp. \(\textbf{Interpret the heterogeneous response:}\) The amount of solid can increase during re-equilibration without directly appearing in the Kp expression. The added NH₃ does not have to be consumed completely. Instead, the two gaseous partial pressures adjust until their product again equals the equilibrium constant. The temperature remains constant, so the original Kp is unchanged. The calculation combines heterogeneous-equilibrium expressions, reaction-quotient direction, gas stoichiometry and rejection of an impossible algebraic root. \(\textbf{Final answer:}\) \[ \boxed{p_{\mathrm{NH_3,eq}}=0.800\,\mathrm{bar},\qquad p_{\mathrm{H_2S,eq}}=0.200\,\mathrm{bar}} \]

Explanation: \(\textbf{Identify the required reaction manipulation:}\) The first reaction produces two B molecules: \[ \mathrm{A\rightleftharpoons2B} \] The second reaction consumes one B molecule together with one C molecule: \[ \mathrm{B+C\rightleftharpoons D} \] To cancel the two B molecules, the second reaction must be used twice. \(\textbf{Double the second equilibrium equation:}\) \[ \mathrm{2B+2C\rightleftharpoons2D} \] Multiplying a reaction equation by two squares its equilibrium constant. Therefore, \[ K_2'=K_2^2 \] \[ K_2'=3.00^2 \] \[ K_2'=9.00 \] \(\textbf{Add the first reaction:}\) \[ \mathrm{A\rightleftharpoons2B} \] Adding it to the doubled second reaction gives \[ \mathrm{A+2C\rightleftharpoons2D} \] The intermediate B species cancel. \(\textbf{Determine the forward composite constant:}\) When reaction equations are added, their equilibrium constants multiply. Therefore, \[ K_{\mathrm{composite}}=K_1K_2^2 \] \[ K_{\mathrm{composite}}=(4.00)(9.00) \] \[ K_{\mathrm{composite}}=36.0 \] \(\textbf{Reverse the composite reaction:}\) The question asks for \[ \mathrm{2D\rightleftharpoons A+2C} \] This is the reverse of the composite reaction. Therefore, \[ K=\frac1{K_{\mathrm{composite}}} \] \[ K=\frac1{36} \] \(\textbf{Determine the sign of standard Gibbs energy:}\) For a consistently defined equilibrium constant, \[ \Delta_rG^\circ=-RT\ln K \] Here, \[ K=\frac1{36}\lt 1 \] Therefore, \[ \ln K\lt 0 \] Consequently, \[ \Delta_rG^\circ\gt 0 \] \(\textbf{Interpret the reaction representation:}\) The equilibrium constant depends on how the balanced overall reaction is written. Doubling a reaction squares K. Adding reactions multiplies their constants. Reversing a reaction takes the reciprocal. The standard Gibbs energy change transforms consistently: it doubles when the equation is doubled, adds when reactions are added and changes sign when the reaction is reversed. The final positive standard Gibbs energy is therefore consistent with an equilibrium constant smaller than one. This conclusion concerns the standard reaction, not the instantaneous direction of every possible mixture. \(\textbf{Final answer:}\) \[ \boxed{K=\frac1{36},\qquad\Delta_rG^\circ\gt 0} \]

Explanation: \(\textbf{Connect average molar mass with dissociation:}\) Suppose the vessel initially contains one mole of N₂O₄. Let its degree of dissociation be \(\alpha\). The equilibrium mole amounts are \[ n_{\mathrm{N_2O_4}}=1-\alpha \] \[ n_{\mathrm{NO_2}}=2\alpha \] Therefore, \[ n_{\mathrm{total}}=1+\alpha \] \(\textbf{Determine the average molar mass:}\) The total gas mass remains equal to the mass of the original mole of N₂O₄. Hence, \[ \overline M=\frac{92.0}{1+\alpha} \] Substituting the measured value, \[ 69.0=\frac{92.0}{1+\alpha_1} \] Therefore, \[ 1+\alpha_1=\frac{92.0}{69.0} \] \[ 1+\alpha_1=\frac43 \] \[ \alpha_1=\frac13 \] \(\textbf{Express the equilibrium partial pressures:}\) At a total pressure P, \[ p_{\mathrm{N_2O_4}} =\frac{1-\alpha}{1+\alpha}P \] \[ p_{\mathrm{NO_2}} =\frac{2\alpha}{1+\alpha}P \] \(\textbf{Construct the pressure equilibrium constant:}\) \[ K_p=\frac{p_{\mathrm{NO_2}}^2}{p_{\mathrm{N_2O_4}}} \] Substituting the partial-pressure expressions, \[ K_p= \frac{\left(\frac{2\alpha}{1+\alpha}P\right)^2} {\left(\frac{1-\alpha}{1+\alpha}P\right)} \] Simplifying, \[ K_p=\frac{4\alpha^2P}{1-\alpha^2} \] \(\textbf{Calculate Kp from the original equilibrium:}\) \[ K_p= \frac{4\left(\frac13\right)^2(0.900)} {1-\left(\frac13\right)^2} \] \[ K_p= \frac{\frac49(0.900)}{\frac89} \] \[ K_p=0.450\,\mathrm{bar} \] \(\textbf{Apply the new equilibrium pressure:}\) The compression is isothermal. Therefore, Kp remains unchanged. At the new total equilibrium pressure, \[ 0.450=\frac{4\alpha_2^2(1.80)}{1-\alpha_2^2} \] Rearranging, \[ 0.450(1-\alpha_2^2)=7.20\alpha_2^2 \] \[ 0.450=7.65\alpha_2^2 \] \[ \alpha_2^2=\frac{0.450}{7.65} \] \[ \alpha_2^2\approx0.05882 \] Taking the physically acceptable positive root, \[ \alpha_2\approx0.2425 \] \(\textbf{Check the direction of change:}\) The degree of dissociation decreases from approximately \(0.333\) to \(0.243\). Dissociation produces two gaseous molecules from one. Higher pressure therefore favours the side with fewer gaseous molecules. The calculated decrease is consistent with this prediction. \(\textbf{Interpret the linked measurements:}\) The average molar mass reveals the original equilibrium degree of dissociation because dissociation changes the total gas mole number without changing total mass. The measured total pressure then permits determination of Kp. That same equilibrium constant controls the degree of dissociation at the new pressure. The problem therefore requires connecting composition-dependent molar mass, gaseous stoichiometry, partial pressures and an isothermal equilibrium disturbance. \(\textbf{Final answer:}\) \[ \boxed{\alpha_1=\frac13,\qquad\alpha_2\approx0.243} \]

Explanation: \(\textbf{Express the equilibrium amounts using dissociation degree:}\) Let the degree of dissociation of PCl₅ be \(\alpha\). Starting from one mole, \[ n_{\mathrm{PCl_5}}=1-\alpha \] \[ n_{\mathrm{PCl_3}}=\alpha \] \[ n_{\mathrm{Cl_2}}=\alpha \] The inert-gas amount is \(1.00\,\mathrm{mol}\). \(\textbf{Recover the degree of dissociation from the pressure ratio:}\) In an ideal-gas mixture, partial-pressure ratios equal mole-number ratios. Therefore, \[ \frac{p_{\mathrm{Cl_2}}}{p_{\mathrm{PCl_5}}} =\frac{\alpha}{1-\alpha} \] The measured ratio is two: \[ \frac{\alpha}{1-\alpha}=2.00 \] \[ \alpha=2-2\alpha \] \[ 3\alpha=2 \] \[ \alpha=\frac23 \] \(\textbf{Determine the total equilibrium mole amount:}\) The reacting gases contribute \[ (1-\alpha)+\alpha+\alpha=1+\alpha \] Including the inert gas, \[ n_{\mathrm{total}}=2+\alpha \] Therefore, \[ n_{\mathrm{total}}=2+\frac23 \] \[ n_{\mathrm{total}}=\frac83 \] \(\textbf{Calculate the reacting-gas partial pressures:}\) For PCl₅, \[ p_{\mathrm{PCl_5}} =\frac{\frac13}{\frac83}(1.60) \] \[ p_{\mathrm{PCl_5}}=0.200\,\mathrm{bar} \] For PCl₃, \[ p_{\mathrm{PCl_3}} =\frac{\frac23}{\frac83}(1.60) \] \[ p_{\mathrm{PCl_3}}=0.400\,\mathrm{bar} \] Similarly, \[ p_{\mathrm{Cl_2}}=0.400\,\mathrm{bar} \] \(\textbf{Calculate the equilibrium constant:}\) \[ K_p=\frac{p_{\mathrm{PCl_3}}p_{\mathrm{Cl_2}}}{p_{\mathrm{PCl_5}}} \] \[ K_p=\frac{(0.400)(0.400)}{0.200} \] \[ K_p=0.800\,\mathrm{bar} \] \(\textbf{Evaluate selective inert-gas removal:}\) For an ideal gas, \[ p_i=\frac{n_iRT}{V} \] The temperature and vessel volume remain unchanged. The amounts of PCl₅, PCl₃ and Cl₂ are also unchanged during inert-gas removal. Therefore, all three reacting-gas partial pressures remain unchanged. \(\textbf{Determine the immediate quotient:}\) \[ Q_p=\frac{(0.400)(0.400)}{0.200} \] \[ Q_p=0.800\,\mathrm{bar} \] Thus, \[ Q_p=K_p \] No net reaction is favoured. \(\textbf{Interpret the pressure distinction:}\) Removing an inert gas decreases the total vessel pressure. However, at constant volume and temperature, it does not change the partial pressures of the reacting gases. The reaction quotient therefore remains unchanged. This differs from changing the vessel volume, which would alter the reacting-gas partial pressures. The presence of an inert gas must be considered through the actual experimental constraint, not merely through the observation that total pressure changes. The result integrates gas-mixture composition, degree of dissociation and the distinction between total pressure and reacting-gas partial pressures. \(\textbf{Final answer:}\) \[ \boxed{\alpha=\frac23,\qquad Q_p=K_p} \]

Explanation: \(\textbf{Determine the gaseous mole-number change for the original reaction:}\) For \[ \mathrm{2SO_2+O_2\rightleftharpoons2SO_3} \] the total gaseous product coefficient is two. The total gaseous reactant coefficient is three. Therefore, \[ \Delta n_g=2-3 \] \[ \Delta n_g=-1 \] \(\textbf{Connect Kp with Kc:}\) For the conventional textbook constants, \[ K_p=K_c(RT)^{\Delta n_g} \] Because \(\Delta n_g=-1\), \[ K_p=\frac{K_c}{RT} \] Substituting, \[ K_p=\frac{5.00}{25.0} \] \[ K_p=0.200\,\mathrm{bar^{-1}} \] This is the pressure constant for the original ammonia-independent sulphur dioxide oxidation reaction as written. \(\textbf{Reverse the reaction:}\) Reversing the original equation gives \[ \mathrm{2SO_3\rightleftharpoons2SO_2+O_2} \] The reverse pressure equilibrium constant is \[ K_{p,\mathrm{reverse}}=\frac1{K_p} \] \[ K_{p,\mathrm{reverse}}=\frac1{0.200} \] \[ K_{p,\mathrm{reverse}}=5.00\,\mathrm{bar} \] \(\textbf{Halve all stoichiometric coefficients:}\) The required equation is \[ \mathrm{SO_3\rightleftharpoons SO_2+\frac12O_2} \] Dividing every coefficient by two takes the square root of the equilibrium constant. Therefore, \[ K_{p,\mathrm{required}} =\sqrt{K_{p,\mathrm{reverse}}} \] \[ K_{p,\mathrm{required}}=\sqrt{5.00} \] \[ K_{p,\mathrm{required}}\approx2.236\,\mathrm{bar^{1/2}} \] \(\textbf{Check the dimensional exponent independently:}\) For the required reaction, \[ \Delta n_g=1+\frac12-1 \] \[ \Delta n_g=\frac12 \] The conventional pressure expression is \[ K_p=\frac{p_{\mathrm{SO_2}}p_{\mathrm{O_2}}^{1/2}}{p_{\mathrm{SO_3}}} \] Its pressure dimension is therefore \[ \mathrm{bar^{1/2}} \] This agrees with the transformed result. \(\textbf{Interpret the sequence of transformations:}\) The conversion from Kc to Kp requires the correct gaseous stoichiometric mole-number change. Reversing the balanced equation requires taking the reciprocal of the equilibrium constant. Halving the equation requires taking its square root. These operations cannot be replaced by simply changing the sign of \(\Delta n_g\) while leaving the numerical constant untouched. The requested pressure-unit exponent provides an independent check on the final expression. The corresponding thermodynamic equilibrium constant is dimensionless when the reaction quotient is constructed using standard-state-normalized activities. \(\textbf{Final answer:}\) \[ \boxed{K_{p,\mathrm{required}}\approx2.236\,\mathrm{bar^{1/2}}} \]

Explanation: \(\textbf{Identify the initial reaction direction:}\) The forward reaction is \[ \mathrm{2A\rightleftharpoons B} \] The vessel initially contains only B. Therefore, the reverse reaction initially produces A. The reverse reaction converts one mole of B into two moles of A. \(\textbf{Construct the conserved stoichiometric inventory:}\) Each B molecule contains two A-derived units. Consequently, the concentration-equivalent conserved quantity is \[ [\mathrm A]+2[\mathrm B] \] Initially, \[ [\mathrm A]_0+2[\mathrm B]_0=0+2(1.00) \] \[ [\mathrm A]_0+2[\mathrm B]_0=2.00\,\mathrm{M} \] At equilibrium, \[ [\mathrm A]+2[\mathrm B]=2.00\,\mathrm{M} \] \(\textbf{Express B through the equilibrium constant:}\) For the written forward reaction, \[ K_c=\frac{[\mathrm B]}{[\mathrm A]^2} \] Therefore, \[ [\mathrm B]=K_c[\mathrm A]^2 \] Substituting, \[ [\mathrm B]=0.500[\mathrm A]^2 \] \(\textbf{Combine equilibrium and conservation:}\) \[ [\mathrm A]+2(0.500[\mathrm A]^2)=2.00 \] Therefore, \[ [\mathrm A]^2+[\mathrm A]-2.00=0 \] Factorising, \[ ([\mathrm A]-1.00)([\mathrm A]+2.00)=0 \] The negative root would imply a negative concentration and is unacceptable. Thus, \[ [\mathrm A]_{\mathrm{eq}}=1.00\,\mathrm{M} \] \(\textbf{Recover B's equilibrium concentration:}\) \[ [\mathrm B]_{\mathrm{eq}}=0.500(1.00)^2 \] \[ [\mathrm B]_{\mathrm{eq}}=0.500\,\mathrm{M} \] \(\textbf{Determine the equilibrium mole fraction of A:}\) For ideal gases in the same volume, \[ y_{\mathrm A}=\frac{[\mathrm A]}{[\mathrm A]+[\mathrm B]} \] Substituting, \[ y_{\mathrm A}=\frac{1.00}{1.00+0.500} \] \[ y_{\mathrm A}=\frac23 \] \(\textbf{Verify both governing constraints:}\) The equilibrium quotient is \[ Q_c=\frac{0.500}{(1.00)^2} \] \[ Q_c=0.500=K_c \] The conserved A-derived inventory is \[ 1.00+2(0.500)=2.00\,\mathrm{M} \] Both requirements are satisfied. \(\textbf{Interpret the inverse equilibrium calculation:}\) The initial vessel contains only the association product B, so the net equilibration direction is reverse. The equilibrium concentrations cannot be determined from Kc alone. The conserved stoichiometric inventory supplies the second equation required to calculate the composition. The physically acceptable concentration root is then converted into a gas mole fraction. The calculation connects equilibrium direction, stoichiometric conservation, a nonlinear equilibrium equation and gas-mixture composition. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm A]_{\mathrm{eq}}=1.00\,\mathrm{M},\qquad y_{\mathrm A}=\frac23} \]

Explanation: \(\textbf{Identify the two different Gibbs-energy quantities:}\) The standard reaction Gibbs energy is related to the equilibrium constant: \[ \Delta_rG^\circ=-RT\ln K \] The instantaneous reaction Gibbs energy depends on the actual composition: \[ \Delta_rG=\Delta_rG^\circ+RT\ln Q \] Combining the relations, \[ \Delta_rG=RT\ln\frac QK \] \(\textbf{Calculate the standard reaction Gibbs energy:}\) \[ \Delta_rG^\circ =-(2.50)\ln(10^{-2}) \] Since \[ \ln(10^{-2})=-\ln100 \] we obtain \[ \Delta_rG^\circ=(2.50)(4.605) \] \[ \Delta_rG^\circ=+11.5125\,\mathrm{kJ\,mol^{-1}} \] \[ \Delta_rG^\circ\approx+11.5\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Calculate the actual reaction quotient ratio:}\) \[ \frac QK=\frac{10^{-4}}{10^{-2}} \] \[ \frac QK=10^{-2} \] \(\textbf{Determine the instantaneous Gibbs energy:}\) \[ \Delta_rG=(2.50)\ln(10^{-2}) \] \[ \Delta_rG=-(2.50)(4.605) \] \[ \Delta_rG=-11.5125\,\mathrm{kJ\,mol^{-1}} \] \[ \Delta_rG\approx-11.5\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Check the reaction direction independently:}\) The equilibrium comparison gives \[ Q\lt K \] Therefore, the forward reaction is thermodynamically favoured. The negative instantaneous Gibbs energy agrees with this conclusion. \(\textbf{Interpret the opposite signs:}\) The equilibrium constant is smaller than one, so the standard Gibbs energy is positive for the written reaction. However, the actual mixture has a reaction quotient even smaller than K. The forward reaction can therefore be thermodynamically favoured under the specified nonstandard composition. A positive standard Gibbs energy does not mean that forward reaction is impossible under every set of conditions. Similarly, a negative instantaneous Gibbs energy does not guarantee rapid reaction. The distinction between standard-state tendency and composition-dependent reaction tendency is essential. The two calculated Gibbs energies have opposite signs without any contradiction. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rG^\circ\approx+11.5\,\mathrm{kJ\,mol^{-1}},\qquad \Delta_rG\approx-11.5\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: Statement I is correct. At constant temperature and pressure, a spontaneous net reaction change proceeds in the direction that decreases the system's Gibbs energy, subject to the relevant constraints. At point P, increasing reaction extent decreases G. Therefore, the forward reaction is thermodynamically favoured. Statement II is correct. At the minimum of the Gibbs-energy curve, the local slope with respect to reaction extent is zero. The corresponding reaction Gibbs energy is \[ \Delta_rG=\left(\frac{\partial G}{\partial\xi}\right)_{T,P} \] At E, the slope vanishes. Therefore, \[ \Delta_rG=0 \] This is the thermodynamic equilibrium condition. Statement III is correct. At Q, increasing the forward reaction extent would increase G. Decreasing reaction extent instead moves the system towards the minimum. Therefore, the reverse reaction is thermodynamically favoured. Statement IV is incorrect. The minimum of G identifies the equilibrium composition. It does not require equal concentrations of reactants and products. Their equilibrium proportions depend on the reaction and its equilibrium constant under the specified conditions. The graph gives the direction of spontaneous net change from different compositions. It does not directly establish the reaction rate. A large Gibbs-energy driving force need not imply rapid equilibration. The minimum also does not require the standard reaction Gibbs energy to be zero. Rather, the actual reaction Gibbs energy becomes zero at the equilibrium composition. The curve therefore distinguishes the thermodynamic reaction direction on either side of equilibrium from both reaction speed and standard-state favourability. Statements I, II and III are correct.

Explanation: \(\textbf{Determine the original pressure equilibrium constant:}\) For \[ \mathrm{A+B\rightleftharpoons C} \] the pressure equilibrium expression is \[ K_p=\frac{p_{\mathrm C}}{p_{\mathrm A}p_{\mathrm B}} \] Substituting, \[ K_p=\frac{0.800}{(0.400)(0.400)} \] \[ K_p=\frac{0.800}{0.160} \] \[ K_p=5.00\,\mathrm{bar^{-1}} \] \(\textbf{Identify the immediate partial-pressure effect of volume change:}\) At constant temperature with fixed reacting-gas mole numbers, \[ p_i\propto\frac1V \] Therefore, doubling volume halves every reacting-gas partial pressure. Halving volume doubles every reacting-gas partial pressure. \(\textbf{Evaluate vessel P after expansion:}\) The immediate pressures are \[ p_{\mathrm A,P}=0.200\,\mathrm{bar} \] \[ p_{\mathrm B,P}=0.200\,\mathrm{bar} \] \[ p_{\mathrm C,P}=0.400\,\mathrm{bar} \] The reaction quotient is \[ Q_{p,P}=\frac{0.400}{(0.200)(0.200)} \] \[ Q_{p,P}=10.0\,\mathrm{bar^{-1}} \] Since \[ Q_{p,P}\gt K_p \] the reverse reaction is favoured. \(\textbf{Evaluate vessel Q after compression:}\) The immediate pressures are \[ p_{\mathrm A,Q}=0.800\,\mathrm{bar} \] \[ p_{\mathrm B,Q}=0.800\,\mathrm{bar} \] \[ p_{\mathrm C,Q}=1.60\,\mathrm{bar} \] Therefore, \[ Q_{p,Q}=\frac{1.60}{(0.800)(0.800)} \] \[ Q_{p,Q}=2.50\,\mathrm{bar^{-1}} \] Since \[ Q_{p,Q}\lt K_p \] the forward reaction is favoured. \(\textbf{Interpret the opposite responses:}\) The forward reaction converts two gaseous molecules into one. Expansion favours the direction producing more gaseous molecules. Compression favours the direction producing fewer gaseous molecules. The reaction-quotient calculations confirm these qualitative predictions. Although both vessels remain at the same temperature, the volume changes disturb the equilibrium compositions. The equilibrium constant therefore remains unchanged while the instantaneous quotients change. The conclusion concerns the immediate reaction tendencies following the mechanical disturbances. The final equilibrium partial pressures would require additional material-balance calculations. \(\textbf{Final answer:}\) Vessel P favours the reverse reaction, while vessel Q favours the forward reaction.

Explanation: \(\textbf{Connect standard Gibbs energy with enthalpy and entropy:}\) The governing relation is \[ \Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ \] The enthalpy and entropy values must be expressed in compatible units. \(\textbf{Convert the entropy unit:}\) \[ \Delta_rS^\circ=100\,\mathrm{J\,mol^{-1}\,K^{-1}} \] Therefore, \[ \Delta_rS^\circ=0.100\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] \(\textbf{Calculate standard Gibbs energy at 300 K:}\) \[ \Delta_rG_1^\circ =40.0-(300)(0.100) \] \[ \Delta_rG_1^\circ=40.0-30.0 \] \[ \Delta_rG_1^\circ=+10.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpret the equilibrium constant at 300 K:}\) The standard Gibbs-energy relationship is \[ \Delta_rG^\circ=-RT\ln K \] Because \[ \Delta_rG_1^\circ\gt 0 \] we have \[ \ln K_1\lt 0 \] Therefore, \[ K_1\lt 1 \] The reactant side is favoured under the standard-state comparison. \(\textbf{Calculate standard Gibbs energy at 500 K:}\) \[ \Delta_rG_2^\circ =40.0-(500)(0.100) \] \[ \Delta_rG_2^\circ=40.0-50.0 \] \[ \Delta_rG_2^\circ=-10.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpret the equilibrium constant at 500 K:}\) Because \[ \Delta_rG_2^\circ\lt 0 \] we obtain \[ \ln K_2\gt 0 \] Therefore, \[ K_2\gt 1 \] The product side is now favoured under the standard-state comparison. \(\textbf{Determine the temperature corresponding to unit equilibrium constant:}\) When \[ K=1 \] the standard Gibbs energy satisfies \[ \Delta_rG^\circ=-RT\ln1=0 \] Thus, \[ \Delta_rH^\circ-T\Delta_rS^\circ=0 \] Rearranging, \[ T=\frac{\Delta_rH^\circ}{\Delta_rS^\circ} \] Substituting, \[ T=\frac{40.0}{0.100} \] \[ T=400\,\mathrm{K} \] \(\textbf{Interpret the temperature-dependent equilibrium favourability:}\) The forward reaction is endothermic because its standard enthalpy change is positive. It also produces a positive standard entropy change. At the lower temperature, the enthalpy contribution dominates the standard Gibbs-energy expression. At the higher temperature, the favourable entropy contribution becomes sufficiently large to make the standard Gibbs energy negative. The corresponding equilibrium constant changes from below one to above one. This does not mean that the reaction is impossible below \(400\,\mathrm{K}\) or complete above it. The actual forward or reverse tendency of a particular mixture still depends on its reaction quotient. The problem connects the standard Gibbs-energy equation with equilibrium favourability and its temperature dependence. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rG_{300}^\circ=+10.0\,\mathrm{kJ\,mol^{-1}},\quad \Delta_rG_{500}^\circ=-10.0\,\mathrm{kJ\,mol^{-1}},\quad T_{K=1}=400\,\mathrm{K}} \]

Explanation: \(\textbf{Infer the reaction stoichiometry from experiment P:}\) The change in A concentration is \[ \Delta[A]=0.400-0.500=-0.100\,\mathrm{M} \] For B, \[ \Delta[B]=0.500-0.700=-0.200\,\mathrm{M} \] For C, \[ \Delta[C]=0.200-0=+0.200\,\mathrm{M} \] The changes have the ratio \[ -\Delta[A]:-\Delta[B]:\Delta[C] =0.100:0.200:0.200 \] Dividing by the smallest value gives \[ a:b:c=1:2:2 \] Therefore, the balanced reaction is \[ \mathrm{A+2B\rightleftharpoons2C} \] \(\textbf{Check the stoichiometry against experiment Q:}\) The second experiment gives \[ \Delta[A]=0.500-0.600=-0.100\,\mathrm{M} \] \[ \Delta[B]=0.300-0.500=-0.200\,\mathrm{M} \] \[ \Delta[C]=0.500-0.300=+0.200\,\mathrm{M} \] These changes independently confirm the same stoichiometric ratio. \(\textbf{Construct the correct equilibrium expression:}\) For the established balanced equation, \[ K_c=\frac{[C]^2}{[A][B]^2} \] The coefficient of B becomes a power of two in the denominator, while the coefficient of C becomes a power of two in the numerator. \(\textbf{Calculate Kc using experiment P:}\) \[ K_c=\frac{(0.200)^2}{(0.400)(0.500)^2} \] \[ K_c=\frac{0.0400}{(0.400)(0.250)} \] \[ K_c=\frac{0.0400}{0.100} \] \[ K_c=0.400\,\mathrm{L\,mol^{-1}} \] \(\textbf{Check the independent equilibrium data:}\) For experiment Q, the concentration changes are \[ \Delta[A]=-0.0100\,\mathrm{M},\quad \Delta[B]=-0.0200\,\mathrm{M},\quad \Delta[C]=+0.0200\,\mathrm{M} \] The change ratio is again \(1:2:2\), consistent with the balanced reaction. The equilibrium constant obtained from the independently measured final concentrations is \[ K_{c,Q}=\frac{(0.0200)^2}{(0.100)(0.100)^2} \] \[ K_{c,Q}=\frac{0.000400}{0.00100} \] \[ K_{c,Q}=0.400\,\mathrm{L\,mol^{-1}} \] \(\textbf{Verify consistency across experiments:}\) Both experiments agree on the simplest stoichiometric coefficients. Both also give the same equilibrium constant at the stated common temperature. The agreement of the concentration-change ratios alone would not suffice to establish equilibrium consistency; the final equilibrium expressions must also agree. Because \(\Delta n\) for the concentration exponents is \(2-(1+2)=-1\), the conventional \(K_c\) units are \(\mathrm{L\,mol^{-1}}\). Option D correctly identifies both the reaction and its equilibrium constant. \(\textbf{Final answer:}\) \[ \mathrm{A+2B\rightleftharpoons2C},\qquad K_c=0.400\,\mathrm{L\,mol^{-1}} \]

Explanation: \(\textbf{Construct the equilibrium expression:}\) For \[ \mathrm{A+2B\rightleftharpoons C} \] the pressure reaction quotient is \[ Q_p=\frac{p_C}{p_Ap_B^2} \] At the original equilibrium, \[ K_p=\frac{(0.160)}{(0.200)(0.400)^2} \] \[ K_p=\frac{0.160}{0.0320} \] \[ K_p=5.00\,\mathrm{bar^{-2}} \] \(\textbf{Evaluate the instantaneous composition in P:}\) Only A is injected. Therefore, \[ p_{A,P}=2(0.200)=0.400\,\mathrm{bar} \] while \[ p_{B,P}=0.400\,\mathrm{bar} \] \[ p_{C,P}=0.160\,\mathrm{bar} \] The new quotient is \[ Q_P=\frac{0.160}{(0.400)(0.400)^2} \] \[ Q_P=\frac{0.160}{0.0640} \] \[ Q_P=2.50\,\mathrm{bar^{-2}} \] Thus, \[ Q_P=\frac12K_p \] The forward reaction is favoured because the quotient is below the equilibrium constant. \(\textbf{Evaluate the instantaneous composition in Q:}\) Here only C is injected: \[ p_{C,Q}=2(0.160)=0.320\,\mathrm{bar} \] The reactant partial pressures remain at their original values. Therefore, \[ Q_Q=\frac{0.320}{(0.200)(0.400)^2} \] \[ Q_Q=\frac{0.320}{0.0320} \] \[ Q_Q=10.0\,\mathrm{bar^{-2}} \] Thus, \[ Q_Q=2K_p \] The reverse reaction is favoured. \(\textbf{Interpret the opposite disturbances:}\) Injecting a reactant increases a denominator term in the reaction quotient. Injecting a product increases the numerator. The effects are opposite even though both operations raise the total gas pressure. Because each injection changes only one reacting-gas partial pressure in its rigid vessel, the composition-based reaction quotient gives the unambiguous direction. The equilibrium constant itself does not change at fixed temperature. The final equilibrium pressures would require separate stoichiometric material balances. \(\textbf{Final answer:}\) \[ \boxed{Q_P=\frac12K_p,\qquad Q_Q=2K_p} \]

Explanation: The Assertion is true. The reaction quotient is \[ Q_p=\frac{p_Cp_D}{p_Ap_B} \] Before compression, the mixture is at equilibrium. Therefore, \[ Q_p=K_p \] At constant temperature, the ideal-gas partial pressure of each species is \[ p_i=\frac{n_iRT}{V} \] Immediately after the volume is halved, the reacting-gas mole numbers have not yet changed. Consequently, each partial pressure doubles. The new quotient is \[ Q_p'=\frac{(2p_C)(2p_D)}{(2p_A)(2p_B)} \] The pressure-scaling factors cancel: \[ Q_p'=\frac{4p_Cp_D}{4p_Ap_B} \] \[ Q_p'=Q_p=K_p \] The Reason is also true. The forward reaction has two moles of gaseous reactants and two moles of gaseous products according to its balanced equation. Thus, \[ \Delta n_g=2-2=0 \] For a uniform pressure-scaling factor f, the quotient transforms as \[ Q_p'=f^{\Delta n_g}Q_p \] With zero gaseous mole-number change, \[ f^{\Delta n_g}=1 \] Therefore, compression does not disturb the equilibrium quotient under the stated ideal-gas conditions. This conclusion does not mean that the individual partial pressures are unchanged. They all increase immediately. It also does not mean that pressure can never influence real equilibria; the given result follows from the balanced gaseous stoichiometry and ideal-gas assumptions. The Reason directly explains why the Assertion is true.