Equilibrium MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: \(\textbf{Represent the concentration changes:}\) The reaction consumes two units of A for each unit of B produced. Let \(x\,\mathrm{M}\) of B form. The equilibrium concentrations are \[ [A]_{\mathrm{eq}}=1.00-2x \] \[ [B]_{\mathrm{eq}}=x \] The physically permissible extent satisfies \[ 0\leq x\leq0.50\,\mathrm{M} \] \(\textbf{Apply the equilibrium expression:}\) For the written reaction, \[ K_c=\frac{[B]}{[A]^2} \] Substituting, \[ 4.0=\frac{x}{(1.00-2x)^2} \] Rearranging, \[ 4(1.00-2x)^2=x \] Expanding, \[ 4(1-4x+4x^2)=x \] \[ 16x^2-17x+4=0 \] \(\textbf{Solve the quadratic equation:}\) Using the quadratic formula, \[ x=\frac{17\pm\sqrt{17^2-4(16)(4)}}{32} \] \[ x=\frac{17\pm\sqrt{33}}{32} \] The roots are approximately \[ x=0.711 \] and \[ x=0.352 \] The larger root exceeds the maximum extent \(0.50\,\mathrm{M}\) and would imply a negative concentration of A. Therefore, it must be rejected. The physically acceptable root is \[ x\approx0.352\,\mathrm{M} \] \(\textbf{Calculate equilibrium concentrations:}\) \[ [B]_{\mathrm{eq}}\approx0.352\,\mathrm{M} \] \[ [A]_{\mathrm{eq}}=1.00-2(0.352) \] \[ [A]_{\mathrm{eq}}\approx0.296\,\mathrm{M} \] \(\textbf{Verify the equilibrium relationship:}\) \[ K_c\approx\frac{0.352}{(0.296)^2} \] \[ K_c\approx4.02\,\mathrm{L\,mol^{-1}} \] The small difference from four arises from rounding. \(\textbf{Verify the equilibrium composition:}\) Using the calculated values, \[ K_c\approx\frac{0.352}{(0.296)^2} \] \[ K_c\approx4.02\,\mathrm{L\,mol^{-1}} \] The small deviation from 4.00 results from rounding the two concentrations. The concentrations also satisfy the material balance: \[ 0.296+2(0.352)=1.000\,\mathrm{M} \] Thus the correct pair is the one given in option C; the other pairs do not satisfy both constraints. \(\textbf{Final answer:}\) \[ \boxed{[A]_{\mathrm{eq}}\approx0.296\,\mathrm{M},\quad[B]_{\mathrm{eq}}\approx0.352\,\mathrm{M}} \]

Explanation: \(\textbf{Relate dissociation to total gaseous moles:}\) Let the initial amount of \(\mathrm{SO_3}\) be \(n_0\). If the fraction dissociated is \(\alpha\), the remaining \(\mathrm{SO_3}\) amount is \[ n_{\mathrm{SO_3}}=n_0(1-\alpha) \] The amount of \(\mathrm{SO_2}\) formed is \[ n_{\mathrm{SO_2}}=n_0\alpha \] From the balanced reaction, two moles of decomposed \(\mathrm{SO_3}\) produce one mole of oxygen. Therefore, \[ n_{\mathrm{O_2}}=\frac{n_0\alpha}{2} \] \(\textbf{Determine total equilibrium moles:}\) \[ n_{\mathrm{total}}=n_0(1-\alpha)+n_0\alpha+\frac{n_0\alpha}{2} \] \[ n_{\mathrm{total}}=n_0\left(1+\frac{\alpha}{2}\right) \] At constant temperature and volume, \[ \frac{P_{\mathrm{eq}}}{P_0}=1+\frac{\alpha}{2} \] \(\textbf{Calculate the dissociation fraction:}\) \[ \frac{2.50}{2.00}=1+\frac{\alpha}{2} \] \[ 1.25=1+\frac{\alpha}{2} \] \[ \frac{\alpha}{2}=0.25 \] \[ \alpha=0.50 \] \(\textbf{Determine equilibrium partial pressures:}\) At constant temperature and volume, each species' pressure contribution is proportional to its mole amount. Using the initial \(\mathrm{SO_3}\) pressure, \[ p_{\mathrm{SO_3}}=P_0(1-\alpha) \] \[ p_{\mathrm{SO_3}}=2.00(0.50) \] \[ p_{\mathrm{SO_3}}=1.00\,\mathrm{bar} \] Similarly, \[ p_{\mathrm{SO_2}}=P_0\alpha \] \[ p_{\mathrm{SO_2}}=1.00\,\mathrm{bar} \] For oxygen, \[ p_{\mathrm{O_2}}=\frac{P_0\alpha}{2} \] \[ p_{\mathrm{O_2}}=0.50\,\mathrm{bar} \] \(\textbf{Calculate Kp:}\) \[ K_p=\frac{p_{\mathrm{SO_2}}^2p_{\mathrm{O_2}}}{p_{\mathrm{SO_3}}^2} \] \[ K_p=\frac{(1.00)^2(0.50)}{(1.00)^2} \] \[ K_p=0.50\,\mathrm{bar} \] \(\textbf{Verify the result against total pressure:}\) The calculated degree of dissociation is \(0.50\), while \(K_p=0.50\,\mathrm{bar}\). The calculated partial pressures also reproduce the measured total pressure: \[ 1.00+1.00+0.50=2.50\,\mathrm{bar} \] The pressure increase must be related to the correct stoichiometric change in total gaseous moles. Treating the reaction as if each \(\mathrm{SO_3}\) molecule produced two complete moles of gas would give an incorrect dissociation fraction. \(\textbf{Final answer:}\) \[ \boxed{\alpha=0.50,\quad K_p=0.50\,\mathrm{bar}} \]

Explanation: \(\textbf{Identify the governing relationship:}\) For ideal gaseous equilibria, the conventional numerical relationship is \[ K_p=K_c(RT)^{\Delta n_g} \] Here, \[ \Delta n_g=\sum n_{\mathrm{gaseous\ products}}-\sum n_{\mathrm{gaseous\ reactants}} \] The exponent depends on the balanced reaction. Therefore, two reactions having the same numerical \(K_c\) do not necessarily have the same numerical \(K_p\). \(\textbf{Evaluate the first reaction:}\) For \[ \mathrm{PCl_5\rightleftharpoons PCl_3+Cl_2} \] one gaseous mole produces two gaseous moles. Hence, \[ \Delta n_{g,1}=2-1 \] \[ \Delta n_{g,1}=1 \] Using the supplied values, \[ K_{p,1}=2.0(25.0)^1 \] \[ K_{p,1}=50.0 \] \(\textbf{Evaluate the second reaction:}\) For \[ \mathrm{H_2+I_2\rightleftharpoons2HI} \] the number of gaseous stoichiometric moles is unchanged. \[ \Delta n_{g,2}=2-(1+1) \] \[ \Delta n_{g,2}=0 \] Therefore, \[ K_{p,2}=2.0(25.0)^0 \] \[ K_{p,2}=2.0 \] \(\textbf{Interpret the different results:}\) The first reaction has a positive gaseous mole-number change, so \(RT\) appears to the first power. The second has no gaseous mole-number change, so the numerical pressure and concentration equilibrium constants coincide. The result does not indicate that one equilibrium constant was measured incorrectly. The two equilibrium expressions contain different net powers of concentration or pressure. Their associated conventional units are also different. Comparing bare numerical values of equilibrium constants without considering the written reactions and conventions can therefore be misleading. \(\textbf{Final answer:}\) \[ \boxed{K_{p,1}=50.0,\quad K_{p,2}=2.0} \]

Explanation: \(\textbf{Determine the initial net direction:}\) The vessel initially contains B and C but no A. For the forward reaction, A is a required reactant. The initial reaction quotient therefore tends towards an infinitely large value because \([A]\) is zero while C is present. Consequently, the reverse reaction is initially favoured. The reverse process converts C into A and B. \(\textbf{Represent the concentration changes:}\) Let \(x\,\mathrm{M}\) of C decompose through the reverse reaction. The equilibrium concentrations are \[ [A]_{\mathrm{eq}}=x \] \[ [B]_{\mathrm{eq}}=1.00+x \] \[ [C]_{\mathrm{eq}}=2.00-x \] The physically acceptable interval is \[ 0\leq x\leq2.00\,\mathrm{M} \] \(\textbf{Apply the equilibrium expression:}\) \[ K_c=\frac{[C]}{[A][B]} \] Substituting, \[ 2.0=\frac{2.00-x}{x(1.00+x)} \] Rearranging, \[ 2x(1+x)=2-x \] \[ 2x+2x^2=2-x \] \[ 2x^2+3x-2=0 \] \(\textbf{Solve the equilibrium equation:}\) Factorising, \[ (2x-1)(x+2)=0 \] The possible roots are \[ x=0.50 \] and \[ x=-2.00 \] The negative root would imply a negative equilibrium concentration of A. It must therefore be rejected. Thus, \[ [A]_{\mathrm{eq}}=0.50\,\mathrm{M} \] \(\textbf{Verify the other equilibrium concentrations:}\) \[ [B]_{\mathrm{eq}}=1.00+0.50=1.50\,\mathrm{M} \] \[ [C]_{\mathrm{eq}}=2.00-0.50=1.50\,\mathrm{M} \] Substitution into the equilibrium expression gives \[ K_c=\frac{1.50}{0.50\times1.50} \] \[ K_c=2.0\,\mathrm{L\,mol^{-1}} \] \(\textbf{Interpret the reverse preparation:}\) The initial absence of A does not prevent equilibrium from being established. A is generated through the reverse reaction. The final mixture contains all three species at positive concentrations. The physically acceptable root must satisfy both the equilibrium relationship and conservation of the reacting material. \(\textbf{Final answer:}\) \[ \boxed{[A]_{\mathrm{eq}}=0.50\,\mathrm{mol\,L^{-1}}} \]

Explanation: Statement I is correct because both reactions have the same stoichiometric form. For the first reaction, \[ K_c=\frac{[Y]_{\mathrm{eq}}}{[X]_{\mathrm{eq}}} \] For the second, \[ K_c=\frac{[Q]_{\mathrm{eq}}}{[P]_{\mathrm{eq}}} \] Since both constants equal ten, both equilibrium product-to-reactant concentration ratios equal ten under the stated convention. Statement II is incorrect because the time required to attain equilibrium depends on reaction kinetics. Different mechanisms can produce different forward and reverse reaction rates even when the equilibrium constants are equal. An equilibrium constant characterises an equilibrium composition relationship, not the duration of the approach to that state. Statement III is therefore correct. A reaction may have a large equilibrium constant while proceeding very slowly because its activation-energy barrier is large. A small equilibrium constant likewise does not imply that every elementary reaction step must be slow. Statement IV is correct for the explicitly stated elementary reversible reaction. For an elementary reaction of the form \(X\rightleftharpoons Y\), the forward and reverse rate laws can be written as \(r_f=k_f[X]\) and \(r_r=k_r[Y]\). At equilibrium, equality of the opposing rates gives \(K_c=k_f/k_r\). Multiplying both rate constants by the same positive factor preserves their ratio. Therefore, the equilibrium constant remains unchanged even though both forward and reverse rates increase for the same concentrations. This is an illustration of the distinction between thermodynamic equilibrium and kinetic behaviour. The same conclusion must not be generalised to arbitrary overall reactions using elementary rate laws without justification. Thus, Statements I, III and IV are correct.

Explanation: \(\textbf{Determine the initial concentration:}\) The initial amount of \(\mathrm{N_2O_4}\) is \(0.100\,\mathrm{mol}\). The vessel volume is \(5.00\,\mathrm{L}\). Therefore, \[ [\mathrm{N_2O_4}]_0=\frac{0.100}{5.00} \] \[ [\mathrm{N_2O_4}]_0=0.0200\,\mathrm{M} \] \(\textbf{Relate nitrogen dioxide formation to dissociation:}\) The balanced reaction is \[ \mathrm{N_2O_4\rightleftharpoons2NO_2} \] If the degree of dissociation is \(\alpha\), the equilibrium nitrogen dioxide concentration is \[ [\mathrm{NO_2}]_{\mathrm{eq}}=2\alpha[\mathrm{N_2O_4}]_0 \] Substituting, \[ 0.0240=2\alpha(0.0200) \] \[ 0.0240=0.0400\alpha \] \[ \alpha=0.60 \] \(\textbf{Determine the remaining N2O4:}\) \[ [\mathrm{N_2O_4}]_{\mathrm{eq}}=[\mathrm{N_2O_4}]_0(1-\alpha) \] \[ [\mathrm{N_2O_4}]_{\mathrm{eq}}=0.0200(1-0.60) \] \[ [\mathrm{N_2O_4}]_{\mathrm{eq}}=0.00800\,\mathrm{M} \] \(\textbf{Calculate the concentration equilibrium constant:}\) \[ K_c=\frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} \] Substituting, \[ K_c=\frac{(0.0240)^2}{0.00800} \] \[ K_c=\frac{0.000576}{0.00800} \] \[ K_c=0.0720\,\mathrm{M} \] \(\textbf{Check the stoichiometric consistency:}\) The dissociated \(\mathrm{N_2O_4}\) concentration is \[ 0.0200-0.00800=0.0120\,\mathrm{M} \] The corresponding nitrogen dioxide concentration should be twice this value. \[ 2(0.0120)=0.0240\,\mathrm{M} \] This agrees with the supplied equilibrium concentration. The calculated degree of dissociation lies between zero and one. Therefore, the result is chemically and quantitatively consistent. \(\textbf{Final answer:}\) \[ \boxed{\alpha=0.60,\quad K_c=0.0720\,\mathrm{mol\,L^{-1}}} \]

Explanation: \(\textbf{Write the original equilibrium condition:}\) For the reaction, \[ K_c=\frac{[C]_{\mathrm{eq}}}{[A]_{\mathrm{eq}}[B]_{\mathrm{eq}}} \] Initially, the mixture satisfies this relationship. \(\textbf{Analyse the first instantaneous operation:}\) Only A is doubled. Therefore, immediately after the first operation, \[ Q_1=\frac{[C]_{\mathrm{eq}}}{(2[A]_{\mathrm{eq}})[B]_{\mathrm{eq}}} \] \[ Q_1=\frac12K_c \] Thus, \[ Q_1\lt K_c \] The forward reaction is favoured because the current quotient is smaller than the equilibrium constant. \(\textbf{Analyse the second instantaneous operation:}\) The concentration of B is now reduced to half its original equilibrium value. A remains twice its original equilibrium value because no appreciable reaction has yet occurred. C remains unchanged. The new quotient is therefore \[ Q_2=\frac{[C]_{\mathrm{eq}}}{(2[A]_{\mathrm{eq}})(\frac12[B]_{\mathrm{eq}})} \] The denominator contains the factors \[ 2\times\frac12=1 \] Consequently, \[ Q_2=\frac{[C]_{\mathrm{eq}}}{[A]_{\mathrm{eq}}[B]_{\mathrm{eq}}} \] \[ Q_2=K_c \] \(\textbf{Interpret the sequential disturbances:}\) The first operation temporarily makes the mixture reactant-rich relative to equilibrium. The second operation removes enough B to restore the original equilibrium concentration ratio. Because the operations are performed before appreciable chemical reaction occurs, the algebraic concentration factors can be combined directly. Immediately after the second operation, the mixture satisfies the equilibrium quotient criterion. Therefore, the net forward tendency caused by the first operation is cancelled by the second operation. The equilibrium constant itself is unchanged throughout because temperature remains constant. The question demonstrates that equilibrium response depends on the complete instantaneous composition, not on the historical sequence of individual additions or removals.

Explanation: \(\textbf{Interpret the three equilibrium constants:}\) Each reaction has the same \(1:1\) stoichiometric form. For P, \[ K_P=\frac{[B]_{\mathrm{eq}}}{[A]_{\mathrm{eq}}}=10^6 \] The large equilibrium constant indicates a very high product-to-reactant concentration ratio at equilibrium. For Q, \[ K_Q=\frac{[D]_{\mathrm{eq}}}{[C]_{\mathrm{eq}}}=10^{-4} \] This indicates that the equilibrium mixture is strongly reactant-favoured. For R, \[ K_R=\frac{[F]_{\mathrm{eq}}}{[E]_{\mathrm{eq}}}=1 \] The equilibrium reactant and product concentrations are equal. \(\textbf{Connect the constants with equilibrium fractions:}\) For a general \(1:1\) transformation initially containing only reactant, \[ K=\frac{x}{1-x} \] where \(x\) is the fraction converted into product. Rearranging, \[ K(1-x)=x \] \[ K=x(K+1) \] \[ x=\frac{K}{1+K} \] Thus, larger K produces a larger equilibrium product fraction for these comparable reactions and initial conditions. \(\textbf{Examine the kinetic limitation:}\) The equilibrium constant provides no direct information about the time required to attain equilibrium. Reaction rates depend on kinetic factors, including activation-energy barriers and reaction mechanisms. A highly product-favoured reaction can proceed slowly. A reaction with a small equilibrium constant can still have rapid forward and reverse elementary processes. The magnitude of K therefore cannot establish which reaction reaches equilibrium first. \(\textbf{Interpret the balanced-rate condition:}\) At equilibrium, the opposing reaction rates are equal for all three systems. This does not require the forward and reverse rate constants to be equal in every reaction mechanism. Equal rates can result from different concentrations and different kinetic constants. The condition \(K_R=1\) specifies an equilibrium concentration ratio, not a universal equality of kinetic rate constants. Thus, P has the greatest equilibrium product fraction, but its equilibrium constant alone does not establish reaction speed.

Explanation: \(\textbf{Identify the governing equilibrium expression:}\) For \[ 2A(g)\rightleftharpoons B(g) \] the reaction quotient is \[ Q_c=\frac{[B]}{[A]^2} \] The equilibrium constant at the specified temperature is \[ K_c=5.0\,\mathrm{L\,mol^{-1}} \] \(\textbf{Evaluate mixture P:}\) \[ Q_{c,P}=\frac{0.40}{(0.40)^2} \] \[ Q_{c,P}=\frac{0.40}{0.16} \] \[ Q_{c,P}=2.5\,\mathrm{L\,mol^{-1}} \] Therefore, \[ Q_{c,P}\lt K_c \] Mixture P favours net forward conversion of A into B. \(\textbf{Evaluate mixture Q:}\) \[ Q_{c,Q}=\frac{0.40}{(0.20)^2} \] \[ Q_{c,Q}=\frac{0.40}{0.04} \] \[ Q_{c,Q}=10.0\,\mathrm{L\,mol^{-1}} \] Thus, \[ Q_{c,Q}\gt K_c \] Mixture Q favours net reverse conversion of B into A. \(\textbf{Interpret the identical product concentrations:}\) Both mixtures initially contain the same concentration of B. However, the concentration of A differs. Because \([A]\) is squared in the denominator, reducing its concentration substantially increases the quotient. Therefore, the two mixtures lie on opposite sides of the equilibrium ratio despite having identical B concentrations. The net reaction direction depends on the complete stoichiometric concentration quotient. It cannot be predicted merely from the amount of product already present. The result illustrates how differences in reactant concentration can reverse the favoured net direction without any difference in product concentration. \(\textbf{Final answer:}\) \[ \boxed{Q_{c,P}=2.5\lt K_c=5.0\lt Q_{c,Q}=10.0} \] Mixture P proceeds forward, whereas mixture Q proceeds reverse.

Explanation: \(\textbf{Determine the equilibrium partial pressures:}\) For an ideal gas mixture, \[ p_i=y_iP_{\mathrm{total}} \] The partial pressure of A is \[ p_A=0.25\times4.00 \] \[ p_A=1.00\,\mathrm{bar} \] For B, \[ p_B=0.50\times4.00 \] \[ p_B=2.00\,\mathrm{bar} \] For C, \[ p_C=0.25\times4.00 \] \[ p_C=1.00\,\mathrm{bar} \] \(\textbf{Construct the pressure equilibrium expression:}\) The balanced reaction is \[ A(g)+2B(g)\rightleftharpoons C(g) \] Therefore, \[ K_p=\frac{p_C}{p_Ap_B^2} \] Substituting, \[ K_p=\frac{1.00}{1.00(2.00)^2} \] \[ K_p=\frac14 \] \[ K_p=0.250\,\mathrm{bar^{-2}} \] \(\textbf{Determine the gaseous stoichiometric change:}\) The products contain one gaseous stoichiometric mole. The reactants contain three. Therefore, \[ \Delta n_g=1-(1+2) \] \[ \Delta n_g=-2 \] \(\textbf{Relate Kp and Kc:}\) For the conventional expressions, \[ K_p=K_c(RT)^{\Delta n_g} \] Therefore, \[ K_p=K_c(RT)^{-2} \] Rearranging, \[ K_c=K_p(RT)^2 \] \(\textbf{Substitute the supplied thermal factor:}\) \[ K_c=0.250(20.0)^2 \] \[ K_c=0.250\times400 \] \[ K_c=100 \] \(\textbf{Verify the conversion directly:}\) For an ideal gas, \[ [i]=\frac{p_i}{RT} \] Thus, \[ [A]=\frac{1.00}{20.0}=0.0500\,\mathrm{M} \] \[ [B]=\frac{2.00}{20.0}=0.100\,\mathrm{M} \] \[ [C]=\frac{1.00}{20.0}=0.0500\,\mathrm{M} \] The concentration equilibrium expression gives \[ K_c=\frac{0.0500}{0.0500(0.100)^2} \] \[ K_c=\frac{0.0500}{0.000500} \] \[ K_c=100\,\mathrm{L^2\,mol^{-2}} \] Both routes agree. The negative value of \(\Delta n_g\) is crucial because it determines whether the conversion multiplies or divides by the appropriate power of \(RT\). The same equilibrium composition produces different numerical \(K_p\) and \(K_c\) values because their expressions involve different pressure and concentration conventions. \(\textbf{Final answer:}\) \[ \boxed{K_p=0.250\,\mathrm{bar^{-2}},\quad K_c=100\,\mathrm{L^2\,mol^{-2}}} \]

Explanation: The Assertion distinguishes the reaction Gibbs energy under actual equilibrium conditions from the standard reaction Gibbs energy. These quantities are related but are not generally equal. The standard reaction Gibbs energy is connected to the equilibrium constant through \[ \Delta_rG^\circ=-RT\ln K \] Because \(K=25\) is greater than one, \(\ln K\) is positive. Consequently, \(\Delta_rG^\circ\) is negative for the reaction as written. This indicates that products are favoured relative to reactants under the corresponding standard-state comparison. However, the reaction Gibbs energy depends on the actual composition of the mixture. Its relationship with the reaction quotient is \[ \Delta_rG=\Delta_rG^\circ+RT\ln Q \] Combining this expression with the standard Gibbs energy relationship gives \[ \Delta_rG=RT\ln\left(\frac{Q}{K}\right) \] At equilibrium, the composition satisfies \(Q=K\). Therefore, \[ \Delta_rG=RT\ln1=0 \] This zero value means that an infinitesimal forward or reverse advancement has no net Gibbs-energy driving force at equilibrium. It does not imply that the total Gibbs energy of the system is zero. It also does not imply that the standard reaction Gibbs energy must be zero. The Reason correctly explains why the actual reaction Gibbs energy vanishes even when the standard reaction Gibbs energy is negative. Thus, both statements are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Identify the instantaneous composition:}\) The equilibrium constant refers to the balanced reaction as written. The actual reaction Gibbs energy depends on the current reaction quotient, not merely on the magnitude of K. \(\textbf{Construct the reaction quotient:}\) For the given reaction, \[ Q=\frac{a_C^2}{a_Aa_B} \] Under the stated ideal concentration convention, the numerical activity ratios correspond to the supplied molarities divided by the standard concentration. Therefore, \[ Q=\frac{(1.00)^2}{(0.50)(0.50)} \] \[ Q=\frac{1.00}{0.25} \] \[ Q=4 \] \(\textbf{Compare the instantaneous and equilibrium ratios:}\) \[ K=16 \] \[ Q=4 \] Since \[ Q\lt K \] the present mixture has a smaller product-to-reactant ratio than required by equilibrium. The forward reaction is consequently favoured. \(\textbf{Calculate the reaction Gibbs energy:}\) \[ \Delta_rG=RT\ln\left(\frac{Q}{K}\right) \] Substituting, \[ \Delta_rG=2.50\ln\left(\frac4{16}\right) \] \[ \Delta_rG=2.50\ln\left(\frac14\right) \] Using \[ \ln\left(\frac14\right)=-\ln4 \] gives \[ \Delta_rG=-2.50(1.386) \] \[ \Delta_rG=-3.465\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpret the sign:}\) The negative reaction Gibbs energy shows that an infinitesimal forward advancement decreases the Gibbs energy under the specified conditions. As A and B are consumed and C is produced, the reaction quotient increases. The magnitude of the forward Gibbs-energy driving force decreases as Q approaches K. At equilibrium, \(Q=K\) and \(\Delta_rG=0\). The calculation therefore links the measured composition, reaction quotient and thermodynamic direction consistently. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rG\approx-3.47\,\mathrm{kJ\,mol^{-1}}} \] The forward reaction is thermodynamically favoured.

Explanation: The Gibbs energy of a reacting system depends on its composition as well as its temperature and pressure. Reaction extent describes the progress of the reaction according to its balanced stoichiometric equation. At constant temperature and pressure, the slope of the total Gibbs-energy curve with respect to reaction extent is the reaction Gibbs energy. The relevant relation is \[ \Delta_rG=\left(\frac{\partial G}{\partial\xi}\right)_{T,P} \] At point P, increasing reaction extent moves the system down the descending part of the curve. The slope is negative, so \(\Delta_rG\) is negative. A small forward advancement therefore decreases the total Gibbs energy. At point Q, the curve reaches its minimum. The slope is zero, indicating \(\Delta_rG=0\). This is the thermodynamic equilibrium condition for the accessible reaction direction. At point R, the curve rises as reaction extent increases. Its slope is positive, so \(\Delta_rG\) is positive. The reverse reaction is therefore thermodynamically favoured from R because decreasing reaction extent lowers G. The total Gibbs energy at Q need not be zero; it is the slope that vanishes. Similarly, the standard reaction Gibbs energy cannot be determined from the existence of this minimum alone. The graph demonstrates that the forward or reverse tendency depends on the composition relative to equilibrium, not merely on the fact that the reaction is reversible. Hence, option C is correct.

Explanation: \(\textbf{Distinguish the two Gibbs-energy quantities:}\) The standard reaction Gibbs energy depends on the equilibrium constant at the specified temperature. The actual reaction Gibbs energy additionally depends on the instantaneous composition. The two quantities can consequently have different signs. \(\textbf{Calculate the standard reaction Gibbs energy:}\) \[ \Delta_rG^\circ=-RT\ln K \] Substituting, \[ \Delta_rG^\circ=-2.50\ln(10^{-2}) \] Using the logarithmic identity, \[ \ln(10^{-2})=-2\ln10 \] Therefore, \[ \Delta_rG^\circ=-2.50[-2(2.303)] \] \[ \Delta_rG^\circ=+11.515\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpret the standard-state sign:}\) The positive value agrees with \(K\lt 1\). The equilibrium is reactant-favoured for the reaction as written. However, this does not yet establish the favoured direction from the specified nonequilibrium mixture. \(\textbf{Calculate the actual reaction Gibbs energy:}\) \[ \Delta_rG=RT\ln\left(\frac{Q}{K}\right) \] Substituting, \[ \Delta_rG=2.50\ln\left(\frac{10^{-3}}{10^{-2}}\right) \] \[ \Delta_rG=2.50\ln(10^{-1}) \] \[ \Delta_rG=-2.50(2.303) \] \[ \Delta_rG=-5.7575\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Verify using the alternative expression:}\) \[ \Delta_rG=\Delta_rG^\circ+RT\ln Q \] \[ \Delta_rG=11.515+2.50[-3(2.303)] \] \[ \Delta_rG=11.515-17.2725 \] \[ \Delta_rG=-5.7575\,\mathrm{kJ\,mol^{-1}} \] Both approaches agree. \(\textbf{Interpret the apparently opposing signs:}\) Although the standard reaction Gibbs energy is positive, the current mixture has \(Q\lt K\). The mixture is therefore sufficiently reactant-rich for the forward reaction to be thermodynamically favoured. Forward reaction continues until the quotient reaches its equilibrium value. A reactant-favoured equilibrium does not imply that the reverse reaction is favoured from every possible initial composition. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rG^\circ\approx+11.52\,\mathrm{kJ\,mol^{-1}},\quad\Delta_rG\approx-5.76\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Combine the successive reactions:}\) The first reaction converts A into B. The second converts B into C. Adding them cancels the intermediate B. The combined reaction is \[ A(g)\rightleftharpoons C(g) \] Standard reaction Gibbs energies are additive when reactions are added. Therefore, \[ \Delta_rG_{A\to C}^\circ=\Delta_rG_1^\circ+\Delta_rG_2^\circ \] Substituting, \[ \Delta_rG_{A\to C}^\circ=-6.0+2.0 \] \[ \Delta_rG_{A\to C}^\circ=-4.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Reverse the combined reaction:}\) The required reaction proceeds from C towards A. Reversing a chemical equation changes the sign of its standard reaction Gibbs energy. Thus, \[ \Delta_rG_{C\to A}^\circ=+4.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Apply the stoichiometric multiplication:}\) The required equation contains two units of C converting into two units of A. Every coefficient in the reverse reaction is therefore multiplied by two. Standard reaction Gibbs energy scales by the same factor. \[ \Delta_rG_{\mathrm{required}}^\circ=2(+4.0) \] \[ \Delta_rG_{\mathrm{required}}^\circ=+8.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Determine the corresponding equilibrium constant:}\) \[ \Delta_rG^\circ=-RT\ln K \] Rearranging, \[ \ln K=-\frac{\Delta_rG^\circ}{RT} \] Substituting, \[ \ln K_{\mathrm{required}}=-\frac{8.0}{2.0} \] \[ \ln K_{\mathrm{required}}=-4 \] \[ K_{\mathrm{required}}=e^{-4} \] \(\textbf{Check the thermodynamic consistency:}\) The required reaction has a positive standard reaction Gibbs energy. Its equilibrium constant must therefore be smaller than one. The result \(e^{-4}\) satisfies this condition. Equivalently, the original combined reaction has a constant greater than one, while reversing it and doubling its coefficients produces the reciprocal squared relationship. The standard energy and equilibrium constant transformations are therefore mutually consistent. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rG_{\mathrm{required}}^\circ=+8.0\,\mathrm{kJ\,mol^{-1}},\quad K_{\mathrm{required}}=e^{-4}} \]

Explanation: \(\textbf{Determine the standard reaction Gibbs energy:}\) The standard reaction Gibbs energy is calculated from the standard molar Gibbs energies of formation. For the balanced reaction, \[ \Delta_rG^\circ=\sum\nu\Delta_fG^\circ(\mathrm{products})-\sum\nu\Delta_fG^\circ(\mathrm{reactants}) \] All stoichiometric coefficients are one. Therefore, \[ \Delta_rG^\circ=\Delta_fG_C^\circ-(\Delta_fG_A^\circ+\Delta_fG_B^\circ) \] Substituting, \[ \Delta_rG^\circ=-40-[(-10)+(-20)] \] \[ \Delta_rG^\circ=-40-(-30) \] \[ \Delta_rG^\circ=-10\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpret the standard-state result:}\) The negative standard reaction Gibbs energy indicates that products are favoured in the standard-state comparison. It also implies that the corresponding equilibrium constant is greater than one. However, the actual direction from the present mixture requires the reaction quotient. \(\textbf{Calculate the composition-dependent contribution:}\) \[ \Delta_rG=\Delta_rG^\circ+RT\ln Q \] The supplied quotient is \[ Q=e^6 \] Therefore, \[ \ln Q=6 \] The composition-dependent contribution is \[ RT\ln Q=2.0\times6 \] \[ RT\ln Q=12.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Calculate the actual reaction Gibbs energy:}\) \[ \Delta_rG=-10+12 \] \[ \Delta_rG=+2.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Verify the reaction direction using K:}\) From the standard Gibbs energy relationship, \[ \ln K=-\frac{-10}{2.0}=5 \] \[ K=e^5 \] Since the actual quotient is \(e^6\), \[ Q\gt K \] The reverse reaction is thermodynamically favoured. The positive reaction Gibbs energy is consistent with this direction. \(\textbf{Physical interpretation:}\) The standard reaction is product-favoured, but the supplied mixture contains an even greater product-to-reactant ratio than equilibrium permits. A reverse adjustment consequently lowers the total Gibbs energy. This illustrates why standard Gibbs energy and actual reaction Gibbs energy can have opposite signs. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rG=+2.0\,\mathrm{kJ\,mol^{-1}}} \] The reverse reaction is thermodynamically favoured.

Explanation: \(\textbf{Identify the linear Gibbs-energy relation:}\) At a fixed temperature, \[ \Delta_rG=\Delta_rG^\circ+RT\ln Q \] This has the mathematical form of a straight line in a plot of \(\Delta_rG\) against \(\ln Q\). The vertical intercept equals \(\Delta_rG^\circ\). The slope equals RT. \(\textbf{Determine the standard reaction Gibbs energy:}\) At \[ \ln Q=0 \] the graph gives \[ \Delta_rG=-12\,\mathrm{kJ\,mol^{-1}} \] Therefore, \[ \Delta_rG^\circ=-12\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Determine the equilibrium quotient:}\) The graph crosses the horizontal axis at \[ \ln Q=3 \] At that point, \[ \Delta_rG=0 \] This is the equilibrium condition. Consequently, \[ \ln K=3 \] \[ K=e^3 \] \(\textbf{Calculate the graph slope:}\) The slope is the change in Gibbs energy divided by the change in \(\ln Q\). \[ RT=\frac{0-(-12)}{3-0} \] \[ RT=4.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Evaluate the Gibbs energy at the requested composition:}\) For \[ Q=e^4 \] we have \[ \ln Q=4 \] Substituting into the linear relation, \[ \Delta_rG=-12+4.0(4) \] \[ \Delta_rG=+4.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Verify the reaction tendency:}\) Since \[ e^4\gt e^3 \] the actual reaction quotient is greater than the equilibrium constant. The reverse reaction is thermodynamically favoured. The positive Gibbs energy obtained from the graph is consistent with this conclusion. The graph therefore permits three connected inferences: standard Gibbs energy from the intercept, equilibrium constant from the zero crossing and actual Gibbs energy from the slope. \(\textbf{Final answer:}\) \[ \boxed{K=e^3,\quad\Delta_rG=+4.0\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: Statement I is correct because the standard reaction Gibbs energy is related to the equilibrium constant by \[ \Delta_rG^\circ=-RT\ln K \] Since \(K=9\gt 1\), the logarithm is positive. Therefore, the standard reaction Gibbs energy is negative. Statement II is correct because the actual reaction Gibbs energy satisfies \[ \Delta_rG=RT\ln\left(\frac{Q}{K}\right) \] When \(Q=9\), the ratio \(Q/K\) equals one. Its natural logarithm is zero, so the reaction Gibbs energy is zero. This is the thermodynamic criterion for equilibrium. Statement III is also correct. When \(Q=27\), \[ \frac{Q}{K}=\frac{27}{9}=3 \] The logarithm of three is positive, so \(\Delta_rG\) is positive for the written forward reaction. Accordingly, the reverse reaction is thermodynamically favoured. The positive actual reaction Gibbs energy does not contradict the negative standard reaction Gibbs energy because the two quantities refer to different compositions. Statement IV is incorrect because an equilibrium constant does not specify the individual reaction rates at every composition. The instantaneous balance of the forward and reverse processes depends on the actual state and reaction kinetics. At equilibrium, the opposing rates are equal despite \(K\gt 1\). A large equilibrium constant therefore does not require the forward rate to exceed the reverse rate universally. Thus, Statements I, II and III are correct.

Explanation: \(\textbf{Recover the equilibrium constant from Gibbs energy:}\) The actual reaction Gibbs energy is related to the reaction quotient by \[ \Delta_rG=RT\ln\left(\frac{Q}{K}\right) \] The measured value is \[ \Delta_rG=RT\ln3 \] Equating the expressions, \[ RT\ln\left(\frac{Q}{K}\right)=RT\ln3 \] At the same positive absolute temperature, RT is nonzero. Therefore, \[ \ln\left(\frac{Q}{K}\right)=\ln3 \] \[ \frac{Q}{K}=3 \] The initial quotient is 12. Hence, \[ \frac{12}{K}=3 \] \[ K=4 \] \(\textbf{Determine the initial direction:}\) Initially, \[ Q_{\mathrm{initial}}=12 \] Since \[ 12\gt K=4 \] the reverse reaction is thermodynamically favoured. This agrees with the supplied positive reaction Gibbs energy. \(\textbf{Analyse the changed composition:}\) After the imposed change, \[ Q_{\mathrm{new}}=3 \] The equilibrium constant remains four because temperature has not changed. Thus, \[ Q_{\mathrm{new}}\lt K \] The forward reaction is now favoured. \(\textbf{Verify the new Gibbs-energy sign:}\) \[ \Delta_rG_{\mathrm{new}}=RT\ln\left(\frac34\right) \] Because the logarithm of a positive fraction smaller than one is negative, \[ \Delta_rG_{\mathrm{new}}\lt 0 \] The reaction therefore has a forward Gibbs-energy driving force after the composition change. \(\textbf{Interpret the reverse inference:}\) The initial Gibbs-energy measurement can be used to infer an unknown equilibrium constant when the reaction quotient is known. That constant can then be applied to a different composition at the same temperature. The reaction tendency reverses because the new quotient lies on the opposite side of K. The inference depends on the sign and magnitude of the logarithmic ratio, not on the concentration of any one species alone. \(\textbf{Final answer:}\) \[ \boxed{K=4} \] The reverse reaction is initially favoured, followed by the forward reaction after the composition change.

Explanation: Both systems initially satisfy their respective equilibrium conditions. Therefore, their initial reaction quotients equal their equilibrium constants. The sudden compression increases every gaseous concentration because the volume is halved while gaseous mole amounts remain instantaneously unchanged. For ideal gases, each species' concentration and partial pressure initially double. The effect on the reaction quotient depends on the stoichiometric coefficients. For system P, \[ Q_P=\frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} \] After compression, the numerator gains a factor of four. The denominator also gains a factor of four. These factors cancel, so the quotient remains equal to its original equilibrium value. Thus, \(Q_P=K_P\), and the reaction Gibbs energy remains zero. For system Q, \[ Q_Q=\frac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2} \] The numerator doubles under the instantaneous compression. The denominator increases by a factor of four. Consequently, the quotient becomes half its original value. Because the system was initially at equilibrium, \[ Q_{Q,\mathrm{new}}=\frac12K_Q \] The reaction Gibbs energy is then \[ \Delta_rG_Q=RT\ln\left(\frac{Q_{Q,\mathrm{new}}}{K_Q}\right) \] \[ \Delta_rG_Q=RT\ln\left(\frac12\right) \] This value is negative. The forward dimerisation of nitrogen dioxide is therefore thermodynamically favoured after compression. The distinction arises because system P has no net gaseous stoichiometric mole-number change, whereas system Q converts two gaseous units into one. The temperature is unchanged in both cases, so neither equilibrium constant changes. The difference lies entirely in how each reaction quotient responds to compression.