Chemical Thermodynamics MCQs with Answers – Part 4 (NEET Chemistry)

Explanation: The first reaction converts A into B. The second converts B into C. Adding these reactions eliminates the intermediate B and produces A converting into C. For compatible equilibrium reactions, addition of reaction equations corresponds to multiplication of their equilibrium constants. Therefore, \[ K_{A\rightarrow C}=K_1K_2 \] \[ K_{A\rightarrow C}=4(9)=36 \] This agrees with the supplied value of \(K_3\). Thus, the first three measurements are thermodynamically consistent. The fourth reaction is the reverse of the third. For a reversed equilibrium equation, the equilibrium constant becomes the reciprocal of the original value. Consequently, \[ K_{C\rightarrow A}=\frac1{K_{A\rightarrow C}} \] \[ K_{C\rightarrow A}=\frac1{36} \] However, the fourth measurement reports \(1/30\). This differs from the required value \(1/36\). The discrepancy cannot be removed by changing the sign of the reaction Gibbs energy while leaving the equilibrium constant unchanged. The equilibrium constant and standard Gibbs-energy change are connected by \[ \Delta_rG^\circ=-RT\ln K \] Reversing a reaction changes the Gibbs-energy sign and reciprocates its equilibrium constant. The positivity of all four equilibrium constants does not by itself establish thermochemical consistency. Only the fourth measurement violates the required relationship.

Explanation: Statement I is correct because a catalyst changes reaction kinetics. It can provide an alternative mechanism with a lower activation-energy barrier. This allows the reacting mixture to approach equilibrium more rapidly. However, the equilibrium constant is a thermodynamic property at the specified temperature for the reaction as written. A catalyst does not change this thermodynamic equilibrium constant. Statement II is incorrect because standard Gibbs energy is not determined by reaction speed. The standard Gibbs-energy change depends on the thermodynamic states of reactants and products. A faster reaction does not require a larger negative Gibbs-energy change. Statement III is correct because chemical equilibrium corresponds to a minimum in the appropriate Gibbs energy at constant temperature and pressure. At this condition, an infinitesimal advancement of the reaction has zero Gibbs-energy change. Forward and reverse microscopic reactions may continue, but there is no net thermodynamic tendency to change the equilibrium composition. Statement IV is correct because Gibbs energy is a state function. Changing the reaction mechanism changes the pathway but not the Gibbs-energy difference between identical initial and final states. The catalyst can alter the rates of the forward and reverse reactions without changing the equilibrium state they approach. Thus, Statements I, III and IV are correct. The essential distinction is between the rate of reaching equilibrium and the thermodynamic position of equilibrium.

Explanation: \(\textbf{Recovering Gibbs energy from the equilibrium constant:}\) The standard Gibbs-energy relation is \[ \Delta_rG^\circ=-RT\ln K \] The equilibrium constant is \[ K=e^6 \] Therefore, \[ \ln K=6 \] Substituting, \[ \Delta_rG^\circ_{400}=-(8.314)(400)(6) \] \[ \Delta_rG^\circ_{400}=-19953.6\,\mathrm{J\,mol^{-1}} \] \[ \Delta_rG^\circ_{400}\approx-19.954\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Recovering standard reaction entropy:}\) Use \[ \Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ \] Rearranging, \[ \Delta_rS^\circ=\frac{\Delta_rH^\circ-\Delta_rG^\circ}{T} \] Substituting, \[ \Delta_rS^\circ=\frac{-30.0-(-19.954)}{400} \] \[ \Delta_rS^\circ=\frac{-10.046}{400} \] \[ \Delta_rS^\circ=-0.025115\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] Therefore, \[ \Delta_rS^\circ\approx-25.1\,\mathrm{J\,mol^{-1}\,K^{-1}} \] \(\textbf{Locating the Gibbs-energy boundary:}\) Under the constant-enthalpy and constant-entropy approximation, the boundary occurs when \[ \Delta_rG^\circ=0 \] Hence, \[ T=\frac{\Delta_rH^\circ}{\Delta_rS^\circ} \] Substituting the compatible kilojoule units, \[ T=\frac{-30.0}{-0.025115} \] \[ T\approx1194\,\mathrm{K} \] Rounding to the precision of the proposed options gives approximately 1200 K. \(\textbf{Checking the thermodynamic trend:}\) The reaction is exothermic and has a negative entropy change. At lower temperatures, the favourable enthalpy contribution can dominate. At sufficiently high temperatures, the negative entropy contribution makes the standard Gibbs-energy change less favourable. The initial equilibrium constant \(e^6\) is greater than one, consistent with the negative standard Gibbs-energy change at 400 K. The inferred threshold is a model prediction based on temperature-independent enthalpy and entropy values. It should not be extrapolated through actual phase transitions without further thermodynamic information. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rS^\circ\approx-25.1\,\mathrm{J\,mol^{-1}\,K^{-1}},\quad T_{\Delta G^\circ=0}\approx1.19\times10^3\,\mathrm{K}} \]

Explanation: Statement I is correct because the complete container is rigid and isolated. No matter or energy crosses the external boundary. With no heat or external work transfer, the first law requires the total internal energy to remain constant. Statement II is correct because an isolated system can undergo irreversible internal changes. These changes can generate entropy even though the system exchanges no heat with its surroundings. For example, spontaneous mixing or free expansion can increase entropy. Statement III is incorrect because Gibbs energy is not identical to internal energy. It is defined by \[ G=H-TS \] The Gibbs-energy change depends on the relevant enthalpy, entropy and temperature conditions. Zero internal-energy change alone is not sufficient to conclude that the Gibbs-energy change is zero. Furthermore, the usual negative-Gibbs-energy spontaneity criterion is specifically associated with constant temperature and pressure. It should not be imposed automatically on every isolated rigid system. Statement IV is correct because the second law permits an isolated system to evolve spontaneously toward states of greater entropy. Entropy production accompanies irreversible progress. At equilibrium, the isolated system has no further spontaneous internal change available under its constraints. The problem distinguishes the energy-conservation criterion from the entropy criterion for isolated systems. Statements I, II and IV are correct.

Explanation: \(\textbf{Reversing the first equilibrium:}\) The original reaction is \[ \mathrm{A\rightleftharpoons B} \] with \[ K_1=25 \] Reversal gives \[ \mathrm{B\rightleftharpoons A} \] Its equilibrium constant is \[ K_{\mathrm{reverse}}=\frac1{25} \] \(\textbf{Doubling the reversed equation:}\) The transformed reaction is \[ \mathrm{2B\rightleftharpoons2A} \] Multiplying every coefficient by two squares the equilibrium constant. Therefore, \[ K'_1=\left(\frac1{25}\right)^2 \] \[ K'_1=\frac1{625} \] \(\textbf{Adding the independent equilibrium:}\) The second equilibrium is \[ \mathrm{C\rightleftharpoons D} \] with \[ K_2=125 \] The combined equation is \[ \mathrm{2B+C\rightleftharpoons2A+D} \] When compatible equilibrium equations are added, their equilibrium constants multiply. Thus, \[ K_{\mathrm{combined}}=K'_1K_2 \] Substituting, \[ K_{\mathrm{combined}}=\frac1{625}(125) \] \[ K_{\mathrm{combined}}=\frac15 \] \(\textbf{Determining the Gibbs-energy sign:}\) The standard Gibbs-energy relation is \[ \Delta_rG^\circ=-RT\ln K \] Here, \[ K_{\mathrm{combined}}=\frac15\lt 1 \] Therefore, \[ \ln K_{\mathrm{combined}}\lt 0 \] The negative sign in the Gibbs-energy relation consequently gives \[ \Delta_rG^\circ_{\mathrm{combined}}\gt 0 \] \(\textbf{Thermodynamic interpretation:}\) The original A-to-B equilibrium is product-favoured. However, the target uses its reversed equation with doubled coefficients, producing a substantial reciprocal-square effect. The second equilibrium partially offsets this effect but does not overcome it completely. The resulting combined reaction has an equilibrium constant below one. Its standard reaction Gibbs-energy change is therefore positive.

Explanation: \(\textbf{Calculating the Gibbs energy for the reference reaction:}\) At constant temperature, \[ \Delta_rG=\Delta_rH-T\Delta_rS \] The reaction enthalpy is \[ \Delta_rH=-24.0\,\mathrm{kJ\,mol^{-1}} \] The entropy change in compatible units is \[ \Delta_rS=-0.060\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] Therefore, \[ \Delta_rG_P=-24.0-(300)(-0.060) \] \[ \Delta_rG_P=-24.0+18.0 \] \[ \Delta_rG_P=-6.0\,\mathrm{kJ} \] This value corresponds to the one-mole reaction extent specified in P. \(\textbf{Scaling the reaction in Experiment Q:}\) Experiment Q proceeds through twice the stoichiometric reaction extent. Gibbs energy is an extensive state function. Under the stated corresponding-state conditions, the total Gibbs-energy change scales with the reaction extent. Therefore, \[ \Delta G_Q=2\Delta G_P \] \[ \Delta G_Q=2(-6.0) \] \[ \Delta G_Q=-12.0\,\mathrm{kJ} \] \(\textbf{Reversing and halving the reaction in Experiment R:}\) Experiment R proceeds in the reverse direction. Reversing a thermochemical transformation changes the sign of its Gibbs-energy change. Therefore, the Gibbs-energy change for one mole of reversed reaction is \[ \Delta G_{\mathrm{reverse}}=+6.0\,\mathrm{kJ} \] Only half that stoichiometric extent occurs in R. Thus, \[ \Delta G_R=\frac12(+6.0) \] \[ \Delta G_R=+3.0\,\mathrm{kJ} \] \(\textbf{Independent enthalpy-entropy check for R:}\) For the reversed half-reaction, \[ \Delta H_R=\frac12(+24.0)=+12.0\,\mathrm{kJ} \] \[ \Delta S_R=\frac12(+60)=+30\,\mathrm{J\,K^{-1}} \] Therefore, \[ \Delta G_R=12.0-(300)(0.030) \] \[ \Delta G_R=+3.0\,\mathrm{kJ} \] \(\textbf{Thermodynamic interpretation:}\) The original forward reaction is favoured under the supplied conditions because its Gibbs-energy change is negative. Doubling its extent doubles the total negative Gibbs-energy change. Reversing it makes the Gibbs-energy change positive, and halving its extent reduces the magnitude accordingly. The calculation requires reaction direction and stoichiometric scaling to be treated separately. \(\textbf{Final answer:}\) \[ \boxed{\left(\Delta G_Q,\Delta G_R\right)=\left(-12.0,+3.0\right)\,\mathrm{kJ}} \]

Explanation: \(\textbf{Applying the first law to the complete cycle:}\) Because the engine returns to its initial state, its internal-energy change over a cycle is zero. The energy received as heat must equal the sum of the work delivered and the heat rejected. \[ Q_h=W_{\mathrm{out}}+Q_c \] The supplied quantities are \[ Q_h=1500\,\mathrm{J} \] \[ W_{\mathrm{out}}=500\,\mathrm{J} \] Therefore, \[ Q_c=1500-500 \] \[ Q_c=1000\,\mathrm{J} \] \(\textbf{Entropy change of the hotter reservoir:}\) The hotter reservoir loses 1500 J at 500 K. \[ \Delta S_h=\frac{-Q_h}{T_h} \] \[ \Delta S_h=\frac{-1500}{500} \] \[ \Delta S_h=-3.000\,\mathrm{J\,K^{-1}} \] \(\textbf{Entropy change of the colder reservoir:}\) The colder reservoir receives 1000 J at 300 K. \[ \Delta S_c=\frac{Q_c}{T_c} \] \[ \Delta S_c=\frac{1000}{300} \] \[ \Delta S_c=+3.333\,\mathrm{J\,K^{-1}} \] \(\textbf{Combining the entropy changes:}\) The working substance undergoes a complete cycle. Since entropy is a state function, \[ \Delta S_{\mathrm{engine}}=0 \] Consequently, \[ \Delta S_{\mathrm{univ}}=\Delta S_h+\Delta S_c \] \[ \Delta S_{\mathrm{univ}}=-3.000+3.333 \] \[ \Delta S_{\mathrm{univ}}=+0.333\,\mathrm{J\,K^{-1}} \] \(\textbf{Checking the thermodynamic possibility:}\) The positive entropy change is compatible with the second law. The engine converts some absorbed heat into work and rejects the remainder. It does not convert all the heat received from a single reservoir into work while completing a cycle. The first law determines the heat rejected, but only the entropy calculation verifies the second-law consistency of the complete operation. \(\textbf{Final answer:}\) \[ \boxed{\left(Q_c,\Delta S_{\mathrm{univ}}\right)=\left(1000\,\mathrm{J},+0.333\,\mathrm{J\,K^{-1}}\right)} \]

Explanation: \(\textbf{Determining the final composition:}\) The total number of moles is \[ n_{\mathrm{total}}=1.00+3.00 \] \[ n_{\mathrm{total}}=4.00\,\mathrm{mol} \] The final mole fractions are \[ x_A=\frac{1.00}{4.00}=\frac14 \] \[ x_B=\frac{3.00}{4.00}=\frac34 \] \(\textbf{Evaluating the entropy-of-mixing relation:}\) The supplied expression is \[ \Delta S_{\mathrm{mix}}=-R\left(n_A\ln x_A+n_B\ln x_B\right) \] Substituting the amounts and mole fractions, \[ \Delta S_{\mathrm{mix}}=-8.314\left[\ln\left(\frac14\right)+3\ln\left(\frac34\right)\right] \] Using logarithmic identities, \[ \ln\left(\frac14\right)=-\ln4 \] \[ \ln\left(\frac34\right)=-\ln\left(\frac43\right) \] Therefore, \[ \Delta S_{\mathrm{mix}}=8.314[1.386+3(0.288)] \] \[ \Delta S_{\mathrm{mix}}=8.314(1.386+0.864) \] \[ \Delta S_{\mathrm{mix}}=8.314(2.250) \] \[ \Delta S_{\mathrm{mix}}\approx+18.71\,\mathrm{J\,K^{-1}} \] \(\textbf{Determining the Gibbs-energy change:}\) The gases are ideal, and the temperature remains constant. Under the supplied model, \[ \Delta H_{\mathrm{mix}}=0 \] Therefore, \[ \Delta G_{\mathrm{mix}}=\Delta H_{\mathrm{mix}}-T\Delta S_{\mathrm{mix}} \] \[ \Delta G_{\mathrm{mix}}=0-(300)(18.71) \] \[ \Delta G_{\mathrm{mix}}\approx-5613\,\mathrm{J} \] \[ \Delta G_{\mathrm{mix}}\approx-5.61\,\mathrm{kJ} \] \(\textbf{Connecting the result to spontaneous mixing:}\) The positive entropy change is consistent with spontaneous mixing of distinguishable ideal gases. Their total internal energy remains unchanged because the enclosure is rigid and insulated, no reaction occurs and ideal-gas internal energy depends on temperature. Nevertheless, the composition distribution changes. The negative Gibbs-energy difference between the specified states is consistent with the favourable direction of mixing at the stated temperature. The entropy calculation must use the unequal mole fractions rather than treating the mixture as equimolar. \(\textbf{Final answer:}\) \[ \boxed{\left(\Delta S_{\mathrm{mix}},\Delta G_{\mathrm{mix}}\right)\approx\left(+18.71\,\mathrm{J\,K^{-1}},-5.61\,\mathrm{kJ}\right)} \]

Explanation: \(\textbf{Locating the equilibrium temperature from energy conservation:}\) Heat lost by the hotter block equals heat gained by the colder block. The blocks have different heat capacities, so the final temperature is not their simple arithmetic mean. \[ C_P(T_{P,i}-T_f)=C_Q(T_f-T_{Q,i}) \] Substituting, \[ 100(400-T_f)=200(T_f-250) \] \[ 40000-100T_f=200T_f-50000 \] \[ 90000=300T_f \] \[ T_f=300\,\mathrm{K} \] \(\textbf{Entropy change of Block P:}\) The hotter block cools from 400 K to 300 K. \[ \Delta S_P=100\ln\left(\frac{300}{400}\right) \] \[ \Delta S_P=100\ln\left(\frac34\right) \] \[ \Delta S_P=100(-0.28768) \] \[ \Delta S_P=-28.768\,\mathrm{J\,K^{-1}} \] \(\textbf{Entropy change of Block Q:}\) The colder block warms from 250 K to 300 K. \[ \Delta S_Q=200\ln\left(\frac{300}{250}\right) \] \[ \Delta S_Q=200\ln\left(\frac65\right) \] \[ \Delta S_Q=200(0.18232) \] \[ \Delta S_Q=+36.464\,\mathrm{J\,K^{-1}} \] \(\textbf{Total entropy change:}\) \[ \Delta S_{\mathrm{total}}=\Delta S_P+\Delta S_Q \] \[ \Delta S_{\mathrm{total}}=-28.768+36.464 \] \[ \Delta S_{\mathrm{total}}=+7.696\,\mathrm{J\,K^{-1}} \] \(\textbf{Checking both thermodynamic laws:}\) Block P releases \[ q_P=100(300-400)=-10000\,\mathrm{J} \] Block Q absorbs \[ q_Q=200(300-250)=+10000\,\mathrm{J} \] The energy transfers cancel, satisfying the first law. Their entropy changes do not cancel because heat is transferred between bodies at unequal temperatures. The entropy of the isolated enclosure increases, satisfying the second law. \(\textbf{Final answer:}\) \[ \boxed{\left(T_f,\Delta S_{\mathrm{total}}\right)\approx\left(300\,\mathrm{K},+7.70\,\mathrm{J\,K^{-1}}\right)} \]

Explanation: \(\textbf{Identifying the required phase correction:}\) The supplied reaction produces two moles of gaseous D. The target reaction produces two moles of liquid D. Therefore, condensation of two moles of D must be added to the gaseous-product reaction. Both enthalpy and entropy change when the physical state changes. \(\textbf{Combining the enthalpy changes:}\) \[ \Delta_rH_l=\Delta_rH_g+2\Delta_{\mathrm{cond}}H_D \] Substituting, \[ \Delta_rH_l=-40+2(-25) \] \[ \Delta_rH_l=-40-50 \] \[ \Delta_rH_l=-90\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Combining the entropy changes:}\) \[ \Delta_rS_l=\Delta_rS_g+2\Delta_{\mathrm{cond}}S_D \] \[ \Delta_rS_l=-100+2(-70) \] \[ \Delta_rS_l=-240\,\mathrm{J\,mol^{-1}\,K^{-1}} \] Converting units, \[ \Delta_rS_l=-0.240\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] \(\textbf{Calculating Gibbs energy for the liquid-product reaction:}\) \[ \Delta_rG_l=\Delta_rH_l-T\Delta_rS_l \] \[ \Delta_rG_l=-90-(300)(-0.240) \] \[ \Delta_rG_l=-90+72 \] \[ \Delta_rG_l=-18\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Independent phase-correction check:}\) For the original gaseous-product reaction, \[ \Delta_rG_g=-40-(300)(-0.100) \] \[ \Delta_rG_g=-10\,\mathrm{kJ\,mol^{-1}} \] For condensation of one mole of D, \[ \Delta_{\mathrm{cond}}G_D=-25-(300)(-0.070) \] \[ \Delta_{\mathrm{cond}}G_D=-4\,\mathrm{kJ\,mol^{-1}} \] Condensing two moles contributes \[ 2(-4)=-8\,\mathrm{kJ} \] Therefore, \[ \Delta_rG_l=-10-8 \] \[ \Delta_rG_l=-18\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Thermodynamic interpretation:}\) Condensation makes the reaction enthalpy considerably more negative. However, it also produces a more negative entropy change. Consequently, the Gibbs-energy correction is smaller in magnitude than the enthalpy correction. The physical-state designation of the products is essential when comparing reaction Gibbs energies. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rG_l=-18\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Separating phase-transition and sensible-heating entropy:}\) Melting and vaporisation occur at fixed equilibrium-transition temperatures. Heating of the liquid and gas occurs over temperature intervals. The entropy of the substance must therefore be calculated as the sum of four distinct contributions. \(\textbf{Stage I — Entropy gained during fusion:}\) \[ \Delta S_1=\frac{\Delta_{\mathrm{fus}}H}{T_m} \] \[ \Delta S_1=\frac{6000}{300} \] \[ \Delta S_1=+20.0\,\mathrm{J\,K^{-1}} \] \(\textbf{Stage II — Entropy gained during liquid heating:}\) For the constant heat capacity, \[ \Delta S_2=C_{p,m}(l)\ln\left(\frac{T_2}{T_1}\right) \] \[ \Delta S_2=100\ln\left(\frac{350}{300}\right) \] \[ \Delta S_2=100(0.15415) \] \[ \Delta S_2=+15.415\,\mathrm{J\,K^{-1}} \] \(\textbf{Stage III — Entropy gained during vaporisation:}\) \[ \Delta S_3=\frac{\Delta_{\mathrm{vap}}H}{T_b} \] \[ \Delta S_3=\frac{35000}{350} \] \[ \Delta S_3=+100.0\,\mathrm{J\,K^{-1}} \] \(\textbf{Stage IV — Entropy gained during vapour heating:}\) \[ \Delta S_4=C_{p,m}(g)\ln\left(\frac{400}{350}\right) \] \[ \Delta S_4=30(0.13353) \] \[ \Delta S_4=+4.0059\,\mathrm{J\,K^{-1}} \] \(\textbf{Total entropy change of the substance:}\) \[ \Delta S_{\mathrm{sys}}=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4 \] \[ \Delta S_{\mathrm{sys}}=20.0+15.415+100.0+4.0059 \] \[ \Delta S_{\mathrm{sys}}=+139.4209\,\mathrm{J\,K^{-1}} \] \(\textbf{Calculating the total heat supplied:}\) The energy supplied for each stage is respectively \[ q_1=6000\,\mathrm{J} \] \[ q_2=100(350-300)=5000\,\mathrm{J} \] \[ q_3=35000\,\mathrm{J} \] \[ q_4=30(400-350)=1500\,\mathrm{J} \] Thus, \[ q_{\mathrm{total}}=6000+5000+35000+1500 \] \[ q_{\mathrm{total}}=47500\,\mathrm{J} \] \(\textbf{Entropy lost by the reservoir:}\) The reservoir remains at 500 K. \[ \Delta S_{\mathrm{surr}}=\frac{-47500}{500} \] \[ \Delta S_{\mathrm{surr}}=-95.0\,\mathrm{J\,K^{-1}} \] \(\textbf{Universe entropy change:}\) \[ \Delta S_{\mathrm{univ}}=139.4209-95.0 \] \[ \Delta S_{\mathrm{univ}}\approx+44.4\,\mathrm{J\,K^{-1}} \] \(\textbf{Physical interpretation:}\) The substance undergoes two phase changes and two sensible-heating stages. The entropy changes of these stages are not obtained by dividing the total heat by one arbitrarily selected temperature. The substance gains more entropy than the constant-temperature reservoir loses. The positive total entropy change is consistent with irreversible heat transfer from the hotter reservoir. \(\textbf{Final answer:}\) \[ \boxed{\Delta S_{\mathrm{univ}}\approx+44.4\,\mathrm{J\,K^{-1}}} \]

Explanation: \(\textbf{Calculating universe entropy per mole of R₁:}\) For the specified reservoir, \[ \Delta S_{\mathrm{univ}}=\Delta S_{\mathrm{sys}}-\frac{\Delta H_{\mathrm{sys}}}{T} \] For R₁, \[ \Delta S_{\mathrm{univ},1}=-80-\frac{-30000}{300} \] \[ \Delta S_{\mathrm{univ},1}=-80+100 \] \[ \Delta S_{\mathrm{univ},1}=+20\,\mathrm{J\,mol^{-1}\,K^{-1}} \] \(\textbf{Universe entropy per mole of R₂:}\) Reaction R₂ absorbs enthalpy and increases system entropy. \[ \Delta S_{\mathrm{univ},2}=20-\frac{12000}{300} \] \[ \Delta S_{\mathrm{univ},2}=20-40 \] \[ \Delta S_{\mathrm{univ},2}=-20\,\mathrm{J\,mol^{-1}\,K^{-1}} \] Thus, the forward direction of R₂ is individually unfavourable at the specified conditions. \(\textbf{Using the measured total entropy increase:}\) Let x be the amount of C converted. The R₁ reaction extent is 0.400 mol. Therefore, \[ \Delta S_{\mathrm{univ,total}}=(0.400)(20)+x(-20) \] Substituting the measured value, \[ 2.0=8.0-20x \] \[ 20x=6.0 \] \[ x=0.300\,\mathrm{mol} \] \(\textbf{Determining the heat transferred by both reactions:}\) For R₁, \[ q_1=(0.400)(-30) \] \[ q_1=-12.0\,\mathrm{kJ} \] For R₂, \[ q_2=(0.300)(+12) \] \[ q_2=+3.6\,\mathrm{kJ} \] The net heat transfer is \[ q_{\mathrm{total}}=-12.0+3.6 \] \[ q_{\mathrm{total}}=-8.4\,\mathrm{kJ} \] \(\textbf{Checking the coupled spontaneity:}\) The favourable reaction contributes \[ \Delta S_{\mathrm{univ},1}=(0.400)(20)=+8.0\,\mathrm{J\,K^{-1}} \] The unfavourable reaction contributes \[ \Delta S_{\mathrm{univ},2}=(0.300)(-20)=-6.0\,\mathrm{J\,K^{-1}} \] Their sum is \[ +8.0-6.0=+2.0\,\mathrm{J\,K^{-1}} \] The positive total entropy change is consistent with the stated coupled transformation. The independently unfavourable second reaction can proceed as part of a favourable overall process if an effective coupling mechanism exists. \(\textbf{Final answer:}\) \[ \boxed{\left(n_{\mathrm{C,converted}},|q_{\mathrm{total}}|\right)=\left(0.300\,\mathrm{mol},8.4\,\mathrm{kJ}\right)} \]

Explanation: Statement I is correct because the expression \[ \Delta_rG=\Delta_rH-T\Delta_rS \] has the form of a straight-line equation when both reaction enthalpy and entropy are constant. The reaction enthalpy determines the extrapolated intercept, while the negative reaction entropy determines the slope. Statement II is incorrect because a phase transition changes the thermodynamic state of the substance. The relevant enthalpy and entropy values generally differ between phases. Therefore, the original constant-property relation cannot automatically be extrapolated across a phase boundary without suitable corrections. The phase transition may alter the temperature dependence of the reaction Gibbs energy. Statement III is correct because the slope of the stated linear relation is \[ \frac{\Delta(\Delta_rG)}{\Delta T}=-\Delta_rS \] A negative slope therefore corresponds to positive reaction entropy. This conclusion depends on the explicitly stated temperature-independent approximation. Statement IV is correct because zero Gibbs-energy change identifies the thermodynamic boundary between favourable and unfavourable reaction directions for the specified states. It does not determine the activation energy or reaction mechanism. A reaction can be thermodynamically favourable yet proceed extremely slowly. Similarly, locating a predicted Gibbs-energy zero crossing does not establish how rapidly equilibrium will be reached. The question requires separating thermodynamic state properties, physical-state changes and kinetic behaviour. Only Statements I, III and IV are correct.

Explanation: \(\textbf{Identifying the relation between the two reactions:}\) Reaction R₁ forms gaseous X. Reaction R₂ forms liquid X. Therefore, R₂ is obtained by adding condensation of X to R₁. The standard Gibbs-energy change for R₂ is consequently the sum of the standard Gibbs energies for gaseous-X formation and condensation. \(\textbf{Calculating the Gibbs energy of condensation:}\) \[ \Delta_{\mathrm{cond}}G=\Delta_{\mathrm{cond}}H-T\Delta_{\mathrm{cond}}S \] Convert the entropy to kilojoule units. \[ \Delta_{\mathrm{cond}}S=-0.100\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] Substituting, \[ \Delta_{\mathrm{cond}}G=-40.0-(300)(-0.100) \] \[ \Delta_{\mathrm{cond}}G=-40.0+30.0 \] \[ \Delta_{\mathrm{cond}}G=-10.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Converting condensation Gibbs energy into its equilibrium contribution:}\) The thermodynamic relation is \[ \Delta_rG^\circ=-RT\ln K \] Therefore, \[ \ln K_{\mathrm{cond}}=\frac{-\Delta_{\mathrm{cond}}G^\circ}{RT} \] \[ \ln K_{\mathrm{cond}}=\frac{10000}{(8.314)(300)} \] \[ \ln K_{\mathrm{cond}}\approx4.01 \] Using the supplied exponential approximation, \[ K_{\mathrm{cond}}\approx e^4 \] \[ K_{\mathrm{cond}}\approx54.6 \] \(\textbf{Combining the reaction equilibria:}\) Adding compatible reactions multiplies their equilibrium constants. Therefore, \[ K_2=K_1K_{\mathrm{cond}} \] Substituting, \[ K_2=(0.20)(54.6) \] \[ K_2=10.92 \] \(\textbf{Interpreting the physical-state effect:}\) Formation of gaseous X has an equilibrium constant below one in the specified standard-state comparison. The condensation step has a negative standard Gibbs-energy change. Its equilibrium contribution therefore increases the equilibrium constant of the combined liquid-product reaction. The product-state correction must be made using Gibbs energy rather than condensation enthalpy alone. The resulting equilibrium constant is approximately 10.9. \(\textbf{Final answer:}\) \[ \boxed{K_2\approx10.9} \]

Explanation: Statement I is correct because the vessel has a rigid boundary and is thermally insulated. Neither heat nor pressure-volume work crosses the specified external boundary. With no other energy transfers, the first law requires the internal energy of the complete contents to remain constant. Statement II is incorrect because internal energy and entropy are different thermodynamic state functions. Conservation of total energy does not require conservation of entropy. Chemical reactions can redistribute energy among molecular states while changing the accessible arrangements of the system. An irreversible spontaneous reaction inside the isolated vessel may therefore increase entropy. Statement III is correct because the familiar negative-Gibbs-energy criterion is associated with constant temperature and pressure. The stated vessel is rigid and insulated. Its temperature need not remain constant during an exothermic or endothermic reaction. Consequently, the specified constraints must be respected when selecting a thermodynamic criterion. The entropy principle for an isolated system is directly relevant here. Statement IV is correct because spontaneous irreversible changes in an isolated system produce entropy. The system approaches equilibrium in the direction permitted by the second law. At equilibrium, no further spontaneous internal change is available under the imposed constraints. This does not imply that all molecular motion or reversible microscopic events stop. The question distinguishes energy conservation, entropy production and the conditions under which different spontaneity criteria apply. Statements I, III and IV are correct.

Explanation: \(\textbf{Constructing the combined pathway:}\) The first reaction produces gaseous C. The second converts gaseous C into liquid C. Adding the two equations cancels the intermediate gaseous C. The resulting equation is the required formation of liquid C from A and B. \(\textbf{Adding compatible Gibbs-energy changes:}\) Standard Gibbs-energy changes are state-function differences. Therefore, they can be added for properly matched reaction equations. \[ \Delta_rG^\circ_{\mathrm{combined}}=\Delta_rG^\circ_1+\Delta_rG^\circ_2 \] Substituting, \[ \Delta_rG^\circ_{\mathrm{combined}}=+6.0-10.0 \] \[ \Delta_rG^\circ_{\mathrm{combined}}=-4.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpreting the result:}\) The gas-product reaction is unfavourable in its specified standard-state comparison. However, condensation contributes a larger negative Gibbs-energy change. Their sum is negative. Thus, the liquid-product reaction is standard-state favoured under the supplied conditions. \(\textbf{Avoiding two common errors:}\) Condensation is not a reversal of the first chemical reaction. It is an additional thermodynamic transformation of the product. Also, phase changes can be included in Gibbs-energy cycles when the states and temperatures are compatible. The final sign depends on the complete specified process, including the product's physical state. The result does not guarantee a fast reaction or establish the direction in every possible nonequilibrium mixture. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rG^\circ_{\mathrm{combined}}=-4.0\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: Statement I is correct because a standard Gibbs-energy change corresponds to a specifically written stoichiometric reaction. Multiplying every reaction coefficient by three multiplies the standard Gibbs-energy change by three. Therefore, the tripled forward reaction has \[ \Delta_rG^{\circ\prime}=3(-5.0) \] \[ \Delta_rG^{\circ\prime}=-15.0\,\mathrm{kJ} \] Statement II is correct because reversing a reaction changes the sign of its Gibbs-energy change. The tripled reverse reaction therefore has \[ \Delta_rG^{\circ\prime\prime}=+15.0\,\mathrm{kJ} \] Statement III is incorrect because the reaction has undergone two transformations: coefficient multiplication and reversal. Reversal alone replaces K with its reciprocal. Multiplying the coefficients by three also raises the equilibrium constant to the third power. Ignoring the coefficient multiplication gives an incomplete result. Statement IV is correct because the transformed equilibrium constant is \[ K'=\frac1{K^3} \] Equivalently, \[ K'=K^{-3} \] The same conclusion follows from the Gibbs-energy relation. Because standard Gibbs-energy changes scale by a factor of negative three for the transformed equation, the logarithm of the equilibrium constant scales by the same factor. These two thermodynamic conventions are consistent with each other. Thus, Statements I, II and IV are correct.

Explanation: \(\textbf{Writing the two Gibbs-energy relations:}\) Under the stated constant-property approximation, \[ \Delta_rG=\Delta_rH-T\Delta_rS \] For Reaction P, \[ \Delta_rG_P=15-0.075T \] For Reaction Q, \[ \Delta_rG_Q=-9+0.030T \] The temperature is measured in kelvin, and both Gibbs-energy values are expressed in kilojoules per mole. \(\textbf{Finding the intersection temperature:}\) At the temperature where their Gibbs-energy changes are equal, \[ \Delta_rG_P=\Delta_rG_Q \] Therefore, \[ 15-0.075T=-9+0.030T \] Rearranging, \[ 24=0.105T \] \[ T=\frac{24}{0.105} \] \[ T=228.571\ldots\,\mathrm{K} \] \(\textbf{Evaluating the common Gibbs-energy change:}\) Substitute into the expression for P. \[ \Delta_rG_P=15-(0.075)(228.571) \] \[ \Delta_rG_P=15-17.143 \] \[ \Delta_rG_P\approx-2.143\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Independent check using Reaction Q:}\) \[ \Delta_rG_Q=-9+(0.030)(228.571) \] \[ \Delta_rG_Q=-9+6.857 \] \[ \Delta_rG_Q\approx-2.143\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpreting the crossing:}\) The two Gibbs-energy changes are equal at approximately 228.6 K. However, equality between the Gibbs-energy changes of two different reactions does not imply that either reaction is at equilibrium. Their common value is negative rather than zero. Therefore, both stated forward reactions are thermodynamically favoured at that temperature under their respective conditions. The intersection identifies equality of reaction Gibbs-energy changes, not a shared equilibrium state. \(\textbf{Final answer:}\) \[ \boxed{\left(T,\Delta_rG_{\mathrm{common}}\right)\approx\left(228.6\,\mathrm{K},-2.14\,\mathrm{kJ\,mol^{-1}}\right)} \]

Explanation: \(\textbf{Selecting the second-law test:}\) The direction of a proposed process must be checked using the total entropy change of the universe. For each process, \[ \Delta S_{\mathrm{univ}}=\Delta S_{\mathrm{sys}}-\frac{\Delta H_{\mathrm{sys}}}{T} \] The reservoir temperature is 300 K. \(\textbf{Process P:}\) The reaction releases 12 kJ. The reservoir receives 12000 J. \[ \Delta S_{\mathrm{surr,P}}=\frac{12000}{300} \] \[ \Delta S_{\mathrm{surr,P}}=+40\,\mathrm{J\,K^{-1}} \] Therefore, \[ \Delta S_{\mathrm{univ,P}}=-30+40 \] \[ \Delta S_{\mathrm{univ,P}}=+10\,\mathrm{J\,K^{-1}} \] The positive result is consistent with spontaneous forward progress. \(\textbf{Process Q:}\) The system absorbs 9 kJ. Therefore, the reservoir loses 9000 J. \[ \Delta S_{\mathrm{surr,Q}}=\frac{-9000}{300} \] \[ \Delta S_{\mathrm{surr,Q}}=-30\,\mathrm{J\,K^{-1}} \] Thus, \[ \Delta S_{\mathrm{univ,Q}}=20-30 \] \[ \Delta S_{\mathrm{univ,Q}}=-10\,\mathrm{J\,K^{-1}} \] The proposed spontaneous forward progress is inconsistent with the second law under the stated assumptions. \(\textbf{Process R:}\) The system absorbs 6 kJ. The reservoir entropy change is \[ \Delta S_{\mathrm{surr,R}}=\frac{-6000}{300} \] \[ \Delta S_{\mathrm{surr,R}}=-20\,\mathrm{J\,K^{-1}} \] Therefore, \[ \Delta S_{\mathrm{univ,R}}=30-20 \] \[ \Delta S_{\mathrm{univ,R}}=+10\,\mathrm{J\,K^{-1}} \] The reported spontaneous progress is consistent with the second law. \(\textbf{Interpreting the consistency test:}\) Process P is spontaneous despite decreasing system entropy. Process R is spontaneous despite absorbing heat. Process Q increases system entropy, but not sufficiently to compensate for the reservoir entropy loss. Consequently, only P and R are consistent with the reported spontaneous direction. The signs of the system enthalpy and entropy must be combined rather than evaluated separately.

Explanation: The overall reaction can be obtained by adding R₁ and R₂. In the combined reaction, B is an intermediate and cancels. For compatible equilibrium constants evaluated at the same temperature, reaction addition corresponds to multiplication of equilibrium constants. However, the two supplied constants are not evaluated at the same temperature. The first belongs to 300 K. The second belongs to 400 K. The requested target constant is to be evaluated at 350 K. An equilibrium constant generally depends on temperature. Therefore, the numerical product \(12\times5\) is not automatically the equilibrium constant of the target reaction at 350 K. Additional information is needed to determine the relevant equilibrium constants at a common temperature. For example, suitable temperature-dependent reaction enthalpy data together with an applicable thermodynamic relation could permit the required correction. The mere fact that reaction equations can be added does not remove the requirement of compatible temperatures. The molecular formulas of A, B and C do not independently establish temperature-invariant equilibrium constants. The important distinction is between algebraic compatibility of reaction equations and thermodynamic compatibility of the numerical data. Thus, the claim is unjustified under the information provided.