Chemical Thermodynamics MCQs with Answers – Part 5 (NEET Chemistry)

Explanation: \(\textbf{Determining the standard reaction Gibbs energy:}\) The supplied enthalpy and entropy refer to the standard reaction. First convert the entropy into kilojoule units. \[ \Delta_rS^\circ=+0.0500\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] Using \[ \Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ \] we obtain \[ \Delta_rG^\circ=25.0-(300)(0.0500) \] \[ \Delta_rG^\circ=25.0-15.0 \] \[ \Delta_rG^\circ=+10.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Recovering the equilibrium constant:}\) \[ \ln K=-\frac{\Delta_rG^\circ}{RT} \] Substituting in joule units, \[ \ln K=\frac{-10000}{(8.314)(300)} \] \[ \ln K\approx-4.01 \] Therefore, \[ K\approx e^{-4.01} \] \[ K\approx0.0181 \] The positive standard Gibbs energy corresponds to an equilibrium constant below one. \(\textbf{Applying the actual reaction quotient:}\) The mixture has \[ \ln Q=-5 \] Its actual reaction Gibbs-energy change is \[ \Delta_rG=\Delta_rG^\circ+RT\ln Q \] Substituting, \[ \Delta_rG=10000+(8.314)(300)(-5) \] \[ \Delta_rG=10000-12471 \] \[ \Delta_rG=-2471\,\mathrm{J\,mol^{-1}} \] \[ \Delta_rG\approx-2.47\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Checking the direction against equilibrium:}\) The initial reaction quotient is approximately \[ Q=e^{-5}\approx0.00674 \] This is smaller than the calculated equilibrium constant. Therefore, \[ Q\lt K \] The forward reaction is thermodynamically favoured in this particular mixture. \(\textbf{Interpreting the apparent contradiction:}\) A positive standard Gibbs-energy change does not prohibit forward reaction from every nonequilibrium mixture. The standard value describes the selected standard-state comparison. The actual value includes the composition-dependent reaction-quotient contribution. Here, the negative contribution from \(RT\ln Q\) exceeds the positive standard Gibbs energy. Consequently, the actual forward reaction is favoured even though \(K\lt 1\). \(\textbf{Final answer:}\) \[ \boxed{K\approx0.0181,\qquad\Delta_rG\approx-2.47\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Recovering the solution enthalpies:}\) The solution enthalpy is the sum of lattice dissociation and ionic hydration contributions. For AX, \[ \Delta_{\mathrm{sol}}H_{\mathrm{AX}}=900-1000 \] \[ \Delta_{\mathrm{sol}}H_{\mathrm{AX}}=-100\,\mathrm{kJ\,mol^{-1}} \] For BX, \[ \Delta_{\mathrm{sol}}H_{\mathrm{BX}}=800-700 \] \[ \Delta_{\mathrm{sol}}H_{\mathrm{BX}}=+100\,\mathrm{kJ\,mol^{-1}} \] AX dissolution is exothermic, whereas BX dissolution is endothermic. \(\textbf{Testing Statement I:}\) For AX, \[ \Delta_{\mathrm{sol}}S_{\mathrm{AX}}=-0.300\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] At 300 K, \[ \Delta_{\mathrm{sol}}G_{\mathrm{AX}}=-100-(300)(-0.300) \] \[ \Delta_{\mathrm{sol}}G_{\mathrm{AX}}=-100+90 \] \[ \Delta_{\mathrm{sol}}G_{\mathrm{AX}}=-10\,\mathrm{kJ\,mol^{-1}} \] The dissolution is exothermic and thermodynamically favoured at 300 K. Statement I is correct. \(\textbf{Testing Statement II:}\) At 400 K, \[ \Delta_{\mathrm{sol}}G_{\mathrm{AX}}=-100-(400)(-0.300) \] \[ \Delta_{\mathrm{sol}}G_{\mathrm{AX}}=+20\,\mathrm{kJ\,mol^{-1}} \] The forward dissolution is not favoured at 400 K. Statement II is incorrect. \(\textbf{Testing Statement III:}\) For BX, \[ \Delta_{\mathrm{sol}}S_{\mathrm{BX}}=+0.400\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] At 200 K, \[ \Delta_{\mathrm{sol}}G_{\mathrm{BX}}=100-(200)(0.400) \] \[ \Delta_{\mathrm{sol}}G_{\mathrm{BX}}=+20\,\mathrm{kJ\,mol^{-1}} \] The positive Gibbs-energy change makes the specified forward dissolution unfavoured. Statement III is correct. \(\textbf{Testing Statement IV:}\) At 300 K, \[ \Delta_{\mathrm{sol}}G_{\mathrm{BX}}=100-(300)(0.400) \] \[ \Delta_{\mathrm{sol}}G_{\mathrm{BX}}=-20\,\mathrm{kJ\,mol^{-1}} \] AX has a Gibbs-energy change of \(-10\,\mathrm{kJ\,mol^{-1}}\) at this temperature. Both are negative. Statement IV is correct. \(\textbf{Thermodynamic interpretation:}\) The two salts have opposite signs of dissolution enthalpy and entropy. AX is favoured at sufficiently low temperatures because its exothermic enthalpy compensates for its negative entropy change. BX becomes favoured at sufficiently high temperatures because its positive entropy contribution overcomes its endothermic enthalpy. The sign of solution enthalpy alone is insufficient to determine dissolution spontaneity. Statements I, III and IV are correct.

Explanation: \(\textbf{Separating chemical dehydration from water vaporisation:}\) The supplied hydrate decomposition produces liquid water. However, the final products contain both liquid and gaseous water. Therefore, a vaporisation correction must be added for the portion of water present as vapour. \(\textbf{Calculating the primary hydrate-decomposition enthalpy:}\) The amount of hydrate is \[ n_{\mathrm{hydrate}}=0.0500\,\mathrm{mol} \] Thus, \[ \Delta H_{\mathrm{dehydration}}=(0.0500)(100) \] \[ \Delta H_{\mathrm{dehydration}}=+5.00\,\mathrm{kJ} \] \(\textbf{Determining total water production:}\) One mole of the hydrate produces seven moles of water. Therefore, \[ n_{\mathrm{H_2O,total}}=7(0.0500) \] \[ n_{\mathrm{H_2O,total}}=0.350\,\mathrm{mol} \] \(\textbf{Determining the vapour amount:}\) The gaseous fraction is 60%. Therefore, \[ n_{\mathrm{H_2O(g)}}=(0.60)(0.350) \] \[ n_{\mathrm{H_2O(g)}}=0.210\,\mathrm{mol} \] The remaining liquid-water amount is \[ n_{\mathrm{H_2O(l)}}=0.350-0.210 \] \[ n_{\mathrm{H_2O(l)}}=0.140\,\mathrm{mol} \] \(\textbf{Calculating the vaporisation correction:}\) Only the 0.210 mol of gaseous water requires the additional vaporisation enthalpy. Therefore, \[ \Delta H_{\mathrm{vap}}=(0.210)(44.0) \] \[ \Delta H_{\mathrm{vap}}=+9.24\,\mathrm{kJ} \] \(\textbf{Calculating the complete enthalpy change:}\) \[ \Delta H_{\mathrm{total}}=\Delta H_{\mathrm{dehydration}}+\Delta H_{\mathrm{vap}} \] \[ \Delta H_{\mathrm{total}}=5.00+9.24 \] \[ \Delta H_{\mathrm{total}}=+14.24\,\mathrm{kJ} \] \(\textbf{Checking the thermochemical endpoints:}\) The first stage forms anhydrous solid and liquid water. The second changes the physical state of only part of that water. No additional water is chemically generated during vaporisation. The product amounts therefore satisfy the original hydrate stoichiometry. Treating all seven waters of crystallisation as gaseous would overestimate the enthalpy required, while omitting vaporisation would underestimate it. \(\textbf{Final answer:}\) \[ \boxed{\Delta H_{\mathrm{total}}=+14.24\,\mathrm{kJ},\qquad n_{\mathrm{H_2O(g)}}=0.210\,\mathrm{mol}} \]

Explanation: \(\textbf{Identifying the required aqueous product state:}\) The supplied elemental reaction forms gaseous hydrogen chloride. The requested final state consists of aqueous hydrogen and chloride ions. The enthalpy of dissolution and complete aqueous ionisation must therefore be included. \(\textbf{Calculating gaseous hydrogen chloride formation per mole:}\) The elemental reaction forms two moles of gaseous HCl. Its enthalpy is \[ \Delta H=-184\,\mathrm{kJ} \] For one mole, \[ \Delta_fH^\circ[\mathrm{HCl(g)}]=\frac{-184}{2} \] \[ \Delta_fH^\circ[\mathrm{HCl(g)}]=-92\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Adding the aqueous-state correction:}\) The dissolution and ionisation stage is \[ \mathrm{HCl(g)\rightarrow H^+(aq)+Cl^-(aq)} \] with \[ \Delta H_{\mathrm{aqueous}}=-75.0\,\mathrm{kJ\,mol^{-1}} \] Therefore, formation of one mole of the specified aqueous HCl from standard elemental states has enthalpy \[ \Delta_fH^\circ[\mathrm{HCl(aq)}]=-92-75.0 \] \[ \Delta_fH^\circ[\mathrm{HCl(aq)}]=-167\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Determining the aqueous HCl amount:}\) The elemental reaction consumes one mole of hydrogen to produce two moles of hydrogen chloride. Thus, \[ n_{\mathrm{HCl(aq)}}=2(0.150) \] \[ n_{\mathrm{HCl(aq)}}=0.300\,\mathrm{mol} \] \(\textbf{Calculating the total enthalpy:}\) \[ \Delta H_{\mathrm{total}}=(0.300)(-167) \] \[ \Delta H_{\mathrm{total}}=-50.1\,\mathrm{kJ} \] \(\textbf{Independent two-stage check:}\) The gaseous formation stage releases \[ q_1=(0.150)(-184) \] \[ q_1=-27.6\,\mathrm{kJ} \] The subsequent dissolution and ionisation of 0.300 mol releases \[ q_2=(0.300)(-75.0) \] \[ q_2=-22.5\,\mathrm{kJ} \] Therefore, \[ q_{\mathrm{total}}=-27.6-22.5 \] \[ q_{\mathrm{total}}=-50.1\,\mathrm{kJ} \] \(\textbf{Chemical interpretation:}\) Gaseous HCl and fully ionised aqueous HCl represent different thermochemical product states. Their formation enthalpies are consequently different. The standard elemental formation reaction must also be normalised to one mole of product before assigning a molar formation enthalpy. The direct one-mole calculation and the full experimental Hess cycle give the same result. \(\textbf{Final answer:}\) \[ \boxed{\Delta_fH^\circ[\mathrm{HCl(aq)}]=-167\,\mathrm{kJ\,mol^{-1}},\qquad\Delta H_{\mathrm{total}}=-50.1\,\mathrm{kJ}} \]

Explanation: \(\textbf{Calculating the slaking enthalpy:}\) The first reaction converts calcium oxide and liquid water into solid calcium hydroxide. Using standard formation enthalpies, \[ \Delta_rH_1^\circ=\Delta_fH^\circ[\mathrm{Ca(OH)_2}]-\{\Delta_fH^\circ[\mathrm{CaO}]+\Delta_fH^\circ[\mathrm{H_2O}]\} \] Substituting, \[ \Delta_rH_1^\circ=-986-[-635-286] \] \[ \Delta_rH_1^\circ=-986+921 \] \[ \Delta_rH_1^\circ=-65\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Calculating the carbonation enthalpy:}\) The second reaction forms calcium carbonate and liquid water from calcium hydroxide and carbon dioxide. Therefore, \[ \Delta_rH_2^\circ=[-1207-286]-[-986-394] \] \[ \Delta_rH_2^\circ=-1493-(-1380) \] \[ \Delta_rH_2^\circ=-113\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Constructing the complete Hess-law reaction:}\) Add \(R_1\) and \(R_2\). The intermediate calcium hydroxide cancels. Water is consumed in the first reaction and regenerated in the second. Therefore, the net equation is \[ \mathrm{CaO(s)+CO_2(g)\rightarrow CaCO_3(s)} \] The overall molar reaction enthalpy is \[ \Delta_rH_{\mathrm{overall}}^\circ=-65-113 \] \[ \Delta_rH_{\mathrm{overall}}^\circ=-178\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Scaling to the calcium oxide amount:}\) The overall net equation consumes one mole of calcium oxide per mole of reaction. For 0.200 mol, \[ \Delta H_{\mathrm{total}}=(0.200)(-178) \] \[ \Delta H_{\mathrm{total}}=-35.6\,\mathrm{kJ} \] \(\textbf{Independent direct-formation check:}\) For the net reaction, \[ \Delta_rH^\circ=-1207-[-635-394] \] \[ \Delta_rH^\circ=-1207+1029 \] \[ \Delta_rH^\circ=-178\,\mathrm{kJ\,mol^{-1}} \] This agrees with the sum of the two stages. \(\textbf{Thermochemical interpretation:}\) Calcium hydroxide is a reaction intermediate rather than a net product. Liquid water also cancels from the overall thermochemical equation. Nevertheless, the individual stages can have different enthalpy changes. The total is fixed by the common initial and final states. The molar enthalpy of the second stage must not be confused with the enthalpy of the entire two-stage process. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rH_2^\circ=-113\,\mathrm{kJ\,mol^{-1}},\qquad\Delta H_{\mathrm{total}}=-35.6\,\mathrm{kJ}} \]

Explanation: \(\textbf{Calculating solid ammonium chloride formation enthalpy:}\) Use the standard formation-enthalpy relation. \[ \Delta_rH^\circ=\sum\nu\Delta_fH^\circ(\mathrm{products})-\sum\nu\Delta_fH^\circ(\mathrm{reactants}) \] For the combination reaction, \[ \Delta_rH_{\mathrm{solid}}^\circ=-314-[(-46)+(-92)] \] \[ \Delta_rH_{\mathrm{solid}}^\circ=-314+138 \] \[ \Delta_rH_{\mathrm{solid}}^\circ=-176\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Including dissolution of the ionic solid:}\) The solid subsequently dissolves with enthalpy \[ \Delta_{\mathrm{sol}}H=+15.0\,\mathrm{kJ\,mol^{-1}} \] Therefore, the complete process from gaseous reactants to the aqueous ionic product has enthalpy \[ \Delta H_{\mathrm{combined}}=-176+15.0 \] \[ \Delta H_{\mathrm{combined}}=-161\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Identifying the limiting gaseous reactant:}\) The combination reaction consumes ammonia and hydrogen chloride in a 1:1 mole ratio. The initial amounts are \[ n_{\mathrm{NH_3}}=0.250\,\mathrm{mol} \] \[ n_{\mathrm{HCl}}=0.180\,\mathrm{mol} \] Hydrogen chloride is therefore limiting. \(\textbf{Calculating the reaction extent:}\) \[ \xi=0.180\,\mathrm{mol} \] The amount of ammonia consumed is also 0.180 mol. Hence, \[ n_{\mathrm{NH_3,remaining}}=0.250-0.180 \] \[ n_{\mathrm{NH_3,remaining}}=0.070\,\mathrm{mol} \] \(\textbf{Calculating the total enthalpy change:}\) \[ \Delta H_{\mathrm{actual}}=\xi\Delta H_{\mathrm{combined}} \] \[ \Delta H_{\mathrm{actual}}=(0.180)(-161) \] \[ \Delta H_{\mathrm{actual}}=-28.98\,\mathrm{kJ} \] \(\textbf{Verifying the two heat contributions independently:}\) The solid-forming reaction releases \[ q_1=(0.180)(-176) \] \[ q_1=-31.68\,\mathrm{kJ} \] Its dissolution absorbs \[ q_2=(0.180)(15.0) \] \[ q_2=+2.70\,\mathrm{kJ} \] Thus, \[ q_1+q_2=-31.68+2.70 \] \[ q_1+q_2=-28.98\,\mathrm{kJ} \] \(\textbf{Interpreting the two stages:}\) The solid-forming reaction is strongly exothermic. Dissolution is endothermic, so it reduces the magnitude of the total heat released. It does not reverse the sign of the complete process. The aqueous product state and the limiting-reactant amount must both be accounted for when calculating the observed enthalpy change. \(\textbf{Final answer:}\) \[ \boxed{\Delta H_{\mathrm{actual}}=-28.98\,\mathrm{kJ},\qquad n_{\mathrm{NH_3,remaining}}=0.070\,\mathrm{mol}} \]

Explanation: \(\textbf{Calculating the decomposition enthalpy:}\) The standard reaction enthalpy follows from the standard formation enthalpies. \[ \Delta_rH^\circ=\sum\nu\Delta_fH^\circ(\mathrm{products})-\sum\nu\Delta_fH^\circ(\mathrm{reactants}) \] The product contribution is \[ \sum\nu\Delta_fH^\circ(\mathrm{products})=-1131-394-242 \] \[ \sum\nu\Delta_fH^\circ(\mathrm{products})=-1767\,\mathrm{kJ} \] The reactant contribution is \[ \sum\nu\Delta_fH^\circ(\mathrm{reactants})=2(-950) \] \[ \sum\nu\Delta_fH^\circ(\mathrm{reactants})=-1900\,\mathrm{kJ} \] Therefore, \[ \Delta_rH^\circ=-1767-(-1900) \] \[ \Delta_rH^\circ=+133\,\mathrm{kJ} \] The decomposition is endothermic. \(\textbf{Identifying gaseous mole production:}\) The reactant sodium hydrogen carbonate is a solid. The products include one mole of carbon dioxide gas and one mole of water vapour. Therefore, \[ \Delta n_g=(1+1)-0 \] \[ \Delta n_g=+2 \] \(\textbf{Converting enthalpy into internal energy:}\) For the stated ideal-gas approximation, \[ \Delta_rH=\Delta_rU+\Delta n_gRT \] Therefore, \[ \Delta_rU=\Delta_rH-\Delta n_gRT \] At 298 K, \[ RT=(8.314)(298) \] \[ RT\approx2477.6\,\mathrm{J} \] \[ RT\approx2.4776\,\mathrm{kJ} \] Thus, \[ \Delta_rU=133-2(2.4776) \] \[ \Delta_rU=133-4.9552 \] \[ \Delta_rU\approx+128.0\,\mathrm{kJ} \] \(\textbf{Determining the sample reaction extent:}\) The balanced equation consumes two moles of sodium hydrogen carbonate. Only 0.100 mol is supplied. Therefore, \[ \xi=\frac{0.100}{2} \] \[ \xi=0.0500\,\mathrm{mol} \] \(\textbf{Calculating constant-pressure heat:}\) At constant pressure with only pressure-volume work, \[ q_P=\xi\Delta_rH \] \[ q_P=(0.0500)(133) \] \[ q_P=+6.65\,\mathrm{kJ} \] \(\textbf{Checking the gas-state correction:}\) The increase in gaseous amount makes the reaction enthalpy larger than the corresponding internal-energy change. The heat required at constant pressure is calculated from enthalpy, not directly from internal energy. The sample heat is also much smaller than the value per balanced equation because only 0.0500 mol of reaction occurs. The water product must remain gaseous throughout this calculation, as specified. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rU\approx+128.0\,\mathrm{kJ},\qquad q_P=+6.65\,\mathrm{kJ}} \]

Explanation: \(\textbf{Constructing the crystalline-hydration pathway:}\) The first dissolution process starts from anhydrous MX. The second starts from crystalline dihydrate. Both lead to the same specified final aqueous solution state after appropriate water accounting. Therefore, the difference between their thermodynamic changes gives the corresponding solid-hydration transformation. \(\textbf{Recovering the enthalpy of dihydrate formation:}\) By Hess's law, \[ \Delta H_{\mathrm{hydrate}}=\Delta H_{\mathrm{anhydrous\ dissolution}}-\Delta H_{\mathrm{hydrate\ dissolution}} \] Substituting, \[ \Delta H_{\mathrm{hydrate}}=8-30 \] \[ \Delta H_{\mathrm{hydrate}}=-22\,\mathrm{kJ\,mol^{-1}} \] The crystalline-hydration process is exothermic. \(\textbf{Recovering the entropy of dihydrate formation:}\) Entropy is also a state function. Using the same thermodynamic pathway, \[ \Delta S_{\mathrm{hydrate}}=\Delta S_{\mathrm{anhydrous\ dissolution}}-\Delta S_{\mathrm{hydrate\ dissolution}} \] Substituting, \[ \Delta S_{\mathrm{hydrate}}=20-80 \] \[ \Delta S_{\mathrm{hydrate}}=-60\,\mathrm{J\,mol^{-1}\,K^{-1}} \] In kilojoule units, \[ \Delta S_{\mathrm{hydrate}}=-0.060\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] \(\textbf{Calculating Gibbs energy at 300 K:}\) \[ \Delta G_{\mathrm{hydrate}}=\Delta H_{\mathrm{hydrate}}-T\Delta S_{\mathrm{hydrate}} \] \[ \Delta G_{\mathrm{hydrate}}(300)=-22-(300)(-0.060) \] \[ \Delta G_{\mathrm{hydrate}}(300)=-22+18 \] \[ \Delta G_{\mathrm{hydrate}}(300)=-4.0\,\mathrm{kJ\,mol^{-1}} \] The negative result favours formation of the dihydrate at 300 K. \(\textbf{Evaluating the higher-temperature condition:}\) At 400 K, \[ \Delta G_{\mathrm{hydrate}}(400)=-22-(400)(-0.060) \] \[ \Delta G_{\mathrm{hydrate}}(400)=-22+24 \] \[ \Delta G_{\mathrm{hydrate}}(400)=+2.0\,\mathrm{kJ\,mol^{-1}} \] The forward hydration process is therefore unfavoured at 400 K under the stated conditions. \(\textbf{Locating the thermodynamic boundary:}\) The Gibbs-energy change becomes zero when \[ T=\frac{\Delta H_{\mathrm{hydrate}}}{\Delta S_{\mathrm{hydrate}}} \] \[ T=\frac{-22}{-0.060} \] \[ T\approx367\,\mathrm{K} \] \(\textbf{Chemical interpretation:}\) The enthalpy decrease favours formation of the hydrated crystal. The negative entropy change opposes the process increasingly strongly as temperature rises. The process is consequently favoured below the calculated boundary but unfavoured above it, within the stated constant-property model. This result combines solution-enthalpy cycles with the Gibbs-energy criterion rather than treating exothermic hydration as automatically spontaneous at every temperature. \(\textbf{Final answer:}\) \[ \boxed{\Delta G_{\mathrm{hydrate}}(300)=-4.0\,\mathrm{kJ\,mol^{-1}},\qquad\Delta G_{\mathrm{hydrate}}(400)=+2.0\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Identifying the bond changes during hydration:}\) In ethene, the two carbon atoms are connected by a double bond. The ethanol product contains a carbon-carbon single bond, an additional carbon-hydrogen bond and a carbon-oxygen bond. One oxygen-hydrogen bond from water is effectively broken, while the other remains in the ethanol molecule. Therefore, the changing bonds are C=C and O–H on the broken-bond side, and C–C, C–H and C–O on the formed-bond side. \(\textbf{Estimating the gas-phase reaction enthalpy:}\) \[ \Delta_rH_{\mathrm{gas}}\approx\sum D(\mathrm{broken})-\sum D(\mathrm{formed}) \] Substituting, \[ \Delta_rH_{\mathrm{gas}}\approx(612+463)-(348+413+358) \] \[ \Delta_rH_{\mathrm{gas}}\approx1075-1119 \] \[ \Delta_rH_{\mathrm{gas}}\approx-44\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Correcting the product's physical state:}\) The bond-energy calculation refers to gaseous ethanol. The experiment instead produces liquid ethanol after condensation. Therefore, \[ \Delta_rH_{\mathrm{liquid}}=\Delta_rH_{\mathrm{gas}}+\Delta_{\mathrm{cond}}H \] \[ \Delta_rH_{\mathrm{liquid}}=-44-38 \] \[ \Delta_rH_{\mathrm{liquid}}=-82\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Identifying the limiting reactant:}\) The chemical reaction consumes ethene and water in a 1:1 mole ratio. Only 0.200 mol of steam is available, compared with 0.300 mol of ethene. Therefore, steam is limiting. The reaction extent is \[ \xi=0.200\,\mathrm{mol} \] \(\textbf{Calculating the total enthalpy change:}\) \[ \Delta H_{\mathrm{total}}=\xi\Delta_rH_{\mathrm{liquid}} \] \[ \Delta H_{\mathrm{total}}=(0.200)(-82) \] \[ \Delta H_{\mathrm{total}}=-16.4\,\mathrm{kJ} \] \(\textbf{Checking the separate energy contributions:}\) The gas-phase reaction releases \[ q_{\mathrm{gas}}=(0.200)(-44)=-8.80\,\mathrm{kJ} \] Condensation releases another \[ q_{\mathrm{cond}}=(0.200)(-38)=-7.60\,\mathrm{kJ} \] Their sum is \(-16.4\,\mathrm{kJ}\). The calculation distinguishes the reaction enthalpy estimated from gaseous bonds, the additional phase-change enthalpy and the actual reaction extent. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rH_{\mathrm{gas}}\approx-44\,\mathrm{kJ\,mol^{-1}},\qquad\Delta H_{\mathrm{total}}=-16.4\,\mathrm{kJ}} \]

Explanation: \(\textbf{Separating oxidation from reduction:}\) Metal M loses two electrons and enters the aqueous phase as \(\mathrm{M^{2+}}\). Aqueous \(\mathrm{N^{2+}}\) accepts two electrons and forms solid N. The supplied atomisation, ionisation and hydration enthalpies can be combined to determine both transformations. \(\textbf{Constructing oxidation of solid M:}\) The thermochemical pathway is \[ \mathrm{M(s)\rightarrow M(g)} \] followed by \[ \mathrm{M(g)\rightarrow M^{2+}(g)+2e^-} \] and then \[ \mathrm{M^{2+}(g)\rightarrow M^{2+}(aq)} \] Therefore, \[ \Delta H_{\mathrm{M,oxidation}}=100+500+1200-1400 \] \[ \Delta H_{\mathrm{M,oxidation}}=+400\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Constructing the corresponding transformation for N:}\) For solid N to form aqueous \(\mathrm{N^{2+}}\), \[ \Delta H_{\mathrm{N,oxidation}}=150+600+1500-1750 \] \[ \Delta H_{\mathrm{N,oxidation}}=+500\,\mathrm{kJ\,mol^{-1}} \] However, N undergoes reduction in the specified displacement reaction. Therefore, this thermochemical pathway must be reversed. \[ \Delta H_{\mathrm{N,reduction}}=-500\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Combining the electron-transfer processes:}\) Adding the oxidation of M and reduction of N cancels the two electrons. The resulting chemical equation is \[ \mathrm{M(s)+N^{2+}(aq)\rightarrow M^{2+}(aq)+N(s)} \] Its molar reaction enthalpy is \[ \Delta_rH=+400-500 \] \[ \Delta_rH=-100\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Scaling to the experimental reaction extent:}\) One mole of M reacts per mole of displacement. Therefore, \[ \xi=0.125\,\mathrm{mol} \] \[ \Delta H_{\mathrm{total}}=\xi\Delta_rH \] \[ \Delta H_{\mathrm{total}}=(0.125)(-100) \] \[ \Delta H_{\mathrm{total}}=-12.5\,\mathrm{kJ} \] \(\textbf{Thermochemical interpretation:}\) Oxidation of M is endothermic in the specified thermochemical cycle. The reverse transformation of aqueous \(\mathrm{N^{2+}}\) releases a larger amount of enthalpy. The net displacement is therefore exothermic. The overall reaction involves two-electron transfer, but that does not mean its enthalpy should be multiplied by an additional factor of two. The electron stoichiometry has already been included in the two successive ionisation steps. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rH=-100\,\mathrm{kJ\,mol^{-1}},\qquad\Delta H_{\mathrm{total}}=-12.5\,\mathrm{kJ}} \]

Explanation: \(\textbf{Choosing a path with known thermochemical data:}\) The initial state is solid X at 300 K. The final state is gaseous X at the same temperature. Simply adding the stated fusion and vaporisation enthalpies would be incomplete because they refer to different transition temperatures. A valid hypothetical pathway can connect the two endpoints using the supplied heat capacities. \(\textbf{Stage I — Fusion at 300 K:}\) \[ \mathrm{X(s,\ 300\,K)\rightarrow X(l,\ 300\,K)} \] The enthalpy contribution is \[ \Delta H_1=+6.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Stage II — Heating the liquid to 400 K:}\) \[ \Delta H_2=C_{p,m}(l)(T_2-T_1) \] \[ \Delta H_2=(100)(400-300) \] \[ \Delta H_2=10000\,\mathrm{J\,mol^{-1}} \] \[ \Delta H_2=+10.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Stage III — Vaporisation at 400 K:}\) \[ \mathrm{X(l,\ 400\,K)\rightarrow X(g,\ 400\,K)} \] The enthalpy contribution is \[ \Delta H_3=+30.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Stage IV — Cooling the vapour to 300 K:}\) \[ \Delta H_4=C_{p,m}(g)(300-400) \] \[ \Delta H_4=(50.0)(-100) \] \[ \Delta H_4=-5000\,\mathrm{J\,mol^{-1}} \] \[ \Delta H_4=-5.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Combining the four stages:}\) \[ \Delta H_{\mathrm{overall}}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4 \] \[ \Delta H_{\mathrm{overall}}=6.0+10.0+30.0-5.0 \] \[ \Delta H_{\mathrm{overall}}=+41.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Calculating the enthalpy for the specified sample:}\) \[ \Delta H_{\mathrm{sample}}=(0.200)(41.0) \] \[ \Delta H_{\mathrm{sample}}=+8.20\,\mathrm{kJ} \] \(\textbf{Checking the physical-state correction:}\) The liquid-heating contribution is larger than the subsequent gas-cooling contribution because the liquid has the greater molar heat capacity. Their difference is \(+5.0\,\mathrm{kJ\,mol^{-1}}\). Consequently, the required enthalpy is 5.0 kJ per mole greater than the uncorrected sum of fusion and vaporisation enthalpies. The result depends only on the specified initial and final thermodynamic states, not on whether the hypothetical intermediate path is the actual experimental route. \(\textbf{Final answer:}\) \[ \boxed{\Delta H_{\mathrm{overall}}=+41.0\,\mathrm{kJ\,mol^{-1}},\qquad\Delta H_{\mathrm{sample}}=+8.20\,\mathrm{kJ}} \]

Explanation: \(\textbf{Recognising why the final states differ:}\) Both samples begin in identical states and undergo the same temperature increase. However, P is heated at constant pressure while Q is heated at constant volume. The final volumes and pressures are therefore not identical. Their entropy changes need not be equal merely because their initial and final temperatures match. \(\textbf{Comparing the entropy expressions:}\) Subtract the supplied relations. \[ \Delta S_P-\Delta S_Q=n(C_{p,m}-C_{V,m})\ln\left(\frac{T_f}{T_i}\right) \] For an ideal gas, \[ C_{p,m}-C_{V,m}=R \] Therefore, \[ \Delta S_P-\Delta S_Q=nR\ln\left(\frac{T_f}{T_i}\right) \] \(\textbf{Evaluating the entropy difference:}\) The temperature ratio is \[ \frac{T_f}{T_i}=\frac{500}{250}=2 \] Substituting, \[ \Delta S_P-\Delta S_Q=(1.50)(8.314)(0.693) \] \[ \Delta S_P-\Delta S_Q\approx+8.64\,\mathrm{J\,K^{-1}} \] \(\textbf{Comparing the heat inputs:}\) For constant-pressure heating, \[ q_P=nC_{p,m}\Delta T \] For constant-volume heating, \[ q_Q=nC_{V,m}\Delta T \] Subtracting, \[ q_P-q_Q=n(C_{p,m}-C_{V,m})\Delta T \] Thus, \[ q_P-q_Q=nR\Delta T \] \(\textbf{Evaluating the heat difference:}\) \[ \Delta T=500-250=250\,\mathrm{K} \] Therefore, \[ q_P-q_Q=(1.50)(8.314)(250) \] \[ q_P-q_Q=3117.75\,\mathrm{J} \] \[ q_P-q_Q\approx+3.12\,\mathrm{kJ} \] \(\textbf{Interpreting both differences:}\) Heating at constant pressure requires more heat because the gas expands and performs pressure-volume work. It also produces a larger gas entropy change because the final gas occupies a greater volume than in the constant-volume experiment. The internal-energy changes are nevertheless identical because both samples undergo the same temperature change. The unequal entropies do not violate state-function behaviour because the two experiments end in different thermodynamic states. \(\textbf{Final answer:}\) \[ \boxed{\Delta S_P-\Delta S_Q\approx+8.64\,\mathrm{J\,K^{-1}},\qquad q_P-q_Q\approx+3.12\,\mathrm{kJ}} \]

Explanation: \(\textbf{Identifying the species entering the reaction quotient:}\) The equation contains solid calcium carbonate, solid calcium oxide and gaseous carbon dioxide. The pure-solid activities are one under the stated conditions. Therefore, the reaction quotient depends on carbon dioxide pressure alone. The amounts of the two solids do not appear as concentration factors in the stated reaction quotient. \(\textbf{Calculating the equilibrium constant:}\) The standard Gibbs-energy relation is \[ \Delta_rG^\circ=-RT\ln K \] Rearranging, \[ \ln K=-\frac{\Delta_rG^\circ}{RT} \] Substituting, \[ \ln K=\frac{-5760}{(8.314)(500)} \] \[ \ln K\approx-1.386 \] Therefore, \[ K\approx e^{-1.386} \] \[ K\approx\frac14=0.25 \] \(\textbf{Evaluating Experiment P:}\) Its reaction quotient is \[ Q_P=\frac{0.100}{1.00} \] \[ Q_P=0.100 \] Because \[ Q_P\lt K \] the forward decomposition direction is thermodynamically favoured. Additional carbon dioxide formation raises the reaction quotient toward its equilibrium value. \(\textbf{Evaluating Experiment Q:}\) Its reaction quotient is \[ Q_Q=\frac{0.500}{1.00} \] \[ Q_Q=0.500 \] Because \[ Q_Q\gt K \] the reverse direction is favoured. Carbon dioxide can react with calcium oxide to produce additional calcium carbonate. \(\textbf{Thermodynamic interpretation:}\) The positive standard Gibbs-energy change indicates an equilibrium constant below one. Nevertheless, decomposition can proceed forward from a mixture whose carbon dioxide pressure is sufficiently low. The actual direction depends on the reaction quotient, not solely on the sign of the standard Gibbs energy. The two vessels have the same equilibrium constant because they are at the same temperature. The initial carbon dioxide pressures place them on opposite sides of equilibrium.

Explanation: \(\textbf{Recovering the standard reaction Gibbs energy:}\) The equilibrium constant is given in exponential form. Therefore, \[ \ln K=2 \] Using \[ \Delta_rG^\circ=-RT\ln K \] we obtain \[ \Delta_rG^\circ=-(8.314)(300)(2) \] \[ \Delta_rG^\circ=-4988.4\,\mathrm{J\,mol^{-1}} \] \[ \Delta_rG^\circ=-4.9884\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Applying formation-Gibbs-energy additivity:}\) For the reaction as written, \[ \Delta_rG^\circ=\Delta_fG^\circ(C)-\Delta_fG^\circ(A)-\Delta_fG^\circ(B) \] Let \[ x=\Delta_fG^\circ(C) \] Substituting, \[ -4.9884=x-(-10.0)-(-20.0) \] \[ -4.9884=x+30.0 \] Therefore, \[ x=-34.9884\,\mathrm{kJ\,mol^{-1}} \] \[ x\approx-35.0\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Recovering the standard reaction entropy:}\) At the specified temperature, \[ \Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ \] Rearranging, \[ \Delta_rS^\circ=\frac{\Delta_rH^\circ-\Delta_rG^\circ}{T} \] Substituting, \[ \Delta_rS^\circ=\frac{-11.0-(-4.9884)}{300} \] \[ \Delta_rS^\circ=\frac{-6.0116}{300} \] \[ \Delta_rS^\circ=-0.020039\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] Converting units, \[ \Delta_rS^\circ\approx-20.0\,\mathrm{J\,mol^{-1}\,K^{-1}} \] \(\textbf{Checking the thermodynamic signs:}\) The reaction is exothermic but has a negative standard reaction entropy. These signs are consistent with an association reaction that reduces the number of gaseous particles. At 300 K, the negative enthalpy contribution is sufficiently large to produce a negative standard Gibbs-energy change. The calculated equilibrium constant is therefore greater than one, consistent with the supplied \(e^2\). This reverse calculation combines formation Gibbs energies, an equilibrium measurement and reaction enthalpy without assuming that formation Gibbs energies equal formation enthalpies. \(\textbf{Final answer:}\) \[ \boxed{\Delta_fG^\circ(C)\approx-35.0\,\mathrm{kJ\,mol^{-1}},\qquad\Delta_rS^\circ\approx-20.0\,\mathrm{J\,mol^{-1}\,K^{-1}}} \]

Explanation: \(\textbf{Distinguishing identical gases from identical initial states:}\) Both experiments contain the same chemical species on either side of the partition. However, this fact alone does not establish that the entropy change upon removing the partition is zero. The pressure and molar volume of the gas on the two sides must also be considered. \(\textbf{Establishing the common final state:}\) Each container contains two moles of the same ideal gas. The final total volume is 20.00 L. Therefore, the final molar volume is \[ V_{m,f}=\frac{20.00}{2.00} \] \[ V_{m,f}=10.00\,\mathrm{L\,mol^{-1}} \] The temperature remains 300 K because the complete system is rigid and insulated and the gas is ideal. \(\textbf{Entropy change in Experiment P:}\) Initially, one mole occupies 5.00 L and the other occupies 15.00 L. The initial pressures are unequal. The change in total entropy can be evaluated from the initial and final ideal-gas state functions. \[ \Delta S_P=R\ln\left(\frac{10.00}{5.00}\right)+R\ln\left(\frac{10.00}{15.00}\right) \] Combining logarithms, \[ \Delta S_P=R\ln\left(\frac{10.00\times10.00}{5.00\times15.00}\right) \] \[ \Delta S_P=R\ln\left(\frac43\right) \] \[ \Delta S_P=(8.314)(0.2877) \] \[ \Delta S_P\approx+2.39\,\mathrm{J\,K^{-1}} \] \(\textbf{Entropy change in Experiment Q:}\) Each compartment initially contains one mole in 10.00 L. Their initial temperatures and pressures are equal. Removing the partition does not alter the thermodynamic state of the uniform gas. Therefore, \[ \Delta S_Q=0 \] \(\textbf{Interpreting the difference:}\) In P, the initially unequal pressures produce spontaneous redistribution and a positive entropy change. In Q, the gas is already uniform at the same temperature and pressure, so no net thermodynamic change occurs upon partition removal. The positive entropy change in P is associated with pressure equalisation, not with mixing chemically distinguishable gases. The example shows why the statement that identical gases have zero entropy of mixing must be applied with attention to their initial thermodynamic states. \(\textbf{Final answer:}\) \[ \boxed{\Delta S_P\approx+2.39\,\mathrm{J\,K^{-1}},\qquad\Delta S_Q=0} \]

Explanation: \(\textbf{Distinguishing internal-energy change from heat at constant pressure:}\) The supplied reaction energy is an internal-energy change. At constant pressure, the heat exchanged equals the reaction enthalpy when only pressure-volume work occurs. Therefore, the gas-mole correction must be applied before calculating the surroundings entropy. \(\textbf{Determining the gaseous mole change:}\) The reaction consumes two moles of gaseous A and produces one mole of gaseous B. Therefore, \[ \Delta n_g=1-2 \] \[ \Delta n_g=-1 \] \(\textbf{Converting internal energy into enthalpy:}\) For ideal gases, \[ \Delta_rH=\Delta_rU+\Delta n_gRT \] At 300 K, \[ RT=(8.314)(300) \] \[ RT=2494.2\,\mathrm{J} \] \[ RT=2.4942\,\mathrm{kJ} \] Substituting, \[ \Delta_rH=-30.0+(-1)(2.4942) \] \[ \Delta_rH=-32.4942\,\mathrm{kJ} \] \(\textbf{Calculating the surroundings entropy:}\) The reaction releases 32.4942 kJ to the reservoir. Therefore, \[ \Delta S_{\mathrm{surr}}=-\frac{\Delta_rH}{T} \] \[ \Delta S_{\mathrm{surr}}=\frac{32494.2}{300} \] \[ \Delta S_{\mathrm{surr}}=+108.314\,\mathrm{J\,K^{-1}} \] \(\textbf{Calculating the universe entropy change:}\) The reaction-system entropy decreases by 100 J K⁻¹. Thus, \[ \Delta S_{\mathrm{univ}}=\Delta S_{\mathrm{sys}}+\Delta S_{\mathrm{surr}} \] \[ \Delta S_{\mathrm{univ}}=-100+108.314 \] \[ \Delta S_{\mathrm{univ}}=+8.314\,\mathrm{J\,K^{-1}} \] \(\textbf{Recovering standard reaction Gibbs energy:}\) \[ \Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ \] Using compatible units, \[ \Delta_rG^\circ=-32.4942-(300)(-0.100) \] \[ \Delta_rG^\circ=-32.4942+30.0 \] \[ \Delta_rG^\circ=-2.4942\,\mathrm{kJ} \] \(\textbf{Checking the second-law connection:}\) At constant temperature and pressure, the corresponding Gibbs-energy and universe-entropy changes satisfy \[ \Delta_rG^\circ=-T\Delta S_{\mathrm{univ}} \] Substituting, \[ -T\Delta S_{\mathrm{univ}}=-(300)(8.314) \] \[ -T\Delta S_{\mathrm{univ}}=-2494.2\,\mathrm{J} \] The two methods agree. The negative Gibbs-energy change is consistent with the positive universe entropy change. Using the constant-volume heat value of 30.0 kJ to calculate the constant-pressure reservoir entropy would incorrectly omit the gas-mole contribution. \(\textbf{Final answer:}\) \[ \boxed{\Delta S_{\mathrm{univ}}\approx+8.31\,\mathrm{J\,K^{-1}},\qquad\Delta_rG^\circ\approx-2.49\,\mathrm{kJ}} \]

Explanation: Statement I is correct because graphite is the specified standard reference state of elemental carbon. The standard enthalpy of formation of an element in its reference state is assigned zero. This is a reference convention for formation enthalpy rather than a claim that the substance contains no internal energy. Statement II is incorrect because diamond and graphite are different allotropes with different thermodynamic properties. Graphite is the standard elemental reference state under the specified conditions. The standard Gibbs energy of formation of diamond is therefore not forced to be zero merely because diamond is composed entirely of carbon. Its formation Gibbs energy describes the thermodynamic difference between diamond and the selected elemental reference state. Statement III is incorrect because standard molar entropy and standard Gibbs energy of formation are different quantities. The standard Gibbs energy of formation of molecular oxygen is zero under the chosen elemental reference convention. However, molecular oxygen possesses molecular motion and accessible microscopic states at 298 K. Its standard molar entropy is consequently not zero. The third law relates zero entropy to an ideal perfect crystal at absolute zero, not to every elemental reference state at ordinary temperature. Statement IV is correct because the standard formation reaction of one mole of gaseous atomic oxygen is \[ \mathrm{\frac12O_2(g)\rightarrow O(g)} \] Dissociation of one mole of molecular oxygen produces two moles of gaseous oxygen atoms. The atomic formation enthalpy is therefore half the corresponding molecular dissociation enthalpy. The question distinguishes elemental reference conventions, absolute entropy and physically different allotropes or atomic states. Only Statements I and IV are correct.

Explanation: \(\textbf{Recognising the two different comparison bases:}\) Heat released per gram of fuel measures energy output relative to fuel mass. Heat released per mole of oxygen measures energy output relative to oxidant consumption. These are different normalisations of the combustion enthalpy. Their rankings need not be identical. \(\textbf{Methane heat release per gram:}\) One mole of methane has mass 16.0 g and releases 890 kJ. Therefore, \[ E_{\mathrm{CH_4,mass}}=\frac{890}{16.0} \] \[ E_{\mathrm{CH_4,mass}}=55.625\,\mathrm{kJ\,g^{-1}} \] \(\textbf{Methanol heat release per gram:}\) One mole of methanol has mass 32.0 g and releases 726 kJ. Therefore, \[ E_{\mathrm{CH_3OH,mass}}=\frac{726}{32.0} \] \[ E_{\mathrm{CH_3OH,mass}}=22.6875\,\mathrm{kJ\,g^{-1}} \] Methane releases more heat per gram of fuel. \(\textbf{Methane heat release per mole of oxygen:}\) The methane combustion equation consumes two moles of oxygen per mole of fuel. Thus, \[ E_{\mathrm{CH_4,O_2}}=\frac{890}{2} \] \[ E_{\mathrm{CH_4,O_2}}=445\,\mathrm{kJ\,mol^{-1}\ of\ O_2} \] \(\textbf{Methanol heat release per mole of oxygen:}\) Methanol consumes 1.5 moles of oxygen per mole of fuel. Therefore, \[ E_{\mathrm{CH_3OH,O_2}}=\frac{726}{1.5} \] \[ E_{\mathrm{CH_3OH,O_2}}=484\,\mathrm{kJ\,mol^{-1}\ of\ O_2} \] Methanol releases more heat per mole of oxygen consumed. \(\textbf{Interpreting the reversal:}\) Methane has the larger mass-specific combustion enthalpy. However, methanol contains oxygen within its molecular formula and requires less external oxygen per mole of fuel. When combustion heat is compared relative to oxygen consumption, methanol has the larger value. The preferred fuel therefore depends on whether fuel mass or oxygen availability is the limiting resource. This comparison does not establish which fuel is preferable under every practical storage, safety or environmental condition. The correct ranking is methane per gram and methanol per mole of oxygen.

Explanation: The Assertion is true because adding water changes the actual composition of the solution. Initially, the saturated solution is at dissolution equilibrium. Its reaction quotient equals the equilibrium constant. For the specified ionic dissolution, the reaction quotient depends on the activities of the dissolved cation and anion. The activity of the pure solid is one and does not enter as a variable concentration factor. When water is first added, both dissolved-ion activities decrease under the stated ideal-dilution approximation. Consequently, the reaction quotient decreases below its former equilibrium value. The equilibrium constant does not change because the temperature remains fixed. The standard reaction Gibbs-energy change also remains associated with the same temperature and standard states. Since the diluted mixture now has a reaction quotient smaller than its equilibrium constant, further dissolution is thermodynamically favoured. Some of the remaining solid can dissolve until equilibrium is restored. The Reason is therefore true and correctly explains the Assertion. The example distinguishes a change in actual Gibbs-energy driving force from a change in the standard reaction Gibbs energy. It also shows why dilution may cause additional solid dissolution without changing the intrinsic equilibrium constant.

Explanation: \(\textbf{Determining the final temperature in direct contact:}\) Experiment P transfers heat directly between blocks of equal heat capacity. No useful work is extracted. Energy conservation requires \[ 100(400-T_P)=100(T_P-300) \] Therefore, \[ 400-T_P=T_P-300 \] \[ T_P=350\,\mathrm{K} \] \(\textbf{Calculating the entropy change in Experiment P:}\) The hotter block cools from 400 K to 350 K. The colder block warms from 300 K to 350 K. Therefore, \[ \Delta S_P=100\ln\left(\frac{350}{400}\right)+100\ln\left(\frac{350}{300}\right) \] Combining the logarithms, \[ \Delta S_P=100\ln\left(\frac{350^2}{400\times300}\right) \] \[ \Delta S_P=100\ln\left(\frac{122500}{120000}\right) \] \[ \Delta S_P=100\ln\left(\frac{49}{48}\right) \] \[ \Delta S_P=100(0.02062) \] \[ \Delta S_P=+2.062\,\mathrm{J\,K^{-1}} \] \(\textbf{Applying reversibility to Experiment Q:}\) The heat engine operates reversibly and returns to its original state after each cycle. The complete arrangement is otherwise insulated. Therefore, the total entropy change of the two blocks is zero in the reversible limit. If their common final temperature is \(T_Q\), \[ 100\ln\left(\frac{T_Q}{400}\right)+100\ln\left(\frac{T_Q}{300}\right)=0 \] Combining logarithms, \[ \ln\left(\frac{T_Q^2}{400\times300}\right)=0 \] Hence, \[ T_Q^2=400\times300 \] \[ T_Q\approx346.4\,\mathrm{K} \] \(\textbf{Determining the work extracted:}\) The engine returns to its initial state, so its net internal-energy change is zero. The work extracted equals the decrease in the total internal energy of the two blocks. Thus, \[ W_{\mathrm{out}}=100(400+300-2T_Q) \] Substituting the unrounded temperature, \[ W_{\mathrm{out}}=100[700-2\sqrt{120000}] \] \[ W_{\mathrm{out}}\approx718\,\mathrm{J} \] \[ W_{\mathrm{out}}\approx720\,\mathrm{J} \] \(\textbf{Comparing the two endpoints:}\) The direct-contact process finishes at 350 K and produces entropy. The reversible engine process finishes at a slightly lower temperature because some of the initial energy is extracted as work. Its final state is therefore different from the direct-contact final state. This difference is essential: entropy remains a state function in both cases. \(\textbf{Final answer:}\) \[ \boxed{\Delta S_P\approx+2.06\,\mathrm{J\,K^{-1}},\qquad W_{\mathrm{out,Q}}\approx720\,\mathrm{J}} \]