Chemical Thermodynamics MCQs with Answers – Part 6 (NEET Chemistry)

Explanation: \(\textbf{Counting bonds broken in the gaseous reactants:}\) The reaction consumes one molecule of nitrogen gas and three molecules of hydrogen gas per balanced reaction. The bonds broken are one N≡N bond and three H–H bonds. Therefore, \[ E_{\mathrm{broken}}=D(\mathrm{N\equiv N})+3D(\mathrm{H-H}) \] \[ E_{\mathrm{broken}}=946+3(436) \] \[ E_{\mathrm{broken}}=946+1308 \] \[ E_{\mathrm{broken}}=2254\,\mathrm{kJ} \] \(\textbf{Counting bonds formed in ammonia:}\) Each ammonia molecule contains three N–H bonds. The balanced reaction produces two ammonia molecules. Therefore, six N–H bonds are formed. \[ E_{\mathrm{formed}}=6D(\mathrm{N-H}) \] \[ E_{\mathrm{formed}}=6(391) \] \[ E_{\mathrm{formed}}=2346\,\mathrm{kJ} \] \(\textbf{Calculating the bond-enthalpy estimate:}\) \[ \Delta_rH_{\mathrm{estimated}}\approx E_{\mathrm{broken}}-E_{\mathrm{formed}} \] \[ \Delta_rH_{\mathrm{estimated}}\approx2254-2346 \] \[ \Delta_rH_{\mathrm{estimated}}\approx-92\,\mathrm{kJ} \] \(\textbf{Calculating the formation-enthalpy result:}\) The reactants are elements in their standard reference states. Their standard formation enthalpies are zero. Therefore, \[ \Delta_rH^\circ=2\Delta_fH^\circ[\mathrm{NH_3(g)}] \] \[ \Delta_rH^\circ=2(-46.0) \] \[ \Delta_rH^\circ=-92.0\,\mathrm{kJ} \] \(\textbf{Determining the discrepancy:}\) \[ |\Delta H_{\mathrm{estimated}}-\Delta H_{\mathrm{formation}}|=|-92-(-92)| \] \[ |\Delta H_{\mathrm{estimated}}-\Delta H_{\mathrm{formation}}|=0 \] \(\textbf{Checking the student's claim:}\) For these particular supplied values, the two calculations agree exactly. Average bond enthalpies nevertheless remain approximate values for bonds in different molecular environments. In other examples, an estimate based on average bond enthalpies need not exactly reproduce a measured reaction enthalpy. Such a discrepancy would not invalidate Hess's law because the law applies to exact enthalpy differences between specified thermodynamic states. The student's general claim is therefore incorrect even though the present numerical comparison happens to agree. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rH_{\mathrm{estimated}}=-92\,\mathrm{kJ},\qquad\text{discrepancy}=0} \]

Explanation: \(\textbf{Identifying the elemental reference state:}\) The problem specifies solid X as the standard elemental reference state. Therefore, its standard formation enthalpy is zero. Gaseous atomic X is a different thermochemical state. The enthalpy required to form gaseous atomic X from solid X is its sublimation enthalpy. Thus, \[ \Delta_fH^\circ[\mathrm{X(g)}]=+100\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Constructing a route through gaseous atoms:}\) The requested transformation starts with two moles of solid X. First convert both moles into gaseous atoms. \[ \mathrm{2X(s)\rightarrow2X(g)} \] Its enthalpy is \[ \Delta H_1=2(100) \] \[ \Delta H_1=+200\,\mathrm{kJ} \] \(\textbf{Forming the gaseous diatomic molecule:}\) The given dissociation equation is \[ \mathrm{X_2(g)\rightarrow2X(g)} \] with \[ \Delta H=+240\,\mathrm{kJ} \] The reverse reaction forms molecular X₂ from two gaseous atoms. Therefore, \[ \mathrm{2X(g)\rightarrow X_2(g)} \] has \[ \Delta H_2=-240\,\mathrm{kJ} \] \(\textbf{Combining the Hess-law stages:}\) \[ \Delta H_{\mathrm{overall}}=\Delta H_1+\Delta H_2 \] \[ \Delta H_{\mathrm{overall}}=200-240 \] \[ \Delta H_{\mathrm{overall}}=-40\,\mathrm{kJ} \] \(\textbf{Checking the direct and reverse paths:}\) Starting with gaseous molecular X₂, dissociation absorbs 240 kJ. Returning the resulting two gaseous atoms to solid X releases 200 kJ. Therefore, \[ \mathrm{X_2(g)\rightarrow2X(s)} \] has enthalpy \[ \Delta H=240-200 \] \[ \Delta H=+40\,\mathrm{kJ} \] Reversing this equation again gives \(-40\,\mathrm{kJ}\) for forming X₂(g) from solid X. \(\textbf{Interpreting the two different quantities:}\) The formation enthalpy of gaseous atomic X is specified per mole of atoms formed. The requested molecular formation equation uses two moles of solid atoms to produce one mole of gaseous diatomic molecules. The bond dissociation enthalpy refers to breaking one mole of gaseous molecules into atoms. These mole bases must be distinguished. The negative result indicates that the molecular gas lies below two moles of solid reference X in enthalpy under the hypothetical supplied conditions. \(\textbf{Final answer:}\) \[ \boxed{\Delta H[2\mathrm{X(s)\rightarrow X_2(g)}]=-40\,\mathrm{kJ},\qquad\Delta_fH^\circ[\mathrm{X(g)}]=+100\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Separating lattice dissociation from hydration:}\) Dissolution of the ionic solid can be represented by formation of separated gaseous ions followed by their hydration. Lattice dissociation absorbs energy. Hydration of the gaseous ions releases energy. The algebraic sum determines the dissolution enthalpy. \(\textbf{Calculating total ionic hydration enthalpy:}\) One mole of MX produces one mole each of M⁺ and X⁻ ions. Therefore, \[ \Delta H_{\mathrm{hydration,total}}=-330-470 \] \[ \Delta H_{\mathrm{hydration,total}}=-800\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Determining the molar dissolution enthalpy:}\) \[ \Delta_{\mathrm{sol}}H=\Delta H_{\mathrm{lattice,dissociation}}+\Delta H_{\mathrm{hydration,total}} \] \[ \Delta_{\mathrm{sol}}H=780-800 \] \[ \Delta_{\mathrm{sol}}H=-20.0\,\mathrm{kJ\,mol^{-1}} \] The dissolution is exothermic under the specified conditions. \(\textbf{Calculating heat released by the sample:}\) The salt amount is 0.150 mol. Therefore, \[ q_{\mathrm{dissolution}}=(0.150)(-20.0) \] \[ q_{\mathrm{dissolution}}=-3.00\,\mathrm{kJ} \] \[ q_{\mathrm{dissolution}}=-3000\,\mathrm{J} \] \(\textbf{Applying calorimetric energy conservation:}\) The solution and calorimeter absorb the heat released by dissolution. Therefore, \[ q_{\mathrm{cal}}=+3000\,\mathrm{J} \] The measured temperature increase is \[ \Delta T=+1.50\,\mathrm{K} \] Using \[ q_{\mathrm{cal}}=C_{\mathrm{eff}}\Delta T \] we obtain \[ 3000=C_{\mathrm{eff}}(1.50) \] \[ C_{\mathrm{eff}}=\frac{3000}{1.50} \] \[ C_{\mathrm{eff}}=2000\,\mathrm{J\,K^{-1}} \] \(\textbf{Checking the physical interpretation:}\) The hydration energy released has a slightly larger magnitude than the energy required to separate the ionic lattice. The net dissolution enthalpy is consequently negative. The positive calorimeter temperature rise is consistent with exothermic dissolution. The lattice dissociation enthalpy must not be confused with lattice formation enthalpy; reversing the lattice process would reverse its sign. The effective heat capacity is recovered from the observed thermal response after the chemical enthalpy is established independently. \(\textbf{Final answer:}\) \[ \boxed{\Delta_{\mathrm{sol}}H=-20.0\,\mathrm{kJ\,mol^{-1}},\qquad C_{\mathrm{eff}}=2000\,\mathrm{J\,K^{-1}}} \]

Explanation: \(\textbf{Checking the reported standard Gibbs energy:}\) Convert the entropy into kilojoule units. \[ \Delta_rS^\circ=-0.0400\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] At 300 K, \[ \Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ \] \[ \Delta_rG^\circ=-18.0-(300)(-0.0400) \] \[ \Delta_rG^\circ=-18.0+12.0 \] \[ \Delta_rG^\circ=-6.00\,\mathrm{kJ\,mol^{-1}} \] The standard Gibbs-energy change is indeed negative. Statement I is correct because \[ \Delta_rG^\circ=-RT\ln K \] A negative standard Gibbs-energy change requires \(\ln K\gt 0\), and therefore \(K\gt 1\). The claimed value \(K\lt 1\) is inconsistent with the stated negative standard Gibbs-energy change. Statement II is correct because both the reaction enthalpy and entropy are negative. As temperature increases, the contribution \(-T\Delta_rS^\circ\) becomes increasingly positive. At the boundary, \[ 0=-18.0+0.0400T \] \[ T=450\,\mathrm{K} \] Above this temperature, the standard Gibbs-energy change becomes positive within the stated approximation. Statement III is incorrect because the actual reaction Gibbs energy also depends on composition. The relation is \[ \Delta_rG=\Delta_rG^\circ+RT\ln Q \] A sufficiently large or small reaction quotient can change the sign of the actual Gibbs-energy change relative to the standard value. Statement IV is correct because reversing a thermochemical reaction exchanges initial and final states. Every corresponding reaction state-function change reverses sign. The reverse equilibrium constant is the reciprocal of the forward equilibrium constant. These operations are consistent with the logarithmic Gibbs-energy relation. Only Statements I, II and IV are correct.

Explanation: \(\textbf{Identifying the condition corresponding to K = 1:}\) The standard Gibbs-energy relation is \[ \Delta_rG^\circ=-RT\ln K \] When \[ K=1 \] then \[ \ln K=0 \] Therefore, \[ \Delta_rG^\circ=0 \] \(\textbf{Applying the enthalpy-entropy relationship:}\) For the specified constant thermodynamic quantities, \[ \Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ \] The entropy in kilojoule units is \[ \Delta_rS^\circ=+0.0600\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] At the boundary, \[ 0=30.0-0.0600T_1 \] \[ 0.0600T_1=30.0 \] \[ T_1=\frac{30.0}{0.0600} \] \[ T_1=500\,\mathrm{K} \] \(\textbf{Determining the high-temperature preference:}\) The reaction is endothermic and has positive reaction entropy. At temperatures below 500 K, \[ \Delta_rH^\circ\gt T\Delta_rS^\circ \] Therefore, \[ \Delta_rG^\circ\gt 0 \] and \[ K\lt 1 \] At temperatures above 500 K, the entropy contribution exceeds the positive enthalpy contribution. Therefore, \[ \Delta_rG^\circ\lt 0 \] and \[ K\gt 1 \] \(\textbf{Checking with a specific temperature:}\) At 600 K, \[ \Delta_rG^\circ=30.0-(600)(0.0600) \] \[ \Delta_rG^\circ=-6.00\,\mathrm{kJ\,mol^{-1}} \] This confirms that K exceeds one above the boundary. \(\textbf{Thermodynamic interpretation:}\) The positive reaction enthalpy opposes the forward reaction. The positive entropy change increasingly favours it as temperature rises. The equilibrium constant therefore crosses one at the temperature where these contributions exactly balance. This condition concerns the standard-state equilibrium preference. The actual reaction direction in a particular mixture would additionally depend on its reaction quotient. \(\textbf{Final answer:}\) \[ \boxed{T_1=500\,\mathrm{K},\qquad T_2\gt T_1} \]

Explanation: \(\textbf{Expressing the Gibbs-energy dependence on temperature:}\) The reaction enthalpy is negative. The reaction entropy is also negative. Therefore, the enthalpy contribution favours the forward reaction, while the entropy term increasingly opposes it at higher temperature. Convert entropy into kilojoule units. \[ \Delta_rS=-0.100\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] Thus, \[ \Delta_rG^\circ=-40.0-T(-0.100) \] \[ \Delta_rG^\circ=-40.0+0.100T \] \(\textbf{Testing observation P:}\) At 250 K, \[ \Delta_rG^\circ_{250}=-40.0+(0.100)(250) \] \[ \Delta_rG^\circ_{250}=-15.0\,\mathrm{kJ\,mol^{-1}} \] The negative value favours the forward reaction. Observation P is consistent. \(\textbf{Testing observation Q:}\) At 400 K, \[ \Delta_rG^\circ_{400}=-40.0+(0.100)(400) \] \[ \Delta_rG^\circ_{400}=0 \] The standard Gibbs-energy difference between forward and reverse states vanishes. Observation Q is consistent. \(\textbf{Testing observation R:}\) At 500 K, \[ \Delta_rG^\circ_{500}=-40.0+(0.100)(500) \] \[ \Delta_rG^\circ_{500}=+10.0\,\mathrm{kJ\,mol^{-1}} \] The positive value indicates that the reverse direction is preferred under the specified standard-state comparison. Observation R is inconsistent. \(\textbf{Identifying the thermodynamic boundary:}\) The sign changes at \[ T=\frac{\Delta_rH}{\Delta_rS} \] \[ T=\frac{-40.0}{-0.100} \] \[ T=400\,\mathrm{K} \] Below 400 K, the forward reaction is favoured. Above 400 K, the standard Gibbs-energy change becomes positive. The two negative thermodynamic quantities do not guarantee forward favourability at all temperatures. The temperature-dependent entropy contribution must be included. Only observations P and Q are consistent.

Explanation: \(\textbf{Evaluating the forward equilibrium constant:}\) For the reaction A to B, \[ K_1=10^3 \] Therefore, \[ \log_{10}K_1=3 \] \(\textbf{Calculating the standard forward Gibbs energy:}\) \[ \Delta_rG_1^\circ=-2.303RT\log_{10}K_1 \] Substituting, \[ \Delta_rG_1^\circ=-(2.303)(8.314)(298)(3) \] \[ \Delta_rG_1^\circ\approx-17108\,\mathrm{J\,mol^{-1}} \] \[ \Delta_rG_1^\circ\approx-17.1\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Reversing the equilibrium reaction:}\) The reverse reaction exchanges reactants and products. Its equilibrium constant is the reciprocal. \[ K_2=\frac1{K_1} \] \[ K_2=10^{-3} \] Therefore, \[ \log_{10}K_2=-3 \] \(\textbf{Calculating the reverse Gibbs energy:}\) \[ \Delta_rG_2^\circ=-2.303RT\log_{10}K_2 \] \[ \Delta_rG_2^\circ=-(2.303)(8.314)(298)(-3) \] \[ \Delta_rG_2^\circ\approx+17.1\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Checking the state-function relationship:}\) The forward and reverse reactions connect identical states in opposite directions. Therefore, \[ \Delta_rG_2^\circ=-\Delta_rG_1^\circ \] This agrees with the logarithmic equilibrium calculations. \(\textbf{Interpreting the signs:}\) The large equilibrium constant favours B under the specified standard-state comparison. The reversed reaction has a small equilibrium constant and positive standard Gibbs-energy change. A positive standard Gibbs energy does not imply that the reverse transformation is impossible from every conceivable mixture. The actual direction depends on composition as well as temperature. The relationship between reciprocal equilibrium constants and opposite standard Gibbs energies is the central thermodynamic consistency requirement. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rG_1^\circ\approx-17.1\,\mathrm{kJ\,mol^{-1}},\qquad\Delta_rG_2^\circ\approx+17.1\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Recovering the entropy of fusion from equilibrium:}\) At the equilibrium melting temperature, solid and liquid have equal Gibbs energies. Therefore, \[ \Delta_{\mathrm{fus}}G=0 \] Using \[ \Delta_{\mathrm{fus}}G=\Delta_{\mathrm{fus}}H-T_m\Delta_{\mathrm{fus}}S \] we obtain \[ \Delta_{\mathrm{fus}}S=\frac{\Delta_{\mathrm{fus}}H}{T_m} \] Substituting, \[ \Delta_{\mathrm{fus}}S=\frac{8000}{320} \] \[ \Delta_{\mathrm{fus}}S=+25.0\,\mathrm{J\,mol^{-1}\,K^{-1}} \] \(\textbf{Determining the reverse phase-transition quantities:}\) Freezing reverses fusion. Therefore, \[ \Delta_{\mathrm{freeze}}H=-8000\,\mathrm{J\,mol^{-1}} \] \[ \Delta_{\mathrm{freeze}}S=-25.0\,\mathrm{J\,mol^{-1}\,K^{-1}} \] \(\textbf{Calculating the Gibbs energy of freezing at 300 K:}\) \[ \Delta_{\mathrm{freeze}}G=\Delta_{\mathrm{freeze}}H-T\Delta_{\mathrm{freeze}}S \] Substituting, \[ \Delta_{\mathrm{freeze}}G=-8000-(300)(-25.0) \] \[ \Delta_{\mathrm{freeze}}G=-8000+7500 \] \[ \Delta_{\mathrm{freeze}}G=-500\,\mathrm{J\,mol^{-1}} \] \(\textbf{Scaling to the sample amount:}\) Exactly 0.100 mol of liquid is considered. Therefore, \[ \Delta G_{\mathrm{sample}}=(0.100)(-500) \] \[ \Delta G_{\mathrm{sample}}=-50\,\mathrm{J} \] \(\textbf{Checking the phase preference:}\) The negative Gibbs-energy change means freezing lowers the Gibbs energy of the sample at 300 K under the stated conditions. This is consistent with the temperature being below the equilibrium melting temperature. At 320 K, the Gibbs-energy difference between the two phases is zero. At 300 K, the solid phase is thermodynamically favoured. The temporary persistence of a metastable liquid does not contradict this conclusion because phase transformation may require nucleation. The problem distinguishes equilibrium phase stability from the kinetic initiation of freezing. \(\textbf{Final answer:}\) \[ \Delta G_{\mathrm{sample}}=-50\,\mathrm{J} \]

Explanation: \(\textbf{Determining the reaction-system entropy change:}\) The reaction entropy is positive. The reaction extent is 0.250 mol. Therefore, \[ \Delta S_{\mathrm{system}}=\xi\Delta_rS \] Substituting, \[ \Delta S_{\mathrm{system}}=(0.250)(50.0) \] \[ \Delta S_{\mathrm{system}}=+12.5\,\mathrm{J\,K^{-1}} \] \(\textbf{Calculating the reaction enthalpy:}\) The molar reaction enthalpy is positive. Thus, the forward transformation absorbs heat at constant pressure. \[ \Delta H_{\mathrm{actual}}=(0.250)(12.0) \] \[ \Delta H_{\mathrm{actual}}=+3.00\,\mathrm{kJ} \] Under the stated conditions, \[ q_P=\Delta H_{\mathrm{actual}} \] \[ q_P=+3000\,\mathrm{J} \] \(\textbf{Calculating the reservoir entropy change:}\) The reservoir supplies 3000 J at 300 K. Therefore, \[ \Delta S_{\mathrm{reservoir}}=\frac{-3000}{300} \] \[ \Delta S_{\mathrm{reservoir}}=-10.0\,\mathrm{J\,K^{-1}} \] \(\textbf{Applying the universe entropy criterion:}\) \[ \Delta S_{\mathrm{univ}}=\Delta S_{\mathrm{system}}+\Delta S_{\mathrm{reservoir}} \] \[ \Delta S_{\mathrm{univ}}=12.5-10.0 \] \[ \Delta S_{\mathrm{univ}}=+2.50\,\mathrm{J\,K^{-1}} \] The positive result is compatible with spontaneous forward transformation under the stated conditions. \(\textbf{Checking with Gibbs energy:}\) Convert molar reaction entropy into kilojoule units. \[ \Delta_rS=+0.0500\,\mathrm{kJ\,mol^{-1}\,K^{-1}} \] Then, \[ \Delta_rG=\Delta_rH-T\Delta_rS \] \[ \Delta_rG=12.0-(300)(0.0500) \] \[ \Delta_rG=12.0-15.0 \] \[ \Delta_rG=-3.00\,\mathrm{kJ\,mol^{-1}} \] The negative Gibbs-energy change supports the same conclusion. \(\textbf{Interpreting the endothermic transformation:}\) The system absorbs heat, and the reservoir consequently loses entropy. However, the system gains more entropy than the reservoir loses. Therefore, the universe entropy increases. The forward transformation is thermodynamically favourable despite being endothermic. This illustrates why reaction enthalpy alone is insufficient for predicting spontaneity. \(\textbf{Final answer:}\) \[ \boxed{\Delta S_{\mathrm{univ}}=+2.50\,\mathrm{J\,K^{-1}},\qquad\Delta_rG=-3.00\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Recovering the separate-liquid volumes:}\) Density is the ratio of mass to volume. For liquid X, \[ V_X=\frac{m_X}{\rho_X} \] \[ V_X=\frac{100}{0.800} \] \[ V_X=125\,\mathrm{mL} \] For liquid Y, \[ V_Y=\frac{100}{1.25} \] \[ V_Y=80.0\,\mathrm{mL} \] \(\textbf{Identifying the volume change on mixing:}\) If the two initial volumes were simply additive, their sum would be \[ V_{\mathrm{sum}}=125+80.0 \] \[ V_{\mathrm{sum}}=205\,\mathrm{mL} \] The measured final volume is only 195 mL. Therefore, \[ \Delta V_{\mathrm{mix}}=195-205 \] \[ \Delta V_{\mathrm{mix}}=-10.0\,\mathrm{mL} \] The negative mixing volume indicates contraction under the specified conditions. \(\textbf{Calculating the mixture density:}\) Mass is conserved during mixing because no substance enters, leaves or reacts. \[ m_{\mathrm{final}}=100+100 \] \[ m_{\mathrm{final}}=200\,\mathrm{g} \] Therefore, \[ \rho_{\mathrm{final}}=\frac{m_{\mathrm{final}}}{V_{\mathrm{final}}} \] \[ \rho_{\mathrm{final}}=\frac{200}{195} \] \[ \rho_{\mathrm{final}}\approx1.0256\,\mathrm{g\,mL^{-1}} \] \(\textbf{Scaling the otherwise identical mixture:}\) In the second experiment, both liquid amounts are doubled while their proportions and intensive conditions remain unchanged. The resulting homogeneous mixture has twice the mass and twice the volume. \[ m_{\mathrm{final,2}}=2(200) \] \[ m_{\mathrm{final,2}}=400\,\mathrm{g} \] \[ V_{\mathrm{final,2}}=2(195) \] \[ V_{\mathrm{final,2}}=390\,\mathrm{mL} \] \(\textbf{Verifying the intensive-property invariant:}\) \[ \rho_{\mathrm{final,2}}=\frac{400}{390} \] \[ \rho_{\mathrm{final,2}}\approx1.0256\,\mathrm{g\,mL^{-1}} \] Density remains unchanged by proportional scaling of the same homogeneous mixture. However, the measured mixture volume cannot be obtained by adding the initial pure-liquid volumes without a mixing correction. This distinguishes extensivity of the total volume from the incorrect assumption that volume is always additive across a change of thermodynamic state. \(\textbf{Final answer:}\) \[ \boxed{\rho_{\mathrm{final}}\approx1.0256\,\mathrm{g\,mL^{-1}},\qquad V_{\mathrm{final,2}}=390\,\mathrm{mL}} \]

Explanation: \(\textbf{Identifying the transfers across each boundary:}\) The same physical event can be analysed using different selected systems. The heat exchange between gas and bath is internal to both P and Q. The mechanical energy transfer from gas to weight crosses the boundary of P but remains internal to Q. The energy balances must therefore depend on the chosen boundary. \(\textbf{Calculating the gas energy change:}\) With the Chemistry work convention, \[ \Delta U_{\mathrm{gas}}=q+w \] The gas absorbs heat: \[ q_{\mathrm{gas}}=+600\,\mathrm{J} \] It performs work on the weight: \[ w_{\mathrm{gas}}=-250\,\mathrm{J} \] Therefore, \[ \Delta U_{\mathrm{gas}}=600-250 \] \[ \Delta U_{\mathrm{gas}}=+350\,\mathrm{J} \] \(\textbf{Determining the bath energy change:}\) The bath supplies the 600 J received by the gas. Therefore, \[ \Delta E_{\mathrm{bath}}=-600\,\mathrm{J} \] \(\textbf{Applying the boundary of system P:}\) P contains the gas and bath. Adding their energy changes, \[ \Delta E_P=\Delta U_{\mathrm{gas}}+\Delta E_{\mathrm{bath}} \] \[ \Delta E_P=350-600 \] \[ \Delta E_P=-250\,\mathrm{J} \] This decrease equals the energy transferred across P's boundary to the excluded weight. \(\textbf{Enlarging the boundary to system Q:}\) Q also includes the weight. The weight gains \[ \Delta E_{\mathrm{weight}}=+250\,\mathrm{J} \] Therefore, \[ \Delta E_Q=\Delta E_P+\Delta E_{\mathrm{weight}} \] \[ \Delta E_Q=-250+250 \] \[ \Delta E_Q=0 \] \(\textbf{Interpreting system classification:}\) For P, work crosses its boundary even though its gas-to-bath heat transfer is internal. For Q, all specified transfers occur between parts of the chosen system. The complete arrangement exchanges neither matter nor energy with the external environment. Thus, Q is isolated under the stated assumptions, while P is not. Enlarging the system boundary changes which transfers are external without changing the physical process itself. \(\textbf{Final answer:}\) \[ \boxed{\Delta E_P=-250\,\mathrm{J},\qquad\Delta E_Q=0} \]

Explanation: Statement I is incorrect because a two-phase mixture contains portions of the same substance in thermodynamically different physical states. Liquid and vapour have different molar enthalpies at the same equilibrium temperature and pressure. The total enthalpy also depends on how much substance occupies each phase. Therefore, temperature, pressure and total amount do not uniquely fix the total enthalpy unless the phase distribution is also specified. Statement II is correct because vaporisation changes liquid into vapour. The enthalpy of vaporisation is positive under the stipulated conditions. For a fixed total amount, increasing the quantity of vapour increases the fraction of material in the higher-enthalpy phase. The total enthalpy consequently increases even though temperature and pressure remain constant. Statement III is correct because enthalpy is extensive. Doubling an otherwise identical two-phase sample doubles the amounts of both phases. Since the phase fractions and intensive conditions remain unchanged, each contribution to the total enthalpy doubles. The total enthalpy therefore doubles. Statement IV is incorrect because constancy of temperature does not require constancy of enthalpy. During equilibrium vaporisation, energy is absorbed as latent heat. The phase distribution changes even though the temperature is unchanged. The molar enthalpy of the combined liquid-vapour sample can consequently change as the vapour fraction increases. The distinction involves intensive variables, extensive state functions and the additional phase-distribution information needed to specify a multiphase thermodynamic state. Only Statements II and III are correct.

Explanation: \(\textbf{Recognising the work represented by the graph:}\) The work depends on the external pressure during expansion. Because the pressure varies linearly with volume, the relevant area is a trapezium rather than a rectangle. The magnitude of expansion work is obtained from the average external pressure multiplied by the volume change. \(\textbf{Calculating the volume change:}\) \[ \Delta V=V_f-V_i \] \[ \Delta V=6.00-2.00 \] \[ \Delta V=+4.00\,\mathrm{L} \] \(\textbf{Recovering the area under the pressure-volume line:}\) \[ |w_1|=\frac{P_{\mathrm{ext},i}+P_{\mathrm{ext},f}}2\Delta V \] Substituting, \[ |w_1|=\frac{3.00+1.00}{2}(4.00) \] \[ |w_1|=8.00\,\mathrm{L\,bar} \] \[ |w_1|=800\,\mathrm{J} \] The gas performs work during expansion. Therefore, \[ w_1=-800\,\mathrm{J} \] \(\textbf{Determining the common internal-energy change:}\) The first-law relation is \[ \Delta U=q+w \] For the first path, \[ \Delta U=1000-800 \] \[ \Delta U=+200\,\mathrm{J} \] \(\textbf{Evaluating work in the alternative path:}\) The second expansion occurs against constant external pressure of 1.00 bar. Therefore, \[ w_2=-P_{\mathrm{ext}}\Delta V \] \[ w_2=-(1.00)(4.00) \] \[ w_2=-4.00\,\mathrm{L\,bar} \] \[ w_2=-400\,\mathrm{J} \] \(\textbf{Recovering the second-path heat:}\) Both paths connect the same thermodynamic states. Consequently, their internal-energy changes are identical. \[ +200=q_2-400 \] \[ q_2=+600\,\mathrm{J} \] \(\textbf{Checking the path dependence:}\) The first path absorbs 1000 J of heat and performs 800 J of work. The second absorbs 600 J of heat and performs 400 J of work. The additional heat absorbed in the first path is exactly balanced by its greater work output. Therefore, both give the same internal-energy change of +200 J. The result combines interpretation of an external-pressure graph with the distinction between path functions and state functions. \(\textbf{Final answer:}\) \[ \boxed{\Delta U=+200\,\mathrm{J},\qquad q_2=+600\,\mathrm{J}} \]

Explanation: \(\textbf{Identifying the two different process constraints:}\) Stage I is isothermal, so the ideal-gas internal-energy change is zero. Stage II is at constant volume, so its pressure-volume work is zero. The heat exchanges must therefore be calculated using different relations for the two stages. \(\textbf{Calculating reversible expansion work:}\) \[ w_I=-nRT\ln2 \] Substituting, \[ w_I=-(2.00)(8.314)(300)(0.6931) \] \[ w_I\approx-3457\,\mathrm{J} \] The gas performs approximately 3457 J of work. \(\textbf{Recovering Stage I heat:}\) For unchanged ideal-gas temperature, \[ \Delta U_I=0 \] Therefore, \[ 0=q_I+w_I \] \[ q_I=-w_I \] \[ q_I\approx+3457\,\mathrm{J} \] \(\textbf{Evaluating constant-volume heating:}\) During Stage II, \[ w_{II}=0 \] The internal-energy change is \[ \Delta U_{II}=nC_{V,m}(T_f-T_i) \] \[ \Delta U_{II}=(2.00)(21.0)(400-300) \] \[ \Delta U_{II}=4200\,\mathrm{J} \] Thus, \[ q_{II}=+4200\,\mathrm{J} \] \(\textbf{Combining the heat transfers:}\) \[ q_{\mathrm{total}}=q_I+q_{II} \] \[ q_{\mathrm{total}}\approx3457+4200 \] \[ q_{\mathrm{total}}\approx7657\,\mathrm{J} \] \[ q_{\mathrm{total}}\approx+7.66\,\mathrm{kJ} \] \(\textbf{Calculating the entropy gained through expansion:}\) \[ \Delta S_I=nR\ln2 \] \[ \Delta S_I=(2.00)(8.314)(0.6931) \] \[ \Delta S_I\approx+11.52\,\mathrm{J\,K^{-1}} \] \(\textbf{Calculating the entropy gained through heating:}\) \[ \Delta S_{II}=nC_{V,m}\ln\left(\frac{400}{300}\right) \] \[ \Delta S_{II}=(2.00)(21.0)(0.2877) \] \[ \Delta S_{II}\approx+12.08\,\mathrm{J\,K^{-1}} \] \(\textbf{Determining total entropy change:}\) \[ \Delta S_{\mathrm{total}}=\Delta S_I+\Delta S_{II} \] \[ \Delta S_{\mathrm{total}}\approx11.52+12.08 \] \[ \Delta S_{\mathrm{total}}\approx+23.61\,\mathrm{J\,K^{-1}} \] \(\textbf{Checking the physical significance:}\) Both stages increase the gas entropy, but for different reasons. The first increases the gas volume at fixed temperature. The second increases temperature at fixed volume. The net internal-energy increase is only 4200 J, even though the total heat absorbed is larger, because part of the heat supplies expansion work. The two different constraints must not be replaced by one constant-volume or constant-pressure heating equation. \(\textbf{Final answer:}\) \[ \boxed{q_{\mathrm{total}}\approx+7.66\,\mathrm{kJ},\qquad\Delta S_{\mathrm{total}}\approx+23.61\,\mathrm{J\,K^{-1}}} \]

Explanation: \(\textbf{Distinguishing constant temperature from constant internal energy:}\) A fixed amount of nonreacting ideal gas has internal energy determined by its temperature. A reacting system can change chemical composition even when its temperature remains unchanged. Consequently, the same isothermal condition does not require the same internal-energy change for both vessels. \(\textbf{Evaluating the nonreacting gas in P:}\) Its temperature and composition remain unchanged. Therefore, \[ \Delta U_P=0 \] The vessel is rigid, so \[ \Delta V_P=0 \] and \[ w_P=0 \] The first law gives \[ \Delta U_P=q_P+w_P \] Thus, \[ q_P=0 \] \(\textbf{Calculating the chemical internal-energy change in Q:}\) The reaction amount is \[ \xi=0.200\,\mathrm{mol} \] Therefore, \[ \Delta U_Q=\xi\Delta_rU \] \[ \Delta U_Q=(0.200)(-10.0) \] \[ \Delta U_Q=-2.00\,\mathrm{kJ} \] \(\textbf{Applying the rigid-vessel work condition:}\) Vessel Q also has constant volume. No pressure-volume work occurs. \[ w_Q=0 \] The first law therefore gives \[ q_Q=\Delta U_Q \] \[ q_Q=-2.00\,\mathrm{kJ} \] \(\textbf{Interpreting the different heat exchanges:}\) The nonreacting gas requires no heat exchange to maintain its unchanged thermodynamic state. The reacting gas releases 2.00 kJ to its thermal reservoir despite having the same initial and final temperature. The energy change in Q arises from chemical transformation rather than sensible heating. The example shows why the ideal-gas relation between internal energy and temperature must be applied with attention to chemical composition. \(\textbf{Final answer:}\) \[ \boxed{q_P=0,\qquad q_Q=-2.00\,\mathrm{kJ}} \]

Explanation: \(\textbf{Identifying the initial direction of heat transfer:}\) Body B is at 400 K while body A is at 300 K. Temperature, not total stored internal energy, determines the spontaneous direction of heat flow between bodies in thermal contact. Thus, heat initially passes from B to A. The fact that A has the larger heat capacity does not reverse this direction. \(\textbf{Applying conservation of energy:}\) The complete arrangement is insulated and performs no work. Therefore, the sensible energy gained by A equals that lost by B. \[ 400(T_f-300)+100(T_f-400)=0 \] Expanding, \[ 400T_f-120000+100T_f-40000=0 \] \[ 500T_f=160000 \] \[ T_f=320\,\mathrm{K} \] \(\textbf{Calculating A's entropy gain:}\) \[ \Delta S_A=400\ln\left(\frac{320}{300}\right) \] \[ \Delta S_A=400(0.06454) \] \[ \Delta S_A=+25.816\,\mathrm{J\,K^{-1}} \] \(\textbf{Calculating B's entropy loss:}\) \[ \Delta S_B=100\ln\left(\frac{320}{400}\right) \] \[ \Delta S_B=100(-0.22314) \] \[ \Delta S_B=-22.314\,\mathrm{J\,K^{-1}} \] \(\textbf{Determining the complete entropy change:}\) \[ \Delta S_{\mathrm{total}}=\Delta S_A+\Delta S_B \] \[ \Delta S_{\mathrm{total}}=25.816-22.314 \] \[ \Delta S_{\mathrm{total}}=+3.502\,\mathrm{J\,K^{-1}} \] \(\textbf{Checking the amount of transferred energy:}\) Body A warms by 20 K. Therefore, \[ q_A=(400)(20) \] \[ q_A=+8000\,\mathrm{J} \] Body B cools by 80 K. \[ q_B=(100)(-80) \] \[ q_B=-8000\,\mathrm{J} \] The energy transfers cancel, as required. \(\textbf{Resolving the student's misconception:}\) The initially colder body may contain more total sensible energy because it has a much larger heat capacity. Nevertheless, spontaneous heat transfer proceeds from higher temperature to lower temperature. The positive entropy change confirms the thermodynamic direction in the insulated arrangement. Total energy is extensive, while temperature is intensive. These properties answer different physical questions. \(\textbf{Final answer:}\) \[ \boxed{T_f=320\,\mathrm{K},\qquad\Delta S_{\mathrm{total}}\approx+3.50\,\mathrm{J\,K^{-1}}} \]

Explanation: \(\textbf{Separating the energy-transfer mechanisms:}\) The gas loses heat to the bath but gains electrical work. It also performs pressure-volume expansion work as it warms at constant pressure. All three transfers must be included in the first-law energy balance. \(\textbf{Calculating the internal-energy increase:}\) For an ideal gas of unchanged composition, \[ \Delta U=nC_{V,m}\Delta T \] Substituting, \[ \Delta U=(2.00)(20.0)(400-300) \] \[ \Delta U=4000\,\mathrm{J} \] \[ \Delta U=+4.00\,\mathrm{kJ} \] \(\textbf{Calculating the pressure-volume work:}\) At constant pressure, \[ P\Delta V=nR\Delta T \] Therefore, \[ w_{PV}=-nR\Delta T \] \[ w_{PV}=-(2.00)(8.314)(100) \] \[ w_{PV}=-1662.8\,\mathrm{J} \] \(\textbf{Recovering the electrical work:}\) The gas releases heat: \[ q=-1000\,\mathrm{J} \] Let \(w_e\) denote the electrical work supplied to the system. The first law gives \[ \Delta U=q+w_{PV}+w_e \] Substituting, \[ 4000=-1000-1662.8+w_e \] \[ w_e=4000+1000+1662.8 \] \[ w_e=6662.8\,\mathrm{J} \] \[ w_e\approx+6.66\,\mathrm{kJ} \] \(\textbf{Calculating the enthalpy change:}\) For the unchanged ideal-gas composition, \[ \Delta H=\Delta U+\Delta(PV) \] \[ \Delta H=\Delta U+nR\Delta T \] \[ \Delta H=4000+1662.8 \] \[ \Delta H=5662.8\,\mathrm{J} \] \[ \Delta H\approx+5.66\,\mathrm{kJ} \] \(\textbf{Checking the constant-pressure heat misconception:}\) The process occurs at constant pressure, but electrical work also enters the gas. Therefore, the heat transfer alone is not equal to the enthalpy change. The appropriate balance is \[ \Delta H=q+w_e \] \[ \Delta H=-1000+6662.8 \] \[ \Delta H=+5662.8\,\mathrm{J} \] The result agrees with the independent ideal-gas calculation. Constant-pressure heat equals enthalpy change only when no additional non-pressure-volume work is involved. \(\textbf{Final answer:}\) \[ \boxed{w_e\approx+6.66\,\mathrm{kJ},\qquad\Delta H\approx+5.66\,\mathrm{kJ}} \]

Explanation: \(\textbf{Connecting the two measured heats:}\) At constant volume, with only pressure-volume work, \[ q_V=\Delta U \] At constant pressure under the stated conditions, \[ q_P=\Delta H \] For a reaction involving ideal gases, \[ \Delta_rH-\Delta_rU=\Delta n_gRT \] \(\textbf{Calculating the observed enthalpy–internal-energy difference:}\) The constant-pressure and constant-volume heats differ by \[ q_P-q_V=8.50-8.00 \] \[ q_P-q_V=+0.500\,\mathrm{kJ} \] \(\textbf{Recovering the implied gaseous mole change:}\) The reaction extent is \[ \xi=0.200\,\mathrm{mol} \] Therefore, \[ q_P-q_V=\xi\Delta n_gRT \] Rearranging, \[ \Delta n_g=\frac{q_P-q_V}{\xi RT} \] Substituting in joules, \[ \Delta n_g=\frac{500}{(0.200)(8.314)(300)} \] \[ \Delta n_g\approx1.002 \] The measured difference is consistent, within rounding, with \[ \Delta n_g=+1 \] \(\textbf{Testing the gaseous stoichiometry of the options:}\) For option A, \[ \Delta n_g=1-2=-1 \] The predicted sign of \(q_P-q_V\) would be negative, so A is inconsistent. For option B, the reactant carbonate and product oxide are solids. Only the carbon dioxide product contributes to gaseous mole change. \[ \Delta n_g=1-0=+1 \] This agrees with the measurement. For option C, \[ \Delta n_g=2-(1+1)=0 \] The two heats would be equal under the ideal-gas approximation. For option D, \[ \Delta n_g=2-(2+1)=-1 \] Its predicted heat difference again has the wrong sign. \(\textbf{Interpreting the reverse inference:}\) The experiment does not merely reveal whether the reaction is endothermic. The difference between two calorimetric conditions gives information about net gaseous mole production. A positive difference indicates a positive \(\Delta n_g\) under the stated assumptions. Among the supplied equations, only carbonate decomposition produces one additional mole of gas per mole of reaction. The measured heats are thermodynamically consistent with option B.

Explanation: The Assertion is true because constant external pressure alone does not guarantee that heat equals enthalpy change. The familiar relation \(q_P=\Delta H\) requires the appropriate restriction that only pressure-volume work is involved. In the described experiment, the motor supplies an additional form of work to the system. The first law requires this stirring work to be included in the energy balance. With the Chemistry sign convention, \[ \Delta U=q+w_{PV}+w_{\mathrm{stirring}} \] For a constant-pressure process involving the usual pressure-volume work, \[ w_{PV}=-P\Delta V \] Using the definition of enthalpy, the balance becomes \[ \Delta H=q_P+w_{\mathrm{stirring}} \] The heat transfer therefore equals the enthalpy change minus the stirring work performed on the contents. The Reason is true because it identifies precisely why the heat and enthalpy change need not agree. For instance, a motor can supply energy while the system releases some heat to its surroundings. The net enthalpy change reflects the combined energy effects, not merely the heat transfer. The Reason correctly explains the Assertion. Both A and R are true, and R correctly explains A.

Explanation: \(\textbf{Establishing the two endpoint invariants:}\) All four experiments connect states A and B. Therefore, every valid path must have the same internal-energy change. \[ \Delta U_{AB}=+400\,\mathrm{J} \] The corresponding enthalpy difference is also fixed. \[ \Delta H_{AB}=\Delta U_{AB}+\Delta(PV) \] \[ \Delta H_{AB}=400+150 \] \[ \Delta H_{AB}=+550\,\mathrm{J} \] \(\textbf{Testing the constant-volume report P:}\) With only pressure-volume work and fixed volume, \[ w_P=0 \] Therefore, \[ \Delta U_P=q_P+w_P \] \[ \Delta U_P=400+0 \] \[ \Delta U_P=+400\,\mathrm{J} \] This agrees with the endpoint data. \(\textbf{Testing the constant-pressure report Q:}\) Because electrical work is also present, the appropriate constant-pressure relation is \[ \Delta H_Q=q_Q+w_{\mathrm{electrical,Q}} \] Substituting, \[ \Delta H_Q=350+200 \] \[ \Delta H_Q=+550\,\mathrm{J} \] This also agrees with the endpoint data. \(\textbf{Testing the general path R:}\) The first law includes all forms of work. \[ \Delta U_R=q_R+w_{PV,R}+w_{\mathrm{electrical,R}} \] \[ \Delta U_R=700-150-150 \] \[ \Delta U_R=+400\,\mathrm{J} \] The internal-energy change is consistent with the reference value. \(\textbf{Testing report S:}\) Experiment S is at constant pressure and includes electrical work. Therefore, \[ \Delta H_S=q_S+w_{\mathrm{electrical,S}} \] \[ \Delta H_S=550+100 \] \[ \Delta H_S=+650\,\mathrm{J} \] This differs from the required endpoint enthalpy change of +550 J. Thus, S is inconsistent. \(\textbf{Interpreting the conflicting report:}\) In S, the reported 550 J of heat alone already equals the entire endpoint enthalpy difference. The additional 100 J of electrical work cannot be ignored. Under the specified assumptions, S would require a different heat transfer or another energy-transfer mechanism. The other three reports have different heat and work values but remain consistent with the same state-function changes. The correct experiment to reject is S.