Thermodynamics: JEE Advanced PYQ, 2007–2026

Answer: (D)

Both solids give the same solution of CuSO₄(aq). Take the dissolution of anhydrous CuSO₄ (ΔH = −66.5) and reverse the dissolution of the pentahydrate (ΔH = −11.7). The solution cancels, leaving CuSO₄ + 5H₂O → CuSO₄·5H₂O. ΔH = −66.5 + (−11.7) = −78.2 kJ mol⁻¹.

Answer: (A)

Bond enthalpies apply to gaseous species, so first convert to H₂O(g): H₂(g) + ½O₂(g) → H₂O(l) is −286 kJ, and H₂O(l) → H₂O(g) is +44 kJ, giving ΔfH(H₂O, g) = −242 kJ mol⁻¹. For H₂(g) + ½O₂(g) → H₂O(g), ΔH = (bonds broken) − (bonds formed) = 436 + ½(498) − 2x. So −242 = 685 − 2x, giving 2x = 927 and x = 463.5 kJ mol⁻¹. Mistakes arise from skipping the liquid-to-gas correction, using a whole O=O instead of half, or flipping the sign of ΔH.

Answer: (C)

Neutralising HCN with NaOH can be split into two steps: (1) HCN → H⁺ + CN⁻ (ΔH_ion) and (2) H⁺ + OH⁻ → H₂O (−57.1). By Hess's law the sum equals the observed −12.1 kJ mol⁻¹. So ΔH_ion = −12.1 − (−57.1) = +45.0 kJ mol⁻¹. Part of the energy released by forming water is spent in ionising the weak acid.

Answer: (A)

Hess's law: ΔH depends only on the initial and final states, so thermochemical equations can be added, reversed (sign of ΔH changes) or multiplied (ΔH scales). The target needs C and CO₂ on the left and 2CO on the right. Equation (i) already has C on the left and one CO₂ on the right. CO₂ must be on the left, so keep equation (i) as written and reverse equation (ii): 2CO₂ → 2CO + O₂, ΔH = +566.0 kJ. Adding: C + O₂ + 2CO₂ → CO₂ + 2CO + O₂. Cancelling O₂ and one CO₂ gives C + CO₂ → 2CO. ΔH = −393.5 + 566.0 = +172.5 kJ.

Answer: (C)

Keep the first equation as written, since Al₂O₃ is a product. Reverse the second, since Fe₂O₃ is a reactant, which changes its ΔH to +824 kJ. Adding them cancels 3/2 O₂, giving Fe₂O₃ + 2Al → Al₂O₃ + 2Fe with ΔH = −1676 + 824 = −852 kJ.

Answer: (A)

Internal energy is a state function, so ΔU depends only on the initial and final states and not on the path. For A→B, q = +60 J and w = −20 J (the system does work), so ΔU = +40 J. For the return B→A, ΔU must be −40 J whatever the path. On path 2, q = −50 J, so w = ΔU − q = −40 − (−50) = +10 J. As a check, over the full cycle ΔU = 0. Net q = 60 − 50 = +10 J and net w = −20 + 10 = −10 J, which sum to 0. So q and w are path-dependent, but their sum for each leg is fixed by the endpoints.

Answer: (C)

Apply the first law, ΔU = q + w. The container is insulated, so q = 0. The gas expands into a vacuum, so p_ext = 0 and w = −p_ext ΔV = 0. Hence ΔU = 0. For an ideal gas U depends only on temperature, so the temperature does not change either. The common errors are assuming an expanding gas always does work, and assuming an expanding gas always cools.

Answer: (A)

Dissolving an ionic solid can be split into two steps. First, the lattice is broken into gaseous ions, which costs the lattice enthalpy (+2200 kJ mol⁻¹). Second, the gaseous ions are hydrated, which releases energy. By Hess's law, ΔsolH = ΔlattH + ΔhydH(cations) + ΔhydH(anions). One formula unit of MX₂ gives one M²⁺ and two X⁻, so ΔsolH = 2200 + (−1650) + 2(−300) = −50 kJ mol⁻¹. For 0.20 mol the enthalpy change is 0.20 × (−50) = −10 kJ, so 10 kJ of heat is released.

Answer: (A)

Heat at constant volume, then expand isothermally: ΔS = nCv ln(T₂/T₁) + nR ln(V₂/V₁) = 2 × (3/2)R × ln 2 + 2 × R × ln 4 = 3R ln 2 + 4R ln 2 = 7R ln 2 = 7 × 8.314 × 0.693 = 40.33 J/K. The answer does not depend on how the change was in fact carried out.

Answer: (A)

A bomb calorimeter works at constant volume, so the heat measured equals ΔU. The relation is ΔH = ΔU + Δng RT, where Δng is the change in moles of gas only. Liquid water is not counted. The gaseous products are 6 mol CO2 and the gaseous reactants are 7.5 mol O2, so Δng = 6 − 7.5 = −1.5. Then ΔngRT = −1.5 × 8.314 × 300 J = −3741 J ≈ −3.74 kJ. So ΔH = −3263 − 3.74 ≈ −3266.7 kJ mol⁻¹. The enthalpy is more negative than ΔU because the surroundings do compression work on the system as the gas moles fall.