Thermodynamics: NEET PYQ, 2018–2026

Answer: (C)

A state function depends only on the current state of the system, so its change depends only on the initial and final states. Internal energy, enthalpy, pressure, volume and temperature are state functions. Heat and work are not: they describe energy transferred during a process, and that depends on the route. Because ΔU is a state function, the first law ΔU = q + w means that q and w can vary along different paths while their sum stays the same.

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: (A)

At constant pressure, work is w = -PΔV, where w is the work done on the system. Compression means ΔV is negative, so w is positive. Substituting P = 2 atm and ΔV = -6 L gives w = 12 L·atm. Converting with 1 L·atm = 101.3 J gives about +1216 J.

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: (A)

Build the target from the given equations. Start with (i) − (ii): C(s) + ½O₂ → CO(g), ΔH = −110.5. The target also needs H₂O → H₂ + ½O₂, which is the reverse of (iii), so ΔH = +242.0. Sum: C(s) + H₂O(g) → CO(g) + H₂(g), ΔH = −110.5 + 242.0 = +131.5 kJ mol⁻¹. Each equation is used once, with the sign set by its direction.

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: (C)

ΔS_surr = −ΔH/T = −(−60 000 J mol⁻¹)/300 K = +200 J K⁻¹ mol⁻¹. ΔS_total = ΔS_sys + ΔS_surr = −150 + 200 = +50 J K⁻¹ mol⁻¹. The total is positive, so the reaction is spontaneous although the entropy of the system falls.

Answer: (A)

Principle: in a Hess cycle the steps are added as written, and any species that appears on both sides cancels. Here 2NO₂ appears as a product in the first step and as a reactant in the second, so the two equations add to N₂ + 2O₂ → N₂O₄. The enthalpies add algebraically: +67.8 + (−57.2) = +10.6 kJ mol⁻¹.