NCERT Exemplar Class 11 Chemistry Chapter 6 MCQ

(c) 37.56Explanation:For gaseous reactants and products, we have a relation between standard enthalpy of vaporization (ΔHvap) and standard internal energy (ΔE) as-ΔHvap = ΔE+ Δng​ RT whereas,Δng = n2​ − n1, i.e., difference between no. of moles of reactant and product.For vaporization of water,H2O (l) → H2O(g)​Therefore, Δ$ ng = 1$ −$0 = 1$Therefore, ΔHvap = ΔE+ Δng RT⇒ ΔE = ΔHvap − Δng​$ RT = 40$.63 − $(1\times8.314\times10-3 \times 373)$ = 37.53 KJ/molHence the value ΔE for this process will be 37.53KJ/mol.