The difference between enthalpies of reaction at constant pressure and constant volume for the reaction: 2(l) + 15(g) - 12(g) + 6(l) at 298 K in kJ is

Question

The difference between enthalpies of reaction at constant pressure and constant volume for the reaction: 2$C_6H_6$(l) + 15$O_2$(g) -> 12$CO_2$(g) + 6$H_2O$(l) at 298 K in kJ is

Prashant Kotian · Accepted Answer

Correct answer: A. The relation between enthalpy of reaction (δ H) and internal energy change (δ E) is: δ H - δ E = δ n_g * R * T. For the reaction: 2(l) + 15(g) -> 12(g) + 6(l), the change in gaseous moles is: δ n_g = n_products(g) - n_reactants(g) = 12 - 15 = -3. Given T = 298 K, R = 8.314 J/(mol K) = 8.314 * 10^-3 kJ/(mol K). Thus: δ H - δ E = -3 * 8.314 * 10^-3 * 298 = -7.43 kJ.

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