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Example · Example 5

Q.For a certain reaction carried out at constant volume and 298 K298\ \text{K}, ΔU∘=−100 kJ mol−1\Delta U^\circ = -100\ \text{kJ mol}^{-1} and the number of moles of gas increases by 11 (Δng=+1\Delta n_g = +1). Calculate ΔH∘\Delta H^\circ for the reaction.

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The relation between enthalpy change and internal energy change for a reaction at constant temperature involving a change in moles of gas is ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT. Here ΔU∘=−100 kJ mol−1=−100,000 J mol−1\Delta U^\circ = -100\ \text{kJ mol}^{-1} = -100{,}000\ \text{J mol}^{-1}, Δng=+1\Delta n_g = +1, R=8.314 J K−1mol−1R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}, and T=298 KT = 298\ \text{K}. First calculate ΔngRT=(1)(8.314)(298)=2477.57 J mol−1≈2.478 kJ mol−1\Delta n_g RT = (1)(8.314)(298) = 2477.57\ \text{J mol}^{-1} \approx 2.478\ \text{kJ mol}^{-1}. Then ΔH∘=ΔU∘+ΔngRT=−100+2.478=−97.52 kJ mol−1\Delta H^\circ = \Delta U^\circ + \Delta n_g RT = -100 + 2.478 = -97.52\ \text{kJ mol}^{-1}. Since more moles o …

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