For the synthesis of ammonia via the Haber process: $\mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} \rightar

Question

For the synthesis of ammonia via the Haber process: $\mathrm{N}_{2(g)} + 3\mathrm{H}_{2(g)} ightarrow 2\mathrm{NH}_{3(g)}$ at 300 K, the enthalpy change ($\Delta H$) is  $-92.4 	ext{ kJ}$. What is the corresponding internal energy change ($\Delta U$) for this reaction? (Given $R = 8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$)

Professor PC Thomas Classes · Answer

(a) : The relationship between enthalpy and internal energy is $\Delta H = \Delta U + \Delta n_g RT$. For this reaction, the change in gaseous moles is $\Delta n_g = 2 - (1 + 3) = -2$. Rearranging for internal energy: $\Delta U = \Delta H - \Delta n_g RT$. Substituting the values: $\Delta U = -92400 	ext{ J} - (-2 	imes 8.314 	imes 300) 	ext{ J} = -92400 + 4988.4 = -87411.6 	ext{ J}$. This is approximately $-87.4 	ext{ kJ}$.