Calculate the heat of reaction at constant volume ($\Delta U^\circ_{298}$ or $q_V$ in $ ext{kcal}$)

Question

Calculate the heat of reaction at constant volume ($\Delta U^\circ_{298}$ or $q_V$ in $	ext{kcal}$) for the hydrogenation of acetylene gas to ethylene gas at $25^\circ	ext{C}$: $	ext{C}_2	ext{H}_2(g) + 	ext{H}_2(g) ightarrow 	ext{C}_2	ext{H}_4(g)$. Given the following standard heats of combustion at constant pressure ($\Delta H^\circ_c$ in $	ext{kcal mol}^{-1}$) at $25^\circ	ext{C}$: Heat of combustion of water ($	ext{H}_2(g) ightarrow 	ext{H}_2	ext{O}(l)$), $\Delta H^\circ_1 = -68.3	ext{ kcal mol}^{-1}$; Heat of combustion of acetylene ($	ext{C}_2	ext{H}_2(g)$), $\Delta H^\circ_2 = -310.6	ext{ kcal mol}^{-1}$; Heat of combustion of ethylene ($	ext{C}_2	ext{H}_4(g)$), $\Delta H^\circ_3 = -337.2	ext{ kcal mol}^{-1}$. (Take $R = 1.987 	imes 10^{-3}	ext{ kcal K}^{-1}	ext{ mol}^{-1}$ and $T = 298.15	ext{ K}$. Report your answer to one decimal place with sign)

Professor PC Thomas Classes · Answer

We first determine the enthalpy of reaction at constant pressure ($\Delta H^\circ$) using the standard heats of combustion of reactants and products: $	ext{C}_2	ext{H}_2(g) + 	ext{H}_2(g) ightarrow 	ext{C}_2	ext{H}_4(g)$. $\Delta H^\circ = \Delta H^\circ_c[	ext{C}_2	ext{H}_2] + \Delta H^\circ_c[	ext{H}_2] - \Delta H^\circ_c[	ext{C}_2	ext{H}_4] = -310.6 + (-68.3) - (-337.2) = -378.9 + 337.2 = -41.7	ext{ kcal}$. To find the heat of reaction at constant volume ($q_V = \Delta U^\circ$), we use the thermodynamic relation: $\Delta H^\circ = \Delta U^\circ + \Delta n_g RT \implies \Delta U^\circ = \Delta H^\circ - \Delta n_g RT$. Here, $\Delta n_g = n_g(	ext{products}) - n_g(	ext{reactants}) = 1 - (1 + 1) = -1	ext{ mol}$. At $T = 298.15	ext{ K}$, the term $\Delta n_g RT = (-1	ext{ mol}) 	imes (1.987 	imes 10^{-3}	ext{ kcal K}^{-1}	ext{ mol}^{-1}) 	imes 298.15	ext{ K} = -0.5924	ext{ kcal}$. Therefore, $\Delta U^\circ = -41.70 - (-0.5924) = -41.1076	ext{ kcal} \approx -41.1	ext{ kcal}$.