The standard molar heats of combustion ($\Delta H^\circ_c$) at $298 ext{ K}$ for acetylene gas ($

Question

The standard molar heats of combustion ($\Delta H^\circ_c$) at $298	ext{ K}$ for acetylene gas ($	ext{C}_2	ext{H}_2$), graphite ($	ext{C}$), and hydrogen gas ($	ext{H}_2$) are $-310.62	ext{ kcal mol}^{-1}$, $-94.05	ext{ kcal mol}^{-1}$, and $-68.32	ext{ kcal mol}^{-1}$, respectively. Calculate the standard heat of formation of acetylene gas ($\Delta H^\circ_f[	ext{C}_2	ext{H}_2(g)]$ in $	ext{kcal mol}^{-1}$) at $298	ext{ K}$. (Report your answer to two decimal places)

Professor PC Thomas Classes · Answer

The synthesis reaction for one mole of acetylene gas from its constituent elements in their standard reference states is: $2	ext{C}(s, 	ext{graphite}) + 	ext{H}_2(g) ightarrow 	ext{C}_2	ext{H}_2(g)$. Using Hess's law, we can compute the enthalpy of formation directly from the combustion enthalpies of the elemental reactants and the product: $\Delta H^\circ_f[	ext{C}_2	ext{H}_2(g)] = 2\Delta H^\circ_c[	ext{C(graphite)}] + \Delta H^\circ_c[	ext{H}_2(g)] - \Delta H^\circ_c[	ext{C}_2	ext{H}_2(g)]$. Substituting the given thermochemical combustion values: $\Delta H^\circ_f[	ext{C}_2	ext{H}_2] = 2(-94.05) + (-68.32) - (-310.62) = -188.10 - 68.32 + 310.62 = -256.42 + 310.62 = +54.20	ext{ kcal mol}^{-1}$.