Calculate the magnitude of the standard enthalpy of combustion ($|\Delta H^\circ_{ ext{comb}}|$ in

Question

Calculate the magnitude of the standard enthalpy of combustion ($|\Delta H^\circ_{	ext{comb}}|$ in $	ext{kJ mol}^{-1}$) of cyclopropane at $298	ext{ K}$. The standard enthalpies of formation of $	ext{CO}_2(g)$, $	ext{H}_2	ext{O}(l)$, and propene($g$) are $-393.5	ext{ kJ mol}^{-1}$, $-285.8	ext{ kJ mol}^{-1}$, and $+20.42	ext{ kJ mol}^{-1}$, respectively. The enthalpy of isomerization of cyclopropane($g$) to propene($g$) is $-33.0	ext{ kJ mol}^{-1}$. (Report your answer to two decimal places)

Professor PC Thomas Classes · Answer

First, we determine the enthalpy of formation of cyclopropane($g$) using the given enthalpy of isomerization: $	ext{cyclopropane}(g) ightarrow 	ext{propene}(g); \Delta H^\circ_{	ext{iso}} = -33.0	ext{ kJ mol}^{-1}$. Since $\Delta H^\circ_{	ext{iso}} = \Delta H^\circ_f[	ext{propene}(g)] - \Delta H^\circ_f[	ext{cyclopropane}(g)]$, we get $-33.0 = 20.42 - \Delta H^\circ_f[	ext{cyclopropane}(g)] \implies \Delta H^\circ_f[	ext{cyclopropane}(g)] = 20.42 + 33.0 = +53.42	ext{ kJ mol}^{-1}$. Now, consider the complete combustion reaction of cyclopropane ($	ext{C}_3	ext{H}_6$): $	ext{C}_3	ext{H}_6(g) + \frac{9}{2}	ext{O}_2(g) ightarrow 3	ext{CO}_2(g) + 3	ext{H}_2	ext{O}(l)$. The enthalpy of combustion is: $\Delta H^\circ_{	ext{comb}} = 3\Delta H^\circ_f[	ext{CO}_2(g)] + 3\Delta H^\circ_f[	ext{H}_2	ext{O}(l)] - \Delta H^\circ_f[	ext{C}_3	ext{H}_6(g)] = 3(-393.5) + 3(-285.8) - (+53.42) = -1180.5 - 857.4 - 53.42 = -2091.32	ext{ kJ mol}^{-1}$. Thus, $|\Delta H^\circ_{	ext{comb}}| = 2091.32	ext{ kJ mol}^{-1}$.