Compute the standard enthalpy of formation of liquid methanol ($ ext{CH}_3 ext{OH}(l)$) in $ ext{

Question

Compute the standard enthalpy of formation of liquid methanol ($	ext{CH}_3	ext{OH}(l)$) in $	ext{kJ mol}^{-1}$ using the following thermochemical data: Enthalpy of vaporization of liquid methanol, $\Delta H^\circ_{	ext{vap}}[	ext{CH}_3	ext{OH}(l)] = +38	ext{ kJ mol}^{-1}$. Standard enthalpies of formation of gaseous atoms from their elements in standard states: $\Delta H^\circ_f[	ext{H}(g)] = +218	ext{ kJ mol}^{-1}$, $\Delta H^\circ_f[	ext{C}(g)] = +715	ext{ kJ mol}^{-1}$, $\Delta H^\circ_f[	ext{O}(g)] = +249	ext{ kJ mol}^{-1}$. Average bond dissociation energies: $E_{	ext{C-H}} = 415	ext{ kJ mol}^{-1}$, $E_{	ext{C-O}} = 365	ext{ kJ mol}^{-1}$, $E_{	ext{O-H}} = 463	ext{ kJ mol}^{-1}$. (Report the magnitude of the exothermic heat of formation as a positive integer)

Professor PC Thomas Classes · Answer

First, let us calculate the standard enthalpy of formation of gaseous methanol, $	ext{C}(s, 	ext{graphite}) + 2	ext{H}_2(g) + \frac{1}{2}	ext{O}_2(g) ightarrow 	ext{CH}_3	ext{OH}(g)$. We can express this reaction using an atomization cycle: $\Delta H^\circ_f[	ext{CH}_3	ext{OH}(g)] = \Delta H^\circ_{	ext{atomization}}(	ext{reactants}) - 	ext{Total Bond Energy of products}$. The energy required to atomize the elements into gaseous atoms is: $\Delta H^\circ_{	ext{atomization}} = \Delta H^\circ_f[	ext{C}(g)] + 4\Delta H^\circ_f[	ext{H}(g)] + \Delta H^\circ_f[	ext{O}(g)] = 715 + 4(218) + 249 = 715 + 872 + 249 = +1836	ext{ kJ mol}^{-1}$. In one molecule of $	ext{CH}_3	ext{OH}(g)$, there are three $	ext{C-H}$ bonds, one $	ext{C-O}$ bond, and one $	ext{O-H}$ bond. The total bond energy released upon forming gaseous methanol is: $	ext{BE}_{	ext{total}} = 3(E_{	ext{C-H}}) + E_{	ext{C-O}} + E_{	ext{O-H}} = 3(415) + 365 + 463 = 1245 + 365 + 463 = +2073	ext{ kJ mol}^{-1}$. Thus, $\Delta H^\circ_f[	ext{CH}_3	ext{OH}(g)] = 1836 - 2073 = -237	ext{ kJ mol}^{-1}$. Finally, to obtain liquid methanol, we condense the gas: $	ext{CH}_3	ext{OH}(g) ightarrow 	ext{CH}_3	ext{OH}(l); \Delta H = -\Delta H^\circ_{	ext{vap}} = -38	ext{ kJ mol}^{-1}$. Therefore, $\Delta H^\circ_f[	ext{CH}_3	ext{OH}(l)] = -237 - 38 = -275	ext{ kJ mol}^{-1}$. The absolute value is $275	ext{ kJ mol}^{-1}$.