The enthalpy of formation of liquid methyl alcohol in kJ/mol, using the following data (in kJ/mol):< — Thermodynamics and Thermochemistry Chemistry Question
Question
The enthalpy of formation of liquid methyl alcohol in kJ/mol, using the following data (in kJ/mol):<br>Heat of vaporization of liquid methyl alcohol = 38.<br>Heat of formation of gaseous atoms from the elements in their standard states: H, 218; C, 715; O, 249.<br>Average bond energies: C–H, 415; C–O, 356; O–H, 463.
💡 Solution & Explanation
Atomization energy of standard elements: $\Delta H_{\text{atom}} = \Delta H_f(\text{C}) + 4\Delta H_f(\text{H}) + \Delta H_f(\text{O}) = 715 + 4(218) + 249 = 1836\text{ kJ/mol}$. Bond energies formed in $\text{CH}_3\text{OH(g)}$: $3\text{BE(C-H)} + \text{BE(C-O)} + \text{BE(O-H)} = 3(415) + 356 + 463 = 2064\text{ kJ/mol}$. $\Delta H_f(\text{CH}_3\text{OH, g}) = 1836 - 2064 = -228\text{ kJ/mol}$. For liquid, $\Delta H_f(\text{l}) = -228 - 38 = -266\text{ kJ/mol}$.