Question

Compute the heat of formation of liquid methyl alcohol in kilojoules per mole using the following data.
The heat of vaporisation of liquid methyl alcohol = $38kJ/mol$.
The heats of formation of gaseous atoms from the elements is their standard states: $H,218kJ/mol$; $C,713kJ/mol$; $O,249kJ/mol$.
Average bond energies: $C-H=415kJ/mol$; $C-O=365kJ/mol$; $O-H=463kJ/mol$.

• A. $-266kJ/mol$
• B. $-488kJ/mol$
• C. Zero
• D. $-356kJ/mol$

Answer 1

The correct option is A $-266kJ/mol$

$C\left(s\right)+2H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to C{H}_{3}OH\left(l\right);\Delta H=?$
So, $\Delta H_f=[\Delta H_{C\left(s\right)\to C\left(g\right)}+2\Delta H_{H-H}+1/2\Delta H_{O=O}]-[3\Delta H_{C-H}+\Delta H_{C-O}+\Delta H_{O-H}+\Delta H_{vap.C{H}_{3}OH}]$
For reactants;
Heat of atomisation of 1 mole of $C=715kJ$
Heat of atomisation of 4 moles of $H=4\times 218kJ$
Heat of atomisation of 1 mole of $O=249kJ$
For products:
Heat of formation of 3 moles of $C-H$ bonds = $-3\times 415kJ$
Heat of formation of 1 mole of $C-O$ bonds = $-356kJ$
Heat of formation of 1 mole of $O-H$ bonds = $-463kJ$
Heat of condenstation of 1 mole of $C{H}_{3}OH$ to liquid $=-38kJ$
So, On adding, we get $\Delta H$ of formation of $C{H}_{3}OH\left(l\right)$:
$\Delta H=715+(4\times 218)+249-(3\times 415)-356-463-38$
$\Delta H=-266kJ/mol$