Question

Compute the enthalpy of formation of liquid methyl alcohol in kJ ${mol}^{-1}$, using the following data. Enthalpy of vaporization of liquid ${CH}_{3}OH=38kJ/mol$.
Enthalphy of formation of gaseous atoms from the elements in their standard states are enthalpy of formation of gaseous atoms from the elements in their standard states are
$H$ $\to$ 218 kJ / mol ; $C$ $\to$ 715 kJ / mol ; $O$ $\to$ 249 kJ / mol
Bond Enthalpies
$C-H$ $\to$ 415 kJ/mol ; $C-O$ $\to$ 356 kJ / mol ; $O-H$ $\to$ 463 kJ/mol

Answer 1

$C+2H_2+\frac{1}{2}{O}_2⟶C{H}_{3}OH\left(g\right)$

Enegry of $C{H}_{3}OH=3(B.E{)}_{C-H}+(B.E{)}_{C-O}+(B.E{)}_{O-H}$

$=3\left(415\right)+\left(356\right)+463$

$=2064KJ/mol$

$\Delta H_{reaction}=(E{)}_{C{H}_{3}OH}-(E{)}_{reactants}$

$=\left(2064\right)-\left[\right(715)+2(218)+\frac{1}{2}\times 2\times 249]$

$=664KJ/mol$

$C{H}_{3}OH\left(l\right)⟶C{H}_{3}OH\left(g\right)$ $\Delta H^{\prime }=38KJ/mol$

$\Rightarrow C+2H_2+\frac{1}{2}{O}_2⟶C{H}_{3}OH\left(l\right)$ $\Delta H$

$\Rightarrow \Delta H=-\Delta H^{\prime }+{\Delta }_{reaction}$

$=664-38=626KJ/mol$ .