Thermodynamics and Thermochemistry 2 Question 19
19. Compute the heat of formation of liquid methyl alcohol in $\mathrm{kJ} \mathrm{mol}^{-1}$, using the following data. Heat of vaporisation of liquid methyl alcohol $=38 \mathrm{~kJ} / \mathrm{mol}$. Heat of formation of gaseous atoms from the elements in their standard states :
$\mathrm{H}=218 \mathrm{~kJ} / \mathrm{mol}, \mathrm{C}=715 \mathrm{~kJ} / \mathrm{mol}, \mathrm{O}=249 \mathrm{~kJ} / \mathrm{mol}$.
Average bond energies:
$(1997,5 \mathrm{M})$
$\mathrm{C}-\mathrm{H}=415 \mathrm{~kJ} / \mathrm{mol}, \mathrm{C}-\mathrm{O}=356 \mathrm{~kJ} / \mathrm{mol}$,
$\mathrm{O}-\mathrm{H}=463 \mathrm{~kJ} / \mathrm{mol}$
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Solution:
- Given : $\mathrm{CH}{3} \mathrm{OH}(g) \longrightarrow \mathrm{CH}{3} \mathrm{OH}(l) ; \quad \Delta H=-38 \mathrm{~kJ}$
$\mathrm{C}(\mathrm{g})+4 \mathrm{H}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})$
$$ \Delta H=-(3 \times 415+356+463) $$
$\because \quad H=H_{1}+H_{2}=-2064 \mathrm{~kJ}$
$\mathrm{C}(\mathrm{g}) \longrightarrow \mathrm{C}(\mathrm{g})$;
$\Delta H=715 \mathrm{~kJ}$
$$ \begin{aligned} 2 \mathrm{H}{2}(g) \longrightarrow & 4 \mathrm{H}(g) ; \ & \Delta H=2 \times 2 \times 218=872 \mathrm{~kJ} \ \frac{1}{2} \mathrm{O}{2}(g) \longrightarrow & \mathrm{O}(g) ; \quad \Delta H=249 \mathrm{~kJ} \end{aligned} $$
Adding : $\mathrm{C}$ (gr) $+2 \mathrm{H}{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)$
$$ \begin{aligned} \Delta H & =-266 \mathrm{~kJ} / \mathrm{mol} \ & =\frac{1 \times 1.25}{0.082 \times 300} \times 20.8 \times(189.86-300) \mathrm{J} \ & =-116.4 \mathrm{~J} \end{aligned} $$
