Thermodynamics And Thermochemistry Result 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:
$\begin{aligned} & \mathrm{C}-\mathrm{H}=415 \mathrm{~kJ} / \mathrm{mol}, \mathrm{C}-\mathrm{O}=356 \mathrm{~kJ} / \mathrm{mol}, \\ & \mathrm{O}-\mathrm{H}=463 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$
(1997, 5M)
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Answer:
Correct Answer: 19. $(-116.4 \mathrm{~kJ})$
Solution:
- Given: $\mathrm{CH}_3 \mathrm{OH}(\mathrm{g}) \longrightarrow \mathrm{CH}_3 \mathrm{OH}(\mathrm{l}) ; \quad \Delta H=-38 \mathrm{~kJ}$
$\begin{aligned} & \mathrm{C}(\mathrm{g})+4 \mathrm{H}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CH}_3 \mathrm{OH}(\mathrm{g}) \text {; } \\ & \Delta H=-(3 \times 415+356+463) \\ & \because \quad H=H_1+H_2=-2064 \mathrm{~kJ} \\ & \mathrm{C}(\mathrm{g}) \longrightarrow \mathrm{C}(\mathrm{g}) ; \quad \Delta H=715 \mathrm{~kJ} \\ & \end{aligned}$
$\begin{aligned} & 2 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{H}(\mathrm{g}) ; \quad \Delta H=2 \times 2 \times 218=872 \mathrm{~kJ} \\ & \frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{O}(\mathrm{g}) ; \quad \Delta H=249 \mathrm{~kJ} \end{aligned}$
Adding : $\mathrm{C}(\mathrm{gr})+2 \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CH}_3 \mathrm{OH}(\mathrm{l})$
$\begin{aligned} \Delta H & =-266 \mathrm{~kJ} / \mathrm{mol} \\ & =\frac{1 \times 1.25}{0.082 \times 300} \times 20.8 \times(189.86-300) \mathrm{J} \\ & =-116.4 \mathrm{~J} \end{aligned}$
BETA
