# Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

Question:

Calculate the standard enthalpy of formation of $\mathrm{CH}_{3} \mathrm{OH}_{(n)}$ from the following data:

$\mathrm{CH}_{3} \mathrm{OH}_{(f)}+\frac{3}{2} \mathrm{O}_{2(g)} \longrightarrow \mathrm{CO}_{2(g)}+2 \mathrm{H}_{2} \mathrm{O}_{(f)} ; \Delta_{r} H^{\theta}=-726 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{C}_{(g)}+\mathrm{O}_{2(g)} \longrightarrow \mathrm{CO}_{2(g)} ; \Delta_{c} H^{\theta}=-393 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{H}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \longrightarrow \mathrm{H}_{2} \mathrm{O}_{(f)} ; \Delta_{f} H^{\theta}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:

The reaction that takes place during the formation of $\mathrm{CH}_{3} \mathrm{OH}_{(f)}$ can be written as:

$\mathrm{C}_{(s)}+2 \mathrm{H}_{2} \mathrm{O}_{(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}_{(f)}$

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:

Equation (ii) + 2 × equation (iii) – equation (i)

$\Delta_{\mathrm{f}} \mathrm{H}^{\theta}\left[\mathrm{CH}_{3} \mathrm{OH}_{(n)}\right]=\Delta_{\mathrm{C}} H^{\theta}+2 \Delta_{f} H^{\theta}\left[\mathrm{H}_{2} \mathrm{O}_{(n}\right]-\Delta_{r} H^{\theta}$

$=\left(-393 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)+2\left(-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)-\left(-726 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$

$=(-393-572+726) \mathrm{kJ} \mathrm{mol}^{-1}$

$\therefore \Delta_{f} H^{\theta}\left[\mathrm{CH}_{3} \mathrm{OH}_{(l)}\right]=-239 \mathrm{~kJ} \mathrm{~mol}^{-1}$