Question

standard enthalpy of combustion of carbon hydrogen and methanol are -393.5, -285.8 and -726 respectively. Calculate the enthalphy of formation of methanol​

Answer 1

The enthalphy of formation of methanol​ is -239 kJ.mol⁻¹

Explanation:

The formation of methanol is given by the equation:

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

In question, the combustion of methanol enthalphy is given and equation for methanol combustion is given below:

$\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\longrightarrow (1)$

$\left.\Delta_{r} H^{0}=-726 \ \mathrm{kJ}\mathrm{mol}^{-1}$

In question, the combustion of carbon enthalphy is given and equation for carbon combustion is given below:

$\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) \longrightarrow (2)$

$\Delta \mathrm{H}^{0}=-393 \ \mathrm{kJ} \mathrm{mol}^{-1}$

In question, the combustion of hydrogen enthalphy is given and equation for hydrogen combustion is given below:

$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\longrightarrow (3)$

$\left.\Delta_{f} H^{0}=-286 \ \mathrm{kJ} \mathrm{mol}^{-1}$

On multiplying equation (3) by 2, we get,

$2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Now, the enthalphy becomes:

$\left.\Delta_{f} H^{0}= 2\times(-286) \ \mathrm{kJ} \mathrm{mol}^{-1} = -572\ \mathrm{kJ} \mathrm{mol}^{-1}$

On adding equation (2) and (3), we get,

$\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Now, the enthalphy becomes:

$\Delta_{r} H^{0}=-393 \ k J \mathrm{mol}^{-1}-572 \ k J \mathrm{mol}^{-1}=-965 \ k J \mathrm{mol}^{-1}$

On subtracting equation (1) from (2), we get,

$\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}-\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})\rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1)-\mathrm{CO}_{2}(\mathrm{g})-2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

which is similar to the equation, $\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})$

Now, the enthalphy becomes:

$\Delta_{r} H^{0}=-965 \ \mathrm{kJ} \mathrm{mol}^{-1}+726 \ \mathrm{kJ} \mathrm{mol}^{-1}=-239 \ \mathrm{kJ} \mathrm{mol}^{-1}$

Thus, enthalphy of formation of methanol​ is -239 kJ.mol⁻¹