Question

enthalpy of combustion of Carbon, Hydrogen and methanol are respectively -300, -250 and - 400 kj/mol. Find enthalpy of formation of methanol.​

Answer 1

$\LARGE\green{\boxed{Answer}}$

$\large\underline{\mathrm {Given}}:-$

$\small C + O_{2}\rightarrow CO_{2},\Delta_{c}H_{Carbon}=-300 KJ/mol$

$\small 2H_{2} + O_{2}\rightarrow{2}O,\Delta_{c}H_{hydrogen}=-250 KJ/mol$

$\small CH_{4} + 2O_{2}\rightarrow CO_{2} + 2H_{2}O,\Delta_{c}H_{methanol}=-400KJ/mol$

$\large\underline{\mathrm {To\:Find}}:-$

$C + 2H_{2}\rightarrow CH_{4} ,\Delta_{f}H_{methanol} = ?$

$\large\bf\underline {\mathrm{Solution}}:-$

$\sf{Let:-}$

Enthalpy of combustion of carbon be x.

Enthalpy of combustion of Hydrogen be y.

Enthalpy of combustion of methanol be z.

$\sf Now:-$

Enthalpy of formation of methanol

= z- x - 2y

= -400 -( -300) - 2×(-250)

= -400 + 300 + 500

= 400 KJ/mol.