Question

The enthalpy of formation $\left(\Delta H_{f}\right)$ of methanol, formaldehyde and water are $-239,-116$ and $-286\, kJ\, mol ^{-1}$ respectively. The enthalpy change for the oxidation of methanol to formaldehyde and water in $kJ$ is

• A. -136
• B. -173
• C. 163
• D. -163

Answer 1

Answer: (D)

Given $H_{f^\circ}$, for $CH _{3} OH =-239\, kJmol ^{-1}$

$H . CHO =-116\, kJmol ^{-1}$

$H _{2} O =-286\, kJmol ^{-1} 0$

Required relation:

$2 CH _{3} OH + O _{2} \longrightarrow 2 HCHO +2 H _{2} O$

$\Delta H_{R}=\frac{1}{2} H_{f^{\circ}}$ of H.CHO. and $H _{2} O$

$\Delta H_{R}=\frac{1}{2} H_{f^{\circ}}[(2 x-116)+(2 x-286)]$

$-1[(2 \times 239)]$

$=\frac{1}{2}[(-232)+(-572)]-[-478]$

$=\frac{1}{2}[-804+478]$

$\Rightarrow \frac{1}{2}[-326]=-163\, kJmol ^{-1}$