The enthalpy of formation $\left(\Delta \mathrm{H}_f\right)$ of methanol, formaldehyde and water are $-239…

The enthalpy of formation $\left(\Delta \mathrm{H}_f\right)$ of methanol, formaldehyde and water are $-239,-116$ and $-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The enthalpy change for the oxidation of methanol to formaldehyde and water in $\mathrm{kJ}$ is
  1. -136
  2. -173
  3. 163
  4. -163

Solution

Given $\Delta H_f$ for $\mathrm{CH}_3 \mathrm{OH}=-239 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\Delta H_f$ of $\mathrm{H} . \mathrm{CHO}=-116 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\Delta H_f$ of $\mathrm{H}_2 \mathrm{O}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Required relation: $ \begin{aligned} & 2 \mathrm{CH}_3 \mathrm{OH}+\mathrm{O}_2 \longrightarrow 2 \mathrm{HCHO}+2 \mathrm{H}_2 \mathrm{O} \\ & \Delta H_R=\frac{1}{2} H_{f^{\circ}}[(2 x-116)+(2 x-286)] \\ & -1[(2 \times 239)] \\ & \Delta_r H=\Sigma\left(\Delta_f H_{\text {Product }}\right)-\Sigma\left(\Delta_f H_{\text {Reactant }}\right) \\ & =(-116+(-286))-(-293+0) \\ & =-163 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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