The enthalpy of formation (Delta H_f) of meth | vedclass

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(D) The chemical equation for the oxidation of methanol is: $CH_3OH(l) + \frac{1}{2} O_2(g) \longrightarrow HCHO(g) + H_2O(l)$The enthalpy change of t

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$-163$

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(D) The chemical equation for the oxidation of methanol is: $CH_3OH(l) + \frac{1}{2} O_2(g) \longrightarrow HCHO(g) + H_2O(l)$The enthalpy change of the reaction $(\Delta_r H)$ is calculated using the formula: $\Delta_r H = \sum \Delta_f H^{\circ}(	ext{products}) - \sum \Delta_f H^{\circ}(	ext{reactants})$Given values:$\Delta_f H^{\circ}(CH_3OH) = -239 \ kJ \ mol^{-1}$$\Delta_f H^{\circ}(HCHO) = -116 \ kJ \ mol^{-1}$$\Delta_f H^{\circ}(H_2O) = -286 \ kJ \ mol^{-1}$$\Delta_f H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state)Substituting the values:$\Delta_r H = [(-116) + (-286)] - [-239 + 0]$$\Delta_r H = -402 - (-239)$$\Delta_r H = -402 + 239 = -163 \ kJ \ mol^{-1}$

The enthalpy of formation $(\Delta H_f)$ 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

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