The standard enthalpy of combustion of C (gra | vedclass

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Question

(B) The formation reaction for methanol is: $C(graphite) + 2H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow CH_3OH_{(l)}$Given combustion reactions:$(i)$ $

Vedclass · Accepted Answer

$-239 \ kJ \ mol^{-1}$

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(B) The formation reaction for methanol is: $C(graphite) + 2H_{2(g)} + \frac{1}{2}O_{2(g)} ightarrow CH_3OH_{(l)}$Given combustion reactions:$(i)$ $C(graphite) + O_{2(g)} ightarrow CO_{2(g)}$, $\Delta H_1 = -393 \ kJ \ mol^{-1}$(ii) $H_{2(g)} + \frac{1}{2}O_{2(g)} ightarrow H_2O_{(l)}$, $\Delta H_2 = -286 \ kJ \ mol^{-1}$(iii) $CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} ightarrow CO_{2(g)} + 2H_2O_{(l)}$, $\Delta H_3 = -726 \ kJ \ mol^{-1}$To get the formation reaction, we perform: $(i) + 2 	imes (ii) - (iii)$$\Delta H_f = \Delta H_1 + 2(\Delta H_2) - \Delta H_3$$\Delta H_f = -393 + 2(-286) - (-726)$$\Delta H_f = -393 - 572 + 726 = -239 \ kJ \ mol^{-1}$

The standard enthalpy of combustion of $C$ (graphite), $H_{2(g)}$ and $CH_3OH_{(l)}$ respectively are $-393 \ kJ \ mol^{-1}$, $-286 \ kJ \ mol^{-1}$ and $-726 \ kJ \ mol^{-1}$. What is the standard enthalpy of formation of methanol?

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