Calculate the standard enthalpy of formation of CH₃OH(l) from the following data: (i) CH₃OH(l) + (3)/(2)O2(g)

Question 98 of 115

CH_3OH_{(l)}

(i)

CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}

\Delta H^\circ = -726\ kJ\ mol^{-1}

(ii)

C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}

\Delta_c H^\circ = -393\ kJ\ mol^{-1}

(iii)

H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}

\Delta_f H^\circ = -286\ kJ\ mol^{-1}

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2022Subjective· 3mImportance★★★★★

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Combining the combustion data by Hess's law gives the standard enthalpy of formation of liquid methanol.

Target reaction: $C(s) + 2H_2(g) + \tfrac12 O_2(g) \rightarrow CH_3OH(l)$ , $\Delta_fH^\circ = ?$

Given:

$CH_3OH(l) + \tfrac32 O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$ , $\Delta H^\circ = -726\ kJ$
$C(s) + O_2(g) \rightarrow CO_2(g)$ , $\Delta_cH^\circ = -393\ kJ$
$H_2(g) + \tfrac12 O_2(g) \rightarrow H_2O(l)$ , $\Delta_fH^\circ = -286\ kJ$ Apply Hess's law: Target = (ii) + 2×(iii) − (i) $\Delta_fH^\circ(CH_3OH) = \Delta H^\circ_{(ii)} + 2\Delta H^\circ_{(iii)} - \Delta H^\circ_{(i)}$ $= (-393) + 2(-286) - (-726) = -393 - 572 + 726 = -239\ kJ\,mol^{-1}$ …

Q.Calculate the standard enthalpy of formation of $CH_3OH_{(l)}$ from the following data: