Question

From the following data $CH _{3} OH (l)+\frac{3}{2} O _{2}(g) \longrightarrow CO _{2}(g)+2 H _{2} O (l)$ $\Delta_{r} H^{\circ}=-726\, k J\, mol ^{-1}$ $H _{2}(g)+\frac{1}{2} O _{2}(g) \longrightarrow H _{2} O (l)$ $\Delta_r H ^{\circ}=-286\, kJ\, mol ^{-1}$ $C (\text { graphite })+ O _{2}(g) \longrightarrow CO _{2}(g)$ $\Delta_{r} H ^{\circ}=-393\, kJ\, mol ^{-1}$ The standard enthalpy of formation of $CH _{3} OH (l)$ in $kJ\, mol ^{-1}$ is

• A. -239
• B. 239
• C. 547
• D. -905

Answer 1

Answer: (A)

-239

Answer 2

$CH _{3}- OH +3 / 2 O _{2} \longrightarrow CO _{2}+2 H _{2} O$,

$\Delta H=-726\, kJ\, / mol\,...(i)$
$2 \times ( H _{2}+ 1 / 2 O _{2} \longrightarrow H _{2} O , \Delta H=-286\, kJ\, /$ mol) ...(ii)

$C + O _{2} \longrightarrow CO _{2}, \Delta H=-393\, kJ\, / mol \, ...(iii)$

From E (ii) $+( iii )- Eq . (i)$
$-286 \times 2-393-(-726)=7$
$=-572-393+726$
$=-965+726$
$=-239\, kJ\, / mol$