QJEEM17Che85d: JEE Main 2017 Chemistry - Class 11 JEEM17Che85d Solution: Given, ; = - 393.5 kJ mol -1 + = - 285.8 kJ mol -1 + + = +…

C(graphite) + O₂(g) $\to$ CO₂ (g);
${\Delta _r}{H^o}$ = - 393.5 kJ mol^-1 .............(1)

${{\rm H}_2}(g)$ + ${1 \over 2}{O_2}(g)$ $\to {{\rm H}_2}{\rm O}(l)$
${\Delta _r}{H^o}$ = - 285.8 kJ mol^-1 .........(2)

$C{O_2}(g)$ + $2{{\rm H}_2}{\rm O}(l) \to$ $C{H_4}(g)$ + $2{O_2}(g)$
${\Delta _r}{H^o}$ = + 890.3 kJ mol^-1 .........(3)

C(graphite) + 2H₂(g) $\to$ CH₄ (g);    $\Delta$ H = ? ........ (4)

You can see,

[Eq. (1) + Eq. (3)] + [2 × Eq. (2)] = Eq. (4)

$\therefore$ [ $\Delta$ _rH₁ + $\Delta$ _rH₃ ] + [2 $\times$ $\Delta$ _rH₂] = $\Delta$ H

$\Rightarrow$ $\Delta$ H = [(-393.5) + (890.3)] + [2(-285.8)] = -74.8 kJ / mol