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Heat of combustion $\Delta H^\circ$ for C(s), $H_2$(g) and $CH_4$(g) are –94, –68 and –213 kcal/mol. Then $\Delta H^\circ$ for $C(s) + 2H_2(g) \rightarrow CH_4(g)$ is : -
A
–17 kcal
B
–111 kcal
C
–170 kcal
D
–85 kcal
Detailed Solution
Given:
(i) $C + O_2 \rightarrow CO_2$, $\Delta H = -94$ kcal
(ii) $H_2 + \dfrac{1}{2}O_2 \rightarrow H_2O$, $\Delta H = -68$ kcal
(iii) $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$, $\Delta H = -213$ kcal
Required reaction = (i) + 2 $\times$ (ii) $-$ (iii):
$\Delta H = \sum\Delta H_c(\text{reactants}) - \sum\Delta H_c(\text{products}) = (-94) + 2(-68) - (-213)$
$= -94 - 136 + 213 = -17$ kcal
(i) $C + O_2 \rightarrow CO_2$, $\Delta H = -94$ kcal
(ii) $H_2 + \dfrac{1}{2}O_2 \rightarrow H_2O$, $\Delta H = -68$ kcal
(iii) $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$, $\Delta H = -213$ kcal
Required reaction = (i) + 2 $\times$ (ii) $-$ (iii):
$\Delta H = \sum\Delta H_c(\text{reactants}) - \sum\Delta H_c(\text{products}) = (-94) + 2(-68) - (-213)$
$= -94 - 136 + 213 = -17$ kcal