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Heat of combustion $\Delta H$ for $C(s), H_2(g)$ and $CH_4(g)$ are -94, -68 and -213 kcal/mol. Then $\Delta H$ for $C(s) + H_2(g) \rightarrow CH_4(g)$ is


(a) -17 kcal
(b) -111 kcal
(c) -170 kcal
(d) -85 kcal
Can you answer this question?
 
 

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Answer: -17 kcal
 
(i) $C(s) + 2H_2(g) \rightarrow CH_4(g)$
(ii) $C(s) + O_2(g) \rightarrow CO_2(g)$ ; $\Delta H = -$94 kcal $mol^{-1}$
(iii) $H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O (l)$; $\Delta H = -$68 kcal $mol^{-1}$
(iv) $CH_4(g) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$ ; $\Delta H = -$213 kcal $mol^{-1}$
To obtain equation (i), We need to add the above equations in the following way
{ $C(s) + O_2(g) \rightarrow CO_2(g)$ } $\times 1$
{ $H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O (l)$ } $\times 2$
{ $CH_4(g) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$ } $\times -1$
Therefore, $\Delta H= -94+2(-68)-(-213) = -94 -136+213$
$\Delta H = -230+213 = -$17 kcal
answered Feb 25, 2014 by mosymeow_1
 

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