The ratio of heats liberated at 298 K from t | vedclass

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(A) One $kg$ of coke contains $n = \frac{1000}{12} = 83.33 \ mol$ of carbon. Heat liberated by burning $1 \ kg$ of coke: $C(s) + O_{2}(g) \rightarrow

Vedclass · Accepted Answer

$0.69$

Vedclass · Answer

(A) One $kg$ of coke contains $n = \frac{1000}{12} = 83.33 \ mol$ of carbon. Heat liberated by burning $1 \ kg$ of coke: $C(s) + O_{2}(g) ightarrow CO_{2}(g) ; \Delta H_{1} = 83.33 	imes 393.5 \ kJ$. Water gas is produced by: $C(s) + H_{2}O(g) ightarrow CO(g) + H_{2}(g)$. Thus,$1 \ kg$ of coke produces $83.33 \ mol$ of $CO$ and $83.33 \ mol$ of $H_{2}$. Heat liberated by burning this water gas: $CO(g) + \frac{1}{2}O_{2}(g) ightarrow CO_{2}(g) ; \Delta H_{CO} = 83.33 	imes 283.5 \ kJ$. $H_{2}(g) + \frac{1}{2}O_{2}(g) ightarrow H_{2}O(l) ; \Delta H_{H_{2}} = 83.33 	imes 285.5 \ kJ$. Total heat $\Delta H_{2} = 83.33 	imes (283.5 + 285.5) = 83.33 	imes 569 \ kJ$. Ratio $= \frac{\Delta H_{1}}{\Delta H_{2}} = \frac{393.5}{569} \approx 0.69 : 1$.

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