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(B) The heat of formation of $CO$ corresponds to the reaction: $C(s) + \frac{1}{2}O_2(g) 	o CO(g)$.Given:$(i) \ C(s) + O_2(g) 	o CO_2(g)$; $\Delta H

Vedclass · Accepted Answer

$-109.5$

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(B) The heat of formation of $CO$ corresponds to the reaction: $C(s) + \frac{1}{2}O_2(g) 	o CO(g)$.Given:$(i) \ C(s) + O_2(g) 	o CO_2(g)$; $\Delta H_1 = -394 \ kJ \ mol^{-1}$$(ii) \ 2CO(g) + O_2(g) 	o 2CO_2(g)$; $\Delta H_2 = -569 \ kJ \ mol^{-1}$Divide equation $(ii)$ by $2$:$(iii) \ CO(g) + \frac{1}{2}O_2(g) 	o CO_2(g)$; $\Delta H_3 = \frac{-569}{2} = -284.5 \ kJ \ mol^{-1}$Subtract equation $(iii)$ from equation $(i)$:$(C + O_2) - (CO + \frac{1}{2}O_2) 	o CO_2 - CO_2$$C(s) + \frac{1}{2}O_2(g) 	o CO(g)$$\Delta H_f = \Delta H_1 - \Delta H_3 = -394 - (-284.5) = -109.5 \ kJ \ mol^{-1}$.

Substance	$H_{2(g)}$	$C_{\text{(graphite)}}$	$C_{2}H_{6(g)}$
$\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$	$-286.0$	$-394.0$	$-1560.0$

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Substance $H_{2(g)}$ $C_{\text{(graphite)}}$ $C_{2}H_{6(g)}$

$\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$ $-286.0$ $-394.0$ $-1560.0$

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