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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}$.

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$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -395 \text{ kJ}$
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From the given diagram, the enthalpy change $\Delta_r H$ for the reaction $C \rightarrow A$ is:

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