If C_{(s)} + O_{2(g)} rightarrow CO_{2(g)},De | vedclass

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(B) To find the enthalpy of formation of $CO_{(g)}$,we need the reaction: $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$.Given equations:

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$Y - X$

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(B) To find the enthalpy of formation of $CO_{(g)}$,we need the reaction: $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$.Given equations:						$(i) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$			$\Delta H = -X$							$(ii) \ CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$			$\Delta H = -Y$			Subtracting equation $(ii)$ from equation $(i)$:$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2} O_{2(g)}) \rightarrow CO_{2(g)} - CO_{2(g)}$$C_{(s)} + \frac{1}{2} O_{2(g)} - CO_{(g)} \rightarrow 0$$C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$The enthalpy change is $\Delta_f H = (-X) - (-Y) = Y - X$.

If $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H = -X$,and $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H = -Y$,calculate $\Delta_f H$ for $CO_{(g)}$ formation.

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