Given C + O_2 rightarrow CO_2 : Delta H = -x | vedclass

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(B) The formation reaction for $CO$ is: $C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)$.Given equations:$(1) \ C + O_2 \rightarrow CO_2$ ; $\Delta H = -x

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$\frac{y - 2x}{2}$

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(B) The formation reaction for $CO$ is: $C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)$.Given equations:$(1) \ C + O_2 \rightarrow CO_2$ ; $\Delta H = -x \ kJ$$(2) \ 2CO + O_2 \rightarrow 2CO_2$ ; $\Delta H = -y \ kJ$Divide equation $(2)$ by $2$:$CO + \frac{1}{2}O_2 \rightarrow CO_2$ ; $\Delta H = -\frac{y}{2} \ kJ$ $(3)$Subtract equation $(3)$ from equation $(1)$:$(C + O_2) - (CO + \frac{1}{2}O_2) \rightarrow CO_2 - CO_2$$C + \frac{1}{2}O_2 \rightarrow CO$Enthalpy change $\Delta H_f = (-x) - (-\frac{y}{2}) = \frac{y}{2} - x = \frac{y - 2x}{2} \ kJ$.

Given $C + O_2 \rightarrow CO_2$ : $\Delta H = -x \ kJ$ and $2CO + O_2 \rightarrow 2CO_2$ : $\Delta H^\circ = -y \ kJ$,find the enthalpy of formation of carbon monoxide.

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