Given that,C_{(s)} + O_{2(g)} longrightarrow | vedclass

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(B) The enthalpy of formation of $CO$ corresponds to the reaction: $C_{(s)} + \frac{1}{2}O_{2(g)} \longrightarrow CO_{(g)}$.Given equations:$(i) \ C_{

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

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(B) The enthalpy of formation of $CO$ corresponds to the reaction: $C_{(s)} + \frac{1}{2}O_{2(g)} \longrightarrow CO_{(g)}$.Given equations:$(i) \ C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H^{\circ} = -x \ kJ \ mol^{-1}$$(ii) \ 2CO_{(g)} + O_{2(g)} \longrightarrow 2CO_{2(g)} ; \Delta H^{\circ} = -y \ kJ \ mol^{-1}$Divide equation $(ii)$ by $2$:$CO_{(g)} + \frac{1}{2}O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H^{\circ} = -\frac{y}{2} \ kJ \ mol^{-1} \ (iii)$Subtract equation $(iii)$ from equation $(i)$:$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2}O_{2(g)}) \longrightarrow CO_{2(g)} - CO_{2(g)}$$C_{(s)} + \frac{1}{2}O_{2(g)} \longrightarrow CO_{(g)}$$\Delta H^{\circ}_{f} = -x - (-\frac{y}{2}) = \frac{y}{2} - x = \frac{y-2x}{2} \ kJ \ mol^{-1}$.

Given that,$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H^{\circ} = -x \ kJ \ mol^{-1}$ and $2CO_{(g)} + O_{2(g)} \longrightarrow 2CO_{2(g)} ; \Delta H^{\circ} = -y \ kJ \ mol^{-1}$. The enthalpy of formation of $CO$ will be:

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