Given 2CO(g)+O2(g)→2CO2(g) ΔH1θ=−x kJ mol−1 C(graphite)+O2(g)→CO2(g) ΔH2θ=−y kJ mol−1 The ΔHθ For the reaction C(graphite)+12O2(g)→CO(g) is:

C(graphite)+O2(g)→CO2(g) ΔH2=−y kJ mol−1 Reversed 12ΔH=x−2y2

Given 2CO(g)+O2(g)→2CO2(g) ΔH1θ=−x kJ mol−1 C(graphite)+O2(g)→CO2(g) ΔH2θ=−y kJ mol−1 The ΔHθ For the reaction C(graphite)+12O2(g)→CO(g) is:

C(graphite)+O2(g)→CO2(g) ΔH2=−y kJ mol−1 Reversed 12ΔH=x−2y2

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Given $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g) Δ H_{1}^{θ} = - x k J m o l^{- 1}$ $C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g) Δ H_{2}^{θ} = - y k J m o l^{- 1}$ The $Δ H^{θ}$ For the reaction $C (g r a p h i t e) + \frac{1}{2} O_{2} (g) \to C O (g)$ is:

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

$2 y - x$

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

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

answer is C.

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Detailed Solution

$\begin{array}{l} C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g) \\ Δ H_{2} = - y k J m o {l^{- 1}}_{}^{} \end{array}$

$\frac{Reversed 1}{2}$

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$Δ H = \frac{x - 2 y}{2}$

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