(B) For the reaction,$CO_{2(g)} + C_{(s)} \rightleftharpoons 2 CO_{(g)}$$K_p = \frac{(p_{CO})^2}{p_{CO_2}} = \frac{(0.6)^2}{0.15} = \frac{0.36}{0.15}

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(B) For the reaction,$CO_{2(g)} + C_{(s)} \rightleftharpoons 2 CO_{(g)}$$K_p = \frac{(p_{CO})^2}{p_{CO_2}} = \frac{(0.6)^2}{0.15} = \frac{0.36}{0.15} = 2.4$Using the relation $K_p = K_c(RT)^{\Delta n}$,where $\Delta n = 2 - 1 = 1$ and $R = 0.08314 \ L \ bar \ K^{-1} \ mol^{-1}$.$K_c = \frac{K_p}{RT} = \frac{2.4}{0.08314 \times 1000} = \frac{2.4}{83.14} \approx 0.02887 \approx 2.89 \times 10^{-2}$.Thus,the correct option is $(B)$.

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Kp and Kc Relationship

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At $1000 \ K$,the partial pressures of $CO_{2(g)}$ and $CO_{(g)}$ for the reaction $CO_{2(g)} + C_{(s)} \rightleftharpoons 2 CO_{(g)}$ in a closed vessel at equilibrium are $0.15 \ bar$ and $0.60 \ bar$ respectively. The $K_c$ for this reaction at the same temperature is approximately

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A
$2.0 \times 10^{-4}$
B
$2.89 \times 10^{-2}$
C
$2.89 \times 10^{-3}$
D
$5.78 \times 10^{-3}$

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1 Equilibrium: Chemical Equilibrium

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Kp and Kc Relationship

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At $1000 \ K$,the partial pressures of $CO_{2(g)}$ and $CO_{(g)}$ for the reaction $CO_{2(g)} + C_{(s)} \rightleftharpoons 2 CO_{(g)}$ in a closed vessel at equilibrium are $0.15 \ bar$ and $0.60 \ bar$ respectively. The $K_c$ for this reaction at the same temperature is approximately

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For the reaction $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$,if the reaction is started with $1 \ mol$ of $H_2$ and $CO_2$ each,and $x \ mol/L$ of $H_2$ is consumed at equilibrium,then the correct expression for $K_p$ will be ........

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