(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 6-1.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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6-1.Equilibrium (Chemical Equilibrium)

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MediumAIIMS 1999

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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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In a $500 \,mL$ flask,the degree of dissociation of $PCl_5$ at equilibrium is $40 \%$ and the initial amount is $5 \,moles$. The value of equilibrium constant in $mol \,L^{-1}$ for the decomposition of $PCl_5$ is

$K_{c}$ for the reaction,$A_{2(g)} \rightleftarrows B_{2(g)}$ is $99.0$. In a $1 \ L$ closed flask,two moles of $B_{2(g)}$ are heated to $T(K)$. What is the concentration of $B_{2(g)}$ (in $mol \ L^{-1}$) at equilibrium?

$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).

The rate constant for forward and backward reactions of hydrolysis of ester are $1.1 \times 10^{-2} \text{ min}^{-1}$ and $1.5 \times 10^{-3} \text{ min}^{-1}$ respectively. The equilibrium constant for the reaction $CH_3COOC_2H_5 + H_2O \rightleftharpoons CH_3COOH + C_2H_5OH$ is: (in $.33$)

For the reactions $(i) \ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ and $(ii) \ \frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$,if their equilibrium constants are $K_c(1)$ and $K_c(2)$ respectively,what is the relation between them?

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$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$
The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).