(C) The equilibrium constant expression for the reaction $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ is given by: $K_c = \frac{[CO]^2}{[CO_2]}$

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(C) The equilibrium constant expression for the reaction $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ is given by: $K_c = \frac{[CO]^2}{[CO_2]}$ Given $K_c = 0.036$ and $[CO_2] = 0.004 \ M$. Substituting the values: $0.036 = \frac{[CO]^2}{0.004}$ $[CO]^2 = 0.036 \times 0.004 = 0.000144$ $[CO] = \sqrt{0.000144} = 0.012 \ M = 1.2 \times 10^{-2} \ mol \ L^{-1}$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Kp and Kc Relationship

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The value of $K_{C}$ for the equilibrium reaction: $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ at $T \ K$ is $0.036$. If the equilibrium concentration of $CO_{2(g)}$ is $0.004 \ M$,the concentration of $CO_{(g)}$ in $mol \ L^{-1}$ is:

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A
$3.6 \times 10^{-2}$
B
$2.0 \times 10^{-2}$
C
$1.2 \times 10^{-2}$
D
$1.2 \times 10^{-3}$

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The value of $K_{C}$ for the equilibrium reaction: $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ at $T \ K$ is $0.036$. If the equilibrium concentration of $CO_{2(g)}$ is $0.004 \ M$,the concentration of $CO_{(g)}$ in $mol \ L^{-1}$ is:

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