In a 500 mL capacity vessel,CO and Cl_2 are | vedclass

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(B) Given,Volume $V = 500 \ mL = 0.5 \ L$. The balanced chemical equilibrium reaction is: $CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)$. At equilibri

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$10$

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(B) Given,Volume $V = 500 \ mL = 0.5 \ L$. The balanced chemical equilibrium reaction is: $CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)$. At equilibrium,the number of moles are: $n(CO) = 0.1 \ mol$,$n(Cl_2) = 0.1 \ mol$,$n(COCl_2) = 0.2 \ mol$. Concentration is calculated as: $[Species] = \frac{n}{V(L)}$. $[CO] = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ mol/L$. $[Cl_2] = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ mol/L$. $[COCl_2] = \frac{0.2 \ mol}{0.5 \ L} = 0.4 \ mol/L$. The equilibrium constant expression is: $K_c = \frac{[COCl_2]}{[CO][Cl_2]}$. Substituting the values: $K_c = \frac{0.4}{(0.2)(0.2)} = \frac{0.4}{0.04} = 10$. Thus,the equilibrium constant $K_c$ is $10$.

In a $500 \ mL$ capacity vessel,$CO$ and $Cl_2$ are mixed to form $COCl_2$. At equilibrium,it contains $0.2 \ mole$ of $COCl_2$ and $0.1 \ mole$ of each of $CO$ and $Cl_2$. The equilibrium constant $K_c$ for the reaction,$CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)$ is:

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