Question

In a $500$ ml capacity vessel $CO$ and $C{l}_2$ are mixed to form $COC{l}_2$ at equilibrium, it contains $0.2$ moles of $COC{l}_2$ and $0.1$ mole each of $CO$ and $C{l}_2$. The equilibrium constant $K_c$ for reaction $CO+C{l}_2\iff COC{l}_2$, is:

• A. 5
• B. 10
• C. 15
• D. 25

Answer 1

Answer: (B)

10

Let us consider the reaction

$CO+{Cl}_2\rightleftharpoons CO{Cl}_{2}Moles0.10.10.2$

Hence,

$K_c=\frac{\left[\frac{0.2}{\frac{1}{2}}\right]}{\left[\frac{0.1}{\frac{1}{2}}\right]\left[\frac{0.1}{\frac{1}{2}}\right]}$

$=\frac{0.4}{0.2\times 0.2}=\frac{0.1}{0.1\times 0.1}$

Hence, the $K_c=10Lmo{l}^{-1}$