Question

In a $500mL$ capacity vessel $CO$ and ${Cl}_2$ are mixed to form $CO{Cl}_2$. At equilibrium, it contains $0.2$ mole of $CO{Cl}_2$ and $0.1$ mole of each of $CO$ and ${Cl}_2$. The equilibrium constant $K_c$ for reaction, $CO+{Cl}_2\rightleftharpoons CO{Cl}_2$ is:

• A. $5$
• B. $10$
• C. $15$
• D. $20$

Answer 1

Answer: (B)

$10$

In a $500mL$ capacity vessel $CO$ and ${Cl}_2$ are mixed to form $CO{Cl}_2$. At equilibrium, it contains $0.2$ mole of $CO{Cl}_2$ and $0.1$ mole of each of $CO$ and ${Cl}_2$. The equilibrium constant $K_c$ for reaction, $CO+{Cl}_2\rightleftharpoons CO{Cl}_2$ is
$K_c=\frac{\left[COC{l}_2\right]}{\left[CO\right]\left[C{l}_2\right]}$
$K_c=\frac{\frac{0.2mol}{0.500L}}{\frac{0.1mol}{0.500L}\times \frac{0.1mol}{0.500L}}$
$K_c=10$
Note: Volume is 500 mL or $\text{500 mL}\times \frac{\text{1 L}}{1000mL}=\text{0.500 L}$