JEE Main 2011 — Chemical Equilibrium Question with Solution

To determine the equilibrium constant $K$, consider the equilibrium reaction:

${\text{CO}}_2+\text{C (graphite)}\rightleftharpoons 2\text{CO}$

Initially, the pressure of ${\text{CO}}_2$ is 0.5 atm, and there is no $\text{CO}$:

Initial Pressure: ${\text{CO}}_2=0.5\text{ atm},\text{CO}=0$

At equilibrium, assume that $x$ atm of ${\text{CO}}_2$ is converted into $\text{CO}$:

Final Pressure: ${\text{CO}}_2=0.5-x\text{ atm},\text{CO}=2x\text{ atm}$

The total pressure at equilibrium is given as 0.8 atm:

$0.5-x+2x=0.5+x=0.8$

Solving for $x$:

$x=0.8-0.5=0.3\text{ atm}$

The equilibrium constant $K_p$ is calculated as:

$K_p=\frac{(P_{\text{CO}}{)}^2}{P_{{\text{CO}}_2}}$

Substitute the equilibrium pressures:

$K_p=\frac{(2x{)}^2}{0.5-x}=\frac{(2\times 0.3{)}^2}{0.5-0.3}=\frac{0.6^2}{0.2}$

$K_p=\frac{0.36}{0.2}=1.8\text{ atm}$