QJEEM25Ch3414: JEE Main 2025 Chemistry - Class 11 JEEM25Ch3414 Solution: A vessel at 1000 K contains with a pressure of 0.5 atm . Some…

The reaction is:

$${\mathrm{CO}}_2(g)+\mathrm{C}(s)\rightleftharpoons 2\mathrm{CO}(g)$$

Initially, the pressure of $\mathrm{CO}_2$ is 0.5 atm, and the pressure of $\mathrm{CO}$ is 0 atm. Let 'x' be the change in pressure of $\mathrm{CO}_2$ at equilibrium. Since the stoichiometric coefficient of CO is twice that of $\mathrm{CO}_2$ , the pressure of CO formed at equilibrium is 2x.

At equilibrium:

Pressure of $\mathrm{CO}_2 = 0.5 - x$

Pressure of $\mathrm{CO} = 2x$

The total pressure at equilibrium is given as 0.8 atm. Therefore:

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

$$0.5+x=0.8$$

$$x=0.8-0.5=0.3$$

So, at equilibrium:

Pressure of $\mathrm{CO}_2 = 0.5 - 0.3 = 0.2 \text{ atm}$

Pressure of $\mathrm{CO} = 2(0.3) = 0.6 \text{ atm}$

The equilibrium constant $K_p$ is given by:

$$K_p=\frac{P_{\mathrm{CO}}^2}{P_{{\mathrm{CO}}_2}}$$

$$K_p=\frac{(0.6{)}^2}{0.2}=\frac{0.36}{0.2}=1.8\text{ atm}$$

Therefore, the value of $K_p$ is 1.8 atm.