A vessel at 1000 K contains $\mathrm{CO}_{2}$ with a pressure of 0.5 atm. Some of $\mathrm{CO}_{2}$ is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then $\mathrm{K}_{\mathrm{p}}$ is :

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Equilibrium

A vessel at 1000 K contains

CO_{2}

with a pressure of 0.5 atm. Some of

CO_{2}

is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then

K_{p}

is :

Answer: (B)

With graphite present,

C O_{2} (g) + C (s) ⇌ 2 C O (g)

.

Starting at

P_{C O_{2}} = 0.5

atm and letting

x

atm of

C O_{2}

convert:

P_{C O_{2}} = 0.5 - x, P_{C O} = 2 x

P_{t o t a l} = 0.5 + x = 0.8 \Rightarrow x = 0.3

At equilibrium

P_{C O_{2}} = 0.2

atm,

P_{C O} = 0.6

atm, so

K_{p} = \frac{P _{C O}^{2}}{P _{C O_{2}}} = \frac{( 0.6 ) ^{2}}{0.2} = 1.8 atm

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option