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 :

Chemistry · Chemical Equilibrium

JEE Main 2025 — 22 January, Morning Shift — Question 36

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 :

Option A:
0.18 atm
Option B:
1.8 atm
Correct
Option C:
0.3 atm
Option D:
3 atm .

Answer: B

Step-by-step solution

$\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})$ 0.5

$0.5-\mathrm{x} \quad 2 \mathrm{x}$

$\mathrm{P}_{\text {total }}=0.5+\mathrm{x}=0.8$

$\mathrm{x}=0.3$

$K_{P}=\frac{(0.6)^{2}}{0.2}=1.8$

Answer key and solution verified before publishing.

Practise Chemical Equilibrium

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 22 January, Morning Shift
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient