JEE Main 2025 Jan 22 Shift 1, Chemistry Q12: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2025Jan 22, Shift 1Chemistry

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 :

A
0.18 atm
B
1.8 atm
C
0.3 atm
D
3 atm

Solution

Reaction: $CO_{2} (g) + C (s) ⇌ 2 CO (g)$ .

Let $x$ atm of CO $_{2}$ react. Then $P_{CO_{2}} = 0.5 - x$ and $P_{CO} = 2 x$ .

Total pressure: $(0.5 - x) + 2 x = 0.5 + x = 0.8$ , so $x = 0.3$ atm.

At equilibrium $P_{CO_{2}} = 0.2$ atm and $P_{CO} = 0.6$ atm.

$K_{P} = \frac{P _{CO}^{2}}{P _{CO_{2}}} = \frac{( 0.6 ) ^{2}}{0.2} = \frac{0.36}{0.2} = 1.8$ atm.

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JEE Main 2025 Jan 22 Shift 1, Chemistry Q12: Law of Chemical Equilibrium And Equilibrium Constant

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