JEE Main 2026 Apr 8 Shift 2, Chemistry Q25: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2026Apr 8, Shift 2Chemistry

Solid carbon, CaO and CaCO $_{3}$ are mixed and allowed to attain equilibrium at $T$ K.

$CaCO_{3} (s) ⇌ CaO (s) + CO_{2} (g)$ , $K_{p_{1}} = 0.08$ atm

$C (s) + CO_{2} (g) ⇌ 2 CO (g)$ , $K_{p_{2}} = 2$ atm

The partial pressure of CO is _____ $\times 1 0^{- 1}$ atm.

Your answer

Solution

Find $P_{CO_{2}}$ from the first equilibrium. Pure solids drop out, so $K_{p_{1}} = P_{CO_{2}} = 0.08$ atm.

Use the second equilibrium for CO.

$K_{p_{2}} = \frac{P _{CO}^{2}}{P _{CO_{2}}} = 2$

$P_{CO}^{2} = 2 \times 0.08 = 0.16 = 16 \times 1 0^{- 2}$

$P_{CO} = 0.4$ atm $= 4 \times 1 0^{- 1}$ atm.

$∴$ The value to fill in the blank is $4$ .

Concept behind this question

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