JEE Main 2026 · 8 Apr · Shift 2 · Chemistry Q75

← JEE Main 2026 · 8 Apr · Shift 2

ChemistryInteger+4 / -1Question 75 of 75

Solid carbon, CaO and $\ce{CaCO3}$ are mixed and allowed to attain equilibrium at T K.

$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$ , $K_{p_1}=0.08$ atm

$\ce{C(s) + CO2(g) <=> 2CO(g)}$ , $K_{p_2}=2$ atm

The partial pressure of CO is ________ $\times10^{-1}$ atm

Your answer

Answer and worked solution

Official answer: 4

Both equilibria together. The first fixes $p_{\ce{CO2}}=0.08$ atm. Then
$p_{\ce{CO}}^2=K_{p_2}\,p_{\ce{CO2}}=2\times0.08=0.16,$
so $p_{\ce{CO}}=0.4$ atm $=4\times10^{-1}$ atm.

Answer: 4

Topics:Law of mass action and Kc, Kp

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