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JEE Main 20268 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
8 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2026 · 8 April · Evening Shift · Chemistry · Chemical Equilibrium

Solid carbon, CaO and CaCO _3 are mixed and allowed to attain...

Solid carbon, CaO and CaCO3_3 are mixed and allowed to attain equilibrium at T K.
CaCO3(s)⇌CaO(s)+CO2(g)Kp1=0.08\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \quad K_{p_1} = 0.08 atm
C(s)+CO2(g)⇌2CO(g)Kp2=2\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g) \quad K_{p_2} = 2 atm
The partial pressure of CO is ________ ×10−1\times 10^{-1} atm

Correct answer

0

Step-by-step solution

For the first equilibrium:

CaCO3(s)⇌CaO(s)+CO2(g)\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)

Kp1=PCO2=0.08 atmK_{p_1} = P_{\text{CO}_2} = 0.08 \text{ atm}

For the second equilibrium:

C(s)+CO2(g)⇌2CO(g)\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)

Kp2=(PCO)2PCO2=2 atmK_{p_2} = \dfrac{(P_{\text{CO}})^2}{P_{\text{CO}_2}} = 2 \text{ atm}

Substituting the value of PCO2P_{\text{CO}_2}:

(PCO)20.08=2\dfrac{(P_{\text{CO}})^2}{0.08} = 2

(PCO)2=0.16(P_{\text{CO}})^2 = 0.16

PCO=0.4 atm=4×10−1 atmP_{\text{CO}} = 0.4 \text{ atm} = 4 \times 10^{-1} \text{ atm}

Answer: 44
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