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JEE MAIN 2026 · Equilibrium

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 75

Updated
Numerical answer+4 / -1Answer matches NTA’s final key

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 atm\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)} \qquad K_{p_1} = 0.08 \text{ atm}

C(s)+CO2(g)⇌2CO(g)Kp2=2 atm\text{C(s)} + \text{CO}_2\text{(g)} \rightleftharpoons 2\text{CO(g)} \qquad K_{p_2} = 2 \text{ atm}

The partial pressure of CO is ______ ×10−1\times 10^{-1} atm

Your answer

Solution

Solution

Both equilibria hold at once. The first fixes the CO2_2 pressure, since the solids do not appear in KpK_p:

pCO2=Kp1=0.08 atm.p_{\text{CO}_2} = K_{p_1} = 0.08 \text{ atm}.

The second then gives

Kp2=pCO2pCO2  ⇒  pCO2=2×0.08=0.16  ⇒  pCO=0.4 atm=4×10−1 atm.K_{p_2} = \frac{p_{\text{CO}}^2}{p_{\text{CO}_2}} \;\Rightarrow\; p_{\text{CO}}^2 = 2 \times 0.08 = 0.16 \;\Rightarrow\; p_{\text{CO}} = 0.4 \text{ atm} = 4 \times 10^{-1} \text{ atm}.

Answer: 4.

Answer checked against the official answer key. The solution is our own working.

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