Q25 / 75 · JEE Main 2026 (08 April Shift 2) · Chemistry
Solid carbon, CaO and CaCO$_3$ are mixed and allowed to attain equilibrium at T K.
$\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \quad K_{p_1} = 0.08$ atm
$\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g) \quad K_{p_2} = 2$ atm
The partial pressure of CO is ________ $\times 10^{-1}$ atm
$\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \quad K_{p_1} = 0.08$ atm
$\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g) \quad K_{p_2} = 2$ atm
The partial pressure of CO is ________ $\times 10^{-1}$ atm
Correct answer: 4
Solution
For the first equilibrium:
$\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$
$K_{p_1} = P_{\text{CO}_2} = 0.08 \text{ atm}$
For the second equilibrium:
$\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)$
$K_{p_2} = \dfrac{(P_{\text{CO}})^2}{P_{\text{CO}_2}} = 2 \text{ atm}$
Substituting the value of $P_{\text{CO}_2}$:
$\dfrac{(P_{\text{CO}})^2}{0.08} = 2$
$(P_{\text{CO}})^2 = 0.16$
$P_{\text{CO}} = 0.4 \text{ atm} = 4 \times 10^{-1} \text{ atm}$
Answer: $4$
$\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$
$K_{p_1} = P_{\text{CO}_2} = 0.08 \text{ atm}$
For the second equilibrium:
$\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)$
$K_{p_2} = \dfrac{(P_{\text{CO}})^2}{P_{\text{CO}_2}} = 2 \text{ atm}$
Substituting the value of $P_{\text{CO}_2}$:
$\dfrac{(P_{\text{CO}})^2}{0.08} = 2$
$(P_{\text{CO}})^2 = 0.16$
$P_{\text{CO}} = 0.4 \text{ atm} = 4 \times 10^{-1} \text{ atm}$
Answer: $4$
Hence, the correct answer is 4.