Question:
Solid carbon, CaO and CaCO$_3$ are mixed and allowed to attain equilibrium at T K.
\[
\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \quad K_{P1} = 0.08 \, \text{atm}
\]
\[
\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2 \text{CO}(g) \quad K_{P2} = 2 \, \text{atm}
\]
The partial pressure of CO is _______ \(\times 10^{-1}\) atm.
Solid carbon, CaO and CaCO$_3$ are mixed and allowed to attain equilibrium at T K.
\[
\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \quad K_{P1} = 0.08 \, \text{atm}
\]
\[
\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2 \text{CO}(g) \quad K_{P2} = 2 \, \text{atm}
\]
The partial pressure of CO is _______ \(\times 10^{-1}\) atm.
Updated On: Apr 12, 2026
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Correct Answer: 0.4
Solution and Explanation
\textcolor{red}{Step 1: Understand the reactions and their equilibrium constants.}
We are given two reactions and their equilibrium constants: 1. \(\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)\), with \(K_{P1} = 0.08 \, \text{atm}\) 2. \(\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2 \text{CO}(g)\), with \(K_{P2} = 2 \, \text{atm}\) We need to find the partial pressure of CO. \textcolor{red}{Step 2: Use the equilibrium constants.}
Since solid substances do not appear in the expression for the equilibrium constant, we only consider the gases. For the first reaction: \[ K_{P1} = \frac{[\text{CO}_2]}{1} = 0.08 \, \text{atm} \] Thus, the partial pressure of \(\text{CO}_2\) is 0.08 atm. For the second reaction: \[ K_{P2} = \frac{[\text{CO}]^2}{[\text{CO}_2]} = 2 \] Let the partial pressure of CO be \(p\). Substituting the values: \[ K_{P2} = \frac{p^2}{0.08} = 2 \] \textcolor{red}{Step 3: Solve for \(p\).}
Rearranging the equation: \[ p^2 = 2 \times 0.08 = 0.16 \] \[ p = \sqrt{0.16} = 0.4 \, \text{atm} \] \textcolor{red}{Step 4: Adjust the answer to the required form.}
The partial pressure of CO is \(0.4 \times 10^0\) atm.
We are given two reactions and their equilibrium constants: 1. \(\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)\), with \(K_{P1} = 0.08 \, \text{atm}\) 2. \(\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2 \text{CO}(g)\), with \(K_{P2} = 2 \, \text{atm}\) We need to find the partial pressure of CO. \textcolor{red}{Step 2: Use the equilibrium constants.}
Since solid substances do not appear in the expression for the equilibrium constant, we only consider the gases. For the first reaction: \[ K_{P1} = \frac{[\text{CO}_2]}{1} = 0.08 \, \text{atm} \] Thus, the partial pressure of \(\text{CO}_2\) is 0.08 atm. For the second reaction: \[ K_{P2} = \frac{[\text{CO}]^2}{[\text{CO}_2]} = 2 \] Let the partial pressure of CO be \(p\). Substituting the values: \[ K_{P2} = \frac{p^2}{0.08} = 2 \] \textcolor{red}{Step 3: Solve for \(p\).}
Rearranging the equation: \[ p^2 = 2 \times 0.08 = 0.16 \] \[ p = \sqrt{0.16} = 0.4 \, \text{atm} \] \textcolor{red}{Step 4: Adjust the answer to the required form.}
The partial pressure of CO is \(0.4 \times 10^0\) atm.
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