Question

CaCO$₃$(s) $\rightleftharpoons$ CaO(s) + CO$₂$(g), \( K_{p1} = 8 \times 10^{-2} \)
C(s) + CO$₂$(g) $\rightleftharpoons$ 2CO(g), \( K_{p2} = 2 \)
If the partial pressure of CO at equilibrium is \( x \times 10^{-1} \) atm, then find the value of \( x \).

Answer 1

Correct Answer: 4

Step 1: Understanding the Question:
This problem involves two simultaneous equilibria. We need to find the equilibrium partial pressure of $CO$ given the equilibrium constants.
Step 2: Key Formula or Approach:
For heterogeneous equilibria involving solids and gases, the equilibrium constant $K_P$ is expressed only in terms of the partial pressures of the gases.
Step 3: Detailed Explanation:
From the first equilibrium:
$CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$
$K_{P_1} = P_{CO_2} = 8 \times 10^{-2}$ atm.

From the second equilibrium:
$C(s) + CO_2(g) \rightleftharpoons 2CO(g)$
$K_{P_2} = \frac{(P_{CO})^2}{P_{CO_2}} = 2$
Substituting the value of $P_{CO_2}$ from the first equation:
$\frac{(P_{CO})^2}{8 \times 10^{-2}} = 2$
$(P_{CO})^2 = 16 \times 10^{-2}$
$P_{CO} = \sqrt{16 \times 10^{-2}} = 4 \times 10^{-1}$ atm.

Comparing with $x \times 10^{-1}$ atm, we get $x = 4$.
Step 4: Final Answer:
The value of $x$ is 4.