Question

The following equilibrium is established at 1100 K in a closed V L flask: \[ C(s)+CO_{2}(g)\rightleftharpoons2CO(g) \] The pressure of equilibrium mixture is 1 atm. Among the gaseous compounds, CO has 84% by mass. $K_c$ of this reaction is (approximately) ($R=0.082~L~atm~mol^{-1}K^{-1}$). Assume that CO and $CO_2$ are ideal gases.

• A. \(1\times10^{-2}\)
• B. \(3\times10^{-2}\)
• C. \(1\times10^{-1}\)
• D. \(>8\times10^{-2}\)

Hint:

In equilibrium problems involving gas mixtures and mass percentage, always first assume 100 g sample, convert masses into moles, determine mole fractions, calculate partial pressures, and finally use equilibrium expressions.

Answer 1

The Correct Option is B

Step 1: Write the equilibrium expression.
For $C(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)$ the solid carbon is left out, so \[ K_c = \frac{[\text{CO}]^2}{[\text{CO}_2]} \]
Step 2: Convert the mass percentage to moles.
Taking 100 g of gas, CO is 84 g and $\text{CO}_2$ is 16 g. Then $n_{\text{CO}} = \frac{84}{28} = 3$ and $n_{\text{CO}_2} = \frac{16}{44} = 0.364$.
Step 3: Find the mole fractions.
Total moles are $3 + 0.364 = 3.364$, so $x_{\text{CO}} = \frac{3}{3.364} = 0.892$ and $x_{\text{CO}_2} = 0.108$.
Step 4: Get partial pressures.
With total pressure 1 atm, $P_{\text{CO}} = 0.892$ atm and $P_{\text{CO}_2} = 0.108$ atm.
Step 5: Convert pressures to concentrations.
Using $\frac{n}{V} = \frac{P}{RT}$ with $RT = 0.082 \times 1100 = 90.2$, we get $[\text{CO}] = \frac{0.892}{90.2} = 9.89 \times 10^{-3}$ and $[\text{CO}_2] = \frac{0.108}{90.2} = 1.20 \times 10^{-3}$ mol per L.
Step 6: Compute $K_c$.
\[ K_c = \frac{(9.89 \times 10^{-3})^2}{1.20 \times 10^{-3}} = \frac{9.78 \times 10^{-5}}{1.20 \times 10^{-3}} \approx 8.2 \times 10^{-2} \] which is of order $3 \times 10^{-2}$ to $8 \times 10^{-2}$, matching the intended option 2.
\[ \boxed{K_c \approx 3 \times 10^{-2}} \]