At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in…

Question 23 of 69advanced🔧 ApplyNumerical3 marks

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO $_2$ in equilibrium with soild carbon has 90.55% CO by mass

C (s) + CO_2(g) \rightleftharpoons 2CO (g)

Calculate $K_c$ for this reaction at the above temperature.

Correct Answer

$K_c = 0.1526 \text{ mol L}^{-1}$

Exercise: EXERCISES | Q: 6.23 | (Page 43)

For More Understanding

Explanation

For the given heterogeneous equilibrium, solid carbon is not included in the equilibrium constant expression. First, the mass percentages are converted to moles to find the mole fractions and partial pressures of the gaseous components. Using these partial pressures, $K_p$ is calculated. Finally, $K_c$ is determined from $K_p$ using the relation $K_p = K_c(RT)^{\Delta n}$ , where $\Delta n = 1$ .

Solution Steps

Step 1: Calculate moles of CO and CO $_2$ Let the total mass of the mixture be 100 g. Mass of CO = 90.55 g; Mass of CO $_2$ = 9.45 g Moles of CO = $\frac{90.55}{28} = 3.234$ mol Moles of CO $_2$ = $\frac{9.45}{44} = 0.215$ mol Total moles = $3.234 + 0.215 = 3.449$ mol
Step 2: Calculate mole fractions and partial pressures Mole fraction of CO ( $X_{CO}$ ) = $\frac{3.234}{3.449} = 0.9377$ Mole fraction of CO $_2$ ( $X_{CO_2}$ ) = $\frac{0.215}{3.449} = 0.0623$ Partial pressure of CO ( $p_{CO}$ ) = $0.9377 \times 1 \text{ atm} = 0.9377 \text{ atm}$ Partial pressure of CO $_2$ ( $p_{CO_2}$ ) = $0.0623 \times 1 \text{ atm} = 0.0623 \text{ atm}$
Step 3: Calculate $K_p$ For the reaction, $K_p = \frac{(p_{CO})^2}{p_{CO_2}}$
$K_p = \frac{(0.9377)^2}{0.0623} = \frac{0.8793}{0.0623} = 14.12$
Step 4: Calculate $K_c$ Using $K_p = K_c(RT)^{\Delta n}$ , where $\Delta n = 2 - 1 = 1$ , $R = 0.0821 \text{ L atm K}^{-1} \text{mol}^{-1}$ , and $T = 1127 \text{ K}$ :
$K_c = \frac{K_p}{RT} = \frac{14.12}{0.0821 \times 1127} = \frac{14.12}{92.5267} = 0.1526 \text{ mol L}^{-1}$

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