Question

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass
$C(s) + CO_2(g) ⇋ 2CO(g)$
Calculate Kc for this reaction at the above temperature.

Answer 1

Let the total mass of the gaseous mixture be 100 g.
Mass of CO = 90.55 g
And, mass of CO2 = (100 - 90.55) = 9.45 g
Now, number of moles of CO, $n_{CO} = \frac {90.55}{28} = 3.234 \ mol$

Number of moles of CO2, $n_{CO_2} = \frac {9.45}{44} = 0.215\ mol$

Partial pressure of CO, $P_{CO} = \frac {n_{CO}}{n_{CO} + n_{CO_2}}×P_{total}$

= $\frac {3.234}{3.234 + 0.215}×1$
= $0.938 \ \text{atm}$
Partial pressure of CO2,
$P_{CO_2} = \frac {n_{CO_2}}{n_{CO} + n_{CO_2}}×P_{total}$

= $\frac {0.215}{3.234 + 0.215}×1$
= $0.062 \ \text {atm}$
Therefore,
$K_p = \frac {[CO]^2}{[CO_2]}$

$K_p$= $\frac {(0.938)^2}{0.062}$
$K_p$= $14.19$
For the given reaction, Δn = 2 - 1 = 1
We know that,
$K_P = K_C(RT)^{Δn}$

⇒ $14.19 = K_C (0.082×1127)^1$

⇒ $K_C = \frac {14.19}{0.082}×1127$
⇒ $K_C$= $0.154$ (approximately)