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 K_c 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)