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Derivation of Kp and Kc

At 1127 K and...

Question

At 1127 K and 1 atm pressure, a gaseous mixture of CO and $C O_{2}$ in equilibrium with solid carbon has 90.55% CO by mass
$C (s) + C O_{2} (g) \leftrightarrow 2 C O (g)$
Calculate $K_{c}$ for this reaction at the above temperature.

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Solution

Let the total mass of the gaseous mixture be 100 g.
Mass of CO=90.55 g
and, mass of $C O_{2} = (100 - 90.55) = 9.45 g$
Now, number of moles of CO, $n_{c o} = \frac{90.55}{28} = 3.234 m o l$
Number of moles of $C O_{2} .$ , $n_{c o_{2}} = \frac{9.45}{44} = 0.215 m o l$
Particular pressure of CO, $P_{c o} = \frac{n_{c o}}{n_{c o} + n_{c o_{2}}} \times P_{t o t a l} = \frac{3.234}{3.234 + 0.215} \times 1 = 0.938 a t m$
Partal pressure of $C O_{2}$ , $P_{c o_{2}} = \frac{n_{c o}}{n_{c o} + n_{c o_{2}}} \times P_{t o t a l} = \frac{0.215}{3.234 + 0.215} \times 1 = 0.062 a t m$
Therefore, $K_{p}$ $= \frac{[C O]^{2}}{[C O_{2}]} = \frac{(0.938)^{2}}{0.062} = 14.19$
For the given reaction, $Δ n = 2 - 1 = 1$
We know that,
$K_{p} = K_{c} (R T)^{Δ n} \Rightarrow 14.19 = K_{c} (0.082 \times 1127)^{1} \Rightarrow K_{c} = \frac{14.19}{0.082 \times 1127} = 0.154 (a p p r o x i m a t e l y)$

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At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass

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