In the reaction, $C (s) + C O_{2} (g) ⇌ 2 C O (g)$ , the equilibrium pressure is $12 a t m$ . If $50$ % of ${C O}_{2}$ reacts, calculate $K_{p}$

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Solution

Let P atm be the initial pressure of $C O_{2}$ .
50% (or $P \times \frac{50}{100} = 0.5 P$ atm) of $C O_{2}$ reacts to reach equilibrium.
$P \times \frac{50}{100} = 0.5 P$ atm of $C O_{2}$ remains at equilibrium.
$2 \times 0.5 P = P$ atm of $C O$ is formed.
Total pressure at equilibrium $= 0.5 P + P = 1.5 P = 12$ atm
$P = \frac{12}{1.5} = 8$ atm
$P_{C O_{2}} = 0.5 P = 0.5 \times 8 = 4$ atm
$P_{C O} = P = 8$ atm
$K_{p} = \frac{P_{C O}^{2}}{P_{C O_{2}}}$
$K_{P} = \frac{8^{2}}{4}$
$K_{P} = 16$ atm.

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