If carbon dioxide is $2$ % dissociated $1800^{o} C$ and $1$ atmospheric pressure,
$2 C O_{2} (g) ⇌ 2 C O (g) + O_{2} (g)$
calculate $K_{p}$ for the reaction

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Solution

Let us assume that initially, 100 moles of $C O_{2}$ are present.
2 moles of $C O_{2}$ decompose to form 2 moles of $C O$ and 1 mole of $O_{2}$ .
$100 - 2 = 98$ moles of $C O_{2}$ remains.
The mole fraction of $C O_{2}$ at equilibrium $= \frac{98}{100} = 0.98$
The partial pressure of $C O_{2}$ at equilibrium $= 0.98 \times 1 a t m = 0.98 a t m$ .
The mole fraction of $C O$ at equilibrium $= \frac{2}{100} = 0.02$
The partial pressure of $C O$ at equilibrium $= 0.02 \times 1 a t m = 0.02 a t m$ .
The mole fraction of $O_{2}$ at equilibrium $= \frac{1}{100} = 0.01$
The partial pressure of $O_{2}$ at equilibrium $= 0.01 \times 1 a t m = 0.01 a t m$ .
$K_{P} = \frac{P_{C O}^{2} P_{O_{2}}}{P_{C O_{2}}^{2}}$
$K_{P} = \frac{(0.02 a t m)^{2} \times 0.01 a t m}{(0.98 a t m)^{2}}$
$K_{P} = 4.12 \times 10^{- 6} a t m$

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