Calculate the equilibrium constant ( $K_{p}$ ) for the reaction, $C (s) + C O_{2} (g) ⇌ 2 C O (g)$ , at $1300 K$ from the following data:
$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g); K_{p} (1300 K) = 3.9 a t m$
$H_{2} (g) + C O_{2} (g) ⇌ C O (g) + H_{2} O (g); K_{p} (1300 K) = 0.623 a t m$

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Solution

$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g); K_{p} (1300 K) = 3.9 a t m$ (1)
$H_{2} (g) + C O_{2} (g) ⇌ C O (g) + H_{2} O (g); K_{p} (1300 K) = 0.623 a t m$ (2)
So, the resultant reaction is forming adding of reactions 1 and 2
The given temperature and condition are the same so the equilibrium constant are multiplying 1 and 2

so $\times 0.623$ atm = 2.42 atm

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