If $450 %$ of $C O_{2}$ converts to CO according to following reaction $C O_{2} (g) + C (s) ⇌ 2 C O (g)$ and the equilibrium presure is 12 atm then $K_{P}$ will be-

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Solution

The correct option is B 16
$C (s) + C O_{2} (g) \to 2 C O (g)$

P(total) = 12 atm
If half the CO2 reacted then CO2 is x/2 and CO is x, and Pt = 12. Then
x/2 + x = 12
1.5x = 12
x = 8
$P_{(} C O_{2})$ = 4 atm
P(CO) = 8 atm

The initial pressure of $C O_{2}$ is 8 atm. When half of it reacts then 4 atm is left and the pressure of CO is twice as great, which is 8 atm.

$K p = P (C O)^{2} / P (C O_{2})$
Kp = 16 / 4
Kp = 16

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