$K_{p}$ for the equilibrium of $F e O (s) + C O (g) ⇌ F e (s) + C O_{2} (g)$ at $1000^{o} C$ is $0.403$ . If $C O (g)$ at a pressure of $1.0 a t m$ with excess of $F e O (s)$ is placed in a container at $1000^{o} C$ . What are the pressures of $C O (g)$ and ${C O}_{2} (g)$ when the equilibrium is attained?

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Solution

f you let x equal the pressure of $C O_{2}$ at equilibrium, then the pressure of CO will equal 1-x. Plugging those into the equation gives:
$0.403 = \frac{x}{1 - x}$
Cross multiplying gives
$0.403 - 0.403 x = x$
$0.403 = 1.403 x$
$x = 0.287$
So, the partial pressure of $C O_{2}$ at equilibrium = 0.287 atm.
The partial pressure of CO at equilibirum is $1 - x$ = 0.713 atm

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