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Characteristics of Equilibrium Constant

The value of ...

Question

The value of $K_{c} = 4.24$ at $800$ K for the reaction
$C O (g) + H_{2} O ⇌ C O_{2} (g) + H_{2} (g)$
Calculate equilibrium concentrations of $C O_{2}, H_{2}, C O$ and $H_{2} O$ at $800$ K, if any $C O$ and $H_{2} O$ are present initially at concentration of $0.10$ M each?

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Solution

$C O (g) + H_{2} O (g) ⟶ C O_{2} (g) + H_{2} (g)$
moles
at $t = 0$ $0.1$ $0.1$ - -
at $E q b^{m}$ $0.1 - x$ $0.1 - x$ $x$ $x$

$K_{C} = \frac{[C O_{2}] [H_{2}]}{[C O] [H_{2} O]} = \frac{x^{2}}{(0.1 - x)^{2}}$
$\Rightarrow 4.24 = \frac{x^{2}}{(0.1 - x)^{2}}$
$\Rightarrow \frac{x}{0.1 - x} = 2.05$
$\Rightarrow x = 0.205 - 2.05 x$
$\Rightarrow x = 0.0675$
Thus concentration at $E q b^{m}$ of
$[C O_{2}] = [H_{2}] = x = 0.0675 M$
$[C O] = [H_{2} O] = 0.1 - x = 0.0324 M$

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