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Reaction Quotient

At 700 K, ...

Question

At $700 K$ , ${C O}_{2}$ and $H_{2}$ react to form $C O$ and $H_{2} O$ . For this process $K_{c}$ is $0.11$ . If a mixture of $0.45$ mole of ${C O}_{2}$ and $0.45$ mole of $H_{2}$ is heated at $700 K$ .
(i) Find out the amount of each gas at equilibrium state.
(ii) After equilibrium is reached another $0.34 m o l e$ of ${C O}_{2}$ and $0.34 m o l e$ of $H_{2}$ are added to the reaction mixture. Find the composition of the mixture at the new equilibrium state.

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Solution

$C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)$
Initial no. of moles: $0.45$ $0.45$ $0$ $0$
No. of moles at $e q^{m}$ : ( $0.45 - x$ ) ( $0.45 - x$ ) $x$ $x$
Applying law of mass action,
$K_{c} = \frac{[C O] [H_{2} O]}{[C O_{2}] [H_{2}]} = \frac{x \times x}{(0.45 - x) (0.45 - x)} = 0.11$
So, $\frac{x}{(0.45 - x)} = 0.33$
$x = 0.11$
At equilibrium, $[C O_{2}] = [H_{2}] = 0.34 m o l e$
$[C O] = [H_{2} O] = 0.11 m o l e$
(ii) $C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)$
Initial moles $0.34$ $0.34$ $0.11$ $0.11$
$+ 0.34$ $+ 0.34$ $-$ $-$
$= 0.68$ $= 0.68$ $0.11$ $0.11$
At equilibrium:( $0.68 - y$ ), ( $0.68 - y$ ), ( $0.11 + y$ ), ( $0.11 + y$ )
$K_{c} = \frac{[C O] [H_{2} O]}{[C O_{2}] [H_{2}]} = \frac{(0.11 + y) (0.11 + y)}{(0.68 - y) (0.68 - y)} = 0.11$
or $y = 0.086$
At equilibrium, $[C O_{2}] = [H_{2}] = 0.11 + 0.086 = 0.196 m o l e$
$[C O] = [H_{2} O] = 0.68 - 0.086 = 0.594 m o l e$

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