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Nitrogen

At a certain ...

Question

At a certain temperature, equilibrium constant ( $K_{c}$ ) is $16$ for the reaction:
$S O_{2} (g) + N O_{2} (g) ⇌ S O_{3} (g) + N O (g)$
If we take $1 m o l e$ of each of the four gases in $1 L$ container, what would be the equilibrium concentration of $N O$ and ${N O}_{2}$ ?

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Solution

$S O_{2} (g) + N O_{2} (g) ⇌ S O_{3} (g) + N O (g)$

Initial concentration: $1$ $1$ $1$ $1$
Equi. Concentration: $1 - x$ $1 - x$ $1 + x$ $1 + x$

Applying law of mass action,

$K_{c} = \frac{[S O_{3}] [N O]}{[S O_{2}] [N O_{2}]} = \frac{(1 + x) (1 + x)}{(1 - x) (1 - x)} = 16$

$\frac{1 + x}{1 - x} = 4$

or $5 x = 3$

$x = \frac{3}{5} = 0.6$

Conc. of ${N O}_{2}$ at equilibrium $= (1 - 0.6) = 0.4 m o l e$
Conc. of $N O$ at equilibrium $= (1 + 0.6) = 1.6 m o l e$

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