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Characteristics of Equilibrium Constant

The equilibri...

Question

The equilibrium constant $K_{p}$ for the reaction, $N_{2} (g) + O_{2} (g) ⇋ 2 N O (g)$ at $1500^{o} C$ is $120$ . If $N_{2}$ and $O_{2}$ at an initial pressure of $0.25 a t m$ each are maintained at $1500^{o} C$ till equilibrium is established, calculate the partial pressure of $N_{2}, O_{2}$ and $N O$ in the equilibrium mixture.

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Solution

The inital partial pressures are
$P_{N_{2}} = P_{O_{2}} = 0.25$ atm
$P_{N O} = 0$ atm
Let us say that x atm of nitrogen reacts with x atm of oxygen to form 2x atm of NO at equilibrium.
The equilibrium partial pressures are
$P_{N_{2}} = P_{O_{2}} = 0.25 - x$ atm
$P_{N O} = 2 x$ atm
$K_{P} = \frac{P_{N O}^{2}}{P_{N_{2}} \times P_{O_{2}}}$
$120 = \frac{(2 x)^{2}}{(0.25 - x) \times (0.25 - x)}$
$10.95 = \frac{2 x}{0.25 - x}$
$0.25 - x = \frac{2 x}{10.95}$
$0.25 - x = 0.18 x$
$0.25 = 1.18 x$
$x = 0.21$
The equilibrium partial pressures are
$P_{N_{2}} = P_{O_{2}} = 0.25 - x = 0.25 - 0.21 = 0.04$ atm

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