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Reaction Quotient

A mixture of ...

Question

A mixture of $0.0373 a t m$ of $N O (g)$ and $0.310 a t m$ of ${C l}_{2} (g)$ is prepared at $500^{o} C$ . The reaction, $2 N O (g) + {C l}_{2} ⇌ 2 N O C l$ , occurs. The total pressure at equilibrium is $0.544 a t m$ . Determine $K_{p}$ of the reaction.

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Solution

The initial pressures are
$P_{N O} = 0.373$ atm
$P_{C l_{2}} = 0.310$ atm
$P_{N O C l} = 0$ atm
The equilibrium pressures are
$P_{N O} = 0.373 - 2 x$ atm
$P_{C l_{2}} = 0.310 - x$ atm
$P_{N O C l} = 2 x$ atm
Total equilibrium pressure $= (0.373 - 2 x) + (0.310 - x) + (2 x)$ atm. But it is equal to 0.544 atm.
$= (0.373) + (0.310 - x) = 0.544$
$x = 0.139$ atm
The equilibrium constant
$K_{p} = \frac{P_{N O C l}^{2}}{P_{N O}^{2} \times P_{C l_{2}}}$
$K_{p} = \frac{(2 x)^{2}}{(0.373 - 2 x)^{2} \times (0.310 - x)}$
$K_{p} = \frac{(2 (0.139))^{2}}{(0.373 - 2 (0.139))^{2} \times (0.310 - 0.139)}$
$K_{p} = \frac{0.0773}{0.00903 \times 0.171}$
$K_{p} = 50.08 a t m^{- 1}$

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