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Standard XII

Chemistry

Characteristics of Equilibrium Constant

At 25oC and...

Question

At $25^{o} C$ and one atmospheric pressure, the partial pressures in an equilibrium mixture of $N_{2} O_{4}$ and ${N O}_{2}$ are $0.7$ and $0.3$ atmosphere, respectively. Calculate the partial pressures of these gases when they are in equilibrium at $25^{o} C$ and at a total pressure of $10$ atmospheres.

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Solution

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
At equi. $0.7$ $0.3 a t m$
$K_{p} = \frac{{(p_{N O_{2}})}^{2}}{p_{N_{2} O_{4}}} = 0.1285 a t m$
Let the degree of dissociation of $N_{2} O_{4}$ be $x$ when total pressure $10$ atmosphere
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
At equi. $(1 - x)$ $2 x$
total number of moles $= 1 - x + 2 x = 1 + x$
$p_{N_{2} O_{4}} = \frac{(1 - x)}{(1 + x)} \times 10; p_{N O_{2}} = \frac{2 x}{(1 + x)} \times 10$
$K_{p} = 0.1285 = \frac{{(\frac{2 x}{(1 + x)})}^{2} \times 10^{2}}{(\frac{(1 - x)}{(1 + x)}) \times 10} = \frac{40 x^{2}}{1 - x^{2}}$
Since, $x$ is very small, $(1 - x^{2}) \to 1$
So, $x = 0.0566$
$p_{N_{2} O_{4}} = \frac{(1 - x)}{(1 + x)} \times 10 = 8.93 a t m$
$p_{N O_{2}} = \frac{2 x}{(1 + x)} \times 10 = 1.07 a t m$

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At 25oC and one atmospheric pressure, the partial pressures in an equilibrium mixture of N2O4 and NO2 are 0.7 and 0.3 atmosphere, respectively. Calculate the partial pressures of these gases when they are in equilibrium at 25oC and at a total pressure of 10 atmospheres.