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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

At 25 and 1...

Question

At $25$ and 1 atmospheric pressure, the partial pressure in equilibrium mixture of $N_{2} O_{4}$ and $N O_{2}$ are $0.7$ are $0.3 a t m$ respectively. The partial pressure of these gases when they are in equilibrium at $25$ and a total pressure of $10 a t m$ is X atom. 100X is

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Solution

Given pressure at equilibrium is $0.7$ and $0.3$ atm
$N_{2} O_{4} \to 2 N O_{2}$

$K_{P} = \frac{P_{N O_{2}}^{2}}{P_{N_{2} O_{4}}}$

$K_{P} = \frac{{0.3}^{2}}{0.7}$

$K_{P} = 0.1286 a t m$

Now due to decomposition at 10 atm
$N_{2} O_{4} 1 (1 - x) \to 2 N O_{2} 0 (2 x)$

$K_{P} = \frac{(P_{N O_{2}})^{2}}{P_{N_{2} O_{4}}}$
$P_{N O_{2}} = X_{N O_{2}} \times P$ , $P_{N_{2} O_{4}} = X_{N_{2} O_{4}} \times P$

Given P= 10 atm
$K_{P} = \frac{(2 x)^{2}}{1 - x} \times \frac{10}{1 + x}$

$0.1286 = \frac{4 x^{2} \times 10}{1 - x^{2}}$

$x = 0.0565$

$P^{1} N O_{2} = \frac{2 x}{1 + x} \times P = 1.07 a t m$

$P^{1} N_{2} O_{4} = [(1 - 0.0565) / (1 + 0.0565)] \times 10 = 8.93 a t m$

$X = 8.93$

$100 X = 893$

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At $25$ and 1 atmospheric pressure, the partial pressure in equilibrium mixture of $N_{2} O_{4}$ and $N O_{2}$ are $0.7$ are $0.3 a t m$ respectively. The partial pressure of these gases when they are in equilibrium at $25$ and a total pressure of $10 a t m$ is X atom. 100X is