$N_{2} O_{4}$ is $25 %$ dissociated at $37^{0}$ C and one atmospheric pressure. Calculate the percentage of dissociation at $0.1$ atmosphere and $37^{0}$ C.

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Solution

$N_{2} O_{4} = 2 N O_{2}$

Initial 1 0
Equilibr. (1-x) 2X

Total moles = (1-x) + 2X = (1+x)

$p_{(} N_{2} O_{4}) = (1 - x) / (1 + x) \times P$

Given X = 0.25 P = 1 atm.
$p_{(} N_{2} O_{4}) = 0.6$ atm
$p_{(} N O_{2}) = 0.4$ atm

$K_{p} = p_{(} N O_{2}) 2 / p (N_{2} O_{4}) = 0.267$ atm
If degree of dissociation of $N_{2} O_{4}$ at 0.1 atm is Y
$p_{(} N_{2} O_{4}) = (1 - Y) / (1 + Y) \times 0.1$

$p_{(} N O_{2}) = 2 Y / (1 + Y) \times 0.1$

$K_{p} = ((2 Y + Y)^{2} \times (0.1)^{2}) / ((1 - Y) / (1 + Y) \times 0.1) = 0.667 Y^{2}$

$0.267 a t m = 0.667 Y^{2}$

$Y = 0.632$ or 63.2%

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N2O4 is 25% dissociated at 370C and one atmospheric pressure. Calculate the percentage of dissociation at 0.1 atmosphere and 370C.

$N_{2} O_{4}$ is $25 %$ dissociated at $37^{0}$ C and one atmospheric pressure. Calculate the percentage of dissociation at $0.1$ atmosphere and $37^{0}$ C.