At 25⁰C and one atmospheric pressure, the partial pressures in equilibrium mixture of N₂O₄ and NO₂ are 0.7 and 0.3 atmosphere respectively. Calculate the partial pressure of these gases when they are in equilibrium at 25⁰C and at a total pressure of 10 atmospheres.

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Solution

Dear Student,

The reaction of equiliirium is:
$N_{2} O_{4} ⇌ 2 N O_{2}$
$K_{p} = \frac{{P_{N O 2}}^{2}}{P_{N 2 O 4}} = \frac{0 . 3^{2}}{0.7} = 0.13$
If total pressure is 10 atm, let pressure of NO2 = x and pressure of N2O4 = 10-x
$K_{p} = \frac{{P_{N O 2}}^{2}}{P_{N 2 O 4}} = \frac{x^{2}}{10 - x}$
$0.13 = \frac{x^{2}}{10 - x} 0.13 (10 - x) = x^{2} 1.3 - 0.13 x = x^{2} x^{2} + 0.13 x - 1.3 = 0$
x = 1.08, x = -1.21
Pressure can not be negative. So, x = 1.08 atm
Equilibirum Pressure of NO2 = 1.08 atm
Equilibirum Pressure of N2O4 = 10-1.08 = 8.92 atm

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