Question

At a particular temperature, Kp=0.22 for the reaction

N2O4(g)=2NO2(g)

A flask containing only NO2 at an initial pressure of 5.8 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases.

Answer 1

first write the equation and then construct the ICE table and substitute the values in the Kp equation and then calculate the equilibrium partial pressures as follows

   N2O4 ---------------------> 2 NO2

I     0                                    5.8

C    +x -2x

E    (x)    (5.8-2x)

Kp = [pNO2]² / [pN2O4]

0.22 = [5.8-2x]² / [x]

0.22 * x = [5.8-2x]²

x = 2.527

where x = [pN2O4] => 2.527 atm

[pNO2] = 5.8 - 2*2.527 => 0.746 atm

answers =>

at equilibrium the partial pressures are as follows

[pN2O4] => 2.527 atm

[pNO2] = 0.746 atm