Question

An equilibrium mixture consists of 0.31 atm NO2(g) and 0.45 atm N2O4(g). 2 NO2(g) <—> N2O4(g) A chemist adds enough NO2 to this mixture to increase the equilibrium partial pressure of NO2(g) to 0.40 atm. What additional pressure of NO2 did she add to the container? Assume that the temperature and volume remain constant.

Answer 1

      2 NO2(g) <—> N2O4(g)

equi   0.31 atm     0.45 atm

add      x            -

change   -2y           +y

final   0.4 atm     0.45+y

Kp = pN2O4/pNO2^2

    = 0.45/(0.31)^2

    = 4.68

final

4.68 = (0.45+y)/(0.4^2)

y = 0.3

so that, 0.31+x-2y = 0.4

         0.31+x-2*0.3 = 0.4

    x = 0.69 atm

additional pressure of NO2 did she add to the container= x = 0.69 atm