Question

The total pressure for a mixture of N₂O₄ and NO₂ is 0.15 atm. If K_p = 7.1 (at 25 degree celsius), calculate the partial pressure of each gas in the mixture

2NO₂(g) <---> N₂O₄(g)

Answer 1

Kp = P-N2O4 / (P-NO2)^2

Kp = 7.1

PT = N2O4 + NO2 = 0.15

assume this mixture is in equilibrium

let "x" be the partial pressure of NO2 and therefore, PT = PN2O2 + PNO2 becomes

PNO2 = x

0.15 = PN2O2 - x

that is:

PN2O4 = 0.15-x

PNO2 = x

then

Kp = P-N2O4 / (P-NO2)^2

7.1 = (0.15-x)/(x^2)

solve for x

7.1*x^2 + x -0.15 = 0

x = 0.09109

then

PN2O4 = 0.15-0.09109 = 0.05891 atm

PNO2 = 0.09109 atm

Proof:

Kp = 7.1

Kp = 0.05891 /(0.09109 ^2) = 7.09982 ; which is pretty close to 7.1

therefore, it is correct