F6 Chem Eqm. constant Kp

At 50℃, the degree of dissociation of N2O4 is 0.5, when the total pressure of the equilibrium mixture is 1 atm.

(a) Calculate the value of Kp.

Let n mol be the initial number of moles of N2O4.

N2O4(g) ⇌ 2NO2(g)
At equilibrium:
Degree of dissociation of N2O4 = 0.5
No. of moles of N2O4 = (n - 0.5n) = 0.5n mol
No. of moles of NO2 = 2 x 0.5n = n mol
Partial pressure of N2O4, PN2O4 = 1 x [0.5n/(0.5n + n)] = 1/3 atm
Partial pressure of NO2, PNO2 = 1 x [n/(0.5n + n)] = 2/3 atm

Kp = PNO2/PN2O4 = (2/3)^2/(1/3) = 1.33 atm


(b) If the total pressure of the system is now increased to 2 atm, what is the degree of dissociation of N2O4?

Let α be the degree of dissociation of N2O4 when total pressure is 2 atm.
Let n mol be the initial number of moles of N2O4.

At equilibrium:
Degree of dissociation of N2O4 = α
No. of moles of N2O4 = n(1 - α) mol = (n - αn) mol
No. of moles of NO2 =2αn mol
Total no. of moles = (n - αn) + 2αn = (n + αn) mol
Partial pressure of N2O4, PN2O4 = 2 x [(n - αn)/(n + αn )] = (2 - 2α)/(1 + α) atm
Partial pressure of NO2, PNO2 = 2 x [2αn/(n + αn)] = 4α/(1 + α) atm

Kp = [4α/(1 + α)]^2 / [(2 - 2α)/(1 + α)] = 1.33
8α² / [(1 + α)(1 - α)] = 1.33
(1 + α)(1 - α) = 8α²/1.33
1 - α^2 = 6α
7α^2 = 1
α = √(1/7)
α = 0.378

Hence, the degree of dissociation = 0.378

2011-02-03 02:01:52 補充：
The last fifth line should be: 1 - α^2 = 6α^2