Question

A mixture containing N₂O₄, N₂O, and O₂, all at an initial partial pressure of 4.60 atm, is allowed to achieve equilibrium according to the equation below. At equilibrium, the partial pressure of O₂ is observed to have decreased by 0.28 atm. Determine K_p for this chemical equilibrium. Report your answer to 3 significant figures in scientific notation.

2N₂O₄(g) ⇌ 2N₂O(g) + 3O₂(g)

Answer 1

2 N2O4 (g) < = > 2 N2O (g) + 3 O2 (g)

Initial partial pressure = 4.60 atm

At equilibrium, Partial pressure of O2 = 0.28

Kp expression of above reaction = [N2O]² [ O2]³ / [N2O4]²

We need to set up ICE chart.

            2N2O4 (g) < == > 2N2O (g) + 3 O2 (g)

I           4.60                    4.60              4.60

C          -2x                         +2x               +3x

E        ( 4.60-2x)            (4.60+2x)            (4.60+3x)

Kc = [ P²(4.60+2x) P³(4.60+3x)/ P²(4.60-2x)

Since equilibrium partial pressure O2 = 0.28 atm.

(4.60+3x ) = 0.28

We now find the value of x.

4.60 +3x = 0.28

3x = 0.28 - 4.60

x = - 1.44

P [N2O4]= 4.60 – 2 x ( -1.44 ) = 7.48

P(4.60+2x) = 4.60 + 2 x (-1.14) = 7.48

Lets plug these value in following kp formula.

Kp = (7.48)²(0.28)³/(7.48)²

Kp = 0.28^3 = 0.02195