1st create an ICE table form the balanced equation & given data:

2 ICl_{(g)} ↔ I_{2(g)} + Cl_{2(g)}

I: 2.16 atm 0 0

C: - 2x +x +x

E: 2.16 - 2x x x

Kp = 1.40 x 10^{-5} = [p_{I2} p_{Cl2}] / p_{ICl} = [(x)(x)] / (2.16 - 2x)

The magnitude of 2x in the denominator is too small to be significant & can be ignored. Thus, 1.40 x 10^{-5} = x^{2}/2.16

Solving for x we get 5.50 x 10^{-3 }atm = p_{I2} = p_{Cl2} and [2.16 - 2(5.50 x 10^{-3})] = approx 2.15 atm = p_{ICl}