What are the partial pressures of all substances involved in the reaction when it comes to equilibrium?

Question

At a temperature below room temperature but with all of the substances still in the gas phase, the equilibrium constant K_p for the decomposition of iodine monochloride (ICl) into I₂ and Cl₂ is 1.40 × 10^–5.

If a sealed vessel initially contains 2.16  atm of ICl but no I₂ or Cl₂, what are the partial pressures of all substances involved in the reaction when it comes to equilibrium?

P_ICI = 

P_I2 =  

P_Cl2 =

Answer 1

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 = x²/2.16

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