If Kp = 0.450 for the reaction below, what is the equilibrium partial pressure of HF?

Question

An empty steel container is filled with 5.00 atm of H₂ and 5.00 atm of F₂. The system is allowed to reach equilibrium. If Kp = 0.450 for the reaction below, what is the equilibrium partial pressure of HF?

H₂ (g) + F₂ (g) ⇌ 2 HF (g)

Answer 1

H₂ (g) + F₂ (g) ⇌ 2 HF (g)

5...........5.................0........Initial

-x..........-x..............+2x......Change

5-x.....5-x.................2x......Equilibrium

Kp = 0.450 = (P_HF)² / (P_H2)(P_F2)

0.450 = (2x)² / (5-x)(5-x)

Solve for x using the quadratic formula

The equilibrium pressure of HF will be 2x