An empty steel container is filled with 4.40 atm of H₂ and 4.40 atm of F₂. Question continues below:

Question

An empty steel container is filled with 4.40 atm of H₂ and 4.40 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) <==> 2HF(g)

4.40.......4.40.............0...........Initial

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

4.4-x......4.4-x...........2x..........Equilibrium

Kp = (HF)² / (H₂)(F₂)

0.450 = (2x)² / (4.40-x)(4.40-x)

0.450 = 4x² / x² -8.6x + 19.36

4x² = 0.45x² - 3.87x + 8.71

x² = 0.1125x² - 0.9675x + 2.178

0.8875x² + 0.968x - 2.18 = 0

x = 1.11 atm

At equilibrium, partial pressure of HF should be 2x = 2 x 1.11 atm = 2.22 atm

(be sure to check the math)