Answer to Question #128351 in General Chemistry for Stefanie Hernandez-Mendez

Question

An empty steel container is filled with 4.6 atm of H₂ and 1.0 atm of F₂. The system is allowed to reach equilibrium according to the reaction below. If Kp = 0.45 for this reaction, what is the equilibrium partial pressure of HF?

H₂ (g) + F₂ (g) ⇌ 2 HF (g)

Answer 1

K_p=$\frac{p^{2}(HF)}{p(H_{2})p(F_{2})}=\frac{(2x)^{2}}{(4.6-x)(1-x)}=0.45$

x=0.487

Partial pressure of HF = 2x = 0.974 atm