Question

An empty steel container is filled with 3.4 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

The equilibrium partial pressure of HF is approximately 1.24 atm.

Step-by-step explanation:

To find the equilibrium partial pressure of HF, we need to use the equilibrium constant expression, which is Kp = [HF]² / ([H₂] * [F₂]²). Given that Kp = 0.45, [H₂] = 3.4 atm, and [F₂] = 1.0 atm, we can solve for [HF].

Substituting the given values into the equilibrium constant expression, we have 0.45 = [HF]² / (3.4 * 1.0²). Rearranging the equation, we get [HF]² = 0.45 * 3.4 * 1.0² = 1.53.

Taking the square root of both sides, we find that [HF] = √1.53 ≈ 1.24 atm. Therefore, the equilibrium partial pressure of HF is approximately 1.24 atm.