Question

n a gas-phase equilibrium mixture of SbCl5, SbCl3, and Cl2 at 500 K, pSbCl5 = 0.17 bar and pSbCl3 = 0.22 bar. Calculate the equilibrium partial pressure of Cl2 given that K = 3.5 × 10−4 for the reaction SbCl5(g) f SbCl3(g) + C

Answer 1

2.7 × 10⁻⁴ bar

Step-by-step explanation:

Let's consider the following reaction at equilibrium.

SbCl₅(g) ⇄ SbCl₃(g) + Cl₂(g)

The pressure equilibrium constant (Kp) is 3.5 × 10⁻⁴. We can use these data and the partial pressures at equilibrium of SbCl₅ and SbCl₃, to find the partial pressure at equilibrium of Cl₂.

Kp = pSbCl₃ × pCl₂ / pSbCl₅

pCl₂ = Kp × pSbCl₅ / pSbCl₃

pCl₂ = 3.5 × 10⁻⁴ × 0.17 / 0.22

pCl₂ = 2.7 × 10⁻⁴ bar