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

2.7 * 10⁻⁴ bar

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