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

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.