See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

5 3 2 PCl Cl Cl 2 P P P 3 6     = 1 atm p 1 1 K 1 1    Let “y” mol of Cl2 were added            5 3 2 PCl g PCl g Cl g Initialmol 2 2 2 y molof equilibrium 2 x 2 x 2 y x        Total mol = 12, because total volume is twice of initial volume Mol of PCl5 + PCl3 = 2 + x +2 –x = 4 Mol of Cl2 = 12 – 4 = 8 5 PCl =   2 x 3 2 x, 12 4     PPCl3 = 2 Cl 2 x 8 3 , P 2 4 12     2 x 2 2 4