Question

2.0 mol each of PCl₅, PCl₃, and Cl₂ are placed in a 2.0L reaction vessel at 620K. K_c for the reaction PCl₅(g) ↔ PCl₃(g) + Cl₂(g) is 0.60 at the given temperature. What is the equilibrium concentration of Cl₂? 1.146 is incorrect.

Answer 1

              PCl₅(g) ↔ PCl₃(g) + Cl₂(g)

    I         2                2               2

   C        +x             -x                -x

    E       2+x          2-x               2-x

           [PCl5]   = 2+x/2

           [PCl3]   = 2-x/2

          [Cl2]     = 2-x/2

        Kc = [PCl3][Cl2]/[PCl5]

       0.6   = (2-x)(2-x)/2(2+x)

      0.6*2(2+x) = (2-x)(2-x)

       x   = 0.328

   [Cl2]     = 2-x/2    = 2-0.328/2    = 0.836M >>>>answer