Question

Consider the reaction: Cl2(g) D 2Cl(g). Exactly 1.000 mole of chlorine gas is placed into a 1.000 L container at 298 K. Determine the number of chlorine atoms in that container at equilibrium, assuming DGof (Cl) = 105.3 kJ/mol.

Answer 1

Cl2(g) -----> 2Cl(g)

G⁰rxn = G⁰f products - G⁰f reactants

                 = 2*105.3-0

                   = 210.6KJ/mole

                    = 210600J/mole

G⁰rxn = -RTlnK

210600   = -8.314*298*2.303logK

210600    = -5705.84logK

logK         = 210600/-5705.84

logK         = -36.9

K              = 10^-36.9    = 1.26*10^-37

            Cl2(g) -----> 2Cl(g)

I            1                    0

C           -x                    2x

E            1-x                  2x

      Kc   = [Cl]^2/[Cl2]

       1.26*10^-37 = (2x)^2/1-x

       1.26*10^-37*(1-x) = 4x^2

         x   = 1.77*10^-19

[Cl]   = 2x   = 2*1.77*10^-19   = 3.54*10^-19 M

no of moles of Cl   = molarity * volume in L

                             = 3.54*10^-19*1 = 3.54*10^-19 moles

no of atoms of Cl = no of moles * 6.023*10^23

                               = 3.54*10^-19*6.023*10^23

                              = 2.13*10^5 chlorine atoms