Question

K_c = 4.15 x 10^-2 at 533°C for PCl_5(g) ↔ PCl_3(g) + Cl_2(g). A closed 2.00 L vessel initially contians 0.100 mol PCl₅. Calculate the total pressure in the vessel (in atm to 2 decimal places) at 533°C when equilibrium is achieved.

Answer 1

                              PCl_5(g) ↔ PCl_3(g) + Cl_2(g)

I                               0.1            0             0

C                             -x               +x           +x

E                            0.1-x             +x           +x

             Kc   = [PCl3][Cl2]/[PCl5]

             4.15*10^-2   = x*x/0.1-x

            4.15*10^-2*(0.1-x) = x^2

                   x = 0.047

          [PCl5]   = 0.1-x   = 0.1-0.047   = 0.053moles

           [PC3]    = x         = 0.047moles

           [Cl2]     = x         = 0.047 moles

total no of moles at equilibrium (n) = 0.053+0.047+ 0.047   = 0.147moles

                                                    V    = 2L

                                                     T   = 533+273   = 806K

                                          PV = nRT

                                          P   = nRT/V

                                               = 0.147*0.0821*806/2   = 4.86atm

total pressure in the vessel   = 4.86atm >>>answer