Question

1)The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PCl5(g) The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PCl5(g) <------ ------> PCl3(g) + Cl2(g) Calculate the equilibrium partial pressures of all species when PCl5(g) is introduced into an evacuated flask at a pressure of 1.14 atm at 500 K. PPCl5 = atm PPCl3 = atm PCl2 = atm 2) The equilibrium constant, Kp, for the following reaction is 55.6 at 698 K: H2(g) + I2(g) <------- -------> 2HI(g) Calculate the equilibrium partial pressures of all species when H2 and I2, each at an intitial partial pressure of 2.22 atm, are introduced into an evacuated vessel at 698 K. PH2 = atm PI2 = atm PHI = atm

Answer 1

          PCl5(g) <-----------> PCl3(g) + Cl2(g)

I        1.14                               0              0

C         -x                                +x              +x

E        1.14-x                          +x             +x

                Kp    = PPCl3 * PCl2/PPCl5

                0.497   = x*x/(1.14-x)

                 0.497*(1.14-x) = x^2

                   x   = 0.544

              PPCl5   = 1.14-x    = 1.14-0.544   = 0.596atm

                PPCl3      = x         = 0.544atm

               PCl2        = x         = 0.544atm

                  H2(g) + I2(g) <-----------> 2HI(g)

I                 2.22     2.22                      0

C                -x           -x                       2x

E              2.22-x     2.22-x                  2x

                      Kp    = P^2HI/PH2*PI2

                      55.6   = (2x)^2/(2.22-x)(2.22-x)

                      55.6   = (2x/2.22-x)^2

                      7.46    = 2x/2.22-x

                      7.46*(2.22-x) = 2x

                     x   = 1.75

              PH2    = 2.22-x       = 2.22-1.75   = 0.47atm

              PI2    = 2.22-x       = 2.22-1.75   = 0.47atm

               PHI    = 2x              = 2*1.75    = 3.5atm