Question

The initial concentration for the compounds involved in the reaction displayed were determined to be [PCl5(g)] = 0.4676 mol/L, [PCl3(g)] = 0.6802 mol/L, [Cl2(g)] = 0.1048 mol/L . Calculate the value of the equilibrium constant ( Kc ) at 292.0 °C if the equilibrium concentration of PCl3(g) was 0.6563 mol/L . PCl5(g) = PCl3(g)+Cl2(g)

Answer 1

Given that initial concentrations are

[PCl5(g)] = 0.4676 mol/L

[PCl3(g)] = 0.6802 mol/L

[Cl2(g)] = 0.1048 mol/L

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

Initial concentration                 0.4676                            0.6802            0.1048

Equilibrium concentration     0.4676-X                             0.6802+X         0.1048+X

Hence,

equilibrium concentration of PCl3 (g) = 0.6802 +X

But given that equilibrium concentration of PCl3(g) = 0.6563 mol/L

Therefore,     0.6802 +X = 0.6563

                            X = -0.0239

Hence, equilibrium concentration are

[PCl5(g)] = 0.4676-X = 0.4676+0.0239 = 0.4915 mol/L

[Cl2(g)] = 0.1048+X = 0.1048-0.0239 = 0.0809 mol/L

[PCl3(g)] = 0.6563 mol/L

Hence,

                      Kc = [PCl3(g)] [Cl2(g)] / [PCl5(g)]

                          = (0.6563 mol/L) ( 0.0809 mol/L) / (0.4915 mol/L)

                          = 0.108 mol/L

                      Kc = 0.108 mol/L

Therefore, the equilibrium constant ( Kc ) = 0.108 mol/L