Question

At a certain temperature, 0.920 mol of SO3 is placed in a 4.00-L container. 2SO3(g) <===>2SO2(g) + O2(g) At equilibrium, 0.130 mol of O2 is present. Calculate Kc.

Answer 1

At a certain temperature, 0.920 mol of SO3 is placed in a 4.00-L container. 2SO3(g) <===>2SO2(g) + O2(g) At equilibrium, 0.130 mol of O2 is present. Calculate Kc

Solution :-

2SO3(g) <===>2SO2(g) + O2(g)

Initial concentration of the SO3 = moles / volume

                                                         = 0.920 mol / 4.00 L

                                                         = 0.230 M

Equilibrium concentration of the O2 = moles / volume

                                                                  =0.130 mol / 4.00 L

                                                                  = 0.0325 M

2SO3(g) <===>2SO2(g) + O2(g)

0.230 M                             0                    0

-2x                                       +2x                 +x

0.230-2x                              +2x                0.0325 M

Using the value of x we can find the equilibrium concentration of the SO3 and SO2

Equilibrium concentration of SO3 = 0.230 M – 2x

                                                              =0.230 M- (2*0.0325)

                                                               = 0.165 M

Equilibrium concentration of the SO2 = 2x

                                                                      =2*0.0325 M

                                                                       = 0.0650 M

Kc= [SO2]^2[O2] /[SO3]^2

Kc= [0.0650]^2[0.0325]/[0.165]^2

Kc= 5.04*10^-3

Therefore the equilibrium constant Kc= 5.04*10^-3