Question

The following reaction wax examined at 250 C

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

At a particular temperature , 14.5 mole SO3 is placed in to a 2.0 L container and dissociates according to the reaction above. At equilibrium 3.0 mole of SO2 is present. Calculate Kc for this reaction

Answer 1

Solution :-

Lets first calculate the initial and equilibrium concentrations of the SO3 and SO2

[SO3] initial = 14.5 mol / 2.0 L = 7.25 M

[SO2] equilibrium = 3.0 mol / 2.0 L = 1.5 M

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

I   7.5 M                                                  0                             0    

C    -2x                                                   +2x                               +x

E 7.5-2x                                                 1.5M                          x

Lets find the value of x

2x = 1.5 M

X= 1.5 M / 2

X= 0.75 M

Equilibrium concentration of SO3 = 7.5 M -2x

                                                             = 7.2 M – (2*0.75 M)

                                                             = 5.75

Equilibrium concentration of the SO2 = 2x = 1.5 M

Equilibrium concentration of O2 = x = 0.75 M

Kc equation is as follows

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

   = [1.5]^2[0.75] /[5.75]^2

   = 0.051

Therefore equilibrium constant Kc for the reaction is 0.051