Question

At a certain temperature, 0.880 mol SO3 is placed in a 2.50 L container.

2SO3(g)↽−−⇀2SO2(g)+O2(g)

At equilibrium, 0.110 mol O2 is present. Calculate ?c.

Answer 1

Initial concentration of SO3 = mol of SO3 / volume in L

= 0.880 mol / 2.50 L

= 0.352 M

Equilibrium concentration of O2 = mol of O2 / volume in L

= 0.110 mol / 2.50 L

= 0.044 M

ICE Table:

[02] [S03] [S02] initial 0.352 0 change -2x +2x +1x equilibrium 0.352-2x +2x +1x

Given at equilibrium,

[O2] = 0.044

+1x = 0.044

x = 0.044

Equilibrium constant expression is

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

Kc = (+2x)^2(+1x)/(0.352-2x)^2

Kc = (+2*0.044)^2(+1*0.044)/(0.352-2*0.044)^2

Kc = 0.00489

Answer: 4.89*10^-3