Question

At a certain temperature, 0.940 mol of SO3 is placed in a 5.00-L container. 2SO3(g)--> 2SO2(g)+O2(g) At equilibrium, 0.110 mol of O2 is present. Calculate Kc.

Answer 1

we know that

conc = moles / volume (L)

so

initally

[S03] = 0.94 / 5= 0.188

at equilibrium

[02] = 0.11 / 5= 0.022

now

consider the given reaction

2 S03 ---> 2 S02 + 02

using ICE table

initial conc of S03 , S02 , 02 are 0.188 , 0 , 0

change in conc of S03 , S02 ,02 are -2x , 2x , x

equilibrium conc of S03 ,S02 ,02 are 0.188-2x , 2x , x

now

given

[02]eq = 0.022

so

x = 0.022

then

[S02]eq = 2x = 0.044

[S03]eq = 0.188-2x = 0.144

now

2 S03 ---> 2 S02 + 02

Kc = [S02]^2 [02] / [S03]^2

Kc = [0.044]^2 [0.022] / [ 0.144]^2

Kc = 2.054 x 10-3

so

the Kc value is 2.054 x 10-3