Question

At a certain temperature, 0.620 mol of SO3 is placed in a 1.50-L container. 2SO3<-->2SO2+O2. At equilibrium, 0.140 mol of O2 is present. Calculate Kc

Answer 1

2SO3 <------> 2SO2 + O2
initially...
0.76 <-------> 0 + 0

at eq.
0.76-2x <-------> 2x + x

given x = 0.16

so no. of moles of SO2 at eq = 2x = 2X0.16 = 0.32
no. of moles of SO3 = 0.76-2x = 0.76-0.32 = 0.44

[SO3] = eq. conc. of SO3 = 0.44/3.5 = 0.126 M
[SO2] = 0.32/3.5 = 0.092 M
[O2] = 0.16/3.5 = 0.046 M

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

Kc = 0.092^2 X 0.046 / 0.126^2

Kc = 8.464 X 10^-3 X 0.046 / 0.016

Kc = 3.893 X 10^-4 / 0.016 = 0.024