Question

At a certain temperature, 0.840 mol SOg is placed in a 5.00 L container. 2SO,(@)2SO,(g)+0,(g) At equilibrium, 0.130 mol O, is present. Calculate Ke Ke

Answer 1

Solution:

For the given reaction, ICE table can be written as,

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

0.840 mol --------0 mol ---------0 mol (initial)

- 2X ---------------- +2X ------------+X  (change)

(0.840 - 2X)--------- 2X ------------- X (equilibrium)

Kc of the given reaction can be calculated from ICE table as,

Given, X = 0.130 mol

Then,

Number of moles of SO3 = 0.840 - 2X = 0.840 -2 x.0.130

= 0.840 - 0.260 = 0.580 mol

[SO3] = 0.580 mol / 5 L = 0.116 M

Number of moles of SO2 = 2 X = 2 x 0.130 = 0.260 mol

[SO2] = 0.260 mol /5L = 0.052 M

[O2] = 0.130 mol / 5 L = 0.026 M

Thus,

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

Kc = (0.052)^2 x ( 0.026) / (0.116)^2

Kc = 0.00523

OR

Kc = 5.23 x 10^-3