Question

At a certain temperature, 0.740 mol of so3 is placed in a 4.00 l container. so_{3}(g)\rightleftharpoons 2so_{2}(g)+o_{2}(g) at equilibruim, 0.190 mol of o2 is present. calculate kc.

Answer 1

Answer : The equilibrium constant kc is 4.76 x 10⁻³

Explanation :

The given equilibrium reaction is

$2SO_{3} (g) \leftrightarrow 2SO_{2} (g) + O_{2} (g)$

Step 1 : Set up ICE table

Let us set up an ICE table for this reaction .

The initial concentration of SO₃ is

$Concentration = \frac{mol}{L} = \frac{0.740mol}{4L} = 0.185 M$

The initial concentrations of products are 0.

Let us assume x is the change .

Please refer the attached picture.

Step 2 : Use the given value to find x

From the ICE table, we can see that at equilibrium, concentration of O₂ is x

But we have been given that , at equilibrium we have 0.190 mol of O₂ .

Let us convert this to concentration unit.

Concentration of O₂ at equilibrium = $\frac{mol}{L} = \frac{0.190mol}{4L} = 0.0475 M$

But concentration of O₂ from the ICE table is x.

Therefore we have x = 0.0475 M

Step 3 : Using x , find equilibrium concentrations

Using this value, let us write the equilibrium concentrations of the given species.

[SO₃]eq = 0.185 M - 2x = 0.185 - 2(0.0475) = 0.09 M

[SO₂]eq = 2x = 0.095 M

[O₂]eq = x = 0.0475 M

Step 4 : Set up equation for kc and solve it

The equilibrium constant kc is calculated as,

$k_{c} = \frac{[SO_{2}]^{2} [O_{2}]}{[SO_{3}]^{2}}$

Let us plug in the above equilibrium values.

$k_{c} = \frac{(0.095)^{2} (0.0475)}{(0.09)^{2}}$

$k_{c} = \frac{0.00042869}{0.09}$

$k_{c} = 4.76 \times 10^{-3}$

The equilibrium constant kc is 4.76 x 10⁻³