Question

SO2 (0.522 mol.L-1) and O2 (0.633 mol.L-1) react and reach equilibrium. Calculate the equilibrium concentrations of the products and reactants given that KC = 5.66 x 10-10 for this reaction: 2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g).

Answer 1

Answer: Hence, the equilibrium concentration of the $SO_2,O_2\text{ and }SO_3$ are 0.521, 0.632 and $10.88\times 10^{-6}M$respectively.

Explanation:

The balanced chemical reaction is:

                  $SO_2(g)+2O_2(g)\rightleftharpoons 2SO_3(g)$

At t = 0          0.522           0.633               0  

At $t=t_{eq}$      0.522-x         0.633-2x      2x

The expression for $K_c$ for the given reaction follows:

$K_c=\frac{[SO_2]^2}{[SO_2]\times [O_2]}$

We are given:

$K_c=5.66\times 10^{-10}$

Putting values in above equation, we get:

$5.66\times 10^{-10}=\frac{(2x)^2}{(0.522-x)\times (0.633-2x)^2}$

$x=5.44\times 10^{-6}M$

Thus equilibrium concentration of $SO_2$ is = (0.522-x )= $(0.522-5.44\times 10^{-6})=0.521$

equilibrium concentration of $O_2$ is = (0.633-2x )= $(0.633-2\times 5.44\times 10^{-6})=0.632$

equilibrium concentration of $SO_3$ is = (2x )= $(2\times 5.44\times 10^{-6})=10.88\times 10^{-6}M$