Respuesta :
Answer: Hence, the equilibrium concentration of the [tex]SO_2,O_2\text{ and }SO_3[/tex] are 0.521, 0.632 and [tex]10.88\times 10^{-6}M[/tex]respectively.
Explanation:
The balanced chemical reaction is:
[tex]SO_2(g)+2O_2(g)\rightleftharpoons 2SO_3(g)[/tex]
At t = 0 0.522 0.633 0
At [tex]t=t_{eq}[/tex] 0.522-x 0.633-2x 2x
The expression for [tex]K_c[/tex] for the given reaction follows:
[tex]K_c=\frac{[SO_2]^2}{[SO_2]\times [O_2]}[/tex]
We are given:
[tex]K_c=5.66\times 10^{-10}[/tex]
Putting values in above equation, we get:
[tex]5.66\times 10^{-10}=\frac{(2x)^2}{(0.522-x)\times (0.633-2x)^2}[/tex]
[tex]x=5.44\times 10^{-6}M[/tex]
Thus equilibrium concentration of [tex]SO_2[/tex] is = (0.522-x )= [tex](0.522-5.44\times 10^{-6})=0.521[/tex]
equilibrium concentration of [tex]O_2[/tex] is = (0.633-2x )= [tex](0.633-2\times 5.44\times 10^{-6})=0.632[/tex]
equilibrium concentration of [tex]SO_3[/tex] is = (2x )= [tex](2\times 5.44\times 10^{-6})=10.88\times 10^{-6}M[/tex]
