Respuesta :
Answer : The correct option is, (A) 0.081 mol/L
Explanation : Given,
Equilibrium constant = 18.0
Initial concentration of [tex]SO_2[/tex] = [tex]\farc{Moles}{Volume}=\frac{2.0}{20}=0.1M[/tex]
Initial concentration of [tex]NO_2[/tex] = [tex]\farc{Moles}{Volume}=\frac{2.0}{20}=0.1M[/tex]
The balanced equilibrium reaction is,
[tex]SO_2(g)+NO_2(g)\rightleftharpoons SO_3(g)+NO(g)[/tex]
Initial conc. 0.1 0.1 0 0
At eqm. (0.1-x) (0.1-x) x x
The expression of equilibrium constant for the reaction will be:
[tex]K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}[/tex]
Now put all the values in this expression, we get :
[tex]4.90=\frac{(x)\times (x)}{(0.1-x)\times (0.1-x)}[/tex]
By solving the term 'x' by quadratic equation, we get two value of 'x'.
[tex]x=0.131M\text{ and }0.081M[/tex]
From this we conclude that, the value of x = 0.131 at equilibrium can not be more than the initial concentration. So, the value of 'x' which is equal to 0.131 M is not consider.
The concentration of [tex]SO_3[/tex] at equilibrium = x = 0.081 M
Therefore, the concentration of [tex]SO_3[/tex] at equilibrium is, 0.081 M
