Respuesta :
Answer : The concentration of [tex]CO_2,CO\text{ and }O_2[/tex] at equilibrium will be, 0.598 M, 0.00216 M and 0.00108 M respectively.
Explanation : Given,
Equilibrium constant = [tex]2.0\times 10^{-6}[/tex]
tex]\text{Concentration of }CO_2=\frac{\text{Moles of }CO_2}{\text{Volume of solution}}=\frac{3.0mole}{5.0L}=0.6mole=0.6M[/tex]
The balanced equilibrium reaction is,
[tex]2CO_2(g)\rightleftharpoons 2CO(g)+O_2(g)[/tex]
Initial conc. 0.6 M 0 0
At eqm. (0.6-2x) M 2x M x M
The expression of equilibrium constant for the reaction will be:
[tex]K_c=\frac{[CO]^2[O_2]}{[CO_2]^2}[/tex]
Now put all the values in this expression, we get :
[tex]2.0\times 10^{-6}=\frac{(2x)^2\times (x)}{(0.6-2x)^2}[/tex]
By solving the term 'x', we get:
[tex]x=0.00108M[/tex]
Concentration of [tex]CO_2[/tex] at equilibrium = [tex](0.6-2x)M=[0.6-2(0.00108)]M=0.598M[/tex]
Concentration of [tex]CO[/tex] at equilibrium = [tex](2x)M=[2(0.00108)]M=0.00216M[/tex]
Concentration of [tex]O_2[/tex] at equilibrium = [tex](x)M=0.00108M[/tex]
