Question

At a particular temperature, K 5.0 x 10-6 for the reaction 2CO2(9) 2CO(9) +02(9) If 3.0 moles of CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species. CO2] CO] O2l [02]-

Answer 1

Answer : The concentration of
`CO_2,CO\text{ and }O_2` at equilibrium will be, 0.598 M, 0.00216 M and 0.00108 M respectively.

Explanation : Given,

Equilibrium constant =
`2.0* 10^(-6)`

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,

`2CO_2(g)\rightleftharpoons 2CO(g)+O_2(g)`

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:

`K_c=([CO]^2[O_2])/([CO_2]^2)`

Now put all the values in this expression, we get :

`2.0* 10^(-6)=((2x)^2* (x))/((0.6-2x)^2)`

By solving the term 'x', we get:

`x=0.00108M`

Concentration of
`CO_2` at equilibrium =
`(0.6-2x)M=[0.6-2(0.00108)]M=0.598M`

Concentration of
`CO` at equilibrium =
`(2x)M=[2(0.00108)]M=0.00216M`

Concentration of
`O_2` at equilibrium =
`(x)M=0.00108M`