Question

At equilibrium, the concentrations in this system were found to be [N21 O20.200 M and [NO]0.500 M. N2(8) 02e) 2NO(g) If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re-established?

Answer 1

Answer : The concentration of
`NO` at equilibrium is 0.9332 M

Solution : Given,

Concentration of
`N_2` and
`O_2` at equilibrium = 0.200 M

Concentration of
`N_2` and
`O_2` at equilibrium = 0.500 M

First we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

`N_2(g)+O_2(g)\rightleftharpoons 2NO(g)`

The expression of
`K_c` will be,

`K_c=([NO]^2)/([N_2][O_2])`

`K_c=((0.500)^2)/((0.200)* (0.200))`

`K_c=6.25`

Now we have to calculate the final concentration of NO.

The given equilibrium reaction is,

`N_2(g)+O_2(g)\rightleftharpoons 2NO(g)`

Initially 0.200 0.200 0

.800

At equilibrium (0.200-x) (0.200-x) (0.800+2x)

The expression of
`K_c` will be,

`K_c=([NO]^2)/([N_2][O_2])`

`K_c=((0.800+2x)^2)/((0.200-x)* (0.200-x))`

By solving the term x, we get

`x=2.6\text{ and }-0.0666`

From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.

And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).

Thus, the concentration of
`NO` at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M

Therefore, the concentration of
`NO` at equilibrium is 0.9332 M