Question

At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.100 M [N2]=[O2]=0.100 M and [ NO ] = 0.600 M . [NO]=0.600 M. N 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 NO ( g ) N2(g)+O2(g)↽−−⇀2NO(g) If more NO NO is added, bringing its concentration to 0.900 M, 0.900 M, what will the final concentration of NO NO be after equilibrium is re‑established?

Answer 1

Answer: The final concentration of NO be after equilibrium is re‑established is 0.825 M.

Step-by-step explanation:

The given balanced chemical equation is as follows.

`N_(2)(g) + O_(2)(g) \rightleftharpoons 2NO(g)`

Now, equilibrium constant for this reaction will be as follows.

`K_(c) = ([NO]^(2))/([N_(2)][O_(2)])`

It is given that concentrations at the equilibrium are:

`[N_(2)] = [O_(2)]` = 0.1 M and [NO] = 0.6 M

Therefore, the value of
`K_(c)` is as follows.

`K_(c) = ((0.6)^(2))/((0.1)(0.1))`

= 36.0

NO concentration of 0.9 M is added to the system. So,

`N_(2)(g) + O_(2)(g) \rightleftharpoons 2NO(g)`

Initial: 0.1 0.1 0.9

Change: x x 2x

Equibm:(0.1 + x) (0.1 + x) (0.9 - 2x)

Now, we will find the value of x as follows.

36.0 =
`((0.9 - 2x)^(2))/((0.1 + x)^(2))`

x = 0.0375

Therefore, final concentration of NO after the equilibrium that is re-established is as follows.

0.9 - 2x

=
`0.9 - (2 * 0.0375)`

= 0.9 - 0.075

= 0.825 M

Therefore, we can conclude that the final concentration of NO be after equilibrium is re‑established is 0.825 M.