At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.500 M.

N2 + O2= 2NO

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?

Substitute the equilibrium concentrations into the Kc expression and solve for Kc. Use this below.


...........N2 + O2 ==> 2NO
I.......0.200..0.200....0.800
C.........+x...+x.......-2x
E.....0.200+x..0.200+x...0.800-2x

Substitute the E line into Kc expression along with the numberical value of Kc determined above and solve for x, then evaluate each of the values in the E line above. Post your work if you get stuck.

To determine the final concentration of NO after equilibrium is re-established, we can use the stoichiometry and the equilibrium constant expression.

The balanced equation for the reaction is N2 + O2 = 2NO.

At equilibrium, the concentrations are given as [N2] = [O2] = 0.200 M and [NO] = 0.500 M.

The equilibrium constant expression for this reaction is Kc = ([NO]^2) / ([N2] * [O2]).

We can use the given information to calculate the value of the equilibrium constant:

Kc = (0.500^2) / (0.200 * 0.200)
Kc = 6.25

Now, if more NO is added, bringing its concentration to 0.800 M, we need to determine the new equilibrium concentrations.

Let's assume that the final concentration of NO is x M. Since 2 moles of NO are produced per mole of N2 and O2, the concentrations of N2 and O2 will decrease by 2x M.

Therefore, after equilibrium is re-established, the new concentrations will be:
[N2] = [O2] = 0.200 - 2x M
[NO] = 0.800 - x M

Using the equilibrium constant expression, we can set up an equation to solve for x:

Kc = ([NO]^2) / ([N2] * [O2])
6.25 = (0.800 - x)^2 / ((0.200 - 2x) * (0.200 - 2x))

Solving this equation will give us the value of x, which represents the final concentration of NO.