At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.500 M.
N2(g) + 02(g) <-> 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?
To determine the final concentration of NO after equilibrium is re-established, we need to understand the concept of Le Chatelier's principle and how it applies to this chemical reaction.
Le Chatelier's principle states that when a system at equilibrium is subjected to a stress, it will respond by shifting in a direction that minimizes the effect of that stress.
In this case, adding more NO to the system increases its concentration, which is a stress. To minimize the effect of this stress, the equilibrium will shift in the opposite direction, towards the reactants (N2 and O2), to consume some of the excess NO.
Since the reaction is 2NO(g) ⇌ N2(g) + O2(g), we expect that the equilibrium will shift to the left, resulting in an increase in the concentrations of N2 and O2 and a decrease in the concentration of NO.
Now, let's calculate the new concentrations:
Initial concentrations:
[N2] = [O2] = 0.200 M
[NO] = 0.500 M
Change in concentration (due to the addition of more NO):
Δ[NO] = 0.800 M - 0.500 M = 0.300 M
According to the stoichiometry of the reaction, for every 2 mol of NO consumed, 1 mol of N2 and 1 mol of O2 are formed. Therefore, the change in concentration of N2 and O2 will both be half of the change in concentration of NO.
Δ[N2] = Δ[O2] = 0.300 M / 2 = 0.150 M
To find the final concentrations, we add the change in concentration to the initial concentrations:
[NO]final = [NO]initial + Δ[NO] = 0.500 M - 0.300 M = 0.200 M
[N2]final = [N2]initial + Δ[N2] = 0.200 M + 0.150 M = 0.350 M
[O2]final = [O2]initial + Δ[O2] = 0.200 M + 0.150 M = 0.350 M
Therefore, the final concentration of NO after equilibrium is re-established will be 0.200 M.