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?
To determine the final concentration of NO after equilibrium is re-established, you can use Le Chatelier's principle, which states that a system at equilibrium will adjust to minimize the effect of any applied stress. In this case, adding more NO to the system is a stress that will cause the equilibrium to shift.
Let's analyze the balanced chemical equation for the reaction: N2 + O2 → 2NO.
The equilibrium expression for this reaction is Keq = [NO]² / ([N2] * [O2]).
At equilibrium, the concentrations are given as [N2] = [O2] = 0.200 M and [NO] = 0.500 M. Plugging these values into the equilibrium expression, we can solve for Keq:
Keq = (0.500 M)² / ((0.200 M) * (0.200 M)) = 6.25
Now, we have to consider what happens when more NO is added, increasing its concentration to 0.800 M. Since the equilibrium expression does not include the concentration of NO2 nor O2, we can compare the new concentration of NO to the initial equilibrium concentration of NO (0.500 M) to determine the direction of the shift.
If the concentration of NO increases, the equilibrium position will shift to the right, favoring the formation of more NO. This means that some of the added NO will react to form N2 and O2, decreasing their concentrations.
We can assume that the change in the concentration of NO is x. Therefore, the equilibrium concentration will be 0.500 M + x.
Using the stoichiometry of the balanced equation, we know that for every 1 mol of NO that reacts, 1 mol of N2 and 1 mol of O2 are formed. So, the decrease in the concentrations of N2 and O2 will also be x.
From the balanced equation, we know that the stoichiometric ratio of NO to N2 and O2 is 2:1. Therefore, the decrease in the concentrations of N2 and O2 will be 2x.
Now we can write the expression for the equilibrium concentrations of the species after the equilibrium is re-established:
[N2] = [N2]initial - 2x
[O2] = [O2]initial - 2x
[NO] = [NO]initial + x (since some of the added NO reacts)
We can substitute the initial concentrations provided and solve for x:
0.200 M = 0.200 M - 2x
0.200 M = 0.200 M - 2x
0 = 2x
Since 2x = 0, x = 0. Therefore, there is no change in the concentrations of N2 and O2, and the concentration of NO will simply be the sum of the initial concentration and the change:
[NO]final = [NO]initial + x = 0.500 M + 0 = 0.500 M
Thus, the final concentration of NO after equilibrium is re-established will remain at 0.500 M.
To determine the final concentration of NO after equilibrium is re-established, we need to use the stoichiometry of the given chemical equation and the concept of Le Chatelier's principle.
1. First, let's analyze the balanced chemical equation: N2 + O2 = 2NO. It states that one molecule of nitrogen gas (N2) reacts with one molecule of oxygen gas (O2) to form two molecules of nitric oxide (NO).
2. Given that at the initial equilibrium the concentrations were [N2] = [O2] = 0.200 M and [NO] = 0.500 M.
3. Now, more NO is added, bringing its concentration to 0.800 M. This disrupts the equilibrium and shifts the reaction to reach a new equilibrium.
4. According to Le Chatelier's principle, when you increase the concentration of one of the reactants, the equilibrium will shift in the direction that consumes that reactant, which is NO in this case.
5. Since the stoichiometry of the reaction is 1:1 for N2 and O2 and 2:1 for NO, for every molecule of NO consumed, one molecule of N2 and one molecule of O2 will also be consumed.
6. Let's assume that x moles of NO are consumed at equilibrium. Therefore, since 1 mole of NO reacts to form 2 moles of NO, the change in concentration for NO would be -2x M.
7. Since the initial concentration of NO was 0.500 M and x moles are consumed, the equilibrium concentration of NO will be 0.500 M - 2x M.
8. Finally, when more NO is added to bring its concentration to 0.800 M, the equilibrium concentration of NO will be 0.800 M - 2x M.
Therefore, the final concentration of NO after equilibrium is re-established will be 0.800 M - 2x M, where x is the number of moles of NO consumed.