At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.400 M.
N2 (g) + O2 (g) --> 2NO (g)
If more NO is added, bringing its concentration to 0.700 M, what will the final concentration of NO be after equilibrium is re-established?
...........N2 + O2 ==> 2NO
E.........0.1....0.1....0.4
add 0.3.................0.7
new I.....0.1....0.1....0.7
C..........+x......+x....-2x
new E....0.1+x..0.1+x...0.7-2x
Use the first E line to calculate Kc. Substitute the equilibrium conditions and solve for kc.
Then substitute the NEW E line into the Kc expression and solve for x, 0.1_x and 0.7-2x
Post your work if you get stuck.
got it!! the answer is 0.6 and it was correct!! Thank you so so much!!
To determine the final concentration of NO after equilibrium is re-established, we can use the information about the initial concentrations and the balanced chemical equation.
Let's set up an ICE (Initial, Change, Equilibrium) table for this reaction:
| | N2 (g) | O2 (g) | 2NO (g) |
|---------|---------|---------|---------|
| Initial | 0.100 M | 0.100 M | 0.400 M |
| Change | | | |
| Equilibrium| | | |
From the balanced chemical equation, we know that the stoichiometric ratio between N2, O2, and NO is 1:1:2. This means that for every 1 mole of N2 and O2, we will have 2 moles of NO.
Since the initial concentrations of N2 and O2 are the same and equal to 0.100 M, and the initial concentration of NO is 0.400 M, we can assume that there is no change in the concentrations of N2 and O2 during the reaction.
Let's calculate the change in concentration of NO:
Change in [NO] = Final concentration of NO - Initial concentration of NO
Since the stoichiometric ratio between NO and O2 is 2:1, we can assume that the change in concentration of NO will be twice the change in concentration of O2.
To find the change in concentration of O2, subtract the initial concentration of O2 from the final concentration of O2:
Change in [O2] = Final concentration of O2 - Initial concentration of O2 = 0.100 M - 0.100 M = 0
Since the change in concentration of O2 is 0, the change in concentration of NO will also be 0.
Therefore, the final concentration of NO after equilibrium is re-established will still be 0.400 M.
To find the final concentration of NO after equilibrium is re-established, we need to use the concept of Le Chatelier's principle. According to Le Chatelier's principle, when a stress is applied to a system at equilibrium, the system will adjust to relieve that stress and re-establish equilibrium.
In this case, the stress is the addition of more NO, which increases its concentration from 0.400 M to 0.700 M. Since the equation shows that 2 moles of NO are consumed for every mole of N2 and O2, increasing the concentration of NO will shift the equilibrium to the left, favoring the formation of N2 and O2.
Let's denote the change in concentration of NO as Δ[NO]. Since the stoichiometry of the reaction is 1:1 for NO:N2 and NO:O2, the change in concentration of N2 and O2 will also be Δ[NO]. Therefore, the final concentrations of N2 and O2 can be written as [N2] = 0.100 M + Δ[NO] and [O2] = 0.100 M + Δ[NO].
Since every mole of NO that forms is consumed in the reaction, the equilibrium concentration of NO is equal to the initial concentration minus the change: [NO] = 0.400 M - Δ[NO].
Now, we can write the equilibrium expression for the given reaction: Kc = ([NO]^2)/([N2][O2]).
Substituting the concentrations we found earlier into the equilibrium expression, we get:
Kc = ([0.400 M - Δ[NO]]^2)/(([0.100 M + Δ[NO]])^2)
Since Kc is a constant value at a given temperature, we can solve this equation to find the value of Δ[NO]. Then, we can substitute this value back into the expression for [NO] to find the final concentration.
This equation can be solved either analytically or numerically using a solver. However, without knowing the value of Kc, we cannot provide a specific numerical answer.