At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.500 M.
The balanced chemical equation is 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, we need to calculate the change in concentration of NO.
Given:
Initial concentrations: [N2] = [O2] = 0.100 M, [NO] = 0.500 M
New concentration of NO: [NO] = 0.800 M
The balanced equation for the reaction is N2 + O2 ⇌ 2NO.
From the stoichiometry of the balanced equation, we can say that for every 1 mole of N2 reacting, 2 moles of NO will be formed. This means that the change in concentration of NO will be twice the change in concentration of N2.
Let's denote the change in concentration of N2 as Δ[N2]. As 1 mole of N2 will react to form 2 moles of NO, the change in concentration of N2 will have a magnitude of -Δ[NO]/2.
So, Δ[N2] = -Δ[NO]/2.
Now, let's calculate the change in concentration of NO.
Δ[NO] = [NO]final - [NO]initial
= 0.800 M - 0.500 M
= 0.300 M
Using the equation Δ[N2] = -Δ[NO]/2, we can find the change in concentration of N2:
Δ[N2] = -Δ[NO]/2
= -(0.300 M)/2
= -0.150 M
To determine the final concentration of NO after equilibrium is re-established, we need to add the change in concentration of NO to the initial concentration of NO:
[NO]final = [NO]initial + Δ[NO]
= 0.500 M + 0.300 M
= 0.800 M
Therefore, the final concentration of NO after equilibrium is re-established will be 0.800 M, which is the same as the new concentration of NO added.
To determine 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 shift in a way that minimizes the effect of that stress. In this case, the addition of more NO is the stress that will cause the reaction to shift.
Since more NO is added, the concentration of NO on the reactant side will increase. Therefore, according to Le Chatelier's principle, the reaction will shift towards the products to alleviate the stress. In other words, the forward reaction will be favored.
Using the balanced chemical equation N2 + O2 ⇌ 2NO, we can determine the change in the concentrations of N2, O2, and NO after the equilibrium is re-established. Let's assume the change in concentration of NO is x.
Initially, we have [N2] = [O2] = 0.100 M and [NO] = 0.500 M. After the reaction shifts to re-establish equilibrium, the final concentration of NO will be [NO] = 0.500 M + x.
Since two moles of NO are formed for every one mole of N2 and O2 consumed, the change in concentration of NO will be twice the change in concentration of N2 and O2. Therefore, the change in concentration of N2 and O2 will be 0.5x.
Now, let's write the equilibrium expression using the concentrations at equilibrium:
Kc = [NO]^2 / ([N2] * [O2])
Since the concentrations of N2 and O2 are initially equal, we can write:
Kc = [NO]^2 / ([N2] * [N2])
Plugging in the concentrations at equilibrium:
Kc = (0.500 M + x)^2 / (0.100 M * 0.100 M)
Now, let's solve for x.
Kc = (0.500 M + x)^2 / (0.010 M^2)
Rearranging and simplifying the equation:
0.010 M^2 * Kc = (0.500 M + x)^2
0.010 M^2 * Kc = 0.250 M^2 + x^2 + 2x * 0.500 M
0.010 M^2 * Kc = 0.250 M^2 + x^2 + x M
0.250 M^2 - 0.010 M^2 * Kc = x^2 + x M
0.010 M^2 * (Kc - 0.250 M^2) = x^2 + x M
Now, we can solve this quadratic equation to find the value of x. The equation will give two possible values for x, but we can discard the negative value since concentrations cannot be negative.
Once we find the value of x, we can calculate the final concentration of NO by adding it to the initial concentration of NO:
[NO]final = [NO]initial + x
[NO]final = 0.500 M + x (where x is the positive value obtained from solving the quadratic equation)
By performing these calculations, you can determine the final concentration of NO after equilibrium is re-established.