A 0.25 mol sample of n204 dissociates and comes to equilibrium in a 1.5 L flask at 100 degrees C. The reaction is N2O4 > 2 NO2. The Kc at 100 degrees C is 0.36. What are the equilibrium concentrations of NO2 and N2O4?
0.25 mol/1.5L = 0.167M which I would round to 0.17. (Note: you may wish to carry an extra place and round the answer at the end to two significant figures. Follow your profs advice on this.)
........N2OP4 ==> 2NO2
I.......0.167.....0
C.......-x........2x
E.....0.167-x....2x
Kc = (NO2)^2/(N2O4)
0.36 = (2x)^2/(0.167-x)
Solve for x and 2x.
To find the equilibrium concentrations of NO2 and N2O4, we can use the given information and the equation for the equilibrium constant (Kc) to set up an ICE table.
Let's start by assuming x is the change in concentration for N2O4 and 2x is the change in concentration for NO2.
The initial concentration of N2O4 is 0.25 mol. Since it all dissociates, the change in concentration for N2O4 is -0.25 mol.
The initial concentration of NO2 is 0 mol, as there is no NO2 initially. The change in concentration for NO2 is 2x.
Using the equation for the equilibrium constant,
Kc = [NO2]^2 / [N2O4]
we can plug in the equilibrium concentrations as [NO2] and [N2O4].
0.36 = (2x)^2 / (-0.25)
0.36 = 4x^2 / (-0.25)
0.36 * (-0.25) = 4x^2
-0.09 = 4x^2
x^2 = -0.09 / 4
x^2 = -0.0225
x = √(-0.0225)
x = 0.15
Since the change in concentration for N2O4 is -0.25 mol and the change in concentration for NO2 is 2x (2 * 0.15 = 0.3), we can calculate the equilibrium concentrations.
[N2O4] = (Initial concentration of N2O4 + Change in concentration)
[N2O4] = 0.25 + (-0.25)
[N2O4] = 0 mol
[NO2] = (Initial concentration of NO2 + Change in concentration)
[NO2] = 0 + 0.3
[NO2] = 0.3 mol
Therefore, at equilibrium, the concentration of N2O4 is 0 mol and the concentration of NO2 is 0.3 mol.
To find the equilibrium concentrations of NO2 and N2O4, we can use the equation for the equilibrium constant (Kc) and the initial moles of N2O4.
The balanced equation for the reaction is: N2O4 ⇌ 2NO2
Let's assume that 'x' represents the change in the concentration of N2O4 (in mol/L) as it dissociates. Therefore, the change in the concentration of NO2 would be '2x' since it is formed in a 2:1 ratio with N2O4.
At equilibrium, the concentration of N2O4 would be: [N2O4] = (0.25 - x) mol/L
At equilibrium, the concentration of NO2 would be: [NO2] = (2x) mol/L
Substituting the values into the equation for Kc:
Kc = ([NO2]^2) / [N2O4]
0.36 = ((2x)^2) / (0.25 - x)
Simplifying further:
0.36 = 4x^2 / (0.25 - x)
Now, we can solve this equation for 'x' using algebra:
0.36(0.25 - x) = 4x^2
0.09 - 0.36x = 4x^2
4x^2 + 0.36x - 0.09 = 0
This is a quadratic equation. We can use the quadratic formula to solve for 'x':
x = (-0.36 ± √(0.36^2 - 4(4)(-0.09))) / (2(4))
Calculating the discriminant (√(b^2 - 4ac)), which is inside the square root:
Discriminant = √(0.36^2 - 4(4)(-0.09))
Discriminant = √(0.1296 + 1.44)
Discriminant = √1.5696
Discriminant ≈ 1.253
Calculating the value of 'x':
x = (-0.36 ± 1.253) / 8
x1 ≈ 0.116, x2 ≈ -0.200
Since the concentration cannot be negative, we can discard the negative root, and we have:
x ≈ 0.116
Now, we can calculate the equilibrium concentrations of NO2 and N2O4:
[N2O4] = (0.25 - 0.116) mol/L ≈ 0.134 mol/L
[NO2] = 2(0.116) mol/L ≈ 0.232 mol/L
Therefore, at equilibrium, the concentrations of NO2 and N2O4 are approximately 0.232 mol/L and 0.134 mol/L, respectively.