When N2O5 (g) is heated it dissociates into N2O3 (g) and O2 (g) according to the following reaction: N2O5 (g) <--> N2O3 (g) + O2 (g) Kc=7.75 at a given temperature. The N2O3 (g) dissociates to give N2O (g) and O2 (g) according the following reaction: N2O3 (g) <--> N2O (g) + O2 (g) Kc= 4.00 at the same temperature. When 4.00 mol of N2O5 (g) is heated in a 1.00-L reaction vessel to this temperature, the concentration of O2 (g) at equilibrium is 4.50 mol/L.
a) Find the concentration of N2O5 in the equilibrium system.
b) Find the concentration of N2O in the equilibrium system.
c) Find the concentration of N2O3 in the equilibrium system.
Get a second opinion on this AND check my thinking..
N2O5 ==> N2O3 + O2
N2O3 ==> N2O + O2
Add the two equations to obtain
N2O5 ==> N2O + 2O2 and k for this reaction is k1*k2 = 7.75*4 = 31.0 = (N2O)(O2)/(N2O5).
The final O2 is 4.5 moles; some came from reaction 1 and some from reaction 2. How much came from each. If we call x the amount from 1, then 4.00/7.75 must have come from rxn 2 or
1+(4.00/7.75) = 4.5 and x = 2.97
The amount from rxn 2 then must be 2.97*(4.00/7.75) = 1.53 (total is still 4.50).
From rxn 2,
N2O3 ==> N2O + O2
If O2 from this rxn is 1.53, that means N2O must be 1/2 that or 0.765 (that is part (c)). From
4.00 = (N2O)(O2)/(N2O3), we solve for N2O3 (substitute 4.5 for O2 because the O2 in the final mixture can't tell where it came from and we know the total is 4.5). That gets part(b).
For part a, concn N2O5 at equilibrium, use K1K2 = 31.0 = (N2O3)(O2)/(N2O5)
You know N2O3 and O2.
Just to check things, if we substitute the values I obtained for N2O3 (0.861) and 4.5 for O2 and 0.500 for N2O5, K1 gives 7.75 which was K in the problem. Again, get a second opinion and check my thinking.
Hi, I'm doing the same assignment. I got to the part where you added the reactions and got the K3 = 31. Then, I got stuck
I don't understand why you're doing the 4.00/7.75 and then stating that 1 +(4.00/7.75) = 4.5? isn't that 1.52 instead of 4.5?
To solve this problem, let's use the given information and the principles of equilibrium to determine the concentrations of N2O5, N2O, and N2O3 in the equilibrium system.
We are given that the concentration of O2 at equilibrium is 4.50 mol/L. Let's denote the concentrations of N2O5, N2O, and N2O3 as [N2O5], [N2O], and [N2O3], respectively.
a) According to the given equilibrium reaction: N2O5 (g) <--> N2O3 (g) + O2 (g)
The stoichiometric coefficient of O2 is 1 in both the forward and backward direction. This means that the change in concentration of O2 is equal to the change in its equilibrium concentration.
Therefore, [O2] = 4.50 mol/L.
b) Now, let's use the equilibrium constant expression for the second reaction: N2O3 (g) <--> N2O (g) + O2 (g)
Kc = [N2O][O2] / [N2O3]
We are given that Kc = 4.00.
Substituting the equilibrium concentrations into the equation:
4.00 = [N2O][4.50] / [N2O3]
Solving for [N2O], we get:
[N2O] = (4.00 * [N2O3]) / [O2]
c) Now, let's use the equilibrium constant expression for the first reaction: N2O5 (g) <--> N2O3 (g) + O2 (g)
Kc = [N2O3][O2] / [N2O5]
We are given that Kc = 7.75.
Substituting the equilibrium concentrations into the equation:
7.75 = [N2O3][4.50] / [N2O5]
Solving for [N2O3], we get:
[N2O3] = (7.75 * [N2O5]) / [O2]
Now, we have two equations that relate the concentrations of N2O and N2O3 to the concentrations of N2O5 and O2.
To find the concentration of N2O5 in the equilibrium system (a), we plug in the given values:
[N2O5] = 4.00 mol
To find the concentration of N2O in the equilibrium system (b), we substitute the values into the equation:
[N2O] = (4.00 * [N2O3]) / 4.50
To find the concentration of N2O3 in the equilibrium system (c), we substitute the values into the equation:
[N2O3] = (7.75 * 4.00) / 4.50
Now, we can solve for the concentrations of N2O5, N2O, and N2O3.
To solve this problem, we can start by setting up an ICE table for each reaction and then use the given equilibrium concentrations and the equilibrium constants (Kc) to calculate the equilibrium concentrations of the reactants and products.
a) Let's set up the ICE table for the first reaction:
N2O5 (g) <--> N2O3 (g) + O2 (g)
Initial: 4.00 0 0
Change: -x +x +x
Equilibrium: 4.00 - x x x
From the given information, we know that at equilibrium the concentration of O2 (g) is 4.50 mol/L. Therefore, we can set up the following equation using the equilibrium expression for the first reaction:
Kc = [N2O3][O2] / [N2O5]
7.75 = (x)(x) / (4.00 - x)
Since the concentration of O2 (g) is the same as x, we can substitute x with 4.50 in the equation above:
7.75 = (4.50)(4.50) / (4.00 - 4.50)
7.75 = 20.25 / (-0.50)
7.75 = -40.50
This equation has no valid solution because the value of Kc cannot be negative. Therefore, there is an error in the information given or the calculations.
b) Since we couldn't find the concentration of N2O5 in part (a), we cannot proceed to find the concentration of N2O in the equilibrium system.
c) Similarly, without the values from part (a) or (b), we cannot find the concentration of N2O3 in the equilibrium system.
It seems like there is a mistake in the information given or the calculation, as the values obtained are not consistent with the equilibrium equation and constants provided.