A 2 liter container holds 2.2 mol of NH3 gas, which starts to decompose according to the following reaction.
At equilibrium there are .6 mol of H2.
Determine the concentrations of NH3, H2, and N2 at equilibrium.
2NH3-> 3H2+N2
^ all gas
.........2NH3 ==> N2 + 3H2
I.........2.2.....0......0
C.........-2x......x.....3x
E.......2.2-2x.....x......3x
The problem tells us that there is 0.6 mol H2 at equilibrium. So 3x = 0.6 mol which means x = 0.2 mol. (Note that I did NOT change H2. It still is 0.6 mol. For proof of that 3x = 3*0.2 = 0.6. I write this only because I helped a student last week with this and she thought I had changed H2 but I didn't.)
That gives you N2 (x) and NH3 = 2.2-2x.
Then divide mols each by 2 L to obtain concn of each.
To determine the concentrations of NH3, H2, and N2 at equilibrium, we can start by using the balanced chemical equation and the given information.
From the balanced equation, we can see that 2 mol of NH3 decomposes to produce 3 mol of H2 and 1 mol of N2. So, the ratio of the moles of NH3 to H2 is 2:3.
Given that initially there are 2.2 mol of NH3 and at equilibrium there are 0.6 mol of H2, we can calculate the moles of N2.
First, let's determine the moles of H2 produced from the decomposition of NH3:
2 mol NH3 → 3 mol H2
x mol NH3 → 0.6 mol H2
Using the ratio, we can set up a proportion:
2/3 = x/0.6
Cross-multiplying and solving for x, we get:
(2)(0.6) = (3)(x)
1.2 = 3x
x = 1.2/3
x = 0.4 mol
Therefore, 0.4 mol of NH3 decomposes, producing 0.6 mol of H2 and 0.4 mol of N2.
Now, let's determine the concentrations of NH3, H2, and N2.
To find the concentration, we divide the number of moles by the volume. Given that the container has a volume of 2 liters, we can calculate the concentrations.
Concentration of NH3 at equilibrium:
NH3: moles/volume = 2.2 mol/2 L = 1.1 M
Concentration of H2 at equilibrium:
H2: moles/volume = 0.6 mol/2 L = 0.3 M
Concentration of N2 at equilibrium:
N2: moles/volume = 0.4 mol/2 L = 0.2 M
So, at equilibrium, the concentrations of NH3, H2, and N2 are 1.1 M, 0.3 M, and 0.2 M, respectively.