A sample of ammonia gas was allowed to come to equilibrium at 400 K.
2NH3(g) --> N2(g) + 3H2(g)
At equilibrium, it was found that the concentration of H2 was 0.0584 M, the concentration of N2 was 0.0195
M, and the concentration of NH3 was 0.430 M. What was the initial concentration of ammonia?
Calculate Kc by substituting the values from the problem into the Kc expression.
.........2NH3 ==> N2 + 3H2
I.........X........0.....0
C........-2Y.......Y.....3Y
E.....X-2Y..........Y....3Y
Eproblem..0.43.....0.0195..0.0584
So you know X-2Y = 0.430. Substitute the value of Y (that's N2 at equil) and solve for X. Interesting problem.
0.218
Well, it seems that ammonia is trying really hard to show off its equilibrium skills. So let's see if we can crack a smile on this chemical reaction:
According to the balanced equation, we have 2 moles of NH3 resulting in 3 moles of H2. That means, for every 2 moles of ammonia consumed, we get 3 moles of hydrogen gas. So the ratio of H2 to NH3 is 3/2.
Now, let's look at the concentrations given at equilibrium. The concentration of H2 is 0.0584 M, and the concentration of NH3 is 0.430 M. Using the ratio we just mentioned, we can set up the following equation:
(0.0584 M) / (0.430 M) = 3/2
Solving this equation will give us the initial concentration of ammonia, which is? Well, that's for you to calculate!
To solve this problem, we'll use the concept of the equilibrium constant (Kc).
The balanced chemical equation is:
2NH3(g) → N2(g) + 3H2(g)
The equilibrium constant (Kc) expression for this reaction is:
Kc = [N2]^1 * [H2]^3 / [NH3]^2
We are given the concentrations of N2, H2, and NH3 at equilibrium:
[N2] = 0.0195 M
[H2] = 0.0584 M
[NH3] = 0.430 M
We'll substitute these values into the Kc expression:
Kc = (0.0195^1) * (0.0584^3) / (0.430^2)
Now we can calculate the value of Kc:
Kc = 4.61 * 10^-2
Since the reaction has reached equilibrium, the value of Kc remains constant at this temperature.
At equilibrium, the ratio of concentrations is given by the stoichiometric coefficients:
[N2] / [H2]^3 = [NH3]^2 / [N2]^1
Substituting the given values, we get:
0.0195 / (0.0584^3) = (0.430^2) / 0.0195
Simplifying the equation, we find:
(0.0195 * 0.0195) / (0.0584^3) = 0.430^2
Solving for the initial concentration of NH3, we have:
[NH3]initial = sqrt(0.430^2 * (0.0584^3) / (0.0195^2))
[NH3]initial ≈ 0.985 M
Therefore, the initial concentration of ammonia (NH3) is approximately 0.985 M.
To find the initial concentration of ammonia (NH3), we need to make use of the stoichiometry and the balanced chemical equation for the given reaction.
The balanced chemical equation is:
2NH3(g) -> N2(g) + 3H2(g)
From the equation, we can see that for every 2 moles of NH3 that react, 3 moles of H2 are produced. This means that the mole ratio between NH3 and H2 is 2:3.
Given that the concentration of H2 at equilibrium is 0.0584 M, we can calculate the concentration of NH3 at equilibrium using the mole ratio. Since 3 moles of H2 are produced for every 2 moles of NH3, the concentration of NH3 at equilibrium can be calculated as:
[ NH3 ] = 0.0584 M / (3/2) = 0.0584 M * (2/3) = 0.0389 M
Therefore, at equilibrium, the concentration of NH3 is 0.0389 M.
Now, to find the initial concentration of NH3, we need to consider the principle of conservation of mass. At equilibrium, the total amount of NH3 that reacted to form N2 and H2 is equal to the concentration of NH3 at equilibrium multiplied by the volume of the container.
So, the initial concentration of NH3 can be calculated using the formula:
[ NH3 ]initial = [ NH3 ]equilibrium * (volume of container)
Since the concentration of NH3 at equilibrium is 0.0389 M and the concentration of NH3 was found to be 0.430 M at equilibrium, we have:
0.430 M = [ NH3 ]initial * (volume of container)
To find the volume of the container, we need additional information. Without the volume, it is not possible to determine the initial concentration of NH3.