I am given the following reaction:

2NH3(g) -------> N2(g) + 3H2(g)

My question is:
6.4 mols of ammonia gas has been put into a 1.7 L flask and has been permitted to reach equilibrium in accordance to the reaction listed above. If the equilibrium mixture has 4.2 mols of nitrogen, what is the value of the equilibrium constant supposed to be?

Thank you

Are you sure of the values for the numbers of moles in the question?

I spent a good deal of time on this and finally came to the conclusion that something was amiss. Either a typo or the original problem was flawed. And the follow up of where did 2.7 = [N2] come from didn't make sense either. Finally, I went on to other posts.

To find the value of the equilibrium constant (K), you need to first determine the concentrations of each species in the equilibrium mixture.

Let's start by calculating the concentrations of the species using the given information:

The initial number of moles of ammonia (NH3) is 6.4 moles, and the volume of the flask is 1.7 L. Therefore, the initial concentration of NH3 is:

[ NH3 ]initial = (6.4 mol) / (1.7 L) = 3.76 M

At equilibrium, there are 4.2 moles of nitrogen (N2). Since the stoichiometric coefficient of nitrogen in the balanced chemical equation is 1, the concentration of N2 can be calculated as:

[ N2 ] = (4.2 mol) / (1.7 L) = 2.47 M

Similarly, the concentration of hydrogen (H2) can be calculated using the stoichiometry:

Since the stoichiometric coefficient of hydrogen in the balanced chemical equation is 3, the concentration of H2 can be calculated as:

[ H2 ] = (3 * [ N2 ]) = (3 * 2.47 M) = 7.41 M

Now that we have the concentrations of NH3, N2, and H2 at equilibrium, we can calculate the value of the equilibrium constant (K) using the expression:

K = ([ N2 ]^1 * [ H2 ]^3) / ([ NH3 ]^2)

Plugging in the calculated values, we get:

K = (2.47^1 * 7.41^3) / (3.76^2)

Now, we can calculate the value of K:

K ≈ 35.79

Therefore, the value of the equilibrium constant (K) for the given reaction is approximately 35.79.