To solve this problem, we need to use the equilibrium constant expression:
Kc = [NH3]^2 / ([N2] x [H2]^3)
where Kc is the equilibrium constant, [NH3], [N2], and [H2] are the molar concentrations of NH3, N2, and H2 at equilibrium, respectively.
At the beginning of the reaction, we have [N2] = 0.1 moles and [H2] = 0.1 moles, and [NH3] = 0 moles. Let's assume that x moles of N2 react to form NH3, so the amount of N2 left at equilibrium will be (0.1 - x) moles. Similarly, 3x moles of H2 react to form 2x moles of NH3, so the amount of H2 left at equilibrium will be (0.1 - 3x) moles, and the amount of NH3 formed will be 2x moles.
Now we can set up an ICE table (Initial, Change, Equilibrium) to keep track of the changes in concentration:
N2 3H2 2NH3
Initial 0.1 0.1 0
Change -x -3x +2x
Equilibrium 0.1-x 0.1-3x 2x
At equilibrium, we know that the reaction quotient Qc is equal to the equilibrium constant Kc:
Qc = [NH3]^2 / ([N2] x [H2]^3) = (2x)^2 / ((0.1-x) x (0.1-3x)^3) = Kc
Simplifying this expression and solving for x, we get:
38,000 = 4x^2 / (0.1-3x)^3(0.1-x)
(0.1-3x)^3(0.1-x) = 4x^2 / 38,000
(0.1-3x)^3(0.1-x) = 1.0526x^2
At this point, we can either use trial and error to solve for x, or we can use a numerical method such as the Newton-Raphson method to iteratively solve the equation. Using the latter method, we can write the equation as:
f(x) = (0.1-3x)^3(0.1-x) - 1.0526x^2 = 0
and its derivative as:
f'(x) = -9(0.1-3x)^2(0.1-x) - (0.1-3x)^3 + 2(1.0526)x
Starting with an initial guess of x = 0.05, we can iterate using the formula:
x1 = x0 - f(x0) / f'(x0)
where x1 is the next approximation, x0 is the current approximation, and f(x0) and f'(x0) are the function and its derivative evaluated at x0. After a few iterations, we find that x ≈ 0.0662 moles, which corresponds to the equilibrium concentrations:
[N2] = 0.1 - x ≈ 0.0338 moles
[H2] = 0.1 - 3x ≈ 0.0016 moles
[NH3] = 2x