A container initially has 0.0186 M ammonia at a certain temperature. When the system reaches equilibrium the concentration of ammonia is 0.0035 M. Calculate Kc for the following reaction as written.

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

Write the Kc expression from the equation..

N2 + H2 ==> 2NH3
Then 0.0186 M NH3 initially.
At equilibrium H2 must be 3x, N2 must be 2x, and NH3 must be 0.0186-2x. Plug those values in and solve for Kc.

Thanks, that's what I got :)

To calculate the equilibrium constant (Kc) for a reaction, we need to use the concentrations of the reactants and products at equilibrium. The equation for the reaction is:

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

Given that the initial concentration of ammonia (NH3) is 0.0186 M and the equilibrium concentration is 0.0035 M, we can find the change in concentration (Δ[NH3]):

Δ[NH3] = [NH3]equilibrium - [NH3]initial
= 0.0035 M - 0.0186 M
= -0.0151 M

Since the stoichiometry of the reaction is 1:2:3 (1 mole of N2 produces 2 moles of NH3, and 3 moles of H2 produce 2 moles of NH3), the change in concentration for N2 and H2 can be calculated as follows:

Δ[N2] = - (1/2) * Δ[NH3] = - (1/2) * (-0.0151 M) = 0.00755 M
Δ[H2] = - (3/2) * Δ[NH3] = - (3/2) * (-0.0151 M) = 0.02265 M

Now we need to express the equilibrium concentrations of the reactants and product in terms of changes and initial concentrations:

[NH3]equilibrium = [NH3]initial + Δ[NH3] = 0.0186 M + (-0.0151 M) = 0.0035 M
[N2]equilibrium = [N2]initial + Δ[N2] = 0 M + 0.00755 M = 0.00755 M
[H2]equilibrium = [H2]initial + Δ[H2] = 0 M + 0.02265 M = 0.02265 M

Now we can calculate the equilibrium constant (Kc) using the equilibrium concentrations:

Kc = ([NH3]equilibrium^2) / ([N2]equilibrium * [H2]equilibrium)
= (0.0035 M)^2 / (0.00755 M * 0.02265 M)
= 0.00001225 / 0.0001709575
= 0.0716 (approximately)

Therefore, the equilibrium constant (Kc) for the reaction is approximately 0.0716.