At 674.5 K the equilibrium concentrations of N2, H2, and NH3 were 0.1614 M, 0.1128 M, and 0.005892 M, respectively. What is the equilibrium constant for the reaction below?

2 NH3(g) → N2(g) + 3 H2(g)

Substitute the equilibrium concentrations given in the problem into the Kc expression and solve for K.

To find the equilibrium constant for the reaction, we need to use the concentrations of the reactants and products at equilibrium. The equilibrium constant is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficient.

In this case, the balanced equation is 2 NH3(g) → N2(g) + 3 H2(g).

At equilibrium, the concentrations of the reactants and products are given as follows:
[N2] = 0.1614 M
[H2] = 0.1128 M
[NH3] = 0.005892 M

To calculate the equilibrium constant (K), we use the following formula:

K = ([N2]^n [H2]^m) / [NH3]^p

Where n, m, and p are the stoichiometric coefficients of N2, H2, and NH3, respectively.

In this case, n = 1, m = 3, and p = 2.

Plugging in the values, we have:

K = ([N2]^1 [H2]^3) / [NH3]^2
= (0.1614^1 * 0.1128^3) / (0.005892^2)

Calculating this expression, we get:
K = 85.6

Therefore, the equilibrium constant (K) for the given reaction is 85.6.