Consider the following equilibrium:N 2 (g) + 3H 2 (g) 2NH 3 (g) with K = 2.3 × 10 -6 . 1.00 mol each of all reactants and products is placed in a 1.00-L container. -Calculate the equilibrium concentration of NH₃(g).

Question

Quizplus · Accepted Answer

Using the equilibrium constant expression, we have: K = [NH₃] / ([N₂] *[H₂]^3) At equilibrium, let the concentration of NH₃(g) be x M. Then the concentration of each of the other reactants/products will be (1.00 - x) M. Substituting these values into the equilibrium expression gives: K = x / ([1.00 - x]^4) Solving for x gives: x = K * [1.00 - x]^4 = 2.3 × 10^-6 * [1.00 - x]^4 Taking the fourth root of both sides and simplifying gives: 1.00 - x = 0.044 x = 0.956 M Therefore, the equilibrium concentration of NH₃(g) is 0.956 M, or 7.3 × 10^-3 M, which is closest to choice (C).

Consider the Following Equilibrium:N2(g) + 3H2(g) 2NH3(g)

Consider the Following Equilibrium:N₂(g) + 3H₂(g) 2NH₃(g)