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 N₂.

Question

Quizplus · Accepted Answer

The equilibrium constant for the reaction is K = [N2]^2/[NH3]^4 = 1.6 x 10^-4. Substituting the given concentrations into this expression, we get:1.6 x 10^-4 = [N2]^2/(0.5 M)^4[N2]^2 = 1.6 x 10^-4 * (0.5 M)^4[N2]^2 = 1.6 x 10^-4 * 0.0625 M^4[N2]^2 = 1.0 x 10^-5 M^4[N2] = sqrt(1.0 x 10^-5 M^4)[N2] = 1.0 x 10^-2.5 M[N2] = 1.5 MTherefore, the equilibrium concentration of N2 is 1.5 M.

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

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