The equilibrium concentrations for the reaction N₂ (g)+ 3H₂ (g)⇔ 2NH₃ (g)at 673 K are [H₂] = 0.63 M, [N₂] = 0.45 M, and [NH₃] = 0.24 M. The value of K_e_q is ________.

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is: K_e_q = [NH₃]^2/([N₂]*[H₂]^3) Substituting the given values, we get: K_e_q = (0.24)^2/((0.45)*(0.63)^3) = 0.51 Hence, the answer is (C) 0.51.

The Equilibrium Concentrations for the Reaction N2 (G)+ 3H2 (G)⇔

The Equilibrium Concentrations for the Reaction N₂ (G)+ 3H₂ (G)⇔