A sample of ammonia gas was allowed to come to equilibrium at 400 K.
2NH₃(g)
N₂(g) + 3H₂(g)
At equilibrium,it was found that the concentration of H₂ was 0.0551 M,the concentration of N₂ was 0.0183 M,and the concentration of NH₃ was 0.383 M.What is K_c for this equilibrium?

Question

A sample of ammonia gas was allowed to come to equilibrium at 400 K.
2NH₃(g)

A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH<sub>3</sub>(g) N<sub>2</sub>(g) + 3H<sub>2</sub>(g) At equilibrium,it was found that the concentration of H<sub>2</sub> was 0.0551 M,the concentration of N<sub>2</sub> was 0.0183 M,and the concentration of NH<sub>3</sub> was 0.383 M.What is K<sub>c</sub> for this equilibrium? A) 3.97 × 10<sup>-3</sup> B) 1.58 × 10<sup>-5</sup> C) 2.10 × 10<sup>-5</sup> D) 2.40 × 10<sup>-1</sup> E) 2.65 × 10<sup>-3</sup>

N₂(g) + 3H₂(g)
At equilibrium,it was found that the concentration of H₂ was 0.0551 M,the concentration of N₂ was 0.0183 M,and the concentration of NH₃ was 0.383 M.What is K_c for this equilibrium?

Quizplus · Accepted Answer

The equilibrium constant expression is: K_c = [N₂][3H₂]/[NH₃]^2 Substituting the given concentrations: K_c = (0.0183)(0.0551)^3/(0.383)^2 K_c = 2.10 × 10^-5 Therefore, the correct answer is choice C.

A Sample of Ammonia Gas Was Allowed to Come to Equilibrium