For the reaction 2 NH 3 (g) - ⇀ ↽ - 3 H 2 (g) + N 2 (g) 2NH3 (g) ↽--⇀ 3H2 (g) + N2 (g) the equilibrium concentrations were found to be [ NH 3 ] = 0.250 M [NH3]=0.250 M, [ H 2 ] = 0.440 M [H2]=0.440 M, and [ N 2 ] = 0.800 M [N2]=0.800 M. What is the equilibrium constant for this reaction?

Question

Dax Arias

Chemistry

11 July, 01:57

For the reaction 2 NH 3 (g) - ⇀ ↽ - 3 H 2 (g) + N 2 (g) 2NH3 (g) ↽--⇀ 3H2 (g) + N2 (g) the equilibrium concentrations were found to be [ NH 3 ] = 0.250 M [NH3]=0.250 M, [ H 2 ] = 0.440 M [H2]=0.440 M, and [ N 2 ] = 0.800 M [N2]=0.800 M. What is the equilibrium constant for this reaction?

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Answers (2)

Know the Answer?

Lexi Flynn · Answer 1

The equilibrium constant Kc for this reaction is 1.09

Explanation:

Step 1: Data given

Concentrations at equilibrium are:

[NH3] = 0.250 M

[H2] = 0.440 M

[N2] = 0.800 M

Step 2: The balanced equation

2NH3 (g) ⇆ 3H2 (g) + N2 (g)

Step 3: Calculate the equilibrium constant Kc

Kc = [N2][H2]³/[NH3]²

Kc = [0.800][0.440]³ / [0.250]²

Kc = 1.09

The equilibrium constant Kc for this reaction is 1.09

Harper Pierce · Answer 2

Equilibrium constant is: 1.09

Explanation:

The proposed equilibrium is:

2NH₃ (g) ⇄ 3H₂ (g) + N₂ (g)

Concentrations are: [NH₃] = 0.250 M; [H₂] = 0.440 M; [N₂] = 0.800 M

Expression for Kc is: [H₂]³. [N₂] / [NH₃]²

We replace data → Kc = (0.440³. 0.800) / 0.250²

Kc = 1.09

Remember that equilibrium constant has no units and it depends on T°