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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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Answer:

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°

Answer:

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

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