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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)
  1. 11 July, 02:16
    0
    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
  2. 11 July, 02:23
    0
    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°
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