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anonymous

  • 4 years ago

Given: N2 + 3H2 <-> 2NH3. At equilibrium [N2] = 0.55M , [H2] = 0.69M, and [NH3] = 0.52M at a certain temperature in a 1.0Lvessel. Some N2 was removed and when equilibrium was established, [H2] = 0.72M. How much N2 was removed

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  1. anonymous
    • 4 years ago
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    so first I found the Ksp , which is 1.4966 and for the values after N2 was removed is it N2 : 0.55-x 3H2 : 0.69 + 0.03 = 0.72 2NH3: 0.52 - 0.02 = 0.50 (3 mol of H2 = 1 mol of N2) but doesnt need to add the 0.01 to N2? cuz when equilibrium is established, N2 will rises too right so shouldnt it be like 0.55 -x + 0.01

  2. Xishem
    • 4 years ago
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    "At equilibrium... [H2] = 0.69M" "...when equilibrium was established, [H2] = 0.72M" What?

  3. anonymous
    • 4 years ago
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    that means the change is 0.03M I 0.69 C 0.03 E 0.72

  4. anonymous
    • 4 years ago
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    the reaction is moving to the reactant side, so the change is +ve

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