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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
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so first I found the Ksp , which is 1.4966 and for the values after N2 was removed is it N2 : 0.55x 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

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.0"At equilibrium... [H2] = 0.69M" "...when equilibrium was established, [H2] = 0.72M" What?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that means the change is 0.03M I 0.69 C 0.03 E 0.72

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the reaction is moving to the reactant side, so the change is +ve
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