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Suppose that you are running this reaction in a sealed flask. N2 (g) + 3 H2 (g) <--> 2 NH3 (g)

If [N2] = 1.4 M, [H2] = 0.80 M, and [NH3] = 0.35 M, what is Keq under the conditions of this flask?
4 years ago

Answers

DrBob222
You can't answer this question unless you know those concentrations prevail at equilibrium. If at equilibrium, then
Keq = (NH3)^2/(N2)(H2)^3
Plug in the numbers and turn the crank.
4 years ago

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