Asked by Rubert
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?
If [N2] = 1.4 M, [H2] = 0.80 M, and [NH3] = 0.35 M, what is Keq under the conditions of this flask?
Answers
Answered by
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.
Keq = (NH3)^2/(N2)(H2)^3
Plug in the numbers and turn the crank.
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