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97lmn
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N2(g) + 2O(g) <--> 2NO2(g)
At a certain temperature, the follwoing are the equilibrium concentrations for the above reaction:
[N2] = 8.0 mol/L
[O2] = 2.0 mol/L
[NO2] = 4.0 mol/L
Determine the equilibrium constant.
My answer..
Ke = [NO2]^2 / [N2][O2]^2
Ke = [4.0]^2 / [8.0][2.0]^2
Ke = 0.5
Correct?
At a certain temperature, the follwoing are the equilibrium concentrations for the above reaction:
[N2] = 8.0 mol/L
[O2] = 2.0 mol/L
[NO2] = 4.0 mol/L
Determine the equilibrium constant.
My answer..
Ke = [NO2]^2 / [N2][O2]^2
Ke = [4.0]^2 / [8.0][2.0]^2
Ke = 0.5
Correct?