1.00 mol of pure NO2 in a 40.0 liter container and we allowed it to react...
1.00 mol of pure NO2 in a 40.0 liter container and we allowed it to react per the reaction listed below at 150°C. At equilibrium, the concetration of NO in the container is found to be 0.0015 M. Use that information to calucalte the equilibrium constant for the following reaction.
2NO2 (g) ? 2NO (g) = O2 (g)
Homework Answers
2 NO2 (g) = 2 NO (g) + O2 (g)
Initial [NO2] = moles / volume in L = 1.00 / 40.0 = 0.0250 M
At equilibrium,
[NO] = 0.00150 M
[NO2] = 0.0250 - 0.00150 = 0.0235 M
[O2] = (1/2)[NO] = 0.00150 / 2 = 0.000750 M
Kc = [NO]2[O2] / [NO2]2
Kc = (0.00150)2(0.000750) / (0.0235)2
Kc = 3.06 * 10-6
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