Initially a 1 L vessel contains 10 moles of NO2 and 6 moles of O2. At equilibrium, the vessel contains 8.8 moles of NO2. Determine Kc.
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3 answers
Determine Kc for which reaction?
I'm not sure, this was all that I was given
I will assume that the reaction is as follows and that the concn of NO2 initially is 10 mol/1L = 10 M and 8.8 mols/L or 8.8 M at the end.
......2xNO2 + O2 ==> NO + O3
I.....10....6......0.....0
C.....-x....-x.....x.....x
E....10-x..6-x.....x.....x
Therefore, x must be 10-8.8 = 1.2 which makes 6-1.2 = 4.8 with both NO and O3 at the E line being 1.2
To find Kc = (NO)(O3)/(NO2(O2) substitute the E line values into the Kc expression and evaluate Kc.
If you find that the equation is different than the one I've assumed, the process will be the same process. Remember to balance the equation first.
......2xNO2 + O2 ==> NO + O3
I.....10....6......0.....0
C.....-x....-x.....x.....x
E....10-x..6-x.....x.....x
Therefore, x must be 10-8.8 = 1.2 which makes 6-1.2 = 4.8 with both NO and O3 at the E line being 1.2
To find Kc = (NO)(O3)/(NO2(O2) substitute the E line values into the Kc expression and evaluate Kc.
If you find that the equation is different than the one I've assumed, the process will be the same process. Remember to balance the equation first.