Consider the following equilibrium process at 700°C:

2H2 + S2 ↔ 2H2S

Analysis shows that there are 2.50 moles of H2, 1.35x10^-5 mole of S2, and 8.70 moles of H2S present in a 12.0-L flask. Calculate the equilibrium constant Kc for the reaction.

4 answers

ASSUMING that the moles indicated are present at equilibrium, then calculate concns of each by mole/L, then plug into the Keq experession. If you need more assistance, please explain what you don't understand about the procedure. Frankly, the numbers don't look right to me but I may have overlooked something.
1.08*10^7
The equilibrium constant can be calculated by: Kc= [product]/[reactants].
Before you know the exact concentration variables necessary to do this you need to calculate the moles present in your reaction. In the example you provided the moles per liter are:
H2S= 8.7M/12.0L= 0.73 moles
H2= 2.5M/12.0 L= 0.21 moles
S2= (1.35 x 10^-5M)/12.0L= 1.13 x 10^-6
Now that you have the exact proportions of moles are are able to insert these variables into the above equation:
Kc= [product]/[reactants]=[0.73]^2/[0.21]^2[1.13 x 10^-6]=9.3 x 10^6
Remember during calculations that you need to place your coefficient from the reaction into the exponent of your Kc equation. That is why I took [0.73]^2 rather than just leaving it at [0.73]. This Kc tells us that the reaction is shifted to the right and we have a high proportion of hydrogen sulfide in this reaction when equilibrium is achieved.
Thinking critically An analysis for copper was performed on two pure solids. One solid was found to contain 43.0% copper. Could these solids be samples of the same copper-containing compound? Explain your answer.