Posted by Laura on Friday, March 8, 2013 at 10:35pm.
0.900mol/2L = 0.45M
.........2SO3(g)<--> 2SO2(g)+O2(g)
I........0.45.........0.......0
C........-2x..........2x......x
E.......0.45-2x.......2x......x
The problem tells you that x = 0.130mol O2 is present which makes (O2) = 0.130/2L = 0.065M and (SO2) is twice that (i.e., 0.260mol/2L) = 0.130M. That makes (SO3) = 0.45-0.260 = 0.190 M
Substitute and solve for Kc. I suspect that you forgot it was in 2L and you didn't correct 0.130 mols O2 to M O2 by dividing by 2 L. Of course that messed up SO2 and SO3 too. Hope this helps.
What is the correct answer?
Most of the tutors here don't work the problem We just help solve it without going through the actual calculation.
I have Kc= [(0.065)(0.130)^2]/(0.190^2)
Kc= 0.030429
This answer was incorrect. Can you please help correct me?
Thanks for showing your work.
I screwed up.
In my calculation I said x = 0.130 mols according to the problem. The problem DID state that O2 was 0.130 mols; however, in my calculation I let x = MOLARITY (not mols). I think the easiest way to explain this is to start over and use mols and not molarity.
.........2SO3 ==> 2SO2 + O2
I........0.9........0......0
C........-2x........x.......x
E.......0.9-2x.......x......x
where 0.9 = mols we started with and x = mols O2 with 2x = mols SO2.
(O2) = mols/L = 0.130/2 = 0.065M (the same number as before).
(SO2) = 2x/2L = 2*0.130/2 = 0.130M (the same number as before).
(SO3) = (0.9-2*0.130)/2 = 0.32 (not the same number as before).
Kc = (0.065)(0.130)^2/(0.32)^2 = ?
Check my work.
Sorry to have given you a bummer. The lesson to learn from this is to check what we do. Sometimes it's late, sometimes we just get in a hurry, sometimes we read the problem too fast (or read it wrong), and sometimes we just screw up as I did in this problem..