# Chemistry

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At a certain temperature, 0.900 mol of SO3 is placed in a 2.00-L container.

2SO3(g)<--> 2SO2(g)+O2(g)

At equilibrium, 0.130 mol of O2 is present. Calculate Kc.

Kc= ?

Please post the answer if possible or the steps in detail; I have attempted this problem many times and have been unsuccessful.

• Chemistry -

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.

• Chemistry -

• Chemistry -

Most of the tutors here don't work the problem We just help solve it without going through the actual calculation.

• Chemistry -

I have Kc= [(0.065)(0.130)^2]/(0.190^2)
Kc= 0.030429

• ooops--Chemistry -

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..

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