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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: !!! on February 13, 2008, 05:53:10 PM

A 0.0040mole sample of S2 (g) is allowed to dissociate in a 0.500L flask at 1000 K. When equilibrium is established, 2.0 x 10^11 moles of S (g) is present. What is the value of Kc for the reaction:
S2 (g) <> 2S (g)
The answer is 2.0 x 10^19, but my calculations are way off from that.
Can someone please show and point out the right steps so I know what I did incorrectly? Thank you.

Can you post your working

S2 at initial concentration: 0.0040 M
2S at initial concentration: 0 M
Change of S2: x
Change of 2S: 2x
S2 at equilibrium concentration: 0.0040 M  x
2S at equilibrium concentration: 2x
[S2]i = 0.0040 M / 0.500 L = 0.008 M
eq = 2.0 x 10^11 mol / 0.500 L = 4 x 10^11 M
2x = 4 x 10^11
x = 2 x 10^11 M
Kc = ^2 / [S2]
= (4 x 10^11)^2 / 2 x 10^11
= 1.6 x 10^21 / 2 x 10^11
Kc= 8 x 10^33 (It's way too off.)

initial [S_{2}] is 0.008M (0.004 mole/0.5L)
At equilibirum, there is 2*10^{11} moles of S  so [ S ] is 4*10^{11}M.
If 2*10^{11} moles of S is formed, that means 1*10^{11} moles of S_{2} reacted  this is a negligible amount so effectively [S_{2}] at equilibirum is still 0.008M.
So K_{c} = (4*10^{11})^{2}/0.008 = 2*10^{19} :)

Thank you!
I understand the concept now.