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### Topic: Determining the Equilibrium Constant  (Read 6617 times)

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#### !!!

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##### Determining the Equilibrium Constant
« on: February 13, 2008, 05:53:10 PM »
A 0.0040-mole sample of S2 (g) is allowed to dissociate in a 0.500-L 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.

#### Sev

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##### Re: Determining the Equilibrium Constant
« Reply #1 on: February 13, 2008, 06:08:16 PM »

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##### Re: Determining the Equilibrium Constant
« Reply #2 on: February 13, 2008, 06:23:00 PM »
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.)

#### Sev

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##### Re: Determining the Equilibrium Constant
« Reply #3 on: February 13, 2008, 06:33:55 PM »
initial [S2] is 0.008M (0.004 mole/0.5L)

At equilibirum, there is 2*10-11 moles of S - so [ S ] is 4*10-11M.

If 2*10-11 moles of S is formed, that means 1*10-11 moles of S2 reacted - this is a negligible amount so effectively [S2] at equilibirum is still 0.008M.

So Kc = (4*10-11)2/0.008 = 2*10-19

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##### Re: Determining the Equilibrium Constant
« Reply #4 on: February 13, 2008, 06:43:28 PM »
Thank you!
I understand the concept now.