hslove142331 Posts: 71, Reputation: 1 Junior Member #1 Mar 3, 2010, 04:13 PM
what will their equilibrium concentrations be?
At 25 °C, Kc= 0.145 for the following reaction in the solvent CCl4.
2BrCl <------------> Br2+Cl2
If the initial concentrations of Br2 and Cl2 are each 0.0195 M, what will their equilibrium concentrations be?

I tried 4 times.. but my answers were all wrong. I have only one chance to get correct.

I don't know I did like this.

(0.0195-x)^2/(2x)^2=0.145

0.0195-x=0.7616x

x=0.011

M=0.0085, right?

I am not sure even I didn't try to put that unsure answer.
And I am so confusing on I/C/E steps.
I know I and E
But I don't know how to figure Out on C that which side have(-)or(+)

 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #2 Mar 4, 2010, 07:15 AM

Where did you get 0.0085?

You worked the problem with concentrations of 0.0195, which is 0.0195 in one litre. So imagine you have only 0.0195 mol each of Br2 and Cl2 in a one litre closed container. You end up with x = 0.011, so 2x (the amount of moles of BrCl) becomes 0.022 moles.

0.022 moles in one litre is 0.022 M, the concentration of your BrCl.

And to check whether your square root should be positive or negative, use the value of x in any of the expressions you used.

To find moles of Br2 (or Cl2) remaining, you use (0.0195 - 0.011) which is positive (that is you must have some Br2 left), as compared to the other way where you get x = 0.081, giving (0.0195 - 0.081) which cannot be since the Br2 cannot get in deficit.
 hslove142331 Posts: 71, Reputation: 1 Junior Member #3 Mar 4, 2010, 08:16 AM
But,, its okay.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #4 Mar 4, 2010, 08:23 AM

Well, you didn't read my post either. I clearly said that 0.022 M is the concentration of the BrCl =/...

Plus, later I said that to have the concentration of Cl2 or Br2, use 0.0195 - 0.011.

I just happened to think that you were looking for the concentration of BrCl.

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