Homework Help: Chemistry

Posted by Saira on Thursday, February 5, 2009 at 9:55pm.

A mixture initially contains A,B, and C in the following concentrations: A1 = 0.300 M, B1 = 1.15M, and C1 = 0.350 M. The following reaction occurs and equilibrium is established:

A+2B <---> C

At equilibrium, A2 = 0.190M and C2 = 0.460M.

Calculate the value of the equilibrium constant, Kc

I am not sure what to do:

This is what i did but i think its wrong:

SO since,

Kc= (C)^C/(A)^a*(B)^b

(0.350)/(0.300)(1.15)=(0.460/(0.190)(B)^2
=(0.350)/(0.345)/(0.460/(0.190)(B)^2

What i did is square root both sides to get rid of B2,then i solve for b i got 1.53.

Then i sub that in the equation:

Kc= 0.460/ (0.190)(1.53)

BUt the answer is wrong

Chemistry - Saira, Thursday, February 5, 2009 at 10:21pm
Never mind i got the right answer worked it out again
Chemistry - DrBob222, Friday, February 6, 2009 at 12:05am
K = 2.80 is what I obtained when I worked the problem. But I think it is
Kc = 0.460/(0.190)(0.930)^2 = 2.80

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