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Topic: rate law  (Read 5415 times)

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Offline chay722

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rate law
« on: February 14, 2008, 05:36:02 PM »
2A+B yields C

                           Initial rate of formation of C
Trial   [A]     
1        .40   .20               8.0*10^-3 M/s
2         .80   .40              1.6*10^-3 M/s
3         .80   .80            3.2*10^-3 M/s
4         .60    .60            2.4*10^-3 M/s

Write the rate law for the reaction above in the form Rate=k[A]^x^y.

Now I understand how to get the value of y by doing the following:
Rate3/Rate2= [3.2*10^-3]/[1.6*10^-3]=(.8M/.4M)^y
2=2^y
y=1

But in order to get the value of x, Im unsure because there are no identical values in B's column. Does anyone know how I would go about doing this? Thanks for any help.

Offline enahs

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Re: rate law
« Reply #1 on: February 14, 2008, 08:28:28 PM »
You can make a value.

You have found that B is first order. Therefor if the concentration doubles the rate double. Correct?
So for instance, take trial 1. Double the concentration of [_B], and double the rate. But do not change [A]. Now you can compare trial 1 and 2 in the same way you did 2 and 3; and theoretically it should work out just fine.

P.S. to fix your post so it looks right, edit it so the concentration for B is written like [_B], other wise it is the bold tag.



 

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