Ishwar S. answered 09/06/18
Tutor
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University Professor - General and Organic Chemistry
Hello Billy
The rate law expression for this reaction is:
Rate = k [A]x [B]y
where x and y are the orders of the reaction with respect to each reactant.
To determine the order for A, look at the data and select the Trials where the concentration of B remains constant. In this case, it is Trials #1 and #2. The concentration of A in Trial #1 is 0.15 M and in Trial #2, it is 0.30 M. Now we can setup a ratio of these 2 trials as:
(Rate2 / Rate1) = ([A]2 / [A]1)x
Plug in the numbers from the data.
(2.6 x 10-2 / 1.3 x 10-2) = (0.30 / 0.15)x
(2) = (2)x
What is the value of the exponent x? x = 1. Therefore, the reaction is 1st order with respect to reactant A.
Now repeat the above process for reactant B. In Trials #1 and #3, the concentration of A remains constant. The concentration of B in Trial #1 is 0.12 M and in Trial #3, it is 0.24 M. Now we can setup a ratio of these 2 trials as:
(Rate3 / Rate1) = ([B]3 / [B]1)y
Plug in the numbers from the data.
(5.2 x 10-2 / 1.3 x 10-2) = (0.24 / 0.12)y
(4) = (2)y
What is the value of the exponent y? y = 2. Therefore, the reaction is 2nd order with respect to reactant B.
Plug in the numbers from the data.
(5.2 x 10-2 / 1.3 x 10-2) = (0.24 / 0.12)y
(4) = (2)y
What is the value of the exponent y? y = 2. Therefore, the reaction is 2nd order with respect to reactant B.
The rate law can now be written as: Rate = k [A] [B]2
To determine the value of the rate constant, k, you can take the data from any of the 3 trials, and plug in the values into the above rate law. Let's choose Trial #1.
[A] = 0.15 M
[B] = 0.12 M
Rate = 1.3 x 10-2 M min-1
1.3 x 10-2 M min-1 = k (0.15 M) (0.12 M)2
1.3 x 10-2 M min-1 = k(2.2 x 10-3 M3)
k = 1.3 x 10-2 M min-1 / 2.2 x 10-3 M3 = 5.9 M-2 min-1