From the data, calculate the rate law expression for the reaction of A with B. I will attach a photo with the data in the comments section as it will be easier to decipher. I have concluded that the rate law would be rate = k[A]^m[B]^n, but how is it calculated given the data provided? They don't give the equation or the rate, just the time and M of each reactant for the three experiments. Please help. Lab is due Monday and this is a prelab question.

Question

hlilly2413 · Answer

Here is a jpg of the data. If you need another format to open the data let me know and i will provide it.

cuanchi · Answer

I assume that A and B are both reactants and the time is the time to produce or consume a certain amount of product or reactant.
then with the time you can calculate the rate of the reaction for each of the three different experiments.

rate = [product]/time
assume that the [product ] = 1 to make the calculation easier
you will have a rate of reaction of 
experiment 1:   1/30= 0.0333 moles/sec
experiment 2:   1/15= 0.0666 moles/sec
experiment 3:   1/60= 0.0166 moles/sec

Then you have to observe that between the experiment 1 and 2 the concentration of B remain constant and the concentration of A increase twice. The rate of reaction between the experiment 1 and 2 also increase twice from 0.033 to 0.066 moles/s. Then we can said that the reaction is the "first" order for the component A because if you increase the concentration of A double you reaction rate also increase double k= [A]^1 [B]^n. If you increase doble the concentration of A and the rate of the reaction doesnt change you will said that the reaction is of "zero order" respect to A. And if you increase the concentration of A to double and the reaction rate increase 4 times you will have a reaction of "second order" for A.

can you figure out what is the order (m) of the reaction for B comparing the rate of the reaction and the change of concentration of B between the experiment 2 and 3?

hlilly2413 · Answer

I was able to deduce that A was 1st order because when the concentration of A was doubled, the time it took for the reaction to occur decreased. Now I can see that the rate doubled as well, thus proving first order. B is where I was having an issue. I could see, ok the concentration of B decreased by half, but the time increased by a factor of 4. Now I can see that the rate decreased by a quarter...is it because 1/2^2 is .25? I believe this would be second order. Please correct me if I am wrong.

hlilly2413 · Answer

I had an issue proving it mathematically because I wasn't aware that we could plug in any number for the concentration of the product. I see that the ratios would come out to be the same as long as you use the same concentration of the product for each calculation. thank you

cuanchi · Answer

Yes you are perfectly right!!

hlilly2413 · Answer

Thank you. Whew.