Question

Help with this rate law question??

Answer 1

Here you need to use the initial rates method. Do you know how that works?

Answer 2

The rate law is given by:

rate = k[A]^n[B]^m, correct?

Answer 3

Yes.

Answer 4

But I don't know how to find the answer @QRAwarrior.

Answer 5

Alright let me think

Answer 6

I have some pictures that I can show you that will help you solve this step-by-step.

Answer 7

Okay. :)

Answer 8

If you can wait just a few minutes, I will upload them and show you

Answer 9

I have instructions instead:
1. Examine the data and find two experiments in which the concentration of one reactant changes, but the other concentrations are the same

Answer 10

Trial 1 and 2? Because A is 0.30M for both of them.

Answer 11

You basically have to do this:
\[\frac{[reactant]_1}{[reactant]_2} = \frac{[rate]_1}{[rate]_2} \]

Answer 12

OHH Now I understand; what you are doing is dividing one rate law equation over another

Answer 13

That's what my instruction tell me to do, but I don't know what to do from there.

Answer 14

So this is what you are doing:
\[\frac{k[reactant_1]^m[reactant_2]^n}{k[reactant_1]^m[reactant_2]^n} = \frac{[rate_1]}{[rate_2]}\]

Answer 15

So instead of reactant_1, imagine it said A, and instead of reactant_2, imagine it said "B"

Answer 16

Suppose you pick trial 1 and trial 2.
Then, you put all the information you know for trial 1 in the numerator, while all the information you know for trial 2 in the denominator expresion.

Answer 17

Then, see what cancels out.

Answer 18

You SHOULD get cancellation

Answer 19

esp. with the k.

Answer 20

I still don't understand. :(

Answer 21

I will do the first one for you

Answer 22

Let's pick trial 1 and trial 2

Answer 23

\[\frac{k[0.30M]^n[0.25M]^m}{k[0.30M]^n[0.50M]^m}=\frac{[1.2*10^-2 M/\min]}{[4.8*10^-2 M/\min]}\]

Answer 24

So it is 0.25^-2M / min?

Answer 25

So what I did here was sub in the information I know for trial 1 into the numerator:
[A] = 0.30M
[B] = 0.25M
rate_1 = 1.2*10^-2 M/min.

Similarly, I did for trial 2 in the denominator.

Answer 26

@JBrightman, I also have to let you know that it does not matter whether we put trial 1 in the numerator and trial 2 in the denominator. We could have correctly put trial 2 in the denominator, and trial 1 in the numerator. Try that

Answer 27

Yes

Answer 28

So do you see cancellation? Can you solve for a variable?

Answer 29

Okay, so it doesn't matter where each of the trials go as long as one is being divided by the other.

Answer 30

Isn't the answer 0.25?

Answer 31

You want to solve for m.

Answer 32

On the left side, it is 0.5, right?

Answer 33

(1/2)^m = 1/4 <-- do you get this?

Answer 34

What? How do you get 1/4?

Answer 35

Let me help you:

Answer 36

\[\frac{k[0.30M]^n[0.25M]^m}{k[0.30M]^n[0.50M]^m}=\frac{[1.2*10^-2 M/\min]}{[4.8*10^-2 M/\min]}\]
^ We have this right?

Answer 37

Yes.

Answer 38

\[\frac{[0.30M]^n[0.25M]^m}{[0.30M]^n[0.50M]^m}=\frac{[1.2*10^-2 M/\min]}{[4.8*10^-2 M/\min]}\]

Answer 39

right?

Answer 40

it is a constant

Answer 41

Okay.

Answer 42

\[\frac{[0.25M]^m}{[0.50M]^m}=\frac{[1.2*10^-2 M/\min]}{[4.8*10^-2 M/\min]}\]

Answer 43

OHHH! I understand, it is 0.5^m = 0.25

Answer 44

Because you have the EXACT same term on the top and bottom: [0.30M]^n is seen in the numerator and the denominator, so it cancels out.

Answer 45

But how do you solve for m from there?

Answer 46

Now, using exponent laws:
\[{(\frac{[0.25M]}{[0.50M]}})^m=\frac{[1.2*10^-2 M/\min]}{[4.8*10^-2 M/\min]}\]