Question

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.

A + B yields products
Trial [A] [B] Rate
1 0.20 M 0.15 M 3.2 × 10-2 M/min
2 0.20 M 0.30 M 6.4 × 10-2 M/min
3 0.40 M 0.30 M 25.6 × 10-2 M/min

Answer 1

Recall the rate law first.

Answer 2

Is it Rf = k[A]^x[B]^y

Answer 3

Yup. Now you look at trials 1 and 2. A is constant and B is twice as much, and the rate is twice faster. So what can you say about the relationship between the rate and the non-constant (B)?

Answer 4

However much the rate goes up, the non-constant goes up the same amount

Answer 5

Well it's actually the reverse since the rate is dependent on the reactants. So you know the rate goes up as many times as B goes up. So y should be 1.

Now compare trials 2 and 3. B is constant, A is doubled. What about the rate?

Answer 6

The rate goes up by four

Answer 7

Right. So when A doubles, the rate quadruples. Meaning the rate is the square of the number A is multiplied by. Can you tell what x is then?

Answer 8

Would it be 2?

Answer 9

Yup! So now you already have Rf=k[A]^2[B]
Now you just plug in values from any trial to get k.

Answer 10

Rf = k[0.2]^2[0.3]

Rf = k[0.04][0.3]

Rf = k[0.012]

Answer 11

Is that all? Or do I have to do something else?

Answer 12

Well, Rf is the rate you are given. So since you're using trial 2, your Rf is 6.4 × 10-2 M/min. Equate that with what you have and solve for k.

Answer 13

So k would equal 5.33?

Answer 14

Yup looks right to me. So now you have k, x, and y, your reaction equation is...?

Answer 15

I don't remember how to do the reaction equation

Answer 16

That's the first thing we talked about :)

Answer 17

Ohh! Rf = 5.33[A]^2[B]^1

Answer 18

Yup! You don't need the power sign for B though since it's just to the power of 1. Good job.

Answer 19

Thank you for helping

Answer 20

No problem.