Identifying Complex Reaction Orders

Sometimes using simple inspection of trials cannot be used to determine reaction rates

Run # [A]0 [B]0 [C]0 v0

1 0.151 M 0.213 M 0.398 M 0.480 M/s

2 0.251 M 0.105 M 0.325 M 0.356 M/s

3 0.151 M 0.213 M 0.525 M 1.102 M/s

4 0.151 M 0.250 M 0.480 M 0.988 M/s

In this case, there are no simple relationships between the trials

Steps to solving complex rates1. Pick two trials where only one variable changes(Use trials 1 and 3 to isolate the effect of C)2. Set up a proportion of the two reaction rates

Trial 1 0.480 M/s =k (0.151 M)a (0.213 M)b (0.398 M)c

1.102 M/s =k (0.151 M)a (0.213 M)b (0.525 M)c

0.4356 = (0.525 M)c

(0.398 M)c

=(0.7581)c

4. To isolate c, take the natural log of each side

ln (0.4356) = ln (0.7581c)

ln (0.4356) = c ln (0.7581)

c = ln (0.4356) ln (0.7581)

c = 3

Trial 3

3. Now we can cancel out some factors that are the same

5.Now use this reaction order to determine another reaction orderWe can use runs 1 and 4 to isolate b

Trial 1 0.480 M/s =k (0.151 M)a (0.213 M)b (0.398 M)3

1.102 M/s =k (0.151 M)a (0.250 M)b (0.480 M)3Trial 4

0.4858 = (0.8520)b (0.5701)

To isolate b, take the natural log of both sides

ln (0.8521) = b ln (0.8520)

b = ln (0.8521) ln (0.8520)

b = 1

(0.8521) = (0.8520)b

6. Now we only need to use two trials where [A] changes (trials 1 and 2)

Trial 1 0.480 M/s =k (0.151 M)a (0.213 M) (0.398 M)3

0.356 M/s =k (0.251 M)a (0.105 M) (0.325 M)3Trial 2

1.3483 = (0.6016)a (2.029) (1.8376)

0.3616 = (0.6016)a

ln (0.3616) = a ln (0.6016)

a = ln(0.3616) ln (0.6016)

a = 2

So the final rate law is

Rate = k [A]2 [B] [C]3

We can now use any trial to figure out k, as we did last time

A. Power that a reactant is raised to Rate = k[A]2 [B] A has a reaction order of 2 and B has a reaction

order of 1(1 isn’t written, but zero would be!)

B. Overall reaction orderSum of all reaction ordersIn our example, the overall reaction order would be

3

C. Units for rate constant Up until now, we have ignored the units for the rate constant (k).

Units depend on the rate orderThe units must cancel with other units in the rate law

to equal the units for the rate (M/s)

First order reactionrate = k [A]M/s = k (M)k should be in 1/sec or s-1

Second order reactionrate = k [A][B]M/s = k (M2) M . = kM2 * sk should be in 1/M*s or M-1s-1

As you add another “order”, you add another “M” in the reaction rate equation. So the rate law must have another “M-1” to cancel it out.

Rate order Units for K

1st

2nd

3rd

4th M-3 s-1

M-2 s-1

M-1 s-1

s-1