Sometimes using simple inspection of trials cannot be used to determine reaction rates Run #[A] 0 [B] 0 [C] 0 v0v M0.213 M0.398 M0.480 M/s M0.105 M0.325 M0.356 M/s M0.213 M0.525 M1.102 M/s M0.250 M0.480 M0.988 M/s In this case, there are no simple relationships between the trials Steps to solving complex rates 1. 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 M/s=k(0.151 M) a (0.213 M) b (0.398 M) c M/s=k(0.151 M) a (0.213 M) b (0.525 M) c = (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 ( c ) 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 order We can use runs 1 and 4 to isolate b Trial M/s=k(0.151 M) a (0.213 M) b (0.398 M) M/s=k(0.151 M) a (0.250 M) b (0.480 M) 3 Trial = (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 M/s=k(0.151 M) a (0.213 M)(0.398 M) M/s=k(0.251 M) a (0.105 M)(0.325 M) 3 Trial = (0.6016) a (2.029) (1.8376) = (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 order Sum of all reaction orders In 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 order The units must cancel with other units in the rate law to equal the units for the rate (M/s) First order reaction rate = k [A] M/s = k (M) k should be in 1/sec or s -1 Second order reaction rate = k [A][B] M/s = k (M 2 ) M. = k M 2 * s k should be in 1/M*s or M -1 s -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 orderUnits for K 1 st 2 nd 3 rd 4 th M -3 s -1 M -2 s -1 M -1 s -1 s -1