The Foundations of Chemistry

Now we know that the rate law is of the form

ratek[A]x[C]

Dependence on[A]: We use experiments 1 and 2 to evaluate x,because [A] is changed,

[B] does not matter, and [C] is unaltered. The observed rate change is due onlyto the

changed [A].

[A] has been multipliedby a factor of 2.0[A] ratio

The rate changes by a factor of

4.0rate ratio

The exponent xcan be deduced from

rate ratio([A] ratio)x

4.0(2.0)x so x 2 The reaction is second order in [A].

From these results we can write the complete rate-law expression.

ratek[A]^2 [B]^0 [C]^1 or ratek[A]^2 [C]

We can evaluate the specific rate constant, k,by substituting any of the four sets of data into

the rate-law expression we have just derived. Data from experiment 2 give

rate 2 k[A] 22 [C] 2

k3.0 10 ^4 M^2 min^1

The rate-law expression can also be written with the value of kincorporated.

rate3.0 10 ^4 M^2 min^1 [A]^2 [C]

This expression allows us to calculate the rate at which this reaction occurs with any known

concentrations of A and C (provided some B is present). As we shall see presently, changes in

temperature change reaction rates. This value of kis valid onlyat the temperature at which the

data were collected.

You should now work Exercises 17 and 18.

9.6 10 ^6 Mmin^1



(0.40 M)^2 (0.20 M)

rate 2



[A] 22 [C] 2

9.6 10 ^6



2.4 10 ^6

0.40



0.20

662 CHAPTER 16: Chemical Kinetics

Problem-Solving Tip:Check the Rate Law You Have Derived

If the rate law that you deduce from initial rate data is correct, it will not matter which

set of data you use to calculate k.As a check, you can calculate kseveral times, once

from each set of experimental concentration and rate data. If the reaction orders in your

derived rate law are correct, then all sets of experimental data will give the same value

of k(within rounding error); but if the orders are wrong, then the kvalues will vary

considerably.