1. Reaction Rate Orders
Chapter 14

2. Rate Laws Review
The overall reaction order can be found by adding the exponents on the reactants in the rate law.
Rate = k[A]m[B]n 2

3. Integrated Rate Laws
Using calculus to integrate the rate law for a first-order process gives us ln
[A]t
[A]0
= −kt
Where
[A]0 is the initial concentration of A.
[A]t is the concentration of A at some time, t, during the course of the reaction. 3

4. Integrated Rate Laws Manipulating this equation produces… ln [A]t [A]0
= −kt
ln [A]t − ln [A]0 = − kt
ln [A]t = − kt + ln [A]0
…which is in the form
y = mx + b 4

5. First-Order Processes
ln [A]t = -kt + ln [A]0
Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k. 5

6. First-Order Processes
Consider the process in which methyl isonitrile is converted to acetonitrile.
CH3NC
CH3CN 6

7. First-Order Processes
CH3NC
CH3CN
This data was collected for this reaction at 198.9°C. 7

8. First-Order Processes
When ln P is plotted as a function of time, a straight line results.
Therefore,
The process is first-order.
k is the negative of the slope: 5.1  10-5 s−1. 8

9. Half-Life
Half-life is defined as the time required for one-half of a reactant to react.
Because [A] at t1/2 is one-half of the original [A],
[A]t = 0.5 [A]0. 9

10. Half-Life For a first-order process, this becomes 0.5 [A]0 [A]0 ln
= −kt1/2
ln 0.5 = −kt1/2
−0.693 = −kt1/2
= t1/2
0.693 k
NOTE: For a first-order process, the half-life does not depend on [A]0. 10

11. Half Life
Since the half life is independent of the original concentration, half life of a first order reaction can be used as a representation of the rate constant because the half life and rate constant are inversely proportional to each other.
The shorter the half life, the faster the reaction rate
Real life examples of first order reactions:
Radioactive decay

12. Second-Order Processes
Similarly, integrating the rate law for a process that is second-order in reactant A, we get 1
[A]t
= −kt +
[A]0
also in the form
y = mx + b 12

13. Second-Order Processes1
[A]t
= −kt +
[A]0
So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k. 13

14. Second-Order Processes
The decomposition of NO2 at 300°C is described by the equation
NO2 (g)
NO (g) + 1/2 O2 (g)
and yields data comparable to this:
Time (s)
[NO2], M
0.0
50.0
100.0
200.0
300.0 14

15. Second-Order Processes
Graphing ln [NO2] vs. t yields:
The plot is not a straight line, so the process is not first-order in [A].
Time (s)
[NO2], M
ln [NO2]
0.0
−4.610
50.0
−4.845
100.0
−5.038
200.0
−5.337
300.0
−5.573 15

16. Second-Order Processes
Graphing ln 1/[NO2] vs. t, however, gives this plot.
Because this is a straight line, the process is second-order in [A].
Time (s)
[NO2], M
1/[NO2]
0.0
100
50.0
127
100.0
154
200.0
208
300.0
263 16

17. Zeroth Order processes
[A] vs time when graphed creates straight line
The negative slope is k
Reaction rate is independent of the concentration of a reactant
Changing its concentration has no effect on the speed of the reaction
Rate stays the constant

18. The number of molecules involved in a collision can tell you the reaction rate order
Unimolecular reactions = first order
Bimolecular reactions = second order