1. The Changes of Concentration with Time
Integrated Rate Laws - 2nd Order Reactions

2. Introduction
A second order reaction is one whose rate depends on the square of the concentration of a single reactant.
A → products
Rate = k [A]2

3. Introduction
A second order reaction is one whose rate depends on the square of the concentration of a single reactant.
A → products
Rate = k [A]2
We also know that the rate of the reaction is equal to the rate of change in [A] with respect to time.
Rate = −
∆[A] ∆t

4. Introduction
Putting these two rate equations together, we get the differential rate law.
Rate = k [A]2 = −
∆[A] ∆t

5. Introduction
Putting these two rate equations together, we get the differential rate law.
k [A]2 = −
∆[A] ∆t

6. Integrated Rate Law
Putting these two rate equations together, we get the differential rate law.
k [A]2 = −
By integrating the equation, we get the relationship between [A] and time, the integrated rate law.
− = kt
∆[A] ∆t 1 1
[A]t
[A]0

7. Integrated Rate Law We can rearrange the integrated rate law.
− = kt ➔ = kt + 1 1 1 1
[A]t
[A]0
[A]t
[A]0

8. Integrated Rate Law We can rearrange the integrated rate law. = kt + 1
This puts it into the familiar form of the slope intercept equation of a line. 1 1
[A]t
[A]0

9. Integrated Rate Law We can rearrange the integrated rate law. = kt + 1
y = mx + b
This puts it into the familiar form of the slope intercept equation of a line. 1 1
[A]t
[A]0

10. Integrated Rate Law We can rearrange the integrated rate law. = kt + 1
This puts it into the familiar form of the slope intercept equation of a line. 1 1
[A]t
[A]0

11. Integrated Rate Law We can rearrange the integrated rate law. = kt + 1
This puts it into the familiar form of the slope intercept equation of a line. 1 1
[A]t
[A]0

12. Integrated Rate Law We can rearrange the integrated rate law. = kt + 1
This puts it into the familiar form of the slope intercept equation of a line.
If we plot [A] vs. time, we get a curved line. 1 1
[A]t
[A]0

13. Integrated Rate Law We can rearrange the integrated rate law. = kt + 1
This puts it into the familiar form of the slope intercept equation of a line.
If we plot [A] vs. time, we get a curved line.
If we plot 1/[A] vs. time, we get a straight line. 1 1
[A]t
[A]0

14. Integrated Rate Law We can rearrange the integrated rate law. = kt + 1
This puts it into the familiar form of the slope intercept equation of a line.
If we plot [A] vs. time, we get a curved line.
If we plot 1/[A] vs. time, we get a straight line.
and the slope of the line is k. 1 1
[A]t
[A]0

15. Integrated Rate Law We can rearrange the integrated rate law. = kt + 1
This puts it into the familiar form of the slope intercept equation of a line.
If we plot [A] vs. time, we get a curved line.
If we plot 1/[A] vs. time, we get a straight line.
and the slope of the line is k.
All we need is one experiment with [A] and time data. 1 1
[A]t
[A]0

16. Sample Problem Such as this one: t (s) [A] (M) 1/[A] 0.0100 100 50
0.0100
100 50
0.0085
118
0.0074
136
150
0.0065
154
200
0.0058
172
250
0.0053
190

17. Sample Problem Such as this one: t (s) [A] (M) 1/[A] 0.0100 100 50
0.0100
100 50
0.0085
118
0.0074
136
150
0.0065
154
200
0.0058
172
250
0.0053
190

18. Sample Problem Such as this one: t (s) [A] (M) 1/[A] 0.0100 100 50
0.0100
100 50
0.0085
118
0.0074
136
150
0.0065
154
200
0.0058
172
250
0.0053
190

19. Sample Problem Such as this one: The 1/[A] vs. time is linear. t (s)
0.0100
100 50
0.0085
118
0.0074
136
150
0.0065
154
200
0.0058
172
250
0.0053
190

20. Sample Problem Such as this one: We calculate k from the slope. t (s)
0.0100
100 50
0.0085
118
0.0074
136
150
0.0065
154
200
0.0058
172
250
0.0053
190

21. Sample Problem Such as this one:
k = ∆(1/[A])/∆t = ([190] - [100])/(250 s - 0 s)
t (s)
[A] (M)
1/[A]
0.0100
100 50
0.0085
118
0.0074
136
150
0.0065
154
200
0.0058
172
250
0.0053
190

22. Sample Problem Such as this one: k = (90)/(250 s) t (s) [A] (M) 1/[A]
0.0100
100 50
0.0085
118
0.0074
136
150
0.0065
154
200
0.0058
172
250
0.0053
190

23. Sample Problem Such as this one: k = 0.360 t (s) [A] (M) 1/[A] 0.0100
0.0100
100 50
0.0085
118
0.0074
136
150
0.0065
154
200
0.0058
172
250
0.0053
190

24. Half-Life
The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.

25. Half-Life
The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.
This means that [A]t½ = ½[A]0.

26. Half-Life
The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.
This means that [A]t½ = ½[A]0.
Substituting this into our equation will give us the half- life with respect to k and [A]0.

27. Half-Life
The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.
This means that [A]t½ = ½[A]0.
Substituting this into our equation will give us the half- life with respect to k and [A]0.
1/[A]t½ = kt + 1/[A]0

28. Half-Life
The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.
This means that [A]t½ = ½[A]0.
Substituting this into our equation will give us the half- life with respect to k and [A]0.
1/[A]t½ - 1/[A]0 = kt

29. Half-Life
The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.
This means that [A]t½ = ½[A]0.
Substituting this into our equation will give us the half- life with respect to k and [A]0.
1/(½[A]0) - 1/[A]0 = kt

30. Half-Life
The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.
This means that [A]t½ = ½[A]0.
Substituting this into our equation will give us the half- life with respect to k and [A]0.
2/[A]0 - 1/[A]0 = kt

31. Half-Life
The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.
This means that [A]t½ = ½[A]0.
Substituting this into our equation will give us the half- life with respect to k and [A]0.
(2 - 1)/[A]0 = kt

32. Half-Life
The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.
This means that [A]t½ = ½[A]0.
Substituting this into our equation will give us the half- life with respect to k and [A]0.
1/[A]0 = kt

33. Half-Life
The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.
This means that [A]t½ = ½[A]0.
Substituting this into our equation will give us the half- life with respect to k and [A]0.
t½ = 1/(k[A]0)

34. Half-Life
The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.
This means that [A]t½ = ½[A]0.
Substituting this into our equation will give us the half- life with respect to k and [A]0.
t½ = 1/(k[A]0)
This means that the half-life of a second order reaction is dependent of the concentration of the reactant.

35. Second Order Reactions
Summary:
The integrated rate law for a first order reaction:
1/[A]t = kt + 1/[A]0
Plotting 1/[A] vs time produces a straight line.
The slope of the line is equal to k
t½ = 1
k[A]0