1. Measurement of Rate Constants
Continuous-Flow:
Stopped-Flow:

2. Measurement of Rate Constants
Quench-Flow:
Flash-Photolysis:

3. Measurement of Rate Constants
Temperature or Pressure Jump:

4. Orders of Reaction Zero-order reactions: A  B
rate = -d[A]/dt = k[A]n,
and since n = 0, [A]0 = 1,
therefore -d[A]/dt = k
or –d[A] = k dt
which upon integration yields: [A]t = A0 – k t -k
[A]
time
[A0]

5. Orders of Reaction First-order reactions: A  B -d[A]/dt = k[A]n,
where n = 1, [A]1 = [A], thus
-d[A]/dt = k[A] or d[A]/[A] = -k dt
which upon integration yields:
In [A]/[A0] = – k t or [A] = [A0] e-kt
time
time
[A]
In [A] / [A0]
- k

6. Orders of Reaction Second-order reactions: 2A  B
-d[A]/dt = k[A]n, where n = 2,
-d[A]/dt = k[A]2 or d[A]/[A]2 = k dt
which upon integration yields:
1 / [A] = kt + 1 / [A0]
time
1 / [A] k
1/[A0]

7. Determination of Half-Lives
Zero-order reactions:
A  B
t = t1/2 when [A] = ½[A0], therefore
[A] = A0 – kt
½ A0 = A0 – k t1/2
k t1/2 = A0 - ½A0
t1/2 = A0 / 2k
t1/2
[A0]
1/2k

8. Determination of Half-Lives
First-order reactions:
A  B
t = t1/2 when [A] = ½[A0], therefore
ln [A0]/2[A0] = - kt1/2
or ln ½ = - kt1/2
and since ln ½ = -ln2,
t1/2 = 1/k ln2 = 0.693/k
t1/2
[A0]
0.693/k

9. Determination of Half-Lives
Second-order reactions:
2A  B
t = t1/2 when [A] = ½[A0], therefore
1 / [A] = 1 / ½[A0] = 2 / [A0], so
-1 / [A0] + 2 / [A0] = k t½ = 1 / [A0]
t½ = 1/k · 1/[A0]
[A0] t½
t1/2
1/[A0]
1/k

10. Determination of Reaction Order
rate constant
half-life -k
[A]
time
[A0]
zero order
t1/2
[A0]
1/2k
time
[A]
t1/2
[A0]
0.693/k
first order
[A0] t½
time
1 / [A] k
1/[A0]
second order

11. LaPlace Transform Method for Solving Integrated Rate Equations
Rate Law Transform
(Diff. Eq.) (p = d/dt)
Integrated Solve for
Solution time variable
Inverse Transform
(from table)

12. LaPlace Transforms k1 define reaction A  B
First-order reaction (irreversible) k1
define reaction
A  B

13. LaPlace Transforms k1 A  B d[A]/dt = -k1[A] define rate law
First-order reaction (irreversible) k1
A  B
d[A]/dt = -k1[A]
define rate law

14. LaPlace Transforms k1 A  B d[A]/dt = -k1[A] let p = d/dt
First-order reaction (irreversible) k1
A  B
d[A]/dt = -k1[A] let p = d/dt
p[A] = -k1[A]
LaPlace transforms assume [A] = 0 at t = 0,
if not then subtract the value of p[A] at t = 0:
p[A] – p[A0] = -k1[A]

15. LaPlace Transforms k1 A  B d[A]/dt = -k1[A] let p = d/dt
First-order reaction (irreversible) k1
A  B
d[A]/dt = -k1[A] let p = d/dt
p[A] = -k1[A]
LaPlace transforms assume [A] = 0 at t = 0,
if not then subtract the value of p[A] at t = 0:
p[A] – p[A0] = -k1[A]
solve for [A] algebraically:
p[A] + k1[A] = (p + k1)[A] = p[A0]

16. LaPlace Transforms k1 A  B d[A]/dt = -k1[A] let p = d/dt
First-order reaction (irreversible) k1
A  B
d[A]/dt = -k1[A] let p = d/dt
p[A] = -k1[A]
LaPlace transforms assume [A] = 0 at t = 0,
if not then subtract the value of p[A] at t = 0:
p[A] – p[A0] = -k1[A]
solve for [A] algebraically:
p[A] + k1[A] = (p + k1)[A] = p[A0]
look up on inverse transform table:
[A] = (p/(p + k1))[A0] = [A0]e-k1t

17. LaPlace Transforms k1 A  B [A] = [A0]e-k1t
First-order reaction (irreversible) k1
A  B
[A] = [A0]e-k1t
time
[A]
where kobs = k1 and Amplitude = [A0]

18. First-order reaction (reversible)
k+1
A  B
k-1

19. First-order reaction (reversible)
k+1
A  B
k-1
1.) write the appropriate rate law:
d[A]/dt = -k+1[A] + k-1[B] where
[B] = [A0] – [A]

20. First-order reaction (reversible)
k+1
A  B
k-1
1.) write the appropriate rate law:
d[A]/dt = -k+1[A] + k-1[B] where
[B] = [A0] – [A]
therefore
d[A]/dt = -k+1[A] + k-1([A0] – [A])
= k-1([A0] - (k+1 + k-1)[A]

21. First-order reaction (reversible)
k+1
A  B
k-1
1.) write the appropriate rate law:
d[A]/dt = -k+1[A] + k-1[B] where
[B] = [A0] – [A]
therefore
d[A]/dt = -k+1[A] + k-1([A0] – [A])
= k-1([A0] - (k+1 + k-1)[A]
2.) rewrite as a LaPlace Transform where
p = d/dt and subtract p[A0] if [A0]  0:
p[A] – p[A0] = k-1[A0] - (k+1 + k-1)[A]

22. k+1
A  B
k-1
3.) combine like terms:
p[A] + (k+1 + k-1)[A] = p[A0] + k-1[A0] or
(p + k+1 + k-1)[A] = (p + k-1)[A0]

23. k+1
A  B
k-1
3.) combine like terms:
p[A] + (k+1 + k-1)[A] = p[A0] + k-1[A0] or
(p + k+1 + k-1)[A] = (p + k-1)[A0]
4.) solve for [A] algebraically:
[A] = [A0]((p + k-1)/(p + k+1 + k-1))

24. k+1
A  B
k-1
3.) combine like terms:
p[A] + (k+1 + k-1)[A] = p[A0] + k-1[A0] or
(p + k+1 + k-1)[A] = (p + k-1)[A0]
4.) solve for [A] algebraically:
[A] = [A0]((p + k-1)/(p + k+1 + k-1))
5.) look up the solution in the inverse
transform table (i.e. for (p + a)/(p + b)):
[A] = [A0]((a/b) – ((a – b)/b)e-bt)
where a = k-1 and b = (k+1 + k-1)

25. k+1
A  B
k-1
3.) combine like terms:
p[A] + (k+1 + k-1)[A] = p[A0] + k-1[A0] or
(p + k+1 + k-1)[A] = (p + k-1)[A0]
4.) solve for [A] algebraically:
[A] = [A0]((p + k-1)/(p + k+1 + k-1))
5.) look up the solution in the inverse
transform table (i.e. for (p + a)/(p + b)):
[A] = [A0]((a/b) – ((a – b)/b)e-bt)
where a = k-1 and b = (k+1 + k-1)
Thus by substitution:
[A] = [A0]((k-1/(k+1 + k-1)) + (k+1/(k+1 + k-1))e-(k+1 + k-1)t

26. [A] = [A0]((k-1/(k+1 + k-1)) – (k+1/(k+1 + k-1))e-(k+1 + k-1)t
A  B
k-1
[A] = [A0]((k-1/(k+1 + k-1)) – (k+1/(k+1 + k-1))e-(k+1 + k-1)t
Thus your observed rate constant (kobs) is a single exponential that is the sum of the forward and reverse rate constants, i.e.
time
[A]
kobs = (k+1 + k-1)
and the observed exponential transient will decay to a final value determined by the relative values of k+1 + k-1 (i.e the final equilibrium concentrations of A and B).

27. k+1
A  B
k-1
kobs = (k+1 + k-1)
To determine k+1 and k-1 explicitly additional information is required:

28. k+1
A  B
k-1
kobs = (k+1 + k-1)
To determine k+1 and k-1 explicitly additional information is required:
1.) Amplitude information from stopped-flow experiment can be used.

29. k+1
A  B
k-1
kobs = (k+1 + k-1)
To determine k+1 and k-1 explicitly additional information is required:
1.) Amplitude information from stopped-flow experiment can be used.
2.) The equilibrium constant for the reaction step determined independently.

30. k+1
A  B
k-1
kobs = (k+1 + k-1)
To determine k+1 and k-1 explicitly additional information is required:
1.) Amplitude information from stopped-flow experiment can be used.
2.) The equilibrium constant for the reaction step determined independently.
3.) If the reaction is actually second order (i.e. a binding step)
but run under psuedo-first order conditions (i.e. [S] >> [E]),
then kobs = (k+1[S] + k-1)
and k+1 + k-1 can be determined from a plot of kobs vs. [S],
where slope = k+1 and y-intercept = k-1.
k+1
E + S  ES
k-1

31. Sequential first-order reactions (irreversible)
k+1
k+2
A  B  C
1.) write the appropriate rate law:
d[A]/dt = -k+1[A]
which is exactly the same rate law that describes a single irreversible step reaction.

32. Sequential first-order reactions (irreversible)
k+1
k+2
A  B  C
1.) write the appropriate rate law:
d[A]/dt = -k+1[A]
which is exactly the same rate law that describes a single irreversible step reaction.
2.) Thus the integrated solution is:
[A] = [A0]e-k+1t
and we can determine k+1 directly from kobs by following the change in [A].

33. Sequential first-order reactions (irreversible)
k+1
k+2
A  B  C
1.) write the appropriate rate law:
d[A]/dt = -k+1[A]
which is exactly the same rate law that describes a single irreversible step reaction.
Thus the integrated solution is:
[A] = [A0]e-k+1t
and we can determine k+1 directly from kobs by following the change in [A].
But what about k+2? We need to be able to examine the change in [B] or [C]!

34. k+1
k+2
A  B  C
1.) write the appropriate rate law:
d[B]/dt = k+1[A] – k+2[B]

35. k+1
k+2
A  B  C
1.) write the appropriate rate law:
d[B]/dt = k+1[A] – k+2[B]
2.) rewrite as a LaPlace Transform where
p = d/dt ( note that now [B0] = 0):
p[B] = k+1[A] – k+2[B]

36. k+1
k+2
A  B  C
1.) write the appropriate rate law:
d[B]/dt = k+1[A] – k+2[B]
2.) rewrite as a LaPlace Transform where
p = d/dt ( note that now [B0] = 0):
p[B] = k+1[A] – k+2[B]
3.) combine like terms:
p[B] + (k+2)[B] = k+1[A]

37. k+1
k+2
A  B  C
1.) write the appropriate rate law:
d[B]/dt = k+1[A] – k+2[B]
2.) rewrite as a LaPlace Transform where
p = d/dt ( note that now [B0] = 0):
p[B] = k+1[A] – k+2[B]
3.) combine like terms:
p[B] + (k+2)[B] = k+1[A]
and since as we already determined:
[A] = [A0](p/(p + k+1))
we can write by substitution:
(p + k+2)[B] = k+1[A0](p/(p + k+1))

38. k+1
k+2
A  B  C
4.) solve for [B] algebraically:
[B] = k+1[A0](p/(p + k+1)(p + k+2))

39. k+1
k+2
A  B  C
4.) solve for [B] algebraically:
[B] = k+1[A0](p/(p + k+1)(p + k+2))
5.) look up the solution in the inverse
transform table (i.e. for (p/((p + a)(p + b))):
[B] = [A0]((1/(b – a)) e-at + (1/(a – b))e-bt)
where a = k+1 and b = k+2

40. k+1
k+2
A  B  C
4.) solve for [B] algebraically:
[B] = k+1[A0](p/(p + k+1)(p + k+2))
5.) look up the solution in the inverse
transform table (i.e. for (p/((p + a)(p + b))):
[B] = [A0]((1/(b – a)) e-at + (1/(a – b))e-bt)
where a = k+1 and b = k+2
Thus by substitution:
[B] = [A0](((k+1/(k+2 - k+1)) e-k+1t + (k+1/(k+1 - k+2))e-k+2t)
Therefore we observe a two exponential decay where the two observed rate constants correspond directly to k+1 and k+2!

41. Determinants of Matrices
Consider a simple 2x2 matrix. The determinant of that matrix is given by:
A B
C D
Det = = A•D – C •B

42. Cramer’s rule P = A •x + B •y Q = C •x + D •y P = A B x Q = C D y P B
Two equations and two unknowns
can be written in matrix form as:
P = A •x + B •y
Q = C •x + D •y
P = A B x
Q = C D y
Where the solutions for x and y can be determined from the determinants of the following matrix equations:
P B
P•D – B •Q
Q D
x = =
A•D – C •B
A B
C D
A P
A•Q – C •P
C Q
y = =
A•D – C •B
A B
C D