1. A B
Reaction Rates
[A]& [B]
determined by monitoring the change in concentration of either reactants or products as a function of time.
[A]
[B]
-[A] t
[B] t
Rate = =
{RxRateIntro}
Spectrometer

2. Reaction Rates t1 -[A2] t2 -[A] t [B] t Rate = = t1 t2
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
butyl chloride
Butanol, but we will ignore that below
- [A1]
Calculateplease
t1
-[A2]
Calculate, please
Answers
Next slide
t2
-[A] t
[B] t
Rate = =
t1
t2

3. Reaction Rates -[C4H9Cl]
C4H9Cl (aq) + H2O (l)  C4H9OH (aq) + HCl (aq)
Notice we change the sign to make all rates positive, even though we are losing reactants
Ave. rate =
-[C4H9Cl] t
Note that the average rate decreases.
As the reaction goes forward, there are fewer collisions between reactant molecules.

4. Reaction Rates Concentration vs. Time Graph
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
The slope of a line tangent to the curve at any point is the instantaneous rate at that time.
Concentration vs. Time Graph
-[C4H9Cl] t
Rate =
Instant at
50 sec
[0.060 final initial ]
210final-0i
= (- and -) or +1.9 x M.sec -1
-[C4H9Cl] t
Rate =
Instant at
600 sec
- [0.017final initial]
800final-400i
= x M.sec -1
= x M.sec -1
Go back to slide 3 to check your work/answer

5. GRAPHICAL DETERMINATION OF RATE
The variation in rate can be investigated by measuring the change in concentration of one of the reactants or products, plotting a graph and then finding the gradients of the curve at different concentrations.
RATE CALCULATION
The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction
Place a rule on the outside of the curve and draw a line as shown on the graph. y x
gradient = y / x
In the reaction…
A(aq) + B(aq) ——> C(aq) + D(aq)
the concentration of B was measured every 200 minutes. The reaction is obviously very slow!

6. GRAPHICAL DETERMINATION OF RATE ½ Lives
The variation in rate can be investigated by measuring the change in concentration of one reactants or product, plotting a graph and then finding the gradients of tangents to the curve at different concentrations.
RATE CALCULATION
The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction
Place a rule on the outside of the curve and draw a line as shown on the graph. y x
gradient = y / x
concentration = mol dm-3
gradient = mol dm-3
1520 min
rate = x 10-3 mol dm-3
The rate is negative because
the reaction is slowing down

7. GRAPHICAL DETERMINATION OF RATE
The variation in rate can be investigated by measuring the change in concentration of one of the reactants or products, plotting a graph and then finding the gradients of the curve at different concentrations.
The gradients of tangents at several other concentrations are calculated.
Notice how the gradient gets less as the reaction proceeds, showing that the reaction is slowing down.
The tangent at the start of the reaction is used to calculate the initial rate of the reaction.
RATE CALCULATION
The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction
Place a rule on the outside of the curve and draw a line as shown on the graph. y x
gradient = y / x

8. Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1 in this
Nucleophillic substitution
Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.
Rate =
-[C4H9Cl] t =
[C4H9OH]

9. Reaction Rates and Stoichiometry
What if the ratio is not 1:1?
2 HI(g)  H2(g) + I2(g)
How do rates compare?
Rate1 = −
[HI] t
Rate2= +
[I2] ≠
Rate = − 1 2
[HI] t =
[I2]
Rate = −
[HI] t =
2 [I2]
To generalize, then, for the reaction
5 A B
4 C + 3 D

10. Practice Problems

12. Measuring rate of reaction
Measure how fast the products are formed
Colour change
Gas formation
Measure how fast the reactants are used up

15. Methods of Measuring Reaction Rates
Property
Observation
Device
Mass
Mass of product increases or reactant decreases
Balance
Temperature
Exothermic reaction
Heat is released and Temperature increases
Endothermic reaction
Heat is absorbed and Temperature decreases
Thermometer
Pressure
If there is a change in number of moles of gas and the reaction is run at a fixed volume then pressure will change.
Manometer
Conductivity
If the reaction involves ions in solution then the conductivity of the solution will change.
Concentration
The concentration of reactant and products changes as the reaction proceeds. So anything that measures concentration can be used.
pH meter

16. RATE EQUATION - SAMPLE CALCULATION
In an experiment between A and B the initial rate of reaction was found for various starting concentrations of A and B. Calculate...
1) the orders for A and B & the overall order
2) the rate equation
3) the value of the rate constant (k)
4) the units of the rate constant
[A]
[B]
Initial rate (r) 1
0.5 1 2 2
1.5 1 6 3
0.5 2 8
r initial rate of reaction mol dm-3 s-1
[ ] concentration mol dm-3
CALCULATING ORDER wrt A
Choose any two experiments where…[B] is KEPT THE SAME
See how the change in [A] affects the rate
As you can see, tripling [A] has exactly the same effect on the rate so...
Compare Experiments 1 & 2
[B] same
[A] 3 x bigger
rate 3 x bigger  rate  [A]
FIRST ORDER with respect to (wrt) A
THE ORDER WITH RESPECT TO A = 1 (it is FIRST ORDER)

17. RATE EQUATION - SAMPLE CALCULATION
CALCULATING ORDER wrt B
Choose any two experiments where...
NOW [A] is KEPT THE SAME
See how a change in [B] affects the rate
As you can see, doubling [B] quadruples the rate so...
THE ORDER WITH RESPECT TO B = 2
It is SECOND ORDER
[A]
[B]
Initial rate (r) 1
0.5 1 2 2
1.5 1 6 3
0.5 2 8
Compare Experiments 1 & 3
[A] same
[B] 2 x bigger
rate 4 x bigger  rate  [B]2
SECOND ORDER wrt B
OVERALL ORDER = THE SUM OF THE INDIVIDUAL ORDERS =
= 3
By combining the two proportionality relationships you can construct the overall rate equation
 rate = k [A] [B]2

18. RATE EQUATION - SAMPLE CALCULATION
Take ANY experiment and substitute into equation
[A]
[B]
Initial rate (r)
 rate = k [A] [B]2
re-arranging k = rate
[A] [B]2 1
0.5 1 2 2
1.5 1 6 3
0.5 2 8
r initial rate of reaction mol dm-3 s-1
[ ] concentration mol dm-3
Chose one experiment (e.g. Expt. 3) and substitute its values into the rate equation
k = mol dm-3 s-1
(0.5mol dm-3) (2 mol dm-3)2
(0.5 mol dm-3) 4 mol2 dm-6
k = mol dm-3 s-1
2.0 mol3 dm-9
= 4 dm6 mol-2 sec-1

19. RATE EQUATION - SAMPLE CALCULATION
SUMMARY
[A]
[B]
Initial rate (r)
[A]
[B]
Initial rate (r) 1
0.5 1 2 1
0.5 1 2 2
1.5 1 6 2
1.5 1 6 3
0.5 2 8 3
0.5 2 8
Compare Experiments 1 & 2
[B] same
[A] 3 x bigger
rate 3 x bigger  rate  [A]
FIRST ORDER with respect to (wrt) A
Compare Experiments 1 & 3
[A] same
[B] 2 x bigger
rate 4 x bigger  rate  [B]2
SECOND ORDER wrt B
 rate = k [A] [B]2
re-arranging k = rate
[A] [B]2
Chose one experiment (e.g. Expt. 3) and substitute its values into the rate equation
k = = 4 dm6 mol-2 sec-1
(0.5) (2)2

20. RATE EQUATION QUESTIONS
No 1
[A] / mol dm [B] / mol dm Rate / mol dm-3 s-1
Expt
Expt
Expt
CALCULATE THE ORDER WITH RESPECT TO A
THE ORDER WITH RESPECT TO B
THE OVERALL ORDER OF REACTION
THE FORMAT OF THE RATE EQUATION
THE VALUE AND UNITS OF THE RATE CONSTANT
Expts 1&2 [A] is constant [B] is doubled Rate is doubled
Therefore rate  [B] 1st order wrt B
Expts 1&3 [B] is constant [A] is doubled Rate is doubled
Rate equation is r = k[A][B]
Value of k Substitute numbers from Exp 1 to get value of k
k = rate / [A][B] = 4 / 0.25 x 0.25 = 64
Units of k rate / conc x conc = dm3 mol-1 s-1

21. Concentration and Rate
Compare Experiments 1 and 2: (done on next slide when [NH4+] doubles, the initial rate doubles.
Likewise, compare Experiments 5 and 6: when [NO2- ] doubles, the rate doubles.

22. How does Concentration affect Rate?
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
The data demonstrates:
Rate  [NH4+]
Rate  [NO2−]
Rate  [NH4+] [NO2−]
Data shows the relationship between the reaction rate and the conc. of reactants.
This equation is the rate law, and k is the rate particular temp. or
Rate = k [NH4+] [NO2−]

23. Concentration and Rate
This equation is called the rate law, and k is the rate constant.
The value of k is determined experimentally.
“Constant” is relative here-
k is unique for each rxn
k changes with T

24. Rate Laws
Exponents tell the order of the reaction with respect to each reactant.
This reaction is
First-order in [NH4+]
First-order in [NO2−]
The overall reaction order can be found by adding the exponents on the reactants in the rate law.This reaction is second-order overall.
Notes : 1)k is a constant that has a specific value for each reaction.
2)The value of k is determined experimentally.
“Constant” is relative here-
k is unique for each rxn
k changes with T

25. Determination of Rate Law using Initial Rate Some Practice, you try and find the rate law
Reaction: S2O82-(aq) + 3I- (aq)  2SO42- (aq) + I3- (aq)
The following data were obtained.

Exp# [S2O82-] [I-] Initial Rate,
(mol/L) (mol/L) (mol/L.s)
x 10-5
x 10-5
x 10-5

26. Determination of Rate Law using Initial Rate
Solution: The rate law = Rate = k[S2O82-]x[I-]y,
here x and y are rate orders.
(a) Calculation of rate order, x: x

27. Determination of Rate Law using Initial Rate
(b) Calculation of rate order, y:
This reaction is first order w.r.t. [S2O82-] and [I-] w.r.t = with respect to
Rate = k[S2O82-]1[I-]1

28. Calculating rate constant and rate at different concentrations of reactants
Rate constant, k =
= 6.6 x 10-3 L.mol-1.s-1

29. Try another?

30. Order of Reaction
Substituting:
Solution:

31. Rate constant, finding k
Using Rate 1, you may use rate 2 (same answer)
7.1 x 10-5 M s-1 = -k[0.01 M]2
k = 0.71 M-1 s-1

32. You try

33. GRAPHICAL DETERMINATION OF RATE
The variation in rate can be investigated by measuring the change in concentration of one of the reactants or products, plotting a graph and then finding the gradients of the curve at different concentrations.
RATE CALCULATION
The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction
Place a rule on the outside of the curve and draw a line as shown on the graph. y x
gradient = y / x
In the reaction…
A(aq) + B(aq) ——> C(aq) + D(aq)
the concentration of B was measured every 200 minutes. The reaction is obviously very slow!

34. GRAPHICAL DETERMINATION OF RATE ½ Lives
The variation in rate can be investigated by measuring the change in concentration of one reactants or product, plotting a graph and then finding the gradients of tangents to the curve at different concentrations.
RATE CALCULATION
The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction
Place a rule on the outside of the curve and draw a line as shown on the graph. y x
gradient = y / x
concentration = mol dm-3
gradient = mol dm-3
1520 min
rate = x 10-3 mol dm-3
The rate is negative because
the reaction is slowing down

35. GRAPHICAL DETERMINATION OF RATE
The variation in rate can be investigated by measuring the change in concentration of one of the reactants or products, plotting a graph and then finding the gradients of the curve at different concentrations.
The gradients of tangents at several other concentrations are calculated.
Notice how the gradient gets less as the reaction proceeds, showing that the reaction is slowing down.
The tangent at the start of the reaction is used to calculate the initial rate of the reaction.
RATE CALCULATION
The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction
Place a rule on the outside of the curve and draw a line as shown on the graph. y x
gradient = y / x

36. FIRST ORDER REACTIONS AND HALF LIFE
One characteristic of a FIRST ORDER REACTION is that it is similar to radioactive decay. It has a half-life that is independent of the concentration.
It should take the same time to drop to one half of the original concentration as it does to drop from one half to one quarter of the original.
The concentration of a reactant falls as the reaction proceeds

37. FIRST ORDER REACTIONS AND HALF LIFE
The concentration of reactant A falls as the reaction proceeds
The concentration drops from
4 to 2 in 17 minutes

38. FIRST ORDER REACTIONS AND HALF LIFE
The concentration of reactant A falls as the reaction proceeds
The concentration drops from
4 to 2 in 17 minutes 2 to 1 in a further 17 minutes

39. FIRST ORDER REACTIONS AND HALF LIFE
The concentration of reactant A falls as the reaction proceeds
The concentration drops from
4 to 2 in 17 minutes 2 to 1 in a further 17 minutes 1 to 0.5 in a further 17 minutes

40. FIRST ORDER REACTIONS AND HALF LIFE
The concentration of reactant A falls as the reaction proceeds
The concentration drops from
4 to 2 in 17 minutes 2 to 1 in a further 17 minutes 1 to 0.5 in a further 17 minutes

41. FIRST ORDER REACTIONS AND HALF LIFE
A useful relationship
k is t½ =
where t½ = the half life
Half life = 17 minutes
k t½ =
k = t½
k = = min-1 17

42. Reaction Mechanisms
The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.
Reactions may occur all at once or through several discrete steps.
Each of these processes is known as an elementary reaction or elementary process.

43. Reaction Mechanisms
The molecularity of a process tells how many molecules are involved in the process.
The rate law for an elementary step is written directly from that step.

44. Multistep Mechanisms
In a multistep process, one of the steps will be slower than all others.
The overall reaction cannot occur faster than this slowest, rate-determining step.

45. Slow Initial Step NO2 (g) + CO (g) NO (g) + CO2 (g)
The rate law for this reaction is found experimentally to be
Rate = k [NO2]2
CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.
This suggests the reaction occurs in two steps.

46. Slow Initial Step A proposed mechanism for this reaction is
Step 1: NO2 + NO NO3 + NO (slow)
Step 2: NO3 + CO NO2 + CO2 (fast)
Overall Reaction is?
The NO3 intermediate is consumed in the second step.
As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

47. rate = k [N2O5] there are no other reactants and the coefficient is 1
RATE DETERMINING STEP
The reaction H2O H3O I¯ I H2O takes place in 3 steps
Step 1 H2O I¯ IO¯ H2O SLOW
Step 2 IO¯ H3O HIO H2O FAST
Step 3 HIO H3O I¯ I H2O FAST
The rate determining step is STEP 1 as it is the slowest
The reaction N2O NO O2 takes place in 3 steps
Step 1 N2O NO NO3 SLOW
Step 2 NO NO NO + NO O2 FAST
Step 3 NO NO NO FAST
The rate determining step is STEP 1
rate = k [N2O5] there are no other reactants and the coefficient is 1

48. Fast Initial Step
The rate of the overall reaction depends upon the rate of the slow step.
The rate law for that step would be