1. OCR A2 F325 5.1.1 Rates of reaction
Dr Carl Thirsk

2. Contents Rate and orders The rate equation
Measuring rates and determining orders
Half lives
Initial rates
Effect of temperature
Rate equations and mechanisms
Selected exam questions

5. How fast? Grade/Level Success Criteria We will be covering:
Rate graphs, orders of reaction and rate equations
Determining rate constants and orders using data
Grade/Level
Success Criteria C
Explain and use the term: rate of reaction and order of reaction
A/B
Deduce reaction rates from initial rate data and concentration–time graphs
A*/A
Relate rate data to reaction mechanisms

6. Rate and orders

7. Starter questions What is meant by rate of reaction?
Which factors affect the rate of reaction?
How do you measure the rate of reaction?

8. Defining rate
Rate of reaction is the change in concentration of a reactant or product in a given time:
Rate= change in concentration (mol dm −3 ) time (s)
Rate has units of mol dm-3 s-1
Square brackets, [   ] are used to represent the concentration, mol dm-3 of a reactant or product.

9. Measuring rate
As a reaction proceeds, the reactants get used up. This means that the concentration decreases.
This explains why reactions:
Slow down: fewer collisions are taking place
Stop: when all the reactant has been used up
The rate can be determined by measuring the concentration of a reactant or product at regular time intervals.

10. Measuring rate
When a graph of concentration of reactant is plotted vs. time, the gradient of the curve is the rate of reaction.
The gradient of a curve is measured by drawing tangents.
The initial rate is the rate at the start of the reaction where it is fastest.
Reaction rates can be calculated from graphs of concentration of reactants or products as they change over time.

11. Concentration change of a reactant during a reaction
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= (𝑦2−𝑦1) (𝑥2−𝑥1)

12. Measuring an initial rate
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= (𝑦2−𝑦1) (𝑥2−𝑥1) 12

13. Measuring the rate from a concentration-time graph
Sulphuryl chloride decomposes producing sulphur dioxide and chlorine:
SO2Cl2(g)  SO2(g) + Cl2(g)
The concentration of SO2Cl2 was measured and plotted

14. Time/s
500
1000
2000
3000
4000
[SO2Cl2]/ moldm-3
0.50
0.43
0.37
0.27
0.20
0.15
Suppose we want to find the initial rate (t = 0) and the rate as t = 3000:
After t = 0
Rate = Change in [SO2Cl2]/
change in time
Rate = (0.5 – 0) / (3300 – 0)
= 1.5 x 10-4 mol dm-3 s-1
blue line = experimental data
After t = 3000
Rate = Change in [SO2Cl2]/
change in time
Rate = (0.38 – 0.14) / (4000 – 0)
= 6.0 x 10-5 mol dm-3 s-1

15. Orders of reaction
What determines whether a collision between two particles will result in a reaction?
Activation energy - Ea
Only particles that have more energy than Ea will react when they collide.
If we increase the concentration of a reactant there will be:
A larger number of collisions per second
A faster reaction
By doing a series of experiments we can work out how concentration of each reactant affects the rate of reaction.
Mathematically this effect is called the order with respect to a reactant.

16. Orders of reaction Take the example reaction: A + B  C
For reactant A, we can say rate is proportional to the concentration of A:
Rate ∝ [A]m
where m is the order of reaction with respect to reactant A
Order is always defined in relation to concentrations.
There are three common reaction orders.

17. Zero order reactions For the reaction: A + B + C  Products
If the order is 0 with respect to reactant A, then;
Rate ∝ [A]0
The rate is not affected by the concentration of A. We can increase the concentration of A as much as we like and the rate of reaction will not change.
This is because any number raised to the power of 0 becomes 1

18. First order reactions For the reaction: A + B + C  Products
If the order is 1 with respect to reactant B, then;
Rate ∝ [B]1
If we double the concentration of B the rate also doubles
This is because any number raised to the power of 1 is itself
If [B] increases by 2 times, rate increases by 21 times = 2 times
If [B] increases by 3 times, rate increases by 31 times = 3 times

19. Second order reactions
For the reaction:
A + B + C  Products
Second Order:
If the order is 2 with respect to reactant C, then;
Rate ∝ [C]2
If we double the concentration of C the rate quadruples
This is because any number raised to the power of 2 …….behaves likes this:
If [C] increases by 2 times, rate increases by 22 times = 4 times
If [B] increases by 3 times, rate increases by 32 times = 9 times

20. Rate ∝ [A]0 , Rate ∝ [B]1 and Rate ∝ [C]2
Combining orders
For the reaction:
A + B + C  Products
Let the orders of reaction with respect to the different reactant be as follows:
A = 0, B = 1, and C=2
So the rate is affected by the concentrations of reactant as follows:
Rate ∝ [A]0 , Rate ∝ [B]1 and Rate ∝ [C]2
Combining these expressions gives: Rate ∝ [A]0[B]1[C]2

21. The rate equation

22. The Rate Equation and The Rate Constant, k
For the reaction:
A + B + C  Products
The rate constant, k, is a proportionality constant that links the rate of reaction with the concentration of reactants.
So if:
Rate ∝ [A]0[B]1[C]2
We can write a rate equation as follows:
Rate = k[A]0[B]1[C]2
Which simplifies to:
Rate = k[B]1[C]2
since [A]0 = 1.

23. Summary For the general reaction: aA + bB →products
The generalised rate equation is:
r = k[A]m[B]n
r is used as the standard symbol for chemical rate
r has units mol dm-3s-1
m, n are called reaction orders
Orders are usually integers 0, 1 ,2
0 means the reaction is zero order with respect to that reactant
1 means first order
2 means second order

24. The Overall Order of Reaction
The overall order of reaction is the sum of the individual orders.
From the example:
Rate = k[B]1[C]2
The overall order is = 3
VERY IMPORTANT:
The orders have nothing to do with the stoichiometric coefficients in the balanced equation.
Orders can only be worked out experimentally

25. The Rate Equation and Orders of Reaction
Questions
The rate of equation for a reaction between P and Q is:
rate = k[P]2[Q]
How will the rate change if:
The concentration of P is doubled
The concentration of Q is tripled
The concentration of P and Q are both tripled.
2) In a reaction between R, S and T:
When the [R] is doubled the rate is unchanged
When the [S] is quadrupled the rate increases by 16 times
When the [T] is halved the rate is halved
Deduce the orders w.r.t R, S and T
Hence write down the rate equation for the reaction
Rate quadruples
Rate triples
Rate multiplies by 27
[R]0 [S]2 [T]1
rate = k [S]2[T] = 3rd order

26. Key points about the rate constant, k
The units of k depend on the overall order of reaction and must be worked out from the rate equation.
The larger the value of k, the faster the reaction
A fast reaction has a large value of k
A slow reaction has a small value of k
The value of k is independent of concentration and time. It is constant at a fixed temperature.
The value of k refers to a specific temperature and it increases if we increase temperature.

27. Finding k and the units of k
The units of the rate constant depend on the overall order in the reaction.
We can determine the units of k by substituting units into the rate equation.
For a first order reaction, the rate equation will be:
rate = k[A]1
The units of this will be:
moldm-3s-1 = k x moldm-3
Hence k for a first order reaction has units s-1

28. Rate constant units
For a 1st order overall reaction the unit of k is s-1
For a 2nd order overall reaction the unit of k is mol-1 dm3s-1
For a 3rd order overall reaction the unit of k is mol-2 dm6s-1

29. The Rate Equation and Orders of Reaction
Question
The reaction between ozone, O3(g), and ethene, C2H4(g), has the rate equation:
rate = k[O3(g)][C2H4(g)]
5 x 10-8 mol dm-3 O3(g) was reacted with 1.0 x 10-8 mol dm-3 C2H4(g).
The initial rate was 1.0 x mol dm-3 s-1
Calculate the rate constant, k, for this reaction and state the units.
k = rate k = x = 2000 mol dm3 s-1
[O3][C2H4] (5 x 10-8)(1 x 10-8)

30. Measuring rates
Half life method
Initial rates

31. Continuous rate experiments
This is an experiment where the concentration of a substance is followed throughout the experiment.
This data can be processed by plotting the data and calculating successive half-lives.
The half life is the time taken for the concentration of a reactant to half.
The half-life of a first-order reaction is independent of the concentration and is constant.
If half-lives rapidly increase then the order is 2nd order.

32. Measuring half-lives
The half life of a reactant is the time it takes for the concentration of that reactant to reduce by half.
The shorter the half life, the faster the reaction is progressing
For a 1st order reaction:
The half-life is constant
Half-life is the same regardless of the concentration
Here half-life is 53 seconds
2N2O(g)  2N2(g) + O2(g)
1st order w.r.t to N2O at a constant temp

33. Half-lives for a zero order reaction
Concentration decreases at a constant rate.
Half-life decreases with time

34. Half-lives for a first order reaction
Concentration halves in equal time intervals
Half life is constant

35. Half-lives for a second order reaction
Concentration decreases rapidly then slows down.
Half-life increases with time

36. Using half-lives Questions
Hydrogen peroxide, H2O2, reacts in a first order reaction with a half-life of 27 seconds. If the initial concentration of hydrogen peroxide is 1.60 mol dm-3 what is the concentration after 81 seconds?
Paracetamol has a biological half-life of 380 s. How long will it take for the level of paracetamol in the body to fall to one 16th of its original value?
0.2 mol dm-3
1520 s or 25.3 min

37. Determining orders from rate-concentration graphs
Orders can also be determined by plotting rate against concentration.
Different orders give different shaped graphs.
The following graphs are of initial rate against concentration.
It is important to understand and learn the shapes of these graphs.

38. Rate is not affected by changes in conc.
Zero order
For zero order: the concentration of A has no effect on the rate of reaction
r = k[A]0 = k
zero-order
Rate is not affected by changes in conc.

39. If [A] is doubled then rate is doubled
First order
For first order: the rate of reaction is directly proportional to the concentration of A:
r = k[A]1
first-order
If [A] is doubled then rate is doubled

40. If [A] is doubled then rate increases by 22 = 4 times
Second order
For second order: the rate of reaction is proportional to the concentration of A squared:
r = k[A]2
second-order
If [A] is doubled then rate increases by 22 = 4 times

41. Relating concentration vs. time and rate vs. concentration graphs
Zero order
Measure half-lives
Measure initial rates

42. Relating concentration vs. time and rate vs. concentration graphs
First order
Measure half-lives
Measure initial rates

43. Relating concentration vs. time and rate vs. concentration graphs
Second order
Measure half-lives
Measure initial rates

44. Determining initial rates
The preceding graphs of initial rate against concentration show the different orders.
The initial rate may have been calculated from taking gradients from concentration vs. time graphs at t=0s
For a rate concentration graph to show the order of a particular reactant, the concentration of that reactant must be varied whilst the concentrations of the other reactants should be kept constant.
Another way to determine initial rates is to use clock reactions

45. Clock reactions
Clock reactions produce some kind of distinct change after a certain period of time:
appearance of a precipitate
disappearance of a solid
a change in colour
During these types of reactions you are measuring the rate for the initial period of the reaction.
This means that there will not be a significant change in concentration so we can assume that the rate is just 1/t.
In clock reactions, experiments are repeated with different concentrations of one reactant, then again with another reactant.
Graphs are plotted for each reactant to determine the order.

46. Sodium thiosulphate and hydrochloric acid reaction
The reaction is:
Na2S2O3(aq) + 2HCl(aq)2NaCl(aq) + S(s) + SO2(aq) + H2O(l)
The experiment is repeated with different [HCl] and then again with different [Na2S2O3]
Graphs are plotted for each
The reaction produces sulfur, which is formed as a suspension
This causes the solution opacity to increase
The rate of reaction is measured by seeing how long it takes for a cross placed under the beaker to vanish

47. Na2S2O3(aq) + 2HCl(aq)  2NaCl(aq) + S(s) + SO2(aq) + H2O(l)
Can you write the rate equation for this reaction?
Results
Na2S2O3(aq) + 2HCl(aq)  2NaCl(aq) + S(s) + SO2(aq) + H2O(l)
Rate = k [Na2S2O3]1 [HCl]0
Rate = k [Na2S2O3]
Na2S2O3
HCl
As the concentration doubles the rate also doubles, as the concentration triples, the rate triples.
The rate and concentration are in direct proportion to each other.
This reactant has no effect on the rate.
The rate continues at a steady pace.

48. Working out orders from experimental initial rate data
Normally, to work out the rate equation we do a series of experiments where the initial concentrations of reactants are changed (one at a time) and measure the initial rate each time.
In questions, this data is normally presented in a table.
To calculate the order for a particular reactant, compare two experiments where only that reactant is being changed
If conc. is doubled and rate stays the same: order= 0 w.r.t that reactant
If conc. is doubled and rate doubles: order= 1
If conc. is doubled and rate quadruples: order= 2

49. Worked example Work out the rate equation for the following reaction,
A + B + 2C  D + 2E, using the initial rate data in the table:
For reactant A compare between experiments 1 and 2
For reactant B compare between experiments 1 and 3
For reactant C compare between experiments 1 and 4

50. Worked example The overall rate equation is r = k[A][B]2
The reaction is 3rd order overall and the unit of the rate constant = mol-2dm6s-1

51. Reactant concentrations are changed simultaneously
In most questions, it is possible to compare between two experiments where only one reactant has its initial concentration changed.
If, however, both reactants are changed then the effect of both individual changes on concentration are multiplied together to give the overall effect on rate.

52. Reactant concentrations are changed simultaneously
In a reaction where the rate equation is:
r = k [A] [B]2
If the [A] is x2, that rate would x2
If the [B] is x3, that rate would x32= x9
If these changes happened at the same time then the rate would x2 x9 = x18
In these questions, you will have to consider the combined effect different reactants have on the rate constant to figure out the orders

53. Worked example
Work out the rate equation for the reaction, between X and Y, using the initial rate data in the table:
For reactant X, compare between experiments 1 and 2
For reactant X as the concentration doubles (Y staying constant) so does the rate
Therefore the order with respect to reactant X is first order

54. Worked example
Work out the rate equation for the reaction, between X and Y, using the initial rate data in the table:
Comparing between experiments 2 and 3:
Both X and Y double and the rate goes up by 8
We know X is first order so that will have doubled rate
The effect of Y, therefore, on rate is to have quadrupled it.
Y must be second order

55. Worked example
Work out the rate equation for the reaction, between X and Y, using the initial rate data in the table:
The overall rate equation is r = k [X] [Y]2
The reaction is 3rd order overall and the unit of the rate constant =mol-2dm6s-1.

56. Calculating a value for k using initial rate data
To do this, choose any one of the experiments and put the values into the rate equation that has been rearranged to give k.
Using experiment 3:
r = k [X][Y]2,
k = r / [X][Y]2
k = 2.40 x 10–6 / 0.2 x 0.22 = 3.0 x 10-4 mol-2dm6s-1

57. Effect of temperature

58. The Effect of Temperature on k
The key factor in the rate of reaction is the number of collisions that exceed the activation energy, Ea.
Rate increases with temperature by much more than can be explained just from an increased frequency of collisions.
rate = k [NO]2[O2]
In the above reaction, the rate increases sharply with increasing T even if the concentrations are unchanged. Obviously the temperature cannot change the concentrations of the reactants, so this shows that k must change with T.
Increasing T speeds up most reactions by increasing the value of k
For many reactions the rate doubles as T increases by 10˚C, demonstrating that a greater number of particles are above Ea

59. The Effect of Temperature on k
Rate of reaction can be measured at different temperatures and k calculated for each temperature. The rate constant, k, increases with increasing temperature
Increasing temperature increases the rate constant k.
The relationship is given by the Arrhenius equation k = Ae-(Ea/RT) where A is a constant, R is the gas constant and Ea is activation energy.
You do not need to know this equation, just be aware that the relationship between k and T is not directly proportional but is as shown on the graph

60. Rate equations and mechanisms
Rate determining step

61. Rate equations and mechanisms
A mechanism is a series of steps through which the reaction progresses, often involving intermediate compounds.
If all the steps are added together they will add up to the overall equation for the reaction.
Each step can have a different rate of reaction. The slowest step will control the overall rate of reaction.
The slowest step is called the rate-determining step (r.d.s.)

62. Rate equations and mechanisms
The molecularity (number of moles of each substance) of the molecules in the slowest step will be the same as the order of reaction for each substance.
r=k[A]m
If m=0, the reaction is zero order w.r.t A and A is not involved in the r.d.s. A must be involved in a fast step.
If m=1, the reaction is first order w.r.t A and 1 mole of A is involved in the r.d.s. of the mechanism.
If m=2, the reaction is second order w.r.t A and 2 moles of A feature in the r.d.s of the mechanism

63. Predicting mechanisms from rate equations
NO2(g) + CO(g)NO(g) + CO2(g)
The overall balanced equation tells us that:
1 mol NO2(g) reacts with 1 mol CO(g) to produce 1 mol NO(g) and 1 mol CO2(g)
The overall equation does not tell you anything about the mechanism. To do this we need to carry out some rate experiments.
If these experiments are carried out the rate equation is found to be:
rate = k[NO2]2
What are the orders of reaction with respect to the two reactants?
[NO2] = second order [CO] = zero order

64. Rate-Determining Step and Molecularity
NO2(g) + CO(g ) NO(g) + CO2(g) r = k[NO2]2
If a reactant appears in the rate equation then it must participate in the r.d.s.
The order with respect to that reactant tells you how many particles of that reactant are involved in the r.d.s. (i.e. its molecularity)
What possible mechanism is consistent with both the rate equation and the overall equation?

65. Rate-Determining Step and Molecularity
Overall Reaction:
NO2(g) + CO(g) → NO(g) + CO2(g)
Mechanism:
Step 1: NO2 + NO2→ NO + NO3 slow
Step 2: NO3 + CO → NO2 + CO2 fast
NO3 is a reaction intermediate.
NO2 appears twice in the slow step so it is second order.
CO does not appear in the slow step so is zero order.
r = k[NO2]2

66. Example 1 Overall reaction: A + 2B + C  D + E Mechanism:
Step 1: A + B  X + D slow
Step 2: X + C  Y fast
Step 3: Y+BE fast
C is zero order as it appears in the mechanism after the slow step
r = k[A][B]

67. Example 2 Overall reaction: A + 2B + C  D + E Mechanism:
Step 1: A + B  X + D fast
Step 2: X + C  Y slow
Step 3: Y + BE fast
We can write from this mechanism:
r = k[X][C]
However, X is an intermediate. We must therefore replace it in the rate equation with the substances it was formed from in the step A + B  X + D
r = k[A][B][C]

68. Example 3
rate equation tell us the slow step (r.d.s.) will involve two molecules of NO and one molecule of H2
Using the rate equation rate = k[NO]2[H2] and the overall equation 2NO(g) + 2H2(g)N2(g) + 2H2O(g), the following three-step mechanism for the reaction was suggested. X and Y are intermediate species.
Step 1: NO + NO  X
Step 2: X + H2  Y
Step 3: Y + H2  N2 + 2H2O
Which one of the three steps is the r.d.s?
Step 2. H2 appears in the rate equation and combining steps 1 and 2 gives the correct molecularity

69. Example 4
Suggest a rate equation for the reaction whose mechanism is shown below:
CH3CH2Br + OH-  CH3CH2OH + Br- slow
The rate equation is r = k[CH3CH2Br] [OH-]
This is called an SN2 reaction. Substitution, Nucleophilic, 2 molecules in the r.d.s

70. The same reaction can also occur via a different mechanism:
Overall Reaction:
(CH3)3CBr + OH–  (CH3)3COH + Br –
Mechanism:
(CH3)3CBr  (CH3)3C+ + Br- slow
(CH3)3C+ + OH–  (CH3)3COH fast
Write the rate equation and draw the ‘curly arrows’ mechanism for this reaction
What would this reaction be called if the other was SN2?
r = k[(CH3)3CBr)]
SN1

71. C3H6O + NaBH4 + H2O  C3H8O + NaBH3.OH
Example 5
Carbonyls can be reduced by hydrides, such as sodium tetrahydroborate, NaBH4
An example reaction is:
C3H6O + NaBH4 + H2O  C3H8O + NaBH3.OH
A student suggests the rate equation for this reaction is:
r = k[H2O][C3H6O]
Using your knowledge of the mechanism for this reaction, state whether or not you agree with this and why.

72. Questions
1) The following reaction is first order w.r.t. H2 and first order w.r.t ICl. H2(g) + 2ICl(g)  I2(g) + 2HCl(g) a) Write the rate equation for this reaction. b) The mechanism for this reaction consists of two steps. i) Identify the molecules that are in the r.d.s. Justify you answer. ii) A chemist suggested the following mechanism: 2ICl(g) I2+Cl2 slow H2(g) + Cl2(g)  2HCl(g) fast Suggest, with reasons, whether this mechanism is likely to be correct. [5 marks overall]

73. Questions
2) The reaction between HBr and oxygen gas occurs rapidly at 700 K. It can be represented by the equation 4HBr + O2  2H2O + 2Br2 The rate equation was found by experiment to be: r = k[HBr][O2] a) Explain why the reaction cannot be a one-step reaction. b) Each of the 4 steps in this reaction involves 1 molecule of HBr. Two of the steps are the same. The r.d.s is the first one and results in the formation of HBrO2. Write equations for the full set of 4 reactions.
Hint: HBrO is another intermediate.

74. Solution to 2b
The rate equation shows that one molecule of HBr is involved in the r.d.s. However, there are 4 molecules in the balanced equations. 3 other molecules of HBr must be involved in other (fast) steps, so the reaction cannot just be one-step. Overall mechanism: Step 1: HBr + O2  HBrO2 slow (r.d.s.) Step 2: HBr + HBrO2  2HBrO fast Step 3: HBr + HBrO  H2O + Br2 fast Step 4: HBr + HBrO  H2O + Br2 fast

75. Further questions

80. Further questions
Orders and rates for simultaneous reactions
Answers

81. Selected exam questions

82. June 2010

85. Jan 2012

88. June 2013