A level

1. Reaction kinetics (A2)

2. Rate of reaction
• Rate of reaction: Change in concentration of
reactants or products per unit time (mol dm-3
s-1
or mol dm-3
min-1
)
TIME
CONCENTRATION
B A
C
•rate at which products are
formed or rate at which
reactants are used up.

3. Instantaneous rate of reaction
y
CONCENTRATION
gradient = y
x
x
TIME
• the rate of change of concentration is found from the slope
(gradient) of the curve
• the slope at the start of the reaction will give the initial rate
• the slope gets less (showing the rate is slowing down) as the
reaction proceeds
The slope of the gradient of
the curve gets less as the
reaction slows down with time

4. •Rate equation shows exactly how the rate is
dependent on the concentration of reactants.
•It relates the rate of reaction to the
concentration of reactants raised to the
appropriate power.
•Rate = kcn
where k is the rate constant
•Can only be obtained experimentally.
•Cannot be deduced theoretically from
stoichiometry equation.
•May not include all the reactants written in
the chemical equation.
Rate equation

5. • Order of reaction shows the exact dependence of
rate of concentration & can only be found
experimentally.
•Defined as the power to which the concentration of
that reactant is raised to in the experimentally
determined rate equation.
•The overall order of reaction is the sum of the
powers of the concentration terms in the
experimentally determined rate equation.
eg: aA +bB  products
Rate = k[A]m
[B]n
m= order with respect to A
n= order with respect to B
m+n = overall order of reaction
Order of reaction

6. • Rate constant k is a constant of
proportionality in the rate equation. It
is constant for a given reaction at a
particular temperature.
•The unit is dependant on the order of
reaction.
Rate constant

7. •The rate of reaction is independent of
the concentration of the reactant, A.
rate  [A]0
•The concentration-time graph is a
straight line showing that rate is
constant.
Zero order

8. •The rate of reaction is proportional to one
concentration term, rate  [A]1
•The concentration-time graph is a curve with
a constant half-life. For other orders of
reaction, the half-life is not constant.
eg: decomposition of hydrogen peroxide:
2H2O2 O2 +2H2O
The rate equation: rate=k[H2O2]
The rate of reaction is directly proportional
to [H2O2].
First order

9. •The half-life of a reaction, t½ is the
time taken for the initial concentration of
a reactant to fall to half its value.
• t½ = ln 2 = 0.693
k k
Half-life (1st
order
independent of concentration)

10. •The rate of reaction is proportional to two
concentration terms rate [A]
 2
•t½ not constant t½ = 1/k[A]
•eg: decomposition of NO2:
2NO2(g)  2NO(g) + O2(g)
The rate equation: rate = k [NO2]2
Therefore doubling [NO2] will increase the
rate of reaction four folds.
Second order

11. Given the variation of initial reaction with
[NO2] for the reaction NO2(g)
+CO(g)NO(g)+CO2(g)
Deduce order by initial rate
Experiment Rate/mol dm-3
s-1
[NO2]/mol dm-3
1 0.010 0.15
2 0.040 0.30
3 0.160 0.60
4 0.360 0.90
On doubling [NO2], the rate increases by
a factor of 4. Thus the rate  to [NO2]2
& the reaction is second order with
respect to NO2.

12. Given the reaction of H2O2 with acidified
I-
ions, liberating I2
Deduce order by initial rate
Exp Initial conc/mol dm-3
Initial rates/mol dm-3
s-1
[H2O2] [I-
] [H+
]
1 0.010 0.010 0.010 2.0x10-6
2 0.030 0.010 0.010 6.0x10-6
3 0.030 0.020 0.010 1.2x10-5
4 0.030 0.020 0.020 1.2x10-5
Comparing exp 1&2
Comparing exp2&3
Comparing exp 3&4

13. •The rate of decomposition of a 3.0 mol dm-3
H2O2(aq)
was measured by withdrawing 10cm3
portions at various
times & titrating with acidified 0.1 mol dm-3
KMnO4(aq).
The results obtain are given.
Calculate rate constant using
initial rate
time/min vol of KMnO4/cm3
0 30.0
5 23.4
10 18.3
15 14.2
20 11.1
25 8.7
30 6.8
Order may be found by plotting the graph
of vol of KMnO4 against time.
Since half life is constant, reaction is 1st
order with respect to H2O2.
Rate equation is: rate=k[H2O2]
From graph, t½ = 14min
Rate constant, k = 0.693/14
= 0.0495min-1

14. • The table given shows the results obtained in a
study of the reaction: RBr+NaOHROH+Na+
Br-
Calculation of initial rates
from concentration-time graph
Time/min [RBr]/mol dm-3
[OH]=0.10 [OH]=0.15
0 0.0100 0.0100
40 0.0079 0.0070
80 0.0062 0.0049
120 0.0049 0.0034
160 0.0038 0.0024
200 0.0030 0.0017
240 0.0024 0.0012
The order may be deduced by
plotting graphs of [RBr]vs
time.
From the graph, t½ = 77min
Therefore, 1st
order with
respect to RBr
From graph1, initial rate
= gradient
= 0.005/90= 5.5x10-5
From graph2, initial rate
= 0.005/60=8.3x10-5
• Hence, when [OH-
] increases by factor of 1.5, initial rate increases
by a factor of 8.3x10-5
/5.5x10-5
=1.5
•Therefore the reaction if 1st
order with respect to NaOH

15. •A reaction mechanism is a sequence of
simple steps proposed in theory to account
for the overall chemical reaction that
takes place& it must be consistent with
the observed kinetics. (satisfy the rate
equation & stoichiometry equation)
Reaction mechanism

16. • The slowest step in the reaction
mechanism is called the rate determining
step. It determines the rate of the
entire multi-step process.
•Therefore, the particles that collide
together in the rate-determining step are
the substance that appear in the rate
equation.
•The order of reaction could be predicted
from the rate-determining step of a given
reaction mechanism.
Rate determining step

17. • CH3CH2CH2CH2Br + OH-
 CH3CH2CH2CH2OH+Br-
•This reaction takes place in one step
•Both CH3CH2CH2CH2Br & OH-
are involved in the slow
step. Hence the rate equation is :
rate=k[CH3CH2CH2CH2Br][OH-
]
•Therefore this reaction is 1st
order with respect to
CH3CH2CH2CH2Br & 1st
order with respect to OH-
•Reaction is bimolecular since 2 reactants are involved
Single-step reaction

18. • (CH3)3C-Br + OH-
 (CH3)3
C-OH + Br-
•This reaction takes place in 2 steps
•1. (CH3)3C-Br  (CH3)3C+
+ Br- slow
•2. (CH3)3C+
+OH-
 (CH3)3C-OH fast
•Only (CH3)3C-Br is involved in the rate determining
step. Hence, rate equation is :
rate = k[(CH3)3C-Br]
•Therefore this reaction is 1st order with respect to
(CH3)3C-Br & zero order with respect to OH-
•Reaction is unimolecular since only one reactant is
involved in the rate-determining step.
Multi-step reaction

19. In the decomposition of hydrogen peroxide,
2H2O2  2H2O + O2, the rate equation is
rate = k[H2O2]. What is the possible
mechanism?
It cannot be a simple 1 step bimolecular
reaction.
Step 1: H2O2  H2O + [O]……..slow
Step 2: [O] + [O]  O2 …….fast
Suggest reaction mechanism from
rate equation

20. (CH3)3C-Cl + OH-
 (CH3)3C-OH +Cl-
Experiment shows that rate of reaction is
1st
order with respect to the organic
chloride & zero order with respect to OH-
.
Suggest a mechanism.
We can conclude that the reaction does not
happen at only 1 step which involves both
reactants.
The rate determining step (slow step)
should involve the organic chloride. The fast
step should involve the OH-
.
(CH3)3C-Cl  (CH3)3C+
+Cl-
…….slow
(CH3)3C+
+OH-
 (CH3)3C-OH ………..fast

21. Predict order from reaction mechanism
Decomposition of hydrogen iodide involves a few steps
HI + HI  H-I ……..slow
H-I
H-I
 H2 + 2I ……..fast
H-I
2I  I2 ………….fast
What is the order?
It is 2nd
order with respect to HI & the stoichiometry
is 2HI  H2 + I2

22. When initial step is not the slow step
Consider this equation & predict the orders:
5Br-
(aq) + BrO3
-
(aq) + 6H+
(aq)  3Br2(aq) + 3H2O(l)
H+
+ Br-
 HBr fast
H+
+ BrO3
-
 HBrO3 fast
HBr + HBrO3  HBrO + HBrO2 slow
HBrO2 + HBr  2HBrO fast
HBrO + HBr  H2O + Br2 fast
Rate = k[Br-
][BrO3
-
][H+
]2

23. • Titrimetric method is useful for reaction
in solutions.
•The concentration of reactants/products
is determined at different times by
withdrawing known volumes of the reaction
mixture & titrating with a suitable
standard solution
Experimental techniques to
study rate of reaction
1.

24. Acid catalysed reaction of propanone with iodine
•CH3COCH3(aq) + I2(aq)  CH2ICOCH3(aq) + HI (aq)
•Unreacted iodine is determined at different times
•Portions of the reaction mixture are pipetted at regular
time intervals into NaHCO3(aq) which “quenches” the
reaction by neutralising the acid catalyst.
•The quenched mixture is then titrated against a
standard solution of Na2S2O3 to determine the amount
of unreacted iodine.
•2S2O3
2-
(aq) + I2 (aq)  S4O6
2-
(aq) +2I-
(aq)
•Vol of S2O3
2-
against time is a straight line showing
rate is constant as [I ] varies. (zero order)
Example 1:

25. Alkaline hydrolysis of an ester
• CH3CO2C2H5(aq) + NaOH(aq)  CH3CO2
-
Na+
(aq) +C2H5OH(aq)
• Unreacted NaOH in the reaction mixture is determined
at different times.
•At various times mixture is withdrawn & run into about
4x its vol of cold water to quench the mixture.
•The NaOH that remains is titrated against a standard
acid (phenolphthalein)
•H+
(aq) + NaOH(aq)  H2O(l) + Na+
(aq)
•Vol of acid against time is a curve with constant half-
life. Rate is α to [OH-
].
Example 2:

26. • Measurement of vol of gaseous product
method is useful for reaction involving
formation of gaseous products.
•Measurement can be done on vol of gas
produced per unit time (with a syringe)
•OR loss of mass per unit time (where gas
is allowed to escape)
2.

27. Reaction between dilute HCl & marble chips
•CaCO3(s) +2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
•CO2(g) is collected in a syringe & its vol measured at
regular time interval
•(Vol of CO2 evolved at the end)- (Vol of CO2 after t) =
amount of unreacted at time t
•Vol of unreacted is plotted against time
•OR loss of mass of conical flask & its content per unit
time is determined.
•Rate of reaction may be found by plotting loss of mass
against time
Example:

28. • Colorimetry method is used when
the colour intensity of a compound is
proportional to its concentration.
3.

29. Acid catalysed reaction of propanone with iodine
•CH3COCH3(aq) + I2(aq)  CH2ICOCH3(aq) + HI (aq)
reddish brown
•The reduction in [I2] (decrease in colour intensity) is
measured in a colorimeter.
•The rate may be found by plotting [I2] against time.
•To study the order, experiment is repeated by varying
the initial concentration of each of the reactants
(propanone, iodine & acid), keeping the other 2 constant
& corresponding [I2] graph against time are plotted.
•Graphs
Example 1:

30. • Clock reaction are used for reactions
accompanied by prominent visual changes.
•The rate may be studied by measuring time
taken for the colour of iodine to disappear.
•Rate α to vol of I2/ time taken
•To study the order, experiment is repeated by
varying the initial concentration of the
reactants, keeping the other 2 constant.
4.

31. Homogeneous catalyst in Fe3+
in
I-
/S2O8
2-
• Homogeneous catalyst - catalyst & reactants same
physical state.
eg1: Oxidation of I2 by S2O8
2-
(all aqueous)
S2O8
2-
+ 2I-
 2SO4
2-
+ I2 Ea too high
step 1: 2I-
+ 2Fe3+
 I2 + 2Fe2+
step2: 2Fe2+
+ S2O8
2-
 2Fe3+
+ 2SO4
2-
•

32. •
Homogeneous catalyst in the
catalytic role of atmospheric NOx
in the oxidation of atmospheric
sulphur dioxide
eg2: Oxidation of atmospheric SO2 by atmospheric
oxides of nitrogen (all gas)
NO2 + SO2  SO3 + NO
NO + ½ O2  NO2
NO2 from car exhaust, SO3 produced dissolves in
water vapour & produce acid rain.

33. •Heterogeneous catalyst – catalyst & reactants
different physical states.
•Usually d-block transition elements, works by
providing a surface for the reactants to be
adsorbed & products are desorbed. (because of
the availability of partially filled 3d orbitals)
•The adsorption weakens the bonds in the
reactant molecules & lowers the Ea & increases
the surface concentration of reactants.
Reactants brought closer together so rate is
increased.
eg: Use of iron as catalyst for the Haber
process.
Heterogeneous catalyst

34. •
Heterogeneous catalyst in Haber
N2(g) + 3H2(g) 2NH3(g) : ∆H = - 92 kJ mol-1
Uncatalysed reaction is very slow at 500°C
Compromise conditions
Conditions Pressure 20000 kPa (200 atmospheres)
Temperature 380-500°C (speed up
reaction but
slightly lower
yield)
Catalyst iron (solid)

35. The 3 catalysts- platinum (Pt), Palladium (Pd) &
rhodium (Rh) –are coated on a ceramic honeycomb
structure (maximise surface area).
2NO(g) + 2CO(g) 2CO2(g) + N2(g) (Rh)
2CO(g) +O2(g)  2CO2(g) (Pt/Pd)
CxHy(g) + (x+y/4)O2(g)  xCO2(g) + y/2H2O (g)
CO, hydrocarbons &NOx
CO2, H2O &N2
heterogeneous catalyst in catalytic
removal of NOx