Doc Brown's
Advanced A Level Chemistry
Advanced A Level Chemistry
Kinetics Part 6
6.3
Experimental techniques i.e. examples of how to obtain rate data and deriving
rate expressions and explaining and deducing orders of reactions with some
exemplar rates questions involving deducing and using orders of reactants and
rate expressions.
Advanced A Level Kinetics
Index
6.3
Obtaining rate data, interpreting rate data, orders of reaction and rate expressions
6.3a.
Examples of obtaining rate data

A BRIEF REVIEW OF METHODS OF COLLECTING
RATE DATA

The speed or rate
of reaction is the rate of removal of reactant or the rate of
formation of product.

Experimental
results can be obtained in a variety of ways depending on the nature
of the reaction e.g.

Collecting a
gaseous product in a gas syringe or inverted burette.

The initial rate is based on the
tangent from the 0,0 origin, over the first few minutes?, which is
usually reasonably linear at the start.

e.g. the
decomposition of hydrogen peroxide by MnO_{2(s)} ,
other transition metal compounds (insoluble–heterogeneous or
soluble–homogeneous) and of course enzymes.

or a metal
reacting with acids, and you can study the effects of a
catalyst e.g. adding Cu^{2+}_{(aq)} ions to a
zinc–acid mixture, though I'm not sure easy it is to get good
quantitative results for advanced level coursework?

i.e. 2H^{+}_{(aq)}
+ M_{(s)} ==> M^{2+}_{(aq)} + H_{2(g)}

Titrating an
acid or alkali where one is produced or used up e.g.

The
hydrolysis of a tertiary chloroalkane produces hydrochloric
acid which can be titrated with standardised alkali NaOH_{(aq)},
or the chloride ion produced can be titrated with
silver nitrate solution, AgNO_{3(aq)}.

R_{3}C–Cl
+ 2H_{2}O ==> R_{3}C–OH + Cl^{–}
+ H_{3}O^{+}

The
hydrolysis of primary halogenoalkanes with sodium
hydroxide can be followed by titrating the remaining
alkali with acid. You may need to use aqueous ethanol as a
solvent since the halogenoalkane is insoluble in water and
a large volume of reactants, so that sample aliquot's can
be pipetted at regular time intervals.

e.g. RCH_{2}Br
+ NaOH ==> RCH_{2}OH + NaBr

or RCH_{2}Br
+ OH^{–} ==> RCH_{2}OH + Br^{–}

Hydrolysis
of an ester with (i) sodium hydroxide or (ii) dil.
hydrochloric acid can be followed by titrating (i) the remaining alkali with standardised acid
HCl(aq) or (ii) using standard alkali to measure the total
acid present – a blank is done to check on the amount of acid
catalyst present and is subtracted from the total titration.

The iodination
of a ketone such as colourless propanone, can be followed by
making colorimetric measurements of the iodine colour intensity
diminishing as the reaction proceeds.

CH_{3}COCH_{3(aq)}
+ I_{2(aq)} ==> CH_{3}COCH_{2}I(aq) +
H^{+}_{(aq)} + I^{–}_{(aq)}

With a
suitably chosen filter, and 'lowish' iodine concentrations,
the absorbance is proportional to concentration, though a
calibration graph should always be produced to check the
relationship.

The oxidation
of iodide to iodine by potassium peroxodisulphate can be followed
by a method known as the 'iodine clock'.

It assumes
the initial rate is reasonably constant in the first few
minutes and only a few % of the reactants are used up during the
'reaction time'.

A small
and constant amount of sodium thiosulphate and starch
solution is added to the reaction mixture.

The first
amount of iodine formed from the reaction by ...

is removed
by reaction with the sodium thiosulphate ...

so the
starch does NOT turn blue, but immediately the next bit of
iodine is formed the starch will turn blue.

Therefore
it is possible to get a reaction time for producing the same
amount of iodine each time. The concentration of iodide,
peroxodisulphate or an added catalyst (e.g. transition metal
ion) can be varied.

How to measure–express initial
rates is introduced in the GCSE Notes on Rates of Chemical Reaction
6.3b.
Rate expressions and orders of reaction

WHAT CAN WE DO WE WITH THE DATA FROM
CHEMICAL KINETICS EXPERIMENTS?

CAN WE DERIVE A MATHEMATICAL EXPRESSION
TO DESCRIBE THE KINETICS OF A GIVEN REACTION?

From experimental
results you need to know how the speed of a
reaction varies with respect to individual reactant concentrations.
Only then is
it possible to derive a rate expression, which summarises what
controls the speed of a particular reaction in terms of the relevant
concentrations, which is not necessarily all the reactants!

The rate at
which a reaction depends on the concentrations of the reactant(s)/catalyst
can be expressed in the form of a rate expression:
Orders of reaction can only be
determined by rate experiments

never from
a balanced
equation, however coincident it may seem, beware!

The orders of
a reaction may or may not be the same as the balancing numbers of
the balanced equations.

The orders of
reaction are a consequence of the
mechanism of the reaction and can only be found from rate
experiments and they cannot be predicted from the balanced
equation.

For example,
many reactions occur via a single bimolecular collision of only
two reactants and no catalyst e.g.

A + B ==> products, giving, by experiment, the rate
expression,

rate =
k[A][B], 1st order for both reactants, 2nd order overall.

Here, the
coincidence is not surprising, the chance of a 'fruitful'
collision is directly dependent on both reactants initially
colliding, its often the slowest step even in a multi–step
mechanism and if there are no other kinetic complications, the
orders of reaction do match the numbers of the balanced
equation

e.g.
the alkaline hydrolysis of esters

RCOOR'
+ OH^{–} ==> RCOO^{–} + R'OH (reactant
ratio 1:1)

rate
= k_{2}[RCOOR'][OH^{–}] (order
ratio 1:1)

For many
other reactions, particularly those involving multi–step
mechanisms and intermediates, things are not so simple, and the orders for each
reactant do not necessarily match completely
the integer ratio of the balanced equation.

e.g.
the acidified oxidation of bromide by bromate(V) ions

BrO_{3}^{–}_{(aq)}
+ 5Br^{–}_{(aq)} + 6H^{+}_{(aq)}
==> 3Br_{2(aq)} + 3H_{2}O_{(l)} (reactant
ratio 1:5:6)

rate
= k[BrO_{3}^{–}_{(aq)}][Br^{–}_{(aq)}][H^{+}_{(aq)}]^{2}, 4th order overall! (order
ratio 1:1:2)

and
sometimes, despite a multi–step mechanism, the rate expression
can be quite simple,

e.g.
the oxidation of iodide by peroxodisulphate goes in at least
two steps with a reactant ratio of 1:2,

S_{2}O_{8}^{2–}_{(aq)} + 2I^{–}_{(aq)}
==> 2SO_{4}^{2–}_{(aq)} + I_{2(aq)},
but the rate expression

rate =
k[S_{2}O_{8}^{2–}_{(aq)}][I^{–}_{(aq)}],
is overall 2nd
order (1+1) for the uncatalysed reaction.

We can examine
theoretically the effect of changing concentration on the rate of
reaction by using a simplified rate expression of the form for a
single reactant ..

rate r = k
[A]^{a}

the constant
k, will include the rate constant k and any other constant
reactant/catalyst concentration
terms,

[A] is
the variable concentration of a reactant A,

a = the
order
with respect to reactant A,

and the
'theoretical results' are shown in the table below for an initial
concentration x.

The numbers in
bold show the factor change in concentration and its effect
on the rate.
Example 
relative
concentration of the reactant [A] 
relative
rate r for orders 0, 1 and 2 (the rate
factor change) 
a =
0 (zero order) 
a =
1 (1st order) 
a =
2 (2nd order) 
Ex.
1 
initially
x 
r =
k (x)^{0} = k 
r =
k (x)^{1} = kx 
r =
k (x)^{2} = kx^{2} 
Ex.
2 
2x 
r =
k (2x)^{0} = k 
r =
k (2x)^{1} =
2kx 
r =
k (2x)^{2} =
4kx^{2} 
Ex.
3 
3x 
r =
k (3x)^{0} = k 
r =
k (3x)^{1} =
3kx 
r =
k (3x)^{2} =
9kx^{2} 
Ex.
4 
x/2 
r =
k(x/2)^{0} = k 
r =
k (x/2)^{1} = kx/2 
r =
k (x/2)^{2} = kx^{2}/4 
Ex.
5 
x/3 
r =
k (x/3)^{0} = k 
r =
k (x/3)^{1} = kx/3 
r =
k (x/3)^{2} = kx^{2}/9 
Example
of a rate
expression 
Examples
of rate r for relative concentrations in terms of x
initially for [A] and y initially for [B]
(the
rate factor change) 
conc^{n}s
of A and B
initial x, y

conc^{n}s
of
A and B x, 2y 
conc^{n}s
of
A and B 2x, y

conc^{n}s
of A and B 2x,
2y

conc^{n}s
of
A and B x, y/2

conc^{n}s
of
A and B x/3, y

conc^{n}s
of
A and B 2x, y/3 
conc^{n}s
of
A and B x/_{2}, y/_{2}

r = k
[A][B]
1st order wrt to both A and B,
overall 2nd order 
'initial'
x, y
r = kxy
'baseline' relative rate

r = kx2y
r =
2kxy
double B, doubles rate

r = k2xy
r =
2kxy
double A, doubles rate

r = k2x2y
r =
4kxy
double both A and B, rate quadrupled

r = kxy/2
r = kxy/2
B halved, rate halved

r = k(x/3)y
r = kxy/3
reduce A by ^{1}/_{3}rd,
rate reduced by ^{1}/_{3}rd

r = k2x(y/3)
r =
^{2}/_{3}kxy
double A, ^{1}/_{3}B,
rate reduced by ^{2}/_{3}rds 
r =
k(x/2)(y/2)
r = kxy/4
halve both A and B, rate reduced by
^{1}/_{4}

r = k
[A][B]^{2}
1st order A and 2nd order, overall 3rd order 
'initial'
x, y r = kxy^{2}
'baseline' relative rate

r =
kx(2y)^{2}
r =
4kxy^{2}
double B, quadruple rate

r = k2xy
r =
2kxy^{2}
double A, double rate

r =
k2x(2y)^{2}
r =
8kxy^{2}
double both A and B, rate inc. 8x

r = kx(y/2)^{2}
r = kxy^{2}/4
halve B, ^{1}/_{4}
rate

r = k(x/3)y^{2}
r = kxy^{2}/3
reduce A by ^{1}/_{3}rd,
rate reduced ^{1}/_{3}

r =
k2x(y/3)^{2}
r =
^{2}/_{9}kxy^{2}
double A, ^{1}/_{3}
B, rate drops by ^{2}/_{9}

r =
k(x/2)(y/2)^{2}
r = kxy^{2}/8
both A and B halved, rate red. by
^{1}/_{8}th

6.3c.
Deducing orders of reaction

HOW DO WE DEDUCE ORDERS OF REACTION?

In rate investigations,
each reactant/catalyst concentration should be individually investigated to
determine its individual order and ultimately to derive the full rate
expression for the reaction. Any other reactant/catalyst concentration must
be kept constant.

The graph below show
typical changes in concentration (or amount of moles remaining) of a
reactant with time, for zero, 1st and 2nd order. The 2nd order graph tends
to be a bit steeper than 1st order BUT that proves nothing! See the second
graph/plot diagram. In the zero order graph the gradient is constant as the
rate is independent of concentration. For the 1st/2nd order graphs, the
initial gradient can be taken as the initial rate as the gradient gradually
decreases as the rate is concentration dependent and the rate decreases as
the concentration decreases.
Connect the graphs with the following:

n=0, zero
order, rate = k(conc^{n})^{0}, rate = k, units
of k = mol dm^{–3 }s^{–1} for
overall zero order, easily obtained from the gradient.

n=1, 1st
order, rate = k(conc^{n}), k = rate/(conc^{n}),
units of k
= mol dm^{–3 }s^{–1} / mol dm^{–3} =
s^{–1} for overall 1st order.

n=2, 2nd
order, rate = k(conc^{n})^{2}, k = rate/(conc^{n})^{2},
units of k
= mol dm^{–3 }s^{–1} / (mol dm^{–3})^{2}
= mol^{–1} dm^{3} s^{–1} for
overall 2nd order.

Some possible graphical
results are shown above.

[1]
shows the results for zero order, where the rate is independent of
concentration.

[2]
shows the results for 1st order, where the rate is directly/linearly
proportional to concentration.

[3]
shows the results for 2nd order, where the rate is proportional to
concentration squared.

[4]
means it cannot be 1st order, order could be <1?

[5]
means it cannot be 2nd order, order could be <2 e.g. 1st? check via
plot [2].

[6]
means it cannot be 1st order, order could be >1 e.g. 2nd? check via
plot [3].

[7]
means it cannot be 2nd order, suggests order >2.

Of
course [4]
to [7]
could simply represent inaccurate data!

There is another
graphical way of showing the order with respect to a reactant is 1st order,
but it requires accurate data showing how the concentration or moles
remaining of a reactant changes with time within a single experiment (apart
from repeats to confirm the pattern).

The rate of radioactive
decay is an example of 1st order kinetics. The decay curve should show
that the time taken for the remaining radioisotope or count rate to
halve is always the same time interval – the
half–life!

A plot of
concentration/moles of reactant remaining versus time, should show the
same pattern as for radioactive decay. The graph below shows what
happens to a reactant with a half–life of 5 minutes.

In other words
100% ==> 50% ==> 25% etc. at 5 minute time intervals, but the
y axis could be concentration or moles left etc.

It is the
constancy of the half–life which proves the 1st order kinetics.

The mathematics of
1st order rate equations (units).

rate = –d[A]/dt =
k_{1}[A]

on integration we
get: k_{1}t = ln[A_{o}] – ln[A]

rate (mol
dm^{–3 }s^{–1})

k_{1}
= 1st order rate constant (s^{–1})

[A_{o}]
= initial concentration of A at t=0 (mol dm^{–3})

[A]
= concentration of A at time t (mol dm^{–3})

A graph of
ln[A] versus t will have a gradient of –k_{1}

The half–life
= t_{1/2} = 0.693/k_{1} (s) and is
independent of the initial concentration.

The above equations
can be expressed in amounts of A e.g. in number of moles of A.

If a =
initial moles of A, x = moles of A reacted,

so (a–x)
= number of moles A remaining at time t

rate = –d(a–x)/dt
= k(a–x)

on integration
we get: kt = ln(a) – ln(a–x)

A plot of
ln(a–x) versus t has a gradient of –k,

and t_{1/2}
= 0.693/k (s)

It is also possible from
the type of graph of concentration versus time, to measure the initial
gradient, which gives the initial rate of reaction (applies to
any order of reaction analysis).
From the point of view
of coursework projects the detailed analysis described above is required,
but quite often in examination questions a very limited amount of data is
given and some clear logical thinking is required. So simplified rate data
questions and their solution is given below.
6.3d.
Simple
exemplar rates questions to derive rate expressions
A FEW EXAMPLES OF PROBLEM SOLVING IN
CHEMICAL KINETICS
From the point of view
of coursework projects the detailed analysis described above is required,
but quite often in examination questions a very limited amount of data is
given and some clear logical thinking is required. So simplified rate data
questions and their solutions avoiding graphical analysis are given below.
Example Q1 The
following rate data was obtained at 25^{o}C for the reaction: A +
2B ==> C
Expt. no 
[A]/mol dm^{–3}

[B]/mol dm^{–3}

rate of
formation of C mol dm^{–3}s^{–1}

1 
0.10 
0.05 
0.02 x 10^{2}

2 
0.10 
0.10 
0.04 x 10^{2}

3 
0.05 
0.10 
0.01 x 10^{2}

4 
0.10 
0.20 
0.08 x 10^{2}

(a) Deduce the
order of reaction with respect to reactant A.
Compare expts.
2 and 3. [B] is kept constant, but on halving A the rate is reduced by a
factor of ^{1}/_{4} (^{1}/_{2} x ^{1}/_{2}) so the rate is proportional to [A]^{2}. Therefore
the order with respect to A is 2 or 2nd order. You can think the
other way round i.e. from expt. 3 to 2, [A] is doubled and the rate
quadruples.
(b) Deduce the
order of reaction with respect to reactant B.
Compare expts. 1 and
2. [A] is kept constant but doubling [B] doubles the rate, so the
reaction is directly proportional to [B]. Therefore the order with
respect to B is 1 or 1st order. Similarly, comparing expts. 1 and
3 (2x [B] gives 2x rate) or comparing expts. 1 and 4 (4x [B] gives 4x
rate).
(c) What is the
overall order of the reaction between A and B?
Total order = 2 + 1
= 3, 3rd order overall.
(d) Write out the
full rate expression.
rate = k [A]^{2}
[B]
(e) Calculate the
value and units of the rate constant.
rearranging rate = k
[A]^{2} [B] gives k = rate / [A]^{2} [B]
and the units will
be mol dm^{–3}s^{–1 }/ (mol dm^{–3})^{2}(mol
dm^{–3}) = s^{–1} / (mol dm^{–3})^{2} =
mol^{–2} dm^{6} s^{–1}
so, using the data
from expt. 1 (or any set, assuming data perfect) gives
k = 0.02 x 10^{2}
/ (0.1^{2} x 0.05) = 4 x 10^{3} mol^{–2}
dm^{6} s^{–1}
or expt. 3 gives
0.01 x 10^{2} / (0.05^{2} x 0.1) = 4 x 10^{3}
etc. check the same for expts. 2 and 4. (hope they work out ok)
Its
not a bad idea to repeat the calculation with another set of data as a
double check!
(f) What will be the
rate of reaction if the concentration of A is 0.20 mol dm^{–3}
and the concentration of B is 0.30 mol dm^{–3}?
You just substitute
the values into the full rate expression:
rate =
k [A]^{2} [B] = 4 x 10^{3} x 0.2^{2} x 0.3 =
0.48 x 10^{2} mol dm^{–3 }s^{–1}
Note: The reacting mole ratio is 2 : 1
BUT that does not mean that the orders are a similar ratio (since here,
it happens to be the other way round for the individual orders). Orders of
reaction can only be obtained by direct experiment and their
'complication' are due to complications of the actual mechanism, which
can be far from simple.
Example Q2
Advanced A Level Kinetics
Index
Advanced Level Theoretical
Physical Chemistry
of chromium – A level Revision notes to help
revise for GCE Advanced
Subsidiary Level AS Advanced Level A2 IB
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(equal to US grade 11 and grade 12 and AP Honours/honors level courses)
Experimental techniques to get rate data and how to derive
rate expressions how to deduce orders of reactions with some
worked out questions involving deducing and using orders of reactants and
rate expressions.
Advanced A Level Kinetics
Index 