Doc Brown's
Advanced A Level Chemistry Advanced A Level Chemistry  KineticsRates
revision notes Part 5
5.3
Rate data, orders of reaction and rate expressions (rate equation) for more advanced kinetics
analysis
Experimental techniques i.e. examples of how to obtain rate
data, explaining and deducing orders of reactants/reactions, calculating rate
constants and deducing their units, with some
exemplar rates questions involving deducing and using orders of reactants and
rate expressions including graphical analysis.
Advanced A Level Chemistry Kinetics
Index
GCSE/IGCSE rates reaction notes INDEX
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5.3
Obtaining rate data, interpreting rate data, orders of reaction and rate expressions
5.3a.
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 peroxodisulfate 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 thiosulfate 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 thiosulfate ...

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,
peroxodisulfate 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
5.3b.
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 (rate equation):
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 peroxodisulfate 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 with respect 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 B, 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

5.3c.
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.

In the zero order graph the gradient is
constant as the rate is independent of concentration, so the graph is of a
linear descent in concentration of reactant.

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.

The 2nd order graph tends
to 'decay' more steeply than 1st order BUT that proves nothing!

The idea is that somehow you test for the
order with an appropriate linear graph ... read on ...
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.
5.3d.
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.
These examples do NOT involve graphs directly, but
a 'graphical' section of examples has been added in section 5.3e. I've made
the numbers quite simple to follow the logic of the argument. I've also shown
how to calculate the rate constant.
These
example calculations below are based on the initial rate of reaction analysis 
so we are assuming the variation of concentration with time for each
experimental run has been processed in some way e.g. by graphical analysis, to
obtain an initial rate of reaction.
The graph on the left illustrates the initial
rate method for the formation of product. The gradients A and B would be for two
different concentrations of a reactant, the concentration for A would be greater
than the concentration of B. The initial rate is taken as the positive tangent 
gradient for the curve at the point 0,0. The same argument applies if you
imagine the graph inverted and you were following the depletion of a reactant.
Then you would get two negative gradients one steeper than the other for the
greater concentration.
Reminder [x] means concentration of x,
usually mol dm^{3}
Example 1.
The table below gives some initial data for
the reaction: A + B ===> products

initial [A]/mol
dm^{3} 
initial [B]/mol dm^{3} 
initial rate/mol dm^{3}
s^{1} 
run (i) 
0.1 
0.1 
3.0 x 10^{4} 
run (ii) 
0.2 
0.1 
6.0 x 10^{4} 
run (iii) 
0.2 
0.2 
6.0 x 10^{4} 
From runs (i) and (ii), keeping [B] constant, by doubling
[A], the rate is doubled, so 1st order with respect to reactant A.
From runs (ii) and (iii), keeping [A] constant, by
doubling [B], the rate is unchanged, so zero order with respect to
reactant B.
Therefore the reaction is 1st order overall, and
the rate expression is ...
rate = k [A] ([B] can be omitted,
anything to the power zero is 1)
To calculate the rate constant, rearrange the rate
expression and substitute appropriate values into it.
Therefore, using run (iii) ...
k = rate constant = rate / [A] = 6.0 x 10^{4}/0.2
= 3.0 x 10^{3} s^{1}
In reality the results would be not this perfect and you
would calculate k for each set of results and quote the average!
Example 2.
The table below gives some initial data for
the reaction: A + B ===> products

initial [A]/mol
dm^{3} 
initial [B]/mol dm^{3} 
initial rate/mol dm^{3}
s^{1} 
run (i) 
0.1 
0.1 
3.0 x 10^{4} 
run (ii) 
0.2 
0.1 
3.0 x 10^{4} 
run (iii) 
0.2 
0.2 
1.2 x 10^{3} 
From runs (i) and (ii), keeping [B] constant, by doubling
[A], the rate is unchanged, so zero order with respect to reactant A.
From runs (ii) and (iii), keeping [A] constant, by
doubling [B], the rate is quadrupled, so 2nd order with respect to
reactant B.
Therefore the reaction is 2nd order overall, and
the rate expression is ...
rate = k [B]^{2} ([A] can be
omitted, anything to the power zero is 1)
To calculate the rate constant, rearrange the rate
expression and substitute appropriate values into it.
Therefore, using run (iii) ...
k = rate constant = rate / [B]^{2} = 1.2 x
10^{3}/0.2^{2} = 3.0 x 10^{2} mol^{1} dm^{3}
s^{1}
In reality the results would be not this perfect and you
would calculate k for each set of results and quote the average!
Example 3.
The table below gives some initial data for
the reaction: A + B ===> products

initial [A]/mol
dm^{3} 
initial [B]/mol dm^{3} 
initial rate/mol dm^{3}
s^{1} 
run (i) 
1.0 
1.0 
2.0 x 10^{3} 
run (ii) 
2.0 
1.0 
4.0 x 10^{3} 
run (iii) 
2.0 
2.0 
8.0 x 10^{3} 
From runs (i) and (ii), keeping [B] constant, by doubling
[A], the rate is doubled, so 1st order with respect to reactant A.
From runs (ii) and (iii). keeping [A] constant, by
doubling [B], the rate is doubled, so 1st order with respect to
reactant B.
Therefore the reaction is 2nd order overall, and
the rate expression is ...
rate = k [A] [B]
To calculate the rate constant, rearrange the rate
expression and substitute appropriate values into it.
Therefore, using run (iii) ...
k = rate constant = rate / ([A][B] = 8 x 10^{3}/(2.0
x 2.0) = 2 x 10^{3} mol^{1} dm^{3} s^{1}
In reality the results would be not this perfect and you
would calculate k for each set of results and quote the average!
Example 4.
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}

initial 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 for the reaction as a whole.
(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.
5.3e
More datagraph work  deducing orders
of a reaction by graphical analysis
Example 1. The inversion (hydrolysis) of
sucrose
sucrose + water ===> glucose +
fructose
In simple terms: C_{12}H_{22}O_{12}
+ H_{2}O ====> C_{6}H_{12}O_{6}
+ C_{6}H_{12}O_{6}
Some rate data for the inversion of sucrose
is given below.
A linear graph, the gradient of the graph of
concentration versus time does not change, therefore a zero order reaction.
Therefore the rate expression is ...
rate = k (mol dm^{–3} s^{–1})
A nonlinear graph of concentration versus
time would suggest first or second order kinetics.
This zero order reaction occurs when the
enzyme (invertase) concentration is low and the substrate (sucrose)
concentration is high. The maximum number of enzyme sites are occupied, which is
itself a constant at constant enzyme concentration.
For more details see section 7.3
Enzyme kinetics
Example 2. Analysing a single set of data to
deduce the order of reaction
The data below are for the hydrolysis of
2–chloro–2–methylpropane in an ethanol–water mixture.
The water concentration is effectively
constant.
The reaction is: 2–chloro–2–methylpropane +
water ===> 2–methyl–propan–2–ol + 'hydrochloric acid'
(CH_{3})_{3}CCl
+ H_{2}O ====> (CH_{3})_{3}COH
+ H^{+} + Cl^{–}
A graph is drawn of (CH_{3})_{3}CCl
concentration versus time.
From the graph the gradient (relative rate)
was measured at 6 points.
Since the gradient (rate) changes with
concentration, it cannot be a zero order reaction.
gradient 
(CH_{3})_{3}C–Cl
concentration 
(conc'n)^{2} 
gradient = rate 
Rate in 
points 
[RX] in mol dm^{–3} 
[RX]^{2} 
rate = d[RX]/dt 
mol dm^{–3} s^{–1} 
(0) 
0.0 
0.000 
0 
0.00000 
(1) 
0.1 
0.010 
0.112/175 
0.00064 
(2) 
0.2 
0.040 
0.2/150 
0.00133 
(3) 
0.3 
0.090 
0.3/177 
0.00169 
(4) 
0.4 
0.160 
0.29/122 
0.00238 
(5) 
0.5 
0.250 
0.2/75 
0.00267 
A graph is then drawn of rate versus
concentration of (CH_{3})_{3}CCl (RX)
The graph is 'reasonably linear'
suggesting it is a 1st order reaction.
Therefore the rate expression is ...
rate = k[(CH_{3})_{3}CCl]
To put this graph in perspective, a 2nd order
plot is done below of rate versus [RX]^{2}. (see data table above)
Wherever you draw a straight line, the data
does not express itself as a linear plot and cannot be a 2nd order reaction.
Nucleophilic substitution by water/hydroxide ion
[S_{N}1 or S_{N}2, hydrolysis to
give alcohols]
for a wider ranging mechanistic and kinetic
analysis of the hydrolysis of haloalkanes.
Example 3. Kinetics of the thermal
decomposition of hydrogen iodide
A single set of reaction rate data at a
temperature of 781K
Reaction: 2HI_{(g)} ==> H_{2(g)}
+ I_{2(g)}

for 1st order plot 
for 2nd order plot 
graph gradient 
time/s 
[HI]/mol dm^{3} 
[HI]^{2}/mol^{2 }dm^{6} 
RATE/mol dm^{3}s^{1} 
0 
0.02000 
0.00040000 

500 
0.01736 
0.00030140 
0.000004640 
1000 
0.01536 
0.00023590 
0.000003600 
1500 
0.01376 
0.00018930 
0.000002890 
2000 
0.01247 
0.00015550 
0.000002360 
2500 
0.01140 
0.00013000 
0.000001980 
3000 
0.01049 
0.00011000 
0.000001680 
3500 
0.00972 
0.00009450 
0.000001430 
4000 
0.00906 
0.00008208 

The concentration of hydrogen iodide was
measured every 500 seconds for 4000 seconds.
A plot of HI concentration versus time
(above) was curved showing it could not be a zero order reaction with respect to
the concentration of HI.
From the [HI] versus time graph (above) the
tangent angle (gradient = rate) was accurately measured from 500 s to 3500 s at
500 second intervals (results tabulated above).
The rate of reaction was is then plotted
against HI concentration to test for 1st order kinetics. This proved to be a
curve  compare the blue rate data curve with the black 'best straight line'
(courtesy of Excel!), so not 1st order either.
The third graph is a plot of HI decomposition
rate versus [HI] squared, and, proved to linear  the blue data line was pretty
coincident with a black 'best straight line'.
This proved that the decomposition of
hydrogen iodide reaction is a 2nd order reaction.
Therefore the rate expression is ...
rate = k [HI]^{2}
For more on this reaction see
KINETICS section 7.4
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