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Rate equationFrom Wikipedia, the free encyclopedia
The rate law orrate equation for a chemical reaction is an equation that links thereaction ratewith
concentrations or pressures of reactants and constant parameters (normally rate coefficients andpartialreaction orders).[1] To determine the rate equation for a particular system one combines the
reaction rate with a mass balancefor the system. [2] For a generic reaction aA + bB C with no
intermediate steps in itsreaction mechanism(that is, an elementary reaction), the rate is given by
where [A] and [B] express the concentration of the species A and B, respectively (usually in moles
per liter (molarity, M));xand yare not the respective stoichiometric coefficients of the balanced
equation; they must be determined experimentally. kis the rate coefficientorrate constantof the
reaction. The value of this coefficient kdepends on conditions such as temperature, ionic
strength, surface area of the adsorbent or light irradiation. For elementary reactions, the rate
equation can be derived from first principles using collision theory. Again, x and y are NOT always
derived from the balanced equation.
The rate equation of a reaction with a multi-step mechanism cannot, in general, be deduced from
the stoichiometric coefficients of the overall reaction; it must be determined experimentally. The
equation may involve fractional exponential coefficients, or it may depend on the concentration of
an intermediate species.
The rate equation is a differential equation, and it can be integratedto obtain an integrated rate
equation that links concentrations of reactants or products with time.
If the concentration of one of the reactants remains constant (because it is a catalystor it is in
great excess with respect to the other reactants), its concentration can be grouped with the rate
constant, obtaining a pseudo constant: If B is the reactant whose concentration is constant,
then r= k[A][B] = k'[A]. The second-order rate equation has been reduced to a pseudo-first-
orderrate equation. This makes the treatment to obtain an integrated rate equation much easier.
Zeroth-order reactions
A Zeroth-order reaction has a rate that is independent of the concentration of the reactant(s).
Increasing the concentration of the reacting species will not speed up the rate of the reaction.
Zeroth-order reactions are typically found when a material that is required for the reaction to
proceed, such as a surface or a catalyst, is saturated by the reactants. The rate law for a zeroth-
order reaction is
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where r is the reaction rate and k is the reaction rate coefficient with units of
concentration/time. If, and only if, this zeroth-order reaction 1) occurs in a closed system, 2)
there is no net build-up of intermediates, and 3) there are no other reactions occurring, it can
be shown by solving amass balance equation for the system that:
If thisdifferential equation is integratedit gives an equation often called the integrated
zero-order rate law.
where represents the concentration of the chemical of interest at a particular
time, and represents the initial concentration.
A reaction is zeroth order if concentration data are plotted versus time and the result is
a straight line. The slope of this resulting line is the negative of the zero order rate
constant k.
The half-life of a reaction describes the time needed for half of the reactant to be
depleted (same as thehalf-life involved in nuclear decay, which is a first-order
reaction). For a zero-order reaction the half-life is given by
Example of a zeroth-order reaction
Reversed Haber process:
It should be noted that the order of a reaction cannot be deduced from the
chemical equation of the reaction.
[edit]First-order reactions
See also Order of reaction.
A first-order reaction depends on the concentration of only one reactant
(a unimolecular reaction). Other reactants can be present, but each will be
zero-order. The rate law for an elementary reaction that is first order with
respect to a reactant A is
kis the first order rate constant, which has units of 1/s.
The integrated first-order rate law is
A plot ofln [A] vs. time tgives a straight line with a slope of k.
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The half-life of a first-order reaction is independent of the starting
concentration and is given by .
Examples of reactions that are first-order with respect to the
reactant:
Further Properties of First-Order Reaction Kinetics
The integrated first-order rate law
is usually written in the form of the exponential decay equation
A different (but equivalent) way of considering first order kinetics is as follows: The
exponential decay equation can be rewritten as:
where tp corresponds to a specific time period and n is an integer corresponding to
the number of time periods. At the end of each time period, the fraction of the reactant
population remaining relative to the amount present at the start of the time period, fRP,will be:
Such that aftern time periods, the fraction of the original reactant population will
be:
where:fBPcorresponds to the fraction of the reactant population that will
break down in each time period. This equation indicates that the fraction of
the total amount of reactant population that will break down in each time
period is independent of the initial amount present. When the chosen time
period corresponds to , the fraction of the population that
will break down in each time period will be exactly the amount present at
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the start of the time period (i.e. the time period corresponds to the half-life of
the first-order reaction).
The average rate of the reaction for the nth time period is given by:
Therefore, the amount remaining at the end of each time period will be
related to the average rate of that time period and the reactant
population at the start of the time period by:
An =An 1 ravg,ntp
Since the fraction of the reactant population that will break down in
each time period can be expressed as:
The amount of reactant that will break down in each time period
can be related to the average rate over that time period by:
Such that the amount that remains at the end of each time
period will be related to the amount present at the start of
the time period according to:
This equation is a recursion allowing for the calculation
of the amount present after any number of time
periods, without need of the rate constant, provided
that the average rate for each time period is known. [3]
[edit]Second-order reactions
A second-order reaction depends on the
concentrations of one second-order reactant, or two
first-order reactants.
For a second order reaction, its reaction rate is given
by:
or or
In several popular kinetics books, the definition of
the rate law for second-order reactions
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is . Conflating the 2 inside
the constant for the first, derivative, form will only
make it required in the second, integrated form
(presented below). The option of keeping the 2 out
of the constant in the derivative form is considered
more correct, as it is almost always used in peer-
reviewed literature, tables of rate constants, and
simulation software.[4]
The integrated second-order rate laws are
respectively
or
[A]0 and [B]0 must be different to obtain
that integrated equation.
The half-life equation for a second-order
reaction dependent on one second-order
reactant is . For a second-
order reaction half-lives progressivelydouble.
Another way to present the above rate
laws is to take the log of both
sides:
Examples of a Second-order reaction
[edit]Pseudo-first-orderMeasuring a second-order reaction rate
with reactants A and B can be
problematic: The concentrations of the
two reactants must be followed
simultaneously, which is more difficult; or
measure one of them and calculate the
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other as a difference, which is less
precise. A common solution for that
problem is the pseudo-first-order
approximation
Ifeither[A] or [B] remains constant as the
reaction proceeds, then the reaction can
be considered pseudo-first-
orderbecause, in fact, it depends on the
concentration of only one reactant. If, for
example, [B] remains constant, then:
where k' = k[B]0 (k' or kobs with units s1)
and an expression is obtained identical to
the first order expression above.
One way to obtain a pseudo-first-orderreaction is to use a large excess of one of
the reactants ([B]>>[A] would work for the
previous example) so that, as the reaction
progresses, only a small amount of the
reactant is consumed, and its
concentration can be considered to stay
constant. By collecting k' for many
reactions with different (but excess)
concentrations of [B], a plot ofk' versus
[B] gives k(the regular second order rate
constant) as the slope.
Example: The hydrolysis of esters by
dilute mineral acids follows pseudo-first-
order kinetics where the concentration of
water is present in large excess.
CH3COOCH3 + H2O CH3COOH + CH3OH
Summary for reaction orders 0, 1, 2, and n
Elementary reaction steps with order 3 (called ternary reactions) are rare and unlikely to occur.
However, overall reactions composed of several elementary steps can, of course, be of any(including non-integer) order.
Zero-Order First-Order Second-Order nth-Order
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Rate Law[4]
Integrate
d Rate
Law [4]
[Except first order]
Units of
Rate
Constant
(k)
Linear
Plot to
determin
e k [Except first order]
Half-life [4
][Except first order]
Where M stands for concentration in molarity(mol L1), tfor time, and kfor the reaction rate
constant. The half-life of a first-order reaction is often expressed as t1/2 = 0.693/k(as ln2 = 0.693).
[edit]Equilibrium reactions or opposed reactions
A pair of forward and reverse reactions may define an equilibriumprocess. For example, A and B
react into X and Y and vice versa (s, t, u, and v are the stoichiometric coefficients):
The reaction rate expression for the above reactions (assuming each one is elementary) can
be expressed as:
where: k1 is the rate coefficient for the reaction that consumes A and B; k2 is the rate
coefficient for the backwards reaction, which consumes X and Y and produces A and B.
The constants k1 and k2 are related to the equilibrium coefficient for the reaction (K) by the
following relationship (set r=0 in balance):
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Concentration of A (A0 = 0.25 mole/l) and B versus time reaching equilibrium kf = 2 min-1 and
kr = 1 min-1
[edit]Simple Example
In a simple equilibrium between two species:
Where the reactions starts with an initial concentration of A, [A]0, with aninitial concentration of 0 for B at time t=0.
Then the constant K at equilibrium is expressed as:
Where [A]e and [B]e are the concentrations of A and B at equilibrium,
respectively.
The concentration of A at time t, [A]t, is related to the concentration of B
at time t, [B]t, by the equilibrium reaction equation:
Note that the term [B]0 is not present because, in this simple
example, the initial concentration of B is 0.
This applies even when time t is at infinity; i.e., equilibrium has been
reached:
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then it follows, by the definition of K, that
and, therefore,
These equations allow us to uncouple the system of
differential equations, and allow us to solve for the
concentration of A alone.
The reaction equation, given previously as:
The derivative is negative because this is the
rate of the reaction going from A to B, and
therefore the concentration of A is decreasing.
To simplify annotation, let x be [A]t, the
concentration of A at time t. Let xe be the
concentration of A at equilibrium. Then:
Since:
The reaction
ratebecomes:
which results in:
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A plot of the
negativenatur
al logarithmofthe
concentration
of A in time
minus the
concentration
at equilibrium
versus time t
gives a straight
line with slope
kf+ kb. By
measurementof Ae and
Be the values
of K and the
two reaction
rate
constants will
be known.[5]
[edit]Generalization of
SimpleExample
If the
concentration
at the time t =
0 is different
from above,
the
simplifications
above are
invalid, and a
system of
differential
equations must
be solved.
However, this
system can
also be solved
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exactly to yield
the following
generalized
expressions:
When the
equilibrium
constant is
close to unity
and the
reaction rates
very fast for
instance
in conformation
al analysis of
molecules,
other methods
are required for
the
determination
of rate
constants forinstance by
complete
lineshape
analysis
in NMR
spectroscopy.
[edit]Consecutivereactions
If the rate
constants for
the following
reaction
are k1 and k2;
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, then the rate
equation is:
For reactant
A:
For reactant
B:
For product
C:
With the
individual
concentrations
scaled by the
total population
of reactants to
become
probabilities,
linear systems
of differential
equations suchas these can
be formulated
as a master
equation. The
differential
equations can
be solved
analytically and
the integrated
rate equations
are
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Thesteady
state approxim
ation leads to
very similar
results in an
easier way.
[edit]Parallel orcompetitivereactions
When a
substance
reacts
simultaneously
to give two
different
products, a
parallel or
competitive
reaction is said
to take place.
Two
first order
reactions:
an
d ,
with
constants k1 an
d k2 and rate
equations
,
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and
The integrated
rate equations
are
then
;
and
.
One important
relationship in
this case
is
One
first order
and one
second
order
reaction:[6]
This can be the
case when
studying a
bimolecular
reaction and a
simultaneous
hydrolysis
(which can be
treated as
pseudo order
one) takes
place: the
hydrolysis
complicates
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the study of the
reaction
kinetics,
because some
reactant is
being "spent"in a parallel
reaction. For
example A
reacts with R
to give our
product C, but
meanwhile the
hydrolysis
reaction takes
away an
amount of A to
give B, a
byproduct:
and
. The rate
equations
are:
and
. Where k1' is
the pseudo first
order constant.
The integrated
rate equation
for the main
product [C]is
, which is
equivalent
to
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. Concentration
of B is related
to that of C
through
The integrated
equations were
analytically
obtained but
during the
process it was
assumedthat
therefeore,
previous
equation for [C]
can only be
used for low
concentrations
of [C]
compared to
[A]0