Rate Equation

7/31/2019 Rate Equation

1/16

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

7/31/2019 Rate Equation

2/16

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.

7/31/2019 Rate Equation

3/16

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

7/31/2019 Rate Equation

4/16

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

7/31/2019 Rate Equation

5/16

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

7/31/2019 Rate Equation

6/16

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

7/31/2019 Rate Equation

7/16

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):

7/31/2019 Rate Equation

8/16

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:

7/31/2019 Rate Equation

9/16

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:

7/31/2019 Rate Equation

10/16

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

7/31/2019 Rate Equation

11/16

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;

7/31/2019 Rate Equation

12/16

, 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

7/31/2019 Rate Equation

13/16

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

,

7/31/2019 Rate Equation

14/16

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

7/31/2019 Rate Equation

15/16

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

7/31/2019 Rate Equation

16/16

. 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