unit 4 rates

8/10/2019 Unit 4 Rates

1/23

Sahan Satharasinghe


2/23

Average rates and initial rates

The average rateof a chemical reaction over a certain interval of time is the rate of change of concentration of a

particular reactant or product:

Average reaction rate = change in concentration of a substance

Time taken

The unitsof reaction rate = (concentration)/ time = moldm-3s-1

Consider a reaction where 1 mole of a reactant (A) produces 1 mole of product (B):

A B

The concentration of the reactant decreases and the

concentration of product increases as time passes:

The amount of product produced between timest3and t4is less than in the equal time period t1to t2.

The rate of reaction decreases as the

reaction proceeds as the reactant is used up.

Clearly, one can only accurately state the

rate of reaction at a certain time.

At time t2, the rate of reaction is given by the

gradient of the tangent to the curve at this point.

This is different to the rate at the start of the reaction at t = 0, the initial rate,

Initial rateis the rate of the reaction when only an infinitesimal amount of the reactant has been used up.

Measuring the rate of a reaction

To measure the rate of a reaction, it is necessary to find the rate of change of concentration of a reactant

or a product.

E.g. in the decomposition of hydrogen peroxide into oxygen and water,

2H2O2(l) 2H2O(l) + O2(g)

the rate of reaction is:

te

E.g. When an ester is hydrolysed, if the concentration of the ester decreases from 1.00 moldm -3to 0.50

moldm-3in 1.00 hour, then the average rate of reaction over the time interval can be calculated:

Average rate = (1.00

0.50) moldm-3 = 1.39 x 10-4moldm-3

(1.00 x 60 x 60) s


3/23

It is easiest to measure the change in concentration of oxygen, especially when realise that, for a gas, the

concentration is proportional to its volume at a constant pressure and temperature. Thus, can write:

Reaction rate = V(O2)

time taken

Apparatus like that in the figure below could be set up, and the volume of oxygen collected could be

measured every 10 seconds or so. If the volume of oxygen given off is plotted against time, a graph

similar the one below would be obtained.

An apparatus for measuring the rate of decomposition

of hydrogen peroxide

Graph of results from an experiment similar to that

shown above

We can measure the rate at which oxygen is evolved at any time by finding out how steep the curve is.

The slope is found by drawing the tangent to a curve at that time. If we measured the slope at time zero,

we would have measured theinitial rateof the reaction.

To provide data on the rate constant and order of the reaction, it is necessary to obtain values for the

rate at various times. It is then necessary to calculate the concentration of specific reactants at those

times.

Finally, plotting rate against the concentration of a reactant will lead to a graph similar to one of the

types discussed earlier.

Inspection of the shape of the graph will show whether the reaction is zero, first or second order with

respect to the reactant


4/23

The study of the factors that affect the rates of chemical reactions is called chemical kinetics.

1. surface area

2. light source (certain reactions only)

3. temperature

4. pressure (gaseous reactions only)

5. catalyst

6. concentration

Collision theory

prticles must COLLIDE before a reaction can take place not ll collisions led to rection rectnts must possess minimum mount of energy - ACTIVATION ENERGY prticles must pproch ech other in certin reltive wy - STERIC EFFECT

to increase the rate one needs, according to collision theory, to have ...

TEMPERATURE

The graph below shows how the energies of particles in a gas are distributed at two different

temperatures (TK and T+10K).

The number of particles with an energy that exceeds EA(a typical value for the activation energy of a

reaction) is proportional to the shaded area beneath the curve.

Notice that only a small fraction of molecules have sufficient energy to react at TK.

However, when the temperature rises by 10K, the fraction with sufficient energy to react roughly

doubles. This causes the reaction rate to double as well.

Maxwell and Boltzmann

distribution curve

Maxwell and Boltzmannderived equations for the distribution

of kinetic energies amongst the

molecules of a gas.

increase the rate

more frequent collisions

increase particle speed more particles present

more successful collisions

give particles more

energylower the

activation energy


5/23

Arrhenius equation

the fraction of molecules with an energy greater than energy EAis given by:

number of molecules = ln )This suggests that the rate of reaction at temperature, T, is proportional to ln(-EA/RT).

Further, since the rate constant, k, is a measure of the rate of reaction, we can write

k = ln )k = ln A ]

factor A(Arrhenius constant) - a collision frequencyandorientation factor.

Thus, the rate of reaction is governed by the energy of reaction, the frequency at which collisions occur and

the orientation geometry of the collision.

A plot of the rate constant of a reaction, k, againsttemperature, T, gives a typically exponential curve

A plot of ln k against 1/T is a straight line graph ofgradient EA/R with an intercept of ln A

The Arrhenius equation can also be stated in alogarithmic form:

k = exp(-EA/RT)

Taking natural logarithms:

ln k = ln A EA/RT = ln A + (-EA/R) (1/T) The average kinetic energy of the molecules at the lower temperature is less than that at the higher

temperature.

This has two effects on the rate of reaction.

1. Fewer of the colliding molecules have a combined energy greater than or equal to the activation energy.

Thus, a smaller proportion of the collisions results in reaction. This means that the rate of reaction is

considerably reduced, even for a small lowering of temperature.

2. The second effect is that the frequency of collision is also slightly reduced. This also lowers the rate of

reaction.

Ea or that there are fewer successful

collisions. Over time, there will be the same number of successful collisions.

As an approximate guide, a 10C reduction in temperature will cause the rate of a reaction to halve.

A decrease in temperature from 25C to 15C will cause the frequency of collision, and hence the rate, to

decrease by less than 2%.


6/23

PRESSURE

When the pressure of a gaseous mixture is increased, the molecules have the same average kinetic

energy, but they are packed together more closely.

This closer contact causes the frequencyof collision to increase.

Ea

remains unaltered.

An increase in pressure produces an increase in frequency of collision, with the same fraction ofcollisions resulting in reaction, so the rate of reaction increases.

Do not state that there are more collisions; it is the frequency (number of collisions per second) that

has increased.

CONCENTRATION

Increasing concentration = more frequent collisions = increased rate

However: Increasing the concentration of some reactants can have a greater effect than increasing others.

Reactions start off at their fastest then slow as the reactant concentration drops.

E.g. In the reaction A + 2B > C the concentrations might change as shown

the steeper the curve the fster the rte of the rection rections strt off quickly becuse of the greter likelihood of collisions rections slow down with time becuse there re fewer rectnts to collide

Reactants A and B)

Concentration decreases with time

Products C)

Concentration increases with time

the rate of change of concentration is found from the slopeor gradient

the slope t the strt will give give the INITIAL ATE the slope gets less (showing the rate is slowing down) asthe reaction proceeds

the vrition in concentrtion of rectnt or product isfollowed with time

method depends on the rection type nd the properties ofreactants/products


7/23

THE RATE EQUATION, ORDERS, AND RATE CONSTANTS

links the rate of reaction to the concentration of reactants

can only be found by doing actual experiments, not by looking at the equation

Consider a reaction between A and B:

A + B C + D + E

The rate of reaction depends on the concentrations of A and B, but we cannot simply say that the rate of

reaction is proportional to the concentration of A and proportional to the concentration of B. The relationship

is:

Reaction rate = [A]m[B]n= k[A]m[B]n

An expression of this kind is called a rate equation (or, sometimes, rate law). The indices mand nare usually

integers, often 0, 1, or 2, and are characteristic of the reaction.

The reaction is of ordermwith respect to A and of ordern with respect to B. The overall orderof the reaction is(m+ n). The constant of proportionality, k, is called the rate constant (or, sometimes, the velocity constant) for

the reaction.

ORDER OF REACTION

The order of a reaction does not follow from its stoichiometric equation. The rate equation for the reaction,

BrO3-(aq) + 5Br-(aq) + 6H+(aq) 3Br2(aq) + 3H2O(l)

bears no relationship to the ratios of specific reactants and products:

Reaction rate = d [r]dt = k [BrO3-] [Br-] [H+] 2

The reaction is first order with respect to bromate(V), first order with respect to bromide ion, second order

with respect to hydrogen ion and fourth order overall. The negative sign in the differential equation means that

[BrO3-] decreases with time.

E.g. The reaction between sodium thiosulphate and hydrochloric acid,

Na2S2O3(aq) + 2HCl(aq) S(s) + SO2(g) + H2O(l) + 2NaCl(aq)

has a rate equation of the type:

reaction rate = k [Na2S2O3]1[HCl]1= k [Na2S2O3][HCl]

The reaction is first orderwith respect to Na2S2O3and is first orderwith respect to HCl. The reaction

is second order overall.


8/23

1. Zero order reactions

The Differential Rate Law

[A] AT ANY INSTANCEA

t

= half-life - a timescale in which each half-life represents the reduction of the initial population to 50% of itsoriginal state.

Consider the reaction,

A + B C + D

The reaction will be zero order with respect to A if it has a rate equation of the type:

Reaction rate = k [A] 0

Reaction rate = k

as [A] 0= 1

te dAdt kA kconstntThe reaction rate of a zero order reaction (w.r.t. A) is independent of the concentration of A, [A]:

Rate vs. time (A) and Concentration vs. time for a zero order reaction.


9/23

The Integrated Rate Law

[A] = A ktR e l a t i o n s h i p B e t w e e n H a lf l i f e a n d Z er o o r d e r R e a c t i on s

[A] = A Using the integrated form of the rate law, we can develop a relationship between zero-order reactions and thehalf-life.

[A] = A - ktSubstitute

A A k t

Solve for time, t

t= A

Notice that, for zero-order reactions, the half-life depends on the initial concentration of reactant and the rate

constant.

Units of the rate constant

Units of the rate constant, k, in a zero order reaction:

Reaction rate = moldm-3s-1= k

Units of k = moldm-3s-1

Examples of zero order react ions

the iodination of propanone which is catalysed by acid. The reaction rate does not change if the

concentration of iodine is changed:

CH3COCH3(aq) + I2(aq) CH3COCH2I(aq) + HI(aq)

The rate equation for this reaction is zero order w.r.t.I2, something which is, perhaps, unexpected:

reaction rate = k [CH3COCH3]1[H+]1[I2]

0= k [CH3COCH3] [H+]

Ammonia decomposes into nitrogen and hydrogen when heated in the presence of a tungsten

catalyst:

2NH3(g) N2(g) + 3H2(g)

The rate equation for this reaction is:

reaction rate = k [NH3]0= k


10/23

2. First order reactions

Th e D i f ferent ia l Rate Law

If the reaction,

A B + C + D

is first order with respect to A, the rate equation will be:Reaction rate = k [A]1= k [A]

te dAdt kA kAThe reaction rate of a first order reaction is directly proportional to the concentration of A, [A]. A plot of rate

against [A] will give a straight line with slope k:

The rate of reaction is proportional to the concentration for a first order reaction


Units of the rate constant, k, in a first order reaction:

Reaction rate = moldm-3s-1= k [A] = k (moldm-3)

k = moldm-3s-1 = s-1

moldm-3

Units of k = s-1


11/23

The Integral Rate Law

The integrated rate equation for a first order reaction is:

[A]

0is the initial concentration of A (at t = 0)

A plot of ln[A]

against tis therefore a straight line with slope kand intercept ln[A]

o

Alternatively, the integrated rate equation can be expressed in the form:

[A] = [A]

0

e

kt

An example of a reaction with first order behaviour is the conversion of cyclopropane into propene:

The rate equation for this reaction is:

reaction rate = k [cylopropane]

For a first order reaction,

reaction rate = dAdt = k [A]1= k [A]

= - kdt

ln[A]=kt + ln[A]o


12/23

Half lives

For a first order reaction,

ln [A] = ln [A] 0- kt

A is the initial concentration of A (at t = 0)k t { }

A A k t

At time t

[A] k t

ln

t

From this, it is apparent that the concentration of A decays exponentially with time:


13/23

The plot of [A] against time for a first order reaction is an exponential decay.

Thus, the half-life of a first order reaction is constant and is independent of the initial concentration of the

reactants.

E. g. An example of a first order reaction is radioactive decay. Radon decays into daughter products with a

constant half-life of 1590 years. How long will it take radon to decay to 10% of its original radioactivity?

t1/2 = 0.693k

k = 0.693 = 0.693 = 4.36 x 10-4year-1

t1/2 1590

The general relationship is:

kt = ln {[A]0/ [A]}

4.36 x 10-4x t = ln {[A]0/ [A]} = ln {100% / 10%} = ln 10 = 2.302

t = 2.302/ 4.36 x 10-4= 5281 years.

Thus, it takes 5281 years for the radon to reach 10% of its original level.


14/23

3. Second order reactions

Th e D i f ferent ia l Rate Law

If the reaction,

A B + C + D

The reaction rate of a second order reaction is proportional to the square of the concentration of A:

if second order with respect to A, the rate equation will be:

Reaction rate = k [A] 2

te dAdt kA

Graph of reaction rate against concentration for a second order reaction


Units of the rate constant, k, in a second order reaction:

Reaction rate = moldm-3s-1 = k [A] 2= k (moldm-3) 2

k = mol dm-3s-1 = s-1

(mol dm-3) 2

Units of k = mol-1dm+3s-1

Reactions may also be second order overall as in the case of sodium thiosulphate and hydrochloric acid:

Na2S2O3(aq) + 2HCl(aq) S(s) + SO2(g) + H2O(l) + 2NaCl(aq)

Reaction rate = k [Na2S2O3]1 [HCl]1= k [Na2S2O3][HCl]

The overall order is given by the sum of the indices of each component of the rate equation. In this case, the

overall order is given by (1 + 1) =2.


15/23

Integrated rate law

For a second order reaction,

Reaction rate = = k [A] 2

dAA The integrated rate equation for a first order reaction is:

[A]0is the initial concentration of A (at t = 0)

Test for a second order reaction: a plot of

against t gives a straight line with slope k:

A plot of 1/[A]against t for a second order reaction. The slope of the line is k and the intercept is at 1/ [A]0

E.g. Examples of second order reactions are:

NO(g) + CO(g) + O2(g) NO2(g) + CO2(g)

Reaction rate = k [NO]2

and, CH3CHO(g) CH4(g) + CO(g)

Reaction rate = k [CH3CHO]2


16/23


17/23

SUMMARY

Plots of rate of reaction against concentration of reactant for zero, first and second order reaction take the

form:

Graphs of rate of reaction against concentration

Order of reaction Rate equation Plot for linearity Variation of half-life with [A]

0 rate = k [A] against t

1 rate = k[A] ln [A] against t

2 rate = k [A]2 1/[A] against t

Rate

Concentration

First order

Zero order

Second order


18/23


19/23

Finding the rate equation of a reaction by inspection of initial rates

E.g. The reaction between hydrogen and nitrogen monoxide at 800C produces water and nitrogen:

2H2(g) + 2NO(g) 2H2O(g) + N2(g)

The following table gives details of an experiment in which the initial concentrations of hydrogen and nitrogen

monoxide were changed and the initial rates of the subsequent reactions were measured.

Experiment number Initial [NO]/ moldm-3 Initial [H2

]/ moldm

-3 Initial rate /moldm-3s-1

1 6 x 10-3 1 x 10-3 3 x 10-3

2 6 x 10-3 2 x 10-3 6 x 10-3

3 6 x 10-3 3 x 10-3 9 x 10-3

4 1 x 10-3 6 x 10-3 0.5 x 10-3

5 2 x 10-3 6 x 10-3 2.0 x 10-3

6 3 x 10-3 6 x 10-3 4.5 x 10-3

In experiments 1, 2, and 3 the concentration of NO is the same (6 x 10-3moldm-3). Between experiments 1 and 2,

when the concentration of H2is doubled, the rate isdoubled. Between experiments 1 and 3, when the

concentration of hydrogen is tripled, the rate is tripled. This suggests that the rate is directly proportional to

[H2]:

reaction rate = [H2]

In experiments 4, 5 and 6, the concentration of H2is the same (6 x 10-3moldm-3). Between experiments 4 and 5,when the concentration of NO is doubled, the rate isquadrupled. Between experiements 4 and 6, when the

concentration of NO is tripled, the rate increases nine-fold. This suggests that the rate is proportional to [NO]2:

reaction rate = [NO]2

We can combine these two results into a single rate equation:

reaction rate = k [H2] [NO]2


20/23

E.g. The following results were obtained in a study of the reaction between peroxodisulphate and iodide ions:

S2O82-(aq) + 2I-(aq) 2SO42-(aq) + I2(aq)

Experiment [S2

O

8

2-

]/ moldm

-3 [I- ]/ moldm-3 Initial rate/ moldm-3s-

1

1 0.040 0.040 9.6 x 10-6

2 0.080 0.040 1.92 x 10-5

3 0.080 0.020 9.6 x 10-6

Find the order of reaction with respect to (a) S2O82-, (b) I-and (c) find the rate constant. What is the initial rate

of the reaction when [S2O82-]0= 0.12moldm-3, and [I-]0= 0.015moldm-3?

(a) By inspection, [I-] is constant (at 0.040moldm-3)in experiments 1 and 2 whereas [S2O82-] doubles from

0.040moldm-3to 0.080moldm-3. The initial rate doubles from 9.6 x 10 -6to 1.92 x 10-5moldm-3s-1as a

consequence, so the reaction must be first order with respect to S2O82-.

(b) Similarly, [S2O82-] is constant (at 0.080moldm-3)in experiments 2 and 3 whereas [I-] halves from 0.040 to

0.020moldm-3. The initial rate also halves from 1.92 x 10 -5 to 9.6 x 10-6 moldm-3s-1as a consequence, so the

reaction must also be first order with respect to I-.

(c) The rate equation takes the form:

initial rate = k [I-] [S2O82-]

Substituting data from experiment 2, we have:

initial rate = k [I-] [S2O82-] = k 0.080 x 0.040 = 1.92 x 10-5moldm-3s-1

k = 1.92 x 10-5/ (0.080 x 0.040) = 6 x 10-3mol-1dm3s-1

The initial rate of the reaction when [S2O82-]0= 0.12moldm-3, and [I-]0= 0.015moldm-3is given by:

initial rate = k [I-] [S2O82-] = 6 x 10-3x 0.12 x 0.015 moldm-3s-1

initial rate = 1.08 x 10-5moldm-3s-1.


21/23

Reaction Mechanisms

Rate studies allow us to interpret reactions on a molecular level. By considering the order of a reaction with

respect to different reactants, we can speculate about the sequence in which bonds break and atoms rearrange.

From these ideas, it is possible to suggest a reaction mechanism.

Iodination of propanone

As an example, consider the reaction between iodine and propanone:

CH3COCH3(aq) + I2(aq) CH3COCH2I(aq) + HI(aq)

The reaction is catalysed by acid, H+(aq). The rate equation for this reaction takes the form:

reaction rate = k [CH3COCH3]1[H+]1[I2]0= k [CH3COCH3] [H+]

The reaction is first order with respect to both propanone and hydrogen ions (the catalyst) but it is zero order

with respect to iodine.

Iodine must be present for the iodination to occur but its concentration does not appear in the rate equation. To

explain this, it is suggested that the reaction takes place in a number of steps. It is the rate of the slowest step

(the rate

determining step orrate-limiting step) that determines the rate of the overall reaction. If iodine is

involved in a step which is too fast to be rate-determining, it will not appear in the rate equation. The suggested

mechanism for the reaction has three steps. The first step, [1], involves protonation of the oxygen of the

carbonyl group on propanone:

This is the slow, rate-determining step. An enol(i.e. a molecule with adjacent C=C and OH groups) is formed.Since hlogens re electrophiles, iodine rects rpidly with the CC double bond through its electrons, :

The reactive intermediateformed has a positive charge on a carbon atom, and fast loses a proton to form the

iodoketone, [3]:

We could not have deduced this reaction precisely from the rate equation but it does fit in with that equation.

Confirmatory evidence is given by the fact that if we carry out the reaction not with iodine but with heavy

water, D2O, a deuterium atom, D, is taken up by the methyl group of propanone at exactly the same rate as

iodine in the first reaction. The two reactions have the same rate-determining step.

Hydrolysis of halogenoalkanes

The hydrolysis of a halogenoalkane is a substitution reaction. The attacking species is a nucleophile. Hydrolyses

are therefore described as SN

reactions. A great deal of work has been done on the mechanisms of hydrolysis

reactions.


22/23

Hydrolysis of primary halogenoalkanes

Experimental evidence:

The alkaline hydrolysis of a primary halogenoalkane:

RX(l) + OH-(aq) ROH(aq) + X-(aq)

is second order overall, following the rate equation:

Reaction rate = k [RX] [OH-]

One can deduce from this evidence that a molecule of RX and an OH-ion must collide before reaction will occur.

Calculations show that the rate of reaction is much less than the rate at which RX and OH -collide. Only a small

fraction of the collisions result in reaction. It must be necessary for the two species not only to collide but also

to collide with enough energy to overcome the repulsion between the hydroxide ion and the electron density in

the halogenoalkane molecule.

Mechanism:

It is known that in order to minimise repulsion between RX and OH-, OH-approaches the carbon atom attached

to X on the opposite side of the molecule from X:

As a bond forms between O and C, the C

X bond weakens, and a transition state is reached in which C is

partially and simultaneously bonded to O and X. Once the transition state has been formed, it is rapidly

converted into the products. A transition state has momentary existence. The reactive intermediates that have

been postulated in other reactions (for an example, see below) are longer-lived.

The proposed reaction mechanism for the alkaline hydrolysis of primary halogenoalkanes is:

The movement of the bonding electrons is often shown by a curly arrow:

The number of reacting species that are involved in the rate-determining step is known as the molecularityof

the reaction. In the mechanism above, two species (i.e. RX and OH-) are involved, hence the mechanism is

known as a bimolecular reaction.

The full name of the reaction mechanism is a bimolecular nucleophilic substitution, or a SN

2 reaction.


23/23

Hydrolysis of tertiary halogenoalkanes

An example of the hydrolysis of a tertiary halogenoalkane is the hydrolysis of

1,1-dimethylchloroethane in aqueous ethanol:

(CH3)3CCl(l) + H2O(l) (CH3)3COH(l) + HCl(aq)

Experimental results show that the reaction is first order with respect to the halogenoalkane:

reaction rate = k [(CH3)3Cl]

It is inferred that the slow, rate-determining step must involve the halogenoalkane alone. It is suggested that

this step is the dissociation of the halogenoalkane into a carbocation (a reactive intermediate) and a halide ion:

The carbocation formed reacts rapidly with water molecules:

This step is fast and the overall reaction is determined by the rate at which carbocations are formed. The

proposed mechanism for the hydrolysis of a tertiary halogenoalkane is:

Only one species, the halogenoalkane, is involved in the rate-determining step so this is a unimolecular reaction.

The full name for the mechanism is a unimolecular nucleophilic substitution, or a SN

1 reaction.

unit 4 rates

Documents