9-Kinetics of the Hydrolysis

8/3/2019 9-Kinetics of the Hydrolysis

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Chemical kinetics

Chemical reactions, reaction rate

Chemical kinetics is the part of physical chemistry that studies reaction rates.

Thereaction rate

orrate of reaction

for a reactant or product in a particular reaction isintuitively defined as how fast a reaction takes place. For example, the oxidation of iron under

the atmosphere is a slow reaction which can take many years, but the combustion of butane in a

fire is a reaction that takes place in fractions of a second.

Consider a typical chemical reaction

aA + bB pP + qQ

The lowercase letters (a, b, p, and q) represent stoichiometric coefficients, while the capital

letters represent the reactants (A andB) and the products (P and Q).

According to IUPAC's Gold Book definition the reaction rate v for a chemical reaction occurringin a closed system under constant-volume conditions, without a build-up of reaction

intermediates, is defined as:

dt

dc

qdt

dc

pdt

dc

bdt

dc

av

QPBA 1111==== (1)

where cI, I=A, B, P, or Q is the concentration of substance. The IUPAC recommends that the

unit of time should always be the second. Reaction rate usually has the units of mol dm-3

s-1

. It is

important to bear in mind that the previous definition is only valid for a single reaction, in a

closed system ofconstant volume.

The quantity

dt

d =& (2)

defined by the equation

dt

dn

qdt

dn

pdt

dn

bdt

dn

a

QPBA 1111====& (3)

where nI designates the amount of substance I (I=A, B, P, or Q) conventionally expressed inunits of mole, may be called the 'rate of conversion' (extent of reaction) and is appropriate

when the use of concentrations is inconvenient, e.g. under conditions of varying volume. In a

system of constant volume, the rate of reaction is equal to the rate of conversion per unit volume

throughout the reaction. For a stepwise reaction this definition of 'rate of reaction' (and 'extent of

reaction', ) will apply only if there is no accumulation of intermediate or formation of sideproducts. It is therefore recommended that the term 'rate of reaction' be used only in cases where

it is experimentally established that these conditions apply.

The rate law or rate equation for a chemical reaction is an equation which links the reaction

rate with concentrations or pressures of reactants and constant parameters (normally rate

coefficients and partial reaction orders). To determine the rate equation for a particular system

one combines the reaction rate with a mass balance for the system. For a generic reaction A + B C the simple rate equation is of the form:

n

B

m

A ckcv = (4)


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Zero-orderreactions are often seen for thermal chemical decompositions where the reaction rate

is independent of the concentration of the reactant (changing the concentration has no effect on

the speed of the reaction). The rate law for a zero-order reaction is

kv = (8)

where v is the reaction rate, and kis the reaction rate coefficient with units of concentration/time.

If, and only if, this zero-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 a mass

balance for the system that:

kdt

dcv A == (9)

If this differential equation is integrated it gives an equation which is often called the integrated

zero-order rate law

0AA cktc += (10)

where cA represents the concentration of the chemical of interest at a particular time, and cA0represents the initial concentration. A reaction is zero order if concentration data are plotted

versus time and the result is a linear function. The slope is the zero order rate constant k.

The half-life of a reaction describes the time needed for half of the reactant to be depleted. For a

zero-order reaction the half-life is given by

k

ct A

20

2

1 = (11)

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.

A Products

The rate law for an elementary reaction that is first order with respect to a reactantA is

AA ck

dt

dcv .== (12)

kis the first order rate constant, which has units of 1/time. The integrated first-order rate law

is

tkc

c

A

A .ln0

= (13)

A plot of ln cA vs. time tgives a linear function with a slope of k. One can easy express the

concentrationcA of reactantA at any time t

tk

AA ecc.

0 .

= (14)

where0Ac is the initial concentration of reactantA.


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The half life of a first-order reaction ( 1 2t ) is independent of the starting concentration and is

given by

1 2

ln 2t

k= (15)

Kinetics of a first-order reaction characterizes also a radioactive decay, the Eq. 15 describes the

time taken for half the radionuclide's atoms to decay.

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:

2

Akcv = (16) or BAckcv = (17)

We will deal with the bimolecular reaction

A +B Products,

supposing the same initial concentration ofA andB reactants,c0A=c0B=c0. The differential ratelaw for the second-order reaction is then

2.ck

dt

dc= (18)

Solving the differential equation, one can obtain

tkcc

.11

0

= (19)

where c is the concentration of reactant at time t(cA=cB=c), and kis the second-order constant,which has dimension of concentration

-1time

-1(eg. dm

3mol

-1s

-1). In this case, a characteristic plot

which will produce a linear function is 1/c vs. time t, with a slope ofk(Fig. 1).

The half-life equation for a second-order reaction dependent on one second-order reactant is

0

2/1.

1

ckt = (20)

Fig. 1 Plot 1/c vs. t.


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Task:

Determine the rate constants and the activation energy of the alkaline hydrolysis of ethyl acetate

using sodium hydroxide.

This experiment illustrates a bimolecular reaction (reacting species are ethyl acetate and sodium

hydroxide):

CH3 COO CH2 CH3 + NaOH CH3 COONa + CH3 CH2 OH (21)

The initial concentrations of the reacting species are the same 0Ac = 0Bc .

Equipments: thermostat, pipettes, burette, volumetric flasks, titrimetric flasks, stop-clock

Chemicals: solution of ethyl acetate ( c = 0,04 mol.dm-3

),

solution of sodium hydroxide ( c = 0,04 mol.dm-3

),

solution of hydrochloric acid ( c = 0,04 mol.dm-3

),

phenolphthalein

Procedure:

1. Transfer 50 ml of the solution ofethyl acetate ( c = 0,04 mol.dm-3

) into the volumetric flask

(V=50 ml) and 50 ml of the solution ofsodium hydroxide ( c = 0,04 mol.dm-3

) into another

volumetric flask (V=200 ml). Both flasks cork down and put them in the thermostated bath

(20 C).

2. Fill the burette with the solution of sodium hydroxide ( c = 0,04 mol.dm-3

).

3. Pipette HClV =5 ml solution ofhydrochloric acid ( c = 0,04 mol.dm

-3

) into a clean anddry titrimetric flask.

4. After 10 minutes, take out the flasks with solutions from thermostated bath and pour the

solution of ethyl acetate to the solution of sodium hydroxide, put the mixture again into the

thermostated bath and start the stop-clock.

5. 5 minutes after mixing, pipette 10 ml of reaction mixture (leave the flask in the bath!) to the

titrimetric flask (with 5 ml of HCl).Remark: HCl stops the reaction given by the Eq. (21).

6. Titrate with the solution of sodium hydroxide adding 1 drop of phenolphthalein as

indicator. When the endpoint of titration has been reached, read the used volume of NaOH

from the burette (VNaOH). Write it down to the Tab.1.

7. Repeat the step 5 and 6 every 5 minutes six times more (in the 10th, 15th, 20th, 25th, 30th

and 35th min. from the moment of mixing).8. Write down to the Tab. 1the temperature of the bath.

9. Repeat the same experiment at 30 C. Because the reaction is faster, the times for titrations

will be in the 5th,10th, 15th, 20th, and 25th min. from the mixing. Write down to the

Tab.1 used volume of NaOH for each titration.


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Table 1. Measured and calculated values

Data treatment

1. Calculate the concentration c (mol dm-3)in the Tab. 1according to

V

cVcVc NaOHNaOHHClHCl

.. =

where: ,HCl NaOH c c are concentrations of HCl a NaOH, respectively (in mol dm-3

)

HClV thevolume of HCl (5 ml)

NaOHV thevolume of NaOH inml

V thevolume of the reaction mixture used in titration (10 ml)

2. Calculate the 1/c values.

3. Use MS Excell to create the dependence )(1 tfc= at given temperature. Fit the experimental

points with a linear function. The slope represents the value of the rate constant kat given

temperature. If the time is in minutes, the unit of the rate constants is dm3

mol-1

min-1

.

4. Calculatethe activation energyEa according to Eq. 7 (R = 8,314 J.K-1

.mol-1

).

The report must include:

Theoretical principles

Equipment and chemicals

Working procedure and measurements

Table of results, calculations, diagrams 1/ ( )c f t= at two temperatures, and the value of the

activation energy.

Literature:

Kopeck F. et al.: Practical and numerical exercises from physical chemistry for students of pharmacy, Faculty of

Pharmacy, Comenius University, Bratislava, 1989, in Slovak

http://en.wikipedia.org/wiki/Reaction_rate

http://www.chm.davidson.edu/ChemistryApplets/kinetics/IntegratedRateLaws.html

http://cnx.org/content/m12728/latest/

J. Oremusov, Manual for laboratory practice in physical chemistry for students of pharmacy, Department of

Physical Chemistry, Faculty of Pharmacy, Comenius University, Bratislava, 2007, in Slovak

Manual written by Doc. RNDr. D. Uhrkov, CSc.

t = 20C t = 30Ct [min]

NaOHV

[ml]

c [mol dm-3]

1/c [dm3 mol-1]

NaOHV

[ml]

c [mol dm-3]

1/c [dm3 mol-1]

5

10

15

20

25

30 -

35 -

9-Kinetics of the Hydrolysis

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