Practical Physical Chemistry Course

8/20/2019 Practical Physical Chemistry Course

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Department of Chemistry

Faculty of Applied Sciences

Umm Al-Qura University

Practical Physical Chemistry

(Physical chemistry II)

Prepared by

Dr. Ahmed Fawzy Saad Prof. Dr. El-Sayed M. Mabrouk

Dr. Hanaa Shokry

Department of ChemistryFaculty of Applied Sciences

Umm Al-Qura University 1434 – 2013


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CONTENTS

Subject Page

1. PREFACE 4

2. I. CHEMICAL KINETICS 5

3. INTRODUCTION 6

4. Catalytic Decomposition of Hydrogen Peroxide. 18

5. Experiment: Determination of the rate of hydrogen peroxide

decomposition catalyzed by manganese dioxide. 18

6. Hydrolysis of Ethyl Acetate in Acid Medium. 20

7. Experiment: Determination of the order of reaction and rate constant of

hydrolysis of ethyl acetate in acid medium. 20

8. Saponification of ethyl acetate in alkaline medium. 23

9. Experiment: Determination of the saponification rate constant of ethyl acetate

in alkaline medium. 23

10. Persulfate-Iodide Reaction. 26

11. Experiment: Determination of the rate constant and half-life period of the

Persulphate - iodide reaction. 26

12. Hydrogen Peroxide - Hydrogen Iodide Reaction. 29

13. Experiment I: Determination of the order of the reaction between hydrogen

peroxide and hydrogen iodide. 30

14. Experiment II: Determination of the rate constant and the energy of activation

of the reaction between hydrogen peroxide and hydrogen iodide. 33

15. Halogenation of Acetone in Solution. 36

16. Experiment: Investigation of the reaction between acetone and iodine. 36

17. Autocatalytic Reaction between Potassium Permanganate and Oxalic Acid. 39

18. Experiment: Investigation of the reaction between potassium permanganate

and oxalic acid.

40


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Subject Page

19. II. THERMOCHEMISTRY 42

20. INTRODUCTION 43

21. Heat Capacity. 46

22. Experiment: Determination of the heat capacity of the calorimeter. 46

23. Experiment II: Determination of the heat capacity of different solutions of NaCl. 48

24. Heat of Solution. 50

25. Experiment: Determination of heat of solution of ammonium chloride

(as endothermic reaction) at infinite dilution. 50

26. Heat of Neutralization. 53

27. Experiment: Determination of the heat of neutralization of hydrochloric acid

and sodium hydroxide. 53

28. Hess's law. 56

29. Experiment: Application of Hess's law. 57

30. III. PHASE EQUILIBRIA 59

31. INTRODUCTION 60

32. Two – Components System. 60

33. Composition – Temperature diagrams. 6034. Experiment: Determination of critical solution temperature of water-phenol system. 62

35. Three Components System. 64

36. Experiment: Construct the binodal curve for the system ethyl acetate ethyl

alcohol-water. 66


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PREFACE

The necessity claimed to originate a simple course of practical physical

chemistry (Physical chemistry II, Code No. 402342, to be studied in the fifth

level) in particularly chemical kinetics, thermochemistry and phase

equilibria. So that the experiments have been chosen to be convenient the

students which starting a laboratory course. The experiments simplified to

can be made by the humble resources and the simple experiment. For

better understanding, every experiment has been supplied with a

theoretical idea to help the student to prepare a short report about the

different subjects.

Our best thanks to Dr. Omer A. Hazazi, Chairman of chemistry

department, faculty of applied sciences, Umm Al-Qura university for the

facilities presented in our department and providing all our requirements.

All members of our department are also acknowledged.

The Staff


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I

CHEMICAL KINETICS


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INTRODUCTION

Chemical kinetics deals with the rates of chemical processes. Any chemical

process may be broken down into a sequence of one or more single-step processes

known either as elementary processes, elementary reactions, or elementary steps.

An important point to recognize is that many reactions that are written as a single

reaction equation in actual fact consist of a series of elementary steps. This will

become extremely important as we learn more about the theory of chemical

reaction rates.

Chemical kinetics is the quantitative study of chemical systems that are

changing with time. (Thermodynamics, another of the major branches of physical

chemistry, applies to systems at equilibrium - those that do not change with time).

Chemical kinetics is conventionally regarded as a topic in physical chemistry.

In this guise it covers the measurement of rates of reaction, and the analysis of the

experimental data to give a systematic collection of information which summarizes

all the quantitative kinetic information about any given reaction. This, in turn,

enables comparisons of reactions to be made and can afford a kinetic classification

of reactions. The sort of information used here is summarized in terms of

the factors influencing rates of reaction,

the dependence of the rate of the reaction on concentration, called the order

of the reaction,

the rate expression, which is an equation which summarizes the dependence

of the rate on the concentrations of substances which affect the rate of

reaction,

this expression involves the rate constant which is a constant of

proportionality linking the rate with the various concentration terms,


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this rate constant collects in one quantity all the information needed to

calculate the rate under specific conditions,

the effect of temperature on the rate of reaction. Increase in temperature

generally increases the rate of reaction. Knowledge of just exactly how

temperature affects the rate constant can give information leading to a

deeper understanding of how reactions occur.

Chemical kinetics is not just an aspect of physical chemistry. It is a unifying

topic covering the whole of chemistry, and many aspects of biochemistry and

biology. It is also of supreme importance in both the chemical and pharmaceuticalindustries. Since the mechanism of a reaction is intimately bound up with kinetics,

and since mechanism is a major topic of inorganic, organic and biological

chemistry, the subject of kinetics provides a unifying framework for these

conventional branches of chemistry. Surface chemistry, catalysis and solid state

chemistry all rest heavily on knowledge of kinetic techniques, analysis and

interpretation. Improvements in computers and computing techniques have resulted

in dramatic advances in quantum mechanical calculations of the potential energy

surfaces and in theoretical descriptions of rates of reaction. Kinetics also makes

substantial contributions to the burgeoning subject of atmospheric chemistry and

environmental studies.

Rate of reaction

When we talk about the rate of a chemical reaction, what we mean is the rate

at which reactants are used up, or equivalently the rate at which products are

formed. The rate, therefore, has units of concentration per unit time, mol dm-3

s-1

(for gas phase reactions, alternative units of concentration are often used, usually

units of pressure – Torr, mbar or Pa). To measure a reaction rate, we simply need


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to monitor the concentration of one of the reactants or products as a function of

time. There is one slight complication to our definition of the reaction rate so far,

which is to do with the stochiometry of the reaction. The stoichiometry simply

refers to the number of moles of each reactant and product appearing in the

reaction equation. For example, the reaction equation for the well-known Haber

process, used industrially to produce ammonia, is:

N2 + 3H2 = 2NH3

N2 has a stochiometric coefficient of 1, H2 has a coefficient of 3, and NH3 has a

coefficient of 2. We could determine the rate of this reaction in any one of three

ways by monitoring the changing 3 concentration of N2

, H2

, or NH3. Say we

monitor N2 and obtain a rate of -d[N2] / d t = x mol dm-3

s-1

. Since for every mole

of N2 that reacts, we lose three moles of H2 , if we had monitored H2 instead of N2

we would have obtained a rate -d[H2] / d t = 3 x mol dm-3

s-1

. Similarly, monitoring

the concentration of NH3 would yield a rate of 2 x mol dm-3

s-1

. Clearly, the same

reaction cannot have three different rates, so we appear to have a problem. The

solution is actually very simple: the reaction rate is defined as the rate of change of

the concentration of a reactant or product divided by its stochiometric coefficient .

For the above reaction, the rate (usually given the symbol ν) is therefore

dt

NH d

dt

H d

dt

N d v

][

2

1][

3

1][322

Note that a negative sign appears when we define the rate using the concentration

of one of the reactants. This is because the rate of change of a reactant is negative

(since it is being used up in the reaction), but the reaction rate needs to be a

positive quantity.


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Rate laws

The rate law is an expression relating the rate of a reaction to the

concentrations of the chemical species present, which may include reactants,

products, and catalysts. Many reactions such as: a A + b B + c C → P, follow a

simple rate law, which takes the form

Rate = ν = k [A] a[B]

b[C]

c

i.e. the rate is proportional to the concentrations of the reactants each raised to

some power. The constant of proportionality, k, is called the rate constant. The

power a particular concentration is raised to is the order of reaction with respect to

that reactant. Note that the orders do not have to be integers. The sum of the

powers is called the overall order. Even reactions that involve multiple elementary

steps often obey rate laws of this kind, though in these cases the orders will not

necessarily reflect the stoichiometry of the reaction equation. For example,

H2 + I2 → 2HI; ν = k [H2] [I2]

Elementary processes always follow simple rate laws, in which the order with

respect to each reactant reflects the molecularity of the process (how many

molecules are involved). For example,

Unimolecular decomposition: A → B; ν = k [A]

Bimolecular reaction: A + B → P; ν = k [A][B]

A + A → P; ν = k [A][A] = k [A]2

We can carry out experiments to determine the orders with respect to each

reactant and then try out various ‘trial’ reaction mechanisms to see which one fits

best with the experimental data. At this point it should be emphasized again that

for multi-step reactions, the rate law, rate constant, and order are determined by

experiment, and the orders are not generally the same as the stoichiometric

coefficients in the reaction equation. A final important point about rate laws is that

overall rate laws for a reaction may contain reactant, product and catalyst


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concentrations, but must not contain concentrations of reactive intermediates

which appear in rate laws for the individual elementary steps.

The units of the rate constant

A point which often seems to cause endless confusion is the fact that the units

of the rate constant depend on the form of the rate law in which it appears i.e. a

rate constant appearing in a first order rate law will have different units from a rate

constant appearing in a second order or third order rate law. This follows

immediately from the fact that the reaction rate always has the same units of

concentration per unit time, which must match the overall units of a rate law in

which concentrations raised to varying powers may appear. The good news is that

it is very straightforward to determine the units of a rate constant in any given rate

law. Below are a few examples:

(i) Consider the rate law; ν = k [H2] [I2]. If we substitute units into the

equation, we obtain

(mol dm-3

s-1

) = [k ] (mol dm-3

) (mol dm-3

)

where the notation [k ] means ‘the units of k . We can rearrange this

expression to find the unit of the rate constant, k , as follows:

[k ] = (mol dm-3

s-1

) / (mol dm-3

) (mol dm-3

) = mol-1

dm3 s

-1

(ii) We can apply the same treatment to a first order rate law, for example

ν = k [CH3 N2CH3].

(mol dm-3

s-1

) = [k ] (mol dm-3

)

[k ] = (mol dm-3 s-1) / (mol dm-3) = s-1

(iii) As a final example, consider the rate law

ν = k [CH3CHO]3/2

(mol dm-3

s-1

) = [k ] (mol dm-3

)3/2

[k ] = (mol dm-3

s-1

) / (mol dm-3

)3/2

= mol-1/2

dm3/2

s-1


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An important point to note is that it is meaningless to try and compare two rate

constants unless they have the same units.

Integrated rate laws

A rate law is a differential equation that describes the rate of change of a

reactant (or product) concentration with time. If we integrate the rate law then we

obtain an expression for the concentration as a function of time, which is generally

the type of data obtained in any kinetic experiment. In many simple cases, the rate

law may be integrated analytically. Otherwise, numerical (computer-based)

techniques may be used. Four of the simplest rate laws are given below in both

their differential and integrated form.

In the above [A]0 and [B]0 represent the initial concentrations of A and B, i.e. their

concentrations at the start of the reaction.


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Half lives

The half life, t1/2 , of a substance is defined as the time it takes for the

concentration of the substance to fall to half of its initial value. We can obtain

equations for the half lives for reactions of various orders by substituting the values

t = t 1/2 and [A] = ½ [A]0 into the integrated rate laws. We obtain

Zeroth order reaction; t 1/2 = [A]0 / 2k

First order reaction; t 1/2 = ln 2 / k

Second order reaction; t 1/2 = 1 / k [A]0

Determining the rate law from experimental data

A kinetics experiment consists of measuring the concentrations of one or more

reactants or products at a number of different times during the reaction. We will

look at the methods that allow us to use the experimental data to determine the

reaction orders with respect to each reactant, and therefore the rate law.

(i) Isolation method

The isolation method is a technique for simplifying the rate law in order to

determine its dependence on the concentration of a single reactant. Once the rate

law has been simplified, the differential or integral methods may be used to

determine the reaction orders. The dependence of the reaction rate on the chosen

reactant concentration is isolated by having all other reactants present in a large

excess, so that their concentration remains essentially constant throughout the

course of the reaction.

As an example, consider a reaction A + B → P, in which B is present at a

concentration 1000 times greater than A. When all of species A has been used up,

the concentration of B will only have changed by 1/1000, or 0.1%, and so 99.9% of

the original B will still be present. It is therefore a good approximation to treat its


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concentration as constant throughout the reaction. This greatly simplifies the rate

law since the (constant) concentrations of all reactants present in large excess may

be combined with the rate constant to yield a single effective rate constant . For

example, the rate law for the reaction considered above will become:

ν = k [A]a

[B]b ≈ k [A]

a[B]0

b= k eff [A]

a with k eff = k [B]0

b

When the rate law contains contributions from a number of reactants, a series of

experiments may be carried out in which each reactant is isolated in turn.

(ii) Differential methods

When we have a rate law that depends only on the concentration of one

species, either because there is only a single species reacting, or because we have

used the isolation method to manipulate the rate law, then the rate law may be

written

ν = k [A]a

(*)

log ν = log k + a log[A]

A plot of log ν against log[A] will then be a straight line with a slope equal to the

reaction order, a, and an intercept equal to log k . There are two ways in which to

obtain data to plot in this way.

1. We can measure the concentration of the reactant [A] as a function of time

and use this data to calculate the rate, ν = - d[A]/d t , as a function of [A]. A

plot of log ν vs. log [A] then yields the reaction order with respect to A.

2. We can make a series of measurements of the initial rate ν0 of the reaction

with different initial concentrations [A]0. These may then be plotted as

above to determine the order, a. This is a commonly used technique known

as the initial rates method .


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(iii) Integral methods

If we have measured concentrations as a function of time, we may compare

their time dependence with the appropriate integrated rate laws. Again, this is most

straightforward if we have simplified the rate law so that it depends on only one

reactant concentration. The differential rate law given in Eq. (*) will give rise to

different integrated rate laws depending on the value of a. The most commonly

encountered ones are:

Zeroth order integrated rate law: [A] = [A]0 – kt

A plot of [A] vs. t will be linear, with a slope of -k .

First order integrated rate law: ln [A] = ln [A]0 – ktA plot of ln [A] vs. t will be linear with a slope of -k .

Second order integrated rate law: 1/[A] = 1/[A]0 + 2kt

A plot of 1/[A] vs. t will be linear with a slope of 2k .

If none of these plots result in a straight line, then more complicated integrated

rate laws must be tried.

(iv) Half lives method

Another way of determining the reaction order is to investigate the behaviour

of the half life as the reaction proceeds. Specifically, we can measure a series of

successive half lives. t = 0 is used as the start time from which to measure the first

half life, t 1/2(1)

. Then t 1/2(1)

is used as the start time from which to measure the

second half life, t 1/2(2)

, and so on.

Zeroth order: t 1/2 = [A]0 / 2 k

Since at t 1/2(1)

, the new starting concentration is ½[A]0, successive half lives

will decrease by a factor of two for a zeroth order reaction.

First order: t 1/2 = ln 2 / k


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There is no dependence of the half life on concentration, so t 1/2 is constant

for a first order reaction.

Second order: t 1/2 = 1 / k [A]0

The inverse dependence on concentration means that successive half lives

will double for a second order reaction.

Experimental techniques

Experimental techniques have been developed to monitor reactions over

timescales varying from hours or days all the way down to a few femto-seconds(1 fs = 10

-15 s). While it is relatively simple to monitor the kinetics of a slow

reaction (occurring over minutes to hours or longer), highly specialized techniques

are required in order to study such fast reactions. Whatever the details of the

experimental arrangement, any kinetics experiment essentially consists of mixing

the reactants and initiating reaction on a timescale that is negligible relative to that

of the reaction, and then monitoring the concentration (s) of one or more reactants

and/or products as a function of time. Because rate constants vary with

temperature, it is also important to determine and control accurately the

temperature at which the reaction occurs.

(i) Techniques for mixing the reactants and initiating reaction

For slow reactions, occurring over minutes to hours, reaction is usually

initiated simply by mixing the reactants together by hand or with a magnetic stirrer

or other mechanical device. For fast reactions, a wide range of techniques have

been developed.


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(ii) Techniques for monitoring concentrations as a function of time

For slow reactions, the composition of the reaction mixture may be analyzed

while the reaction is in progress either by withdrawing a small sample or by

monitoring the bulk. This is known as a real time analysis. Another option is to use

the quenching method , in which reaction is stopped a certain time after initiation

so that the composition may be analyzed at leisure. Quenching may be achieved in

a number of ways such as: sudden cooling, adding a large amount of solvent, rapid

neutralization of an acid reagent, removal of a catalyst and addition of a quencher.

The composition of the reaction mixture may be followed in any one of a

variety of different ways by tracking any chemical or physical change that occurs

as the reaction proceeds as follows:

For reactions in which at least one reactant or product is a gas, the reaction’s

progress may be followed by monitoring the pressure, or possibly the

volume.

For reactions involving ions, conductivity or pH measurements may often be

employed.

If the reaction is slow enough, the reaction mixture may be titrated.

If one of the components is colored then colourimetry may be appropriate.

Absorption or emission spectroscopy is common (more on these later).

For reactions involving chiral compounds, polarimetry (measurement of

optical activity) may be useful.

Other techniques include mass spectrometry, gas chromatography,

NMR/ESR, and many more.


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(iii) Temperature control and measurement

For any reaction with a non-zero activation energy, the rate constant is

dependent on temperature. The temperature dependence is often modelled by the

Arrhenius equation,

k = A exp (- E a / RT )

where k is the rate constant (min-1

) , E a is the activation energy for the reaction (cal.

/ mole), T is the absolute temperature (K), R is the gas constant (cal. / mole. K),

and A is a constant known as the pre-exponential factor . This temperature

dependence means that in order to measure an accurate value fork , the temperature

of the reaction mixture must be maintained at a constant, known value. If

activation energies are to be measured as part of the kinetic study, rate constants

must be measured at a series of temperatures. The temperature is most commonly

monitored using a thermocouple, due to its wide range of operation and potential

for automation; however, standard thermometers are also commonly used.

There are numerous ways in which the temperature of a reaction mixture may

be controlled. For example, reactions in the liquid phase may be carried out in a

temperature-controlled thermostat, while reactions in the gas phase are usually

carried out inside a stainless steel vacuum chamber, in which thermal equilibrium

at the temperature of the chamber is maintained through collisions of the gas

molecules with the chamber walls. High temperatures up to 1300 K may be

obtained using conventional heaters. Low temperatures may be achieved by

flowing cooled liquid through the walls of the reaction vessel, and very low

temperatures may be reached by using cryogenic liquids such as liquid nitrogen

(~77 K) or liquid helium (~ 4 K). Extremely low temperatures (down to a few

Kelvin) such as those relevant to reactions in interstellar gas clouds may be

obtained by preparing the reactant gases in a supersonic expansion.


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Catalytic Decomposition of Hydrogen Peroxide

Theory :

Inhibitor free hydrogen peroxide solutions decompose spontanouosly librating

oxygen in accordance with the following equation:

H2O2 (aq) = H2O (aq) + ½ O2 (g)

The decomposition rate is markedly accelerated by solids such as manganese

dioxide or colloidal platinum, which act as catalysts. The course of reaction may be

followed either by titrating the peroxide with potassium permanganate in acid

medium, or by collecting the oxygen gas evolved.

Experiment:

Determination of the rate of hydrogen peroxide decomposition catalyzed by

manganese dioxide.

Procedure :1. Prepare 250 ml of 0.1 N KMnO4 and 100 ml of 0.1 N H2O2 solution (Ten-

volume hydrogen peroxide is approximately 3%). Thermostat the peroxide

solution at 25oC. Add about 0.03 g manganese dioxide and record the time.

2. After about 3 minutes pipette out 10 ml of the decomposing mixture into a

flask containing about 10 ml of about 2.0 N sulfuric acid and titrate rapidly

with potassium permanganate recording the main time of titration.

3. Repeat the above step at increasing time intervals extending for about 80

minutes, i.e. 5, 8, 13, 20, 30, 45, 60 and 80 minutes.


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Calculations:

1. Tabulate the results in the following order:

t, (a-x), log (a-x)

where, t is the main time of titration in minutes and (a-x) the amount of

undecomposed peroxide expressed in volume (ml) of KMnO4.

2. Plot log (a-x) against t to identify the order of reaction, then deduce the

value of k (reaction rate constant) and t1/2 (half-live time).

Results:

t (a-x) log (a-x)

3

5

8

12

20

30

45

60

80


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Hydrolysis of Ethyl Acetate in Acid Medium

Theory:

Ethyl acetate hydrolyzes in water to give ethanol and acetic acid in

accordance with

CH3COOC2H5 + H2O = CH3COOH + C2H5OH

The reaction does not proceed at any measurable rate in pure water but is highly

catalyzed by hydrogen ions. Although, more than one molecular species are

involved, the reaction is kinetically of the first order since water is usually present

in a large excess. For a first order reaction, the rate constant k may be expressed as

)(log

303.2

xa

a

t k

where a is the original concentration and (a-x) the concentration after time t.

Experiment:

Determination of the order of reaction and rate constant of hydrolysis of ethyl

acetate in acid medium.

Procedure :

1. Prepare 100 ml of 0.1 N HCl and 250 ml of 0.1 N NaOH solutions.

2. Transfer the whole amount of acid into a bottle + about 10 ml ethyl acetate.

3. After about 5 minutes withdraw 10 ml of the solution + 10 ml distilled

water. Titrate the solution rapidly with 0.1 N NaOH using phenolphethalin

as an indicator.

4. Repeat at increasing time intervals making a total about ten determinations

over a period extending for about 80 minutes.


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5. Carry out one similar determination after a sufficiently long time, say 24

hours when hydrolysis is complete; or alternatively, run out 10 ml of the

ester solution into a small flask containing some water and fitted with an air

condenser. Boil in a water bath for about one hour to affect complete

hydrolysis. Wash the condenser into the flask and titrate with 0.1 N NaOH.

Calculations :

1. In any titration the amount (expressed in volume, ml) of alkali consumed

(Vt) is equivalent to the amount of HCl in 10 ml (VHCl) plus the amount of

acetic acid librated ( x). After complete hydrolysis, the amount of alkali

consumed (V∞) is equivalent to the amount of HCl in 10 ml (VHCl) plus the

amount of acetic acid equivalent to all ester ( a). The remaining ester

concentration after time t is therefore (a-x).

2. Plot log (a-x) against t and identify the order of reaction.

3. From the graph, calculate the rate constant k and the half-live time t1/2.


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Saponification of Ethyl Acetate in Alkaline Medium

Theory :

In the presence alkali ethyl acetate undergoes saponification in accordance

with

CH3COOC2H5 + OH- = CH3COO

- + C2H5OH

The rate of saponification is directly proportional to both concentrations of ester

and alkali, and the reaction therefore, is a second order one, i.e.

)(.

1

xaa

x

t k

where a and b are the initial concentrations of ester and alkali, respectively (a = b

in this case) and k is the reaction rate constant.

Experiment:

Determination of the saponification rate constant of ethyl acetate in alkaline

medium.

Procedure :

1. Prepare the following solutions:

100 ml of exactly 0.1 N Na2CO3,

100 ml of about 0.1 N HCl,

100 ml of exactly 0.1 N NaOH.

2. Standardize the acid against the carbonate and the hydroxide against theacid.

3. By appropriate dilution prepare 100 ml of exactly 0.025 N HCl and 100 ml

of each of exactly 0.05 N and 0.025 N NaOH.

4. Prepare 100 ml of exactly 0.05 N ethyl acetate (density is 0.901 g/ml).


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Results:

t x (a-x) x / (a-x)

20

30

40

50

60

70

80

90


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ml of 0.01 N K 2Cr 2O7 , leave for sometimes, then titrate the librated iodine

against the thiosulphate solution using starch as indicator).

3. In a dry bottle mix up quickly the iodide and persulphate solutions; register

the time and thermostat. The reacting mixture therefore, is initially 0.1 M

with respect to KI and 0.0125 M with respect to K 2S2O8 .

4. After about 3 minutes pipette out 25 ml of the mixture into a conical flask

containing some distilled water and starch solution. Record the time of

dilution and titrate the librated iodine with the thiosulphate.

5. Follow the reaction by carrying out in the same way about eight titrations

over a period extending for about 90 minutes.

Calculations :

1. Calculate the amount of persulphate (x) decomposed at the end of each

interval expressing it as millimole of K 2S2O3 per liter of mixture.

2. Tabulate the results in the following order:

t, x, a-x, log(a-x),

where, t is the time in minutes, x is the amount of persulphate decomposed

(= millimoles of I2 librated = millimoles of K 2S2O3 consumed) and a is the

initial concentration per liter of the original mixture (0.0125 M).

3. Plot log (a-x) against t, and deduce the values of k and t1/2 of the reaction.


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Results:

t x (a-x) log (a-x)

3

5

8

12

20

30

45

60

80


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Hydrogen Peroxide - Hydrogen Iodide Reaction

Theory:

The overall reaction between H2O2 and HI which represented by the equation

H2O2 + 2 HI = 2 H2O + I2

is kinetically of second order -not, as might be expected, third order. The suggested

mechanism is probably as

H2O2 + I- = H2O + IO

- (slow),

IO

-

+ 2 H

+

+ I

-

= H2O + I2 (fast)The rate-determining step is the slow stage, Rate α [H2O2][I

-] (first order with

respect to both [H2O2] and [I-]). The order of the reaction with respect to H2O2 can

be studied conveniently by choosing conditions such that there is practically

constant excess of HI. Then, the rate of the reaction depends only on [H2O2] and

temperature and, hence, the kinetics then follows the first order law, Rate α

[H2O2].

This is achieved experimentally by continually adding small volumes of

sodium thiosulphate solution to remove the iodine as soon as it is liberated and to

regenerate iodide according to the reaction

2S2O3--

+ I2 = S4O6--

+ 2I-

Note that : by using a large volume of solution and adding small amounts of

concentrated thiosulphate solution, one can neglect the small increase of volume of

the solution and take the concentration of I- ions as constant.

The course of the reaction can readily be followed by timing the appearance

of iodine (indicated by starch solution) after the addition of a small known volume

of thiosulphate solution. The amount of iodine librated by the reaction at a series of

times corresponds to the volume of thiosulphate added. The total amount of iodine


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librated at infinite time can be determined from a standardization of the hydrogen

peroxide used. Thus, it is possible to determine the concentration of hydrogen

peroxide at any time, since 1 mol of iodine is librated for every mol of hydrogen

peroxide destroyed.

The order of the reaction with respect to HI can be determined by determining

the first order velocity constant of the reaction with different concentrations of HI.

Generally, the rate equation of the overall reaction is

Rate = k [H2O2] a [HI]

b

where, k is the rate constant, a and b are the order with respect to H2O2 and HI,

respectively.

The rate of this reaction can be determined by allowing the reaction to proceed

in the presence of thiosulphate and determining the time taken between mixing of

the reactants and the appearance of iodine. The reciprocal of the time interval is a

measure of the rate of reaction.

Experiment I:Determination of the order of the reaction between hydrogen peroxide and

hydrogen iodide.

Procedure :

1. Prepare the following solutions

500 ml of 1.0 N H2SO4 ,

250 ml of 0.10 M KI ,

100 ml of 0.01 N Na2S2O3 . 5 H2O,

50 ml of 1 vol. H2O2 ,

Freshly prepared starch solution.


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To get the order with respect to H2O2:

2. In dry small flask bottom flasks make up the following series of mixtures

(volumes in ml):

Solution

Mixture

I II III IV V

1.0 N H2SO4 25 25 25 25 25

0.10 M KI 25 25 25 25 25

0.01 N Na2S2O3 . 5 H2O 5 5 5 5 5

Starch + distilled water 4 3 2 1 --

3. To each mixture add respectively and separately: 1, 2, 3, 4 and 5 ml of H2O2

(making a total volume of 60 ml), mix thoroughly and meanwhile start a

clock on. Determine the time period (t, sec.) indicated by the sudden

appearance of the blue color. The results are tabulated as follows:

VH2O2 (ml), t, 1/t , log (1/t), log VH2O2 .

4. At constant [I-], the rate law of the reaction can be written as follos:

Rate = = k [H2O2] a

Then, log Rate = log k + a log [H2O2]

Considering the rate of reaction is measured as 1/t and [H2O2] is represented

by VH2O2

log (1/t) = log k + a log VH2O2

Plot log (1/t) against log VH2O2 . This gives a straight line, the slope of which

is a and the intercept is log k.


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To get the order with respect to HI:

5. In dry small flask bottom flasks make up the following series of mixtures

(volumes in ml):

Solution

Mixture

I II III IV V

1.0 N H2SO4 25 25 25 25 25

0.01 N Na2S2O3 . 5 H2O 5 5 5 5 5

H2O2 2 2 2 2 2

Starch + distilled water 23 18 13 8 3

6. To each mixture add respectively and separately: 5, 10, 15, 20 and 25 ml of

0.1 M KI solution (making a total volume of 60 ml), mix thoroughly and

meanwhile start a clock on. Determine the time period (t, sec.) between

addition of KI solution and the sudden appearance of the blue color. The

results are tabulated as follows:

VKI

(ml), t, 1/t , log (1/t) , log VKI

.

7. Similarly,

log (1/t) = log k + b log VKI

Plot log (1/t) against log VH2O2 . This gives a straight line, the slope of which

is b and the intercept is log k.


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Experiment II:

Determination of the rate constant and the energy of activation of the reaction

between hydrogen peroxide and hydrogen iodide.

Procedure:

1. Prepare 250 ml of the following solutions:

2.0 N H2SO4 ,

0.4 M KI solution ,

0.1 M Na2S2O3 . 5 H2O ,

1 vol. H2O2 ,


2. Into a clean dry conical flask introduce 20 ml of the hydrogen peroxide

solution, add about 2 g solid potassium iodide and 10 ml of sulfuric acid.

Leave the mixture for about five minutes in a dark place, then titrate the

librated iodine with standard sodium thiosulphate solution, and, hence,

standardize the hydrogen peroxide solution.

Note that : the consumed volume of Na2S 2O3 is equivalent to the librated I 2 ,

equivalent to [H 2O2] destroyed and equals a.

3. To 50 ml of potassium iodide solution in a dry bottle, add 20 ml of the

diluted sulfuric acid and place in a thermostat at about 25oC. At the same

time have 20 ml of hydrogen peroxide solution and 10 ml of the starchsolution (in separate boiling tubes) at the same temperature.

4. Arrange the 50 ml burette containing the sodium thiosulphate solution above

the flask in the thermostat so that it will deliver directly into the solution,

add the starch and hydrogen peroxide solutions (with vigorously shaking)


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and record the time at which a blue color appears, then add immediately 1

ml of the thiosulphate solution from the burette to discharge the color.

5. Continue the addition of 1 ml aliquots of sodium thiosulphate until the blue

color appears until the blue color takes five to six times the initial time to

reappear (it is essential that the reaction mixture be shaken continuously).

6. Repeat the experiment at various temperatures in the range 25 to 50 oC.

7. To determine the order of the reaction with respect to hydrogen iodide,

repeat the experiment (at 25oC) now using half and double the amounts of

sulfuric acid and potassium iodide in the same total volume of the reaction

mixture.

Treatment of the experimental data and discussion :

1. The total amount of librated iodine at infinite time can also be determined

from the preliminary standardization. Thus, it is possible to determine the

concentration of hydrogen peroxide in mole per liter, [H2O2] at each time,

(a-x). Note that: x is the volume of Na2S 2O3 added. 2. Plot a graph of ln [H2O2] (ln (a-x)) against t, at each temperature to obtain a

straight line with a slope of k (rate constant) at constant iodide ion

concentration (first order reaction) according to the equation:

ln [H2O2] = - k t + B

3. From the values of the rate constant at different temperatures, calculate the

activation energy of the reaction from the equation:

ln k = - E a /RT + const.

by plotting the values of ln k against 1/T to get a straight line with a slope

of E a /R, then calculate E a (activation energy).


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From the values of the rate constant at different iodide concentrations

determine the order of the reaction with respect to hydrogen iodide. The

overall reaction will be of a second order.


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Halogenation of Acetone in Solution

Theory :

General acid catalysis was first established for the reaction between acetone

and iodine which can be represented by the equation

CH3 CO CH3 + I2 = CH3 CO CH2I + HI

The rate of such reaction is independent of the concentration of iodine (zero-

order reaction).

Experiment:

Investigation of the reaction between acetone and iodine.

Procedure:

1. Prepare the following solutions:100 ml of 1.0 M H2SO4 ,

2.0 M acetone,

100 ml of 0.05 M I2 in 10% KI solution,

1.0 M sodium acetate,

250 ml of 0.02 M Na2S2O3 . 5 H2O ,


2. Prepare the following four mixtures in four dry bottles (volumes in ml) kept

in a large water bath at room temperature


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Solution

Mixture

I II III IV

1.0 M H2SO4 20 20 25 20

0.05 M I2 20 25 20 20

Distilled water 50 45 45 45

3. To the first bottle add 10 ml of acetone and start (record) the time.

4. Remove 20 ml of the sample from the reaction mixture after 5 minutes

interval and run into 10 ml of sodium acetate solution to stop the reaction

and titrate the residual iodine against 0.01 M sodium thiothlfate using starch

indicator (till the solution becomes colorless).

5. Repeat the above titration at constant time intervals (5 min.).

6. Repeat the last three steps for the other mixtures.

:Calculations

1. Tabulate the volume of consumed sodium thiosulphate at different time

intervals for each.

2. Plot a graph of titre volume of sodium thiosulphate against time for each

mixture (you should obtain straight lines parallel to each other, i.e. k is

constant indicating that the reaction is zero order ).

3. Determine the order of the reaction with respect to acetone, iodine and

sulfuric acid.

4. Investigate the effect of changing the concentration of: a) I2 , b) acetone

and c) acid on the rate of reaction.


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Results:

Rate = k [acetone] [acid] [iodine]

Volume of consumed sodium thiosulphateTime,

min IVIIIIII

5

10

15

20

25

30

35

40

45

50


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Autocatalytic Reaction between Potassium Permanganate and

Oxalic Acid

Theory:

The reaction between potassium permanganate and oxalic acid in presence of

acid takes place according to the following equation:

2 KMnO4 + 5 H2C2O4 + 3 H2SO4 = K 2SO4 + 2 MnSO4 + 10 CO2 + 8 H2O

Such reaction is catalyzed by manganese ions formed during the reaction. The

progress of the reaction can be followed by estimating the residual permanganate

concentration from time to time. This is done by running the samples of the

mixture to excess of KI and then titrating the liberate iodine with standered

Na2S2O3 solution. A plot of titre value of sod thiosulphate against the time reveals

the autocatalytic effect of Mn2+

ions.

The order of the reaction with respect to potassium permanganate can be

determined from these curves by determining the time required for half change

(half-life method). If the order of the reaction is n, then the half-life of the reaction

t1/2 depends on the initial concentration [A]0 of the reactant according to the

equation:

0

1

21 ]log[)1()1(

12loglog An

k nt

A

n

A plot of log t1/2 versus log [A]0 gives a straight line of slope (n-1) then calculate n.


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Experiment:

Investigation of the reaction between potassium permanganate and oxalic acid

Procedure:

1. Prepare the following solutions

1.0 M H2SO4 ,

0.1 M oxalic acid,

0.02 M potassium permanganate,

0.2 M manganese sulphate,

2. Prepare the following mixtures (volumes in ml) kept in a large water bath at

room temperature.

Solution

Mixture

I II III IV

0.2 N Oxalic acid 80 80 80 80

2.0 M H2SO4 10 10 10 10

0.4 M MnSO4 0 0 2 2

Distilled water 15 25 13 138

0.1 N KMnO4 20 10 20 20

Total volume, ml 125 125 125 250

3. At time t0 add the required volume KMnO4 to the mixture and start

stopwatch.

4. Withdraw 10 ml of the reaction mixture at regular time intervals and run into

5 ml of 5% KI taken in a conical flask. For reaction mixtures I and II, the

time interval can taken as 60 sec., while for reaction mixtures III and IV, it

can be taken as 30 sec.


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5. Cover the mouth of the conical flask and keep it in the dark for 5 min till the

iodine is fully liberated. Titrate the liberated iodine with stander sodium

thiosulphate solution using starch solution as the indicator.

6. The titre value of sodium thiosulphate is equivalent to the concentration of

potassium permanganate in the reaction mixture at that time.

Calculations:

1. Plot a graph of the time against the titre of sodium thiosulphate for each

mixture.

The addition of Mn2+

in mixtures III and IV causes the autocatalytic part of

the curve, evident in I and II, to discharge. Finally, using the curves of III

and IV, determine the order of the reaction.


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Results:

Time min

Titre value (volume)

I II III IV


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II

THERMOCHEMISTRY


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INTRODUCTION

Thermochemistry is a part science deals with the heat changes accompanying

chemical reactions. This subject is based largely on the law of conservation of

energy.

.

Heat changes in chemical reactions

Since different substances have different amount of energy, the total energy

of the products of a chemical reaction may be different from the total energy of the

reactants, so that, the process may be accompanied by an absorption or liberation

of energy in the form of heat. If at ordinary temperature heat is liberated in the

reaction the process is said to be exothermic reaction, but if heat is absorbed it is

described as an endothermic reaction.

Heat of reaction

Since many reactions normally occur at constant (atmospheric) pressure the

heat changes are usual record in terms of Q p. So that the heat of reaction is

defined as the change in the heat content (enthalpy) accompanying the reaction at

constant pressure. For exothermic reaction (Q p = - ∆H) and for endothermic

reaction ( - Q p = ∆H).

As examples:

1) C(s) + O2 (g) = CO2 (g) + 94030 cal (exo.)

or C(s) + O2 (g) = CO2 (g) ; ∆H = 94.03 kcal.

this means that the quantity of heat (Q p) evolved is equal to 94.03 kcal.


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2) H2 (g) + I2 (g) = 2 HI (g) - 11900 cal. (endo.)

or H2 (g) + I2 (g) = 2 HI (g); ∆H = 11.90 kcal.

this means that the quantity of heat (Q p) absorbed is equal to 11.90 kcal.


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Heat Capacity

Theory:

The heat capacity (C) of a substance (or specific heat ) is defined as the

amount of heat needed to raise the temperature of one gram of a pure substance by

1oC or 1 K.

The rnolar heat capacity (C') is defined as the amount of heat (by joules)

needed to raise the temperature of one mole (or molar solution) of a pure substance

by IoC or 1 K.

Experiment I:

Determination of the heat capacity of the calorimeter.

Equipment:

Calorimeter - Bekmann thermometer - graduated cylinder - beaker (250 ml) -

distilled water.

Procedure:

1. Weigh the empty calorimeter W1 .

2. Introduce exactly 100 ml of hot distilled water (60 "C.) in the calorimeter

and record tire temperature quickly as possible T1 .

3. At steady state (constant temperature over two minutes) record T2 .

4. Weigh the calorimeter with its contents W2 .

You can make a simple calorimeter as shown in the Figure below.


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Calculations:

1. Calculate the weight of distilled water W3 = W2 -W1 .

2. Where Q the amount of heat absorbed by the calorimeter: the amount of heat

lost by water, hence;

Q = W3 x Cwater x ∆T

where, ∆T = (T1 – T2).

3. The heat capacity of the calorimeter;

Cc = Q / ∆T

Results:

W1 W2 W3 = W2 -W1 T1 (K) T2 (K) ∆T = T1 – T2


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Experiment II:

Determination of the heat capacity of different solutions of NaCl.

Equipment:

Calorimeter - Bekmann thermometer - graduated cylinder - beaker (250 ml) -

distilled water – solid NaCl.

Procedure:

1. Prepare (10 N) NaCl solution [585 g in 1000 m1].

2. Use the dilution law: N x V = N' x V' to prepare (2,4,6, and 8 N) NaCl

solutions.


4. Introduce exactly 100 ml of hot (5N) NaCl solution (50oC) in the

calorimeter and record the temperature quickly as possible T1 .



7. The weight of NaCl solution, W3 = W2 - W1 .

8. Repeat steps (2 - 4) by using different concentrations of NaCl.

Recording Data:

Cs W1 W2 W3 = W1 –W2 T1 (K) T2 (K) ∆T = T1 – T2


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Calculations:

1. Where Q the amount of heat absorbed by the calorimeter = the amount of

heat lost by solution, hence;

Qc = QNaCl

So that;

Cc x ∆T = W3 x CNaCl

Cc = the heat capacity of calorimeter (from the previous experiment) and

∆T = T1 – T2

2. Calculate C NaCl, where;

CNaCl

= Cc x ∆T/W

3

3. Plot Q NaCl against concentration, comment.

4. Calculate the molar heat capacity of NaCl from the relation

C’ = ∆H' = Q x 1000/N x 100

it must be the same for all concentration, (calculate the average).

5. Plot log Q NaCl against concentration and extrapolate the straight line to zero

concentration, then calculate C, at infinite dilution from the intercept.

6. Determine the heat capacity of 5 N NaCl.

7. Repeat this experiment by using NH4CI.

Results:

Item Calc. Result

Q = Cc x ∆T =

CNaCl = Q / W3 =

C’NaCl = ∆H’ = (Q x 1000) / (N x 100) =


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Heat of Solution

Theory:

When a solid solute dissolved in a suitable solvent an amount of heat evolved

or absorbed as a result of formation of solution. The quantity of heat depends on

the types of the two components and changes as the concentration of solution

change. Heat of solution may be defined according to the condition of the reaction

(temperature, pressure, saturated solution, dilutes solution, etc…).

Heat of solution for saturated solution is the change in heat content when one

mole of solute dissolved in a volume of water (solvent) of the nearly saturated

solution. Heat of solution for infinite dilution is the change in heat content when

one mole of solute dissolved in a large amount of water (solvent) till infinite dilute

solution.

Experiment:

Determination of heat of solution of ammonium chloride (as endothermic

reaction) at infinite dilution.

Using suitable calorimeter to calculate the heat of solution for different

concentrations. The extrapolation of the straight line obtained from the plot of the

relation between the concentration and the heat of solution to Y axis, gives an

intercept which equals to the heat of solution at infinite dilution.

Equipment:

Calorimeter - Bekmann thermometer - graduated cylinder - beaker (250 ml)-

distilled water - ammonium chloride.


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Procedure:


2. Introduce exactly 100 ml of distilled water in the calorimeter and record the

temperature T1 .

3. Weigh the calorimeter with its water W2 , hence;

W3 (water) = W2 - W1

4. Add W4 of NH4Cl (5 g), hence;

W5 (solution) = W3 + W4

5. Shake or stir the closed calorimeter to dissolve the solid.


7. Repeat steps (2 - 6) with different weights of NH4CI as (8, 12, 16, 20 g).

Recording Data:

Cc M Csol T1 T2 ∆T W1 W2 W3 W4 W5

Calculations:

1. Calculate the molality of solution from the relation;

m = (W4 /W3) x (1000 / M) molal

2. Calculate the absorbed heat by the solution (∆H1), where

∆H1 = Csol (NH4Cl, aq) x W5 x ∆T ;

consider Csol = 3.975 cal/g for all concentrations.

3. Calculate the absorbed heat by calorimeter (∆H2),

∆H2 = Cc x ∆T

4. The total heat absorbed: ∆H3 = ∆H1 + ∆H2 .


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5. Heat of solution: ∆H = ∆H3 / n = ∆H3 x M/ W4 .

6. Plot the relation between m and ∆T; comment.

7. Plot the relation between m and ∆H and determinate ∆H at infinite dilution

and the molar heat of solution (∆H') at m = 1 .

Results:

Item Calc. Result

M, molal = (W4 / W3) x (1000 /M) =

∆H1 Csol x W5 x ∆T =

∆H2 = Cc x ∆T = molal

∆H3 = ∆H1 + ∆H2 =

∆H = ∆H3 x M / W4 =

For exothermic reaction use NaOH instead of NH4Cl.


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Heat of Neutralization

Theory:

Heat of neutralization is defined as; the constant quantity of heat liberated

when one mole of strong acid is neutralized with one mole of strong base to

produce one mole of water. As an example;

HCl (aq) + NaOH (aq) = NaCl (aq) + H2O + Q

The reaction can be written as:

H+

(aq) + Cl-(aq) + Na

+(aq) + OH

-(aq) = Na

+(aq) + Cl

-(aq) + H2O; ∆H = -13.733 cal/mol.

this indicates that the process is an exothermic reaction. Strictly speaking (aq)

implies that the reaction is taking place in such dilute solution that the addition of

further solvent causes no heat change, i.e. the heat of dilution is zero; this is

exactly true, however, only at infinite dilution. The constant quantity of heat a is

equal to the heat of formation of one mole of undissociated water from

combination of one mole of H+ and one mole of OH

-, where

H

+

(aq) + OH

-

(aq) = H2O ; ∆H = - 13.733 cal/mol.The neutralization reactions of week acid or week base evolved not only heat

of neutralization but also heat of ionization. So the quantity of heat of reaction Q

will be either lower or higher than 13.733 cal/mol, according to the endothermic or

the exothermic ionization process.

Experiment:Determination of the heat of neutralization of hydrochloric acid and sodium

hydroxide.


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Equipment:

Calorimeter - Bekmann thermometer - graduated cylinder - beaker (250 rnl) -

distilled water - burette - pipette - conical flask - funnel - spatula - ph.ph. indicator.

Procedure:

1. Prepare 500 ml of 5N HCI and 500 ml of NaOH, thermostat the two solution

at room temperature T1 .

2. Titrate 50 ml NaOH (in a flask) against 5N HCI by using ph. Ph. as

indicator, determine V1 (HCl).

3. Weigh the empty calorimeter W1.

4. Place 50 ml of 5N NaOH in the calorimeter and record the temperature at

steady state T2.

5. Introduce quickly V1 (HCl) and justly record the temperature T3.


7. The weight of NaCl solution, W3 = W2 – W1.

Recording Data:

Cc CNaCl V1 W1 W2 W3 T1 T2 T3

Calculations:

1. Calculate ∆T = [T3 - (T1 + T2) / 2]; T1 mal be = T2 .

2. Calculate the heat absorbed by 5N NaCl solution where,

∆H1 = CNaCl x W3 x ∆T

consider (C NaCl = 3.895 cal), or use the heat capacity of 5N NaCl solution


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(obtained from previous experiment).

3. The heat absorbed by the calorimeter;

∆H2 = Cc x ∆T

where, Cc from Previous experiment.

4. Calculate heat of neutralization where;

∆H = ∆H1 + ∆H2

5. Calculate the number of water produced from neutralization reaction

(undissociated water) where;

n = (5 x 36.5 x V1) / 1000

6. Calculate the molar heat of neutralization;

∆H' = ∆H / n

Results:

Item Calc. Result

∆T = T3 - (T1 + T2) / 2 =

∆H1 = CNaCl x W3 x ∆T =

∆H2 = Cc x ∆T =

∆H = ∆H1 + ∆H2 =

n = (5 x 36.5 x V1) / 1000 =

∆H’ = ∆H / n =


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Hess’s Law

Theory:

Hess's law is a law stating that the heat given off or absorbed in a chemical

reaction is the same regardless of whether the reaction occurs in a single step or in

many steps. It is often called the law of constant heat summation and it is direct

consequence of the law of conservation of energy. The amount of heats of reaction

is dependent only on the initial and final states. As an example;

A -------- ∆HI -----→ B

∆H reaction = ∆H1+ ∆HII = ∆HIII + ∆HIV ↓ ∆HIII ∆HII ↓

C ------- ∆HIV -----→ D

Experiment:

Application of Hess's law.

The reaction between HCl, H2O and NaOH can be runs in two different ways

as following;

I] a) H2O (L) + HCl conc. = HCl (aq) ; ∆H1

b) HCl (aq) + NaOH (s) = NaCl (aq) + H2O (L) ; ∆HII

II] a) H2O (L) + NaOH (s) = NaOH (aq) ; ∆HIII

b) NaOH (aq) + HCl conc. = NaCl (aq) + H2O (L) ; ∆HIV

Equipment:

HCl conc. – NaOH (s) - distilled water – thermometer.


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Recording Data:

Run I Run II

T1 T2 T3 T4 T5 T1 T2 T3 T4 T5

Calculations and Results (Run I):

Item Calc. Result

∆H1

∆T1 = T2– T1 = - =

∆HW = ∆T1 x VW (ml) x CW = x x =

∆T2 = T2 – T3 = - =∆Hc = ∆T2 + Cc = x =

∆HI = ∆Hw + ∆Hc = + =

∆HII

∆T3 = T4 – T3 = - =


∆T4 = T4 – T5 = - =

∆Hc = ∆T4 + Cc = x =

∆HII = ∆Hw + ∆Hc = + =

∆H ∆H1 + ∆H2 = + =

Calculations and Results (Run II):

Item Calc. Result

∆H1

∆T1 = T2– T1 = - =


∆T2 = T2 – T3 = - =

∆Hc = ∆T2 + Cc = x =

∆HI = ∆Hw + ∆Hc = + =

∆HII

∆T3 = T4 – T3 = - =∆HW = ∆T3 x VW (ml) x CW = x x =

∆T4 = T4 – T5 = - =

∆Hc = ∆T4 + Cc = x =

∆HII = ∆Hw + ∆Hc = + =

∆H ∆H1 + ∆H2 = + =

Interpret your results


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III

PHASE EQUILIBRIA


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INTRODUCTION

Two Components System

When two liquids are mixed together, one of the following cases may arise:

(a) The two liquids are completely miscible at all proportions yielding one

homogeneous liquid pha5e, for example alcohol and water.

(b) The two liquids are partially miscible yielding either one or two liquid

phases, depending on the conditions, for example phenol and water.

(c) The two liquids are practically immiscible yielding always two distinct

phases under ordinary conditions, for example carbon disulphide and water.

The mutual solubility of partially miscible liquids usually increases with

temperature. In this case, the solubility curve exhibits a maximum at the critical

solution temperature above which the two liquids become completely miscible at

all proportions. For Some liquid pairs such as ether and water, however, the mutualsolubility decreases with temperature, and the solubility curve shows a minimum at

the critical solution temperature below which the two liquids become completely

miscible at all proportions.

Composition-Temperature Diagrams

The temperature-composition diagram of the water-phenol system as shown

in the figure. Outside the area bounded by the curve ABC, there occurs one

unsaturated homogeneous liquid phase. Within that area, two liquid phases in

equilibrium with each other coexist; one is water saturated with phenol and the

other phenol saturated with water. Any point on the curve represents one saturated


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Experiment:

Determination of critical solution temperature of water-phenol system.

Procedure:

1. In a clean dry test tube (phenol tube) weigh accurately about 1 g phenol.

2. From a burette, add 0.5 ml of distilled water. Cover the tube with a cork

stopper carrying a thermometer and a stirrer, then place in a beaker

containing water to serve as bath.

3. Heat (or cool) gradually while the mixture is constantly stirred until the two

layers disappear forming one homogeneous layer. The two temperatures (t1

and t2), at which this occurs on passing from a lower (t1) to a higher (t2)

temperature and the reverse, are recorded. These two temperatures should be

nearly the same, and their mean gives the miscibility temperature of the

mixture used.

4. To the same mixture add the necessary volumes of water (0.5, 1, 1.5,…) and

heat gradually, then determine the miscibility temperature of the new

mixture as described above.

Calculations:

1. Record the volume of water used for each composition.

2. Plot miscibility temperature against percentage water or phenol for the

various mixtures.

3. From the curve obtained find out the critical solution temperature and the

corresponding composition.


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Results:


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Three Components System

Theory:

With a system of three components, four variables are possible; the pressure,

temperature and two concentration terms (the concentration of the third component

will be automatically fixed). To simplify the study of the phase properties, the

system is treated as a condensed one at a constant temperature, in which case the

pressure and temperature as variables, are dispensed with. The phase rule

expression reduces hence to: F = C - P. Under such conditions, the phase

properties are best defined by using the triangular diagram. On an equilateral

triangle (as shown in figr.rre) the apexes A, B and C represent pure components

(100% each). A point on any one side represents a mixture of two components

only. A point P inside the triangle represents the composition of a mixture of A, B

and C of percentages respectively proportional to the lengths of the lines Pa, Pb

and Pc drawn parallel to the sides of the triangle. This contention is borne out by

the fact that in an equilateral triangle the sum of Pa + Pb + Pc (also for any point

other than P) is always equal to the length of any one side (corresponding to a total

of 100%).

If components A, B and C are completely miscible with each other at all

proportions, then any point inside the triangle represents a system of three

components and one phase; therefore F, the number of degrees of freedom, is 2,

which means that the concentrations of any two components can be varied with

respect to one another while that of the third component is maintained constant. If

components B and C form a pair of partially miscible liquids, a mixture of the two

composition X (as shown in the figure) will separate into two layers having the

composition determined by points b and a (compare the phenol-water system). If


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component A is gradually added to the above mixture, the composition, will vary

following points on the line Ax. If A is completely miscible with both B and C, it

will distribute itself between the two layers and eventually a composition

corresponding to point M is reached where complete miscibility is observed.

Similarly, for any other mixture of A and B of composition between a and b, a

point is reached where there is complete miscibility. The line connecting the

various points is the bimodal curve b M a. Within this curve (immiscibility region)

any point represents a system with C = 3 and P = 2; therefore: F = l, which means

that changing the concentration of one component, the concentrations of both other

components must change. Outside the curve, is the region of complete miscibility

and is treated as explained before.


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Experiment:

Construct the binodal curve for the system ethyl acetate ethyl alcohol-water.

Procedure:

1. Into five dry stopper bottles introduce the following volumes of ethyl acetate

and water.

Bottle I II III IV V

Ethyl acetate, ml 10 8 6 4 2

Water, ml 2 4 6 8 10

The mixtures will appear turbid. Thermostat the bottles at 30oC.

2. From a burette, run in ethyl alcohol in small quantities at a time into each of

the bottles in turn. shake well after each addition and thermostat again.

3. Continue addition of ethyl alcohol till turbidity just disappears on shaking.

(The final additions should be made drop by drop).

Calculations:

1. Calculate the percentage composition by weight of each mixture at the stage

when turbidity lust disappears.

Densities at 30 oC are

Ethyl acetate = 0.894 g/c.c.

Ethyl alcohol = 0.789 g/c.c.

Water = 0.996 g/c.c.

2. Tabulate the results and plot them on a triangular diagram. Join the points by

a smooth curve to obtain the binodal curve.


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Results:

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Practical Physical Chemistry Course