Indian Institute of Science

Education and Research Kolkata

Advanced Physical Chemistry Laboratory

CH3105

Laboratory Manual

2015

Page | 2

CH3105

3rd

year Practical

Autumn Semester, 2015

GENERAL INSTRUCTIONS

1. Attendance is mandatory. In case of illness etc. the student must contact the instructor and fix a

schedule for making up the missed lab. All labs must be completed in order to get a passing

grade.

2. All data and results should be recorded directly in the lab notebook. The recording should

include, title of the experiment, date of experiment, working formula, data in tabulated forms,

results and calculations.

3. The instructor should sign the data before the student leaves the lab.

4. Graph papers and computer print-outs may be directly pasted on the lab notebook.

5. The lab notebook must be submitted to the instructor with the completed work before the student

leaves the lab.

Grading:

The marking scheme in the lab will be as follows:

1. Mid-semester examination …………………………….. 30

2. Lab notebook …………………………….. 10

3. Attendance …………………………….. 10

4. Continuous assessment by teacher …………………………... 10

5. Final examination …………………………….. 40

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Lab Safety is not-negotiable!

To help you realize how important it is, 2 marks will be deducted everytime you are found not

wearing the labcoat, proper footwear, and so on.

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List of Experiments

Experiment

No.

Experiments Page

No.

1. Critical Solution Temperature of Phenol-Water System 5

2. Verification of Langmuir Adsorption Isotherm 7

3. Determination of Eº value of Ag/AgCl Electrode and hence f+ by

Potentiometric Method

10

4. Determination of Fluroescence Quantum Yield 13

5. Measurement of Rate-Constant of Fluorescence Quenching: Stern

Volmer Method

15

6. Potentiometric Titration of Strong and Weak Acid Mixture Using a

Strong Base

19

7. Verification of Ostwald’s Dilution Law by Conductometric Method 21

8. Determination of activation energy of a clock reaction 24

9. Estimation of Partition Coefficient of Succinic Acid in n-Butanol and

Water Biphasic System

26

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EXPERIMENT 1

Critical Solution Temperature of Phenol-Water System

THEORY:

Two partially miscible liquids may become completely miscible at a higher temperature since solubility

increases with temperature generally. This miscibility temperature is different for different compositions

of the mixture. The highest miscibility temperature is called the critical solution temperature or CST.

Above this temperature, all compositions of this mixture are completely miscible.

PROCEDURE:

(1) Take a clean boiling tube fitted with a cork carrying a sensitive (110 oC) thermometer and stirrer.

(2) Measure 5 mL of phenol into the boiling tube.

(3) Fill a burette with water and add 2 mL water into the boiling tube containing phenol.

(4) Fix the thermometer and stirrer and heat slowly on a water bath, stirring continuously.

Note the temperature at which turbidity just disappears.

(5) Take out from the water bath and allow the tube to cool slowly while stirring. Note the

temperature at which turbidity reappears.

(6) Then add 2 mL more of water into the same boiling tube and repeat the experiment.

(7) Continue like this till a total of 30 mL water is added.

(8) Plot the average miscibility temperatures against percentage of phenol on a graph paper. The

maximum point in the curve is the CST of phenol-water system.

RESULTS:

Vol. of phenol

(mL)

Vol. of water

(mL)

% of phenol Miscibility temperature

Disappear Appear Average

5

5

…

2

4

…

Page | 6

CST of phenol-water system = ____ oC.

Critical composition = ____ % phenol, ____ % water.

PRECAUTIONS: • Use gloves. • DO NOT ingest phenol.

Follow-up experiment: EFFECT OF ADDING NaCl ON THE CST OF PHENOL-WATER SYSTEM AND

DETERMINATION OF UNKNOWN CONCENTRATION

THEORY:

The CST of a system is usually increased when a solute soluble in only one of the component liquids is

added. Thus when NaCl or KCl, soluble only in water and not in phenol, is added, the CST of phenol-

water system is found to increase with concentration.

PROCEDURE:

(1) Prepare 250 cm3 of 0.2 molar NaCl solution. Transfer 10, 20, 30, 40 and 50 cm

3 of this solution

into five different 100 cm3 volumetric flasks and make up with distilled water to get 0.02, 0.04,

0.06, 0.08 and 0.10 molar solutions of NaCl respectively.

(2) Determine the CST of phenol-water system using each of these solutions in the place of water.

(3) Also determine the CST using the given solution of unknown concentration.

RESULTS: Concentration of the given NaCl solution = ____ M.

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EXPERIMENT 2

Verification of Langmuir Adsorption Isotherm

THEORY:

In this laboratory exercise, we will generate a Langmuir Isotherm for the adsorption of acetic acid on the

surface of activated charcoal. The surface area of the activated charcoal available for adsorption of

acetic acid will be determined from the isotherm data and other assumptions. Activated charcoal is

particularly useful in purification processes involving impurity adsorption because of its high porosity,

giving it a very high surface to mass ratio.

Surface adsorption to a solid falls into two broad categories; physisorption and chemisorption.

Physisorption is a non-specific loose binding of the adsorbate to the solid via van der Waals type

interactions. Multilayered adsorption is possible and it is easily disrupted by increasing temperatures.

Chemisorption involves a more specific binding of the absorbate to the solid. It is a process that is more

akin to a chemical reaction and hence, only monolayer adsorption is possible

Irving Langmuir was awarded the Nobel Prize in 1932 for his investigations concerning surface

chemistry. Langmuir’s isotherm describing the adsorption of adsorbate (A) onto the surface of the

adsorbant (S) requires three assumptions:

• The surface of the adsorbant is in contact with a solution containing an adsorbate which is strongly

attracted to the surface.

• The surface has a specific number of sites where the solute molecules can be adsorbed.

• The adsorption involves the attachment of only one layer of molecules to the surface, i. e. monolayer

adsorption.

The chemical reaction for monolayer adsorption can be represented as follows:

A + S ⇌ AS

where AS represents a solute molecule bound to a surface site on S. The equilibrium constant

Kads for this reaction is given by: SA

ASK ads (Eq. 1)

[A] denotes the concentration of A, while the other two terms [S] and [AS] are two-dimensional analogs

of concentration and are expressed in units such as mol/cm2. The principle of chemical equilibrium

holds with these terms. The complete form of the Langmuir isotherm considers (Eq. 1) in terms of

surface coverage θ which is defined as the fraction of the adsorption sites to which a solute molecule has

become attached. An expression for the fraction of the surface with unattached sites is therefore (1- θ).

Given these definitions, we can rewrite the term [AS]/[S] as

1S

AS (Eq. 2)

Now we express [A] as C and rewrite (Eq. 1) as:

1CK ads (Eq. 3)

Page | 8

Rearranging, we obtain the final form of the Langmuir adsorption isotherm:

CK

CK

ads

ads

1 (Eq. 4)

If we define Y as the amount of adsorption in units of moles adsorbate per mass adsorbant, and

Ymax and the maximal adsorption, then:

maxY

Y (Eq. 5)

and the isotherm can be expressed as:

(Eq. 6)

This is the form of the isotherm we will use for our Charcoal-Acetic Acid system.

PROCEDURE:

We will prepare our acetic acid-charcoal isotherm by allowing acetic acid solutions of various

concentrations to equilibrate for about an hour with a given mass of activated charcoal. The amount of

acetic acid (HAc) not adsorbed to the activated charcoal will be determined by titration with sodium

hydroxide (NaOH).

HAc (aq) + NaOH (aq) NaAc (aq) + H2O (Eq. 7)

This will allow us to easily determine the amount of acetic acid that has adsorbed to a given mass of the

charcoal.

Prepare about 1L of 0.1M NaOH from a 6M stock solution and store it in a tightly stoppered plastic

bottle. Standardize the NaOH against standard oxalic acid.

Prepare 0.4 N Acetic Acid and standardize against NaOH solution.

Page | 9

Weigh about 1.5g charcoal into 6 stopper 250 mL Erlenmeyer flasks.

Prepare a series of acetic acid solutions according to:

For each sample above, add 100 mL of the solution to a charcoal sample. Swirl the flasks vigorously and

then place them into the shaker bath at 25oC. Let them agitate until about an hour.

Filter your charcoal solutions (discard the first 10 ml of filtrate to clean the filter flask of any

contaminants and to saturate with acid any adsorption sites which might be on the filter paper) and

titrate a suitable aliquot with the standardized NaOH. Perform your titrations in triplicate.

Data Analysis

1. Determine the concentration of the NaOH solution from your standardization data.

2. Using your titration data, for each sample:

Calculate the number of moles of acetic acid in the solution before adsorption.

Calculate the number of moles of acetic acid in the solution after adsorption.

3. Determine Y for each sample. Include appropriate error estimates for each Y value.

4. Plot your isotherm; Y vs. C. Include appropriate error bars for Y.

5. Plot the isotherm in the form of (Eq. 6) and determine the Langmuir parameters Ymax and

Kads.

7. Plot your isotherm; Y vs. C.

8. Assume the surface area occupied by one acetic acid molecule on the surface of the

charcoal is 21Å2. Determine the surface area of 1 g of charcoal. Express your result in

m2/g.

Is your number reasonable? Compare your result with that of the literature.

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EXPERIMENT 4

Determination of Fluorescence Quantum Yield

Theory

When a fluorophore absorbs a photon of light, an energetically excited state is formed. The fate of this

species is varied, depending upon the exact nature of the fluorophore and its surroundings, but the end

result is deactivation (loss of energy) and return to the ground state. The main deactivation processes

which occur are fluorescence (loss of energy by emission of a photon), internal conversion and

vibrational relaxation (non-radiative loss of energy as heat to the surroundings), and intersystem

crossing to the triplet manifold and subsequent non-radiative deactivation.

The fluorescence quantum yield (Φ) is the ratio of photons absorbed to photons emitted through

fluorescence. In other words the quantum yield gives the probability of the excited state being

deactivated by fluorescence rather than by another, non-radiative mechanism. The most reliable method

for recording Φ is the comparative method, which involves the use of well characterised standard

samples with known Φ values. Essentially, solutions of the standard and test samples with identical

absorbance at the same excitation wavelength can be assumed to be absorbing the same number of

photons. Hence, a simple ratio of the integrated fluorescence intensities of the two solutions (recorded

under identical conditions) will yield the ratio of the quantum yield values. Since Φ for the standard

sample is known, it is trivial to calculate the Φ for the test sample.

Page | 14

Procedure

1. Prepare 0.1 M aq. NaOH solution.

2. Prepare a solution of fluorescein in 0.1 M NaOH so that the solution absorbance at λmax is ~0.05.

3. Dilute OD = 0.05 solution using 0.1 M NaOH soln. to prepare 4 different solutions with ODs at λmax

in 0.05-0.01 range.

4. Prepare four solutions of anthracene in cyclohexane with ODs at 356 nm in 0.05-0.01 range.

5. Prepare four different solutions of Rhodamine 6G in ethanol with OD at λ (= λmax of fluorescein) in

0.05-0.01 range, and a 5th

solution with OD = 1.

6. Record the absorbance of fluorescein and Rhodamine 6G solutions at λ (= λmax of fluorescein).

7. Record the absorbance of anthracene solutions at 356 nm.

8. Record the fluorescence spectrum of fluorescein and Rhodamine 6G solutions by exciting at λ (=

λmax of fluorescein).

9. Record the fluorescence spectrum of anthracene solutions upon exciting at 356 nm.

10. Calculate the integrated fluorescence intensity of each solution.

11. Plot together the integrated fluorescence intensity as a function of their respective absorbance in the

same plot. Determine the fluorescence quantum yield of Rhodamine 6G using fluorescein and

anthracene as reference standards, and compare the two values.

The ratio of slopes is proportional to the ratio of the respective quantum yields, given by the relation:

denotes the quantum yield and refers to the refractive index of the solvent. At 493 nm, (EtOH)

= 1.3656 and (H2O) = 1.335. At 350 nm, (cyclohexane) = 1.4266.

Rhodamine 6G = 0.95

Fluroescein, aq. NaoH = 0.91

Anthracene = 0.36

2

2

ref

sample

ref

sample

ref

sample

Slope

Slope

Page | 15

EXPERIMENT 5

Measurement of Rate-Constant of Fluorescence Quenching: Stern Volmer Method

Aim: To determine the rate constant of fluorescence quenching of a fluorophore by a quencher in a

solvent.

Introduction: The relative fluorescence intensity of a fluorophore depends on the number of

fluorophores, natural lifetime of its first excited singlet state and on the rate that the first excited state is

deactivated by nonradiative processes. Fluorescence quenching refers to any process that decreases the

fluorescence intensity of a sample. A variety of molecular interactions can result in quenching. These

include ground-state complex formation, excited-state reactions, molecular rearrangements, energy

transfer, and collisional quenching.

Much of these nonradiative deactivations come about because of collisions of the excited

molecule with solvent molecules. However, some substances are particularly efficient at deactivating, or

quenching, excited states. These tend to be substances with heavy atoms (for example halogens) or

paramagnetic compounds. Oxygen is a particularly good quencher; therefore it is often necessary to

remove the oxygen from solution before measuring fluorescence spectra. Fluorescent probes, such as

dansyl chloride, are used in biochemistry to study the various binding sites in large macromolecules

through the difference of the quenching rates of the bound verses free probe.

Theory: The emission intensity of the fluorophore can be quenched by a quencher (Q) by complex

formation in the ground state (static quenching) or in excited-state fluorophore quencher reaction

(dynamic quenching). The following general scheme illustrates the nature of processes that deactivate an

electronically excited state of a molecule.

If quenching happens in ground state then the equation for quenching reads as:

Page | 16

If quenching happens in excited state then the equation for quenching reads as:

In these equations F0 and F are the integrated fluorescence intensities in the absence and presence of

quencher, respectively; KS and KD are static and dynamic quenching constant. For both static and

dynamic quenching a linear plot is obtained. kq is the bimolecular quenching constant; τ0 is the lifetime

of the fluorophore in the absence of quencher, and [Q] is the concentration of quencher. The Stern-

Volmer quenching constant is given by KD = kqτ0. If the quenching is known to be dynamic, the Stern-

Volmer constant will be represented by KD. Otherwise this constant will be described as KS or KSV.

Quenching data are usually presented as plots of F0/F versus [Q]. This is because F0/F is expected to be

linearly dependent upon the concentration of quencher. A plot of F0/F versus [Q] yields an intercept of

one on the y-axis and a slope equal to KS or KD. From the linear plot of F0/F versus [Q], it is not possible

to say whether the quenching mechanism is static and dynamic. Time resolved fluorescence studies are

necessary to identify whether the quenching mechanism is static or dynamic. For static quenching

excited state fluorescence lifetime does not change but for dynamic quenching fluorescence lifetime

decreases with increase in [Q]. Therefore, for static quenching τ/τ0 = 1, whereas, for dynamic quenching,

F0/F = τ0/τ.

Examples of Quencher: One of the best-known collisional quenchers is molecular oxygen, which

quenches almost all known fluorophores. It is frequently necessary to remove dissolved oxygen to

obtain reliable measurements of the fluorescence yields or lifetimes. Known quenchers include halides,

heavy metals, different amines etc.

Generally linear Stern-Volmer plot is observed however, there are several reasons which make the Stern

– Volmer plot non-linear.

Page | 17

Reference: Principles of Fluorescence Spectroscopy, 3rd

Edition, J. R. Lakowicz.

Chemicals Required: Anthracene, CCl4, ethanol (95%).

Equipments Required: Fluorescence cuvette, Volumetric flasks, Pipettes; Spectrophotometer (to check

absorbance of the stock anthracene solution); Spectrofluorimeter (to measure fluorescence of different

solutions).

Procedure:

1. Make a ~5.0 x 10-4

M solution of anthracene by dissolving about 1 mg of anthracene in 10mL of

ethanol in volumetric flask.

2. Dilute this solution by 50 times by transferring 1 mL of this solution to a 50 mL volumetric flask

thus making a 1.0 x 10-5

M stock solution.

3. Calculate how much amount of CCl4 is required to make the final concentration as 0.02, 0.04,

0.12, 0.16, 0.20 moles/litre.

4. There will be total six solutions, one without CCl4 and other five solutions will be with different

concentrations of CCl4. Keep it in mind concentration of anthracene should be exactly same in

all solutions.

5. Measure the fluorescence of the stock solution of anthracene (i.e. without CCl4) and note down

the integrated intensity (F0) (area under the fluorescence curve).

6. Measure the fluorescence of all other solutions with equal concentration of anthracene but with

different concentration of CCl4 and note down the integrated intensity (F) for each five solutions.

7. Make a table of F0/F against the quencher concentration [Q].

Page | 18

8. Plot F0/F against the six different quencher concentrations.

9. Fit the plot with fixed intercept at 1.

10. Calculate the slope of the linear plot.

11. Comment on the goodness of the linear fit.

12. Write down the sources of error that could creep into the result.

13. If fluorescence lifetime of anthracene at 250C is 5 nanosecond then calculate the value of kq.

(Assume quenching to be dynamic)

14. Write the experiment report and submit the notebook to the instructor.

Cautions:

1. It is important that the absorbance (optical density) of the solution is not more than about 0.2 so

that the fluorescence intensity is uniform throughout the solution.

2. Rinse the fluorescence cuvette several times with the new solution before taking its fluorescence

spectrum.

3. Experimental conditions for fluorescence measurements, i.e. excitation wavelength, slits etc.

should be same for all measurements.

Page | 19

EXPERIMENT 6

Potentiometric Titration of Strong and Weak Acid Mixture Using a Strong Base

Aim: To determine the concentrations of a strong acid and a weak acid present in a mixture by

potentiometric titration.

Apparatus: Potentiometer assembly, KCl salt bridge, Pt electrode and saturated calomel electrode

Principle: When a electrode is dipped in an electrolyte solution, its potential depends on the

concentration of the ions to which the electrode is reversible. In an acid-base titration, quinhydrone

(QH) which is an equimolar mixture of quinone and hydroquinone is used as an H+ reversible system,

and the following cell is constructed:

Pt | acid mixture+QH || calomel

The reduction potential of QH and EMF of the cell will depend on the pH of the solution. On adding

small aliquots of base the EMF of the cell drops gradually; but near the equivalence point a large

decrease in EMF occurs. Beyond the equivalence point, the change in EMF again becomes gradual. For

a mixture of strong and weak acids, first the strong acid will be neutralized followed by the weak acid.

Consequently, two equivalence points will be seen.

Procedure:

1. Prepare 100 ml ~0.1 N NaOH solution and standardize against 0.1 N oxalic acid.

2. Pipette out 20 ml of the acid mixture (provided) into a clean beaker and add a pinch of QH to it.

3. Place a Pt electrode in the solution and connect it to a calomel electrode through a salt bridge.

4. Connect the Pt to the positive and SCE to the negative terminals of the potentiometer and

measure the initial EMF.

5. Keep adding 1 ml of standardized NaOH solution from the burette to the acid mixture and record

the cell EMF after every addition.

Page | 20

6. The EMF decreases gradually, and then dips suddenly, continues to decrease gradually again.

Note the volume when the decrease is sudden as VA.

7. The points when the EMF measurement becomes difficult, reverse the polarity, add few more 1

ml aliquots of NaOH and measure the EMF. Note the volume at which pole reversal occurs as

VB.

8. Repeat the experiment by taking fresh 20 ml of the acid mixture solution and titrating with 0.1

ml of NaOH in the neighborhood of VA and VB.

9. Plot E vs volume of NaOH and dE/dV vs. volume of NaOH for both runs. VA corresponds to the

neutralization of the strong acid, and VB VA corresponds to the that of the weak acid. Use these

values to calculate the respective acid concentrations.

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EXPERIMENT 7

Verification of Ostwald’s Dilution Law by Conductometric Method

Objective : Verification of Ostwald’s Dilution Law using acetic acid as the weak acid

Discussion : Equivalent conductance (Λ) of an eletrolyte solution is related to the specific conductance

(κ) according to the relation: Λ = 1000 κ /c

Where c is the concentration of the electrolyte in g equivalent/L

Thus, if the specific conductance and the concentration of a solution are known, the equivalent

conductance of the solution can easily be calculated.

The dissociation of a weak binary electrolyte (AB) with degree of dissociation α at a concentration c mol

L-1

may be represented as,

AB A+ + B

-…………………………………………………………(1)

c(1- α) cα cα

The dissociation constant Ka is given by,

Ka = (aA+

. aB-)/ aAB………………………………………………………………………(2)

Where ai represents activity of the ith species. For the dissociation of a weak acid (HA) equation (1) can

be written as follows.

HA + water H+(aq) + A

- (aq)……………………………………………………….(3)

Ionisation constant (Ka) of the weak acid, HA, may be defined according to,

Ka = [H+][A

-]/[HA].fH

+.fA

-/fHA………………………………………………………. (4)

here, terms in the square bracket represent concentrations c in gmolL-1

or gionL-1

. For a dilute solution

of weak acid, the ionic strength of the medium will be very low, and the numerical value of the activity

coefficients f’s are very close to unity (Debye-Huckel limiting law). Under this condition, equation (4)

may be written as:

Ka = [H+][A

-]/[HA]

= c2

/(1-)………………………………………………………(5)

The degree of ionization (), at a particular concentration (c) of the week electrolyte, HA, may be well

approximated by the ratio, /o, where is the equivalent conductance of the weak acid HA at

concentration c and o is its equivalent conductance at infinite dilution.

Substituting, = /o, one obtains from equation

Page | 22

Ka = (2 c)/[(o - )o]…………………………………………………………….(7)

Which on rearrangement yields,

1/ = 1/o + 1/(Ka. o2) c………………………………………………………….(8)

Thus, a plot of 1/ against c at different concentration should give a straight line, values of Ka and o

can be obtained from the slope and intercept.

Equipments/Glass Apparatus: Conductivity bridge, burette, pipette, conical flask, beaker.

Reagents: oxalic acid (C2H2O4, 2H2O), sodium hydroxide, acetic acid (CH3COOH), potassium chloride.

Procedure :

1. Prepare 100 mL standard 0.1N oxalic acid solution and 0.1 N (exact) KCl solution (100 mL). Prepare

solutions of 0.5N NaOH (100 mL) and 0.1N acetic acid (100 mL).

2. Prepare 100 mL 0.01N standard KCl solution by proper dilution of 0.1N solution.

3. Rinse a 100 mL beaker and the conductivity cell with the exact 0.01N KCl solution thoroughly and

then pour sufficient volume of this solution into the beaker so that the electrodes of the cell are

completely immersed in the solution. Record the conductance. Repeat this procedure with the exact

0.1N KCl solution. Calculate the mean cell constant from the measured conductance values of these two

solutions, using the literature values of the specific conductance of KCl solutions at these concentrations

at the same temperature. Find the mean cell constant.

4. Take 60 mL of the water in a 100 mL beaker. Dip the clean, dry conductivity cell into it. Add 0.5 mL

acetic acid, to water (in the conductivity cell), stir well and record the conductance. Repeat the

procedure of adding 0.5 mL acetic acid for at least eight times, and measure the conductivity of the

solution after each addition.

5. Calculate the concentration and equivalent conductance values of all the acetic acid solutions.

Results:

Table 1. Preparation of 100 mL standard 0.1N oxalic acid solution

Weight taken (g) Weight to be taken (g) Strength of oxalic acid solution

Table 2. Standardization of NaOH solution using standard oxalic solution

Page | 23

Sl.

No.

Vol. of oxalic

acid solution/ mL

Burette Reading /mL Average

volume/mL

Strength of NaOH

solution Initial Final Difference

Table 3. Standardization of Acetic acid solution against NaOH solution

Sl.

No.

Volume of acetic

acid solution/ mL

Burette reading/ mL Average

volume/mL

Strength of Acetic

acid solution Initial Final Difference

Table 4. Determination of specific conductance for KCl solution

Sl. No. Strength of KCl Conductance (S) Specific conductance

Table 5. Cell constant determination

Sl.

No.

Concentration

of KCl

Specific

conductance

Resistance Cell

constant

Mean cell constant

Table 6. Conductance of acetic acid solution of different concentration

Volume of water = …….mL

Sl.

No.

Volume of acetic acid

solution added (mL)

Concentration of

acetic acid solution

Conductance

(S)

Equivalent

conductance

Calculation : Calculate Ka and Λo from the slope and the intercept.

Page | 24

EXPERIMENT 8

Determination of Activation Energy of a Clock Reaction

Theory: In this experiment we will determine the effect of temperature on the rate of a chemical

reaction. A reaction is chosen which proceeds conveniently slowly near room temperature and which

can be measured easily by a dramatic color change. This reaction is the oxidation of iodide ion (I-) to

molecular iodine (I2) by hydrogen peroxide (H2O2) in acidic medium. The reaction is as follows:

2I- + H2O2 + 2H

+ = I2 + 2H2O (slow) (1)

As this reaction proceeds, the colorless reactants gradually develop a brown color due to the product I2.

Because of the difficulty of timing the appearance of the I2, we make use of another much faster reaction

in the same solution to mark the progress of the slow reaction:

I2 + 2 S2O32-

=2 I- + S4O6

2- (fast) (2)

Reaction (2) is so fast that I2 produced by reaction (1) is consumed instantaneously by the thiosulfate

(S2O32-

), so that the I2 color cannot develop. Because both S2O3

2- and S4O6

2- are colorless, the solution

remains colorless. However, we do not add enough thiosulfate to react with all the I2 that will be formed

from reaction (1). By this device the reaction solution stays colorless until the instant at which all the

thiosulfate is consumed, and then free I2 begins to appear. We time the reaction from the initial mixing

until the appearance of I2. In order to help see this appearance we add starch indicator which forms an

intensely blue dark complex with I2 and signals the appearance of I2

by a dramatic color change.

We know the initial concentration of S2O32-

and measure the time interval for this S2O32-

to be consumed.

These quantities determine the rate of the slow reaction (1). By repeating the experiment with different

temperatures, we can determine the effect of temperature on the rate of reaction (1). In this procedure

the concentration of I- is kept fixed and the reaction becomes pseudo-unimolecular. Thus,

kcdt

dc ; c = concentration of H2O2 (3)

So, 2

1

12

0 ln1

ln1

TT

TT

ttc

c

tk

(4)

Where T = titre value at infinite time, T1 = titre value at time t1, and T2 = titre value at time t2.

At a fixed acid concentration, the reaction is studied at different temperatures. Actual specific rate

constant k is

]['

H

kk

(5)

Now, RT

EAk a ln'ln

(6)

From which the activation energy, Ea, is determined graphically.

Page | 25

Procedure: Prepare the following solutions.

(1) 500 mL 0.4% KI

(2) 100 mL 12 N H2SO4

(3) 100 mL 2 vol H2O2

(4) 100 mL 0.2 N Na2S2O3

(5) For each set kept in a bath, take 100 mL 0.4% KI + 6 mL 12 N H2SO4 + 2 mL starch + 1 mL 0.2

N Na2S2O3 + 4 mL 2 vol H2O2 (stop-watch is started during H2O2 addition). At the time of

appearance of iodine color, note the time and add 1 mL 0.2 N Na2S2O3 and shake well. Take 5-6

such readings.

Plot iTT

TT

1ln versus (ti-t1). A straight line would be obtained according to Eq (4). Determine

the value of rate constant from the slope.

(6) Determine T, i.e., the titre value at infinite time (i.e., complete reaction) as follows: Take 100

mL 0.4% KI + 6 mL 12 N H2SO4 + 2 mL starch + 1 mL of 1% ammonium molybdate solution +

4 mL 2 vol H2O2 and titrate the liberated I2 against 0.2 N Na2S2O3 solution.

(7) Perform the experiment at 20, 25 30, and 35 oC.

(8) Plot ln k’ versus 1/T and determine the activation energy from the slope. (slope = −Ea/R)

Note:

Ea ~ 16 k cal/mol.

Calculate S# from the plot.

Dependence of the Reaction Rate on the Presence of Catalyst: Some ions have a pronounced catalytic

effect on the rates of many reactions in water solution. Homogenous catalyst [NH4]2MoO4, ammonium

molybdate is used in the reaction. The activation energy of the catalyzed reaction is smaller than the Ea

of an uncatalyzed reaction. Since the function of the catalyst is to lower the activation energy, the

reactions moves forward a lot quicker in comparison to the uncatalyzed reaction. Although the catalyst

doesn’t get consumed in the reaction, it creates new transition states and elementary steps to allow the

reaction to precede a lot faster. Hence, the catalyst lowers the Ea for the sodium thiosulfate and iodine

reaction and this experiment demonstrated the function of a catalyst in chemical kinetics.

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EXPERIMENT 9

Estimation of Partition Coefficient of Succinic Acid in n-Butanol and Water Biphasic System

Aim: To determine the partition coefficient of succinic acid in system of two highly immiscible liquids:

1-butanol-water.

Theory: Consider a system consisting of highly immiscible liquids. If another substance which is

soluble in both of these liquids is added then the substance is found to distribute itself into these two

liquids in a definite manner. At constant temperature and at equilibrium the ratio of the concentration of

the added substance in these two layers has a definite value and assuming no chemical reactions occur or

no chemical or physical changes happen, for dilute solutions that value is independent of the actual

amount of the substance added.

If the concentration of this substance in solvent 1 and solvent 2 are c1 and c2, then the partition

coefficient (K) is given by

K = c1/c2

When two phases are in equilibrium, then the chemical potential of the dissolved/partitioned substance

will be the same in both phases. The chemical potential of the dissolved substance in these two phases is

given by:

µ1 = µ10+ RT ln(a1) and µ2 = µ20+ RT ln (a2)

where a1 and a2 are the activities of the partitioned substance in the two phases.

When the two phases are at equilibrium then µ1 and µ2 must be equal, so

µ20 - µ10 = RT ln a1/a2

At a constant temperature the standard potentials, i.e. µ10 and µ20 are constant, and since ‘R’ is also a

constant, therefore, a1/a2 is also a constant, which is exact form of the distribution law.

For systems which are ideal or do not deviate from the ideal behavior, the ratio of activities may

be replaced by the ratio of mole fractions, which should also be constant. Further for dilute solution of

for gases the ratio of mole fractions may be replaced by concentrations. Thus under all these

considerations c1/c2 should be constant in agreement with the simple form of the distribution law.

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Chemicals Required: Succinic acid (aqueous solution c = 20mg/dm3), NaOH (c = 0.05 mol/dm

3), 1-

Butanol, Phenolphthalein, Distilled water

Equipments Required: 4 separation funnels (100 mL), 4 titrimetric flasks, 2 burettes, pipettes, 8

beakers

Procedure:

1. Prepare 4 mixtures of butanol, water and succinic acid into separation funnels following the

Table 1.

2. Shake gently the separating funnels with prepared solutions for 40 min.

3. Put them to the holders and wait a few minutes until the mixture is apparently separated into two

clear layers: lower layer is water (Wat), and upper layer is butanol (Org).

4. Pour carefully the water phase into a beaker using the funnel’s valve.

5. Transfer the water solution from the beaker into two titrimetric flasks, 5 mL per flask, using

pipette (V = 5 mL).

6. Add 2-3 drops of phenolphthalein.

7. Fill a burette with a standard solution of NaOH (titrant) and perform titration.

8. When the endpoint of titration has been reached, read the used volume of NaOH from the burette

(VT). Write it down to the Table 2.

9. Repeat the procedure for funnels 2 – 4.

10. Proceed in the same way the upper layer (butanol) of the mixtures in the funnels No. 1-4.

Data treatment:

1. Calculate the average volume of titrant (VT) from 2 titrations of the same solution in the Table 2.

2. Succinic acid is a diprotic acid. Process of titration is based on the neutralization reaction:

3. Calculate the concentration of succinic acid in the water and organic phase, cWat, cOrg,

respectively, for each composition of the partition mixture. Remember, an analyte is the succinic

acid, the volume of VA = 5 mL.

4. Calculate the partition coefficient k of succinic acid at each composition of the partition mixture.

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5. Express the average value of the partition coefficient k from four experiments.

Table 1: Composition of partition mixtures

Table 2: Experimental data

Write down the report including the following and submit the notebook to the instructor:

• Theory (partition coefficient)

• Working procedure and measurements

• Tables of results and calculation

• In conclusion, provide the obtained average value of partition coefficient from 4 experiments.