1. s
E M a n u a l
To be a value-driven
Institute providing
quality education in
science which thrives
for producing scholars
with concern and care
for environment and
society that will best
serve the nation in the
21st
century.
PHYSICAL CHEMISTRY
M.Sc I SEMESTER
SESSION 2019-20
GOVT. MADHAV SCIENCE
P.G.COLLEGE UJJAIN M.P.
2. DECLARATION BY STUDENT
I ……………………………………….Son/Daughter of shri…………………………………….. Class
M.Sc. I Semester Certify that the Practical report has been prepared by me
or my group as the case may be after working for stipulated hours in the
laboratory under guidance of Dr Kalpana Singh Professor of Chemistry Govt.
Madhav Science P.G. College Ujjain
Date ___________
Name :-
Class :-M.Sc,I Semester
Place Ujjain
Roll No. ……………………………………
3. DECLARATION BY THE INSTRUCTOR
I Dr. Kalpana Virendra Singh Professor of Chemistry certify that the
practical report has been prepared by the student and his/her group after
working for stipulated hours in the laboratory. Student has taken care of
optimum green practices during the practical sessions and has helped in
diminishing the harmfulchemical impact thereon.
Date
Dr. Kalpana Virendra Singh
Professor Chemistry
Govt. Madhav Science P.G.College Ujjain M.P.
4. HOUSE KEEPING
Before starting an experiment, wash, and if necessary dry, any glassware,
before use. Remember, the last person to use the glassware was a student.
At the end of the experiment be sure to clean and wash any glassware. If
you found it on the bench, in the cupboard, the draws or outside on the
platform return it there on the rack. If it came from a stock cupboard , put
it on the drying rack. Make sure you leave the balances clean. If any
equipment was disassembled when you got it, disassemble it again and return
to where you found it. Leave any equipment or chemicals that the
instructor/technician gave you on the bench. Clean up any spills, paper
towels, or any other mess you make. Report any breakages or depleted
reagents so we can make replacements. Failure to do this housekeeping may
result in marks being deducted.
Instructor
Dr. Kalpana V.Singh
5. INDEX
Sr.
No.
Name of the Experiment Date Page
No.
Remark and
Signature of
Instructor
1. Errors , Types of Errors, minimization
of errors distribution curves,
precision accuracy and combination.
Numericals based upon this
01-07
2. Calibration of Volumetric burette ,
Pipette and Standard Flask.
Preparation of Standard Solutions and
equalization of strengths of two acids
titrimetrically
08-14
3. Determine the Order of reaction for
Acid catalyzed Hydrolysis of Methyl
Acetate
15-18
4. Determine the Activation Energy and
Temperature Coefficient at two given
Temperatures for Acid catalyzed
Hydrolysis of Methyl Acetate
19-24
5. Determine the Congruent Composition
and Temperature of Binary Mixture
Diphenylamine Benzophenone mixture
25-27
6. 01
Experiment No. 01
Errors , Types of Errors, minimization of errors distribution curves, precision
accuracy and combination. Numericals based upon this
In qualitative analysis, when numerical data and numerical results are measured with the greatest
correctness that the instrumental method and observer are capable of, it has been observed that the
results of successive determination differ among themselves to a greater or lesser extent. These are
considered as errors.
Types of errors-
Errors in any set of measurements can be divided into the following two categories-
(1) Determinate or constant errors
(2) Indeterminate or random errors
[1] Determinant Errors:- A determinate error is a tangible error, the magnitude of which can be
determined and the measurements thereby corrected. A determinate error is characterized by the fact
that it ordinarily affects the se degree the results of a series of determination. If the determinate error
cannot be controlled, it may become an indeterminate errors.
Type of Determinate Error:-
There are basically four types of determinate error:-
(1) Instrumental and Reagents Errors:-
Instrumental errors may be caused by balance arms of unequal lengths, weight, that are uncalibrated
or improperly calibrated and the burettes that are incorrectly graduated. They also arise due to foreign
material deposition upon glass ware porcelain etc. and by loss of weight in platinum crucible when
strongly heated.
Similarly, reagent error arise by the utilisation of reagents containing impurities, if pure
reagents are not available, the degree of impurity must be measured, in order that a suitable correction
may be made.
7. 02
(2) Personal Errors and Operation Errors:-
Personal errors are largely the results of improper techniques and carelessness, e.g. errors in reading
burette, mechanical loss of material in various steps of an analysis, failure to wash and ignite a
precipitate properly etc.
(3) Methodical Errors:-
These errors are the most serious errors encountered in chemical analysis. These errors can be
eliminated or reduced to very small magnitude by the proper technique, because they are mostly
physical in nature e.g. Co-precipitation and post precipitation, occurrence of induced or side reaction
etc.
(4) Additive and Proportional Errors:-
An additive error is one which has the same absolute value, no matter what is the account of the
constituent being determined. Error in weight are well known example of additive errors.
A proportional error is one whose absolute magnitude depends upon the quantity of the
constituent e.g. it may arise due to an impurity present in a standard substance, which leads to an
incorrect value for the normality of a standard solution.
[2] Indeterminate or Random Errors:- These errors are accidental and more or less intangible errors
over which the operator has little, if any control. These errors are difficult to define, but their existence
is revealed by the small differences in the successive values of a measured quantity when the
measurements are made by the same person will great care under conditions, as nearly the same as
possible. These errors can be divided into two classes:-
(1) Variation within Determinate Error:-
The temperature changes and variation in humidity in a balance room also give rise to determinate
errors, but if they are not controlled they become indeterminate.
(2) Erratic Errors:- It is very difficult to pinpoint such types of errors and in fact the analysis is
unaware of their presence, e.g. accidental loss of material during an analysis etc.
8. 03
Removal or Minimization of Errors:-
The determinate errors often be minimized by one of the following methods:-
(1)Running a blank determination:- Errors arising from the introduction of impurities through the
reagents and vessel and sometimes from other sources are often accomplished by running a blank
such a procedure involves going through all the analysis, using the same solvent and reagent in the
same quantities but emitting the unknown component or sample.
(2) Calibration of apparatus and application of corrections:- All instruments such as burettes, pipettes,
weight measuring flask etc. Must be properly calibrated and the appropriate corrections are applied to
the original measurements.
(3) Running a Controlled Determination:- This method of minimisation of errors consists in carrying
out a determination under identical experimental conditions upon a quantity of a standard substance
which contains the same weight of the constituent as is contained in the unknown sample. The weight
of the constituent X in the unknown can be calculated by making use of the following relationship.
Result found for standard = Weight of constituent in standard
Result found for unknown.
(4) Running a parallel determination:- The results obtained is running parallel determination serve as
a check on the results of a single determination and indicate only the precision of the analysis.
(5) Standard Addition:- This method is specially applicable to physico-chemical procedures such a
paleography and spectrophotometer. A known amount of the constituent which is to be determined is
added to the sample which is then analyzed for the total amount of constituent present. The recovery
of the amount of added constituent will be equal to the difference between the analytical results for
samples with and without the added constituent.
(6) Isotopic Dilution:- This method consists in adding a known amount of the pure component
containing radioactive isotope to be unknown and throughoutly mixed with it.
Measurement of Errors (Accuracy and Precision)
The difference between the observed or measured value and the true or most probable value is known
as the absolute error of the determination. The absolute error is the measure of the accuracy of the
9. 04
quantity measured. The accuracy has been described as the degree of arrangement between a
measured value and the most probable or the true value.
In analytical chemistry, the term precision of a numerical value is defined as the
degree of agreement between it and other values, obtained under substantially the same conditions. In
other words, precision is used to describe the responsibility of the results, while accuracy has been
used to express the correctness of measurement.
Precision Measures:-
“The mean deviation or the relative mean deviation is a measure of the precision. The mean is the
numerical value obtained by dividing the sum of a set of measurements by the number of individuals
results in the set. Thus the mean m is given by.
m= ΣMn/n
Where M is the individual measurement and n is the total number of measurements.
Q.1 calculates the result given by replicate determination of chloride in a metal chloride.
X1=32.22, X2=32.64, X3=36.52, X4=32.46
Sol. Authentic mean=32.22+32.64+.36.52+32.46/4
=133.84/4=33.46
Mean of a discrete frequency distribution.
1. Direct method-
Q.2 Find the mean of the following distribution
X=4, 6, 9, 10, 15
f=5, 10, 10, 7, 8
Sol. Using formula
X =(f1.x1+f2.x2+f3.x3+.....fnxn)/( f1+f2+f3+.............fn)
10. 05
X F XiFi
X1 4 F1 5 20
X2 6 F2 10 60
X3 9 F3 10 90
X4 10 F4 7 70
X5 15 F5 8 120
Median- The median is a value about which all the other are equally distributed. Half the value is
scalar and other half are larger than the median value. Let the total no. Observations be n.
If n is odd ((n+1)/2) th Observations.
If n is even median is the arthenetic means of the value of (n/2)than ((n+2)/2)th Observations.
Let - the data be-
38, 70, 48, 34,42,55,63,46,54,44
Arrange in ascending order.
34,38,42,44,46,48,54,55,63,70
n=10 even
Median = value of ( 10/2)th term + value of ((10+1)/2)th term
2
= 5th term + 6th term
2
= (46+48)/2 = 94/2= 47
Median is 47.
Q.3 calculate median of the data for the liberation of ferrous avvoniueesulphate potassium
permanganate.
19.4CC , 19.5CC ,19.6CC ,20.3CC, 19.8CC ,20.1CC ,20.5CC
Ans. Arrange in increasing order.
19.4, 19.5, 19.6, 20.3, 19.8, 20.1, 20.5
n=7 (odd)
Median = ((n+1)/2 )th term
11. 06
= ((7+1)/2)the term
=4th term i.e. 19.8
Median =19.8
Q.4 Find the median of the following data
25, 34, 31, 23, 22, 26, 35, 28, 20, 32
Sol. Arrange in increasing order
n=10 even
Median =( (n/2)th term + ((n+1)/2+1)th term)/2
= ((10/2)the term + (12/2)th term)/2
= (5th term + 6th term)/2
= (26+28)/2
Median = 27.
Mean deviation - the mean deviation of a single measurement is the mean of the deviation of all the
individual measurements.
1. This can be calculated by determining the arithmetical mean of the results.
2. Then calculating the deviation of each individual measurement from the mean.
3. Finally dividing the sure of the deviations regardless of sign , by the nuclear of measurements.
The mean deviation is calculated by-
d= Ʃ(|Mn-m|)/n
d =Mean deviation
|Mn-m| = Absolute value of the deviation if
Mn is the nuclear from the mean.
n = no. Of total item's.
ƩXi = Sun of individual deviation from the mean.
12. 07
Q.5 calculate the mean deviation of following values.
46.62 ,46.47, 46.64, 46.76, 46.53, 46.60, 46.71, 46.60, 46.71
And:- Mean=(46.62+46.47+46.64+46.76+46.53+46.60+46.71+46.6+46.71)/9
M=(419.64)/9= 46.627
The difference between any one of the values and the mean is the deviation XL of that value
from the mean.
Mn M =(Mn-M) (Mn-M)*(Mn-M)
1 46.62 46.627 =0.007 0.000049
2 46.47 46.627 =0.157 0.02464
3 46.64 46.627 =0.013 0.000169
4 46.76 46.627 =0.133 0.01768
5 46.53 46.627 =0.097 0.009409
6 46.60 46.626 =0.027 0.000729
7 46.71 46.627 =0.083 0.00688
8 46.60 46.627 =0.027 0.000729
9 46.71 46.627 =0.083 0.00688
Total 0.627 0.06716
Now the average or deviation
d=0.627/9=0.070
Relative mean deviation - if is the mean deviation divided by the mean ,experssed in terms of
percentage or parts per thousand.
Relative mean deviation =(d/m)*100
= (0.070/46.627)*100
=.150%
Standard deviation- the standard deviation is obtained by extracting the square root of the quotient
obtained by dividing the sun of the square of the individual deviation by of the number of
measurement made. Thus
б=[ (1/n) Ʃ (Mn-M)*(Mn-M)]
б=[(1/n)*((0.007)*(0.007)+(0.157)*(0.157)+(0.013)*(0.013)+(0.133)*(0.133)+(0.97)*(0.97)+(0.027)*
(0.027)+(0.083)*(0.083)+(0.027)*(0.027)+(0.083)*(0.083)]
б =√(0.06717)/9 =0.086
13. 08
Experiment No. 02
Calibration of Volumetric burette, Pipette and Standard Flask. Preparation of
Standard Solutions and equalization of strengths of two acids titrimetrically
PRINCIPLE
Volumetric flask, burette, and pipette are used to deliver or to contain a known volume of a dilute
solution of the analyte or the reagent. Their calibration is done by accurately weighing the mass of the
water contained in or delivered by these apparatus. These weights are converted to volumes by using
accurately known density of water at the temperature of measurement. The formula given in Eq.1.1 is
used for the purpose,
V= m/p
Where, V, m, and p. are the volume, mass and density of water respectively.
Water serves as the ideal reference material because it has similar viscosity and rate of drainage as the
liquids to be contained in or delivered by these apparatus. Further, the ready availability and cost add
to its choice as the reference material. It may also be noted here that the volumes measured would also
depend on the way apparatus is used. In this context, three aspects need to be mentioned; these are
the way the apparatus is filled, the way it is drained and the way the meniscus is read. Therefore, the
apparatus should be used in exactly the same way as it is used during calibration.
You would recall that any measurement is accompanied by systematic as well as random errors. If
sufficient care is taken to avoid the systematic errors the random errors would still affect the precision
of the data. The precision of a given set of measurements can be as curtained from the value of
standard deviation. However, probability distributions are quite important for judging there liability of
analytical data. If we have sufficient number of observations we can apply the laws of probability to
estimate the uncertainty introduced by random errors. According to the laws of probability:
i) Small errors occur with high frequency,
ii) Large errors occur with low frequency and
iii) Positive and negative errors occur with equal frequency ,so that the arithmetic mean is the most
probable value.
14. 09
These rules can be verified graphically by plotting the frequency of errors as a function of the is
magnitudes. Such a plot’s known as normal error curve.
Fig1: a) A schematic representation of a normal error curve and b) Characteristics of normal error curve
The characteristics of the Normal Error Curve given in Fig 1.1(b) can be summarized as follows:
• The highest point of the normal error curve occurs at x=µ, the mean value.
• The curve is symmetric about the mean,µ.
• The total area under the curve equals unity.
• One half of the curve is a mirror image of the other half, i.e ., the area under the curve to the right to
fµ is equal to the area under the curve to the left of µ equals one half (½).
• The curve has inflection points at µ-σ and µ+σ where б is the standard deviation
• The curve becomes asymptotic to the horizontal axis at the extremes.
You would be plotting one such normal error curve after acquiring a set of data.
REQUIREMENTS:
Volumetric flasks (100 cm3) 1
Burette (50cm3) 1
Pipette (10 cm3) 1
Pipette(1 cm3) 1
Weighing bottle 1
Burette stand with clamp 1
Chemicals:
Double Distilled water
REQUIREMENTS:
15. 10
1.1 PROCEDURE
Having learnt about the general principle of calibration of volumetric apparatus let us now perform the
calibration of these apparatus.
1.1.1 Calibration of Burette
A burette is designed to transfer definite but variable volumes of a liquid into another container. It is a
long glass tube, commonly of 50.0 cm3 capacity with a scale etched on it in 0.1 cm3 unit graduation
marks and has a stop cock at the lower end to control the amount of solution drained. The scale can be
used to precisely measure the volume of the solution delivered.
Follow the steps given below to calibrate the burette:
1. Take a clean and dry burette of 50 cm3 capacity and rinse it thoroughly with distilled water.
2. Fill the burette upto the 0.0 mark and mount it upright on the burette stand; ensure that the flow
of water is smooth and there are no air bubbles in the burette.
3. Wipe off the drop of water, if hanging, from the tip of the nozzle and wait for sometime to see that
the burette does not leak.
4. Take a clean and dry conical flask of 100 cm3 capacity and accurately weigh it on an electronic
balance to the nearest milligram. Record the mass of conical flask in the Observation Table 1.1.
5. Deliver exactly 5 cm 3 of water in it from the burette; touch the tip of the nozzle of burette with
the wall of the flask so as to transfer the adhering drop into the flask.
6, Accurately weigh the flask containing water; record the reading in Observation Table 1.1.
7. Repeat step 5 successively delivering 5cm 3 of water each time till the burette is emptied to 50.0
mark and record the readings in Table 1.1.
16. 11
S.
No.
Vol. of
water
delivere
d (in
cm3)
(I)
Mass of
the flask
+ water
(in g)
(ii)
Mass of
empty
flask (in
g)
(iii)
Mass of
water
(in g)
(iv)
Density*of
water
attempt. of
Measurem
ent
(v)
Calibrated
volume
(incm3)(vi)
Error
(I)-(vi)
1 5
2 10
3 15
4 20
5 25
6 30
7 35
8 40
9 45
10 50
1.1.1 Calibration of Pipette
A pipette is an accurate piece of glass apparatus used for measuring volume. It is of two types, a fixed
volume pipette or transfer pipette and a graduated or measuring pipette. A fixed volume pipette is
used to measure and transfer a fixed volume of liquid. It is a long tube with a bulb in the middle. A
horizontal line is etched on the narrow upper part of the pipette. This mark indicates the level to which
the pipette has to be filled to deliver the liquid equal to the volume indicated on the bulb when used in
the proper way.
Follow the steps given below to calibrate the pipette:
1. Take a clean and dry fixed volume pipette of 10cm3 capacity and rinse it thoroughly with distilled
water. For this, take water in a clean dry beaker and dip the pipette into it with the tip nearly touching
the bottom. Suck a little of the water in the pipette. Immediately close the upper end of the pipette with
your index finger (the one next to the thumb). Tilt the pipette and rotate it so as to rinse the inside of
the pipette.
2. Take a clean and dry conical flask of 50cm3 capacity (or you may use a weighing bottle) and
accurately weigh it on an electronic balance (if possible) to the nearest milligram. Record the mass of
conical flask (or weighing bottle) in the Observation Table 1.2.
17. 12
3. Fill the pipette up to the etched mark. For it suck the water in the pipette, till a little above the
etched mark and close the upper end of the pipette with moist index finger. Release the water back in
to the beaker till its level (as ascertained by the lower meniscus) reaches the etched mark.
4. Insert the tip of the pipette in to the weighed conical flask (or weighing bottle and remove the
finger to discharge the solution. Do not blow or force the solution out, let it flow on its own. Also do not
blow out the last drop in the pipette. Touch the tip of pipette o the inner wall of the container, apart of
the solution passes into the container. Neglect the remaining liquids till at the tip of the pipette.
5. Accurately weigh the flask or weighing bottle econ taining water; record the reading in
Observation Table 1.2.
6. Repeat step 3-5 at least five times and record your observation sin the Observation Table 1.2.
Observation Table1.2: Calibration of a pipette.
S.
No.
Vol. of
water
delivered
Mass of
flask +
water
Mass of
empty flask
Mass of
water
delivered
Density*of water
at temp. of
Calibrated
volume
(in cm3
)
Error
(in cm3
)
(I)
(in g)
(ii)
(in g)
(iii)
(in g)
(iv)
measurement
(v) (vi) (I)-(vi)
1 10
2 10
3 10
4 10
5 10
6 10
Volumetric analysis
1. Solution -Homogeneous mixture of solution water or any solution is called solution.
Normally water is used as solvent in most of the reactions.
2. Titration – The quantitative analysis is performed with solutions of the substances.
The quality of the substance is determined in specific volume. It is done by measuring the volumes of
reagents of known as well as unknown concentration. Therefore it is called the volumetric analysis or
trimetric analysis and the process is called titration.
3. Standard solutions – A solution of known concentration is called standard solutions.
Standard solutions are of two types
18. 13
Primary standard substances :- These are the chemicals which can be weighed correctly and the
solution of known strength Leo standard solutions may be prepared directly. The chemical composition
of such substances is constant under working conditions e.g. oxalic acid, alcohol solution, anhydrous
solution carbonate, K2 Cr2O7 etc.
Secondary standard substances:- The substances which do not full fill the requirement of primary
standard substances but are used in titrations are called secondary standard substances
There Chemical composition may undergo change during working period or in solutions. These may not
available in pure form or may absorb moisture or loose water of crystallization in atmosphere or react
with O2 or CO2 of atmosphere e.g. HCl, NaOH, Na2CO3.10 H2O KMnO4 can only be used as secondary
standard. To make standard solutions of such chemicals a solution of little higher concentration is
prepared. It is higher concentration is prepared. It is titrated with standard solutions of proper primary
standard chemical and exact concentration calculated. This is called standardization.
(1) Concentration of solution
The concentration or strength of a solution is written in several ways.
1. Grams per liter (gm/l) - Mass in grams of a substance (solute) present in one liter of a
solution.
Q. 1 calculate the strength of solution prepared by dissolving 0.45 g in 100 ml
(a) In g/liter
Strength = 0.45 / 100 × 1000 =4. 5 g/ liter
(b) by dissolving 0.45g in 250 ml.
Strength = 0.45 /250 × 1000 =1.8g/liter
(2) Percent (%)
(a) By weight W/w: - Mass of substance present in 100 g of solution.
(b) By volume V/v: - when solute is liquid V/v representative is desirable. It is volume of solute in 100
ml of solution.0.20ml alcohol present in 100 my solution is 20% V/v
Q. 2 what is 10% NaOH solutions 10 % of alcohol.
Ans-10% g of NaOH in 100 ml alcohol in 100 ml solutions.
(3) Normality – It is the number of gram equivalents of solution present in one liter of solution. It is
denoted by litter N e.g. 1 N solutions of oxalic acid (eq. wt 63) contains 1 × 63=63 g of the acid in liter of
the solution.
Q.3 How will you prepare KMnO4 solutions where eq. wt is 158.03 of strength.
(a) 1 N = by dissolving 158.03 g in 1 liter solutions.
(b) .1 N or N/10 = 15.8 g in one liter solutions.
(c) 2N = 2 × 158.03 =316.06 g per liter.
(d) 5N =5 × 158.03 =790.15 g liter
19. 14
Q. 4 For making 1 N solutions of oxalic acid 63g oxalic acid is dissolved in 1 liter solutions. How
you will prepare –
(a) N/10 oxalic acid – 63/10 =6.3g in 1 liter .
(b) N/10 in 250 ml – 63/10 × 250/1000 = 1.57g per 250ml.
(4) Molarity: - It is the number of gram moles present in one liter of solution. It is denoted by M, .e.g. 20
gm NaOH in one liter of a solution is 0.5 M
Relation between molarity and normality
N = nM
Where n is number of H+ ions given or accepted by the acid or base.
(5) Molality: - The number of gram moles of solute in 1000 gm (1kg) of solvent is called morality .
Calculations of titration
Calculations of titrations are done or the bases of equivalent mass let the normality of acid solution
(x) be N1 it’s volume required V2 ml of alkyls solutions (y) of normality N2 then.
1000ml of solution x contains N1 gm eq. acid
: V1 ml shall contain = N1 × v1/1000 g. eq. acid
Similarity =N2 V2/1000 g eq. of alkali is used for neutralization.
On completion of the reaction the gm equivalents of the reactants shall be same or equal.
I.e. N1 V1/1000=N2 V2/1000
I.e. N1V1 =N2 V2
20. 15
Experiment No. 03
Determine the Order of reaction for Acid catalyzed Hydrolysis of Methyl
Acetate
Aim:
To find out the order of reaction in case of acid hydrolysis of Methyl acetate.
Materials required:
1. 100ml 0.1N HCl
2. 0.1N NaOH
3. 10ml Methyl acetate
4. Burette (50ml), Pipette (10ml)
5. Standard flask, Beaker, Conical flask
6. Phenolphthalein indicator & Stop watch
Procedure:
1. Prepare 0.1N NaOH solution and standardize it.
2. Prepare 0.1N HCl solution.
3. Fill the burette up to zero with NaOH solution.
4. Mix 100ml 0.1N HCl and 10ml Methyl acetate in a beaker.
5. Pipette out 10ml mixture and transfer it in conical flask.
6. Add 5ml cold water (to slow down the reaction) to it and 1-2 drops of phenolphthalein to it and
titrate it against NaOH solution.
7. Note down the volume of NaOH required to neutralize 10ml of mixture name this reading as V0.
8. Take five more readings V1,V2,V3,V4 and V5 at intervals of 10 minutes.
9. After V5 heat the remaining mixture for some time, then cool it and again pipette out 10ml and
note the last reading V∞ .
10. Find out the rate constant for each step and compare.
21. 16
Reaction involved:
CH3COOCH3 + H2O → CH3COOH + CH3OH
Formula used :
Hydrolysis of methyl acetate is a first order reaction so rate will be given by-
K = 2.303/t log10 (a/a-x)
Where –
a = (V∞-V0)
(a-x) = (V∞-Vt)
V∞= Volume of HCl and CH3COOH formed at the end of reaction in terms of volume of NaOH.
V0= Volume of HCl only in term of NaOH.
Vt = Volume of HCl and CH3COOH for used at only time t of the reaction in term of volume of NaOH.
K = 2.303/t log10 (V∞-V0/V∞-Vt)
Theory:
The rate of reaction Methyl acetate hydrolysis in presence of on acid which acts as a catalyst to give
acetic acid, methyl alcohol is given by.
dx/dt= K (CH3COOCH3) (H2O)
Since water is present in large excess, so
dx/dt = K (CH3COOCH3)
Hence rate of reaction is determined by the first power of the concentration of the order ester, so the
reaction is of first order and pseudo uni-molecular reaction.
22. 17
Observation Table:
Temperature of reaction - 25◦C (Room temperature)
S.No. Time in
minutes
Volume of NaOH
in ml
V∞-V0 V∞- Vt K= 2.303/t log V∞-V0/V∞-Vt
in min-1
01 0 min 3.5 4.7 4.7 -----
02 10 min 4.5 4.7 3.7 0.02390
03 20 min 5.1 4.7 3.1 0.04159
04 30 min 5.5 4.7 2.7 0.0554
05 40 min 6.0 4.7 2.2 0.0759
06 50min 6.7 4.7 1.5 0.1142
07 60 min 7.7 4.7 0.5 0.2241
08 ∞ 8.2 4.7 ---- -----
Observation:
The amount acetic acid formed at the end of the reaction is equivalent to the initial volume of methyl
acetate. The values of velocity constant are calculated at different intervals of time t. we see the values
are nearly constant.
Result:
Rate constant or velocity constant for the studied reaction = 0.089191 min-1
Precautions:
1. The zero time should only be noted when ice is added is the first 10ml of mixture.
2. Distilled water used for preparing the solution must be free from CO2 as the end points of the
titration are very sharp when no CO2is present. Similarly for save reason NaOH should also be free
from CO2.
23. 18
Graph: Log (V∞-V0/V∞-Vt) versus to time
60, 0.2241
0
0.05
0.1
0.15
0.2
0.25
0 10 20 30 40 50 60 70
Log(V∞-V0/V∞-Vt)
Time in min.
Log (a/a-x)
Log (a/a-x)
24. 19
Experiment –4
Determine the Activation Energy and Temperature Coefficient at two given
Temperatures for Acid catalyzed Hydrolysis of Methyl Acetate
Aim: - To find out the order of reaction in case of acid hydrolysis of Methyl acetate at 300 C and
determine the energy of activation and temperature coefficient.
Material required –
1.100ml N/2 HCl and N/5 NaOH .
2.10ml Methyl acetate.
3.Burette and pipette (10ml)
4. Titration flask ,beakers ,conical flasks .
5. Phenolphthalein indicator and stop watch.
Theory : - The rate of reaction of Methyl acetate hydrolysis in presence of an acid which acts as a
catalyst to give acetic acid and Methyl acetate alcohol is given by
dx/dt = k (CH3COOCH3) (H2O)
Since water is present in large excess ,so
dx/dt=k (CH3COOCH3)
Hence rate of reaction is determined by the first power of the concentration of the ester and so the
reaction is of first order and pseudo uni-molecular reaction.
25. 20
Observation Table- 1
Temperature of reaction-300C
S.No. Time of
Minutes
Volume of
NaOH in ml
V∞-Vo
(72-7)
in ml
V∞-Vt
in ml
K=2.303/t log
V∞-V0/ V∞-Vt
In ml
1 0 7 65 65 log(V∞-Vt)
2 10 16 65 56 1.7481 2.3×10‐³ 0.2388
3 20 26 65 46 1.6627 2.5×10-³ 0.2511
4 30 36.5 65 35.5 1.5502 2.6×10‐³ 0.2693
5 40 47.5 65 24.5 1.3891 3.0×10‐³ 0.3005
6 50 59 65 13 1.1139 3.7×10‐³ 0.3798
7 ∞ 72 65 0
K 300C= 2.303/t×1.4345/5
K 300C = 0.2869
K 300C = 2.86×10-³ min-¹
26. 21
Observation Table- 2
Temperature of reaction= 400C
S.No. Time of
Minutes
Volume of
NaOH
V∞-Vt
(15-134)
in ml
V∞-Vt
in ml
log(V∞-Vt) K=2.303/t log
V∞-V0/ V∞-Vt
In ml
1 0 15 119
2 10 32 102 2.008 2.3×10-
³ 0.2379
3 20 50 84 1.923 2.4×10-
³ 0.2483
4 30 69 65 1.8129 2.6×10-
³ 0.2637
5 40 89 45 1.6532 2.8×10-
³ 0.2890
6 50 110 24 1.3802 3.4×10-
³ 0.3460
7 ∞ 134 0
K 400C = 0.27695
K 400C = 27.6×10-³ min-¹
Calculation:-
1. Temperature Coefficient: - Increase in temperature almost in all cases increases the velocity
of a reaction to a marked extent . It has been seen the general the specific rate is approximately
doubled or tripled for an increase in temperature by 10⁰C .
K40/K30 = 0.27695/0.2869
= 0.9653789
2. Activation Energy : - The measurement of specific rates of a reaction at two temperature
enables E to be calculated.
If on the other hand, the values of specific rates of a series of temperature are known as energy of
activation E, can be plotting log K values:
Formula =logK2/K1= -E/2.303R [1/T2 -1/T1]
27. 22
Log K2/K1= E/19.147 [T2 - T1/T1T2]
Where K2 = Rate constant at t = 400C
K1= Rate constant at t = 300C
log 0.27695/0.2869 = E/19.147 [313-303/313×303]
-0.0153 = E/19.147 [10/94839]
-0.0153 = E×10/19.147×94839
-0.0153 = E×10/1815882.333
E = -0.0153×1815882.333/10
E = - 27782.99919/10
E = - 0.27×10⁴
Procedure : -
1 Prepare N/5 NaOH solution and N/2 HCl Solution and standardize it it.
2 Fill the burette up to Zero mark with NaOH solution.
3 Keep HCl solution and Methyl acetate on water bath to heat up to 400 c
4 Mix 100ml HCl solution and 10 ml Methyl acetate in conical flask. Maintain the temperature of the
mixture at 400c .
5 Pipette out 10 ml mixture and transfer it in titration flask
6 Add 5 ml cold water (to arrest the reaction) to it and 1-2 drop phenolphthalein to it and titrate it
against NaOH solution
7 Note done the volume of NaOH required to neutralise 10 ml of mixture. NaOH the reading as Vo.
8 Take five more reading V1, V2, V3, V4, V5 at intervals of 10 minutes
9 After V5 heat the remaining mixture for some time, then cool it and again pipette out 10 ml and
note the last reading V∞
10 Find out the rate constant for each step and compare .
11 Find the temperature coefficient and activation energy for values of rate constants at 30O c and
40Oc by substituting values in formula.
28. 23
Observation:-
The amount of acetic acid formed at the and of the reaction is equivalent to the initial volume of
Methyl acetate. The values of velocity constant are calculated at different intervals of time (t) and
values are nearly constant.
Result –
1 - The Rate constant for studied reaction at 40oc is27.6 ×10-³ minute -1
2- The temperature coefficient is 0.9653
3- The activation energy is-0.27×10⁴ Calories.
Precaution –
1- The zero time should only be noted when ice is added to first 10 ml of mixture.
2- Distilled water used for preparing the solution should be free form CO2 as the end points of
titration are very sharp when no CO2 is present.
40˚C Temperature
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0 10 20 30 40 50 60
Log(a/a-x)
Times in min.
Log (a/a-x)
30. 25
Experiment - 05
Determine the Congruent Composition and Temperature of Binary Mixture
Diphenylamine Benzophenone mixture
Aim: -
To determine the congruent composition and Temperature of Binary Mixture (Diphenylamine
Benzophenone mixture).
Material Required:-
11 Test Tubes with Cork, Ice, Thermometer, Beaker, Stirrer, Boiling Tubes.
Theory:-
Solid and liquid can exist in equilibrium at the freezing point. When a homogeneous binary liquid
mixture is cooled. Eventually a temperature (freezing point) is reached when the solid phase separates
out. The composition of the solid phase depends upon the nature of the two components. In a system
like Diphenylamine and Benzophenone, a component of the two components separates out when the
compound has the same composition as that of liquid phase. No change in the composition of the liquid
occurs during freezing and the freezing point remains constant. Such a compound which on melting
gives the liquid of its own composition is called a congruently melting compound.
Procedure:-
1) Prepare the following mixtures of the components by taking weighed amount in properly labelled test
tubes.
Diphenylamine - 10 9 8 7 6 5 4 3 2 1 0
Benzophenone- 0 1 2 3 4 5 6 7 8 9 10
2) Using a cork fit up a thermometer and a ring stirrer in one of the test tubes. Place the test tube in hot
water contained in a beaker and melt the contents completely.
3) Remove the test tube from the bath, wipeit clean and place it in the large boiling tube as an air jacket.
Clamp the boiling tube in a beaker containing crushed ice to ensure uniform cooling.
4) Allow the temperature to fall, stir thoroughly and record the temperature every half minute.
5) Note the temperature at which crystals begin to separate from the molten mass.
31. 26
6) Continue to record the temperature until it becomes constant and the whole mass solidifies.
7) Similarly, repeat the determination with other mixtures. Determine the maximum temperature of first
arrest and the eutectic temperature.
8) Then plot the freezing point diagram.
Observation Table:-
Test
Tubes
1 2 3 4 5 6 7 8 9 10 11
Diphenyl
Amine(g)
10 9 8 7 6 5 4 3 2 1 0
Benzophenone
(g)
0 1 2 3 4 5 6 7 8 9 10
Time
30 sec
60 sec
90 sec
120 sec
150 sec
180 sec
210 sec
240 sec
49
44
44
35
34
31
28
48
42
40
38
36
36
34
50
48
47
43
40
38
38
35
45
44
35
40
40
40
40
35
45
44
44
35
40
22
22
21
26
26
25
26
26
26
26
21
23
25
25
25
25
24
23
14
16
21
22
22
27
25
27
25
25
20
29
28
27
43
40
40
39
35
29
23
40
48
48
45
Observation:-On plotting the freezing point diagram the maximum of the curve CDE is obtained at
(6:4) composition of Diphenylamine and Benzophenone.
Result:- The phase diagram of two component system of Diphenylamine and Benzophenone is as
shown in diagram .The composition of the given compound formed is 6:4 and formula will be A3B2
where
