chem 311 lab manual - uoh.edu.sa€¦ · 13-instrumentation can only be used after instructor...

College of Engineering Dept. of Chemical

Engineering

قسم الھندسة الكیمیائیة جامعة حائل

KINGDOM OF SAUDI ARABIA

Ministry of Higher Education University of Hail

Dr. F. Khlissa

Department of

Chemical Engineering

Laboratory Manual

Physical Chemistry II – Chem 311 2017/2018

Dr. F. KHLISSA


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Dr. F. Khlissa

Laboratory safety policy and rules

1- Approved eye protection (safety glasses/goggles) must be worn at all times

in the laboratory. Even if you have completed you lab work, eye protection

must be remain on in the lab. 2- No eating, drinking any liquids (including water), chewing gum or smoking in

the lab.

3- Skin is less susceptible to chemical exposure when properly covered. As

such, the use of full-legged pants and a laboratory coat is necessary in the

laboratory when working with chemicals.

4- Shoes Sandals, flip-flops, and other open-toed shoes expose the feet to

chemicals as well as sharp objects and are not permitted.

5- Long hair must be kept tightly in place. Hair and loose clothes can catch fire easily.

6- Do not enter the lab until the instructor or the supervisor is present.

7- Any accident or injury must be reported to the supervisor at once. 8- Chemicals are generally to kept in designated access area. Caps are to be

replaced promptly on any reagents used in lab. Use care when transferring

or dispersing chemicals. Any spills must be cleaned up promptly and completely.

9- If any material is spilled on the skin, wash it off immediately with a large

volume of water. Notify your instructor.

10- Reading labels and using the exact chemical in its proper concentration is in

the responsibility of the student. For some reagents, only the instructor is allowed to disperse them (these will be announced in the lab introduction).

11- Do not use cracked or chipped glassware. Dispose of it in a proper manner as

indicated by your instructor. 12- Proper clean up and maintaining a well-ordered work space is essential to lab

safety. Bench tops and the weighing areas are to be wiped clean. Clean all

shared equipment. Improper cleaning can damage expensive equipment and render it useless.

13- Instrumentation can only be used after instructor approval or according to any instructions given in the lab instruction.

14- Proper disposal of waste chemical is the student's responsibility. If there

is a question, ask the supervisor. 15- Read the entire experiment and complete any pre-assignments before

entering the laboratory.

16- Inform the instructor immediately of any broken thermometers. If a mercury thermometer is broken, any spills must be cleaned up by the

instructor.


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Dr. F. Khlissa

17- Any special health factors such as an allergic reaction to a chemical must be

reported to the instructor. As soon as possible.

18- Always use the common sense in the laboratory. If something is unclear, be sure to ask the instructor before proceeding.


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Dr. F. Khlissa

Good Lab Practices

Proactively make your work environment a safer place.

Housekeeping

• Keep your work area well-organized and clean. • Avoid storing empty cardboard boxes in the lab.

• Keep needles capped or stuck into a silicone rubber stopper to prevent

accidental pricks.

Food and Drink

Food and drink should not be present in the laboratory.

Solvent and Reagent Bottles

• Do not place chemical or solvent bottles on the floor where they can be kicked over.

• Transport solvent and acid bottles in rubber carriers.

• All 4L bottles of solvents or acids must be plastic or plastic-coated glass. • Store bulk solvent containers (4L and above) in flammable material cabinets.

Label Everything

• Ensure that you clearly label all wash bottles, base baths, solvent bottles, and

chemicals. • Label research samples with chemical names and/or chemical structures and/or

the relevant laboratory notebook number and page number.

Safety Equipment

• Know the location and operation of eye wash stations, safety showers, fire

extinguishers, and emergency exits. • Maintain unfettered access to safety equipment.

Running a Reaction in Lab

1) Plan ahead of time and read procedures in their entirety.

2) Understand the hazards of the chemicals and procedures you are about to employ.

3) Store and transport the chemicals properly.

3) Choose the appropriate equipment and reaction vessel sizes. 4) Use the appropriate Personal Protective Equipment (PPE).

5) Dispose of hazardous waste properly.


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Content

Laboratory safety policy and rules……………………………………………………………………………….2

Good Laboratory Practices…………………….………………………………………………………….………….4

Experiment 1: Boyle and Mariotte’s law……………………………………………………………………….6

Experiment 2: Determination of the enthalpy of neutralization………………………….12

Experiment 3: Determination of the mixing enthalpy of binary fluid mixtures…17

Experiment 4: Determination of the hydration enthalpy of an electrolyte……….22

Experiment 5: Determination of the enthalpy of vaporization of liquids……………27

Experiment 6: Viscosity of gases Estimation of molecular diameter………………….33

Experiment 7: Mutual Solubility Curve for Phenol and Water……………………………..38

Experiment 8: Chemical Equilibrium and Le Chatelier's Principle………………………..41

Experiment 9: Determination of the surface tension by capillary rise method.43

Experiment 10: Ternary Phase Diagram: Water-Toluene-Ethanol…………………..46


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Dr. F. Khlissa

Experiment 1: Boyle and Mariotte’s law

Principle

The state of a gas is determined by temperature, pressure and amount of

substance. For the limiting case of ideal gases, these state variables are linked via

the ideal gas law. In the case of isothermal process control this equation converts

to Boyle and Mariotte’s law.

Tasks

Experimentally investigate the validity of Boyle and Mariotte’s law for a constant

amount of gas (air). From the resulting relationship calculate the universal gas

constant.

Set-up and procedure

Set up the experiment as shown in Fig. 1. Install the gas syringe in the glass jacket

as described in the operating instructions supplied with the glass jacket.

Figure 1: Experimental Set-up

Pay particular attention to the air-tightness of the gas syringe. As an exception

here, because no air is to be allowed to leak out even at higher pressures,


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Dr. F. Khlissa

lubricate the plunger with a few drops of multigrade motor oil, so that the glass

plunger is covered with an uninterrupted clear film of oil throughout the entire

experiment; but avoid excess oil. Fill the glass jacket with water via the funnel.

The water avoids fluctuations of the temperature connected with compression

and expansion because of its high heat capacity. Insert the thermocouple and

place it as close to the syringe as possible. After adjusting the initial volume of

the gas syringe to exactly 50 ml, connect the nozzle of the gas syringe to a

reducing adaptor via a short piece of silicone tubing, whereby the reducing

adaptor should directly abut on the glass tubular sleeve after the tubing has been

slipped over it. Secure the tubing on both the gas syringe’s nozzle and on the

reducing adaptor with hose clips. Connect the reducing adaptor to the measuring

module with a short piece of silicone tubing (di = 2 mm). Keep the tubing

connections as short as possible.

Connect the measuring module to the Cobra3 Basic-Unit using a module converter

and a data cable. Call up the “Measure” programme in Windows and enter <Ideal

gas law> as gauge. Set the measuring parameters as shown in Fig. 2. Under <Start /

Stop> choose <Get value on key press>. Under <Other Settings> select Digital

Display 1 for <Pressure p>, Digital Display 2 for <Temperature> and Diagram 1 with

Channel <Pressure p>, X bounds <from 1 to 20> and Mode <no auto range>.

Now calibrate your sensors under <Calibration> by entering temperature and

pressure values measured with a thermometer respectively manometer and

pressing <Calibrate>. After having made these settings, press <Continue> to reach

the field for the recording of measured values. Arrange the displays as you want

them. Subsequently expand the enclosed quantity of air in 1 ml steps to a volume

of approximately 70 ml. Record the volume for each step by pressing <Save value>.

Terminate measurement by pressing <Close>. Save the measurement with <File>


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Dr. F. Khlissa

Figure 2: Measurement Parameters

<Save the measurement as…>. To have the plot of Pressure versus Volume carried

out, make the following alterations. Under the menu prompt <Measurement>

<Channel manager>, choose Volume for the x-axis and Pressure for the y-axis. Fig.

3 shows the graph as it is then presented by the programme. If you choose 1/V as

x-axis, you get the graph as it is shown in Fig. 4. Under menu prompt Analysis you

can let the programme show the slope.

Theory and evaluation

The state of a gas is a function of the state variables temperature T, pressure p

and the amount of substance n, which reciprocally determine one another. Thus,

the dependence of pressure on the temperature, volume and amount of substance

variables is described by the total differential

�� ,� �� ,� ��

��, �� (1.1)


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Dr. F. Khlissa

Analogously, the following is true for the change of pressure with T, V and n:

�� ,� �� ,� ��

��,� �� (1.2)

This relationship simplifies for a given amount of substance (n = const., dn = 0;

enclosed quantity of gas in the gas syringe) and isothermal change of state (T =

const., dT = 0) to

�� ,� �� (2.1)

and

�� ,� �� (2.2)

The partial differential quotient (∂V/∂p)T,n resp. (∂p/∂V)T,n corresponds

geometrically to the slope of a tangent to the function V = f(p) or p = f(V) and

therefore characterizes the mutual dependence of pressure and volume. The

degree of this dependence is determined by the initial volume or the initial

pressure. One thus defines the cubic compressibility coefficient x0 by referring

it to V or V0 at T0 = 273.15 K.

��

��,� (3)

For the limiting case of an ideal gas (sufficiently low pressures, sufficiently high

temperatures), the correspondence between the state variables p, V, T and n is

described by the ideal gas law:

PV = nRT (4)

R: Universal gas constant

For cases of constant quantity of substances and isothermal process control this

equation changes into the following equations:

PV = const (4.1)

and

P = const . 1/V (4.2)


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Dr. F. Khlissa

According to this correlation, which was determined empirically by Boyle and

Mariotte, a pressure increase is accompanied by a volume decrease and vice versa.

The graphic representation of the functions V = f(p) or p = f(V) results in

hyperbolas (Fig. 3). In contrast, plotting the pressure p against the reciprocal

volume 1/V results in straight lines where p = 0 at 1/V = 0 (Fig. 4). From the slope

of these linear relationships,

� ��,� � �� (5)

it is possible to determine the gas constant R experimentally when the enclosed

constant quantity of air n is known. This is equal to the quotient of the volume V

and the molar volume Vm,

� � �� (6)

which is V0 = 22.414 l · mol-1 at T0 = 273.15 K and p0 = 1013.25 hPa at standard

conditions. A volume measured at p and T is therefore first reduced to these

conditions using the relationship obtained from (4):

��

� (7)

Data and results

Figs. 3 and 4 confirm the validity of Boyle and Mariotte’s law. From the slope

obtained for n = 2.086 mmol and T = 295.15 K, (∂p/∂V-1)T,n = 4.6464 kPa/m-3 =

4.6464 Nm of the linearized correlation between p and 1/V (Fig. 4), the universal

gas constant can be calculated to be R = 7.547 Nm · K-1 · mol-1. The deviation from

the literature value (R = 8.31441 Nm · K-1 · mol-1 = 8.31441 J · K-1 · mol-1) is due to

the unavoidable lack of gas-tightness with increasing deviation from atmospheric

pressure through compression or expansion, whereby the condition dn = 0 is

violated and the observed slope (∂p/∂V-1)T is diminished in comparison with the

value measurable with a constant quantity of substance.


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Dr. F. Khlissa

Figure 3: Correlation between the

volume V and the pressure p at

constant temperatures (T = 295.15 K) and constant quantity of substance (n

= 2.086 mmol)

Figure 4: Pressure p as a function of

the reciprocal volume 1/V at constant temperature (T = 295.15 K)

and constant quantity

of substance (n = 2.086 mmol)


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Dr. F. Khlissa

Experiment 2: Determination of the enthalpy of neutralization

Principle

When a strong acid is neutralized with a strong base in dilute solution, the same

amount of heat is always released. If the reaction takes place under isobaric

conditions, this heat is known as the enthalpy of neutralization. The chemical

reaction which generates this heat is the reaction of protons and hydroxyl ions to

form undissociated water. It therefore correlates to the enthalpy of formation

of water from these ions.

Tasks

1. Measure the temperature change during the neutralization of a dilute potassium

hydroxide solution with dilute hydrochloric acid.

2. Calculate the enthalpy of neutralization.

Figure 1. Experimental set-up.


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Dr. F. Khlissa


Set up the experiment as shown in Fig. 1 but for the time being do not connect the

heating coil to the work and power meter. Connect the temperature probe to T1 of

the measuring module. Call up the Measure programme in Windows and enter

<Temperature> as measuring instrument. Set the measuring parameters as shown

in Fig. 2. Under <Diagram 1> select Temperature T0a, the appropriate range for the

temperature and the X bounds and ‘auto range‘. Now calibrate your sensor under

<Calibrate> by entering a temperature value measured with a thermometer and



them. Pour approximately 750 g water and 60 g of the 2 M potassium hydroxide

solution (both weighed to an accuracy of 0.1 g) into the calorimeter. Using a

delivery pipette and a pipettor, draw around 50 ml of the 2 M hydrochloric acid

from a small glass beaker.

Figure 2: Measurement parameters


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Dr. F. Khlissa

The exact mass of the hydrochloric acid contained in the delivery pipette is

calculated from the difference between the masses of the filled and the empty

delivery pipette (accuracy 0.1 g). The 600 ml beaker is used as a pipette stand.

Place the filled calorimeter on the magnetic stirrer, put in the oval magnetic

stirrer bar and switch on the stirrer (Caution: Do not switch on the heating unit by

mistake!). Push the delivery pipette through the cap of the calorimeter from

below and mount the lid on the calorimeter vessel. Now attach the pipette to the

support rod using a clamp in such a manner that the opening is above the level of

the liquid and that the stirrer bar can rotate unhindered. Insert the heating coil

and the temperature probe into the lid of the calorimeter and fix them in position.

When the temperature equilibrium has been reached (after approximately 10 min)

start the measurement by pushing <Start measurement>. Wait 3 to 4 minutes,

then blow the hydrochloric acid out of the delivery pipette into the potassium

hydroxide solution in the calorimeter. To do this, first clamp a pinchcock onto the

tube of the rubber bulb, blow up the air reservoir of the rubber bulb and quickly

release the pinchcock. Continue to measure until a new equilibrium has been

reached. Subsequently perform electrical calibration to determine the total heat

capacity of the calorimeter. Supply 10 V AC to the work and power meter for the

electric heating. Push the <Reset> button and then put the free ends of the

heating coil connection cables into the output jacks. The system is now

continuously heated and the supplied quantity of energy is measured. The

temperature increase in the system should be approximately 2 K. When this value

has been reached, switch off the heating and read the exact quantity of electrical

energy supplied. After a further three minutes, stop the recording of

temperature. Fig. 3 shows the graph as it is presented by the programme when the

measurement is stopped. If you use <survey> from the toolbar you can read the

temperature difference data.


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Dr. F. Khlissa

Figure 3: Temperature-time curve of neutralization and determining

the heat capacity of the system

- and evaluation

The value of the enthalpy of neutralization ∆RH for the reaction of strong acids

with strong bases is independent of which strong acid or base is used, because the

heat of reaction is generated by the reaction of hydrogen and hydroxyl ions to

form water: H+ + OH- --> H2O ∆RH = -57.3 kJ · mol-1

In the case of the neutralization of weak acids and bases, additional heat effects

arise from dissociation, hydration and association of molecules, so that the value

of the enthalpy of neutralization will differ to that given above. The heat capacity

of the system must be determined in order to be able to determine the system

change in enthalpy ∆H. This is undertaken, after completion of the neutralization

reaction, by introducing a specific amount of heat into the filled calorimeter using

electrical heating. The electrical energy Wel = I · U · t which is converted into heat

Q causes an increase in temperature ∆Tcal. From this the heat capacity of the

system Csys can be calculated using equation (1).

Q = I · U · t = Csys . ∆Tcal = Wel (1)


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Dr. F. Khlissa

Using the heat capacity of the system, the enthalpy of neutralization ∆RH can be

calculated from the temperature increase ∆T of the neutralization reaction for a

known amount n of converted hydrochloric acid.

∆�� ∆�� !"#$

�"#$%"#$� &'('∆� (2)

n Amount of hydrochloric acid introduced

cHCl Concentration of hydrochloric acid (= 2 mol/l)

mHCl Mass of hydrochloric acid introduced

ρHCl Density of hydrochloric acid (= 1.0344 g/ml for 2 M HCl at 20°C)

∆�� Enthalpy of neutralization

Csys Heat capacity of system

For reasons of simplification it is assumed that the heat capacity of the dilute

salt solution differs only negligibly from that of water.

Data and results

Enthalpy of neutralization:

∆�� = -57.3 kJ · mol-1


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Dr. F. Khlissa

Experiment 3: Determination of the mixing enthalpy

of binary fluid mixtures

Principle

When two miscible liquids are mixed, a positive or negative heat effect occurs,

which is caused by the interactions between the molecules. This heat effect is

dependent on the mixing ratio. The integral mixing enthalpy and the differential

molar mixing enthalpy can be determined by calorimetric measurements of the

heat of reaction.

Tasks

1. Measure the integral mixing enthalpy of 7 different water-acetone mixtures.

2. Plot the molar integral mixing enthalpy versus the quantity of substance (mole

fraction) and determine the molar mixing enthalpy.

3. Discuss the results on the basis of the interactions in the mixture.

Figure 1: Experimental set-up.


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Dr. F. Khlissa


Set up the experiment as shown in Figure 1 but for the time being do not connect

the heating coil with the work and power meter.

Weigh out the individual components of these mixtures with an accuracy of 0.1 g in

accordance with the values given in Table 1.

Table 1: Preparation of the seven test mixtures

Substance

quantity Calorimeter

Erlenmeyer

flask

Mixture number

X = 0.1 432 g Water 154 g Acetone 1

X = 0.2 Mixture 1 194 g Acetone 2

X = 0.9 464 g Acetone 16 g Water 7

X = 0.9 Mixture 7 20 g Water 6




Connect one of the temperature probes to T1 of the measuring module, the other

to T2. Call up the ‘Measure’ program in Windows and enter <Temperature> as

measuring instrument.

Set the measuring parameters as shown in Fig. 2. Under <Digital display 1> choose

Temperature T0a as channel, under <Digital display 2> Temperature T0b. Under

<Diagram 1> select Temperature T0a, the appropriate range for the temperature

and the X bounds and ‘auto range‘. Now calibrate the sensors under <Calibrate>

each by entering the temperature value measured with a thermometer and



them.


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Dr. F. Khlissa

Figure 2: Measurement Parameters.

For the first measurement, fill 432 g water into the calorimeter. Insert the oval

magnetic stirrer bar in the calorimeter and switch the magnetic stirrer on

(Caution: Do not switch on the heating unit by mistake!). Insert the heating coil


Weigh 154 g of acetone in a 250 ml Erlenmeyer flask. Cut a rubber stopper with

hole lengthwise, put the second temperature probe through the hole and close the

Erlenmeyer flask before hanging it into the temperature-controlled bath. Adjust

the immersion thermostat to the temperature of the water in the calorimeter and

wait until the temperature difference between the acetone in the bath and the

calorimeter does not exceed 0.02 K.

Start the measurement with <Start measurement>. Wait a few minutes, then pour

the acetone into the water in the calorimeter. After a new temperature

equilibrium has been reached, perform electrical calibration for the

determination of the total heat capacity of the calorimeter. To do this, supply 10

V AC to the work and power meter for the electric heating. Push the <Reset>


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Dr. F. Khlissa

button and then put the free ends of the heating coil connection cables into the

output jacks. The system is now continuously heated and the supplied quantity of

energy is measured. When the temperature increase in the calorimeter induced by

the electrical heater is approximately equal in size to the temperature change

resulting from mixing the two liquids, switch off the heating and read the exact

quantity of electrical energy supplied.

Continue to measure for another three minutes, then stop temperature recording

by <Stop measurement>. Fig. 3 shows the graph as it is presented by the program

when the measurement is stopped. If you use <survey> from the toolbar you can

read the temperature difference data. In a second experiment, add an additional

portion of acetone (194 g) to the mixture in the calorimeter (see Table 1). Perform

the experiment completely analogously to the first measurement and pay

attention that the temperature of the mixture in the calorimeter and of the

acetone is the same. In a further series of experiments, successively add the 5

portions of water listed in Table 1 to the 464 g of acetone in the calorimeter.

Carry out this series in the same manner as in the first set of measurements,

after carefully cleaning and drying the calorimeter. It is important that the

calorimeter is recalibrated after each addition, as the heat capacity of the

system is different after each temperature change.


The change in enthalpy observed when two liquids are mixed is the sum of the

changes in enthalpy which occur during the mixing process. The mixing enthalpy

∆MH is influenced by the interactions of the molecules involved, which in turn are

a function of the mixing ratio. The mixing enthalpy is zero if there are no

interactions between the molecules (so-called ideal mixtures). The interactions

between two liquids can cause endothermic effects (decreasing supramolecular


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Dr. F. Khlissa

assemblies) or exothermic effects (formation of supramolecular assemblies of

different molecules).

The quantity of heat exchanged by mixing nA moles of the component A with nB

moles of component B is termed the integral mixing enthalpy ∆MHI. If a substance

is successively added to another one until a certain mixing ratio is reached, the

integral mixing enthalpy is obtained by adding the individual enthalpy values:

∆MhI =∑ ∆*+,- (1)

with

∆*+-, = ./0 � .�12 ∆�345∆�#6$ �7/2∆�345∆�#6$ (2)

Figure 3: Temperature-time curve of the mixing enthalpy of mixture 1

Figure 4: Integral molar mixing enthalpy as a function of the molar fraction


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Dr. F. Khlissa

Experiment 4: Determination of the hydration enthalpy of an

electrolyte

Principle When a solid electrolyte dissolves in water, a positive or negative heat effect

occurs as a result of the destruction of the crystal lattice and the formation of

hydrated ions. The enthalpy of hydration of copper sulphate can be calculated

from the different heats of reaction measured when anhydrous and hydrated

copper sulphate are separately dissolved in water.

Tasks

1. Record temperature-time curves for the dissolution of anhydrous copper

sulphate and hydrated copper sulphate in water.

2. Calculate the hydration enthalpy of anhydrous copper(II)sulphate.


Set up the experiment as shown in Figure 1 but for the time being do not connect

the heating coil to the work and power meter.

Figure 1: Experimental Set-up


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Dr. F. Khlissa

Prepare the two copper salts by grinding each of them separately to a fine powder

in a mortar. Make sure that the anhydrous copper sulphate really is anhydrous by

heating it in a porcelain dish over a Butane burner until it is completely white and

allowing it to cool in a desiccator. Weigh 24.97 g (0.1 mol) of copper(II) sulphate

and 15.96 g (0.1 mol) of anhydrous copper(II) sulphate in two separate beakers

(weighing accuracy 0.01 g). Fill the calorimeter with 900 g of distilled water

(weighing accuracy 0.1 g).

Connect the temperature probe to T1 of the measuring module. Call up the

‘Measure’ programme in Windows and enter <Temperature> as measuring

instrument. Set the measuring parameters as shown in Figure 2. Under <Diagram 1>

select Temperature T0a, the appropriate range for the temperature and the X

bounds and ‘auto range‘.

Figure 2: Measurement Parameters

Now calibrate your sensor under <Calibrate> by entering a temperature value

measured with a thermometer and pressing <Calibrate>. After having made these

settings, press <Continue> to reach the field for the recording of measured values.

Arrange the displays as you want them. Place the filled calorimeter on the


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Dr. F. Khlissa

magnetic stirrer, insert the oval magnetic stirrer bar and switch on the stirrer

(Caution: Do not switch on the heating unit by mistake!). Insert the heating coil


When temperature equilibrium has been reached (after approximately 10 min)

start the measurement by pushing <Start measurement>. Wait 3 to 4 minutes,

then add the first copper salt to the water by pouring it through the powder

funnel which has been inserted in the opening of the lid. Make sure that the entire

quantity of salt is added to the water without any loss. Continue to measure until a

new equilibrium has been reached. Subsequently perform electrical calibration to

determine the total heat capacity of the calorimeter. Supply 10 V AC to the work

and power meter for the electric heating. Push the <Reset> button and then put

the free ends of the heating coil connection cables into the output jacks. The

system is now continuously heated and the quantity of energy supplied is

measured. When the work and power meter shows approximately 4000 Ws, switch

off the heating and read the exact quantity of electrical energy supplied. After a

further three minutes, stop recording the temperature.

Figure 3: Temperature-time curve of solution

of copper(II) sulphate and determining the

heat capacity of the system

Figure 4: Temperature-time curve of solution

of anhydrous copper (II) sulphate and

determining the heat capacity of the system


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Dr. F. Khlissa

Figures 3 and 4 show the graphs as they are presented by the program when the

measurements are stopped. If you use <survey> from the toolbar you can read the

temperature difference data.

Repeat the experiment to determine the enthalpy of solution of the second

copper salt. At least two measurements for each salt should be performed to

avoid errors and to calculate a mean value.


The dissolving of a solid electrolyte in water is primarily determined by two

simultaneously occurring processes: the destruction of the crystal lattice and the

hydration of the ions. The destruction of the crystal lattice is an endothermic

process because energy is required to break down the chemical bonds, whereas

the hydration of the ions is exothermic. Depending on the type of lattice, and on

both the radius and the charge of the ions (charge density), the resulting enthalpy

of solution can be either endothermic or exothermic.

When a salt exists in both hydrated and dehydrated forms, and on assuming that

when the hydrated salt dissolves only the degradation of the crystal lattice

occurs, the enthalpy of hydration can be calculated using Hess’s theorem.

Diagram of Hess’s theorem

∆8� � ∆9:� (1)

∆;��<=>? � ∆8��<=>? � ∆8��<=>?.A;B>


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Dr. F. Khlissa

∆;� Enthalpy of hydration

∆8� Molar enthalpy of solution

∆8+ Integral enthalpy of solution

The integral enthalpy of solution can be calculated according to equation (3).

∆8� � C345� (2)

./0 � ./0 ∆�345∆�D6$ (3)

Qexp Heat of solution of a salt

Qcal Electrical work for calibration

∆�/0 Temperature difference during the dissolution of the salt

∆��12 Temperature difference during the calibration

n Quantity of salt (CuSO4)

Data and results

MCuSO4 = 159.6 g · mol-1

MCuSO4•5H2O = 249.68 g · mol-1

∆8��<=>?= - 66.2 kJ · mol-1

∆8��<=>?.A;B> = + 11.5 kJ · mol-1

∆;��<=>?= - 77.7 kJ · mol-1


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Dr. F. Khlissa

Experiment 5: Determination of the enthalpy of vaporization of liquids

Principle

The vaporization of a liquid occurs with heat absorption. To determine the

enthalpy of vaporization, a known mass of the liquid which is to be investigated is

vaporized in a special vaporization vessel in a current of air. The quantity of heat

absorbed, which corresponds to the enthalpy of vaporization, can be

calorimetrically determined.

Figure 1: Experimental Set-up.

Tasks

1. Determine the molar enthalpy of vaporization of diethyl ether and methanol.

2. Calculate the molar entropies of vaporization and discuss the results under

consideration of Trouton’s rule.


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Dr. F. Khlissa

Set-up and procedure Set up the experiment as shown in Figure1 but for the time being do not connect

the heating coil to the work and power meter. Connect the temperature probe to

T1 of the measuring module. Call up the ‘Measure’ program in Windows and enter

<Temperature> as measuring instrument. Set the measuring parameters as shown

in Figure 2. Under <Diagram 1> select Temperature T0a, the appropriate range for

the temperature and the X bounds and ‚auto range‘.

Figure 2: Measurement parameters

Now calibrate your sensor under <Calibrate> by entering a temperature value

measured with a thermometer and pressing <Calibrate>. After having made these

settings, press <Continue> to reach the field for the recording of measured values.

Arrange the displays as you want them.

Place the clean and dry evaporation vessel in an Erlenmeyer flask and fill it

through the straight inlet tube with 15 ml of the liquid to be evaporated using a

syringe with a cannula. Following this, attach a 5 cm length of rubber tubing to the


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Dr. F. Khlissa

air inlet tube and connect it to the air control valve. Only then, close off the

right-angled air outlet tube with a rubber cap to prevent loss of substance due to

vaporization. Subsequent to this, determine the mass of the evaporation vessel,

which has been thus prepared (weighing accuracy: 0.0001 g). The Erlenmeyer flask

serves merely as a support for the evaporation vessel.

Fill the calorimeter with 900 g of distilled water which is at room temperature

(weighing accuracy = 0.1 g). Put the oval magnetic stirrer bar into the calorimeter

and switch the magnetic stirrer on (Caution: Do not mistakenly switch on the

heating unit!). Insert the heating coil, the temperature probe and the evaporation

vessel into the lid of the calorimeter and fix them in position. Take off the rubber

cap and connect the evaporation vessel to the filter pump via the safety bottle.

Wait until a temperature equilibrium has been reached (approximately 10 min).

Start the measurement with <Start measurement>.

Ensure that the equilibrium temperature has been reached in the calorimeter, i.e.

the temperature remains constant or shows only a slight drift, then turn on the

water jet pump and start the vaporization process by carefully opening the air

control valve on the air inlet. Avoid vigorous sputtering and delayed boiling. When

the temperature of the water has decreased by approx. 1°C, close the air control

valve and shut off the water jet pump (let air into the safety bottle!).

Immediately remove the vacuum tubing from the air outlet tube of the

evaporation vessel and close this outlet tube with the rubber cap. Continue to

measure and record the temperature of the system until a new equilibrium has

been reached or until the alteration only occurs slowly and linearly. Now, perform

electrical calibration to determine the total heat capacity of the calorimeter. To

do this, supply 10 V AC to the work and power meter for the electric heating. Push

the <Reset> button and then put the free ends of the heating coil connection


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Dr. F. Khlissa

cables into the output jacks. The system is now continuously heated and the

supplied quantity of energy is measured.

When the work and power meter shows approximately 4000 Ws, switch off the

heating and read off the exact quantity of electrical energy that has been

supplied. Continue to measure for another three minutes, then stop temperature

recording with <Stop measurement>.

Figure 3 shows the graph as it is presented by the program when the measurement

is stopped. If you use <survey> from the toolbar you can read the temperature

difference data. Subsequently remove the closed evaporation vessel, carefully

dry it, then weigh it in the Erlenmeyer flask which was previously used. From the

two weightings, i.e. before and after vaporization, the quantity of vaporized

substance can be calculated. Always thoroughly clean and dry the evaporation

vessel before performing a new measurement.

Figure 3: Temperature- time curve of the vaporization of diethyl ether and for

determining the heat capacity of the system.

Theory and evaluation When liquids are heated at constant pressure, a certain temperature is reached

at which the substance is converted from the liquid to the vapor phase. This


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Dr. F. Khlissa

boiling point remains constant under further addition of heat until the liquid phase

has been completely vaporized. If a mole of liquid is vaporized, its molar enthalpy

H increases by the amount of heat absorbed QV. This difference in enthalpies is

termed the enthalpy of vaporization ΔVH and is explained by the energy required

to break the bonds between the molecules in the liquid and the constantly

changing pressure that the vapor formed must exert against the existing external

pressure. If the vapor is cooled, condensation occurs at the same temperature –

the process is reversible.

Consequently, the increase in entropy ΔVS which accompanies the vaporization can

be formulated by applying the second principle of thermodynamics:

∆EF � ∆G;�G (1)

In many liquids the molar enthalpy of vaporization lies in the range between 80

and 90 J · mol-1 · K-1. This rule was named after Trouton and is especially valid for

non-associated liquids.

Strong interactions between the molecules of liquid, such as hydrogen bonds,

result in higher entropies of vaporization. In the present experiment, the liquid is

vaporized in an airstream below the boiling point of the liquid. In doing so, the

equilibrium between the liquid and the vapor phase is disturbed by the continuous

removal of the vapor phase. The enthalpy of vaporization is equal to the amount of

heat which is withdrawn from the calorimeter.

∆E� � ∆EF. �E (1.1)

∆E� � ∆G:� (2)

∆EF � �&H . ∆�E (3)

7/2 � &HF. ∆�/2 (4)

&H � I3$�3$ (4.1)

∆E� � �&H . ∆�E %* (5)

∆E� � �7/2 . ∆�G∆�3$ .%* (5.1)


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Dr. F. Khlissa

ΔTV Temperature difference during vaporization

Tel Temperature difference during calibration

Wel Electrical work

CK Heat capacity of the calorimeter

n Quantity of vaporized substance (m/M)

Data and results Molar mass of methanol: 32.04 g · mol-1

Molar mass of diethyl ether: 74.12 g · mol-1

Boiling point of methanol: 337.7 K

Boiling point of diethyl ether: 307.7 K

Enthalpy of vaporization of methanol: 33.8 kJ · mol-1

Enthalpy of vaporization of diethyl ether: 25.6 kJ · mol-1

Entropy of vaporization of methanol: 100.0 J · mol-1 · K-1

Entropy of vaporization of diethyl ether: 83.2 J · mol-1 · K-1


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Dr. F. Khlissa

Experiment 6: Viscosity of gases Estimation of molecular diameter

Principle

Expressed most simply, the viscosity of a fluid (liquid or gas) relates to its

resistance to flow. The viscosity of a gas is determined in particular by the rate of

transfer of the flow momentum from faster moving layers (laminas) to slower

ones. The so-called transpiration methods provide a convenient way of measuring

gas viscosities. In the approach used here, the flow rate of the gas (which is

inversely proportional to its viscosity) is recorded by monitoring the evacuation of

a vessel through a capillary tube under a constant pressure differential.

Tasks

Measure the viscosities of the gases nitrogen, carbon dioxide, hydrogen and

helium. Use simple gas kinetic theory to estimate the molecular diameter of each

of the gases measured.


Set up the experiment as shown in Figure 1, but first ensure that the syringe

plunger stop is positioned to prevent the plunger from being fully removed from

the barrel, and determine the exact length of the capillary tube with the Vernier

calliper before integrating it in the measuring arrangement.

Disconnect the syringe from the flow line, turn the 3-way stopcock to connect the

gas bottle to the syringe and fill the syringe with nitrogen. Turn the stopcock to

connect the syringe to the exit tube and lightly press the plunger to expel

nitrogen. Turn the stopcock to reconnect the syringe to the gas bottle. Rinse the

syringe in this manner three times, then leave it filled and connected to the gas

line. Switch on the pump and evacuate the flow line up to the stopcock by pumping

for ten minutes.


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Dr. F. Khlissa

Figure 1: Experimental Set-up.

Once the flow line is evacuated and with the pump still on, turn the stopcock to

connect the syringe to the flow line. Start the stopwatch when the syringe volume

has reached 90 ml and stop it as soon as the syringe is empty. Record the ambient

laboratory temperature and atmospheric pressure. Repeat the measurement

twice. Carry out the same rinsing and measuring procedure for the gases carbon

dioxide, hydrogen and helium.


It is a property of fluids (liquids and gases) that an external force will result in a

steady flow rate. The force applied is resisted by the internal frictional forces

between adjacent layers of the fluid moving with differential velocities. Consider

a fluid undergoing ideal parallel-plane laminar flow in the x-direction, as

illustrated in Figure 2. The flow velocity varies linearly from the stationary

bottom layer to the uppermost layer moving with velocity vx.

With the distance between the two layers dz, and the difference of the velocities

dv, the force F resisting the relative motion of the plane layers is found to be

proportional to the area of contact A between the adjacent layers of the fluid and

the flow velocity gradient in the fluid dvx/dz.

J � KL ME4MN (1)

where η is the constant of proportionality, the viscosity.


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Dr. F. Khlissa

F/A is the tangential or shear stress across the boundary of a layer and, acc. to

Newton’s second law, this is equal to the rate of change of flow momentum

between adjacent layers.

In the experiment performed here it is assumed that the gas flows laminarly along

the cylindrical capillary tube. The flow motion in a gas is transmitted by collisions

of molecules from one layer to the next. As the velocity is almost zero at the wall,

but increases with increasing distance from the wall and reaches its maximum in

the center of the capillary tube, a molecule moving from an outer layer to the

center slows down the inner layers. The reverse direction causes acceleration of

the slower layers. After a short while a stationary state is established where the

relative velocities of the layers are constant. This phenomenon is known as viscous

flow.

The rate of gas flow along the tube can be calculated using Poiseuille’s equation.

M�MO � PQBR��BS�BBT

�U2V (2)

M�M� Volume rate of flow

p1, p2 Pressure at the high resp. low pressure ends of the capillary

r Radius of the capillary tube

l Length of the capillary tube

p Pressure at which the volume is measured (here p1)

From the constant pressure differential used in this experiment and the known

dimensions of the capillary, the evacuation rate data measured can be used to

calculate absolute values of the gas viscosities. Due to the r4 term in (2) minor

irregularities in the radius of the capillary can have a significant effect on the

viscosity values determined. It is therefore better to calibrate the dimensions of

the capillary by measuring the volume flow rate for a reference gas of known

viscosity.


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Dr. F. Khlissa

Viscosity is a temperature dependent quantity. It increases with increasing

temperature in the case of gases, and decreases with increasing temperature with

liquids.

In an ideal gas, the velocity of the gas molecules increases with increasing

temperature. The viscosity increases to the same extent, proportionally to √�. A

greater velocity results in a greater braking effect being exerted on a molecule

when it enters a neighboring layer.

The pressure of the gas in the syringe p1 corresponds to the pressure measured

with the weather monitor, while the pressure p2 at the other end of the capillary

is determined by the filter pump, which produces a vacuum of about 18 mbar (≈

1824 Pa). The Poiseuille equation is derived on the assumption that the gas

behaves as an incompressible fluid undergoing laminar flow. This is strictly valid

only if the pressure differential (p1 – p2) is small compared to p1 resp. p2, which is

not the case here. Nevertheless, the method enables realistic estimates of gas

viscosities to be obtained.

Figure 2: Ideal plane-parallel laminar flow.


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Dr. F. Khlissa

In terms of the kinetic theory of gases, these values can be further interpreted

on a molecular level. The gas kinetic expression for the viscosity η of a gas

undergoing laminar flow is the following:

η =�XYZ[\ (3)

m Mass of a molecule

λ Mean free path

N Number of molecules per unit volume

\ Mean molecular speed

The mean free path for an ensemble of molecules having a Maxwell distribution of

molecular velocities is

Z � �√]PMB^ (4)

N Number of molecules per unit volume

d Hard sphere collision diameter

Substituting (4) into (3) the following is obtained

K � %_X√]PMB (5)

For a Maxwellian distribution of molecular velocities the mean molecular speed c is

\ � `ab�P% (6)

k Boltzmann constant

T Temperature

Substituting (6) into (5) and rearranging leads to

�] � ]XV`b�%Pc (7)


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Dr. F. Khlissa

Experiment 7: Mutual Solubility Curve for Phenol and Water

Introduction

A few liquids are miscible with each other in all proportion, for example: ethanol

and water. Others have miscibility in limited proportions in other liquids, for

example: ether-water, phenol-water.

Generally both liquids become more soluble with rising temperatures until the

critical solution temperature is attained, and above this point the liquids

completely miscible. There is a big possibility that any pair of liquids can form a

closed system where both upper and lower critical solution temperatures exist,

however it is not easy to determine both the temperatures (before the substance

freezes or evaporates) except for nicotine and water.

At any temperature below critical solution temperature, the composition for two

layers of liquids in equilibrium state is constant. The mutual solubility for a pair of

partially miscible liquids in general is extremely influenced by the presence of a

third component.

Tasks

1) to determine the solubility of two partially liquids (phenol – water solution)

2) to construct a mutual solubility for the pair

3) to determine their critical solution temperature.

Procedure

1. Seven boiling tubes were prepared and label by A,B,C,D,E,F and G.

2. The boiling tubes was filled with different amount of phenol and water.

Boiling tube A filled with 8%, B 11%, C 35%, D 50%, E 63%, F 75% and G 80% of

phenol and water was added until 50mL of solution reach.


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Dr. F. Khlissa

3. Heat the boiling tubes with water bath. The boiling tubes were swirled

and shaken well.

4. Once the turbid becomes clear, the tubes were observed and

temperatures were recorded.

5. The tubes were removed from the hot water and the temperature was

allowed to reduce. Then the temperature were recorded when the liquid become

turbid and two separate layers were form.

6. By using the temperature obtained, the average temperature for each

tubes at which the two phases were no longer seen or at which two phases were

exist were determined.

Theory

Phase rule is a useful device to relate the effect of least number of independent

variables, such as concentration, pressure and others upon the various phases that

can exist in an equilibrium system containing a given number of components. The

phase rule is expressed as followed: F = C – P + 2

in which F is the number of degrees of freedom in the system, C the number of

components and P the number of phases present.

For two-component system contains one liquid phase, for instance, when phenol

and water are miscible, we need at least three intensive variables to define the

system where

F = C – P + 2

= 2 – 1 + 2

= 3

in which the three variables are temperature, pressure and concentration of

phenol.


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Dr. F. Khlissa

For two-component system contains two liquid phases, for instance, when phenol

and water are immiscible, we need at least two intensive variables to define the

system where

F = C – P + 2

= 2 – 2 + 2

= 2

in which the two variables are temperature and concentration of phenol and the

pressure is fixed. In another words, the system has 2 degrees of freedom.

A system comprising of liquid, for example water is in equilibrium with its vapor.

By stating the temperature, the system is completely defined as the pressure

under which liquid and vapor coexist is also defined. This agrees the phase rule as

F= C-P+2

=1-2+2

=1

The state of 3 phases ice-water-vapor system is completely defined and the rule

is F=C-P+2

=1-3+2

=0

This means there is no degree of freedom.


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Dr. F. Khlissa

Experiment 8: Chemical Equilibrium and Le Chatelier's Principle

Theory Some reactions involving chromium compounds are suitable for illustrating Le

Chatelier’s Principle because they involve clear color changes. One such

equilibrium is:

&d]ef]S � �]e ↔ 2&dei]S � 2�j

Orange yellow

This experiment will be used to demonstrate the effects of concentration

changes on an equilibrium mixture. Adding an acid will increase the concentration

of H+, and adding a base will reduce it.

Procedure NB: Wear your safety glasses.

Quarter fill a test tube with the solution of sodium dichromate provided. This

should have an orange color. The following equilibrium exists:

&d]ef]S � �]e ↔ 2&dei]S � 2�j

Orange yellow

Since the orange colour predominates, the equilibrium must lie on the left hand

side of the equation. Keep a second sample of the sodium dichromate solution in a

test tube as a control.

Carefully add some bench dilute sodium hydroxide solution until the orange color

changes to yellow.


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Dr. F. Khlissa

The sodium hydroxide removes the H+ ions giving rise to a stress and therefore, in

keeping with Le Chatelier's Principle, the forward reaction predominates to

produce more H+ ions.

Carefully add dilute hydrochloric acid until the yellow color changes back to

orange. The added hydrochloric acid creates an excess of H+ ions that causes the

equilibrium reaction to be shifted to the left in order to absorb this excess of H+

ions.

Student Questions When sodium hydroxide solution is added to a solution of potassium dichromate, a

color change occurs. Describe the color change, and explain why it happens.

The color changes from orange to yellow. In the equilibrium

&d]ef]S � �]e ↔ 2&dei]S � 2�j

Orange yellow

Adding sodium hydroxide solution removes H+, and so shifts the equilibrium to the

right, according to Le Chatelier’s Principle. Therefore the color changes to yellow.

Why does adding hydrochloric acid reverse the color change referred to in

question 1?

The color changes from yellow to orange. In the equilibrium

&d]ef]S � �]e ↔ 2&dei]S � 2�j

Orange yellow

Adding acid increases the concentration of H+ ions, and therefore shifts the

equilibrium to the left, according to Le Chatelier’s Principle. Therefore the color

changes to orange.


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Dr. F. Khlissa

Experiment 9: Determination of the surface tension

by capillary rise method

Objective

To determine the surface tension of a liquid by capillary rise method.

Apparatus

(i) Capillary tubes, (ii) Experimental liquid (water, acetone,..), (iii) Beaker, (iv)

Ruler, (v) Support stands and clamps.

Theory

A molecule well within a liquid is surrounded by other molecules on all sides. The

surrounding molecules attract the central molecule equally in all directions,

leading to a zero net force. In contrast, the resulting force acting on a molecule at

the boundary layer on the surface of the liquid is not zero, but points into the

liquid. This net attractive force causes the liquid surface to contract toward the

interior until the repulsive collisional forces from the other molecules halt the

contraction at the point when the surface area is a minimum. If the liquid is not

acted upon by external forces, a liquid sample forms a sphere, which has the

minimum surface area for a given volume. Nearly spherical drops of water are a

familiar sight, for example, when the external forces are negligible.

The surface tension γ is defined as the magnitude F of the force exerted

tangential to the surface of a liquid divided by the length l of the line over which

the force acts in order to maintain the liquid film.

γ =F/l (1)

In this experiment we will determine the surface tension of water by capillary

rise method. Capillarity is the combined effect of cohesive and adhesive forces


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Dr. F. Khlissa

that cause liquids to rise in tubes of very small diameter. In case of water in a

capillary tube, the adhesive force draws it up along the sides of the glass tube to

form a meniscus. The cohesive force also acts at the same time to minimize the

distance between the water molecules by pulling the bottom of the meniscus up

against the force of gravity.

Consider the situation depicted in Figure 1, in which the end of a capillary tube of

radius, r, is immersed in a liquid of density ρ. For sufficiently small capillaries, one

observes a substantial rise of liquid to height, h, in the capillary, because of the

force exerted on the liquid due to surface tension. Equilibrium occurs when the

force of gravity on the volume of liquid balances the force due to surface tension.

The balance point can be used to measure the surface tension.

Figure 1 Figure 2

Thus, at equilibrium force of gravity is given as,

Jk � l+Rmd]Tn (2)

where g is the acceleration due to gravity.

Force due to surface tension (see Figure 2) is along the perimeter of the liquid.

Let θ be the angle of contact of the liquid on glass. The vertical component of the

force (upward) at equilibrium is given as

J � o p 2md p \qrs (3)


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Dr. F. Khlissa

Assuming θ to be very small and neglecting the curvature of liquid surface at the

boundaries, one can obtain surface tension by equating Eqs. 2 and 3 as follows:

o � !kQ:] (4)

It should also be noted that surface tension of a liquid depends very markedly

upon the presence of impurities in the liquid and upon temperature. The SI unit

for surface tension is N/m.

Procedure

- When working with the capillaries, keep one just for pure water and one for

acetone solution.

- Dip the capillary into the liquid, and note that surface tension will keep the

liquid in the tube after you remove it.

- Let both acetone and pure water be drawn into the two capillaries provided.

Measure the height with a ruler; measure the capillary diameter and the

angle of contact using the magnifier as illustrated in Figure 3.

Figure 3

Analysis and Report

Compare the values of the surface tension measured for the two liquids. In

particular, note the calculated fractional error in each case.


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Dr. F. Khlissa

Experiment 10: Ternary Phase Diagram: Water-Toluene-Ethanol

Objective

To determine the phase diagram for the ethanol/ toluene/ water system

Introduction

In three components system at constant temperature and pressure, the

composition can be expressed in the form of coordinates of an equilateral triangle

diagram.

Diagram 1: Triangular diagram for 3 component system

In the above diagram, every corner of the triangle represents a pure component,

which are 100%A, 100%B and 100%C. Each side represents one binary mixture and

area in this triangular diagram represents ternary components. Rules Relating to

Triangular Diagrams are as below:


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Dr. F. Khlissa

1. Each of three corners or apexes of the triangle represent 100% by weight of

one component (A, B, or C). As a result, that same apex will represent 0% of the

other two components.

2. The three lines joining the corner points represent two-component mixtures

of the three possible combinations of A, B and C. Thus the lines AB, BC and CA are

used for two-component mixtures of A and B, B and C, and C and A, respectively

3. The area within the triangle represents all the possible combinations of A, B,

and C to give three-component systems

4. If a line is drawn through any apex to a point on the opposite side (e.g. line DC

in Diagram 1) then all systems represented by points on such a line have a constant

ratio of two components, in this case A and B

5. Any line drawn parallel to one side of the triangle, for example, line HI ,

represents ternary systems in which the proportion (or percent by weight) of one

component is constant. In this instance, all systems prepared along HI will contain

20% of C and varying concentrations of A and B

Addition of the third component into one pair of miscible liquids can change their

solubility. If this third component is more soluble in either one from the two

components, the solubility of both components will reduce. But if the third

component is soluble in both components at the same time, the solubility

increases. Thus, when ethanol is added into a mixture of benzene and water, the

solubility of these two components will increase until a point is reached, where

the mixture become homogenous. This application can be used in formulations of

solutions. Examples of three-component liquid system that have been tested, are

castor oil/ alcohol/ water; peppermint oil/ propylene glycol/ water; peppermint

oil/ polyethylene glycol/ water.

The advantages in preparing an oily substance as a homogeneous aqueous liquid are

obvious. However, we need to know that what will happen to a system like this,


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Dr. F. Khlissa

when it is diluted, and this can be explained through the understanding of

triangular phase diagram.

Procedures

1. Ethanol/ toluene mixtures of different compositions were prepared and

placed in sealed conical flasks.

2. Each mixture contained different % volume of ethanol in 50 ml: 10, 25,

35, 50, 65, 75, 90, 95% v/v.

3. A burette was filled with distilled water.

4. The mixtures were titrated with water, accompanied by vigorous shaking

of the conical flask.

5. Titration was stopped when a cloudy mixture was formed.

6. The volume of the water used was recorded.

7. Steps 1-6 were repeated to do a second titration. The volume of water

required for complete titration of each mixture was recorded.

8. Average volume of water used was calculated.

9. % volume of each component of the ternary system for when a second

phase became separated was calculated.

10. These values were plotted on a graph paper with triangular axes to produce

a triple phase diagram.


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Dr. F. Khlissa

Results

% ethanol (v/v) Volume of Water Used (mL)

Average Titration I Titration II

10 1.3 1.1 1.2

25 1.6 1.0 1.3

35 1.1 1.9 1.5

50 1.9 2.1 2.0

65 2.7 2.7 2.7

75 4.3 3.9 4.1

90 10.0 10.4 10.2

95 16.8 14.2 15.5

Calculation

Total

volume Water

Toluene

Ethanol

(x+20mL) Volume (mL) % Volume (mL) % Volume (mL) %

21.2 1.2 5.7 18.0 84.9 2.0 9.4

21.3 1.3 6.1 15.0 70.4 5.0 23.5

21.5 1.5 7.0 13.0 60.4 7.0 32.6

22.0 2.0 9.1 10.0 45.45 10.0 45.45

22.7 2.7 11.9 7.0 30.8 13.0 57.3

24.1 4.1 17.0 5.0 20.7 15.0 62.3

30.2 10.2 33.8 2.0 6.6 18.0 59.6

35.5 15.5 43.7 1.0 2.8 19.0 53.5

Graph :


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Dr. F. Khlissa

Discussion

A ternary phase diagram has three components. The three components are usually

compositions of elements, but may include temperature or pressure also. This type

of diagram is three-dimensional but is illustrated in two-dimensions for ease of

drawing and reading. Instead of being a rectangular plot, it is a triangle. It is

called triangular diagram.

In the case of toluene, ethanol, and water which water and toluene are usually

form a two-phase system because they are only slightly miscible. The heavier of

the two phases consists of water saturated with toluene, while the lighter phase

is toluene saturated with water. However, ethanol is completely miscible with both

toluene and water. Thus, the addition of sufficient amount of ethanol to the

toluene-water system would produce a single liquid phase (upper region in the

diagram, region B) in which all the three components are miscible and the mixture


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Dr. F. Khlissa

is homogenous. This is shown in the triple phase diagram that has been plotted on

the triangular diagram.

The curve of the plotted graph is termed a binodal curve or binodal. The region

bounded by this curve, which is marked A, represent the two-phase region.

Mixture with composition contained within region A are cloudy in appearance due

to the phase separation. In other words, the amount of ethanol is not sufficient

for homogenous mixture to be produced. The region of the graph that is not

bounded by the binodal curve represents the one–phase region. It is marked B on

the diagram. Mixture with composition that falls into this region is clear and they

are homogenous. For these mixtures, the amount of ethanol is sufficient to

produce a single liquid phase.

The points that are at both ends of the curve are the limits of solubility of

toluene in water and water in toluene. Along the toluene-water line, which

represents a binary mixture of toluene and water, the liquids are able to form a

homogenous mixture as long as the first point is not exceeded. However, the

second point must be exceeded for a homogenous mixture to be formed. The

length of line between the two points represents the mixture of toluene and water

with such composition that they cannot form a homogenous mixture. This may be

due to insolubility of toluene in water or water in toluene.

In this experiment, some errors may be happened and influence the accuracy of

the result formed. Our eyes must be parallel to the meniscus position when taking

reading on burette or pipette. It can make sure the volume taken and recorded is

accurate. Besides, the conical flask must be shaked well after each addition of

water. Furthermore, judgement of the cloudy solution formed depends on

personal judgement. Hence, different group may vary in results Thus, precautions

should be taken to minimize it.


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Dr. F. Khlissa

Practice

1. Will a mixture containing 70% ethanol, 20% water and 10% toluene remain

clear or form two phases?

From the graph plotted, at these concentrations, appear as one liquid phase, the

solution is appear clear.

2. What will happen if you dilute 1 part of the mixture with 4 parts of (a)

water; (b) toluene; (c) ethanol?

1 part x 70% ethanol = 1 part x 70/100 = 0.7 part of ethanol

1 part x 20% water = 1 part x 20/100 = 0.2 part of water

1 part x 10% toluene = 1 part x 10/100 = 0.1 part of toluene

There are 0.7 part of ethanol; 0.2 part of water; 0.1 part of toluene in the

mixture.

(a) 1 part of mixture + 4 parts of water:

Ethanol = (0.7/1+4) x 100% =14%

Water = (0.2+4 / 1+4) x 100% = 84%

Toluene = (0.1 /1+4) x 100% =2%

From the phase diagram, this mixture is outside the area of the binodal curve.

Therefore, a clear single liquid phase of solution is formed.

(b) 1 part of mixture + 4 parts of toluene

Ethanol = (0.7 / 1+4) x 100% =14%

Water = (0.2 / 1+4)x 100% = 4%

Toluene = (0.1+4 / 1+4) x 100% =82%

From the phase diagram, this mixture is within the area of the bimodal curve.

Therefore, a two liquid phase will form and the mixture is cloudy.

(c) 1 part of mixture + 4 parts of ethanol

Ethanol = (0.7+4 / 1+4) x 100% =94%


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Dr. F. Khlissa

Water = (0.2/ 1+4)x 100% = 4%

Toluene = (0.1 / 1+4) x 100% =2%

From the phase diagram, this mixture is outside the area of the bimodal curve.

Therefore, a clear single liquid phase of solution is formed

Conclusion:

Ethanol,toluene and water system is a ternary system with one pair of partially

miscible liquid ( toluene and water). The addition of sufficient amount of ethanol

to the toluene-water system would produce a single liquid phase in which all the

three components are miscible and the mixture is homogenous.

p

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