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PHYSICAL CHEMISTRY (LAB MANUAL)
EXPERIMENT -3
AIM: Determination of integral enthalpy of solution of potassium nitrate in
1:200 molar ratio.
LEARNING OBJECTIVES
After performing this experiment student will be able to

define integral and differential enthalpies of solution and make a distinction
between them.

identify the major energy factors that govern the enthalpy of solution and
explain on their basis why dissolution of some solutes is exothermic process
while for others it is endothermic process.

explain the significance of molar ratio associated with each value of integral
enthalpy of solution.

correlate the molar ratio of solute and solvent to the mass composition of
solution.
REQUIREMENTS
(a) Chemicals
Potassium nitrate
(b) Apparatus
Weighing bottle, test tube, rubber cork, 250 mL beaker, Dewar/ Thermos flask,
rubber cork (3” dia.), glass stirrer, 0.1oC thermometers – 2, stop watch / stop clock,
a glass rod fitted with a rubber policeman.
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PHYSICAL CHEMISTRY (LAB MANUAL)
THEORY
Integral enthalpy of solution, solnH, is defined as the change in enthalpy when one
mole of the solute is dissolved in a specified amount (number of moles) of solvent at
a constant temperature and under a constant pressure (1 bar for standard enthalpy
of solution, solnHo). It corresponds to the process:
X (s) + n H2O(l) 
X .nH2O (aq)
solnH
Where solnH is the integral enthalpy of solution of the solute X in water in 1 : n molar
ratio.
It is observed that dissolution process for some solutes is exothermic while for others
it is endothermic. The reason for this difference can be broadly understood in terms
of two major energy factors. The process of dissolution can be assumed to occur in
two steps.
i.
Breaking of crystal lattice to make the constituent particles of solute free.
These particles are ions in case of ionic solutes and molecules in case of
molecular solutes. This process requires energy and is endothermic in nature.
The energy required is equal to the lattice energy of the solute, HL, Stronger
the lattice more is the value of lattice energy.
ii.
Solvation of solute particles whereby each solute particle is surrounded by
certain number of solvent molecules. During this process energy is released
i.e. it is exothermic in nature. The change in enthalpy during the process is
called the enthalpy of hydration of solute hydH. The amount of energy
released depends upon the extent of solvation (hydration in case of water
being used as solvent) which further depends upon the concentration of the
solution. Greater the dilution of a solution, greater is the extent of solvation
and hence more energy is released.
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PHYSICAL CHEMISTRY (LAB MANUAL)
The heat lost or gained during the dissolution process is roughly the sum of
these two energy factors and may be written as
solnH = HL + hydH
Overall sign of left hand side depends upon the relative magnitudes of the two terms
on the right hand side.
solH = -ve when hydH > HL
and solH = +ve when HL > hydH
PROCEDURE
This experiment is performed in the following two steps.
I.
Determination of heat capacity of calorimeter
It is carried out as mentioned in experiment 1(a). If the same set of calorimeter
apparatus is being used which includes 250 mL beaker, thermos flask, 0.1 oC
thermometer and glass stirrer and if the volume of water being used is also the same
(200 mL) then the heat capacity obtained in experiment 1(a) may be used.
II.
Determination of Enthalpy of Solution
It can be done by the following procedure:
1.
Calculate the mass of KNO3 required for preparing a solution in 200 mL water
in which its molar ratio is 1:200
Mass of water to be used = 200 g
Molar mass of KNO3 = 101 g mol-1
Molar mass of water = 18 g mol-1
Mass of 200 moles of water = 200 x 18 g
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PHYSICAL CHEMISTRY (LAB MANUAL)
Mass of KNO3 to be dissolved in 200 x 18 g water = 101 g
Mass of KNO3 to be dissolved in 200 g water =
101 200
 5.61g
200 18
2. Powder KNO3 in a mortar to a very fine powdery state.
3. Take a clean and dry test tube and place about 6 g of potassium nitrate in it.
Fix rubber cork and weigh it accurately.
4. Take 200 mL distilled water in the calorimeter (250 mL beaker placed in
thermos flask) and partially dip the closed test tube with potassium nitrate in
it. Start the stop watch/clock. Record the temperature every 30 seconds 4 to 5
times.
5. Perform the following operation as quickly as possible. Lift the test tube out of
the water in calorimeter, touch it with sides of calorimeter and take it out.
Wipe off the water from it with a dry cloth or filter paper. Pour its contents into
the water in calorimeter, note down time of mixing from the stop watch/ stop
clock and stir the water briskly with the help a glass rod fitted with a rubber
“policeman” (a ring of rubber tubing slipped over one end of the glass rod) till
the salt dissolves completely.
6. Withdraw the glass rod with policeman immediately and replace the lid/ cover
of the thermos flask fitted with a thermometer and a glass stirrer.
7. Start recording the temperature of the solution every 30 seconds for next 3-4
minutes.
8. Weigh the test tube with rubber cork again to determine the mass of
potassium nitrate dissolved.
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PHYSICAL CHEMISTRY (LAB MANUAL)
9. Plot a graph of temperature versus time . Mark the time of mixing. Join the
points of temperature of
water (before dissolving potassium nitrate) by a
straight line and extend it to the time of mixing to obtain the initial
temperature (tinitial ). Join the points of temperature of the solution (after
dissolving potassium nitrate) by a smooth curve and extend it to the time of
mixing to obtain the final temperature ( tfinal ) at the time of mixing as shown
in the figure 4.
Fig. 4: Temperature versus time curve for determination of enthalpy of
solution
10.Calculate the enthalpy of solution of KNO3 as given later.
11.Repeat the experiment once more.
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PHYSICAL CHEMISTRY (LAB MANUAL)
III.
SOURCES OF ERROR AND PRECAUTIONS
A.
Handling of Chemicals and Apparatus
In addition to the sources of errors and precautions given in experiment 1, the
following are the additional ones:
1. Allow the temperature of water in calorimeter to attain a steady valve while
the test tube containing salt is kept in it. Otherwise record the temperature
every 30 seconds and obtain the initial temperature also from the graph as in
earlier experiments.
2. The salt should be finely powdered so that it dissolves quickly completely.
3. Dissolution of salt should be completed as fast as possible, since, during this
period there is no lid/cover on thermos flask and there would be some
exchange of heat with surroundings.
4. When test tube containing salt or the glass rod with policeman is taken out of
the beaker kept in thermos flask, it should be touched with side of the beaker
so that minimum quantity of solution sticks to it.
B.
Errors in Measurements and Calculations
5. Actual molar ratio of salt and solvent should be finally calculated from the
mass of salt dissolved and reported along with enthalpy of solution.
OBSERVATIONS
I.
Mass of salt dissolved
1. Mass of test tube + KNO3 = m1 g
2. Mass of test tube + KNO3 after transfer = m2 g
3. Mass of KNO3 dissolved = m = (m1 - m2) g
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PHYSICAL CHEMISTRY (LAB MANUAL)
II.
Mass of water taken in calorimeter
Volume of water taken = 200 mL
Density of water  1g mL-1

Mass of water taken = (200 x 1) = 200 g
III.
Initial temperature of water in the calorimeter and salt = t1 oC
IV.
Temperature – time data
Temperature / oC
Time / min
CALCULATIONS
Final temperature of solution at the time of mixing from graph = t2 oC
Change in temperature t = (t2 – t1) oC
In the dissolution process of KNO3 heat is absorbed by the system from
the
calorimeter and the water in the solution whose temperature falls
Total heat gained, q = - (heat lost by calorimeter + heat lost by water)
Let the heat capacity of calorimeter be C, then the heat lost by it = C x t
Mass of water in the solution = 200 g
(taking density as 1 g mL-1
Heat capacity of water = 4.18 J oC-1g-1
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PHYSICAL CHEMISTRY (LAB MANUAL)
Change in temperature = t
Heat lost by water = m x s x t
= (200 g) x (4.18 J oC-1) x t
Total heat lost = (C x t) + (200 x 4.18 x t) = (C+200 x 4.18) t J
q = - (C+200 x 4.18) t J
Mass of KNO3 dissolved = m g
Molar mass of KNO3 = 101 g mol-1
Number of moles of KNO3 dissolved = n =
3
mol
101
Heat absorbed when n moles of KNO3 are dissolved = q
heat absorbed when 1 mole of KNO3 is dissolved
= solnHo =
q
(C 
 
n
200  4.18) t
n
J mol 1
The values are often reported in kJ mol-1
Calculation of molar ratio
Since n moles of KNO3 are dissolved in 200 g water, moles of water =
1 mole of KNO3 is dissolved in = n’ =
200
mol
18
200
mol
18  n
and the molar ratio is 1 : n’
Calculate solnHo for both the sets and report the average valve.
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PHYSICAL CHEMISTRY (LAB MANUAL)
Result : The standard integral enthalpy of solution of KNO3 in molar ratio 1 : n’ was
found to be …….. kJ mol-1.
EXTENSION
1. A major source of error in calorimetric measurements is the loss of heat by
radiation. How can the procedure be modified to minimize it while using the
same apparatus.
2. Suggest alternatives to Dewar/ thermos flask.
3. Suggest alternative method for determination of heat capacity.
4. Determine the enthalpy of neutralization by using 0.5 M HCl and 0.5 M NaOH
solutions. What changes are expected in the observations? How would you
modify the calculations?
TROUBLE SHOOTING
1. Normally it is desirable to heat the water by 8o-10oC for determination of heat
capacity of calorimeter. Supposing you forget to stop heating and temperature
rise is much higher. Allow the water to cool down to the desired temperature
and then start the stop clock and begin recording of data.
2. If you forget to start the mixing process at the pre-decided time of mixing,
start the mixing process and note down the time when one-half of mixing has
been done. Use this time as the time of mixing and continue with the
experiment normally.
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