SCH 4U0
Name: _______________________
Louise Arbour Secondary School
SCIENCE DEPARTMENT
GRADE 12 UNIVERSITY CHEMISTRY
24 T/I
10 C
ENERGY CHANGES & RATES OF REACTIONS: MEDICAL COLD PACKS LAB
(VIRTUAL)
We can study physical changes that involve liquids and aqueous solutions, using a polystyrene
calorimeter like that one shown in Figure 1. Such materials can be applied to practical
thermochemical systems. For example, when athletes are injured, they may immediately hold an
“instant cold pack” against the injury. The medical cold pack operates on the principle that certain
salts dissolve endothermically in water. The amount of heat per unit mass involved in the dissolving
of the compound is a characteristic property of the substance. It is called the enthalpy of solution.
Table 1 lists some compounds that might be possible candidates for a cold pack because they absorb
energy when they dissolve.
Figure 1- A simple laboratory calorimeter consists of an insulated container made of three nested
polystyrene cups, a measured quantity of water, and a thermometer. The chemical system is placed
in or dissolved in the water of the calorimeter. Energy transfers between the chemical system and
the
surrounding water are monitored by measuring changes in the temperature of
the water.
For this lab you will be using a simulation that provides the ability to
manipulate variables of mass (or volume), temperature and type of
substances. You will observe how changing these variables affects the initial
and/or final temperature and the amount of heat exchanged in a constant
pressure calorimeter experiment.
Purpose
The purpose of this investigation is to use calorimetry to determine the enthalpy of solution of an unknown salt, and to
use that value to identify the salt in a medical cold pack.
Table 1: Enthalpies of Solution for Compounds in a Medical Cold Pack
Salt
Ammonium chloride, NH4Cl
Enthalpy of Solution
0.277 kJ/g
Calcium fluoride, CaF
0.119 kJ/g
Potassium nitrate, KNO3
0.345 kJ/g
Ammonium nitrate, NH4NO3
0.311 kJ/g
Sodium chloride, NaCl
0.064 kJ/g
Potassium chloride, KCl
0.225 kJ/g
Pre Lab Questions:
What is the difference between the system and the surroundings when measuring heat transfer in
a laboratory? Use an example to illustrate your answer. (2 marks C)
SCH 4U0
Name: _______________________
The system is the sample that is being observed while the surroundings is everything around the
system that is able to absorb or release thermal energy. An example of this is if we are looking at a
cell of the human body and looking at this reaction C6H12O6 + 6O2 → 6H2O +6CO2 + energy, the
molecules we are observing in the cell would be the system while the extracellular fluid in your
cells would be the surroundings.
Procedure:
VISIT THE FOLLOWING WEBSITE:
https://media.pearsoncmg.com/bc/bc_0media_chem/chem_sim/calorimetry/Calor.php
Click on Experiment (Tab) → Run Demonstration (Optional)
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Run through the steps and questions of the demonstration to become familiar with how the
simulation works.
Click on Experiment (Tab)→ Run Experiment
Part 1: NH4NO3 in Water (H2O)
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For the first table (beaker and electronic balance), select “solid” from the tab. Using the scroll
down menu select, ammonium nitrate, NH4NO3.
Increase the mass of ammonium nitrate to 3 g and ensure the temperature is set at 20°C. Click
“Next”.
For the second table (calorimeter), select “liquid” from the tab. Using the scroll down menu
select, water, H2O.
Increase the mass of water to 100 g and ensure the temperature is set at 20°C. Record this
value as T1 in the table below.
Click on “Show Graph View”, “Show Microscopic View” and then Start.
Record the final temperature T2 located above the calorimeter in the table below.
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Calculate the heat of reaction, Q (kJ) and enthalpy of solution, ∆Hsol (kJ/g). (2 marks T/I)
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work shown here:
ΔT = T2 - T1
= 17.77 - 20
= -2.23
Q = mcΔT
= (100g)(4.18 J/g°C)(-2.23)
= -932.14
ΔH = -Q/massNH4NH3
= 0.932kJ/3g
= 0.312kJ/g
SCH 4U0
Part 2: Unknown Compound III
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Name: _______________________
For the first table (beaker and electronic balance), select “solid” from the tab. Using the scroll
down menu select, unknown compound III.
Increase the mass of unknown compound III to 3 g and ensure the temperature is set at 20°C.
Click “Next”.
For the second table (calorimeter), select “liquid” from the tab. Using the scroll down menu
select, water, H2O.
Increase the mass of water to 100 g and ensure the temperature is set at 20°C. Record this
value as T1 in the table below.
Click on “Show Graph View”, “Show Microscopic View” and then Start.
Record the final temperature T2 located above the calorimeter in the table below.
Complete the discussion questions below.
Observations
Table 2: Quantitative Data of the Endothermic Reaction when 3.0 g of the salt dissolves in 100.0 mL of water
NH4NO3
3g
Unknown Compound III
3g
mass of surroundings (g)
100g
100g
Initial temperature (°C) of H2O
20°C
20°C
17.77°C
18.07°C
Change in temperature (°C) ΔT
-2.23°C
Heat of Reaction (kJ) Q
-0.932kJ
ΔT = T2 - T1
= 18.07 - 20
= -1.93°C
Q = mcΔT
= (100g)(4.18 J/g°C)(-
Mass of system (g)
T1
Final temperature (°C) of H2O
T2
Enthalpy of the Solution (kJ/g)
0.312kJ/g
1.93°C)
= -806.74J
= -0.806kJ
m ∆Hsol = - Q
∆Hsol = -Q/m
= 0.806kJ/3g
= 0.269kJ/g
QUESTIONS:
1. Calculate the heat of reaction for the unknown compound III. Show all your work. (3 marks
T/I)
SCH 4U0
Name: _______________________
Q = mcΔT
= (100g)(4.18 J/g°C)(-1.93°C)
= -806.74J
= -0.806kJ
2. Calculate the enthalpy of solution (kJ/g) for the unknown compound III. (2 marks T/I)
Use m ∆Hsol = - Q
m ∆Hsol = - Q
∆Hsol = -Q/m
= 0.806kJ/3g
= 0.269kJ/g
3. Determine the identity of the unknown compound III using Table 1 and your calculations. (1
mark T/I)
The identity of the unknown compound III using table 1 and my calculations is Ammonium Chloride
NH4Cl since the enthalpy of solution I got was 0.269kJ/g which is closest to the enthalpy given of
NH4CL, 0.277kJ/g, out of all the salts on the table.
4. Calculate a percentage difference between your experimental value and the accepted
value for enthalpy of solution of the identified salt. (2 marks T/I)
Percent Error = (Theoretical Yield - Actual Yield/Theoretical Yield) x 100%
= (0.277 - 0.269/0.277) x 100%
= 2.89%
The percentage difference between the experimental value and the accepted value was
2.89%.
SCH 4U0
Name: _______________________
5. Why is it important to control how much of the unknown compound or an ionic salt is used in a
medical hot or cold pack? (2 marks T/I)
It is important to control the amount of the unknown compound or ionic salt used in a medical hot or
cold pack because the amounts of each will determine the temperature that they cool down or
heat up to which directly affects the usability of a hot or cold pack. To make sure that it doesn't get
too hot or too cold and is just at the temperature they want, it is important to control the amount that
is in the hot or cold pack.
6. Suggest three sources of experimental error that could have occurred in this experiment if
conducted using a styrofoam calorimeter (three nested polystyrene cups). Do not include
human errors. (3 marks C)
One source of error with using a styrofoam calorimeter is that since the contents within the
calorimeter are not as insulated as the contents within the calorimeter used in this experiment, there
are more surroundings that can absorb or release thermal energy other than just the water. Also, we
are not sure if the process taking place is under constant pressure using a styrofoam cup. Lastly, the
temperature of the contents within the cup can be easily altered by the outside temperature since
the styrofoam calorimeter isn’t as well closed off as the calorimeter used in this experiment. These are
three sources of error that could possibly arise from using a styrofoam calorimeter.