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Experiment
9
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Heats of
Reaction and Hess's Law
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Overview
Whenever a chemical reaction takes
place, there is a change in the energy of the system. For many
reactions, this energy change takes the form of a transfer of
heat energy either into (endothermic) or out of (exothermic) the
system. In this experiment, you determine the heat flows for three
processes: the reaction of an acid (HCl) with a base (NaOH); the
reaction of an active metal (Mg) with an acid (HCl); and the
dissolving of a salt (KBr).
The write up for this experiment is
very long, and may seem confusing. The steps in the calculations are
very systematic and straightforward, however, and are based in
sequence on the data you have recorded.
Temperature changes (DT's)
are determined from the graphs of your data. These temperature
changes are needed at many places in the calculations, so it is
important that you make your graphs first before attempting
the calculations.
Watch the Pre-Lab video for this experiment. (7 Mb file:
Broadband connections only; Real Player required)
Data:
This is a very long lab report. To
simplify this help file, the data for each section of the experiment
will be presented separately along with the calculations.
Calculations:
Part A: Neutralization of HCl and NaOH
Data: Page 89, Part IA
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Concentration of
Reagents |
NaOH
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HCl
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0.950 M
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0.981 M
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Time
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NaOH
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HCl
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1.00 |
20.1 |
20.2 |
2.00 |
20.2 |
20.3 |
3.00 |
20.3 |
20.4 |
4.00 (mix)
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mix solutions
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mix solutions
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5.00 |
27.4 |
same (mixed)
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6.00 |
27.3 |
same (mixed)
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7.00 |
27.2 |
same (mixed)
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Based on your data, you prepare a graph
for Trial 1 (example shown on Page 53), and determine from your
graph the change in temperature during the reaction (DT).
For the data above, DT
would be 7.1oC for my Trial 1.
Calculations, Page 91, Part IIA
1. Temperature from graphs
This is the information you read off
your graphs for the 4.00 minute point where the solutions were
mixed. You need the extrapolated (from the graph) temperatures of
the separate solutions as well as the extrapolated (from the graph)
temperature of the mixture.
These extrapolations allow for the fact
that the solutions before mixing were very gradually warming up
with time, whereas the combination after mixing was very gradually
cooling off with time. The extrapolations try to estimate
what the temperatures would have been at 4.00 minutes when the
solutions were actually mixed.
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Temperature from
Graphs |
Trial 1
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Trial 2 (no data
shown) |
NaOH and HCl at
mixing |
20.4 |
21.2 (no data shown
above) |
Reaction mixture
at mixing |
27.5 |
28.2 (no data shown
above) |
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For Trial 1,
DT is 7.1oC.
For Trial 2, DT
is 7.0oC.
2. Heat Change of Mixed Solutions
(Trial 1)
For a transfer of heat energy to or
from a material, the quantity of heat transferred is dependent on
the mass of material, the temperature change of the
material, and the specific heat capacity of the material (a
constant).
Since we measured out the solutions
used in this part of the experiment by volume (using a
graduated cylinder), we will have to convert from the volume
of solution taken to the mass of the solution, using the
density of the solution (1.02 g/mL) given on Page 54. The specific
heat capacity is also given on Page 54 as 3.93 Joules/goC
For Trial 1, calculate the Heat Change
for the Mixed Solutions, then click here to
check your answer.
3. Heat Change of Calorimeter
It is noted in the discussion for the
experiment that no calorimeter is completely inert as regarding its
ability to transfer heat to and from itself. Although plastic foam
is a good insulator, it is not a perfect insulator.
When the reaction occurs in the calorimeter, a small portion of the
heat generated by the reaction is absorbed by the calorimeter.
The amount of heat absorbed by a
calorimeter is a function of the calorimeter itself and is called
the heat capacity of the calorimeter, Ccal. We
have previously measured the heat capacities of plastic foam
calorimeters such as the one you used in this experiment and have
found the average value of such heat capacities to be 41.7 Joules/oC.
This number means that anytime a reaction is performed in such a
calorimeter, for each degree increase in the temperature of the
contents of the calorimeter, the calorimeter itself absorbs 41.7
Joules. For Trial 1 of my data, in which the temperature of the
reaction mixture increased by 7.1 oC, calculate the
quantity of heat absorbed by the Calorimeter, then click
here to check your result.
4. Heat of Reaction
The heat of reaction represents the
total quantity of heat energy transferred by the reaction. This
energy occurred in two parts: the heat change that we were able to
measure for the solutions, as well as the heat energy that was lost
to warming up the calorimeter itself by 7.1 oC. The
equation given in this part on Page 59 of the lab manual takes some
explanation. The equation reads
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Heat of reaction +
Heat Change of Solutions + Ccal(DTcal)
= 0 |
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This is nothing more than a
mathematical statement of the First Law of Thermodynamics: during
any process, the total energy of the universe is constant. In other
words, whatever changes in energy occurred in the chemicals
themselves, in the solutions, and in the calorimeter must total to a
net change of zero. Whatever amount of heat energy is given
off by the actual chemicals when they react with each other is
exactly equal to the total amount of heat energy transferred to the
water in which the chemicals were dissolved and to the
calorimeter. Rearranging the equation above to solve for the
Heat of Reaction gives the following:
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Heat of Reaction = 0
- [Heat Change of Solutions + Ccal(DTcal)]
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This rearranged equation says that the
Heat of Reaction is just the negative of the sum of the heat
change of the solutions and the heat change of the calorimeter.
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Heat of Reaction =
-[2.8 kJ + 0.30 kJ] = -[3.1 kJ] = -3.1 kJ |
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We took 50.0 mL of 0.950 M NaOH
and combined it with 50.0 mL of 0.981 M HCl. The reactant
which is present in the lesser amount on a mole/stoichiometric basis
is the limiting reactant. For our experiment, the NaOH is the
limiting reactant. Some of the HCl is present in excess and does not
react. For the data above, calculate the number of moles of NaOH
used, then click here to check your result.
6. Molar Heat of Neutralization (per
mole of water formed)
The heat of reaction calculated in Part
4 above (-3.1 kJ) was for the specific amounts of materials used
in the experiment. Heats of Reaction are most commonly tabulated
in handbooks in terms of the Molar heat of reaction: the heat
of reaction when 1 mole of a particular reactant or product is
involved.
We want here to calculate the Molar
Heat of Reaction per mole of water formed in the reaction. In Part 5
above, we calculated that the limiting reactant was the NaOH used.
Since 0.0475 moles of NaOH reacted completely, from the balanced
chemical equation, 0.0475 moles of water must have been produced.
For the data above for Trial 1, calculate the Molar heat of reaction
(per mole of water formed), and then click here
to check your answer.
Part B: Reaction of Mg and HCl
Data: Page 90, Part IB
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Concentration of
reagents: HCl = 0.981 M |
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Mass Data
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Trial 1
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Trial 2
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Mass beaker + Mg
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11.0385 g
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10.9163 g
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Mass beaker empty
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10.7891 g
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10.6675 g
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Time
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Trial 1
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Trial 2
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1.00 |
20.1oC
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21.2oC
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2.00 |
20.2oC
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21.3oC
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3.00 |
20.3oC
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21.4oC
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4.00 |
add Mg to HCl
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add Mg to HCl
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5.00 |
26.2oC
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27.3oC
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6.00 |
26.7oC
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27.8oC
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7.00 |
27.2oC
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28.3oC
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8.00 |
26.5oC
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27.6oC
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9.00 |
26.0oC
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27.1oC
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Calculations: Page 92, Section IIB
1. Temperature from Graphs
Since this experiment was first
written, we have discovered that it is not possible to determine the
DT in
the same manner as in Section IIA. In Part A, the temperature change
was essentially instantaneous at the time of mixing: your
graphs for Part A show an immediate increase at the 4.00 minute
mark. Consider the data for Part B above, however. The temperature
continues to rise past the 4.00 minute mark. This indicates
that it must take a few minutes for the magnesium to react
completely with the acid and dissolve. To calculate
DT for this
Part of the reaction, you should use the following two temperatures
for each trial:
- the temperature of the HCl at the
4.00 minute mark (extrapolated from your graph). For Trial 1 in
the data table above, this temperature would be 20.4oC.
- The highest temperature reached
after the magnesium has been added. For Trial 1 in the data table
above, this temperature would be the temperature at the 7.00
minute point: 27.2oC. At this point, the greatest
amount of heat has been transferred from the reactants to the
water and the calorimeter.
- Therefore
DT for Trial
1 above would be calculated as
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DT = (27.2oC
- 20.4oC) = 6.8oC |
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2. Heat Change of Reaction Mixture
As mentioned earlier, for a transfer of
heat energy to or from a material, the quantity of heat transferred
is dependent on the mass of material, the temperature
change of the material, and the specific heat capacity of
the material (a constant).
Since we measured out the solutions
used in this part of the experiment by volume (using a
graduated cylinder), we will have to convert from the volume
of solution taken to the mass of the solution, using the
density of the solution (1.02 g/mL) given on Page 54.
The specific heat capacity is also
given on Page 54 as 3.93 Joules/goC.
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Heat change Trial 1
= (50.0 mL)(1.02 g/mL)(3.93 Joules/goC)(6.8oC)
= 1363 Joules = 1.4 kJ |
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3. Heat Change of Calorimeter
This is the same consideration we had
in Part A: the calorimeter itself absorbs part of the energy change.
The specific heat capacity of the calorimeter is given as 41.7
Joules/oC.
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Heat Change of
Calorimeter = (41.7 J/oC)(6.8oC) = 283.6
Joules = 0.28 kJ |
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4. Heat of Reaction for Trial 1
The equation given in this section is
again written as a statement of the conservation of energy
principle: that the total of all the energy changes in the universe
when a process takes place is zero. Solving this equation for the
Heat of Reaction gives
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Heat of reaction =
-[Heat Change of Reaction Mixture + Heat Change of Calorimeter]
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Heat of
Reaction = -[1.363 kJ + 0.284 kJ] = -1.647 kJ = -1.6 kJ
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5. Moles of Limiting Reactant
The Magnesium is the limiting reactant.
For Trial 1, the mass of Mg used is (see data table)
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Mass of Mg = 11.0385
g - 10.7891 g = 0.2494 g Mg |
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Moles of Mg =
(0.2494 g)(1 mole/24.31 g) = 0.01026 moles Mg |
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6. Molar Heat of Reaction (per mole of
Mg reacted)
The Heat of Reaction calculated in Part
4 is for the specific amounts of chemicals taken for the experiment.
Heats of Reaction are usually tabulated on a mole basis
(i.e., per mole of substance reacting). So we want to calculate the
Heat of Reaction per mole of Mg.
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Molar Heat of
Reaction = (-1.647 kJ/0.01026 mol) = -160.5 kJ/mole = -1.6 X 102
kJ/mole |
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Notice that the heat of reaction is a
negative number, which means the process is exothermic. The
temperature of the calorimeter and the liquid in it increased when
the reaction took place as energy was transferred from the reactant
substances.
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Mass Data
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Trial 1
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Trial 2
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Mass of beaker +
salt |
16.172 g
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16.572 g
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Mass of beaker
empty |
11.051 g
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11.523 g
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Time
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Trial 1
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Trial 2
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1.00 |
21.2oC
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22.1oC
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2.00 |
21.3oC
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22.2oC
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3.00 |
21.4oC
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22.3oC
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4.00 |
mix |
mix |
5.00 |
15.4oC
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16.3oC
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6.00 |
14.9oC
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15.1oC
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7.00 |
15.7oC
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16.6oC
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As mentioned in the discussion, similar
to the magnesium/HCl reaction earlier, there is a time delay in the
maximum temperature change because of the time required for the salt
to dissolve. As the final temperature, we take the minimum
temperature reached by the system. For the data above:
Minimum temperature
reached: Trial 1 - 14.9oC Trial 2 - 15.1oC
Calculations: Page 94, Part IIC
1. Formula of Salt
KBr was used.
2. Mass of Salt Dissolved
For Trial 1
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Mass of Salt for
Trial 1 = (16.172 g - 11.051 g) = 5.121 g |
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Moles of KBr for
Trial 1 = (5.121 g)(1 mole/119.0 g) = 0.04030 mole KBr
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4. Temperature of Solution at Mixing
(extrapolated from graph)
This temperature is read from the
extrapolation of the left hand portion of your graph to the 4.00
minute point. Since the temperature of the liquid in the calorimeter
in my data is rising slightly, the temperature of the liquid at the
4.00 minute point is extrapolated to be 21.5oC.
5. DT
between Minimum Temperature Reached and Temperature Before Mixing
For Trial 1, the liquid in the
calorimeter was at 21.5oC at the 4.00 minute mark, and
then decreased to a minimum temperature of 14.9oC. This
means that DT
for the process is
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DT = 14.9oC
- 21.5oC = -6.6oC |
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6. Heat Change of Solution
As mentioned earlier, for a transfer of
heat energy to or from a material, the quantity of heat transferred
is dependent on the mass of material, the temperature
change of the material, and the specific heat capacity of
the material (a constant).
Since we measured out the solutions
used in this part of the experiment by volume (using a
graduated cylinder), we will have to convert from the volume
of solution taken to the mass of the solution, using the
density of the solution (1.02 g/mL) given on Page 54.
The specific heat capacity is also
given on Page 54 as 3.93 Joules/goC.
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Heat Change Solution
= (50.0 mL)(1.02 g/mL)(3.93 Joules/goC)(-6.6oC)
= -1323 Joules = -1.3 kJ |
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7. Heat Change of Calorimeter
This is the same consideration we had
in Parts A and B: the calorimeter itself is involved in part of the
energy change. The specific heat capacity of the calorimeter is
given as 41.7 Joules/oC.
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Heat Change of
Calorimeter = (41.7 Joules/oC)(-6.6oC) =
-275.2 Joules = -0.28 kJ |
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8. Total Heat of Solution
Let's rearrange the equation given to
solve for the heat of solution (as we did above in Parts A and B)
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Heat of Solution =
-[Heat Change of Contents + Heat Change of Calorimeter]
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Heat of Solution =
-[(-1.323 kJ) + (-0.275 kJ)] = -[-1.598 kJ] = 1.598 kJ = 1.6 kJ
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Molar Heat of
Solution = (1.598 kJ/0.04030 mole KBr) = 39.65 kJ/mole = 40
kJ/mole |
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