Thermodynamics—Enthalpy of Reaction and Hess’s Law
November 16, 2011
Ryan Miller and Molly Brinser
Purpose
The purpose of this experiment is to verify Hess’s Law.
Theory
Four main topics were covered during this experiment including enthalpy
of reaction, heat of formation, Hess’s Law, and calorimetry.
Enthalpy of reaction, ΔHrxn, is the heat or enthalpy change for a chemical
reaction. This energy change is equal to the amount of heat transferred, at
constant pressure, in the reaction. This change represents the difference in
enthalpy of the products and the reactants and is independent of the steps in
going from reactants from products.
Heat of Formation (ΔH°f), which is also known as standard enthalpy of
formation, is defined as the change in enthalpy that accompanies the
formation of one mole of a compound from its elements with all substances in
their standard states. The enthalpy change for a given reaction can be
calculated by subtracting the enthalpies of formation of the reactants from the
enthalpies of formation of the products. This can be illustrated in this equation:
𝛥𝐻°𝑟𝑥𝑛 = Σ𝑛𝑝 𝛥𝐻°𝑓 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − Σ𝑛𝑟 𝛥𝐻°𝑓 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠).
During this experiment, Hess’s Law was a major concept that the lab
concentrated on. Hess’s Law states that if a reaction can be carried out in a
series of steps, the sum of the enthalpies for each step equals the enthalpy
change for the overall reaction. For example, the three chemical equations
used throughout the experiment were:
Equation 1: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Equation 2: NH4Cl(aq) + NaOH(aq) → NH3(aq) +NaCl(aq) + H2O(l)
Equation 3: NH3(aq) + HCl(aq) → NH4Cl(aq)
So, Equation 1 plus Equation 2(reversed) will equal Equation 3. In this case,
Hess’s Law gave the ΔH for Equation 3.
The last important concept covered in this lab is calorimetry. Calorimetry
is the science of measuring heat, and is based on observing the temperature
change when a body absorbs or discharges energy as heat. A calorimeter is
the device used experimentally to determine the heat associated with a
chemical reaction. During this experiment, a coffee cup calorimeter was used.
The equation 𝑞 = 𝑚𝑐𝛥𝑇 is associated with a calorimeter because it discovers the
amount of heat produced by multiplying the mass of the substance by the
specific heat and by the change in temperature. Since the coffee cup
calorimeter will absorb heat, the data must be corrected for that by solving for
qcal and ccal. The value of qcal will be the opposite of 𝑞𝐻2 𝑂 , and in order to find
the value of ccal, the equation 𝑞𝑐𝑎𝑙 = 𝛥𝑇𝑐𝑐𝑎𝑙 must be used.
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ΔHrxn & Hess’s Law
Page 2
Procedure
For Part 1, 50.0 mL of distilled water at room temperature will need to be
measured, its initial temperature recorded, and placed into the calorimeter.
Then, 50.0 mL of heated water will need to be measured, its initial temperature
record, and placed into the calorimeter with the room temperature water. With
the motor on and the magnet spinning, immediately insert the thermometer
probe and collect the data. For Part 2, get 50.0 mL of HCl and NaOH and
record their initial temperatures. Then add them to the calorimeter, cover the
calorimeter, insert the temperature probe, and start collecting the data. Do this
procedure for Reaction 2 and Reaction 3. Then, the heat change and enthalpy
of reaction are to be calculated. The value of ΔH is to be found two ways,
through experimental data and using Hess’s Law. These values are then
compared to find the percent error of the experiment.
Data
Part 1: Determination of the Heat Capacity of the Calorimeter
Initial Temperature
50.0 mL H2O—room temperature:
22.4 °C
50.0 mL H2O—heated:
68.8 °C
Mixing Data
Time (seconds) Temperature (°C)
20
44.1
40
43.9
60
43.7
80
43.5
100
43.4
120
43.4
140
43.4
160
43.2
180
43.0
Calculated Values
Tmix = 44.04 °C
Tavg = 45.6 °C
qcal = +652 J
Ccal = 30.1 J/°C
Part 2: Determination of Heats of Reaction
Reaction 1
Initial Temperature
50.0 mL 2.0 M HCl
22.9 °C
50.0 mL 2.0 M NaOH
22.7 °C
Mixing Data
Time (seconds) Temperature (°C)
20
35.8
40
35.8
60
35.8
80
35.8
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ΔHrxn & Hess’s Law
100
35.7
120
35.7
140
35.6
160
35.6
180
35.6
Calculated values
Tmix = 35.87 °C
qrxn = -6020 J
Reaction 2
Initial Temperature
50.0 mL 2.0 M NH4Cl
23.0 °C
50.0 mL 2.0 M NaOH
22.8 °C
Mixing Data
Time (seconds) Temperature (°C)
20
23.8
40
23.8
60
23.8
80
23.8
100
23.8
120
23.8
140
23.8
160
23.8
180
23.8
Calculated Values
Tmix = 23.77 °C
qrxn = -401 J
Reaction 3
Initial Temperature
50.0 mL 2.0 M NH3
22.8 °C
50.0 mL 2.0 M HCl
22.4 °C
Mixing Data
Time (seconds) Temperature (°C)
20
44.1
40
43.9
60
43.7
80
43.5
100
43.4
120
43.4
140
43.3
160
43.2
180
43.0
Calculated Values
Tmix = 27.42 °C
qrxn = -2160 J
Page 3
ΔH = -60 kJ/mol
ΔH = -4.0 kJ/mol
ΔH = -22 kJ/mol
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ΔHrxn & Hess’s Law
Page 4
Calculations
Part 1
1. Tmix = 44.04 °C (from graph)
22.4 °𝐶+68.8 °𝐶
2. 𝑇𝑎𝑣𝑔 =
= 45. 6 °𝐶
2
3. 𝑞𝐻2 𝑂 = 𝑚𝑐(𝑇𝑚𝑖𝑥 − 𝑇𝑎𝑣𝑔 )
𝐽
𝑞𝐻2 𝑂 = (100.0 𝑔)(4.18 ⁄𝑔 °𝐶 )(44.04 °𝐶 − 45.6 °𝐶)
𝑞𝐻2 𝑂 = −652 𝐽
𝑞𝑐𝑎𝑙 = +652 𝐽
4. 𝑞𝑐𝑎𝑙 = 𝛥𝑇𝑐𝑐𝑎𝑙
𝑞𝑐𝑎𝑙 = (𝑇𝑚𝑖𝑥 − 𝑇𝑐𝑜𝑙𝑑 )𝑐𝑐𝑎𝑙
+652 𝐽 = (44.04 °𝐶 − 22.4 °𝐶)𝑐𝑐𝑎𝑙
+652 𝐽 = (21.64 °𝐶)𝑐𝑐𝑎𝑙
𝐽
30.1 ⁄°𝐶 = 𝑐𝑐𝑎𝑙
Part 2
Reaction 1
1. Tmix = 35.87 °C (from graph)
22.9 °𝐶 + 22.7 °𝐶
2. 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒𝑠 =
= 22.8 °𝐶
2
𝑞𝑟𝑥𝑛 = −[𝑚𝑐(𝑇𝑚𝑖𝑥 − 𝑇𝑖 ) + 𝑐𝑐𝑎𝑙 (𝑇𝑚𝑖𝑥 − 𝑇𝑖 )]
𝐽
𝐽
𝑞𝑟𝑥𝑛 = − [(103 𝑔) (4.18 ⁄𝑔 °𝐶 ) (35.87 °𝐶 − 22.8 °𝐶) + (30.1 ⁄°𝐶 ) (35.87 °𝐶 − 22.8 °𝐶)]
𝑞𝑟𝑥𝑛 = −6020 𝐽
1 𝑘𝐽
−6.02 𝑘𝐽
𝑘𝐽
3. 𝛥𝐻 = −6020 𝐽 𝑥
=
= −60. ⁄𝑚𝑜𝑙
1000 𝐽
.10 𝑚𝑜𝑙𝑒𝑠
Reaction 2
*See calculation answers in the Data section
Reaction 3
*See calculation answers in the Data section
Part 3
1. The net ionic equations of the three chemical equations used throughout
the experiment
Equation 1:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l)
*OH-(aq) + H+(aq) → H2O(l)*
Equation 2:
NH4Cl(aq) + NaOH(aq) → NH3(aq) +NaCl(aq) + H2O(l)
NH4+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → NH3(aq) + Na+(aq) + Cl-(aq) + H2O(l)
*NH4+(aq) + OH-(aq) → NH3(aq) + H2O(l)*
Equation 3:
NH3(aq) + HCl(aq) → NH4Cl(aq)
NH3(aq) + H+(aq) + Cl-(aq) → NH4+(aq) + Cl-(aq)
*NH3(aq) + H+(aq) → NH4+(aq)*
***Equation 1 – Equation 2(reversed) = Equation 3
RRM
ΔHrxn & Hess’s Law
2. Equation 1 ΔH = -60. kJ/mol
Equation 2 ΔH = -4.0 kJ/mol
OH-(aq) + H+(aq) → H2O(l)
NH3(aq) + H2O(l) → NH4+(aq) + OH-(aq)
NH3(aq) + H+(aq) → NH4+(aq)
Page 5
ΔH = -60. kJ/mol
ΔH = +4.0 kJ/mol
ΔH = -56 kJ/mol
Error Analysis
|𝐴𝑐𝑡𝑢𝑎𝑙 − 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙|
|−22 − (−56)|
𝑥 100 =
𝑥 100 = −61 %
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙
−56
There were many errors that could have affected our results in this
experiment but three main errors were all of the initial temperatures were
different, the group had to assume the densities of the solutions, and the group
had to assume that the molarity of all solutions was 2.0 M.
The first error was that the initial temperature measurements for all
solutions were not the same. For example, in the first reaction of Part 2, the initial
temperature of hydrochloric acid was 22.9 °C and the initial temperature of
sodium hydroxide was 22.7 °C. In order to find the initial temperature, the
average of these two numbers had to be taken, which could have caused the
calculations to be slightly higher or slightly lower than what they should have
been. However, this error would have a negligible effect on the data since
there is not a large difference between the numbers.
The second error that occurred in the experiment was that the group had
to assume the densities of the solutions. In the calculation selection of the lab
handout, it stated to assume the density of the solutions is 1.03 g/mL. Instead of
assuming this value, the group could have calculated it to have a more
accurate answer. Like the error before, it would have a small effect on the data
that was produced.
The third error that occurred in the experiment was that the group
had to assume that the molarity of all of the solutions were 2.0 M.
All of these solutions were student prepared and could have been
made improperly since they were the first solutions the class made.
Mixing solutions that are not close to 2.0 M would have reduced the
overall molarity of the mixed solutions. If the solutions’ molarity was
less than 2.0 M, then it would increase the final answer, but if the
solutions’ molarity was greater than 2.0 M, then the final answer
would have decreased. This error would have been the biggest
concern during the lab since it had the largest effect on the data.
% 𝑒𝑟𝑟𝑜𝑟 =
Conclusion
The purpose of this experiment was not met because the group was
unable to verify Hess’s Law. From the calculations, ΔH3 was found to be -22
kJ/mol, but when Hess’s Law was used, ΔH3 was found to be -56 kJ/mol. The
percent error calculated from these values ended up being -61%. Since the
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ΔHrxn & Hess’s Law
Page 6
calculated answers were not nearly as close as they should have been, the
group was unable to verify Hess’s Law.
Questions
1. The direct method worked better to find ΔH3 than the indirect method
because in order to use Hess’s Law, the data had to be nearly error free.
If the group does have an error in the calculations, than the Hess’s Law
answer will not match the answer the group calculated. The Hess’s Law
answer had a -61% error compared to the calculated answer. Therefore,
the direct method worked better than the indirect method.
2. Hess’s Law states that if a reaction is carried out in a number of steps, ΔH
for the overall reaction is equal to the sum of the ΔH’s from each
individual step.
3. ΔH means the heat or enthalpy change for a chemical reaction. This
energy change is equal to the amount of heat transferred, at constant
pressure, in the reaction. This change represents the difference in
enthalpy of the products and the reactants and is independent of the
steps in going from reactants to products.
4. The true initial temperature is found by using a modified linear fit because
when excluding any points that are skewed, it is giving the results a more
accurate reading. The first couple points that were skewed are due to
incomplete mixing and lack of equilibrium with the thermometer.
5. All of the solutions did not have the same initial temperature because
they were not contained in the same area of the room. These areas
could have a slightly different temperature causing the different initial
temperatures. Also, transfer of heat from holding the graduated cylinder
could have increased the initial temperature, and depending on how
long the graduated cylinder was held, could have determined how high
the initial temperature rose.
Calculations:
Part 1
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ΔHrxn & Hess’s Law
Page 7
5. Tmix = 44.5 °C (from graph)
22.7 °𝐶+62.6 °𝐶
6. 𝑇𝑎𝑣𝑔 =
= 42.65 °𝐶
2
7. 𝑞𝐻2 𝑂 = 𝑚𝑐(𝑇𝑚𝑖𝑥 − 𝑇𝑎𝑣𝑔 )
𝐽
𝑞𝐻2 𝑂 = (100.0 𝑔)(4.18 ⁄𝑔 °𝐶 )(44.5 °𝐶 − 42.65°𝐶)
𝑞𝐻2 𝑂 = −773.3 𝐽
𝑞𝑐𝑎𝑙 = +773 𝐽
8. 𝑞𝑐𝑎𝑙 = 𝛥𝑇𝑐𝑐𝑎𝑙
𝑞𝑐𝑎𝑙 = (𝑇𝑚𝑖𝑥 − 𝑇𝑐𝑜𝑙𝑑 )𝑐𝑐𝑎𝑙
+773 𝐽 = (44.5 °𝐶 − 22.7 °𝐶)𝑐𝑐𝑎𝑙
+773 𝐽 = (21.8°𝐶)𝑐𝑐𝑎𝑙
𝐽
35.5 ⁄°𝐶 = 𝑐𝑐𝑎𝑙
Part 2
Reaction 1
4. Tmix = 36.82 °C (from graph)
24.5 °𝐶 + 23.9 °𝐶
5. 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒𝑠 =
= 24.2 °𝐶
2
𝑞𝑟𝑥𝑛 = −[𝑚𝑐(𝑇𝑚𝑖𝑥 − 𝑇𝑖 ) + 𝑐𝑐𝑎𝑙 (𝑇𝑚𝑖𝑥 − 𝑇𝑖 )]
𝐽
𝐽
𝑞𝑟𝑥𝑛 = − [(103 𝑔) (4.18 ⁄𝑔 °𝐶 ) (36.82 °𝐶 − 24.2 °𝐶) + (30.1 ⁄°𝐶 ) (36.82 °𝐶 − 24.2 °𝐶)]
𝑞𝑟𝑥𝑛 = −6020 𝐽
6. 𝛥𝐻 = −6020 𝐽 𝑥
1 𝑘𝐽
1000 𝐽
=
−6.02 𝑘𝐽
.10 𝑚𝑜𝑙𝑒𝑠
= −60.
𝑘𝐽⁄
𝑚𝑜𝑙
Reaction 2
*See calculation answers in the Data section
Reaction 3
*See calculation answers in the Data section
Part 3
3. The net ionic equations of the three chemical equations used throughout the experiment
Equation 1:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l)
*OH-(aq) + H+(aq) → H2O(l)*
Equation 2:
NH4Cl(aq) + NaOH(aq) → NH3(aq) +NaCl(aq) + H2O(l)
NH4+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → NH3(aq) + Na+(aq) + Cl-(aq) + H2O(l)
*NH4+(aq) + OH-(aq) → NH3(aq) + H2O(l)*
Equation 3:
NH3(aq) + HCl(aq) → NH4Cl(aq)
NH3(aq) + H+(aq) + Cl-(aq) → NH4+(aq) + Cl-(aq)
*NH3(aq) + H+(aq) → NH4+(aq)*
***Equation 1 – Equation 2(reversed) = Equation 3
4. Equation 1 ΔH = -60. kJ/mol
RRM
ΔHrxn & Hess’s Law
Equation 2 ΔH = -4.0 kJ/mol
OH-(aq) + H+(aq) → H2O(l)
ΔH = -60. kJ/mol
NH3(aq) + H2O(l) → NH4+(aq) + OH-(aq) ΔH = +4.0 kJ/mol
NH3(aq) + H+(aq) → NH4+(aq)
ΔH = -56 kJ/mol
Page 8