Lab Report 5
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Experiment 5: Qualitative and Quantitative Aspects of Equilibrium
Report Sheet
You are welcome to seek guidance on the content of this lab report from your TA during your lab session
and from the lab coordinator during lab office hours. You are also welcome to work constructively with
your peers on the general content and understanding of the material. However, the work submitted here
in this report sheet must be your own. For more details on academic integrity and potential penalties see
the
Code of Behaviour on Academic Matters
.
I certify that this submitted laboratory report represents entirely my own efforts. I have read and
understand the University of Toronto policies regarding, and sanctions for, academic honesty.
I regret that I violated the Code of Behaviour on this assessment and would like to admit that
now so that I can take responsibility for my mistake.
A. Heat capacity of calorimeter
1.
Insert your two plots of temperature vs. time for the calibration of the calorimeter and
complete the table.
Plot #1
Plot #2
Run #
Initial Temperature of HCl (
o
C)
Initial Temperature of NaOH (
o
C)
Extrapolated Final Temperature (
o
C)
1
18.6
20.4
26.6
2
19.5
20.2
26.3
2. Was your solution acidic or basic after the run was complete? What does this tell you about
the reaction?
After the run was complete the solution was highly basic at pH 12. This means the reaction was a strong
acid-base reaction and is exothermic which can be seen from the rise in the temperature at the beginning.
3. Show your calculation of the heat capacity of the calorimeter (for run #1).
C = n/V
∴
n = CV
n
HCl = (1.0 M) (0.75)
n
NaOH = (1.1 M) (0.75)
= 0.075 mol
= 0.0825 mol
(HCl = limiting reagent)
qReaction =
nΔH
0
H3O + OH
= (0.075 mol) (-58985 J/mol)
= 4423.88
(
-58985 J/mol is found using the table in lab manual page 5-7 for the initial temperature 18.6
o
C )
HEAT CAPACITY OF CALORIMETER -
nΔH
0
H3O + OH = - (C
CAL
ΔT
CAL
+
C
WATER
V
ACID
ΔT
ACID
+
C
WATER
V
BASE
ΔT
BASE
∴
- (4.13
J/
o
C
-1
mL
-1
* 28.1
o
C + 4.13
J/
o
C
-1
mL
-1
* 75 mL* 28.1
o
C + 4.183
J/
o
C
-1
mL
-1
* 75 mL * 28.1
o
C
= - 5431.91
nΔH =
- C
CAL
ΔT - 2C
ACID
VΔT
∴
- 4423.88 = - C
CAL
(6.9) - 2(4.18)(75 mL)(6.9)
∴
- 4423.88 = - C
CAL
(6.9) - 4326.3
∴
C
CAL
(6.9) / (6.9) = - 4326.3 + 4423.88 / 6.9
∴
C
CAL
= 14.14 J/
o
C
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3. Complete the table:
Run #
Heat Capacity of Calorimeter
1
14.14 J/
o
C
2
33.1 J/
o
C
Average
23.62 J/
o
C
B. Enthalpy of formation of magnesium oxide
4. Insert your plot of temperature vs time for the Mg trial.
5. Complete the table:
Reactant
Mass (g)
Initial Temperature
of HCl (
o
C)
Extrapolated Final
Temperature (
o
C)
ΔH∞ for the reaction
with HCl
Mg
0.28
20.5
26.9
- 2707.42 J mol
-1
6. Show your calculation of ΔH∞ for the reaction Mg + 2HCl
→
MgCl
2
+ H
2
Limiting reagent - Mg, nMg = 0.0012 mol
0.012*ΔH = -((28.1*4.13) + (4.183*150*4.13))
ΔH = - 2707.42 J mol
-1
7. Insert your plot of temperature vs time for the MgO trial.
8. Complete the table:
Reactant
Mass (g)
Initial
Temperature
of HCl (
o
C)
Extrapolated Final Temperature (
o
C)
ΔH∞ for the reaction
with HCl
MgO
0.28
20.5
26..9
- 43487.94 J mol
-1
9. Show your calculation of ΔH∞ for the reaction, MgO + 2HCl
→
MgCl
2
+ H
2
O
Limiting reagent - MgO,
nMgO = 0.0018mol
0.018*ΔH = -((28.1*6.9) + (4.183*150*6.9))
ΔH = - 43487.94 J mol
-1
10. Use your experimental data to calculate the enthalpy of formation, ΔH
f
∞, of MgO.
To help with the calculation, first, summarize the relevant ΔH∞ values in the table below. Note
that some are experimental (meaning you determined them using your data) and some are from
the literature.
Reaction
Δ
H
∞
(kJ/mol)
Mg(s) + 2HCl(aq) → MgCl
2
(aq) + H
2
(g)
- 2707.42 J mol
-1
MgO(s) + 2HCl(aq) → MgCl
2
(aq) + H
2
O(l)
- 43487.94 J mol
-1
H
2
(g) + ½ O
2
(g) H
2
O (l)
- 285826 J mol
-1
11. Then write the chemical equation for the enthalpy of formation of MgO:
Mg (s) + ½ O
2
(g) = MgO (s)
12. Show your calculation of the enthalpy of the formation of MgO:
We need Mg and ½ O
2
on the left side and MgO on the right -
It is possible to deduce the enthalpy of formation by adding the sum of the enthalpy reaction between Mg
and HCl and the negative enthalpy of the reaction between MgO and HCl, as we need MgO towards the
right and the enthalpy of formation of water.
ΔH
f
= (- 2707.42) + (-285826) - (-43487.94)
= -245045.48 J mol
-1
≈ -245045 J mol
-1
≈ 254.045 KJ mol
-1
13. How does your experimental value for the enthalpy of formation of MgO compare with the
literature value? How can you account for any discrepancies?
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The enthalpy of formation for MgO in the literature is -601 kJ mol, which is rather similar to the value we
discovered. This mismatch may be the result of random error brought on by erroneous temperature
readings, the heat lost to the environment when adding ingredients or setting up the next step of the
experiment, or both.
14. List any references used.
“Thermodynamics Enthalpy of Formation Magnesium Oxide - Lab Report Experiment
Thermodynamics:”
StuDocu
,
https://www.studocu.com/row/document/quaid-i-azam-university/chemistry/thermodynamics-enthalpy-of-
formation-magnesium-oxide/17338623.
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