
H
2
O
(l)
-285.83
OH
-
(aq)
-229.99
The following chemical reactions are witnessed in the experiment in the same
order as below:
NH
3
(aq)
+ H
+
(aq)
NH
4
+
(aq)
CH
3
COOH
(aq)
+ OH
-
(aq)
OAc
-
(aq)
+ H
2
O
(l)
NH
3
(aq)
+ CH
3
COOH
(aq)
NH
4
+
(aq)
+ OAc
-
(aq)
H
+
(aq)
+ OH
-
(aq)
H
2
O
(l)
2 H
+
(aq)
+ Mg
(s)
Mg
2+
(aq)
+ H
2(g)
Mg
(s)
+ 2 CH
3
COOH
(aq)
Mg
2+
(aq)
+ 2OAc
-
(aq)
+ H
2(g)
Cu
2+
(aq)
+ Zn
(s)
Zn
2+
(aq)
+ Cu
(s)
Ca
2+
(aq)
+ CO
3
2-
(aq)
CaCO
3
(s)
The purpose of this experiment is to determine the heat evolved or absorbed from
the given reaction and compare it to the theoretical. It is also done to practice the students
in the process of calorimetry.
ANSWERS TO QUESTIONS
From Manual:
1. Errors could arise from the improper set-up of the calorimeter. The Styrofoam ball
could be damaged which could cause improper insulation, therefore, improper
temperature readings. Wrong preparation of solutions would also cause discrepancies
from the experimental and theoretical heat of reaction. Using a wet test tube would also
yield incorrect results because it would lead to the misreading of the temperature.
Residue from chemicals used in the calibration or previous trials that are left from
improper washing would also affect experimental results.
2. a.) Because of the set-up used. We used an 8” or larger Styrofoam ball and a 6” test
tube. A lesser amount of solution might cause the improper reading of the temperature
because the bulb would not reach the solution. A big difference from 15ml might fill the
test tube and interfere with proper temperature readings.
b.) We need to know the concentration of reactants in order to solve for their moles and
further for the limiting reagent in the reaction. The moles of limiting reagent would then
be used to compute for the enthalpy of reaction.
c.) For the same reason of knowing the concentration of liquid reactants, knowing the
weight of metals would let us know the number of moles that we would use in the
reaction. Using stoichiometry, we could compute for the moles of the limiting reagent
which we could use for computing the enthalpy of reaction.
3. a.) mol = 0.2 L (0.5 M) = 0.1 moles HA
Enthalpy = -6 kJ/0.1 moles = -60 kJ/mol