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1st RDR - Chem 17TRANSCRIPT
EXPERIMENT NO. 1
Calorimetry
Avila, Romar Angelo M.
Department of Mining, Metallurgical and Materials Engineering, College of Engineering, University of the Philippines, Diliman, Quezon City, Philippines
Introduction
In chemistry, scientists often encounter the problem of having to measure heat changes of
systems. A calorimeter is used in these instances. By knowing either the mass and specific heat or
the heat capacity of the calorimeter, plus the change in temperature, one can solve for the heat of
reaction, qrxn [1]. There are two types of calorimeter: the constant-pressure or bomb calorimeter
used for combustion reactions and the constant-volume or ‘coffee cup’ calorimeter used to for
noncombustion reactions such as neutralization and oxidation-reduction (redox) reactions [2].
In the experiment, a constant-volume calorimeter is used to measure heat changes in four types of
reactions: neutralization, acid-active metal reaction, metal displacement, and precipitation
reaction. The reactions involving acids and bases will also vary in terms of the relative strengths
of the compounds in solution.
[1] Petrucci, R.H.; Herring, F.G.; Madura, J.D.; Bissonnette, C. General Chemistry: Principles
and Modern Applications, 10th ed; Pearson Education Inc: Philippines, 2012; pp 248-251
[2] Chang, R. Chemistry, 10th ed; McGraw-Hill Publishing: New York, NY, USA, 2010; pp
247-249
Answers to Questions Posted on the Bulletin Board
Calibration of the Calorimeter
1. Give the net ionic equation of the reaction used to calibrate the calorimeter.
H(aq) + OH(aq) H2O(l)
a. Is the reaction endothermic or exothermic?
The reaction is exothermic, since heat was released to the surroundings by the system
during the reaction.
b. Which is the limiting reactant?
(10mL NaOH)(1L/1000mL)(1mol/1L) = 0.010mol NaOH
(5mL HCl) (1L/1000mL)(1mol/1L) = 0.005mol HCl
HCl is the limiting reactant, since there are only 0.005mol of it compared to 0.01mol of
NaOH.
c. How much (in moles) limiting reactant was used?
As seen from the calculations above, the number of moles of HCl used in calibration is
0.005mol.
d. How much heat was generated (or absorbed) by the reaction?
(-55.8kJ/mol)(0.005mol)(1000J/1kJ) = 297J
279J were released by the system to its surroundings.
2. Relate the sign of the ΔT to the ΔH of the reaction used for calibration.
The sign of ΔT determines the sign of the qcal, which is the opposite sign of qrxn:
qcal = mcΔT; qcal = -qrxn
So, -qrxn = mcΔT. In turn, -qrxn = -nΔH. We can form another relation, which is:
-nΔH = mcΔT
Since n (number of moles) is always positive, the negative sign is multiplied to ΔH.
Therefore, we can conclude that the sign of ΔT will be directly proportional but opposite the
sign of ΔH.
3. What is the heat capacity of the calorimeter? Relate its sign to the sign of the ΔT.
The heat capacity of the calorimeter (Ccal) is the amount of heat energy needed to raise the
temperature of the calorimeter by 1ºC. This takes into account both the specific heat and the
mass of the calorimeter. In the experiment, the author of this paper got a value of 79.7J/ ºC.
The Ccal and ΔT are used to calculate the qcal:
qcal = CcalΔT
There is no straightforward relationship between the signs of Ccal and ΔT. However, we do
know that Ccal is always a positive value, since its factors (mass and specific heat) are always
positive.
4. In the appendix, show the derivation to obtain the equation used to calculate the heat capacity
of the calorimeter.
Determination of Heats of Reaction
5. Give the net ionic equation for each reaction.
a. NH3(aq) + H(aq) NH4(aq)
b. CH3COOH(aq) + OH(aq) H2O(l) + CH3COO(aq)
c. NH3(aq) + CH3COOH(aq) NH4(aq) + CH3COO(aq)
d. H(aq) + OH(aq) H2O(l)
e. Mg(s) + 2H(aq) Mg(aq) + H2(g)
f. Mg(s) + CH3COOH(aq) Mg(aq) + CH3COO(aq)
g. Zn(s) + Cu(aq) Zn(aq) + Cu(s)
h. CO3(aq) + Ca(aq) CaCO3(s)
6. Determine the limiting reactant and the amount of the limiting reactant in each of the reactions
performed.
7. Calculate for the theoretical and experimental enthalpy of each reaction.
8. Relate the sign of ΔT to the sign of the experimental ΔH.
Like in the explanation above in item number 2, the sign of ΔT will be directly proportional
but opposite the sign of ΔH.
9. For reactions 1-4, which pair gave the most and the least exothermic (or endothermic)
reaction? Explain the observation. (Use both the theoretical and experimental values for this)
The theoretical values of the neutralization reactions are close to one another, as shown below.
Table 1. Theoretical and Actual Enthalpies for Neutralization in Experiment 1
Reaction Theoretical ΔHrxn Actual ΔHrxn % ErrorNH3 + HCl -52.2kJ/mol -55.8kJ/mol 6.93%
NaOH + CH3COOH -56.1kJ/mol -55.8kJ/mol 0.445%NH3 + CH3COOH -52.5kJ/mol -55.8kJ/mol 6.35%
NaOH + HNO3 -55.8kJ/mol -55.8kJ/mol 0.00%
The students were limited to two significant figures due to the calibration of the thermometers,
so the values are rounded off. This could explain why they recorded the same ΔHrxn for all
four reactions. The fourth reaction has a 0.00% error because its net ionic
equation is the same as the reaction used for calibration, which is:
H(aq) + OH(aq) H2O(l)
10. For reactions 5 and 6, which gave the most exothermic reaction? Why? (Use both the
theoretical and experimental values for this)
The active metal and acid reactions produce hydrogen gas as a product. This caused the test
tube to pop periodically while performing the experiment. Since the popping transferred heat
and matter to the surroundings, the solution can no longer be called adiabatic and isobaric as
the system did work through expansion. This led to a large deviation in the theoretical and
actual enthalpy values in the experiment.
Table 2. Theoretical and Actual Enthalpies for Active Metal and Acid Reactions in Experiment 1
Reaction Theoretical ΔHrxn Actual ΔHrxn % ErrorMg + HCl -466.85kJ/mol -387kJ/mol 17.1%
Mg + CH3COOH -467.35kJ/mol -271kJ/mol 42.0%
11. For reactions 7 and 8, what are the solid products of the reaction? What is the theoretical yield
of each?
To get the theoretical yield, we must multiply the moles of limiting reagent (after applying
their stoichiometric ratios) with the molecular weight of the precipitate.
Table 3. Table of Molecular Weights for Precipitation Reactions in Experiment 1
Precipitate Moles LR Molecular WeightCu(s) 0.00764mol (Zn) 63.5g/mol
CaCO3(s) 0.005mol (none) 100.1g/mol
After multiplying we get 0.485g Cu and 0.501g CaCO3.
12. The net ionic of reaction 8 is a synthesis (combination) reaction. Using the theoretical ΔH of
the reaction, deduce the relative magnitude of the energy of bond breaking and bond forming
during the reaction.
The reaction between CaCl2 and Na2CO3 is endothermic. This means that it absorbed energy
from its surroundings during the course of the reaction. The reaction can be broken down into
3 basic steps:
CaCl2(aq) Ca(aq) + 2Cl(aq) ΔH1 = positive
Na2CO3(aq) 2Na(aq) + CO3(aq) ΔH2 = positive
Ca(aq) + CO3(aq) CaCO3(aq) ΔH3 = negative
ΔH1 and ΔH2 are processes that involve bond breaking, thus they require energy from the
surroundings. On the other hand, ΔH3 is a bond forming process, which releases energy in
order to gain stability. Observing the theoretical ΔHrxn for reaction 8, we see that it is a
positive value, and this shows that the energy gained from dissolution (ΔH1 and ΔH2) is
greater than the energy gained from precipitate formation (ΔH3).
|ΔH1 +ΔH2| > |ΔH3|
13. In the appendix, show the equation used to determine the heats of reaction for reactions 1-6
and reactions 7 and 8.
14. Tabulate the possible sources of errors and their effect to the following parameters:
ΔT, Ccal, and ΔH. Accompany each with a reason.
Source of Error Effect on ΔT Effect on Ccal Effect on ΔHMg ribbon not cleaned well
Decrease Not Affected Decrease
Entire test tube not covered by Styrofoam
Decrease Increase Decrease
Excess HCl used in calibration
Increase Decrease Increase
Excess NaOH used in calibration
Decrease Increase Decrease
One pipette was used for transferring all
compounds
Decrease Increase Decrease
Excess reactant mistaken for limiting
reactant during calibration
Not Affected Increase Decrease
Stopper pops due to gas buildup
Decrease Not Affected Decrease
Mg ribbon not cleaned well
The MgO surface of the ribbon does not allow complete dissociation in acid. Small amounts
of the impurities will remain, absorbing some of the heat released during the process. Thus, the
ΔT will be lower than the actual value. Ccal will be unaffected since its value was calculated in
the calibration part of the experiment, while the ΔH will decrease as it is directly proportional to
ΔT at constant Ccal.
Entire test tube not covered by Styrofoam
Styrofoam insulates the calorimeter, preventing the escape of heat. Exposing even a small part
of the calorimeter will allow heat to flow out of that area, giving a lower ΔT. Ccal increases
because it is inversely proportional to ΔT, while ΔH decreases because it is directly proportional
to ΔT.
Excess HCl used in calibration
HCl is the limiting reactant, so an increase in the volume used will increase the ΔT, since more
moles of HCl and NaOH will be reacting. Since Ccal is inversely proportional to ΔT, the Ccal will
decrease. ΔH is constant in the calibration part (-55.8kJ/mol), but in the other reactions it will
be higher since it is directly proportional to ΔT.
Excess NaOH used in calibration
Since NaOH is the excess reactant, an increase in the volume used will not increase the number
of moles that will react. It will instead lower ΔT because the extra volume absorbs some of the heat
of the reaction compared to the reaction with 10ml 1M NaOH. The Ccal increases because it is
inversely proportional to ΔT, while the ΔH will decrease, except in calibration where it is
constant.
One pipette was used for transferring all compounds
Using only one pipette for different compounds will affect the actual number of moles of the
compounds used. The pipette may, for example, be used to transfer NaOH first then HCl. We may
not see it, but small amounts of NaOH will be left in the pipette, and when it is used to transfer
HCl, the excess NaOH will react thereby reducing the moles of limiting reagent. Thus, we are left
with a lower ΔT and ΔH because of the decrease in moles, and a higher Ccal.
Excess reactant mistaken for limiting reactant during calibration
Human error is a crucial factor that affects how the overall result of the experiment appears. If
NaOH is mistaken to be the limiting reactant instead of HCl, the moles to be used in calibration
will be 10mmol, which is twice the correct value of 5mmol. ΔT will not be affected since its
value depends on the actual moles of limiting reagent and is calculated experimentally. Ccal and
ΔH, however, are calculated theoretically, and substituting the moles of excess reagent for the
moles of limiting reagent we get a higher Ccal and a lower ΔH.
Stopper pops due to gas buildup
The calorimeter is supposed to be at a constant volume at all times. In the reactions involving
magnesium, one of the products is hydrogen gas, which causes pressure to build inside the test
tube. When the stopper pops, the system does work on its surroundings, causing a loss of energy.
Thus, some of the energy that can be used to increase the temperature is lost, causing a decrease in
ΔT. Ccal is unaffected since its value was calculated in the calibration part, while the ΔH will
decrease since it is directly proportional to the temperature change.
Answers to Questions in the Laboratory Manual
1. After obtaining experimental values of ΔHrxn, explain any discrepancy of the values to the
theoretical. Give some possible sources of errors.
There are many possible causes to the discrepancies in the value of ΔHrxn. First, the system is
imperfectly adiabatic; some heat is absorbed by the Styrofoam and some heat is supposed to
leave the system through the thermometer, which gives the reading. The instruments
themselves such as the thermometer and balance are calibrated to only a specific decimal
place, which means that small differences will not be detected. Rounding off is also a possible
cause, since we follow significant figures in solving for ΔHrxn. Lastly, there are the
problems with the preparations and equipment stated earlier in number 14 of the answers to
questions posted on the bulletin board. Here is the list again, which gives seven examples.
Source of Error Effect on ΔT Effect on Ccal Effect on ΔHMg ribbon not cleaned well
Decrease Not Affected Decrease
Entire test tube not covered by Styrofoam
Decrease Increase Decrease
Excess HCl used in calibration
Increase Decrease Increase
Excess NaOH used in calibration
Decrease Increase Decrease
One pipette was used for transferring all
compounds
Decrease Increase Decrease
Excess reactant mistaken for limiting
reactant during calibration
Not Affected Increase Decrease
Stopper pops due to gas buildup
Decrease Not Affected Decrease
2. In the procedure for the determination of ΔH, explain why it is important:
a. that the total volume of the resulting solution be 15ml?
The calorimeter used is a constant-volume calorimeter. The volume used to calibrate the
calorimeter must be the same in all reactions to give the same initial volume and pressure,
which affect the energy changes in the reaction.
b. to know the exact concentrations of the reactants?
The concentration of reactants are vital in thermochemical calculations. If the experimental
values deviate from the real values, there will be many errors, especially in equations
involving the limiting reagent, especially qrxn and ΔHrxn.
c. to know the exact weight of the metal solids used?
The solids present at the end of the reaction absorb some of the heat released by the
reaction. We must take this into account so that the values of qrxn and ΔHrxn will become
more accurate.
3. The neutralization of 200mL 0.5M HA by sufficient amount of NaOH evolves 6.0kJ of heat.
a. Calculate the enthalpy change for the neutralization of 1 mole HA
(200mL)(1L/1000mL)(0.5mol/1L) = 0.1mol
ΔHrxn = qrxn/nlr = -6kJ/0.1mol = -60kJ/mol
The ΔHrxn is quite close to the theoretical value, -56.1kJ/mol.
b. Is HA a weak or strong acid? Justify your answer using thermochemical equations.
Acetic acid is known to be a weak acid because it does not completely ionize in solution.
Strong acids have this net ionic equation when reacted with strong bases:
H(aq) + OH(aq) H2O(l)
The first ionization of all strong acids follow this; the anions do not participate in the
reaction and are merely dissolved.
Using the standard enthalpies of formation of the product and reactants, we can solve for
the value of ΔHrxn for all strong acids in their first ionization with a strong base, which
turns out to be -55.8kJ/mol. However, the theoretical value for the ΔHrxn of acetic acid and
a strong base is -56.1kJ/mol.Therefore, acetic acid is not a strong acid.
c. Write the net ionic equation for the reaction between HA and NaOH.
CH3COOH(aq) + OH(aq) H2O(l) + CH3COO(aq)
4. A calorimeter similar to your Styrofoam-ball calorimeter was used to determine the enthalpy
change associated with the reaction between Cu and Zn(s). The reaction between 20mL of
0.450M CuSO4 and 0.264g Zn(s) resulted in a temperature change of 8.83ºC.
Prior to the analysis, the calorimeter was calibrated using the reaction between 15mL of 2.0M
HCl and 5mL of 2.0M NaOH, which brought about a temperature change of 5.60ºC.
a. Write the net ionic equation for the calibration reaction
H(aq) + OH(aq) H2O(l)
b. Write the net ionic equation for the displacement reaction
Zn(s) + Cu(aq) Zn(aq) + Cu(s)
c. Calculate Ccal
d. Calculate the enthalpy change (per mole) for the displacement reaction.
5. Given that the standard enthalpy of formation of liquid water, ΔHºf,H2O is -285kJ/mol, calculate
the ΔHºf of OH(aq).
Appendix
Equation used to calculate the heat capacity of the calorimeter
We start with three basic equations:
(1) qcal = -qrxn
(2) qcal = CcalΔT
(3) qrxn = nlrΔHrxn
Where qcal = heat of the calorimeter ΔT = change in temperature
qrxn = heat of reaction nlr = moles of limiting reactant
Ccal = heat capacity of the calorimeter ΔHrxn = enthalpy per mole of reaction
Using equation 1, we can equate equation 2 and 3 by substitution. We are left with:
CcalΔT = -nlrΔHrxn
We isolate Ccal on one side, and we now have the final equation:
Ccal = (-nlrΔHrxn) ÷ ΔT
Equation used to determine the heats of reaction for reactions 1-6 and reactions 7 and 8
For reactions 1-6:
qrxn = -CcalΔT
For reactions 7 and 8:
qrxn = -(mc + Ccal)ΔT
Where qrxn = heat of reaction m = mass of precipitate
Ccal = heat capacity of the calorimeter c = specific heat of precipitate
ΔT = change in temperature