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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:


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


One pipette was used for transferring all

compounds


Excess reactant mistaken for limiting

reactant during calibration

Not Affected Increase Decrease

Stopper pops due to gas buildup


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.


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.


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.


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.


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





Increase Decrease Increase



One pipette was used for transferring all

compounds


Excess reactant mistaken for limiting

reactant during calibration

Not Affected Increase 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:


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


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

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