Calorimeter

Experiment 1: Calorimetry

C.L. Mercado institute of chemistry, college of science university of the philippines, diliman quezon city, philippines date performed: November 16, 2012 instructor’s name: Irina Diane V. CastaÑo

_________________________________________________________________________________________________________________________

introduction Chemical reactions absorb or release heat. One way to measure this is by using a homemade calorimeter. The calorimeter used in this experiment is made up of a Styrofoam ball with a hole that can fit a test tube. The heat capacity of the styro-ball calorimeter is measured using a neutralization reaction of know enthalpy change. This heat capacity is then used to calculate the amount of heat absorbed or released by a chemical reaction. This will be used to measure the heat of reaction of different reaction systems such as neutralization reactions, displacement, precipitation and active metals and acids. After the experiment and recording of raw data, the heat of reactions can be calculated and compared with known literature values. Having a comparison of the results to known values will verify if this experiment works, or if there is something wrong with the experiment done. Calorimetry is used to directly measure physical and chemical processes and reactions. This can be done even at a molecular level when the right set-up is used.

Results and Discussion
Calculations
(see attached papers)

Answers to questions (lab manual) 1. Only 2 reaction systems have an error below 10%. The errors that have occurred in the experiment may be due to the limitation of the thermometer’s scale. The errors could also be due to the escape of heat when the temperature reading is taken because the system has to be uncovered. Also see table of possible errors and its effects below. 2. A. It is important the solution be 15 mL so that the use of the test tube is maximizded, allowing less errors because of a larger sample compared to a smaller volume. Also because it has the same volume as the solution used to calibrate the calorimeter.
B. Knowing the exact amounts of concentration will allow for the calculation of the heat released or absorbed because the number of reactants will be known.
C. Knowing the exact weight of the metal is important because it is a reactant that will allow for the computation of the heat of reaction and enthalpy. 3. A. 6.0KJ/[(200mL)(0.5 moles HA/1000ml)] = -60 KJ
B. it is a strong acid because its enthalpy value is near -56 KJ
C. OH-(aq) + H+(aq) H2O(L)

4. A. OH-(aq) + H+(aq) H2O(L)
B. Cu2+(aq) + Zn(s) Zn2+(aq) + Cu(s)
C. Ccal = (55.8 KJ)(nLR)/(ΔT) = 55.8KJ(5mL x 2M/1000mL)/(5.6) = 0.0998 KJ
D. ΔH = -(Ccal + mc)(ΔT)/nLR = -(99.8J + (0.264g)(0.3884J))(8.83)/(0.264g/63.55g/mol) = - 218.1 KJ
Answers to questions (Additional)
Calibration of the calorimeter: 1. OH-(aq) + H+(aq) H2O(L) a. Exothermic b. 5mL 1M HCl c. 0.005 moles d. -0.279 2. ΔT has a positive sign while ΔH has a negative sign.

3. For our calorimeter, the average heat capacity was 116.25 J. It has a positive sign like ΔT.

4. See appendix for derivation to obtain the equation used to calculate the heat capacity of the calorimeter.

Determination of heats of reactions: 5. 1. NH3(aq) + H+(aq) NH4+(aq)
2. OH-(aq) + H+(aq) H2O(l)
3. NH3(aq) + H+(aq) NH4+(aq)
4. OH-(aq) + H+(aq) H2O(l)
5. 2H+(aq) + 2Cl-(aq) + Mg(s) MgCl2(aq) + H2(g)
6. 2CH3COO-(aq) + 2H+(aq) + Mg(s) H2(g) + Mg(CH3COO)2(aq)
7. Cu2+(aq) + Zn(s) Zn2+(aq) + Cu(s)
8. CO3 2-(aq) + Ca2+(aq) CaCO3(s)

6. 1. HCl, 0.005 moles
2. CH3COOH 0.005 moles
3. CH3COOH 0.005 moles
4. HNO3 0.005 moles
5. Mg 0.00206 moles
6. Mg 0.00206 moles
7. Zn 0.00765 moles
8. Ca 0.005 moles

7. Table for enthalpies Rxn no. | ExperimentalEnthalpy(kJ/mol) | TheoreticalEnthalpy(kJ/mol) | type | %error | 1 | -69.75 | -52.22 | exo | 33.57% | 2 | -51.15 | -55.8 | exo | 8.33% | 3 | -33.45 | -52.22 | exo | 34.94% | 4 | -65.1 | -55.8 | exo | 16.6% | 5 | -64 | -307.4 | exo | 79.1% | 6 | -447 | -467.35 | exo | 4.35% | 7 | -29.32 | -218.66 | exo | 86.59a% | 7 | -9.8 | -218.66 | exo | 95.48% | 8 | - | 13.07 | endo | - |

8. All reactions gave a positive sign for ΔT, while all were negative for ΔH. 9. Reactions 1 and 2 are the least exothermic based on theoretical data. But according to experimental data, the least exothermic reaction is reaction 3, followed by 2, 4 and 1. 10. Reaction 6 is more exothermic than reaction 5 by 159.95 KJ per mol. This is also true for the experimental data. 11. The solid products were Cu(s) and CaCO3(s). Their theoretical yields are 0.49 grams and 0.50 grams respectively. 12. Bond breaking = 13.07KJ/mol
Bond formation = -13.07KJ/mol 13. See appendix for equation used to determine the heats of reaction for reactions 1 – 6 and reactions 7 and 8. 14. Table of possible errors and its effects Source of Error | Effect on ΔT | Effect on Ccal | Effect on ΔH | Wrong weight of metal samples | Decrease or increase, because there is more or less of the sample to react with | none | Decrease or increase if the metal sample is the limiting reactant | Spaces in the styrofoam calorimeter | Decrease, because heat can escape | Increase, because it will have a lower temperature | No effect | Wet test tube | none | none | Increase because the molarity will decrease | Taking time looking at the thermometer | Decrease, because heat is escaping | none | Decrease, because ΔT decreased | Solution not covered immediately | Decrease , because heat is escaping | Increase because heat escapes | Decrease, because ΔT decreased |

References:
General Chemistry II Laboratory Manual. Quezon City: Institute of Chemistry, University of the Philippines - Diliman, 2011 June.
Petrucci, R., Harwood, W., Herring, F. G. General Chemistry: Principles and Modern Application, 10th ed.; Prentice Hall, Inc, 2010
Standard enthalpy change of formation (data table) http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table)

Thermochemistry http://wikieducator.org/The1stLawo fThermodynamicsLesson5

Calorimetry—an important tool in solution chemistry http://www.sciencedirect.com/scienc e/article/pii/S0040603104002187

Appendix: A. Derivation for Ccal
Given: qcal = (Ccal) (ΔT) qrxn = -qcal ΔHrxn = qrxn/nLR
ΔHrxn(neutralization)(H2O) = -55.8 KJ
ΔHrxn = -55.8 KJ = -qcal/nLR = - (Ccal) (ΔT)/nLR
Therefore, Ccal = (55.8 KJ)(nLR)/(ΔT)

B. Equation used for reactions 1-6
ΔHrxn = - (Ccal) (ΔT)/nLR C. Equation used for reaction 7
ΔHrxn = - (Ccal+mc) (ΔT)/nLR D. Equation used for reaction 8
ΔHrxn = - (Ccal+mc) (ΔT)/nLR

References: General Chemistry II Laboratory Manual. Quezon City: Institute of Chemistry, University of the Philippines - Diliman, 2011 June.