Manuel Matthew P. Chanco VBS Chemical Engineering2010-01334

I. DiscussionThe experiment was all about Calorimetry. Basically, it is the process of measuring the enthalpy change of a given chemical reaction. A calorimeter was first set up, specifically Styrofoam Calorimeter and calibrated using Sodium Hydroxide (NaOH) and Hydrochloric Acid (HCl). After the calibration, different chemical reactions were made and the Enthalpy changes of each were measured using the information from the calibration. The results were compared to the theoretical values which made a very large percentage error. A possible cause of this error is the mistake in measurements of reagents that were used. It will have a great change on the value of the Enthalpy change because the amount of the limiting reactant will change. The experiment may be considered as failed because of the large discrepancy of the values to the experimental.

II. Calculations

Hrxn = (qrxn)/nLR qcal = (Ccal)(T) qcal = -qrxn

Sample: T1 = 3.5C; T2 = 3.0C; nLR = 0.005mol; Ccal = 279J/C

qcal = (Ccal)( T)qcal = (279J/C)(3.5C); (279J/C)(3.0C)qcal = 976.5J; 837.0J

qcal = -qrxnqrxn = -976.5J; -837.0J

Hrxn = (qrxn)/nLR Ave. Hrxn = (-195.3kJ/mol + -167.4kJ/mol)/2Hrxn = (-976.5J)/ (0.005mol); (-837.0J)/ (0.005mol) Ave. Hrxn = -181.4kJ/molHrxn = -195.3kJ/mol; -167.4kJ/mol

Theo. Hrxn =Hf,products Hf,reactants = (-486.01 + -132.51) (-80.29 + -485.76) kJ/mol = -52.47kJ/mol

% error = abs ([theoretical experimental]/ theoretical) x 100 = abs ([(-52.47kJ/mol) (-181.4kJ/mol)]/ -52.47) x 100 = 245.7%

III. Answers to questions

1. a. If the resulting solution is not equal to 15mL, then there might be an increase or a decrease of amount of reagent that was made to react. If that happened, it will change the amount of limiting reactant (or even change the species of the limiting reactant) thereby causing the change of the calculated Enthalpy change of the reaction.b. It is important so that the computed H is more accurate and precise.c. Just like in b, it will make the calculated H more accurate and precise. Moreover, it also affects its magnitude compared to reactions with only liquid reagents.

2. a. HA + NaOH NaA + H2O ; H = -6.0kJ, 200mL of 0.5M HAmol HA = (0.200L)(0.5M) = 0.1mol(1mol)(-6000J/0.1mol) = -60000J/mol or -60kJ/mol; Hrxn = -60kJ/molb. Since the energy that was released in comparison to the formation of water is almost equal, then it can be assumed that HA is a strong acid because of how ions are neutralized by water.c. H+ + OH- H2O

3. a. H+ + OH- H2Ob. Cu2+ + Zn Zn2+ + Cuc. Given: 20mL 0.450M CuSO4; 0.264g Zn T = 8.83C 0.00404mol = nLR 15mL 2.0M HCl; 5mL 2.0M NaOH T = 5.60C 0.01mol = nLR

H+ + OH- H2O H = -55.8kJ-55.8kJ/mol = (qrxn)/ 0.01mol qrxn = -0.558kJ qcal = 0.558kJ

qcal = CcalT 0.558kJ = Ccal(5.60C) Ccal = 99.6J/C

d. qcal = (99.6J/C)(8.83C) qrxn = -879.8J

Hrxn = qrxn/nLR = -879.8J/0.00404mol = -218.0kJ or -220kJ

4. Hfo = -285kJ/molH+ + OH- H2O H = -55.8kJ

Hrxn =Hf,products Hf,reactants -55.8kJ/mol = (-285kJ/mol) (Hf OHo + Hf Ho)

-55.8kJ/mol + 285kJ/mol = Hf OHo Hf OHo = -229.2kJ/mol

IV. References

Petrucci, R.H., Harwood, W.S., & Herring, F.G. (1997). General Chemistry: Principles and Modern Applications. New Jersey: Prentice-Hall, Inc. 8th ed. Redmore, F.H. (1980). Fundamentals of Chemistry. Quezon City: Apson Enterprises.