# Experiment 1: Calorimetry

Published: 2021-09-14 11:15:09  Category: Physics, Heat, Thermodynamics

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GET MY ESSAY Experiment 1: Calorimetry Nadya Patrica E. Sauza, Jelica D. Estacio Institute of Chemistry, University of the Philippines, Diliman, Quezon City 1101 Philippines Results and Discussion Eight Styrofoam ball calorimeters were calibrated. Five milliliters of 1M hydrochloric acid (HCl) was reacted with 10 ml of 1M sodium hydroxide (NaOH) in each calorimeter. The temperature before and after the reaction were recorded; the change in temperature (? T) was calculated by subtracting the initial temperature from the final temperature. The reaction was performed twice for every calorimeter.
The heat capacity (Ccal) of each calorimeter was calculated using the formula, C_cal=(-?? H? _rxn^o n_LR)/? T where ? Horxn is the total heat absorbed or evolved for every mole of reaction and nLR is the number of moles of the limiting reactant. The ? Horxn used was -55. 8kJ per mole of water while the nLR was 0. 005 mole. Table 1. Average Ccal from recorded ? T values. Trial? T, (oC)Ccal, (J)Ave Ccal, (J) 112. 2126. 82202. 91 21. 0279. 00 213. 093. 00108. 50 22. 3124. 00 310. 5558. 00558. 00 20. 5558. 00 412. 0139. 50244. 13 20. 8348. 75 513. 093. 0081. 38 24. 069. 75 612. 0139. 50209. 25 21. 0279. 00 712. 111. 60111. 60 22. 5111. 60 813. 093. 00116. 25 22. 0139. 50 Different heat capacities were calculated for each calorimeter (Table 1). After calibration, a reaction was performed in a calorimeter by each pair. A total of eight reactions were observed by the whole class. The temperature before and after the reaction were recorded. Then the change in temperature was calculated. Each reaction was performed twice to produce two trials. The experimental ? Horxn for each reaction was solved using the formula, ?? H? _rxn^o=(-C_cal ? T)/n_LR  where Ccal is the heat capacity previously calculated for each calorimeter.
The percent error for each reaction was computed by comparing the computed experimental ? Horxn to the theoretical ? Horxn using the formula, % error=|(computed-theoretical)/theoretical|? 100%  Table 2. Comparison of calculated ? Horxn and theoretical ? Horxn. RxnLRTrial? T, (oC)? Horxn, (kJ/mol)Ave ? Horxn, (kJ/mol)Theo ? Horxn, (kJ/mol)% Error 1HCl13. 5-142. 04-131. 89-132. 510. 47 23. 0-121. 75 2HOAc11. 3-26. 34-41. 61-56. 0924. 65 22. 7-56. 89 3HOAc11. 8-189. 61-203. 16-52. 47287. 18 22. 0-216. 70 4HNO311. 5-73. 24-70. 80-55. 8426. 78 21. 4-68. 36 5Mg13. 0-118. 67-138. 45-466. 8570. 34 24. 0-158. 23 6Mg15. 5-559. 4-635. 72-953. 1133. 30 27. 0-712. 01 7Zn13. 0-43. 80-43. 80-218. 6679. 97 23. 0-43. 80 8CaCl210. 00. 00-5. 8113. 07144. 47 20. 5-11. 63 There were differences in experimental and theoretical values of ? Horxn as shown by the percent error for each reaction (table 2). The discrepancies were caused by many factors. One factor was the loss of heat. The heat may have been released when the thermometer was pushed or pulled during the reaction. The heat may also have been lost because the calorimeter is not totally isolated. Another factor was the dilution of the solution. The pipette or test tube may still have been wet when used.

However, the concentration used in solving for values was the concentration of the undiluted solution. Another factor that may have contributed to the difference in the experimental and theoretical values was human error. It was manifested when reading the thermometer or measuring chemicals with different instruments. The factors aforementioned are the limitations of this experiment. References Petrucci, R. H. ; Herring, F. G. ; Madura, J. D. ; Bissonnette, C. General Chemistry, 10th ed. ; Pearson Education: Canada, 2011; Chapter 7. Appendices Appendix A Comparison of Observed and Theoretical Heats of Reactions RxnLRTrial? TnLRqrxn?
HorxnAve ? HorxnTheo ? Horxn% Error 1HCl13. 500. 00500-710. 19-142. 04-131. 89-132. 510. 47 23. 000. 00500-608. 73-121. 75 2HOAc11. 250. 00515-135. 63-26. 34-41. 61-56. 0924. 65 22. 700. 00515-292. 95-56. 89 3HOAc11. 750. 00515-976. 50-189. 61-203. 16-52. 47287. 18 22. 000. 00515-1116. 00-216. 70 4HNO311. 500. 00500-366. 19-73. 24-70. 80-55. 8426. 78 21. 400. 00500-341. 78-68. 36 5Mg13. 000. 00206-244. 13-118. 67-138. 45-466. 8570. 34 24. 000. 00206-325. 50-158. 23 6Mg15. 500. 00206-1150. 88-559. 44-635. 72-953. 1133. 30 27. 000. 00206-1464. 75-712. 01 7Zn13. 000. 00764-334. 80-43. 80-43. 80-218. 6679. 97 23. 000. 00764-334. 80-43. 0 8Na2CO3/ CaCl210. 000. 005000. 000. 00-5. 8113. 07144. 47 20. 500. 00500-58. 13-11. 63 Appendix B Sample Calculations Calibration of Calorimeter 10ml 1M NaOH + 5ml 1M HCl n. i. e. : OH-(aq) + H+(aq) ? H2O(l)? Horxn= -55. 8kJ LR: HCLnLR= 0. 005mol Grp 1 Trial 1 ?T= 2. 2oC Sol’n: C_cal=(-?? H? _rxn^o n_LR)/? T C_cal=(-(-55. 8kJ)(0. 005mol))/(? 2. 2? ^o C)? 1000J/1kJ ?(C_cal=126. 82 J) Determination of Heats of Reaction Neutralization Reaction Rxn 4 Trial 1: 10ml 1M NaOH + 5ml 1M HNO3 n. i. e. : OH-(aq) + H+(aq) ? H2O(l) LR: HNO3nLR= 0. 005mol ?T= 1. 5oCCcal= 244. 125 J Sol’n ?? H? _rxn^o=(-C_cal ? T)/n_LR ?? H? _rxn^o=(-(244. 25J)(? 1. 5? ^o C))/0. 005mol? 1kJ/1000J ? (?? H? _rxn^o=-73. 24kJ) Reaction between an Active Metal and an Acid Rxn 5 Trial 1: 15ml 1M HCl+ 0. 05g Mg n. i. e. : 2H+(aq) + Mg(s) ? Mg+2(aq) + H2(g) LR: MgnLR= 0. 00206mol ?T= 3oCCcal= 81. 375 J Sol’n ?? H? _rxn^o=(-C_cal ? T)/n_LR ?? H? _rxn^o=(-(81. 375J)(3^o C))/0. 00206mol? 1kJ/1000J ?(?? H? _rxn^o=-118. 67kJ) Displacement of One Metal by Another Rxn 7 Trial 1: 15ml 1M CuSO4 + 0. 5g Zn n. i. e. : Cu+2(aq) + Zn(s) ? Zn+2(aq) + Cu(s) LR: ZnnLR= 0. 00764mol ?T= 3oCCcal= 111. 6 J Sol’n ?? H? _rxn^o=(-C_cal ? T)/n_LR ?? H? _rxn^o=(-(111. 6J)(3^o C))/0. 00764mol? 1kJ/1000J ?(?? H? rxn^o=-43. 80kJ) Precipitation Reaction Rxn 8 Trial 1: 10ml 0. 5M Na2CO3 + 5ml 1M CaCl2 n. i. e. : CO3-2(aq) + Ca+2(aq) ? CaCO3(s) LR: Na2CO3/ CaCl2nLR= 0. 005mol ?T= 0. 5oCCcal= 116. 25 J Sol’n ?? H? _rxn^o=(-C_cal ? T)/n_LR ?? H? _rxn^o=(-(116. 25J)(? 0. 5? ^o C))/0. 005mol? 1kJ/1000J ? (?? H? _rxn^o=-11. 63kJ) Appendix C Answers to the Questions in the Lab Manual There are many possibilities that explain the discrepancy of the experimental and theoretical values of ? Horxn. First, heat might have been lost to the surroundings. This is possible whenever the thermometer is pulled out or pushed in the calorimeter during the reaction.
Also, the calorimeter might not have been thoroughly isolated. Second, the solution might have been diluted in the test tube or pipette. They might have been wet when used with the solution. Lastly, the discrepancies might have occurred due to human error. The students might have misread the thermometer when taking the temperature or the pipette when measuring the solutions. a. It is important to keep the total volume of the resulting solution to 15ml because any more or any less than that of the volume can contribute to the absorption or release of additional heat therefore affecting the ? Horxn. b.
It is important to know the exact concentrations of the reactants to solve for their number of moles and to find out the limiting reactant. c. It is important to know the exact weight of the metal solids used to solve for their number of moles and to find out whether one of them is a limiting reactant. Also, the weight is needed to solve for the heat capacity of the solid when the specific heat is given. 200ml 0. 5M HA + NaOH ? -6. 0kJ LR: HAnLR= 0. 1mole ?? H? _(rxn,mol)^o= (-6. 0 kJ)/(0. 1 mol) ?(?? H? _(rxn,mol)^o= -60 kJ) HA is a strong acid. OH-(aq) + H+(aq) ? H2O(l)? Horxn= -60 kJ/mole Calibration:15ml 2. M HCl + 5ml 2. 0M NaOH? T=5. 60oC LR: NaOHnLR= 0. 01mole Reaction:20ml 0. 450M CuSO4 + 0. 264g Zn? T=8. 83oC LR: ZnnLR= 0. 00404mole n. i. e. : OH-(aq) + H+(aq) ? H2O(l) n. i. e. : Cu+2(aq) + Zn(s) ? Zn+2(aq) + Cu(s) C_cal=(-?? H? _rxn^o n_LR)/? T C_cal=(-(-55. 8kJ)(0. 01mol))/(? 5. 60? ^o C)? 1000J/1kJ ?(C_cal=99. 6 J) ?? H? _rxn^o=(-C_cal ? T)/n_LR ?? H? _rxn^o=(-(99. 6J)(? 8. 83? ^o C))/0. 00404mol? 1kJ/1000J ? (?? H? _rxn^o=-218. 0 kJ) OH-(aq) + H+(aq) ? H2O(l)? Horxn= -55. 8kJ ?Hof,H2O= -285 kJ ?Hof,OH-= ? ?Horxn= ? Hof,product - ? Hof,reactant -55. 8 kJ = ? Hof,OH- - (-285 kJ) ?(?? H? _(f,? OH? ^-)^o=-218. 0 kJ)

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