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Trent University Introductory Chemistry II CHEM 1010H, 2011 Winter

Lab Report#1 Calorimetry

Professor: Dr. Matthew Thompson Lab. Coordinator: Sue Landry

Y.C. Matthew Ho Trent#: 0473584

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Due: 31-01-2012 Name: Y.C. Matthew Ho

ID: 0473584

CHEM1010 – Report 1 (Be sure to state references for any cited value at the end of this report) 1. Raw data: Part 1 Mass of weighing bottle + paper towel + ice (g): 30.7741g. Mass of weighing bottle + wet paper towel (g): 22.0178g. Time (min.)

Temperature (oC)

Time (min.)

Temperature (oC)

Time (min.)

Temperature (oC)

1 (15:07)

18.7

6 (15:12)

9.6

11 (15:17)

10.5

2 (15:08)

18.7

7 (15:13)

9.7

12 (15:18)

10.7

3 (15:09)

18.7

8 (15:14)

9.9

13 (15:19)

10.8

4 (15:10)

10.7

9 (15:15)

10.0

14 (15:20)

11.0

5 (15:11)

9.8

10 (15:16)

10.3

15 (15:21)

11.2

Part 2 Weighing paper + Mg (g): 0.4832g. Weighing paper after transferred (g): 0.2813g.

Time (min.)

Temperature (oC)

Time (min.)

Temperature (oC)

Time (min.)

Temperature (oC)

1 (13:42)

19.8

6 (13:47)

30.7

11 (13:52)

29.3

2 (13:43)

19.8

7 (13:48)

30.5

12 (13:53)

29.0

3 (13:44)

19.8

8 (13:49)

30.2

13 (13:54)

28.8

3

4 (13:45)

31.1

9 (13:50)

29.9

14 (13:55)

28.7

5 (13:46)

30.9

10 (13:51)

29.6

15 (13:56)

28.5

Part 3 Time (min.)

Temperature (oC)

Time (min.)

Temperature (oC)

Time (min.)

Temperature (oC)

1 (14:07)

19.7

6 (14:12)

26.3

11 (14:17)

25.4

2 (14:08)

19.7

7 (14:13)

26.1

12 (14:18)

25.2

3 (14:09)

19.7

8 (14:14)

25.9

13 (14:19)

25.0

4 (14:10)

26.7

9 (14:15)

25.7

14 (14:20)

24.9

5 (14:11)

26.4

10 (14:16)

25.6

15 (14:21)

24.7

Part 4

Weighing paper + NH4NO3 (g): 1.2871g. Weighing paper after transferred (g): 0.2798g.

Time (min.)

Temperature (oC)

Time (min.)

Temperature (oC)

Time (min.)

Temperature (oC)

1 (14:36)

18.9

6 (14:41)

16.8

11 (14:46)

17.6

2 (14:37)

18.9

7 (14:42)

17.0

12 (14:47)

17.7

3 (14:38)

18.9

8 (14:43)

17.2

13 (14:48)

17.8

4 (14:39)

16.6

9 (14:44)

17.4

14 (14:49)

17.9

5 (14:40)

16.5

10 (14:45)

17.5

15 (14:50)

18.0

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2. Part 1: 



Prepare a plot of temperature vs. time for Part 1. You must prepare this graph manually using the mm-scale graph paper available as a .pdf in the folder for experiment 1 in myLearningSystem, and not using an electronic resource to receive credit. Use your prepared graph to show T1 and T2 using the extrapolation process described in the theory section of the instructions for the experiment. Be sure to include a copy of your plot with your report submission, and please pay attention to axis labels and titles.  T1 = 18.7o C, T2 = 9.2oC.



Using the known enthalpy of fusion (Hfus) of ice (do not forget to include your references), calculate the heat capacity of the calorimeter with Eq. (5) from the laboratory instructions. To use this equation, you will need to calculate the mass of the water initially present using the volume of water and the density of water at the initial temperature, T1 ANSWER:  Mass of ice = 30.7741g - 22.0178g = 8.7563g.  T1 = 18.7oC, T2 = 9.2oC.  Volume of water = 75.00mL.  Density of water = 1g/mL.  Swater = 4.18Jg-1oC-1. 

of water = 333Jg-1.  By: Density = Mass / Volume 1g/mL = Mass(g) / 75.00(mL) Mass (g) = 75.00g.

 Cp = {[(333Jg-1)(8.7563g) + (4.18Jg-1oC-1)( 8.7563g)(9.2 oC - 0 oC)] / (18.7 oC -9.2 oC)} – (4.18Jg-1oC-1)(75.00g) = [(2915.8479J + 336.7322728J) / 9.5oC] -313.5JoC-1 = 342.3768603JoC-1 -313.5JoC-1 = 28.87686029oC-1 = 28.88JoC-1 . 3. Part 2:

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

Prepare a plot of temperature vs. time for Part 2. You must prepare this graph manually using the mm-scale graph paper available as a .pdf in the folder for experiment 1 in myLearningSystem, and not using an electronic resource to receive credit. Use your prepared graph to show T1 and T2 using the extrapolation process described in the theory section of the instructions for the experiment. Be sure to include a copy of your plot with your report submission, and please pay attention to axis labels and titles. Use Eq. (6) from the laboratory instructions to calculate the ∆Hrxn of Mg reacting with the HCl solution. (Note: state any assumptions that have been made here)  T1 = 19.8o C, T2 = 31.2oC. ANSWER: 

mm = 0.4832g - 0.2813g = 0.2019g.



T1 = 19.8o C, T2 = 31.2oC. = 31.2oC – 19.8oC = 11.4oC.





Cp = 28.88JoC-1



Sm = 4.18Jg-1oC-1

= -(4.18 Jg-1oC-1*0.2019g + 28.88 JoC-1)(31.2oC - 19.8oC) = -(29.723942JoC-1)(11.4oC) = -338.8529388J = -338.8529J. 

o

Using your data, calculate the heat of formation of Mg2+ (aq), i.e. H f (Mg2+(aq)), and the the reaction that is given below. Remember, elements in their standard states, as well as the free proton (H+ (aq)) have + H of = 0, therefore H of for Mg (s), H (aq) and H2 (g) are zero by definition. To do this, you will find equation 6.15 in the textbook helpful. ANSWER:

-338.8529J = [(1*Mg2+) + (1*H2)] – [(1*Mg) + (2*H+)] -338.8529J = [Mg2+ + 0] – [0 + 0] Mg2+ = -338.8529J 4. Part 3:

Mg (s) + 2 H+ (aq)  Mg2+ (aq) + H2 (g)

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

Prepare a plot of temperature vs. time for Part 3. You must prepare this graph manually using the mm-scale graph paper available as a .pdf in the folder for experiment 1 in myLearningSystem, and not using an electronic resource to receive credit. Use your prepared graph to show T1 and T2 by the extrapolation process described in the theory section of the instructions for this experiment. Be sure to include a copy of your plot with your report submission, and please pay attention to axis labels and titles.  T1 = 19.7oC, T2 = 26.8oC.



Using your data, calculate the enthalpy of the overall reaction, ∆Hrxn , for the neutralization of HCl with NaOH, per 1 mole of HCl neutralized. Note: You are calculating the enthalpy of the reaction listed below by doing this (recall: strong acid, strong base definitions?). Please remember: as you have not used exactly 1 mole of HCl, you will need to account for this in your calculations. H3O+ + OH-  2 H2O ANSWER:  Mass of HCl and NaOH are 37.50g.  HCl and NaOH: By n = C/V = 1.00M / 37.50mL = 0.026666666mole = 26.7 * 10-3 mole

= [2(H2O)] – [0.0267(HCl) + 0.0267(NaOH)] = [2(-285.8kJ/mol)] – [0.0267(-167.21kJ/mol) + 0.0267(-470.1kJ/mol)] = [-571.6kJ] – [(-4.464507kJ) + (-12.55167kJ)] = -554.583823kJ = -554.5838kJ. 

Given the definitions of strong acid and strong base, if you were to repeat this experiment using HNO 3 as the acid, should you get the same value? Explain your reasoning. ANSWER: 

5. Part 4:

If I repeat this experiment using HNO3 as the acid, I will not get the same value. It is because HNO3 and HCl are having different values of the standard enthalpy of formation, therefore, the values and result will not be the same.

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



Prepare a plot of temperature vs. time for Part 4. You must prepare this graph manually using the mm-scale graph paper available as a .pdf in the folder for experiment 1 in myLearningSystem, and not using an electronic resource to receive credit. Use your prepared graph to show T1 and T2 by the extrapolation process described in the theory section of the instructions for this experiment. Be sure to include a copy of your plot with your report submission, and please pay attention to axis labels and titles. 



T1 = 18.9o C, T2 = 16.7oC.

Using your data, calculate the ∆Hrxn for the aqueous dissociation of NH4NO3 into NH4+(aq) and NO3-(aq), per mole of NH4NO3 that dissociates. Again, since you did not use exactly 1 mol of the ammonium nitrate compound, you will need to account for this in your calculations. ANSWER: 

Mass of NH4NO3 (g): 1.2871g - 0.2798g = 1.0073g.



NH4NO3 → NH4+ + NO3-



n = mass / molar mass = 1.0073g / 80.0486 g/mol = 0.012583605mol = 12.58 * 10-3 mol.

= [(12.58 * 10-3mol)(NH4+) + (12.58 * 10-3mol)(NO3-)] – [(12.58 * 10-3mol)(NH4NO3)] = [(12.58*10-3mol)(-133.26 kJ/mol) + (12.58*10-3mol)(-206.85 kJ/mol)] – [(12.58*10-3mol)(365.6 kJ/mol)] = [(-1.676891264kJ) + (-2.602918789kJ)] – [-4.600565988kJ] = 0.320755935kJ = 32.0756*10-2J. 

Is this process exothermic or endothermic? Based on this answer, what molecules must be supplying the necessary thermal energy in order for the dissociation reaction to occur? ANSWER:  This process is endothermic reaction. It is because the system absorbs energy during reaction. Water molecules are supplying the necessary thermal energy in order for the dissociation reaction to occur.

Reference

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1) Nivaldo J. Tro(2011). Pearson Chemistry: A Molecular Approach, Second Edition. Location: Pearson. Appendix II, B. Standard Thermodynamic Quantities for Selected Substances at 25o C....