4 1 Fundamentals of Calorimetry Lab Report Dr Z Edits PDF

Preview

Summary

week 4 assignment for fundamentals of chemistry Lab 4 1 Fundamentals of Calorimetry Lab Report Dr Z Edits...

Description

The Fundamentals of Calorimetry Student Name Josue Merino Date September 26, 2021

Time (min) 1 2 3 4 5 6 7 8 9 10

Data Table 1A Trial 1 Temp. °C Trial 2 Temp. °C 48 46 45 44 44 43 42 42 41 41

46 45 44 43 43 42 41 41 40 40

Question 1: Create a spreadsheet and graph of the data from Data Table 1A, plotting Temperature vs. Time. Insert a trendline. Use the y intercept to find the temperature at time 0, (T0), when the two volumes of water are mixed. Insert a copy of your graph below. Be sure to display the equation of the line on the graph. Your name and date in the title on your graph. Be sure the equations of the lines are displayed clearly and are legible.

Temp °C

Josue Merino Zavala 9/25/2021 49 48 47 46 45 44 43 42 41 40 39

f(x) = − 0.73 x + 47.6 f(x) = − 0.67 x + 46.2

0

2

4

6

Time(min.)

© 2016 EW5 Carolina Biological Supply Company

8

10

Trial 1 Linear (Trial 1) 12 Trial 2 Linear (Trial 2)

Data Table 1B: You must include your work, with units, in the Data Table. Calculation #

Trial 1

Trial 2

50g

50g

Initial temperature of cold water = Tcold (°C)

24.0°C

22.0°C

Initial temperature of warm water = Thot (°C)

75.0°C

75.0°C

(24.0 + 75.0)/2= 49.5°C

(22.0+75.0)/2 = 48.5°C

47.6°C

46.2°C

47.6°C – 24.0°C=

46.2°C-22.0°C =

23.6°C

24.2°C

47.6°C – 75.0°C=

46.2°C - 75°C =

-27.4°C

-28.8°C

4.18J/°Cg x 23.6°C x 50g =

4.18J/°Cg x 24.2°C x 50g =

4932.4 J

5057.8 J

Mass of water

1

Average temperature, Tave(°C) Temperature at time 0 from graph, T0, (°C) Change in Temp Cold Water (ΔTCold= T0- Tcold)

2

3

Change in Temp Hot Water (ΔTHot = T0- THot)

Change in q Cold Water (final unit should be J) 4 ∆q(cold water) = Cwater x ΔTCold x mass of water Change in q Hot Water (J) 5

∆q(hot water) = Cwater x ΔTHot x mass of water

Heat (Energy) gained by the Calorimeter (final unit should be J) 6

4.18J/°Cg x (-27.4°C) x 50g = 4.18J/°Cg x (-28.8°C) x 50g = -5726.6 J

-6019.2 J

| 4932.4 J + (-5726.6 J) | =

|5057.8J + (-6019.2J| =

794.2 J

961.4 J

33.65 J/°C

39.727 J/°C

∆qcal = |∆q(cold water) + ∆q(hot water)| **∆qcal should be a positive value**

7

Heat capacity of calorimeter Ccalorimeter (final unit should be J/°C) ∆qcal = Ccalorimetry x |ΔTCold|

8

Average Heat Capacity of Calorimeter, (CAVE) in J/°C

© 2016 EW5 Carolina Biological Supply Company

(33.65 + 39.727)/2 = 36.6885 J/°C

Data Table 2a You must include your work, with units, in the Data Table. Calculation #

9

Mass of water (g) mw Mass of salt (g) ms Moles of salt solution (ms x mol/g)

5g CaCl2 100g

10g CaCl2 100g

15g CaCl2 100g

5.01g

10.07g

15.08g

(5.01g/110.98 mol/g) = 0.045 mol

(10.07g/110.98 mol/g) = 0.091 mol

(15.08g/110.98 mol/g) = 0.136 mol

Initial Temperature, Ti (°C)

24°C

24°C

25°C

Final Temperature, Tf (°C)

31°C

37°C

43°C

10

Change in Temperature (°C) ∆T = Tf - Ti

31°C - 24°C = 7 °C

37°C - 24°C = 13°C

43°C - 25°C = 18°C

11

Heat absorbed by the solution (J) qw = -[4.18 x 100 x 7] qw = -[4.18 x 100 x 18] = -2926 J qw = -[4.18 x 100 x 13] qw = -[cw x mw x ∆T] = -7524 J = -5434 J

8

12

13

14

15

Average Heat Capacity of the Calorimeter, CAVE. From Data 36.6885 J/°C Table 1B Heat absorbed by the calorimeter (J) qc = -[ 36.6885 x 7] = -256.8195 J qc = -[ CAVE x ∆T] Enthalpy of solution (J) ∆H = qw + qc Enthalpy of solution (kJ) *Note: 1 kJ = 1000 J

Enthalpy/mole of solution (kJ/mol) ∆H/moles of salt

© 2016 EW5 Carolina Biological Supply Company

∆H = -2926 J + (256.8195 J) = -3182.8195 J

36.6885 J/°C qc = -[ 36.6885 x 13] =-476.95

∆H = -5434 J + (476.95 J) = -5910.95 J

36.6885 J/°C

qc = -[ 36.6885 x 18] =-660.393 J ∆H = -7524 J + (-660.393 J) = -8184.393 J

-31.82.8195 J /1000J / -5910.95 J / 1000 J /1kJ -8184.393 J /1000 J /1kJ 1 kJ = -3.1828195 kJ = -5.91095 kJ = -8.18439 kJ

-2926 J / 0.045 mol = -5434 J /0.091 mol = -7524 J / 0.136 mol = -65022.22 J/mol/ 1 -59714.29/1000 J/ 1 -55323.53 / 1000 J / 1 kJ = kJ/ 1000 J = kJ= -59.71 kJ/mol -55.32 kJ/mol -65.02 kJ/mol

Data Table 2b You must include your work, with units, in the Data Table. Calculation #

Mass of water (g) mw

16

Mass of salt (g) ms Moles of salt solution (ms x mol/g) Initial Temperature, Ti (°C)

5g NH4Cl

10g NH4Cl

15g NH4Cl

100g

100g

100g

5.03g

10.09g

15.08g

(5.03g/53.491 g/mol) = (10.09g/53.491 g/mol) (15.08g/53.491 g/mol) 0.094 mol = = 0.189 mol 0.282 mol 25°C 25°C 25°C

Final Temperature, Tf (°C)

21°C

19°C

15°C

17

Change in Temperature (°C) ∆T = Tf - Ti

21°C - 25°C = -4°C

19°C - 25°C = -6°C

15°C - 25°C = -10°C

18

Heat absorbed by the solution (J) qw = -[cw x mw x ∆T]

8

19

20 21 22

Average Heat Capacity of the Calorimeter, CAVE. From Data Table 1B Heat absorbed by the calorimeter (J) qc = -[ CAVE x ∆T]

Enthalpy of solution (J) ∆H = qw + qc Enthalpy of solution (kJ) *Note: 1 kJ = 1000 J Enthalpy/mole of solution (kJ/mol) ∆H/moles of salt

© 2016 EW5 Carolina Biological Supply Company

qw = -[4.18 x 100 x (-4)]qw = -[4.18 x 100 x (-6)] qw = -[4.18 x 100 x (= 1672 J = 2580 J 10)] = 4180 J 36.6885 J/°C

36.6885 J/°C

qc = -[ 36.6885 x (-4)] qc = -[ 36.6885 x (-6)] = 146.75 J = 220.13 J

36.6885 J/°C

qc = -[ 36.6885 x (10)] = 366.89 J

∆H =1672 J + 146.75 J∆H= 2580 J + 220.13 J ∆H= 4180 J + 366.89 J = 1818.75 J = 2800.13 J = 4546.89 J 1818.75 J /1000J /1 kJ 2800.13 J /1000J /1 kJ 4546.89J /1000J /1 kJ = = = 4.54689 kJ 1.81875 kJ 2.80013 kJ 1672 J /0.094 mol = 2580 J / 0.189 mol = 4180 J / 0.282 mol = 17787.23 J/mol / 1000 13650.79 J/mol /1000 J/ 14822.70 J/mol /1000 J/ 1 kJ= 1 kJ= J / 1 kJ =

17.79 kJ/mol

13.65 kJ/mol

14.82 kJ/mol

Question 2: Create a spreadsheet and graph for CaCl2 and NH4Cl, using the data from Data Table 2a and b. Plot mass on the X axis and change in temperature on the Y axis for both graphs. The slope will be the change in temperature per gram of salt dissolved. Insert a trendline and display the equation on the graph. Include your name and date in the titles. Insert graphs below the Data Tables 3 and 4. Data Table 3 Mass CaCl2 5.01 10.07 15.16

∆T

7.0 13.0 18.0

Data Table 4

© 2016 EW5 Carolina Biological Supply Company

Mass NH4Cl 5.03 10.09 15.08

∆T

-4.0 -6 -10

Copy and Paste Question 2 graph(s) here:

© 2016 EW5 Carolina Biological Supply Company

Josue Merino 9/25/2021 20 15

f(x) = 1.08 x + 1.74

10

∆T

5 0 -5

4

6

8

10

12

14

16

CaCl2 Linear (CaCl2) NH4Cl Linear (NH4Cl)

f(x) = − 0.6 x − 0.66

-10 -15

Mass (g)

Question 3: What sort of relationship exists between the temperature change and the mass of the solid? Explain why that relationship exists. Hint: See your lab manual Background. Heat transmission is equal to internal energy change. Because heat is directly proportional to mass, Thus, temperature change and solid mass are directly related.

Question 4: How do the calculated molar heats (∆H/moles of salt) of solution for calcium chloride compare to one another? How do the calculated molar heats of solution for ammonium chloride compare to one another? Hint: Compare the values of the determined Enthalpy/mole of solution for each compound at the different masses. Include numerical information in your response. Are they similar? How similar? The biggest temperature increase was caused by 15g of CaC12, which was 5 degrees higher than the second highest temperature increase, resulting in a final temperature of 18 degrees Celsius, which was the most significant change. While this was happening, the high temperature change caused by the NH4CI was 10 degrees colder than the initial temperature.

Question 5: The actual molar enthalpy of solution (∆H/moles of salt) for calcium chloride is -81.3 kJ/mol, whereas the molar enthalpy of solution of ammonium chloride is 14.8 kJ/mol. Calculate the average molar enthalpy of solution for each compound based on your data, and then calculate the percentage error for calcium chloride and ammonium chloride using your calculated average. Do not calculate the percent error for each mass. Percent error is (|Actual – Experimental|/Actual) * 100% CaCl2 Actual: -81.3 kJ/mol CaCl2 Experimental avg: -59.02 kJ/mol

-81.3 – (-59.02)/ -81.3 * 100% = 27.4047% error for CaCl2 NH4Cl Actual: 14.8 kJ/mol NH4Cl Experimental avg: 15.42 kJ/mol © 2016 EW5 Carolina Biological Supply Company

14.8 – 15.42 / 14.8 * 100% = 4.18919% error for NH4Cl

Activity 4 Question 6: Based on the data and graphs for calcium chloride and ammonium chloride (Question 2), determine which compound to use and what quantity of each compound will be needed to make a chemical hot pack and cold pack. Both packs should be calculated based on using 100 g (100 mL) of water. The hot pack should reach 60 °C, and the cold pack should go down to 3.0 °C from a room temperature of 25 °C. This is a 4-part question! You must show work to earn credit for this question. Hot Pack: Compound needed to reach 60 °C? CaCl2

How many grams are needed? Show your work for this calculation. To reach 60°C, 24.75g of CaCl2 is needed. (See below for attached picture of the calculations)

Cold Pack: Compound needed to reach 3.0 °C? NH4Cl How many grams are needed? Show your work for this calculation. To reach 3.0°C, 28.89g of NH4Cl is needed. (See below for attached picture of the calculations)

© 2016 EW5 Carolina Biological Supply Company

Question 7: What were some potential sources of error in this investigation? Inaccurate water measurements for compound comparison, contaminants in the water if not using bottled water, and inaccuracies in calculations when graphing. Question 8: Suggest some ways in which the calorimeter or lab protocol could be improved to have lower percentage of errors. It is necessary to place a lid on the calorimeter in order to prevent heat from escaping. And Making use of purified water and proper measuring techniques.

© 2016 EW5 Carolina Biological Supply Company...