Calorimetry- Choice I, II, and III lab report PDF

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Calorimetry- Choice I, II, and III Chemistry 1412 PO5- General Chemistry II South Texas College Spring- 2020 For: Dr. Ludivina Avila By: Lydia Michel Duenas, Vanessa Medrano, Eliza Vela

A) Objective

Chemical and physical changes always occur with change in energy. It is known energy cannot be created nor destroyed but can be redistributed. This experiment was detrimental in finding the calorimetry constant for a simple coffee-cup calorimetry. This was then used to determine the quantity of heat flowing in and out of the system in multiple physical and chemical processes. B) Materials ● Goggles and Lab Coat ● A Coffee-cup Calorimeter #85 ● Thermometer ● 600 mLBeakers ● 100 mL Graduated cylinders ● Copper strip ● Hot plate ● Tongs/Towel ● Glass Beads ● Ring Stand ● Boiling stones C) Summary of Procedures a) Choice I For this experiment, a coffee-cup calorimeter was obtained along with two 100 mL graduated cylinders. 50 mL of cold water was put into the calorimeter tool and recorded the temperature constant. 150 mL of water was heated up to 90°C to 100°C. After the water reached this temperature, the water was place onto the side to cool for 2-3 minutes. The temperature was recorded and 50 mL was placed into the calorimeter and stirred together with the cold water for 30 seconds. The highest temperature reached by the water was then recorded. The temperature change, final temperature and calorimetry constant were determined and recorded. b) Choice II The same calorimetry assembly was used for this experiment to determine the temperature change when a hot metal is added to cold water. In this experiment an unknown metal was obtained and its identification code was recorded. A 600 mL beaker was filled with water along with tamer tabs to control the boiling water. 75 mL of cold water was obtained and added to the calorimeter. The temperature was then recorded. The water in the 600 mL beaker was brought to a boil at 90°C-100°C. Using a ring stand, the unknown metal was hung over with a string attached and lowered for the entire metal piece to be submerged. This stayed like this for 10 minutes and then transferred to the cold water when the metal was dry. The final temperature and temperature change was determined. After finding this the specific heat of the substance was determined along with the mean value of substance, literature value, reference and percent error in specific heat determination. c) Choice III

Like Choice I and II, the same coffee-cup calorimeter was used in the assembly. This time this experiment involved two different chemicals, HCl and NaOH. In this experiment 50 mL of NaOH was obtained and measured in a graduated cylinder. 50 mL HCl was also measured in a different graduated cylinder. Two thermometers were used in this experiment to prevent the accidental mixing of chemicals. The temperature was recorded for both and then both were mixed together when the temperatures were at a constant. The constant temperature after the mixture was then recorded. The final temperature, total volume of the mixture, temperature change, heat flow (J), moles of water produced, ΔH, the mean value of ΔH and the literature value was determined. D) Pre-laboratory Questions a) Choice I 1. What is the definition of the joule in terms of the basic SI units? A joule is the SI unit of heat and energy. One calorie equals 4.184 joules. 2. A calorimeter is to be calibrated: 51.203 g of water at 55.2°C is added to a calorimeter containing 49.783 g of water at 23.5°C. After stirring and waiting for the system to equilibrate, the final temperature reached is 37.6°C. Calculate the calorimeter constant. Cold water

Hot water

m=39.78 g

m= 51.2 g

T(i)= 23.5 Celsius

T(i)= 55.2 Celsius

T(f)= 37.6 Celsius

T(f)= 35.6 Celsius

s=4.18

s=4.18

(49.783)(4.18)(37.6-23.5)+(51.203)(4.18)(37.6-55.2)+c(37.6-23.5)=0 =59.1 J/C b) Choice II 1. Using the handbook chemical data, look up the specific heat capacities of the following materials, Give your references. Substance Specific heat Reference Al(s) 0.900 J/g°C CRC Handbook C(s) 0.170 J/g°C CRC Handbook Fe(s) 0.444 J/g°C. CRC Handbook H2O(l) 4.184 J/g°C CRC Handbook Pb(s) 0.128 J/g°C CRC Handbook 2. What are the units of specific heat capacity? J/g°C c) Choice III 1. The acids and bases to be determined in the experiment in this experiment are all classified as strong acids or bases. Use your

textbook to find what is meant by the word strong in this context. Acids and bases that are completely ionized when dissolved in water are called strong acids and strong bases 2. What is meant by neutralization? Neutralization reaction is a chemical reaction where an acid and a base react with each other in order to create salt and water; with strong acids and bases the essential reaction is the combination of hydrogen ions with hydroxyl ions to form water E) Data tables a) Choice I. Determination of a Calorimeter Constant Trial 1

Trial 2

Trial 3

Mass(volume) of cold water

50 ml

50 ml

50 ml

Temperature of Cold Water

23.9°C

25.3°C

25.3°C

Mass (volume) of hot water

50 ml

50 ml

50 ml

Temperature of hot water

75°C

72.4°C

77.2°C

Final temperature reached

45.1°C

46.7°C

47.6°C

Temperature change, ΔH

ΔT(hot) 29.9°C ΔT(cold) 21.2°C

ΔT(hot) 25.7°C ΔT(cold) 21.4°C

ΔT(hot) 29.6°C ΔT(cold) 22.3°C

Calorimetry constant

85.77

40.83

68.42

Mean value of Calorimeter constant

65.01

b) Choice II. Specific Heats of a Metal and Glass Specific heat of a metal Metallic substance used G

Trial 1

Trial 2

Trial 3

Initial heat of metal

100°C

100°C

100°C

Mass of substance taken

58.03 g

58.03 g

58.03 g

Mass (volume) cold water used

50 ml

50 ml

50 ml

Initial temperature of water

26.3 °C

26.3 °C

26.3 °C

Final temperature of water

33.5 °C

33.2 °C

27.2°C

Temperature change, ΔT

7.2°C

6.9°C

4.7°C

Specific heat of substance

.511

.48

.30

Mean value of specific heat of substance .39 =.388 Literal value .388 Reference CRC 86th Edition Percent error in specific heat determination .51 c) Choice III. Heat of Acid/Base Reaction Reaction of HCl and NaOH Trial 1

Trial 2

Volume of 2 M NaOH used

30 ml

30 ml

Initial temperature of NaOH

21.5°C

21.8°C

Volume of 2 M HCl

30 ml

30 ml

Initial temperature of HCl

21.6°C

21.8°C

Final temperature reached

35.3°C

35.5°C

Total mass (volume) of mixture

60 ml

60 ml

Temperature change, ΔT

13.7

13.7

Heat Flow, Joules

2571.91 J

2516.8 J

Moles of water produced

.06 mol

.06 mol

ΔH (kj/mole water)

-135 kj/mol

-135 kj/mol

F) Calculations Choice I: Determination of a Calorimeter Constant Mass of Unknown Metal: G Metal 58.03g Mass of water: 50.0g Initial Temperature of water: 23.9 ℃ Final Temperature of water: 45.1 ℃

We need to find change in temperature and calorimeter constant Trial 1 Final temperature 45.1℃- 23.9℃= 21.2℃ (50.0g) (4.184 J/g ℃ ) (45.1℃ -75.0℃ )+50.0g (4.184 J/g ℃ ) (45.1℃ -24℃ )+ C(45.1℃ 24℃)=0 Calorimeter Constant= 85.7 ℃ Trial 2: Mass of Unknown Metal: G Metal 58.03g Mass of water: 50.0g Initial Temperature of Water: 25.3℃ Final Temperature of water: 46.7℃ Find change in temperature and calorimeter constant Final Temperature 46.7℃-25.3℃= 21.4℃ (50.0g)(4.184 J/g ℃ )(46.7℃ -25.3℃ )+50.0g(4.184J/g℃)+C(46.7℃ -25.3℃)= 40.8℃ Trial 3: Mass of unknown metal: G Metal 58.03 g Mass of water 50.0g Initial Temperature: 25.0℃ Final Temperature of water: 47.6℃ Find change in temperature and calorimeter constant

Final Temperature: 47.6℃-25.0℃= 22.6℃ (50.0g)(4.184J/g℃ )(47.6℃ -25.0℃ )+50.0g(4.184J/g℃ )(47.6℃ -72.2℃ ) +C(47.6℃-25.0℃)=68.4℃ Mean of calorimeter constant (85.7℃ + 40.8℃ + 68.4℃) / 3 = 64.9℃ Choice II Specific Heats of Metals and Glass Trial I: Mean value of specific heat of substance (.511 + .48 +.30)/3= .39 Percent error in specific heat determination ((.39-.388)/.388) *100 =51%

Choice III HCl + NaOH à NaCl + H2O (-92) (-470) (-411) (-286) (1*-411 + 1*-286)- (1*-92 + 1* -470)= -135 kj/mol

G) Post-laboratory Questions 1. What effect on the calorimeter constant calculated would be observed if the calorimeter cup were made of a conducting material (such as metal) rather that plastic foam? The plastic foam is a better insulator than a metal so its able to keep the temperature from changing drastically. In the metal the temperature is lost to the surroundings faster and that would cause the calorimeter constant to be greater. 2. Why is water typically used as the heat-absorbing liquid in calorimeters? The specific heat in water is large which is beneficial when adding hot liquids because water can take high heats before its temperature starts to change. 3. What is the significance of the minus sign in Equation 8-2 in the Introduction of Choice I of this experiment? The - sign signifies that the water is losing energy. 4. A unit of heat energy that was formerly used frequently was the calorie.

Look up the definition of the calorie in your textbook or a handbook and record it here. The amount of heat required to raise the temperature of one gram of water. 5. Calculate the calorimeter constant for your calorimeter in cal/℃. Mean value of calorimeter constant 85.7℃ + 40.8℃ +68.4℃ = 64.9℃ Choice 2 post lab 1. In the pre-lab questions, you were asked to look up the specific heat of ice [i.e., H2O(s)]. Why is the value for ice not the same as for liquid of ice? The molecules in a solid such as ice are closely packed as opposed to the molecules in a liquid form. For this reason ice has a lower specific heat. Choice III Post lab 1. The heat flows measured in this experiment were actually not for the simple neutralization of a proton and hydroxide ion (as indicated in the Introduction of Choice III); rather they include contributions based on the fact that these species are hydrated in aqueous solution. What does it mean to say that a proton is hydrated, and how will the heat of hydration affect the measured heat flow for a neutralization reaction? Adding H2O to a H+ molecule is an example of hydrating a proton. This hydration affect will cause the heat flow to be higher.

H) Conclusion...