Hess\'s Law Lab Report PDF

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lab report on Hess's Law ...

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Thermodynamics and Calorimetry

Objective: The purpose of the lab is to combine two reactants in the calorimeter and measure the heats of reaction in solution, eventually allowing an individual to prove or disprove Hess’s law.

Introduction: Thermodynamics is the study of energy exchanged between the system and the surroundings. The heat of a reaction, change in enthalpy, or ΔH refer to the energy transferred from the surroundings in the form of heat in the course of a reaction occurring at constant pressure. If ΔH is greater than zero, the reaction is referred to as being endothermic, which mean heat is taken from the surrounding and the surroundings get colder. If ΔH is less than zero, the reaction is exothermic, meaning that heat is released, and the surroundings get warmer. Another important aspect of this experiment is that an object’s or solution’s temperature will rise in proportion to how much heat it has absorbed. The increase in temp that is directly proportional to the amount of heat absorbed is considered the constant of heat capacity. Therefore, the larger the heat capacity of the object, the smaller the temperature rise will be. The lab is a calorimetry experiment which measures the heat transfer associated with a process. A calorimeter is a device used to measure the heat of a reaction. For this lab, Styrofoam cups work because they are insulated, which prevents heat exchange within the environment. After determining the changes in temperature in reactions, it is evident that the total quantity of energy remains constant. Using Hess’s law, which states that change in enthalpy for a stepwise process is the sum of the enthalpy

changes of the steps, if the reaction is changed, ΔH reflects those changes, it is clear that the law of conservation of energy is applicable.

qrxn=m× C × ∆ T

KO H (aq )+ H Cl(aq ) → KCl( aq ) +H 2 O(l) ∆ H 1 -56.96 KOH ( s) → KOH ( aq ) ∆ H 2 -55.29 KOH ( s) + H Cl (aq ) → KCl(aq )+ H 2 O(l ) ∆ H 3 -112.25

ΔH1+ ΔH2= ΔH3

Methods: For the first week, set up a calorimeter from two Styrofoam coffee cups, one inside of the other one. Put a lid on the calorimeter and push a temperature probe through the lid hole. Calibrate the probe with a thermometer. Put 25.0mL of room temperature water in the calorimeter, then 75.0mL of ~40oC water in the other Styrofoam cup. Record the temperatures and the combine them in the calorimeter. Use the thermometer in the warm water, and the temperature probe in the calorimeter. Pour the warm water into the calorimeter then put the lid on and stir. Once the temperature stops increasing, the final temperature has been reached. Repeat the procedure twice. The second experiment includes the same process, but with 50 mL of 2 M HCl (aq) and 50 mL of 2M KOH. The third reaction is between HCl and KOH, which involves reacting 100 mL of 1M HCl with around 4 g KOH pellets. To determine the amount of heat that was produced, multiply the change in temperature by the sum of the heat capacity of the final solution and the Calorimetry

constant. The last reaction during the second week, set up the calorimeter and calibrate the temperature probe like the first week. The reactants are 100 mL of distilled water and 4 g KOH. Repeat both procedures twice and find the average values for each reaction.

Results and Discussion:

Table 1. First Calorimetry Experiment

Initial Temperature of the “cold” water – 25mL

21.8°C

Initial Temperature of the “hot” water – 75mL

43°C

Temperature after the waters mixed

31.7°C

Temperature change of the “hot” water

37.1°C – 43°C = - 5.9°C

Temperature change of the “cold” water

37.1°C – 21.8°C = 15.3°C

Q (heat) of the ‘hot” water

-1851.42 J

Q (heat) of the “cold” water

1600.38 J

Q of the calorimetry

16.41 J/°C

Table 2. Second Calorimetry Experiment

Initial Temperature of the KOH 1.55M – 50mL 22.1°C

Initial Temperature of the HCl 2M – 50mL

22.1°C

Temperature after the substances were mixed

33.2°C

Qcal

4.826 kJ

Qrxn

-4.826 kJ

Number of moles

0.0755 mol

ΔH

-62.3 J/mol

Week 1: q hot : (75 g ) × ( 4.184 J / g ∙℃ ) × ( 37.1℃−43 ℃ )=−1851.42 q cold : ( 25 g ) × ( 4.184 J / g ∙℃ ) × (37.1 ℃−21.8 ℃) =1600.38

Heat capacity of calorimeter (Ccal): J J +(4.184 J /g ∙ ℃ ) ∙100 g=434.81 16.41 ℃ ℃

Table 3. Second week Calorimetry values

Trial Number Which ΔH

Mass of KOH Change in Temp ΔH Value

1

ΔH

3

4.053 g

18.78°C

-113.046 kJ

2

ΔH

3

4.003 g

17.92°C

-109.218 kJ

3

ΔH

2

4.044 g

10.15°C

-61.234 kJ

4

ΔH

2

4.038 g

8.57°C

-51.779 kJ

Week 2: 434.81 J /℃ × ∆ T ¿ mol

J ×18.78 ℃ ℃ =−113.046 kJ 1 mol 4.053 g × 56.11 g

434.81 trial1 :

J ×17.92 ℃ ℃ =−109.22 kJ 1 mol 4.003 g × 56.11 g

434.81 trial2 :

J ×10.15 ℃ ℃ =−61.23 kJ 1 mol 4.044 × 56.11 g

434.81 trial3 :

J ×8.57 ℃ ℃ trial 4 : =−51.78 kJ 1 mol 4.038 g × 56.11 g 434.81

H2O and KOH heat of reaction 30

Temperature (C)

25 20 15 10 5 0 0

10

20

30

40

50

60

Time (seconds)

Figure 1. Graph of change of heat in reaction between water and KOH, which displays the heat of the reaction.

HCl and KOH Heat of Neutralization 20.5 20

Temperature (C)

19.5 19 18.5 18 17.5 17 16.5 0

10

20

30

40

50

60

Time (seconds)

Figure 2. Graph of change of heat in reaction between HCl and KOH, which displays the heat of neutralization.

Conclusion: ¿−111.132 kJ −(−112.25 kJ ) ∨

¿ ×100=0.99 % −112.25 kJ

% error :¿

In order to ensure that all of the heat goes into a medium with a known heat capacity, it is important to keep the heat from leaving the calorimeter. As well, it is essential to know the heat capacity of the calorimeter and the solutions in it. At the end, when the solution in the calorimeter is well-mixed, the reaction is complete. Throughout the reaction, if the mixture is not stirred fast enough, the resulting line is not vertical, meaning the transition will be faster if NaOH is added rapidly and stirred well, which leaves a nearly vertical temp. rise. After the rise, the reaction is complete. If the temperature probe is not placed into the calorimeter fast enough, it will result in an incorrect initial temperature, varying the change in temperature in the reaction. Since the ΔH for the reactions is negative, there is a release in energy. The graph scale should be 1 Hz instead of 10 Hz in order to accurately demonstrate the release in heat. Overall, the ΔH of the reactions should mirror the values of Hess’s Law....