The specific heat capacity and enthalpy of dissolution Lab PDF

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The Specific Heat Capacity and Enthalpy of Dissolution DATA Part 1: Enthalpy of Dissolution

Table 1: Calculations for determining the Enthalpy of dissolution

Trial

Mass (g)

Volume Water (mL)

KNO3

1

2.7674 g

200 mL

25.00⁰C

23.13⁰C

S.S

2

2.4667 g

200 mL

25.00 ⁰C

22.57 ⁰C

3

2.0204 g

200 mL

25.01 ⁰C

23.31 ⁰C

KOH

1

2.4748 g

200 mL

25.00 ⁰C

30.42⁰C

S.S

2

2.5161 g

200 mL

25.01 ⁰C

31.66⁰C

3

2.0379 g

200 mL

25.00 ⁰C

31.08 ⁰C

Tinitial  (⁰C)

Tfinal  (⁰C)

Figure 1: Temperature curve for KNO3

Figure 2: Temperature curve for KOH

Part 2: Specific Heat Capacity

Table 2: Calculations for determining the Specific Heat Capacity

Metal

Mass Metal (g)

Tinitial,metal (⁰C)

Tfinal,metal (⁰C)

Volume Water (mL)

Tinitial,wate r (⁰C) 

Tfinal,water (⁰C)

Zn

18.9416 g

200 ⁰C

27.87 ⁰C

100 mL

24.99 ⁰C

27.87⁰C

Cu

24.7173 g

200 ⁰C

28.79⁰C

100 mL

24.99⁰C

28.79⁰C

DATA ANALYSIS Part 1 1. Use the equation below to calculate the amount of heat energy gained by the solution (qsol). In determining the mass of water, m, use the density of water at 25o C = 0.998 g/mL. Use 4.184 J/(g•°C) as the specific heat, Cs, of the solution. qsoln = Cs ´ msolution ´ ∆Twater C= 4.184 J/(g•°C) D= 0.998 g/mL msolution= density (0.998 g/mL) x volume of solution (200 mL) ∆Twater = 25 oC

KNO3 trial 1: msolution= (0.998 g/mL) X (200mL) + 2.7674g =202.37 g qsoln = 4.184 J/(g•°C) X (202.37 g) X (23.13 °C - 25.00°C) = -1582.2 J --------------------------------------------------------------------------------------------------------------------KOH trial 1: msolution= (0.998 g/mL) X (200 mL) + 2.4748 g = 202.07 g qsoln = 4.184 J/(g•°C) X (202.07 g) X ( 30.42 °C - 25.00 °C) = 45783 Using the equation qsoln = Cs ´ msolution ´ ∆Twater plug in 4.184 J/(g•°C) as C, msolution is going to be (0.998 g/mL x 200 mL) then ∆Twater will be the temperature of the water in this case 25.00 °C. Solve for each reaction inorder to determine the amount of heat gained 2. The heat calculated above represents the heat gained by the solution (the solution being predominantly water). Since we are interested in the enthalpy of dissolution, we need to remember that it will be equal in magnitude but opposite in sign to the head gained or lost by the solution. This relationship can be expressed by the following equation: qdissolution = -qsoln KNO3 trial 1: qdissolution = -qsoln = 1582.2 J = - 1582.2 J

--------------------------------------------------------------------------------------------------------------------KOH trial 1: qdissolution = -qsoln = 45783 J = -45783 J 3. Determine the number of moles of salt that were dissolved. Use the moles of the sald along with qdissolution to determine the enthalpy change, ∆Hdissolution in terms of kJ/mol of dissolved salt. This is your experimental value of ∆Hdissolution . ∆Hdissolution = qrxn/moles salt KNO3 trial 1 2.7674 g / 101.11= 0.027370 mol Qdissolution = 1.5822 kJ/ 0.027370 mol = 57.808 kJ/mol --------------------------------------------------------------------------------------------------------------------KOH trial 1 2.4748 g/ 56.1056 g/mol = 0.044110 mol Qdissolution = 4.5782 kJ / 0.044110 mol = 103.81 kJ/mol The quantity of kJ/mols of dissolved salt should be determined. Take the mass (g) / molar mass (g/mol) to find the moles of salt. In order to provide the Qdissolution in kJ/mol, take the amount of kJ for each component produced and divide by the mol. 4. Look up the enthalpy of dissolution for each potassium nitrate and potassium hydroxide. Use those values to determine the % error. Be sure to provide a citation for your source of the accepted values.

Sample Calculations: KNO3 trial 1: msolution= (0.998 g/mL) X (200mL) + 2.7674g =202 g qsoln = 4.184 J/(g•°C) X (202 g) X (23.13 °C - 25.00°C) = -1582 J 2.7674 g / 101.11= 0.027370 mol Qdissolution = 1.5822 kJ/ 0.027370 mol = 57.808 kJ/mol (( 34.89 kj/mol - 57.808 kJ/mol ) / 34.89 kj/mol)) x 100= 65.69 %

--------------------------------------------------------------------------------------------------------------------KOH trial 1: msolution= (0.998 g/mL) X (200 mL) + 2.4748 g = 202 g qsoln = 4.184 J/(g•°C) X (202 g) X ( 30.42 °C - 25.00 °C) = 45783 J qdissolution = -qsoln = 45783 J = -45783 J 2.4748 g/ 56.1056 g/mol = 0.044110 mol Qdissolution = 4.579221084 kJ / 0.044110 mol = 103.81 kJ/mol ((-57.61kj/mol - 103.81 kJ/mol) / -57.61 kj/mol)) x 100= 280.0% Citation for source of enthalpy of dissolution values:

Libretexts. “Chapter 9.5: Enthalpies of Solution.” Http://Chem.libretexts.org/, Libretexts, 11 Aug. 2020 “Potassium Nitrate.” ChemSpider, www.chemspider.com/. Part 2 1. For this, it is important to remember that the heat gained by the water is equal to the heat lost by the metal. Don’t forget the change in sign. qmetal = -qwater This can be restated as this Cs, metal ´ mmetal ´ ∆T metal = - Cs, water ´ mwater ´ ∆Twater Zinc (Cs ) ( 18.9416 g) (200 °C - 27.87 °C) = ( 4.184 J/(g•°C) (99.8) (27.87 °C x 24.99 °C) Copper (Cs ) ( 24.7174 g) (200 °C - 28.79 °C) = ( 4.184 J/(g•°C) (99.8) (28.79 °C x 24.99 °C)

2. Solve for Cs. In determining the mass of water, m, use the density of water at 25oC = 0.998 g/mL. Use 4.184 J/(g•°C) as the specific heat, Cs, of water. Zinc (Cs ) ( 18.9416 g) (200 °C - 27.87 °C) = ( 4.184 J/(g•°C) (99.8) (27.87 °C - 24.99 °C) 3260.417608 = 1202.582016 (Cs ) = 0.369 J/(g•°C) Copper (Cs ) ( 24.7174 g) (200 °C - 28.79 °C) = ( 4.184 J/(g•°C) (99.8) (28.79 °C - 24.99 °C) 4231.866054 = 1586.74016 (Cs ) = 0.375 J/(g•°C) Using this equation (Cs ) (mass g) x (Tinitial,metal C - Tfinal,metal C ) = - (4.184 J/(g•°C) x (density) x ( Tfinal,metal C - Tinitial,water C), then take the right side and divide by the left which gives the answer of what (Cs ) would be. 3. Look up the specific heat capacities for each of the metals. Use those values to determine the % error of your results. Be sure to provide a citation for your source of the accepted values. Sample calculations: Zinc → 0.39 (Cs ) ( 18.9416 g) (200 °C - 27.87 °C) = ( 4.184 J/(g•°C) (99.8) (27.87 °C - 24.99 °C) 3260.417608 = 1202.582016 (Cs ) = 0.369 J/(g•°C) ((0.39 J/(g•°C) - 0.36884294 J/(g•°C)) / 0.39 J/(g•°C)) x 100% = 5.4% error Copper → 0.385 (Cs ) ( 24.7174 g) (200 °C - 28.79 °C) = ( 4.184 J/(g•°C) (99.8) (28.79 °C - 24.99 °C) 4231.866054 = 1586.74016 (Cs ) = 0.375 J/(g•°C) ((0.385 J/(g•°C) - 0.3749504686 J/(g•°C) ) / 0.385 J/(g•°C)) x 100% = 2.61 % error

Citation for source of specific heat capacities: Specific Heat Capacity Table, www2.ucdsb.on.ca/tiss/stretton/database/Specific_Heat_Capacity_Table.html.

RESULTS Table 3: Using Empirical (kJ/mol) and Accepted (kJ/mol) to calculate percentage error for KNO3 and KOH .

Salt

∆Hdissolution Empirical (Kj/mol)

∆Hdissolution Accepted (Kj/mol)

% error

KNO3

57.808 kJ/mol

34.89 (Kj/mol)

65.69 %

KOH

103.81 kJ/mol

-57.61 (Kj/mol)

280.0%

Was the addition of potassium nitrate to water an endothermic or exothermic reaction? What is the sign of ΔH for this process? In addition to adding potassium nitrate to water this caused an endothermic reaction. The reaction of absorbing heat during the endothermic reaction was the process of ΔH being positive.

Was the addition of potassium hydroxide to water an endothermic or exothermic reaction? What is the sign of ΔH for this process? In addition to adding potassium hydroxide to water this caused an exothermic reaction. The reaction of heat being released during the exothermic reaction was the process of ΔH being negative.

Table 4: Using Cs Empirical (J/g· C) and Cs Accepted (J/g· C) to calculate percentage error for Zn and Cu .

Metal

Cs Empirical (J/g·⁰C)

Cs Accepted (J/g·⁰C)

% error

Zn

0.369 J/(g•°C)

0.39 (J/g·⁰C)

5.4%

Cu

0.375 J/(g•°C)

0.385 (J/g·⁰C)

2.61%

In the equation, qmetal = -qwater, what does the negative sign (-) represent? Could it also be written as -qmetal = qwater? The negative sign means that the water absorbs the amount of heat emitted by the metal. The negative sign shows one direction of heat flow....