DEPARTMENT OF CHEMISTRY

FACULTY OF SCIENCE & MATHEMATICS

UNIVERSITI PENDIDIKAN SULTAN IDRIS

CHEMISTRY 2 SKU 3023

LABORATORY REPORT

NAME

MATRIC NUMBER

Aqmarollah bin Mohd Nasip

D20141066992

Muhammad Zul Afiq bin Anuar

D20141066978

Muhammad Nurakmal bin Abdul Rahman

D20141066961

DATE : 15/4/2015

LECTURER : Prof. Madya Dr. Ismail bin Zainol

EXPERIMENT : Entalphy

EXPERIMENT 4 : ENTHALPY

OBJECTIVES :

1.Determine the enthalpy of neutralization of strong acid and weak acid

2.Determine the quantity and direction of the heat transfer in the dilution of a salt

INTRODUCTION :

Enthalpy is the amount of heat content used or released in a system at constant pressure. Enthalpy is usually expressed as the change in enthalpy. The change in enthalpy is related is related to a change in internal energy (U) and a change in the volume (V) which is multiplied by the constant pressure of the system.

Enthalpy of neutralization is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of base undergo neutralization reaction to form water and salt. When a reaction is carried out under standard conditions at the temperature of 298 K and 1atm of pressure and one mole of water is formed,it is called the standard enthalpy of neutralization.

Enthalpy of solution is produced when a salt dissolve in water. It can be defined as the enthalpy change associated with the dissolution of a substance in a solvent at constant pressure resulting in infinite dilution. The enthalpy of solution is most often expressed in kJ/mol at constant temperature.

PROCEDURE :

4a) Heat of Neutralization of Acid-Base

1.50.0 mL 1.0 M HCL was obtained and the temperature was measured

2.50.0 mL 1.0 M NaOH standard solution was obtained and put into a dry stiroform cup and the temperature is measured. The molar concentration for the base was recorded.

3.Acid was added to the base in the stiroform cup quickly. The cup was covered by placing a lid on the top to prevent heat lost. The cup was stirred and temperature recorded for 10-20 seconds.

4.Temperature-Time graph was plotted and the maximum temperature for neutralization reaction was determined.

5.Experiment repeated to get accurate data.

6.Steps 1-5 repeated by using HNO3 and both of the Hn was compared.

4b) Heat of Salt Solution

1.5.0 g salt was weight.

2.Mass of dry cup was measured. Then, 20 mL distilled water was put into it and temperature was recorded.

3.Salt added into the stiroform and covered with lid. The cup was swirled until salt completely dissolved. Temperature recorded for every 10-20 seconds.

4.Temperature-Time graph was plotted and maximum temperature for neutralization of the reaction was determined.

5.Experiment repeated to get accurate data.

RESULT

Enthalphy of neutralization for acid-base reaction

HCl + NaOH -> NaCl + H2O

HCl +HNO3 -> NOCl + H2O

Trial 1

Trial 2

Trial 1

Trial 2

oC

Time(s)

oC

Time(s)

oC

Time(s)

oC

Time(s)

26

20

26

20

22

20

22

20

26

40

26

40

22

40

22

40

26

60

26

60

22

60

22

60

26

80

26

80

22

80

22

80

26

100

26

100

21.8

100

21.8

100

26

120

26

120

21.8

120

21.8

120

26

140

26

140

21.5

140

21.5

140

25.8

160

25.8

160

21.5

160

21.5

160

25.5

180

25.5

180

21.5

180

21.5

180

25.5

200

25.5

200

21.5

200

21.5

200

25

220

25

220

21.5

220

21.5

220

Enthalphy of neutralization for salt solution

Trial 1

Trial 2

oC

Time(s)

oC

Time(s)

36

20

36

20

36

40

36

40

36.3

60

36.3

60

36.9

80

36.9

80

36.9

100

36.9

100

37.3

120

37.3

120

37.4

140

37.4

140

37.3

160

37.3

160

37.5

180

37.5

180

37.5

200

37.5

200

37.5

220

37.5

220

Experiment 4a : Heat of neutralization of acid-base

Item

HCl + NaOH

HN + NaOH

Trial 1

Trial 2

Trial 1

Trial 2

Acid volume ( mL)

50

50

50

50

Acid temperature (

24.6

24.6

22

22

NaOH volume ( mL)

50

50

50

50

NaOH temperature ()

23

24

23

23

NaOH concentration (mole/L)

1.0

Maximum temperature from graph()

i. Calculation of the Heat of neutralization

Item

HCl + NaOH

HN + NaOH

Trial 1

Trial 2

Trial 1

Trial 2

Average initial temperature of acid and base ()

26

27

22

21

Temperature change,T ()

28

29

24

23

Volume of final mixture (mL)

500

500

500

500

Mass of final mixture (mL)

100

100

100

100

Specific heat of the mixture

4.18 J/g.

4.18 J/gC

Yielded heat (J)

836

836

836

836

Amount of O reacted (mol)

0.05

0.05

0.05

0.05

Amount of produced (mol)

0.05

0.05

0.05

0.05

Yielded heat per mole ,

1106.48

1106.48

1106.48

1106.48

Average ( kJ/mole H2O

)

22.13

22.13

22.13

22.13

Calculation

*Density of solution is 1.0 g/mL

HCl + NaOH

1. Temperature of calorimeter and NaOH = 26 C

2. T determined from curve after adding HCl to the NaOH = 28C

3. Heat gained by solution = q= Msolution Cwater t

= (100 g)( 4.18 J/g. C)(28-26) C = 836 J

4. Heat gained by calorimeter = q = Ccalt

= (135.24 J/C)(28-26) C

= 270.48 J

5. Total joules released by reaction [(3) + (4)]= 836 J + 270.48 J

= 1106.48 J

6. HCl + NaOH NaCl + H2O

7. The number of moles of HCl in 50 mL of 1.0 M HO = molarity x liters of solution

= 1.0 M x 0.05 L

= 0.05 mol HCl

8. The number of moles of H2O produced in reaction of : 50 mL 1.0 M HCl and 50 mL 1.0 M NaOH

= number of moles of HCl ratio of moles of H2O to HCl

= 0.05 mol HCl (1mol of H2O /1mol of HCl)

= 0.05 mol water

9. Joules released per mole of water formed

= total joules released / number of moles of water produced

= 1106.48 J / 0.05 mol H2O

= 22.13 kJ

HNO3 and NaOH

1. Temperature of calorimeter and NaOH = 22 C

2. T determined from curve after adding HC2H3O2 to the NaOH = 24 C

3. Heat gained by solution = q= Msolution Cwater t

= (100 g)( 4.18 J/g. C)(24-22) C

= 836 J

4. Heat gained by calorimeter = q = Ccalt

= (135.24 J/C)(24-22) C

= 270.48 J

5. Total joules released by reaction [(3) + (4)]= 836 J + 270.48 J

= 1106.48 J

6. HC2H3O2 + NaOH NaC2H3O2 + H20

7. The number of moles of H2O produced in reaction of: 50 mL 1.0 M HC2H3O2 and 50 mL 1.0 M NaOH

= number of moles of HC2H3O2 ratio of moles of H2O to HC2H3O2

= 0.05 mol HC2H3O2 (1mol H2O/1mol HC2H3O2)

= 0.05 mol H2O

8. Joules released per mole of water formed

= total joules released / number of moles of water produced

= 1106.48 J / 0.05 mol H2O

= 22.13 kJ

Experiment 4b : Heat of salt solution

Salts :

Items

Trial 1

Trial 2

Mass of salt (g)

5.000

5.125

Mole of salt (mole)

0.005

0.005

Mass of cup and water(g)

22.147

23.150

Mass of stiroform cup (g)

2.110

2.100

Mass of water (g)

20.037

21.050

Initial temperature of water(g)

34

33

Initial temperature of water () from graph

ii. Calculation of Heat of Salt Solution

Items

Trial 1

Trial 2

Temperature change , T ()

30

35

Water heat release (J)

-335.02

-87.99

Salt heat released (J)

-540.96

-135.24

Total enthalpy change

-875.98

-223.23

Amount of O recorded

0.005

0.005

Amount of produced

0.005

0.005

( J/mole salt)

-875.98

-223.23

Average Hs

-549.61

Calculation

1. Temperature of salt and water= 34 C

2. T determined from curve of water= 30C

3. Heat released by water = q= Mwater Cwater t

= (20.037 g)( 4.18 J/g. C)(30-34) C = -335.02 J

4. Heat released by salt = q= Csalt t

= (135.24 J/C)(30-34) C

= -540.96 J

5. Total joules released by reaction [(3) + (4)]= -(335.02+540.96) J

= -875.98 J

6. HCl + NaOH NaCl + H2O

7. The number of moles of salt in 0.005 L of 1.0 M H2O = molarity x liters of solution

= 1.0 M x 0.005 L

= 0.005 mol salt

8. The number of moles of H2O produced in reaction of : 5 mL 1.0 M salt and 5 mL 1.0 M water

= number of moles of salt ratio of moles of H2O to salt

= 0.005 mol salt (1mol of H2O /1mol of salt)

= 0.005 mol water

9. Joules released per mole of water formed

= total joules released / number of moles of water produced

= -875.98 J / 0.005 mol H2O

= -4.38 x 103 kJ

DISCUSSIONS

The basic principle of heat transfer applied to obtain the calorimeter heat capacity and heat of neutralization is when heat is evolved, the reaction is exothermic. If heat is absorbed, the reaction is endothermic. In this experiment, two exothermic reactions will be observed, and the heat of reaction for each will be found. The basic principle for the heat of neutralization, which is the enthalpy change produced when an acid and a base react to form water.

The first three readings temperature were recorded for every 5 seconds because in order to detect if there are small change of temperature as the solution does not mix well yet. This time was used to wait for the solution to well mixed and reach their equilibrium temperature. At this time, the transfer of heat is in action. It is also to get the average temperature of the calorimeter. This average temperature reading would portray the stable equilibrium reading of the calorimeter.

The value obtain are experimental data that must be compared to the true value in order to gauge the accuracy of this experiment. The theoretical value of the joules released per mole of water for the HCl reaction is 58.2 kJ/mol while for acetic acid is 55.8 kJ/mol.

When the magnitude of heat of neutralization for weak and strong acid is compared the number of moles of NaOH acting on acid in both experiment are same. While initial any final temperature may vary, the change in temperatures should be the same. Thus, the two heat of neutralization should be the same.

A few suggestions appropriate apparatus should be used in order to get more accurate result is a pipet should have been used for maximum accuracy because using a graduated cylinder leaves room for a crucial error in volume determination, which would then lead to errors in determination of mass, molar content of the solution, and every other derivative formula. In addition, the thermometer had to be calibrated, which improves accuracy but is itself an imprecise technique. The thermometer and the calorimeter should wipe till it completely dry to prevent the errors in reading. The calorimeter also should run in closed area to prevent the heat easily lost to the surrounding.

CONCLUSION

For experiment part A, heat capacity of calorimeter is equal to heat gained by the calorimeter / temperature increases and the answer is 270.48 J.

For experiment part B, we calculated heat of neutralization of HCI-NaOH. At first, temperature of calorimeter and NaOH is 26.0 C. While we can see that T determined from curve after adding HCl to the NaOH is .Heat gained by calorimeter is 270.48 J. Heat gained by the calorimeter for heat neutralization of HCI NaOH is 1106.48 J.

For experiment part C, we can conclude that there are slightly different in heat gained by calorimeter of strong acid of HCI and weak acid, HC2H3O2. The heat gained by calorimeter of weak acid is low than heat gained by calorimeter of strong acid.

At the end,we can say that the experiment was successful as the procedure was carried out correctly. The calorimeter prove as a good heat insulator for this experiment, as it prevent the heat lose as the solution is put in it. Even the result is not too accurate, we can conclude that neutralization process releases energy in the form of heat and it is exorthermic processes. Moreover, accurate apparatus can lead to a improvement in the result we got.

REFERENCES

Raymond,C. Chemistry Tenth Edition. (2010). McGraw.Hill International Edition

Laboratory Manual SKU 3023 (Chemistry II)