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2011

Edited by: Faridah Yusoff

Ramli Hitam

Norlinda Daud

SKU 3023: Chemistry II


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SAFETY AND RULES IN THE CHEMISTRY LABORATORY

1. Do not enter the laboratory without the presence of the laboratory instructor.2. No experiment to be performed without the permission of the laboratory instructor.3. Safety glasses must be worn at all the times in the laboratory.4. Laboratory coat must be worn at all the times in the laboratory.5. Long hair and scarf must be properly manageable.6. Do not wear slippers, sandals, heels or sides open shoes.7. Do not wear contact lenses.8. No smoking, drinking and eating at all the times in the laboratory.9. Do not throw waste organic liquids into the sink.10. Know the location of all safety equipment.11. Read the label on the container of a chemical twice.12. Use a fume hood for poisonous or irritating fumes.13. Throw all unused or contaminated chemical properly.14. Do not return use reagent to the stock bottle.15. Do not send your product with your report.16. Evacuate the laboratory when a fire alarm sounds.17. Turn off the flame and switch off the hot plate before leaving the laboratory.

18.Always practice goods housekeeping.

19.Always add acids to water.20. Do not aim the opening of any glassware at yourself or anyone else.21. Do not use cracked or chipped glassware.


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22. Never use mouth suction when using a pipette.23. Report any accident in the laboratory.24. Do not make fun and joke in the laboratory.25. Do not leave any heating, vigorous or rapid reaction unattended.


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EXPERIMENT 1: ENTHALPY

1.1: OBJECTIVES

1. To determine the enthalpy of neutralization of strong acid and weak acid.2. To determine the quantity and direction of the heat transfer in the dilution of a salt.

1.2: INTRODUCTION

Thermochemistry deals with the heat involved in the chemical and physical change and especially with

the concepts of enthalpy. Enthalpy is a thermodynamic variable for reactions at constant pressure. The

enthalpy of a system is defined as the internal energy plus the product of the pressure and volume.

The change in the enthalpy () equals the heat gained or lost at constant pressure. It dependsonly on the difference between and. The enthalpy change of reaction also called the heatof reaction, , always refers to . An exothermic process releases heat and resultsin a decrease in the enthalpy of the system. An endothermic process absorbs heat and results in an

increase in the enthalpy of the system.

Heat capacity is the quantity of heat required to change its temperature by 1 with the unitJ

. Specific heat capacity, s, is the quantity of heat required to change the temperature of 1g of a

substance by 1.Specific heat =

A calorimeter is a device used to measure the heat released or absorbed by the physical or

chemical process. The coffee-cup calorimeter has been used in this experiment to measure the heat at

constant pressure. The enthalpy of neutralization involves acid base reaction and for strong acid and

base, the reaction is exothermic. Hn is determined by

Salt dispersion enthalpy, HS is determined experimentally with the combination of heat loss from the salt

and water whenever both of them are mixed.

Hn = specific heat H2O x total mass acid-base x T

Hs = (heat loss H2O ) + (heat loss salt)


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1.4: RESULTS AND DATA

A: Table of Data to Plot Graph

Enthalpy (heat) of neutralization for acid-base reactionEnthalpy (heat) of solution for

the salt

HCl NaOH HNO3 NaOHTrial 1 Trial 2

Trial 1 Trial 2 Trial 1 Trial 2

t

(sec)

T

(C)

t

(sec)

T

(C)

t

(sec)

T

(C)

t

(sec)

T

(C)

t

(sec)

T

(C)

t

(sec)

T

(C)

0 31 0 29 0 30 0 30 0 23 0 23

20 31 20 29 20 30 20 30 20 23 20 23

40 31 40 29 40 29 40 30 40 23 40 23

60 31 60 29 60 29 60 30 60 23 60 23

80 31 80 29 80 29 80 30 80 23 80 23

100 31 100 29 100 29 100 30 100 23 100 23

120 31 120 29 120 29 120 30 120 23 120 23

140 31 140 29 140 29 140 30 140 23 140 23

160 31 160 29 160 29 160 30 160 23 160 23

180 30 180 29 180 29 180 30 180 23 180 23

200 30 200 28 200 29 200 29 200 23 200 23

220 30 220 28 220 29 220 29 220 23 220 23

240 30 240 28 240 29 240 29 240 23 240 23

260 30 260 28 260 29 260 29 260 23 260 23

280 30 280 28 280 29 280 29 280 23 280 23


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B: Heat of Neutralization

ItemHCl + NaOH HNO3 + NaOH


1. Acid volume (mL) 50.0 50.0 50.0 50.0

2. Acid temperature (C) 23.0 23.0 24.0 24.0

3. NaOH volume (mL) 50.0 50.0 50.0 50.0

4. NaOH temperature (C) 23.0 23.0 24.0 24.0

5. NaOH concentration (molL-1) 1.00 1.00

6. Maximum temperature from graph (C) 30 29 30 30

Section C: Calculation of the Heat of Neutralization

ItemHCl + NaOH HNO3 + NaOH


1. Average initial temperature of acid and base (C) 23.0 23.0 24.0 24.0

2. Temperature change, T (C) 6.6 5.7 5.1 5.7

3. Volume of final mixture (mL) 100 100 100 100

4. Mass of final mixture (g) 100 100 100 100

5. Specific heat of the mixture 4.18 Jg-1oC 4.18 Jg-1oC

6. Yielded heat (J) -2758.8 -2382.6 -2131.8 -2382.6

7. Amount of OH- reacted (g) 0.85 0.85 0.85 0.85

8. Amount of H2O produced (g) 0.90 0.90 0.90 0.90

9. Yielded heat per mole H2O, Hn (kJmol-1 H2O) -55.176 -47.652 -42.636 -47.652

10. Average Hn (kJmol-1 H2O) -51.414 -45.144


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Section D: Heat of Salt Solution (e.g: NaCl, Na2SO4, KOH, Na2S2O3)

Salt: NaCl

Item Trial 1 Trial 2

1. Mass of salt (g) 5.023 5.008

2. Mole of salt (mole) 0.0859 0.0857

3. mass of cup and water (g) 21.378 21.116

4. Mass of Styrofoam cup (g) 2.126 2.182

5. Mass of water (g) 19.252 18.934

6. Initial temperature of water (C) 25.0 25.0

7. Initial temperature of water (C) from graph 23.0 23.0

Section E: Calculation for Heat of Salt Solution

Item Trial 1 Trial 2

1. Temperature change, T (C) 2.0 2.0

2. Water heat released (J) 160.945 158.288

3. Salt heat released (J) 8.680 8.654

4. Total enthalpy change -169.625 -166.942

5. Amount of OH- reacted (g) 1.4603 1.4569

6. Amount of H2O produced (g) 1.5462 1.5426

7. HS (kJ/mol salt) -0.251 -0.251

8. Average HS (kJ/mol salt) -0.251


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1.5: QUESTIONS

1. Show your calculation for Trial 1 (in sec. C)

HCl + NaOH

i) Average initial temperature of acid and base for HCl + NaOH:Average initial =

= 23Average final temperature of acid and base for HCl + NaOH:

Average final = 29.6ii) Temperature change, = 29.6 - 23.0

= 6.6

iii) Volume of final mixture = 100.0 mLiv) Mass of final mixture =100.0 gv) Specific heat of mixture = 4.18 Jvi) Yielded heat (J)

x total mass acid base x T= - 4.18 x 100g x 6.6=

vii) Amount of

reacted

+

1 mol 1 mol x mol 0.05 mol

x = 0.05 mol

mass of = 0.05 mol x 17.0 g/mol = 0.85 g


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viii)Amount of produced +

1 mol 1 mol 0.05 mol 0.05 mol mass of = 0.05 mol x 18.0 g/mol

= 0.9 g

ix) Yielded heat per mole , =

= -55176 J

= -55.176 kJ

x) Average Trial 1 + trial 2= -55.176 + -47.652

2

= -51.414

+ NaOHi) Average initial temperature of acid and base for HCl + NaOH:

Average initial = = 24

Average final temperature of acid and base for HCl + NaOH:

Average final = 29.1ii) Temperature change, = 29.1 - 24.0

= 5.1iii) Volume of final mixture = 100.0 mLiv) Mass of final mixture =100.0 gv) Specific heat of mixture = 4.18 Jvi) Yielded heat (J)

x total mass acid base x T= - 4.18 x 100g x 5.1=


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vii) Amount ofreacted +

1 mol 1 mol x mol 0.05 mol

x = 0.05 mol mass of = 0.05 mol x 17.0 g/mol

= 0.85 g

viii)Amount of produced +

1 mol 1 mol 0.05 mol 0.05 mol mass of = 0.05 mol x 18.0 g/mol

= 0.9 g

ix) Yielded heat per mole , =

= -42636 J

= -42.636 kJ

x) Average = -42.636 + -47.652

2

= -45.144


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2. Compare both Hn values and give your opinion.

The average Hn of the reaction between HCl and NaOH was -51.414 while thereaction ofand NaOH was -45.144. HCl yielded higher heat per mole water because it wasstronger acid compared to . The acidity constant () of HCl is higher. It ionizes more .

A strong acid like hydrochloric acid, (HCl) is completely ionized.

A strong acid will have an extremely large value for Ka, whereas a weak acid will have a small value.

A concentrated acid has a high initial value for [HA] and a dilute acid has a low initial [HA].

pH depends on [H+] which depends on concentration and acid strength.

3. Write down the balance equation for the neutralization.

4. Show your calculation for Trial 1 ( in sec. E)

i) Temperature change, = final initial

= 23

- 25

= - 2= 2

ii)Water heat released x total mass x T

= -4.18 J x 19.252 g x 2= -160.945 J

iii)Salt heat released x total mass x T

= -0.864 J x 5.023 g x 2= -8.680 J


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iv)Total enthalpy changeEnthalpy salt + enthalpy water = -8.68 J + -160.945 J

= -169.625 J

v)Amount of reacted 1 mol NaCL 1 mol NaOH0.0859 mol NaCL 0.0859 mol NaOH

Mass = mol x molar mass = 0.0859 mol x 17 g/mol

= 1.4603 g

vi)Amount of producedMass

= 0.0859 mol x 18 g/mol

= 1.5462 g

vii) Salt dispersion enthalpy,

(

)

=

= -150.474 J + -101.05

= -251.523 J

= -0.251 kJ

viii) average =

=

=


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5. Instead of using specific heat, what other formula can you use to calculate the heat of reaction?

Explain in what condition that formula can be applied.

can use the molar heat capacity.

It is the energy required to raise one mole of any substance by one degree Celsius.

molar heat capacity = _ J_ _mole oC

A useful relationship is J X g = Jg oC mole mol oC

It can be used when the value of specific heat capacity was not given and there was no value oftemperature differences, T.

6. What is the appropriate name should be given to the Styrofoam cup as an apparatus to measure

heat of reaction?Coffee-cup calorimeter


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1.6: DISCUSSION

Experiment A

This experiment is carried out to determine the enthalpy of neutralization of strong acid,

hydrochloric acid HCl and weak acid, . Both of the acids reacted with the strong basesodium hydroxide, NaOH. Both resulted with different enthalpy of neutralization. Through this experiment,

the strong acid, HCl produced higher enthalpy. This happened due to its ionization property. It ionized

completely, thus higher energy required and more heat was produced.

Both reactions were exothermic reactions which release heat and resulted in the negative values of

heat enthalpies. of the HCl-NaOH reaction was -51.414 kJ/mol which is near to the theoretical value of-57.0kJ/mol. Whereas the -NaOH reaction came out with lower value of -45.144kJ/mol. The valueswere obtained by the calculation using the formula of specific heat capacity. All the acids and base were

assumed to have the same specific heat capacities as same as water, 4.18kJ/mol. All the equations of the

reactions must be balanced first to get the correct ratio to each other compound.

Most of the trials have decrements on the temperature during the reaction (plotted on graph). This

happened due to the surrounding temperature which is much lower than in the system experimented. Thus,

it lowers the temperatures of the solutions through the period of time. Although the styrofoam cup has been

used, the heat transfer between the system and the surrounding still happened due to the errors on the

apparatus. The lid might not been covered completely.

Experiment B

Sodium chloride salt has been used in this experiment to react with water to produce acid and base. Both

trials shows decrement of 2 of the temperature which means both reactions undergo exothermic process.There was no much difference in the result for both trials which ended up with salt dispersion enthalpies of

-0.251kJ/mol salt. Balanced equation must be written first to avoid mistake in determining mol and mass of

any compound.


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1.7: CONCLUSION(S)

All the reactions experimented undergo exothermic process. Thus, all of the reactions give

negative values of enthalpy. Strong acid has higher enthalpy of neutralization compared to weak acid. All

the acids and base were assumed to have same specific heat capacity as water, 4.18kJ/mol. Coffee cup is

the apparatus used to measure the enthalpy of reaction.

1.8: REFERENCE(S)

Silberberg M. S. (2009). Chemistry: the molecular nature of matter and change. New York: The McGraw Hill

Companies, Inc.

Brown, LeMay (2009). Chemistry: the central science. New Jersey, Pearson Education, Inc.

Eng S. C. (2009). Chemistry for Matriculation 1. Kuala Lumpur: Oriental Academic Publication.

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