LAB Report AS1202I CHM 271 2021 PDF

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Results/Data/Observatio n / Data Analysis/Validation / Discussion / Conclusion / Questions / Format / Total /FACULTY OF APPLIED SCIENCESUiTM PAHANG (JENGKA CAMPUS)CHM271 PRINCIPLES OF PHYSICAL CHEMISTRYLAB REPORTNAME :STUDENT ID :GROUP : AS1202ILECTURER :DATE OF SUBMISSION : Declaration of Academic ...

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Results/Data/Observatio n Data Analysis/Validation Discussion Conclusion Questions Format Total

FACULTY OF APPLIED SCIENCES UiTM PAHANG (JENGKA CAMPUS)

CHM271 PRINCIPLES OF PHYSICAL CHEMISTRY

LAB REPORT NAME

: NURFARAH DIYANA BINTI DAUD

STUDENT ID

: 2020876972

GROUP

: AS1202I

LECTURER

: MADAM SHAHIDA HANUM KAMARULLAH

DATE OF SUBMISSION

: 23 JULY 2021

--------------------------------------------------------------------------------------------------------------------------Declaration of Academic Honesty Academic honesty or academic integrity is a very important virtue that all students should uphold at all times. I/We declare that the lab report submitted is not plagiarised and is entirely my/our own work, and that no part of it has been copied from any work produced by other person(s)/ source(s) or provided by any other student(s). I/We understand that issuing a false declaration can result in severe penalties and I/we am/are willing to be penalized if any form of copying found valid.

___________________________ (STUDENT’S NAME) NURFARAH DIYANA BINTI DAUD 2020876972

/2 /2 /2 /2 /1 /1 /10

Format (writing quality) for report writing/Lab Report: 1. 2. 3. 4. 5. 6. 7. 8. 9.

Objective(s) List of apparatus List of chemicals Procedure Results Data/ observation Discussion Conclusion References

FACULTY OF APPLIED SCIENCES DIPLOMA IN SCIENCE (AS120) GRADING RUBRIC FOR WRITTEN PRACTICAL REPORT

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EXPERIMENT 1: Thermochemistry - Hess's Law (DATASHEET + QUESTION)

QUESTION 1 Calculate △ 𝐇2 and △H3. △ Q = (ma × C ×△ T) ÷ 1000kJ =(30g x4.25g-1℃-1x47℃)÷ 1000kJ =5.922kJ

△ H2 =

△Q n

0.60

N= 24.305 =0.025 mol =

5.922𝑘𝐽 0.025𝑚𝑜𝑙

△ 𝐻2 =236.88 kJ/mol △ Q = (30g x4.2Jg −1℃−1 x 24℃) ÷ 1000kJ =3.024kJ △𝑄

△ H3 =

𝑛 1.00

n=24.305+15.999 =0.025 mol 3.024

=0.025 △ 𝐻3 =120.96 kJ/mol QUESTION 2 State whether the reaction are exothermic or endothermic.

Reaction of magnesium crystal are exothermic while reaction for magnesium oxide crystals are exothermic. Overall both reaction are exothermic.

QUESTION 3 Calculate △ 𝐻1 , the standard enthalpy of formation of magnesium oxide. Given that the standard enthalpy of formation of water (value of △ 𝐻4 is -286kJ/mol). △ 𝐻2 = 236.88 𝑘𝐽/𝑚𝑜𝑙 △ 𝐻3 = 120.96 𝑘𝐽/𝑚𝑜𝑙 △ 𝐻4 = −286𝑘𝐽/𝑚𝑜𝑙 △ 𝐻1 = △ 𝐻2 + (− △ 𝐻3 ) +△ 𝐻4 =236.88kJ/mol+(-120.96kJ/mol)+(-286kJ/mol) = -170.08kJ/mol

CHM271 LABORATORY

OCTOBER 2020 - FEBRUARY 2021

INSTRUCTIONS TO STUDENTS: Complete the datasheet for the following experiment DATASHEET EXPERIMENT 1: THERMOCHEMISTRY – HESS’S LAW Table: Mass and Temperature of Substances Magnesium (Mg)

Magnesium oxide (MgO)

Mass,m (g)

0.60

1.00

Initial temperature (°C)

27.0

27.0

Final temperature (°C) (HCl + Mg or MgO)

74.0

51.0

+ 47℃

+ 24℃

Temperature change (∆T)

Volume of acid is used = 30 .......................................... mL Mass of acid, ma = 30 ....................................... g (assume density of acid is 1g mL-1)

EXPERIMENT 2: pH Determination of Acid, Base and Buffer Solutions (QUESTION) QUESTION 1

State whether the following solutions is acidic, basic or neutral and explain your answer. A) 0.5M Na2CO3 is a basic solution. This can be proved by calculation below. (H+) = 0.5M POH = -log(0.5) = 0.301 pH = 14-0.301 = 13.7 (base) Na2CO3→Na ++3CO32CO32-+H2O⇌ H2CO3+OH0.1M CH3COONa is a basic solution. When you add NaCO3 to the water, it dissociates leaving Na + ion and CO32- ion. When Na+ reacts with water, NaOH will formed. This is a strong base. When CO32- reacts with water, it will produce week acid H2CO3. When strong base meet weak acid, the result of the solution will be basic. B) 0.1M CH3COONa is a basic solution. This can be proven by calculation below. (OH-) = 0.1M POH = -log (0.1) =1 Ph = 14-1 = 13(base) NaOH + CH3COOH → CH3COONa NaOH is a strong basic while CH3COOH is a weak acid. When strong base meet weak acid the solution will remained basic. C) 0.1M AlCl3 is an acidic solution. This can be proven by calculation below. (H+) = 0.1M pH = -log(0.1) = 1(acid) HCl+Al(OH) 3=AlCl3+H2O AlCl3 is produced by mixing strong acid and weak base. When strong acid was added to the weak base, the solution will remained acidic.

QUESTION 2 Explain your observation in procedure part II in terms of the effect of addition of a strong acid and strong base to the buffer solution. When strong acid is added to buffer solution, the conjugate present in the buffer consume the hydromium ion and converting it into weak acid of the conjugate base and water. This result is in the increase amount of conjugate base present and an increase amount of week acid.

EXPERIMENT 3: Electrochemical Cell (DATASHEET+DATA TREATMENTS + QUESTION)

DATA TREATMENT 1 FOR EACH ELECTROCHEMICHAL CELL ,(a) DRAW A COMPLETE CELL DIAGRAM AND (b) WRITE THE CELL SYMBOL .

(A)

(B)

DATA TREATMENT 2 CALCULATE THE STANDART CELL POTENTIAL IN THE SAME MANNER SHOW IN THE INTRODUCTION SECTION OF THIS LABORATORY EXPERIMENT. (I) Anode = Zn → Zn2++2e (oxi) Cathode = Cu2++2e → Cu (red) E°= -0.76 V E°= +0.34 V E°cell = E °red - E°oxi = 0.34 V - (-0.76 V) = 1.1 V (II) Anode = Fe → Fe2++2e (oxi) Cathode = Cu2++2e → Cu (red) E°= -0.44 V E°= +0.34 V E°cell = E °red - E°oxi = 0.34 V - (-0.44 V) = 0.78 V (III) Anode = Zn → Zn2++2e (oxi) Cathode = Fe2++2e → Fe (red) E°= -0.76 V E°= -0.44 V E°cell = E °red - E°oxi = -0.44 V - (-0.76 V) = 0.32 V DATA TREATMENT 3 COMPARE THE CALCULATED CELL POTENTIAS TO THE MEASURED VOLTAGES OF YOUR ELECTROCHEMICALS CELLS. DESCRIBE THE POSSIBLE SOURCES OF ERROR. (I) Zn-Cu |𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 − 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠|

Percentage difference = 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 E°exp = 1.15V E°theory = 1.10V |1.15 𝑉− 1.10 𝑉 | Percentage difference = × 100 1.10 𝑉 = 4.545 %

× 100

(II) Fe-Cu |𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 − 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠|

Percentage difference = 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 E°exp = 0.73 V E°theory = 0.78 V |0.73 𝑉− 0.78 𝑉 | Percentage difference = × 100 0.78 𝑉 = 6.410%

× 100

(III) Fe-Zn |𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 − 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠|

Percentage difference = 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 E°exp = 0.44 V E°theory = 0.32 V |0.44 𝑉− 0.32 𝑉 | Percentage difference = × 100 0.32 𝑉 = 37.5 %

× 100

The possible sources of error are surface area of the electrode used in the cell. More over concentration and temperature of the solution used as electrolyte also can gives changes in the current produced by the cell.

QUESTION 1 WHAT ARE THE FUNCTION OF SALT BRIDGE? Salt bridge keeps the solution in two half-cells electrically neutral. In anodic half cell, positive ion pass into the solution and there shall be accumulation of extra positive charge in the solution around the anode which will prevent the flow of electrons from anode. This does not happened because negative ions are provided by salt bridge. In cathodic half-cell negative ions will accumulate around cathode due to deposition of positive ions by reduction. To neutralize these negative ions, sufficient number of positive ions is provided by salt bridge. Basically salt bridge maintains electrical neutrality. QUESTION 2 CALCULATE THE STANDARD CELL POTENCIAL OF A CELL CONSTRUCTED FROM Mg2+/Mg AND Ni2+/Ni. WHICH HALF-CELL IS ANODE AND WHICH HALF-CELL IS CATHODE? Mg2+ + 2e → Mg

( E° = -2.37 V )

Ni2+ + 2e → Ni

( E° = -0.25 V )

A) Anode (oxidation) = Mg → Mg2+ + 2e Cathode (reduction) = Ni2+ + 2e → Ni B) Cell potential , E °cell E°cell = E °red - E°oxi = - 0.25 V - (- 2.37 V) = 2.12 V C) Overall cell reaction Anode (oxidation) = Mg → Mg2+ + 2e Cathode (reduction) = Ni2+ + 2e → Ni Overall reaction = Mg + Ni2+ → Mg2+ + Ni

QUESTION 3 USING THE NERNST EQUATION, WHAT WOULD BE THE POTENTIAL 0F CELL [Ni2+]=[Mg2+]=0.10M.

= Mg → Mg2+ + 2e

( E° = -2.37 V )

Cathode (reduction) = Ni2+ + 2e → Ni

( E° = -0.25 V )

Anode (oxidation)

Overall reaction = Mg + Ni2+ → Mg2+ + Ni

E°cell = 2.12 V E= E° -

0.0592

E= 2.12 -

𝑛

0.0592 2

𝑝𝑟𝑜𝑑𝑢𝑐𝑡

𝑙𝑜𝑔 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 0.10

𝑙𝑜𝑔 0.10

= 2.12 V E > 0; spontaneous reaction

(n=2)

CHM271 LABORATORY

MARCH 2021 - AUGUST 2021

INSTRUCTIONS TO STUDENTS: Complete the datasheet for the following experiment

DATASHEET EXPERIMENT 3: ELECTROCHEMICAL CELL Table: Voltage Readings and Identify of Metals as Anode and Cathode

Electrochemical Cell

Anode (-)

Cathode (+)

Measured Voltage (V)

Calculated Voltage, E°cell (V)

(I) Zn – Cu

Zinc

Copper

1.15

1.10

(II) Fe – Cu

Iron

Copper

0.73

0.78

(III) Fe – Zn

Zinc

Iron

0.44

0.32

EXPERIMENT 4: Kinetics - Determination Order of Reaction (DATASHEET + QUESTION)

QUESTION 1 PLOT A GRAPH OF CONCENTRATION OF H2O 2 VERSUS TIME.

QUESTION 2 DETERMINE FROM THE GRAPH, THE FIRST, SECOND AND THIRD HALFLIFE OF THIS REACTION.

QUESTION 3 HOW DOES THE CONCENTRATION AFFECTS THE RATE OF DISSOCIATION OF HYDROGEN PEROXIDE? The rate of reaction will increase as the hydrogen concentration increase. The dissociation of hydrogen peroxide is catalysed by enzyme catalase. The rate of dissociation increase by the intra-cellular enzyme catalyse. More reactant particles moving allow more collisions to happen. As a result the rate of dissociation increased. The higher the concentration of the reactants, the faster the rate of dissociation will be. QUESTION 4 DETERMINE THE ORDER OF THE ABOVE REACTION. The order of the above reaction is the first order reaction. QUESTION 5 WRITE THE RATE EQUATION FOR THE ABOVE REACTION. Rate = K[H2O2]

CHM271 LABORATORY

MARCH 2021 - AUGUST 2021

INSTRUCTIONS TO STUDENTS: Complete the datasheet for the following experiment

DATASHEET EXPERIMENT 4 : KINETICS – DETERMINATION OF ORDER OF REACTION Table: Volume of KMnO4 Used and Concentration of H2O2 Reaction Time (min)

3

6

9

12

15

18

Volume of KMnO 4 (Vfinal )(mL)

16.0

50.0

10.0

14.8

22.0

25.0

Volume of KMnO 4 (Vinitial)(mL)

0

16.0

0

10.0

14.8

22.0

Volume of KMnO 4 needed (Vfinal - Vinitial)(mL)

16.0

34.0

10.0

4.8

7.2

3.0

0.050

0.038

Concentration of H 2O2 (mol dm-3 ) remained 0.160 0.075 unreacted at time t (using Eq. 1)

0.020 0.0125

EXPERIMENT 5: Freezing Point Depression (LAB REPORT)

INTRODUCTION The freezing point of a solvent rely on the concentration of the dissolved solute and the nature of the solvent. If the dissolved solute is a non electrolyte, then the reduction in the freezing point, ∆T is corresponding to the molality, m, (moles of solute per kg of solvent) of a weaken solution according to the equation: ∆T =Kf x m Where Kf is the molal freezing-point depression constant unique for each solvent. In this experiment you will be given cyclohexane, C6H12 as a solvent and a solid and a solid whose molar mass you will determine from the observed freezing point depression. OBJECTIVES 1) To determine the freezing point of a pure solvent and a solution of an unknown 2) To determine molar mass of the unknown solid using freezing point depression method. APPARATUS Boiling tube Thermometer Water bath Ice Glass rod Stopwatch CHEMICALS Cyclohexane, C6H12 A unknown solid

PROCEDURE

PART 1: FREEZING POINT OF THE SOLVENT 1) 30 ml of cyclohexane was poured into a boiling tube. 2) The tube was placed in a 250 ml beaker full of ice and a thermometer was placed inside. 3) When temperature was drop to 10℃ , the temperature sampling started for every 30 second until the crystal begin to appear and the temperature became constant. Temperature was recorded as the freezing point of pure cylcohexane. Graph was prepared for the data obtained. Solution from procedure Part 1 was kept for further experiment in Part 2. PART 2 : SAMPLE PREPARATION AND FREEZING POINT OF THE MIXTURE (SOLUTION) 1) O.5g weight of unknown was solute into a small dry test tube.

2) The tube with the solvent and thermometer from procedure 1 was placed back into the beaker of boiling water (water bath). 3) Thermometer was removed once the solvent melts and the solute sample prepared above was added. The solution was stirred gently with thermometer. 4) Heat was removed and cool at the room temperature for five minutes. 5) The tube was placed into an ice-filled beaker. 6) When the temperature was drop to 10℃, the temperature sampling started for every 30 second until the crystal begin to appear and the temperature became constant. Temperature was recorded as the freezing point of the mixture. Graph was prepared from the data.

RESULT Part I: Freezing point of solvent

Time (s) Temp (°C)

0 10.0

30 9.3

60 8.8

90 8.4

Time (s) Temp (°C)

270 6.5

300 6.5

330 6.2

360 6.0

120 8.2

150 7.5

180 7.0

210 6.5

240 6.5

120 6.6

150 6.0

180 5.4

210 4.6

240 3.8

Part II: Freezing point of solution

Time (s) Temp (°C)

0 10.0

30 8.5

60 7.8

90 7.0

Time (s) Temp (°C)

270 3.5

300 3.5

330 3.5

360 3.2

DATA ANALYSIS

Diagram 1 shows graph of temperature vs time (solution)

Diagram 2 shows graph of temperature versus time (solvent) Freezing point for the solution is at 3.5℃. freezing point for the solvent is at 6.5℃.

DISCUSSION Freezing point depression is the phenomena that describes why adding a solute to a solvent results in the lowering of the freezing point of the solvent.

CONCLUSION 1. The freezing point of a pure solvent is 6.5℃ and the freezing point of the solution is 3.5℃ 2. The molar mass of the unknown solid using freezing point dispersion method is 0.024kg.

QUESTION 1) What is the freezing point depression, ∆T for the solution? ∆Tf = T°f (solvent) -T°f(solution) = 6.5℃ − 3.5℃ = 3℃

2) the freezing point depression constant, Kf for the cyclohexane is 20.0℃/m and the density of cyclohexane is 0.799 g/mL.n calculate the molality of your solution. ∆ T = Kf m 3℃ = 20.0 (m) m= 0.15 molal 3) How many kilogram of solvent were used? 30mL of solvent were used. 1mL =0.799g/mL 30mL x 23.97𝑔 1000

0.799𝑔 𝑚𝐿

= 23.97g

= 0.024kg

4) What is the number of moles of solutes ? Moles of solvent = molality x mass of solution Moles of solvent = 0.15 mol/kg x (0.024kg) Moles of solvent = 3.6 x10-3 = 0.0036 mol 5) What is the molar mass of the solute? Molar mass = Molar mass =

𝑚𝑎𝑠𝑠 𝑚𝑜𝑙𝑒𝑠 0.5 0.0036

Molar mass = 138.89 g/mol 6) Why is the curve for the freezing of a solution different in slope from the freezing point of the solution? When a pure solvent freezes, it’s particles become more arranged as the intermolecular forces act between the atoms become permanent. By dissolving a solute into the solvent, the arrangement is disrupted and prevent the solvent from going into solid phase. This required more energy to be taken out in order to freeze. To solidify the solution, the temperature must drop more due to the solute-solvent interactions. That is why the curve for the freezing of a solution different in slope from the freezing point of the solution. 7) What are the natural random uncertainties in reading your thermometer and balance ? The natural random uncertainties of reading a thermometer is due to the position of eyes are not perpendicular to the reading scale on thermometer. The uncertainty of a measuring instrument is calculated as half of the smallest scale division plus or minus. The uncertainty for a thermometer with a mark every 1.0 ℃ is ± 0.5℃ and the balance is ± 0.01....