aristo book 5 experiment solution essay

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HKDSE CHEMISTRY – A Modern Look at (Chemistry)

Experiment Workbook 5

Suggested answers

Chapter 52 Importance of professional processes

Chapter 53 Rate equation

Test 53. you Determining the speed equation of a reaction using method of first rate (A microscale experiment) 1

Chapter fifty four Activation strength

Test 54. you Determining the activation strength of a reaction 3

Chapter fifty-five Catalysis and industrial techniques

Try things out 55. one particular Investigating the action of a catalyst

6

Experiment fifty five. 2 Examining homogeneous catalysis

almost 8

Test 55. 3Investigating ways to change the rate of your reaction which has a suitable catalyst 9

Experiment 55. 3 Sample laboratory statement

13

Experiment 55. 4Preparing ethanol simply by fermentation

16

Chapter 56 Industrial procedures

Section 57 Green chemistry pertaining to industrial operations

Chapter 53Rate equation

Research 53. 1Determining the rate equation of a effect using method of initial level (A microscale experiment)

7. and eleven. (a)

Well number

one particular

a couple of

3

four

5

6

six

almost 8

Volume of drops of 0. a few M Na2S2O3(aq)

twelve

being unfaithful

almost 8

7

6

your five

some

a few

Period, t (s)

13. 2

15. eight

18. 8

20. 5

3. 8

28. six

35. 7

47. 6th

(s1)

0. 070

0. 063

zero. 056

0. 049

0. 042

0. 035

zero. 028

0. 021

10. and 12. (a)

Well number

1

2

3

4

5

6

7

8

Number of drops of 1. zero M H2SO4(aq)

twelve

9

eight

7

6th

5

four

several

Time, t (s)

59. 4

59. six

60. 0

60. 7

fifty nine. 9

60. zero

61. 0

60. a few

(s1)

0. 017

0. 017

0. 017

0. 016

zero. 017

0. 017

zero. 016

0. 017

11. (a)inversely

(b)

(c)1

12. (b)0

(c)From the results in Table 53. a couple of, the blood pressure measurements of time are close, proving the fact that the reaction features zeroth buy with respect to H+(aq).

13. Price = t[S2O32(aq)]

14. S2O32(aq) + 2H+(aq)  S(s) + SO2(g) + H2O(l)

15. With this experiment, the time for the organization of a fixed, but tiny amount of insoluble sulphur precipitate is definitely measured. The shorter the time, the quicker is the reaction. It is assumed which the extent of reaction remains to be small if the time is usually recorded, in order that the time registered can be used like a measurement of initial rate of the response. Chapter 54Activation Energy

Test 54. 1Determining the service energy of a chemical reaction

five.

Temperature of the response mixture (°C)

12-15

twenty-five

thirty five

forty-five

fifty-five

Coming back the appearance of dark blue color (s)

679

(at 11°C)

232

(at 27°C)

112

(at 37°C)

70

(at 43°C)

33

(at 56°C)

6. (a)rate constant; service energy; Universal gas regular; temperature;

(b)

log ()

2. 83

2. 37

2. 05

1. 80

1. 52

several. 52

3. thirty-three

a few. 23

3. 16

a few. 04

(c)

(d)2750

(e)slope sama dengan 2750 sama dengan

Tool = 2750 × 2 . 3 × 8. 314 J mol1

sama dengan 52 586 J mol1

sama dengan 52. six kJ mol1

six. Arrhenius equation; log k = journal A

8. direct line;

on the lookout for. S2O82(aq) + 2I(aq)  2SO42(aq) + I2(aq)

15. To monitor the formation of iodine from your reaction of S2O82(aq) ions and I(aq) ions.

11. The moment all S2O82(aq) ions have got reacted, any iodine formed will change the starch solution dark blue. The time for this coloring change is known as a measure of the rate of reaction shown involved 9. (Note: The reaction charge is inversely proportional towards the time used for the starch answer to turn dark blue. )

12. The amount of reactants used in each experiment may not be the identical.

There may be an error in measuring or reading the temperatures from your thermometers.

As along with change from the solution mix is not really a sudden one, especially in low temps, there may be a mistake in documenting the time of colour alter.

Chapter 55Catalysis and commercial processes

Try things out 55. 1Investigating the actions of catalyst

1 . (b)No.

5. (b)

Time (s)

10

20

30

40

50

60

Volume of O2(g) released (cm3), with the addition of zero. 5 g MnO2(s) 35

60

eighty-five

95

ninety six

ninety six

Time (s)

70

80

90

100

110

120

Volume of O2(g) released (cm3), with the addition of 0. 5 g MnO2(s) ninety six

6. (b)

Time (s)

10

20

30

40

50

60

Volume of O2(g) released (cm3), with the addition of 1 . 5 g MnO2(s) 70

85

ninety five

ninety six

ninety six

Time (s)

75

80

85

100

128

120

Amount of O2(g) released (cm3), with the addition of 1 . five g MnO2(s)

8.

9. Manganese(IV) o2

MnO2(s)

15. 2H2O2(aq)  2H2O(l) + O2(g)

10. (a)The addition of manganese(IV) oxide greatly increases the level of decomposition of hydrogen peroxide.

(b)(i)The first rate of reaction is higher.

(ii)The total time of reaction is shorter. (Note: increasing the amount of catalyst would improve the reaction price. )

(c)No.

doze. Add more H2O2(aq) to the reaction mix, rapid spirit shows that manganese(IV) oxide will not be used up in the reaction. The catalytic home of manganese(IV) oxide remains present. Research 55. 2Investigating homogeneous catalysis

5. Mixture ‘y’.

It has a characteristic sweet smell like certain glues or nail enhance removers.

six. ethyl ethanoate; concentrated sulphuric acid

six. CH3COOH(l) & CH3CH2OH(l) ⇌ CH3COOCH2CH3(l) + H2O(l)

almost 8. Homogeneous catalyst. This is because every species will be in the same phase in the reaction, we. e. the liquid stage.

9. Sodium carbonate option reacts with any unreacted ethanoic acid solution left in the reaction combination. The strong vinegar smell of ethanoic acid can be thus taken out. The salt sodium ethanoate created has no smell. Besides, the ester is definitely insoluble in water and floats boating surface. Can make us simpler to detect the smell of ester. Experiment 55. 3Investigating ways to replace the rate of the reaction with a suitable catalyst

1 . Device:

Protection spectacles

Protective safety gloves

Cone-shaped flask (100 cm3)

5 computing cylinders (10 cm3)

Dropper

Stopwatch

Boiling tube

White colored tile

Chemicals:

Ammonium peroxodisulphate solution (0. 020 M)

Potassium iodide solution (0. 60 M)

Sodium thiosulphate solution (0. 010 M)

zero. 2% starch solution

Iron(II) chloride solution (~0. 010 M)

Unadulterated water

2 .

What you will keep constant

(Controlled variable)

What you will change

(Independent variable)

What you should measure

(Dependent variable)

volume of ammonium peroxodisulphate solution

volume of potassium iodide option

amount of sodium thiosulphate solution

amount of 0. 2% starch solution

with or without needing iron(II) answer

the time for the appearance of the darker blue color

3.

Figure 1

4.

(1)Using a measuring cyndrical tube, add 15 cm3 of ammonium peroxodisulphate solution to a conical flask. (2)Using diverse measuring cyl, add your five cm3 of potassium iodide solution, five cm3 of sodium thiosulphate solution, 1

cm3 of iron(II) chloride answer and installment payments on your 5 cm3 of starch solution to a boiling tube. (3)Pour the contents in the boiling conduit into the conical flask. (4)Immediately start the stopwatch.

(5)When a dark blue color of the starch-iodine complex shows up in the answer, stop the stopwatch. (6)Record the time intended for the appearance of the dark green colour in Table 1 ) (7)Repeat measures (1) to (6), but replace iron(II) chloride remedy with you cm3 of distilled water.

5.

Risk evaluation form

6th.

Time for seen the dark blue color

With Fe2+(aq) ions (as a catalyst) added

fifty nine s

Without any catalyst added

3 a few minutes and 52 s

Table one particular

7. The response involves the collision of two negatively charged ions, S2O82(aq) ions and I(aq) ions, which usually actually repel each other.

8. S2O82(aq) & 2Fe2+(aq)  2SO42(aq) + 2Fe3+(aq)

2Fe3+(aq) & 2I(aq)  2Fe2+(aq) + I2(aq)

on the lookout for. Referring to both the equations showcased 8, the S2O82(aq) ions oxidize the Fe2+(aq) ions to Fe3+(aq) ions. Concurrently, the S2O82(aq) ions are reduced to SO42(aq) ions. The Fe3+(aq) ions are strong oxidizing agents that oxidize I(aq) ions to I2(aq). Concurrently, Fe3+(aq) ions are decreased back to Fe2+(aq) ions (i. e. the catalyst is usually regenerated).

Both equations shown in question almost 8 involve the collision among positive and negative ions. This will become much more likely to achieve success

than the collision among two bad ions inside the uncatalysed effect. Thus, the activation energy of this pathway will be decrease and the reaction rate is likewise higher.

10. The reaction can be speeded up by addition of iron(II) ions, which behave as a homogeneous catalyst of this reaction.

14. It can be regenerated after the reaction. OR It is particular in action. OR POSSIBLY A small amount of catalyst is usually enough for the catalytic action.

12. Homogeneous catalyst can be one which provides the same period as the reactants and products. Sample laboratory survey

Title: Investigating ways to change the rate of the reaction using a suitable catalyst

Objective

To design and carry out an test to investigate approaches to change the level of a effect – by the use of a suitable catalyst.

Apparatus and materials

Safety spectacles

Protective gloves

Conical flask (100 cm3)

five measuring cylinders (10 cm3)

Dropper

Stop watch

Boiling tube

White tile

Ammonium peroxodisulphate remedy (0. 020 M)

Potassium iodide solution (0. 50 M)

Sodium thiosulphate answer (0. 010 M)

0. 2% starch option

Iron(II) chloride option (~0. 010 M)

Distilled normal water

Chemical reactions involved

S2O82(aq) + 2Fe2+(aq)  2SO42(aq) + 2Fe3+(aq)

2Fe3+(aq) + 2I(aq)  2Fe2+(aq) + I2(aq)

Procedure

1 . By using a measuring tube, 10 cm3 of ammonium peroxodisulphate answer was added to a cone-shaped flask. 2 . Using distinct measuring cyl, 5 cm3 of potassium iodide option, 5 cm3 of salt thiosulphate solution, 1 cm3 of iron(II) chloride remedy and 2 . 5 cm3 of starch solution had been added to a boiling pipe. 3. The contents inside the boiling conduit were poured into the conical flask. four. The stop watch was began immediately.

five. When a dark blue colour of the starch-iodine complex made an appearance in the remedy, the stop watch was ended. 6. Enough time for the appearance of the darker blue shade was recorded in Table 1 . 7. Measures (1) to (6) were repeated, nevertheless iron(II) chloride solution was replaced with 1 cm3 of distilled drinking water.

Results

Coming back the appearance of the dark green colour

With Fe2+(aq) ions (as a catalyst) added

59 t

Without any catalyst added

3 mins and 52 s i9000

Stand 1

After mixing every one of the chemicals inside the conical flask, the reaction blend with Fe2+(aq) ions will require a short time for the dark blue colour to show up.

Analysis

1 . In the absence of Fe2+(aq) ions, the response between S2O82(aq) ions and I­­(aq) ions is gradual. As the two reactant ions are negatively charged, they have a tendency to repel each other. However , when Fe2+(aq) ions will be added, the

response becomes faster. Fe2+(aq) ions have the same phase (i. at the. aqueous phase) as the reactants and products, thus they are homogeneous catalyst on this reaction. 2 . Fe2+(aq) ions is a lowering agent which can reduce S2O82(aq) ions to SO42(aq) ions. The Fe3+(aq) ions created act as a great oxidizing agent, which oxidize I­­(aq) ions to I2(aq) ions and regenerate Fe2+(aq) ions once again. Being a catalyst, Fe2+(aq) ions are not consumed in the catalytic process.

Discussion

1 . Either Fe2+(aq) ions happen to be Fe3+(aq) ions is a good selection of catalyst for this reaction since the interconversion among Fe2+ and Fe3+ facilitates the reaction among S2O82(aq) ions and I­­(aq) ions to occur. 2 . The catalytic real estate of Fe2+(aq) ions may be due to the fact that it is easier for the adversely charged S2O82(aq) ions to approach the positively recharged Fe2+(aq) ions. The same holds true when the favorably charged Fe3+(aq) ions formed can procedure the in a negative way charged I­­(aq) ions easier. 3. The experiment is a simple evaluation tube experiment but the end result (colour change) is quite evident and easy to detect.

Conclusion

The chemical reaction may be speeded up by the addition of Fe2+(aq) ions, which will act as a homogeneous catalyst of this effect.

Answers to questions for additional thought

11. It can be regenerated following the reaction. Another problem is that it can often be specific in action. OR A small amount of catalyst is generally enough to get the catalytic action.

12. Homogeneous catalyst is one which has the same phase because the reactants and products. Experiment fifty-five. 4Preparing ethanol by fermentation

3. (b)

Glucose answer

with yeast

Glucose answer

without yeast

Appearance from the glucose solution

a pale brown suspension

a clear solution

Findings in the limewater

crystal clear and colourless

clear and colourless

6. (d)Acidified potassium dichromate solution improvements colour via orange to green.

(f)No shade change for the acidified potassium dichromate solution.

7.

Glucose solution

with yeast

Glucose answer

without yeast

Appearance from the glucose remedy

over cast; a soft brown suspension system

very clear, no noticeable change

Observations inside the limewater

milky

remains obvious and colourless

Smell of the sugar solution

a smell of liquor

no characteristic smell

8. catalyst

9. ethanol; carbon dioxide

10. The solution transforms milky. It indicates that co2 is produced during fermentation.

11. Inside the presence of yeast, blood sugar is converted to ethanol. The existence of ethanol can be indicated by colour transform of the response with acidified potassium dichromate solution. Ethanol is a lowering agent. It reduces dichromate ions to chromium(III) ions.

one particular

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