Aristo Book 5 experiment answer

HKDSE CHEMISTRY – A Modern View (Chemistry)
Experiment Workbook 5
Suggested answers

Chapter 52 Importance of business processes
Chapter 53 Rate equation
Experiment fifty three.1 Determining the speed equation of a reaction using technique of preliminary rate (A microscale experiment) 1
Chapter fifty four Activation energy
Experiment fifty four.1 Determining the activation vitality of a chemical reaction 3
Chapter fifty five Catalysis and industrial processes
Experiment 55.1 Investigating the action of a catalyst
6
Experiment 55.2 Investigating homogeneous catalysis
8
Experiment 55.3Investigating methods to alter the speed of a response with an appropriate catalyst 9
Experiment 55.three Sample laboratory report
13
Experiment 55.4Preparing ethanol by fermentation
16
Chapter 56 Industrial processes
Chapter 57 Green chemistry for industrial processes

Chapter 53Rate equation

Experiment 53.1Determining the speed equation of a response using method of preliminary price (A microscale experiment)

7. and eleven.(a)
Well number
1
2
3
4
5
6
7
8
Number of drops of zero.5 M Na2S2O3(aq)
10
9
8
7
6
5
4
3
Time, t (s)
14.2
15.8
17.8
20.4
23.8
28.6
35.7
47.6
(s1)
0.070
zero.063
0.056
zero.049
0.042
zero.035
zero.028
0.021

10. and 12.(a)
Well number
1
2
3
4
5
6
7
8
Number of drops of 1.0 M H2SO4(aq)
10
9
8
7
6
5
4
3
Time, t (s)
fifty nine.4
fifty nine.7
60.0
60.7
fifty nine.9
60.0
sixty one.0
60.5
(s1)
0.017
zero.017
0.017
zero.016
0.017
0.017
zero.016
zero.017

11.(a)inversely
(b)

(c)1

12.(b)0
(c)From the leads to Table 53.2, the readings of time are close, indicating that the reaction is of zeroth order with respect to H+(aq).

13.Rate = k[S2O32(aq)]

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

15.In this experiment, the time for the formation of a hard and fast, but small quantity of insoluble sulphur precipitate is measured. The shorter the time, the sooner is the reaction. It is assumed that the extent of reaction is still small when the time is recorded, so that the time recorded can be used as a measurement of initial price of the response.

Chapter 54Activation Energy

Experiment 54.1Determining the activation power of a chemical reaction

5.
Temperature of the response combination (°C)
15
25
35
45
55
Time for the looks of darkish blue color (s)
679
(at 11°C)
232
(at 27°C)
112
(at 37°C)
80
(at 43°C)
33
(at 56°C)

6.(a)rate constant; activation power; Universal fuel constant; temperature;
(b)
log ()
2.83
2.37
2.05
1.90
1.52

3.52
3.33
3.23
three.16
three.04

(c)

(d)2750
(e)slope = 2750 =
Ea = 2750 × 2.three × 8.314 J mol1
= fifty two 586 J mol1
= fifty two.6 kJ mol1
7.Arrhenius equation; log k = log A
eight.straight line;

9.S2O82(aq) + 2I(aq)  2SO42(aq) + I2(aq)

10.To monitor the formation of iodine from the response of S2O82(aq) ions and I(aq) ions.

11.When all S2O82(aq) ions have reacted, any iodine shaped will flip the starch resolution darkish blue. The time for this color change is a measure of the speed of reaction shown in question 9. (Note: The reaction rate is inversely proportional to the time taken for the starch resolution to turn darkish blue.)

12.The amount of reactants used in every experiment may not be precisely the identical.
There may be an error in measuring or studying the temperatures from the thermometers.
As the colour change of the answer mixture is not a sudden one, particularly at low temperatures, there could also be an error in recording the time of color change.

Chapter 55Catalysis and industrial processes

Experiment 55.1Investigating the motion of catalyst

1.(b)No.

5.(b)
Time (s)
10
20
30
40
50
60
Volume of O2(g) launched (cm3), with the addition of zero.5 g MnO2(s) 30
60
85
95
96
96
Time (s)
70
80
90
100
110
120
Volume of O2(g) released (cm3), with the addition of zero.5 g MnO2(s) 96

6.(b)
Time (s)
10
20
30
40
50
60
Volume of O2(g) launched (cm3), with the addition of 1.5 g MnO2(s) 70
90
95
96
96

Time (s)
70
80
90
100
110
120
Volume of O2(g) launched (cm3), with the addition of 1.5 g MnO2(s)

8.

9.Manganese(IV) oxide
MnO2(s)
10.2H2O2(aq)  2H2O(l) + O2(g)

11.(a)The addition of manganese(IV) oxide tremendously increases the speed of decomposition of hydrogen peroxide.
(b)(i)The initial rate of response is higher.
(ii)The complete time of response is shorter. (Note: increasing the quantity of catalyst would enhance the response fee.)
(c)No.

12.Add more H2O2(aq) to the reaction combination, speedy effervescence exhibits that manganese(IV) oxide has not been used up within the response. The catalytic property of manganese(IV) oxide is still present. Experiment 55.2Investigating homogeneous catalysis

5.Mixture ‘y’.
It has a characteristic candy odor like certain glues or nail polish removers.

6.ethyl ethanoate; concentrated sulphuric acid

7.CH3COOH(l) + CH3CH2OH(l) ⇌ CH3COOCH2CH3(l) + H2O(l)

8.Homogeneous catalyst. This is as a outcome of all species are in the same part within the reaction, i.e. the liquid phase.

9.Sodium carbonate answer reacts with any unreacted ethanoic acid left in the response combination. The sturdy vinegar odor of ethanoic acid is thus eliminated. The salt sodium ethanoate shaped has no scent. Besides, the ester is insoluble in water and floats on the water floor. This makes us easier to detect the scent of ester. Experiment fifty five.3Investigating ways to alter the rate of a reaction with a suitable catalyst

1.Apparatus:
Safety spectacles
Protective gloves
Conical flask (100 cm3)
5 measuring cylinders (10 cm3)
Dropper
Stopwatch
Boiling tube
White tile
Chemicals:
Ammonium peroxodisulphate resolution (0.020 M)
Potassium iodide solution (0.50 M)
Sodium thiosulphate resolution (0.010 M)
0.2% starch solution
Iron(II) chloride solution (~0.010 M)
Distilled water
2.
What you’ll maintain constant
(Controlled variable)
What you’ll change
(Independent variable)
What you’ll measure
(Dependent variable)
volume of ammonium peroxodisulphate solution
quantity of potassium iodide solution
volume of sodium thiosulphate solution
quantity of zero.2% starch solution
with or with out utilizing iron(II) solution
the time for the looks of the dark blue colour

3.

Figure 1
four.
(1)Using a measuring cylinder, add 10 cm3 of ammonium peroxodisulphate solution to a conical flask. (2)Using different measuring cylinders, add 5 cm3 of potassium iodide answer, 5 cm3 of sodium thiosulphate resolution, 1
cm3 of iron(II) chloride solution and a pair of.5 cm3 of starch solution to a boiling tube. (3)Pour the contents in the boiling tube into the conical flask. (4)Immediately begin the stopwatch.

(5)When a darkish blue colour of the starch-iodine complicated appears within the solution, stop the stopwatch. (6)Record the time for the appearance of the darkish blue color in Table 1. (7)Repeat steps (1) to (6), however substitute iron(II) chloride solution with 1 cm3 of distilled water.

5.
Risk evaluation form

6.

Time for the appearance of the darkish blue colour
With Fe2+(aq) ions (as a catalyst) added
59 s
Without any catalyst added
3 mins and 52 s
Table 1

7.The reaction involves the collision of two negatively charged ions, S2O82(aq) ions and I(aq) ions, which actually repel one another.

8.S2O82(aq) + 2Fe2+(aq)  2SO42(aq) + 2Fe3+(aq)
2Fe3+(aq) + 2I(aq)  2Fe2+(aq) + I2(aq)

9.Referring to the 2 equations in query 8, the S2O82(aq) ions oxidize the Fe2+(aq) ions to Fe3+(aq) ions. At the identical time, the S2O82(aq) ions are decreased to SO42(aq) ions. The Fe3+(aq) ions are strong oxidizing agents that oxidize I(aq) ions to I2(aq). At the identical time, Fe3+(aq) ions are decreased again to Fe2+(aq) ions (i.e. the catalyst is regenerated).

Both the equations shown in question 8 involve the collision between constructive and unfavorable ions. This shall be much more likely to be successful
than the collision between two adverse ions within the uncatalysed reaction. Thus, the activation vitality of this pathway might be decrease and the reaction fee may even be greater.

10.The chemical reaction can be accelerated by the addition of iron(II) ions, which act as a homogeneous catalyst of this response.

11.It can be regenerated after the reaction. OR It is particular in action. OR A small quantity of catalyst is normally sufficient for the catalytic motion.

12.Homogeneous catalyst is one which has the same part as the reactants and merchandise. Sample laboratory report

Title: Investigating methods to vary the speed of a reaction with an acceptable catalyst

Objective
To design and perform an experiment to research ways to alter the rate of a response – by means of a suitable catalyst.

Apparatus and materials
Safety spectacles
Protective gloves
Conical flask (100 cm3)
5 measuring cylinders (10 cm3)
Dropper
Stopwatch
Boiling tube
White tile
Ammonium peroxodisulphate resolution (0.020 M)
Potassium iodide solution (0.50 M)
Sodium thiosulphate solution (0.010 M)
zero.2% starch solution
Iron(II) chloride solution (~0.010 M)
Distilled water

Chemical reactions involved
S2O82(aq) + 2Fe2+(aq)  2SO42(aq) + 2Fe3+(aq)
2Fe3+(aq) + 2I(aq)  2Fe2+(aq) + I2(aq)

Procedure
1.Using a measuring cylinder, 10 cm3 of ammonium peroxodisulphate resolution was added to a conical flask. 2.Using different measuring cylinders, 5 cm3 of potassium iodide resolution, 5 cm3 of sodium thiosulphate answer, 1 cm3 of iron(II) chloride solution and 2.5 cm3 of starch solution had been added to a boiling tube. three.The contents within the boiling tube were poured into the conical flask. four.The stopwatch was began immediately.

5.When a darkish blue colour of the starch-iodine complex appeared in the solution, the stopwatch was stopped. 6.The time for the appearance of the dark blue colour was recorded in Table 1. 7.Steps (1) to (6) had been repeated, however iron(II) chloride answer was replaced with 1 cm3 of distilled water.

Results

Time for the appearance of the dark blue colour
With Fe2+(aq) ions (as a catalyst) added
fifty nine s
Without any catalyst added
3 minutes and 52 s
Table 1

After mixing all the chemicals within the conical flask, the reaction combination with Fe2+(aq) ions will take a shorter time for the dark blue colour to look.

Analysis
1. In the absence of Fe2+(aq) ions, the reaction between S2O82(aq) ions and I­­(aq) ions is slow. As each reactant ions are negatively charged, they have a tendency to repel one another. However, when Fe2+(aq) ions are added, the
reaction turns into sooner. Fe2+(aq) ions have the identical phase (i.e. aqueous phase) as the reactants and merchandise, so they are homogeneous catalyst of this reaction. 2. Fe2+(aq) ions is a decreasing agent which might reduce S2O82(aq) ions to SO42(aq) ions. The Fe3+(aq) ions shaped act as an oxidizing agent, which oxidize I­­(aq) ions to I2(aq) ions and regenerate Fe2+(aq) ions once more. Being a catalyst, Fe2+(aq) ions are not consumed in the catalytic process.

Discussion
1.Either Fe2+(aq) ions are Fe3+(aq) ions is an efficient selection of catalyst for this reaction because the interconversion between Fe2+ and Fe3+ facilitates the reaction between S2O82(aq) ions and I­­(aq) ions to occur. 2.The catalytic property of Fe2+(aq) ions may be due to the truth that it’s easier for the negatively charged S2O82(aq) ions to approach the positively charged Fe2+(aq) ions. The identical is true when the positively charged Fe3+(aq) ions shaped can approach the negatively charged I­­(aq) ions easier. three.The experiment is just a simple check tube experiment but the outcome (colour change) is sort of obvious and easy to detect.

Conclusion
The chemical response could be sped up by the addition of Fe2+(aq) ions, which act as a homogeneous catalyst of this reaction.

Answers to questions for additional thought
eleven.It may be regenerated after the response. OR It is particular in action. OR A small quantity of catalyst is usually enough for the catalytic action.

12.Homogeneous catalyst is one which has the same part as the reactants and merchandise. Experiment fifty five.4Preparing ethanol by fermentation

3.(b)

Glucose solution
with yeast
Glucose solution
with out yeast
Appearance of the glucose solution
a pale brown suspension
a transparent solution
Observations in the limewater
clear and colourless
clear and colourless

6.(d)Acidified potassium dichromate answer modifications colour from orange to green.
(f)No color change for the acidified potassium dichromate resolution.

7.

Glucose solution
with yeast
Glucose solution
without yeast
Appearance of the glucose solution
cloudy; a pale brown suspension
clear, no visible change
Observations in the limewater
milky
remains clear and colourless
Smell of the glucose solution
a smell of alcohol
no characteristic smell

8.catalyst

9.ethanol; carbon dioxide

10.The resolution turns milky. It signifies that carbon dioxide is produced during fermentation.

11.In the presence of yeast, glucose is converted to ethanol. The presence of ethanol is indicated by the color change of the response with acidified potassium dichromate solution. Ethanol is a lowering agent. It reduces dichromate ions to chromium(III) ions.