GCSEScienceWJECScience (Double Award)ChemistryExam QuestionsChemical Substances, Reactions & Essential ResourcesRate Of Chemical Change

Rate Of Chemical Change (WJEC GCSE Science (Double Award): Chemistry): Exam Questions

Exam code: 3430

2 hours12 questions

1 mark

A class investigated the rate of the reaction between magnesium and hydrochloric acid by measuring the volume of gas produced in 10 seconds using different concentrations of acid.

Tick (✓) the name of the piece of apparatus they used to measure the volume of gas produced.

beaker	$□$
thermometer	$□$
gas syringe	$□$
conical flask	$□$

How did you do?

6 marks

The table shows the results of their experiment.

Concentration of acid (M)	Volume of gas produced (cm³)
Concentration of acid (M)	Test 1	Test 2	Test 3	Mean
0.2	16	14	15	15
0.4	31	33	30	32
0.6	47	49	29	48
0.8	63	64	65	64
1.0	82	83	79	81

(i) Circle the result in the table which was not used to calculate a mean.

Give the reason why this result was not used.

Reason ..................................................................................

[2]

(ii) Plot the mean volume of gas produced against the concentration of acid. Draw a suitable line.

Figure: Blank grid for a graph with x-axis "Concentration of acid (M)" ranging from 0 to 1.0 in steps of 0.2, and y-axis "Mean volume of gas produced (cm3)" ranging from 0 to 100 in steps of 10

[3]

(iii) Use the information to find the volume of gas produced in 10 seconds using acid of concentration 0.5 M.

........................................................................ cm³

[1]

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2 marks

In another investigation the class found that as the temperature of the acid increases the reaction is faster.

Tick (✓) the two statements which explain why the reaction is faster at a higher temperature.

there are more particles of acid	$□$
the acid particles are moving faster	$□$
there is a higher surface area	$□$
there are more collisions per second	$□$
the acid particles have less energy	$□$

[2]

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1 mark

Hydrogen peroxide solution, H₂O₂, is used in commercial stain removers.

A GCSE class investigated how effective four stain removers are at removing stains. Stain removers A, B, C and D contain different concentrations of hydrogen peroxide.

The students tested how effective each one is at removing an identical oil stain from four towels.

Their findings are outlined below.

Figure: Four information cards, one per stain remover. Stain remover A: working temperature 50°C, cost per 100 cm³ = 99p, time to remove stain = 40 min, volume needed = 20 cm³. Stain remover B: working temperature 30°C, cost per 100 cm³ = £1.99, time to remove stain = 40 min, volume needed = 10 cm³. Stain remover C: working temperature 20°C, cost per 100 cm³ = £2.49, time to remove stain = 20 min, volume needed = 5 cm³. Stain remover D: working temperature 30°C, cost per 100 cm³ = £1.49, time to remove stain = 30 min, volume needed = 10 cm³.

In carrying out this investigation, which variables were kept the same in order to get valid results? Tick ( $✓$ ) the correct answer.

type of oil used, towel material and volume of hydrogen peroxide	$□$
type of oil used, towel material and temperature of stain remover	$□$
type of oil used and towel material	$□$
type of oil used, towel material and cost of stain remover	$□$

How did you do?

1 mark

Tick ( $✓$ ) all of the statements which could explain why stain remover A has to be heated to 50°C before it removes the stain.

it is the cheapest stain remover	$□$
it is heat resistant	$□$
it has a low concentration of hydrogen peroxide	$□$
it takes a long time to work	$□$

How did you do?

1 mark

The students found that stain removers B and D used the same volume and worked best at the same temperature.

Assuming that they have the same hydrogen peroxide concentration, suggest a possible reason why D removes the stain more quickly than B.

...................................................................................................................................................................

How did you do?

5 marks

One student went on to investigate the decomposition of hydrogen peroxide.

The equation for the reaction is as follows.

2H₂O₂ → 2H₂O + O₂

The student investigated the effect of changing the concentration of the hydrogen peroxide solution on the rate of the reaction. She used manganese dioxide as a catalyst in each experiment.

This is the method she used.

Pour 50 cm³ of hydrogen peroxide solution of concentration R into a conical flask on a digital balance.
Add 1 g of catalyst and place some cotton wool loosely in the neck of the flask. Record the balance reading and immediately start a stopwatch.
Record the balance reading every minute until the mass no longer changes.
Carry out the experiment twice more using hydrogen peroxide of different concentrations, S and T.

Her results are plotted on the grid below.

Figure: Line graph. X-axis "Time (minutes)" from 0 to 10. Y-axis "Balance reading (g)" from 163.0 to 164.0 (decreasing mass as oxygen gas escapes). Three curves are plotted, labelled R, T and S, each starting near 164.0 g at t=0 and decreasing at different rates before levelling off — R falls fastest and levels off soonest, T falls at an intermediate rate, and S falls most slowly. A tangent line is drawn touching curve T at t=2 minutes, with a right-angled triangle construction beside it showing a rise of 0.4 g over a run of 2.8 minutes.

(i) Using the tangent shown on the graph, calculate the rate of reaction for concentration T at 2 minutes. Show your working.

Rate at 2 minutes = ....................................................... g / minute

[2]

(ii) The initial rate for concentration S is half the initial rate for concentration T.

Explain this difference in rate in terms of the particle theory.

...................................................................................................................................................................

[3]

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11a

3 marks

Catalysts are used to speed up the reaction. The graphs show the volume of oxygen formed over 120 seconds when 1.0 g of different metal oxides were added to 50 cm³ of hydrogen peroxide solution.

Figure: Graph showing volume of oxygen (cm³, y-axis 0 to 140) against time (s, x-axis 0 to 120), with three curves labelled lead oxide, iron(III) oxide and zinc oxide. Lead oxide rises fastest and reaches the highest volume; iron(III) oxide rises at a moderate rate; zinc oxide rises slowest and reaches the lowest volume, all levelling off before 120 s.

(i) Use the graph to find the time taken to form 44 cm³ of oxygen using iron(III) oxide as the catalyst.

..................................................... s

[1]

(ii) State which metal oxide is the best catalyst. Give a reason for your answer.

[1]

(iii) After 120 seconds, the contents of each beaker were washed into a filter paper and funnel. The catalyst left over on each filter paper was dried and weighed.

Tick ( $✓$ ) the box next to the statement which is correct.

the same mass of all catalysts is left over	$□$
more zinc oxide is left over than lead oxide	$□$
about 80 % of the iron(III) oxide is left over	$□$
no lead oxide is left over	$□$

[1]

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11b

6 marks

Exhaust gases from car engines contain harmful molecules, for example, carbon monoxide and nitrogen oxides. All cars are fitted with catalytic converters that split up these harmful molecules.

The catalysts are made from platinum (Pt) and palladium (Pd) or rhodium (Rh). A mesh structure is used that exposes the maximum surface area of catalyst to the exhaust gases, while also reducing the amount of catalyst required. Platinum, palladium and rhodium are extremely expensive.

As the exhaust gases from the engine pass over the catalysts, chemical reactions take place on their surfaces. The harmful molecules are broken up and converted into other gases that are "safe" to enter the air. These gases include carbon dioxide, nitrogen, oxygen and water.

Figure: Diagram of a catalytic converter — a cylindrical honeycomb/mesh structure coated with platinum (Pt), palladium (Pd) and rhodium (Rh), with exhaust gases entering from the engine at one end. Labels indicate that reduction occurs at one region of the mesh and oxidation occurs at another, with "safe" gases exiting the other end.

There are two different types of catalyst, a reduction catalyst and an oxidation catalyst. The reduction catalyst removes oxygen from nitrogen oxides to make nitrogen and oxygen. The oxidation catalyst adds oxygen to carbon monoxide to make carbon dioxide.

One of the biggest drawbacks of the catalytic converter is that it only works effectively at high temperatures. The average running temperature of a car engine is between 90 and 110 °C. It takes nearly 30 minutes for these temperatures to be reached.

The table below shows the percentages of carbon monoxide and nitrogen oxides converted to safe gases by a catalytic converter at different temperatures.

Temperature (°C)	Carbon monoxide converted (%)	Nitrogen oxides converted (%)
25	16	25
50	19	28
75	26	35
100	60	72
125	91	92
150	93	94
175	95	95
200	97	98

(i) Using your knowledge of particle theory suggest why a catalyst in mesh form works better than a lump of catalyst.

[2]

(ii) Tick ( $✓$ ) the box that best describes the adverse effect that gases leaving the exhaust would have on the environment.

they have no effect on the environment	$□$
they reduce oxygen levels	$□$
they deplete the ozone layer	$□$
they cause global warming	$□$

[1]

(iii) Tick ( $✓$ ) the box that best describes the conversion of carbon monoxide and nitrogen oxides into "safe" gases at different temperatures.

equal amounts of carbon monoxide and nitrogen oxides are converted at every temperature	$□$
more carbon monoxide is converted than nitrogen oxides up to 100 °C	$□$
more nitrogen oxides are converted than carbon monoxide up to 100 °C	$□$
40 % more nitrogen oxides are converted than carbon monoxide up to 100 °C	$□$

[1]

(iv) In your opinion how effective are catalytic converters? Explain your answer.

[2]

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Time (s)	Volume of gas produced (cm³)
0	0
5	12.0
10	20.5
15	29.0
20	33.5
25	35.0
30	35.0

Concentration of sodium thiosulfate (g / dm³)	Time 1 (s)	Time 2 (s)	Time 3 (s)	Mean time (s)
0.2	114	113	112	113
0.4	74	70	72	72
0.6	40	38	57	39
0.8	21	23	22	22
1.0	14	16	15	15

Time (s)	Volume of gas (cm³)
0	0
10	34
20	64
30	78
40	94
50	98
60	100
70	100

Graph stops at 60 s	$□$
Graph is still rising at 60 s	$□$
Graph reaches a maximum temperature of 56 °C	$□$

The particles collide with less energy so less chance of successful collisions	$□$
The particles move slower so less chance of successful collisions	$□$
The particles have less surface area so less chance of successful collisions	$□$
The particles get used up so less chance of successful collisions	$□$

from 0 – 0.5 minutes	$□$
from 1 – 1.5 minutes	$□$
from 2 – 2.5 minutes	$□$
from 3 – 3.5 minutes	$□$

Time (minutes)	Volume of hydrogen produced (cm³)
0	0
1	15
2	28
3	32
4	45
5	49
6	50
7	50