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GCSE · Chemistry

C6a Rates of Reaction

AQA 8464 · Calculator allowed · about 130 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Why Do Reactions Happen at Different Speeds?

Original text written for Revision Library.

A firework explodes in well under a second, but an iron gate left out in the rain can take years to rust away completely. Both are chemical reactions, yet their rates, the speed at which reactants turn into products, are wildly different. Chemists explain this using collision theory: reacting particles must collide with each other, and they must collide with enough energy to react. Anything that increases how often particles collide, or how much energy those collisions carry, will speed up a reaction. This is why food is kept in a refrigerator (a lower temperature slows down the reactions that cause spoilage), why a powdered stock cube dissolves faster than the same mass in one solid block, and why the catalytic converter in a car exhaust is built with a fine honeycomb structure, to expose as much surface area as possible to the exhaust gases passing through it. In the laboratory, chemists measure rate of reaction by tracking how quickly a reactant is used up or a product is formed over time, often using a gas syringe, a mass balance, or by timing how long it takes for a marker to become obscured by a forming precipitate. This topic covers collision theory, the factors that affect rate of reaction (concentration, pressure, surface area, temperature and catalysts), the required practicals for measuring rate using a change in turbidity and a change in gas volume, and calculating mean rate and rate from the gradient of a graph.

Rate of reaction describes how quickly reactants are converted into products, and can be explained using collision theory.

(a)State what is meant by the term 'rate of reaction'.(1)

(b)According to collision theory, state two conditions that must be met for a reaction to occur between two particles.(2)

(Total for Question 1 is 3 marks)

Both concentration and surface area affect the rate of a reaction between a solid and a solution.

(a)Increasing the concentration of a solution increases the rate of reaction. Explain why, in terms of particles.(2)

(b)Breaking a solid reactant into smaller pieces also increases the rate of reaction. Explain why, in terms of particles.(2)

(Total for Question 2 is 4 marks)

Increasing the temperature of a reaction mixture increases the rate of reaction more than most students expect, because temperature affects particles in two separate ways.

(a)Explain, using ideas about collision theory, why increasing the temperature of a reaction mixture increases the rate of reaction.(5)

(Total for Question 3 is 5 marks)

A cube of solid reactant of side length 2 cm is cut up into 8 identical smaller cubes, each of side length 1 cm, so that the total volume of solid stays the same.

Figure (to be drawn): Diagram: a single cube of side length 2 cm, alongside the same cube cut into 8 identical smaller cubes each of side length 1 cm (a 2 x 2 x 2 arrangement).

(a)Calculate the total surface area of the single cube before it is cut up.(2)

(b)Calculate the total surface area of all 8 smaller cubes added together, and state whether it is greater or smaller than your answer to part (a).(2)

(c)Explain, in terms of collision theory, why cutting the reactant into smaller pieces increases the rate of reaction with a surrounding acid.(1)

(Total for Question 4 is 5 marks)

Pressure is a factor that affects the rate of reactions involving gases.

(a)Increasing the pressure of a gaseous reaction mixture increases the rate of reaction. Explain why, in terms of particles.(3)

(b)Give one everyday or industrial situation where controlling the rate of a reaction is important, and explain whether a faster or slower rate is more desirable in that situation.(2)

(Total for Question 5 is 5 marks)

A catalyst is a substance that changes the rate of a chemical reaction.

(a)Define the term 'catalyst'.(2)

(b)State one reason why using a catalyst is beneficial in an industrial chemical process, in terms of cost.(2)

(Total for Question 6 is 4 marks)

The reaction profile below shows the energy change during a reaction, both without a catalyst and with a catalyst present. The activation energy without a catalyst is 92 kJ/mol; with a catalyst present, the activation energy is 55 kJ/mol. The products have less energy than the reactants in both cases.

Figure (to be drawn): Reaction profile diagram: energy (y-axis) against reaction progress (x-axis), showing reactants at a higher energy level than products (exothermic), with two activation energy humps shown: an uncatalysed pathway peaking at 92 kJ/mol above the reactants, and a lower, dashed catalysed pathway peaking at 55 kJ/mol above the reactants.

(a)State whether this reaction is exothermic or endothermic, and justify your answer using the reaction profile.(2)

(b)Calculate the reduction in activation energy caused by the catalyst.(2)

(c)Explain, in terms of activation energy and successful collisions, why a catalyst increases the rate of reaction without being used up.(2)

(Total for Question 7 is 6 marks)

Required practical: a student investigates the effect of the concentration of sodium thiosulfate solution on the rate of its reaction with dilute hydrochloric acid, using the disappearing cross method.
Na2S2O3(aq) + 2HCl(aq) -> 2NaCl(aq) + S(s) + SO2(g) + H2O(l)

(a)Describe how the student could carry out this experiment. Include reference to the apparatus used and how the end-point is judged.(4)

(b)Give one safety precaution the student should take during this experiment, and explain why it is necessary.(2)

(c)Identify the independent variable and the dependent variable in this investigation.(1)

(Total for Question 8 is 7 marks)

The student repeats the experiment in Question 8 with three different concentrations of sodium thiosulfate solution, keeping everything else the same. Rate of reaction is calculated using rate = 1 / time.
Concentration of Na2S2O3 (g/dm³): 10, time for cross to disappear (s): 60, rate of reaction (s^-1): 0.0167
Concentration of Na2S2O3 (g/dm³): 20, time for cross to disappear (s): 30, rate of reaction (s^-1): to be calculated
Concentration of Na2S2O3 (g/dm³): 40, time for cross to disappear (s): 15, rate of reaction (s^-1): 0.0667

(a)Calculate the missing rate of reaction for the experiment using 20 g/dm³ sodium thiosulfate solution. Give your answer to 3 significant figures.(2)

(b)Describe the relationship between the concentration of sodium thiosulfate solution and the rate of reaction shown by these results.(2)

(c)Using the relationship you described in part (b), predict the time it would take for the cross to disappear if the student used a sodium thiosulfate concentration of 80 g/dm^3.(2)

(Total for Question 9 is 6 marks)

Required practical: a student investigates the effect of the concentration of hydrochloric acid on the rate of its reaction with small marble chips (calcium carbonate).
CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)

(a)Describe how the student could use a gas syringe to measure the rate of this reaction.(3)

(b)Suggest one reason why a gas syringe is used to measure the volume of carbon dioxide produced, rather than collecting the gas in an upside-down measuring cylinder filled with water.(1)

(c)State two variables that must be controlled for this to be a fair test.(2)

(d)Explain why the marble chips are used in small pieces of a similar size, rather than one large lump.(1)

(Total for Question 10 is 7 marks)

The student's results for the reaction in Question 10 are shown below.
Time (s): 0, 10, 20, 30, 40, 50, 60
Volume of CO2 collected (cm³): 0, 18, 32, 42, 48, 50, 50
The reaction was complete (no more gas produced) at 50 seconds.

Figure (to be drawn): Graph: volume of gas produced (cm³, y-axis, 0 to 60) against time (s, x-axis, 0 to 60), plotted from the data table given. The curve rises steeply at first, then its gradient decreases, becoming horizontal from t = 50 s onward. A tangent line is shown touching the curve at t = 10 s.

(a)Calculate the mean rate of reaction during the first 30 seconds of the experiment.(2)

(b)Calculate the mean rate of reaction for the whole reaction, using the time at which the reaction was complete.(2)

(c)A tangent drawn to the curve at t = 10 s passes through the points (0 s, 5 cm³) and (20 s, 29 cm³). Use this tangent to calculate the rate of reaction at t = 10 s.(2)

(d)Explain why the rate of reaction calculated in part (c) is greater than the mean rate calculated in part (b).(2)

(Total for Question 11 is 8 marks)

A different student investigates the same reaction (marble chips and dilute hydrochloric acid) by measuring the loss in mass of the reaction flask over time, using a balance. The flask is loosely plugged with a small piece of cotton wool. The initial mass of the flask and its contents was 150.00 g. After 80 seconds, the reaction was complete and the final mass was 149.12 g.

(a)Explain the purpose of the cotton wool plug in this experiment.(2)

(b)Calculate the total loss in mass during the reaction.(1)

(c)Calculate the mean rate of reaction, in g/s, over the 80 seconds. Give your answer to 2 significant figures.(2)

(d)Explain why the total mass of the flask and its contents decreases during the reaction, even though mass is conserved overall in the chemical reaction.(1)

(Total for Question 12 is 6 marks)

A student reacts an excess of magnesium ribbon with 50 cm³ of dilute hydrochloric acid, collecting the hydrogen gas produced using a gas syringe.
Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

(a)Explain, in terms of collision theory, why the gradient of the graph of volume of gas against time decreases as the reaction proceeds.(2)

(b)Explain why the graph becomes horizontal (flat) before all of the magnesium ribbon has reacted.(2)

(c)State what would happen to the final (maximum) volume of gas collected if a larger excess of magnesium ribbon were used, with the same volume and concentration of hydrochloric acid.(1)

(Total for Question 13 is 5 marks)

A student compares the rate of reaction between magnesium ribbon and excess dilute hydrochloric acid at two different temperatures, using the same mass of magnesium and the same volume and concentration of acid in each case. A tangent drawn to the graph for the experiment carried out at 40 degrees C, at t = 4 s, passes through the points (0 s, 10 cm³) and (8 s, 58 cm³).

Figure (to be drawn): Graph: volume of gas (cm³) against time (s) for the reaction at 40 degrees C, with a tangent line drawn at t = 4 s passing through the points (0 s, 10 cm³) and (8 s, 58 cm³).

(a)Use the tangent to calculate the rate of reaction at t = 4 s for the experiment carried out at 40 degrees C.(2)

(b)Convert your answer to part (a) into units of dm³ per minute.(2)

(c)The same tangent method applied to the experiment carried out at 20 degrees C, at the same point on its graph, gives a rate of only 2 cm³/s. Explain, using collision theory, why the rate of reaction is greater at 40 degrees C than at 20 degrees C.(2)

(Total for Question 14 is 6 marks)

Manganese(IV) oxide catalyses the decomposition of hydrogen peroxide solution.
2H2O2(aq) -> 2H2O(l) + O2(g)
A student measures the volume of oxygen gas produced over time using a gas syringe. A tangent to the graph at t = 15 s passes through the points (5 s, 12 cm³) and (25 s, 68 cm³).

Figure (to be drawn): Graph: volume of oxygen gas (cm³) against time (s), with a tangent line drawn at t = 15 s passing through the points (5 s, 12 cm³) and (25 s, 68 cm³).

(a)Calculate the rate of reaction at t = 15 s. Give your answer to 3 significant figures.(3)

(b)The student repeats the experiment without the manganese(IV) oxide catalyst. State and explain the effect this would have on the rate of reaction, and on the total volume of oxygen gas eventually produced.(3)

(Total for Question 15 is 6 marks)

A student carries out the same reaction between magnesium ribbon and excess dilute hydrochloric acid at 20 degrees C (Graph A) and 50 degrees C (Graph B), using identical masses and concentrations in each case.

Figure (to be drawn): Two graphs of volume of gas (cm³) against time (s): Graph A (20 degrees C) and Graph B (50 degrees C), for the same reaction and reactant amounts, drawn on the same axes.

(a)State and explain which graph, A or B, would show the steeper gradient at the start of the reaction.(2)

(b)State, with a reason, whether graphs A and B would reach the same final (maximum) volume of gas.(2)

(c)A student claims that increasing the temperature from 20 degrees C to 50 degrees C will exactly double the rate of reaction, because the temperature (in degrees C) has more than doubled. Evaluate this claim.(3)

(Total for Question 16 is 7 marks)

A student wants to investigate how the concentration of hydrochloric acid affects the rate of its reaction with magnesium ribbon, using the gas volume (gas syringe) method.

Plan a method for this investigation. In your answer you should refer to: the equipment and procedure you would use; how you would ensure your results are valid, accurate and reproducible; and how you would use your results to calculate the rate of reaction for each concentration.

(Total for Question 17 is 6 marks)

The Boltzmann distribution shows the spread of energies of particles in a gas or solution at a particular temperature, and can be used to explain the effect of temperature on rate of reaction in more detail.

Figure (to be drawn): Two diagrams: (1) a Boltzmann distribution graph of number of particles (y-axis) against particle energy (x-axis), with the activation energy, Ea, marked on the x-axis; (2) a volume of gas (cm³) against time (s) graph with a tangent drawn at t = 8 s passing through the points (2 s, 6 cm³) and (14 s, 42 cm³).

(a)Higher tier only. On a Boltzmann distribution graph (particle energy on the x-axis, number of particles with that energy on the y-axis, with the activation energy, Ea, marked on the x-axis), state what happens to the shape of the curve when the temperature of a reaction mixture is increased. In your answer, refer to the total area under the curve.(3)

(b)A tangent to the volume-time graph for this reaction at t = 8 s passes through the points (2 s, 6 cm³) and (14 s, 42 cm³). Calculate the rate of reaction at t = 8 s, in cm³/s, and then convert your answer into mm³/s (1 cm³ = 1000 mm³).(3)

(c)The student judges the end-point of a similar disappearing-cross experiment by eye. Evaluate this method, and suggest one improvement.(2)

(Total for Question 18 is 8 marks)

Mark scheme · C6a Rates of Reaction

Question 1

(a) B1 the rate of reaction is a measure of how fast a reactant is used up, or how fast a product is formed (oe: change in amount of reactant/product per unit time)
(a) Answer: The rate of reaction is how fast a reactant is used up or a product is formed (change in amount per unit time).
(b) B1 the particles must collide with each other (oe)
(b) B1 the collision must have energy greater than or equal to the activation energy (oe: a minimum amount of energy needed to react)
(b) Answer: The particles must collide, and the collision must have energy greater than or equal to the activation energy.

Question 2

(a) B1 there are more particles of the reactant in the same volume, so the solution is more crowded (oe)
(a) B1 this leads to more frequent collisions between reacting particles per unit time (oe), so the rate of reaction increases
(a) Answer: A higher concentration means more particles in the same volume, so collisions between reacting particles are more frequent, increasing the rate.
(b) B1 breaking the solid into smaller pieces increases its surface area (exposed to the other reactant) for the same total mass/volume (oe)
(b) B1 this leads to more frequent collisions between reacting particles per unit time (oe), so the rate of reaction increases
(b) Answer: Smaller pieces have a greater surface area, so more particles are exposed and collisions happen more frequently, increasing the rate.

Question 3

(a) B1 at a higher temperature, particles have more kinetic energy and move faster (oe)
(a) B1 faster-moving particles collide more frequently (more collisions per second) (oe)
(a) B1 a greater proportion of particles have energy greater than or equal to the activation energy (oe)
(a) B1 so a greater proportion of collisions are successful (result in a reaction) (oe)
(a) B1 both effects (more frequent collisions and a greater proportion being successful) combine to increase the rate of reaction (oe, linking mark)
(a) Answer: Particles move faster and collide more often, and a greater proportion of collisions have energy above the activation energy, so more collisions are successful. Both effects together increase the rate of reaction.

Question 4

(a) M1 area of one face = 2 x 2 = 4 (cm²), and a cube has 6 faces
(a) A1 24 cm² (cao)
(a) Answer: 24 cm²
(b) M1 area of one small cube = 6 x 1² = 6 cm², and there are 8 small cubes
(b) A1 48 cm² (cao), which is greater than the answer to (a) (ecf)
(b) Answer: 48 cm², which is greater than in part (a).
(c) B1 more surface area is exposed to the acid, so there are more frequent collisions between reacting particles per unit time (oe)
(c) Answer: More surface area is exposed to the acid, so collisions between reacting particles are more frequent, increasing the rate.

Question 5

(a) B1 increasing the pressure means the same number of gas particles occupy a smaller volume (oe)
(a) B1 so the gas particles become more crowded together, increasing their (effective) concentration (oe)
(a) B1 this increases the frequency of collisions between reacting particles per unit time, increasing the rate of reaction (oe)
(a) Answer: Higher pressure squeezes the same number of gas particles into a smaller volume, increasing their concentration, so collisions between reacting particles are more frequent.
(b) B1 a valid situation identified, e.g. the catalytic converter in a car exhaust, food spoilage, setting of a bone-repair cement, or an industrial reaction vessel (any one, oe)
(b) B1 correct reasoning about why a faster or slower rate is desirable for the chosen situation (e.g. a fast rate is needed in a catalytic converter so harmful gases fully react in the short time they are in contact with the catalyst; a slow rate is desirable for food spoilage so food lasts longer)
(b) Answer: Example: a catalytic converter needs a fast rate so that harmful exhaust gases fully react before leaving the exhaust system.

Question 6

(a) B1 a substance that increases the rate of a chemical reaction (oe)
(a) B1 without being used up (chemically unchanged) at the end of the reaction (oe, cao)
(a) Answer: A catalyst is a substance that increases the rate of a reaction without being used up (chemically unchanged) at the end.
(b) B1 catalysts allow reactions to be carried out at a lower temperature and/or pressure than otherwise needed (oe)
(b) B1 this reduces the energy costs (and equipment costs) of the process, and since the catalyst is not used up it can be reused, making it cost-effective (oe, any one linked reason)
(b) Answer: Catalysts allow lower operating temperatures and pressures, reducing energy costs, and since they are not used up they can be reused.

Question 7

(a) B1 exothermic (cao)
(a) B1 because the products have less energy than the reactants, so energy is transferred to the surroundings overall (oe)
(a) Answer: Exothermic, because the products have less energy than the reactants, so energy is released to the surroundings.
(b) M1 92 - 55
(b) A1 37 kJ/mol (cao)
(b) Answer: 37 kJ/mol
(c) B1 a catalyst provides an alternative reaction pathway with a lower activation energy (oe)
(c) B1 so a greater proportion of particle collisions have energy greater than or equal to this lower activation energy, meaning more collisions are successful per second; the catalyst itself takes no part in the overall reaction, so it is chemically unchanged at the end (oe)
(c) Answer: A catalyst provides a lower-activation-energy pathway, so more collisions have enough energy to react; it is not consumed because it is not part of the overall reaction.

Question 8

(a) B1 measure a fixed volume of sodium thiosulfate solution into a conical flask, placed on top of a cross drawn on a piece of paper or card (oe)
(a) B1 add a fixed volume of dilute hydrochloric acid to the flask and immediately start a stop-clock/timer (oe)
(a) B1 look down through the solution from above and stop the timer at the moment the cross can no longer be seen, as the sulfur precipitate that forms makes the solution turbid/cloudy (oe)
(a) B1 repeat using different concentrations of sodium thiosulfate solution (diluting with water, keeping the total volume the same), keeping the volume/concentration of hydrochloric acid and the temperature the same as controlled variables (oe)
(a) Answer: Add a fixed volume of sodium thiosulfate over a cross, add a fixed volume of dilute hydrochloric acid and start timing, then stop timing when the cross disappears from view; repeat with different thiosulfate concentrations, keeping the acid volume, concentration and temperature the same.
(b) B1 wear eye protection/safety goggles (oe)
(b) B1 because the sulfur dioxide gas produced is toxic/irritating and the hydrochloric acid is corrosive/irritant, so the experiment should also be carried out in a well-ventilated area/fume cupboard (oe, any suitable linked reason)
(b) Answer: Wear eye protection, and work in a well-ventilated area, because sulfur dioxide gas produced is toxic and the acid is irritant/corrosive.
(c) B1 independent variable: concentration of sodium thiosulfate solution; dependent variable: time taken for the cross to disappear (both needed for the mark)
(c) Answer: Independent variable: concentration of sodium thiosulfate solution. Dependent variable: time for the cross to disappear.

Question 9

(a) M1 rate = 1 / 30
(a) A1 0.0333 s^-1 (awrt 0.0333, cao)
(a) Answer: 0.0333 s^-1
(b) B1 as the concentration of sodium thiosulfate solution doubles, the rate of reaction also doubles (oe: rate is directly proportional to concentration)
(b) B1 supported with reference to the data, e.g. concentration doubling from 10 to 20 g/dm³ doubles the rate from 0.0167 to 0.0333 s^-1, and doubling again from 20 to 40 g/dm³ doubles the rate again to 0.0667 s^-1
(b) Answer: Rate of reaction is directly proportional to concentration: doubling the concentration of sodium thiosulfate doubles the rate.
(c) M1 identifies that doubling the concentration again (40 to 80 g/dm³) would double the rate again, to 0.133 s^-1 (ecf from their stated relationship in (b), even if that relationship is incorrect, provided it is applied consistently)
(c) dM1 dependent on M1: time = 1 / their rate, giving 7.5 s (awrt 7.5, ecf)
(c) Answer: 7.5 s

Question 10

(a) B1 add a fixed mass of marble chips to a conical flask, then add a fixed volume of dilute hydrochloric acid and quickly attach the gas syringe/bung, starting the stop-clock at the same time (oe)
(a) B1 record the volume of gas (carbon dioxide) collected in the gas syringe at regular time intervals (e.g. every 10 seconds), until no more gas is produced/the volume stops changing (oe)
(a) B1 repeat using hydrochloric acid of different concentrations (diluted with water to keep the total volume the same), keeping the mass/size of marble chips, volume of acid and temperature the same as controlled variables (oe)
(a) Answer: Add a fixed mass of marble chips and a fixed volume of acid to a flask, quickly seal it with a gas syringe and time how the volume of gas collected changes; repeat with different acid concentrations, keeping other variables constant.
(b) B1 carbon dioxide is (slightly) soluble in water, so some gas would dissolve and the measured volume would be an underestimate; a gas syringe avoids this source of error (oe)
(b) Answer: Carbon dioxide is slightly soluble in water, so using a measuring cylinder filled with water would give an underestimate of the gas volume; a gas syringe avoids this.
(c) B1 temperature of the acid/reaction mixture
(c) B1 total volume of acid used (any two of: temperature, volume of acid, mass/size of marble chips)
(c) Answer: Temperature of the acid, and the total volume of acid used (mass/size of marble chips also acceptable).
(d) B1 to control/standardise the surface area of the marble chips exposed to the acid, so that surface area does not become an uncontrolled variable affecting the rate (oe)
(d) Answer: Using pieces of a similar size controls the surface area of marble exposed to the acid, so it does not become an uncontrolled variable.

Question 11

(a) M1 rate = volume / time = 42 / 30
(a) A1 1.4 cm³/s (cao)
(a) Answer: 1.4 cm³/s
(b) M1 rate = 50 / 50 (using the volume and time at completion, not 60 s)
(b) A1 1.0 cm³/s (cao)
(b) Answer: 1.0 cm³/s
(c) M1 gradient = (29 - 5) / (20 - 0) = 24 / 20
(c) A1 1.2 cm³/s (cao)
(c) Answer: 1.2 cm³/s
(d) B1 the rate of reaction decreases over time as the reaction proceeds, because the concentration of acid decreases as it is used up, so collisions become less frequent (oe)
(d) B1 the tangent in (c) gives the (instantaneous) rate near the start of the reaction, when it is fastest, whereas the mean rate in (b) is averaged over the whole reaction, including the much slower later stages, so it is lower (ecf from their answers to (b) and (c))
(d) Answer: Rate decreases over time as acid concentration falls, so the instantaneous rate early on (part c) is higher than the rate averaged over the whole, slower-finishing reaction (part b).

Question 12

(a) B1 it allows the gas (carbon dioxide) produced to escape from the flask, so the mass loss can be measured (oe)
(a) B1 while stopping any acid spray/splashes from escaping (or preventing loss of solid/liquid), so the measured mass loss is due only to the gas escaping (oe)
(a) Answer: The cotton wool lets carbon dioxide gas escape (so mass loss can be measured) while stopping acid spray from escaping, so the mass loss recorded is due only to gas leaving.
(b) A1 0.88 g (cao)
(b) Answer: 0.88 g
(c) M1 rate = 0.88 / 80 (ecf from (b))
(c) A1 0.011 g/s (awrt 0.011, ecf)
(c) Answer: 0.011 g/s
(d) B1 the carbon dioxide gas produced escapes into the surrounding air (through the cotton wool), so the mass remaining inside the (open) flask decreases, even though no mass is actually created or destroyed overall (oe)
(d) Answer: Carbon dioxide gas escapes into the air, reducing the mass remaining in the open flask, even though total mass is conserved overall.

Question 13

(a) B1 the concentration of hydrochloric acid decreases as it is used up during the reaction (oe)
(a) B1 so the frequency of collisions between reacting particles decreases, meaning fewer successful collisions occur per second, and the rate decreases (oe)
(a) Answer: As the acid is used up, its concentration falls, so collisions between reacting particles become less frequent and the rate decreases.
(b) B1 the hydrochloric acid is the limiting reactant/is fully used up (there is excess magnesium remaining) (oe)
(b) B1 once all of the acid has reacted, no more collisions can occur between (the remaining) reacting particles, so no more gas can be produced and the reaction stops (oe)
(b) Answer: The acid (the limiting reactant) is fully used up, so no more reacting collisions can occur, even though excess magnesium remains, so the reaction stops.
(c) B1 the final volume of gas would stay the same (unchanged), because the amount of hydrochloric acid (the limiting reactant) determines the maximum amount of gas produced (oe)
(c) Answer: It would stay the same, because the acid (the limiting reactant) determines the maximum amount of gas produced, not the amount of magnesium.

Question 14

(a) M1 gradient = (58 - 10) / (8 - 0) = 48 / 8
(a) A1 6 cm³/s (cao)
(a) Answer: 6 cm³/s
(b) M1 6 x 60 = 360 cm³/min (ecf from (a))
(b) A1 360 / 1000 = 0.36 dm³/min (awrt 0.36, ecf)
(b) Answer: 0.36 dm³/min
(c) B1 at the higher temperature, particles have more kinetic energy and move faster, so they collide more frequently (oe)
(c) B1 a greater proportion of collisions have energy greater than or equal to the activation energy, so a greater proportion of collisions are successful, increasing the rate (oe)
(c) Answer: At 40 degrees C, particles have more energy and move faster, colliding more often and with more collisions exceeding the activation energy, so more collisions are successful.

Question 15

(a) M1 gradient = (68 - 12) / (25 - 5) = 56 / 20
(a) M1 = 2.8
(a) A1 2.80 cm³/s (cso, 3 s.f.)
(a) Answer: 2.80 cm³/s
(b) B1 without the catalyst, the rate of reaction would be slower (a lower gradient/rate at any given time) (oe)
(b) B1 because there is no alternative, lower-activation-energy pathway, so fewer collisions have sufficient energy to react, meaning fewer successful collisions per second (oe)
(b) B1 the total (maximum) volume of oxygen gas produced would be unchanged/the same, because the same amount of hydrogen peroxide reacts overall; a catalyst does not affect the amount of product formed, only how quickly it forms (oe)
(b) Answer: The rate would be slower without the catalyst, because collisions need to overcome a higher activation energy, but the total volume of oxygen produced would be the same, since the catalyst only affects rate, not the amount of product.

Question 16

(a) B1 Graph B (the experiment at 50 degrees C) (cao)
(a) B1 because particles have more kinetic energy at the higher temperature, so they collide more frequently and with more energy, resulting in more successful collisions per second at the start of the reaction (oe)
(a) Answer: Graph B (50 degrees C), because particles have more energy and collide more frequently and successfully at the higher temperature.
(b) B1 yes, the same final volume (cao)
(b) B1 because the same amount/mass of hydrochloric acid (the limiting reactant) is used in each experiment, so the same amount of product (hydrogen gas) is eventually formed; temperature only affects how quickly the maximum volume is reached, not the total amount (oe)
(b) Answer: Yes, the same final volume, because the same amount of the limiting reactant (acid) is used each time; temperature changes only how quickly this volume is reached.
(c) B1 the claim is not correct: the relationship between Celsius temperature and rate is not simply proportional (a doubling of Celsius temperature does not cause a doubling of rate) (oe)
(c) B1 rate depends on the proportion of particles with energy greater than or equal to the activation energy, and this proportion increases sharply (non-linearly) with a rise in temperature, rather than in direct proportion to the Celsius value (oe)
(c) B1 collision frequency and collision energy both change together in a complex way as temperature rises, so a simple doubling claim based on Celsius values is not a scientifically valid comparison (oe)
(c) Answer: The claim is incorrect: rate does not scale directly with Celsius temperature. It depends on the (non-linearly increasing) proportion of particles with energy above the activation energy, and on collision frequency, so a simple doubling of temperature value does not mean a doubling of rate.

Question 17

Level 3 (5-6): A logically sequenced method is described that would lead to a valid, reproducible outcome. The answer includes relevant equipment, how variables are controlled, how measurements are taken and repeated, and explains how the rate of reaction is calculated from the results.
Level 2 (3-4): A method is described with some steps in a logical order. The answer includes some relevant equipment and some reference to controlling variables or repeating measurements, and some reference to calculating rate, but is not fully complete or has minor omissions.
Level 1 (1-2): Simple statements are made about the equipment or method, with little or no logical structure. Little or no reference to controlling variables, repeats, or calculating rate.
Level 0 (0): No relevant content.
Indicative content:
- Use a conical flask, bung and gas syringe (or delivery tube to an inverted measuring cylinder) to collect the gas produced.
- Use a fixed length/mass of magnesium ribbon in each repeat, cleaned with emery paper/sandpaper to remove the surface oxide layer.
- Use the same total volume of hydrochloric acid each time, varying only its concentration (e.g. by diluting with water using a measuring cylinder or burette, keeping the total volume constant).
- Add the acid to the flask, quickly insert the bung/gas syringe, and start the stop-clock at the same moment.
- Record the volume of gas produced at regular time intervals (e.g. every 10 seconds) until no more gas is produced.
- Control the temperature (e.g. using a water bath) so it does not affect the comparison between concentrations.
- Repeat each concentration at least twice (or three times), calculate a mean, and discard anomalous results.
- Plot a graph of volume of gas against time for each concentration.
- Calculate the mean rate of reaction using rate = volume of gas produced / time taken, or use the gradient of a tangent to find the rate at a specific time.
- Compare the rate at different concentrations to draw a conclusion about the relationship between concentration and rate.
- Reference to safety: wear eye protection, and keep naked flames away, since hydrogen gas is flammable.

Question 18

(a) B1 the peak of the curve moves to the right (towards higher energies) and becomes lower, so the curve becomes flatter and more spread out (oe)
(a) B1 the total area under the curve stays the same, because the total number of particles has not changed (cao)
(a) B1 the area under the curve to the right of the activation energy, Ea, increases, showing that a greater proportion of particles now have energy greater than or equal to the activation energy (oe)
(a) Answer: The peak shifts right and the curve flattens, but the total area under the curve stays the same (same total number of particles); the area beyond Ea increases, showing more particles now have enough energy to react.
(b) M1 gradient = (42 - 6) / (14 - 2) = 36 / 12
(b) A1 3 cm³/s (cao)
(b) dM1 dependent on the previous A1: 3 x 1000 = 3000 mm³/s (ecf)
(b) Answer: 3 cm³/s, which is 3000 mm³/s
(c) B1 judging by eye is subjective: different people may judge the point at which the cross disappears slightly differently (or the same person may be inconsistent), reducing the reproducibility/accuracy of the results (oe)
(c) B1 improvement: use a light sensor/colorimeter (data logger) to measure the light transmitted or absorbed by the solution, recording when it reaches a fixed, objective threshold value, removing human judgement from the measurement (oe)
(c) Answer: Judging by eye is subjective and can vary between observers, reducing reproducibility. Using a light sensor/colorimeter with a fixed threshold value would give a more objective, reproducible measurement.

Rates of Reaction - Worksheets, Questions and Revision