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Questions on measuring the rate of a reaction
Following a reaction by gas volume in a gas syringe, by mass loss on a balance, by colour in a colorimeter and by a precipitate hiding a cross; choosing a method for a given reaction; the reaction of sodium thiosulfate with hydrochloric acid and its sulfur dioxide hazard; colorimetry and the calibration curve; the resolution of a balance and the uncertainty in a mass loss; varying the thiosulfate concentration by dilution; why 1/t measures the rate; the iodine clock and a rate in mol dm⁻³ s⁻¹ from a fixed amount of product; and required practical 3, the effect of temperature on rate, with its method, results, uncertainties and improvements.
20 original questions · 65 marks · the measuring the rate of a reaction notes · Kinetics · Flashcards for this unit
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Describe how to investigate the effect of temperature on the rate of the reaction between sodium thiosulfate solution and dilute hydrochloric acid.
Mark scheme
Indicative content
- measured volumes of sodium thiosulfate solution, for example 50 cm³, in a conical flask and of hydrochloric acid, for example 10 cm³, in a separate tube
- both are warmed separately in a water bath to the chosen temperature, checked with a thermometer
- the flask stands on paper marked with a cross; the acid is added and a stopwatch started at the moment of mixing
- the watch is stopped when the cross can no longer be seen looking down through the liquid, and the temperature of the mixture is recorded
- the experiment is repeated at about five temperatures, for example 20 to 60 °C, with repeats at each, keeping volumes, concentrations, flask, cross and observer the same
- 1/t is calculated as a measure of rate and plotted against temperature, in a well-ventilated room because sulfur dioxide is given off
Levels
- Level 3 (5–6 marks) the volumes, the separate warming of both solutions, the cross and the timing from mixing to disappearance, the range of temperatures with repeats, the variables kept the same and the processing as 1/t against temperature are all present, with the ventilation precaution
- Level 2 (3–4 marks) the cross method and the timing are described over a range of temperatures; the two solutions are not warmed separately, the controls are not named, or the results are not processed into 1/t
- Level 1 (1–2 marks) the solutions are mixed and timed at different temperatures, with the time itself taken as the rate and no controls or repeats
Calculate the rate of formation of iodine in run 2 of the iodine clock results in the table, in mol dm⁻³ s⁻¹, and deduce how the rate depends on the iodide concentration. Every run held 5.0 cm³ of 0.0040 mol dm⁻³ sodium thiosulfate with starch, and hydrogen peroxide and acid in excess, in a total volume of 40.0 cm³. The thiosulfate reacts as I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq).
Run KI(aq) / cm³ Water / cm³ Time for the blue-black colour / s 1 5.0 15.0 120 2 10.0 10.0 61 3 20.0 0.0 30 Mark scheme
- M1 amount of thiosulfate = 0.0050 dm³ × 0.0040 mol dm⁻³ = 2.0 × 10⁻⁵ mol
- M1 iodine formed at the end point = 2.0 × 10⁻⁵ ÷ 2 = 1.0 × 10⁻⁵ mol, a concentration of 1.0 × 10⁻⁵ ÷ 0.0400 = 2.5 × 10⁻⁴ mol dm⁻³
- B1 doubling the iodide volume, at the same total volume, halves the time and quadrupling it quarters the time, so the rate is proportional to the iodide concentration
- M1 rate in run 2 = 2.5 × 10⁻⁴ ÷ 61
- A1 rate = 4.1 × 10⁻⁶ mol dm⁻³ s⁻¹ to two significant figures; allow error carried forward from the amount of iodine
Calculate 1/t for two runs of the disappearing-cross experiment in which the cross disappeared after 47 s and after 18 s, and the factor by which the rate increased, explaining why 1/t can be used as a measure of the rate.
Mark scheme
- B1 the cross disappears when a fixed amount of sulfur has formed, the same in every run, so rate = fixed amount ÷ t is proportional to 1/t
- M1 1/t = 1 ÷ 47 and 1 ÷ 18
- A1 1/t = 0.021 s⁻¹ and 0.056 s⁻¹, to two significant figures, with the unit s⁻¹
- A1 factor = 47 ÷ 18 = 2.6, a ratio with no unit, to two significant figures; allow 0.056 ÷ 0.021 = 2.7
Evaluate a student's conclusion that 'the rate doubles for every 10 °C rise' from the results for the thiosulfate reaction given in the table.
Temperature / °C 1/t / s⁻¹ 20 0.0110 30 0.0205 40 0.0370 50 0.0640 Mark scheme
- B1 the ratios for each 10 °C rise are 1.9, 1.8 and 1.7
- B1 so the rate nearly doubles for each 10 °C rise, which supports the conclusion approximately
- B1 the ratio falls as the temperature rises, so 'doubles' is not exact and the rule does not hold at every temperature
- B1 the results cover only 20 to 50 °C for one reaction, with no repeats shown, so the conclusion holds only for that range and that reaction
Explain why, when the effect of thiosulfate concentration is investigated by the disappearing-cross method, water is added so that the total volume is the same in every run, and give the volumes of 0.050 mol dm⁻³ sodium thiosulfate solution and of water needed to make 50 cm³ of solution of concentration 0.020 mol dm⁻³.
Mark scheme
- B1 the same volume in the same flask gives the same depth of liquid over the cross
- B1 so the cross is hidden by the same amount of sulfur in every run, and 1/t stays proportional to the rate
- B1 the concentration of the acid in the mixture is also the same, so thiosulfate concentration is the only variable changed
- B1 0.020 × 50 ÷ 0.050 = 20 cm³ of thiosulfate solution, made up with 30 cm³ of water
Calculate the initial rate of the reaction of marble chips with excess hydrochloric acid in an open flask on a balance, using the mass lost in the table, first in g s⁻¹ and then as the rate of production of carbon dioxide in mol s⁻¹. Mr of CO₂ = 44.0.
Time / s Mass lost / g 0 0.00 20 0.26 40 0.44 60 0.56 90 0.67 120 0.73 180 0.78 Mark scheme
- M1 mass lost plotted against time with a smooth curve from the origin, and a tangent ruled to it at t = 0
- A1 gradient of the tangent = 0.016 g s⁻¹; accept 0.013 to 0.019 g s⁻¹
- M1 the mass lost is carbon dioxide, so divide by its Mr: 0.016 ÷ 44.0
- A1 initial rate = 3.6 × 10⁻⁴ mol s⁻¹ to two significant figures; allow error carried forward from the gradient
Identify the precipitate and the hazardous gas formed when sodium thiosulfate solution reacts with dilute hydrochloric acid, and give one precaution against the gas.
Mark scheme
- B1 the precipitate is sulfur, S, which forms as a pale yellow solid and makes the mixture cloudy
- B1 the gas is sulfur dioxide, SO₂, which is toxic and irritates the eyes and lungs
- B1 work in a well-ventilated room or fume cupboard and pour each mixture away as soon as the time is recorded; eye protection alone is not credited, as it answers the acid
Explain why the reaction of 0.10 g of magnesium with excess hydrochloric acid cannot be followed by mass loss on a balance that reads to 0.01 g. Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
Mark scheme
- B1 amount of magnesium = 0.10 ÷ 24.3 = 4.1 × 10⁻³ mol, so 4.1 × 10⁻³ mol of hydrogen forms
- B1 the mass of hydrogen is 4.1 × 10⁻³ × 2.0 = 0.0082 g
- B1 that is less than the 0.01 g resolution of the balance, so the whole reaction changes the reading by less than one division
Calculate the volume of carbon dioxide, at room temperature and pressure, given off when 0.500 g of calcium carbonate reacts with excess hydrochloric acid, and state whether it fits in a 100 cm³ gas syringe. Mr of CaCO₃ = 100.1; molar volume 24.0 dm³ mol⁻¹.
Mark scheme
- M1 amount of CaCO₃ = 0.500 ÷ 100.1 = 5.00 × 10⁻³ mol, which gives the same amount of CO₂
- M1 volume = 5.00 × 10⁻³ × 24 000 cm³ mol⁻¹
- A1 1.20 × 10² cm³, to three significant figures, which is more than the 100 cm³ the syringe holds
Calculate the mean time and its percentage uncertainty for the three runs at 30 °C in the table, taking the uncertainty as half the range.
Run Time for the cross to disappear / s 1 35 2 33 3 37 Mark scheme
- M1 mean time = (35 + 33 + 37) ÷ 3 = 35.0 s
- M1 uncertainty = half the range = (37 − 33) ÷ 2 = 2 s
- A1 percentage uncertainty = 2 ÷ 35.0 × 100 = 5.7%, to two significant figures
Explain why, in the thiosulfate practical, both solutions are brought to temperature separately before mixing, and why the temperature of the mixture is measured at the end of each run.
Mark scheme
- B1 if either solution were colder or warmer, the mixture would start at a temperature different from the one being investigated
- B1 the mixture loses heat once it stands on the bench, most quickly in the hottest runs
- B1 recording the starting and final temperatures and using the mean gives the temperature at which the reaction ran
Suggest a method for following how the rate of reaction between zinc granules and dilute sulfuric acid changes with time, naming the apparatus and the measurements.
Mark scheme
- B1 hydrogen is given off, so collect it in a gas syringe connected to a bunged conical flask
- B1 start a stopwatch as the zinc meets the acid and read the gas volume at regular intervals, such as every 15 s
- B1 plot volume against time; the gradient of a tangent gives the rate at any time, since hydrogen is too light to follow by mass loss
Compare apparatus A, a gas syringe, with apparatus B, an upturned measuring cylinder of water, for collecting the carbon dioxide from marble and hydrochloric acid.
The two ways of collecting the gas, set up on the same flask of marble chips and dilute hydrochloric acid. Mark scheme
- B1 carbon dioxide dissolves slightly in water, so the measuring cylinder records less gas than was produced, whereas the syringe collects it all
- B1 the syringe reads directly and does not need filling and inverting, but its plunger can stick and then jump
- B1 the syringe holds only 100 cm³, whereas a larger measuring cylinder can collect more gas
Give three variables that must be controlled when investigating the effect of temperature on the rate of the thiosulfate and acid reaction.
Mark scheme
- B1 the volume and concentration of the sodium thiosulfate solution
- B1 the volume and concentration of the hydrochloric acid
- B1 the depth of liquid over the cross, through using the same flask, the same cross and the same observer
Outline the difference between continuous monitoring and a method timed to an end point, giving an example of each.
Mark scheme
- B1 continuous monitoring takes readings at intervals through a single run, so one experiment gives a whole curve
- B1 a method timed to an end point gives one time per run, so the run is repeated at each value of the variable
- B1 gas volume in a syringe is an example of continuous monitoring; the disappearing cross is an end-point method
Show that the mean rate of production of hydrogen is 5.0 × 10⁻⁵ mol s⁻¹ when 48.0 cm³ of the gas, measured at room temperature and pressure, is collected in the first 40 s of the reaction of magnesium with hydrochloric acid. Molar gas volume = 24.0 dm³ mol⁻¹.
Mark scheme
- M1 n(H₂) = 48.0 ÷ 24 000 = 2.00 × 10⁻³ mol, with the molar gas volume converted to cm³ first; allow 0.0480 ÷ 24.0
- M1 mean rate = amount formed ÷ time = 2.00 × 10⁻³ ÷ 40
- A1 = 5.0 × 10⁻⁵ mol s⁻¹, with the unit, and the working shown in full because the answer is given
Suggest an improvement to the disappearing-cross method that removes the need to judge the end point by eye, and explain how it works.
Mark scheme
- B1 a light sensor connected to a data logger, placed under the flask or beside it, or a colorimeter
- B1 it records the light passing through the mixture, so the end point is a set value of transmitted light, the same in every run
Suggest why the reaction between bromine water and methanoic acid, in which the orange colour of bromine fades, could be followed with a colorimeter.
Mark scheme
- B1 bromine is the only coloured species, so the absorbance of the mixture falls as the bromine is used up
- B1 readings can be taken at intervals through one run, and a calibration curve turns absorbance into bromine concentration
Explain why a loose plug of cotton wool is placed in the neck of the flask when following the reaction of marble with acid by mass loss.
Mark scheme
- B1 the plug lets the carbon dioxide escape, so the mass falls as the gas is given off
- B1 it stops droplets of acid thrown up by the bubbles leaving the flask, which would add to the mass lost and make the rate too high
Calculate the percentage uncertainty in a mass loss of 0.18 g found as the difference between two readings on a digital balance that reads to 0.01 g.
Mark scheme
- M1 each reading is uncertain by one in the last digit, ±0.01 g, so the difference of two readings is uncertain by ±0.02 g
- A1 0.02 ÷ 0.18 × 100 = 11%, to two significant figures; allow 5.6% from ±0.005 g a reading
Practise measuring the rate of a reaction one question at a time