Questions on finding orders from initial rates, half-lives and graphs

Question 1

Describe how to find the order with respect to propanone in the acid-catalysed reaction of propanone with iodine, using a colorimeter.

Accepted Answer

Indicative content

calibrate: measure the absorbance of iodine solutions of known concentration with a blue filter and plot a calibration graph
bring the propanone, acid and water mixture and the iodine solution to the same temperature in a water bath before mixing
mix, start the timer at the moment of mixing, and record the absorbance at regular intervals such as every 30 s
convert absorbance to iodine concentration, plot it against time and find the gradient, which is the rate
repeat with a different volume of propanone, made up with water to the same total volume so that the other concentrations and the temperature are unchanged
compare the factor on the propanone concentration with the factor on the rate: halving propanone halves the rate, so it is first order

Levels

Level 3 (5–6 marks) the calibration graph, the temperature control before mixing, the timing and regular absorbance readings, the conversion to concentration and the rate from the gradient, and the repeat with a different propanone concentration at the same total volume are all present, with the order deduced from the factors on concentration and rate
Level 2 (3–4 marks) the colorimeter follows the iodine concentration with time and the run is repeated with a different propanone concentration; the calibration, the constant concentrations of iodine and acid, or the step from rates to order is not there
Level 1 (1–2 marks) the colorimeter is said to measure the colour of iodine; the propanone concentration is often changed along with the total volume, so two things vary at once

Question 2

Calculate the rate constant, with its units, for the reaction of A with B, after deducing the order with respect to each from the initial rates in the table.

Accepted Answer

B1 runs 1 and 2: [A] doubles and the rate is multiplied by 4, so the reaction is second order in A B1 runs 1 and 3: [B] triples and the rate triples, so the reaction is first order in B M1 rate = k[A]²[B], so k = 3.0 × 10⁻⁵ ÷ (0.010² × 0.020) = 3.0 × 10⁻⁵ ÷ 2.0 × 10⁻⁶ M1 units: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ for an overall order of three A1 k = 15 mol⁻² dm⁶ s⁻¹, to two significant figures; accept dm⁶ mol⁻² s⁻¹, and allow error carried forward from wrong orders

Question 3

State what is meant by the half-life of a reaction, and identify the order shown by each set of successive half-lives in the table.

Accepted Answer

B1 the half-life is the time taken for the concentration of a reactant to fall to half of its starting value B1 P is first order, since its half-life is constant B1 Q is zero order, since each half-life is half the one before B1 R is second order, since each half-life is double the one before

Question 4

Compare following the iodine concentration in the propanone reaction with a colorimeter and by titration of quenched samples.

Accepted Answer

B1 the colorimeter gives many readings without disturbing the mixture, whereas titration removes samples and gives only a few points B1 the colorimeter reading is almost instant, whereas each titration sample must be quenched and any delay adds a timing error B1 titration needs no calibration, whereas the colorimeter needs a calibration graph to turn absorbance into concentration B1 titration end points are judged by eye with starch, whereas the colorimeter gives an objective reading

Question 5

Evaluate a student's conclusion that a reaction is first order overall, drawn from the concentration–time graph of reactant X, off which she read half-lives of 52 s and 49 s. The other reactant, Y, was in large excess.

Accepted Answer

B1 the graph gives successive half-lives of 50 s each (0.080 to 0.040 to 0.020 to 0.010 at 50, 100 and 150 s); the student's 52 s and 49 s agree within the precision of reading a graph, consistent with first order in X B1 two values are a small sample; the third half-life, 100 to 150 s on the graph, would make the conclusion more secure B1 Y was in large excess, so its concentration hardly changed and the graph says nothing about the order in Y B1 the conclusion should be first order with respect to X; the overall order needs further runs with a different concentration of Y

Question 6

Deduce the order with respect to hydrogen peroxide from the iodine clock results in the table. Each run used the same volumes of potassium iodide, acid, thiosulfate and starch, and the total volume was the same in every run.

Accepted Answer

B1 with the total volume fixed, the volume of H₂O₂ is proportional to its concentration in the mixture M1 1/t is taken as the rate, because each run forms the same fixed amount of iodine before the colour appears M1 halving the volume from 20.0 to 10.0 cm³ changes 1/t from 0.031 to 0.016 s⁻¹, a factor of 0.5 A1 the rate is directly proportional to [H₂O₂], so the reaction is first order with respect to H₂O₂

Question 7

Calculate the rate constant of a first-order reaction whose half-life is 34 s, and the time taken for the reactant's concentration to fall to one eighth of its starting value.

Accepted Answer

M1 k = ln 2 ÷ t½ = 0.693 ÷ 34 A1 k = 0.020 s⁻¹, to two significant figures A1 one eighth is three half-lives, 3 × 34 = 102 s, so 1.0 × 10² s to two significant figures

Question 8

Calculate the initial rate from the tangent drawn at t = 0 on the concentration–time graph, using the two points circled on the tangent.

Accepted Answer

M1 the rate is the gradient of the tangent, change in concentration ÷ change in time M1 gradient = (0.010 − 0.050) ÷ (120 − 0) = −0.040 ÷ 120 A1 initial rate = 3.3 × 10⁻⁴ mol dm⁻³ s⁻¹, to two significant figures, quoted as a positive value

Question 9

Calculate the rate of reaction in mol dm⁻³ s⁻¹ from the colorimeter readings in the table. The absorbance falls steadily and is directly proportional to the iodine concentration.

Accepted Answer

M1 the absorbance falls at 0.36 ÷ 240 = 1.5 × 10⁻³ s⁻¹ M1 the calibration factor is 0.90 ÷ 0.0030 = 300 dm³ mol⁻¹, so divide the absorbance rate by 300 A1 rate = 5.0 × 10⁻⁶ mol dm⁻³ s⁻¹, to two significant figures

Question 10

Explain why 1/t, where t is the time for the colour to appear in a clock reaction, can be used as a measure of the initial rate.

Accepted Answer

B1 the colour appears when a fixed amount of product has formed, the same amount in every run B1 rate is amount formed ÷ time, so with a fixed amount the rate is proportional to 1/t B1 the amount is small, so little reactant has been used and the rate over that time is close to the initial rate

Question 11

Explain why propanone and hydrochloric acid are used in large excess when the reaction of propanone with iodine is followed with a colorimeter.

Accepted Answer

B1 with both in large excess their concentrations hardly change during the run B1 the only concentration that changes is that of iodine, which is the coloured species the colorimeter measures B1 so any change in rate during the run is due to iodine alone, and the graph shows the order with respect to iodine

Question 12

Explain why a run of the acid-catalysed reaction of propanone with iodine, in which the iodine concentration falls by the same amount in every minute until the iodine is used up, shows that the reaction is zero order with respect to iodine.

Accepted Answer

B1 an equal fall in every minute is a constant rate (a straight line of constant gradient on a concentration–time graph), so the rate is the same throughout the run B1 the iodine concentration falls during the run while the rate stays constant B1 so the rate does not depend on the iodine concentration, which is zero order

Question 13

Suggest how the decomposition of hydrogen peroxide, catalysed by manganese(IV) oxide, could be followed by continuous monitoring to find the order with respect to hydrogen peroxide.

Accepted Answer

B1 collect the oxygen in a gas syringe and record its volume at regular time intervals B1 the volume still to be released is proportional to the hydrogen peroxide remaining, so a graph of it against time is a concentration–time graph B1 read successive half-lives from the graph: if they are constant, the reaction is first order in hydrogen peroxide

Question 14

Suggest why a student who doubled [A] and [B] together in a second run could not deduce either order from those two runs alone, and what further run is needed.

Accepted Answer

B1 the change in rate is the product of the effects of both changes, so it cannot be split between A and B B1 a factor of 4, for example, could be second order in one and zero in the other, or first order in each B1 a further run changing only one of the two concentrations, the other kept as in the first run, separates the two effects

Question 15

Calculate the rate constant k in rate = k[CH₃COCH₃][H⁺] at 35 °C, given that a run with propanone at 1.00 mol dm⁻³ and H⁺ at 0.500 mol dm⁻³, both in large excess, loses iodine at a steady 2.5 × 10⁻⁵ mol dm⁻³ s⁻¹.

Accepted Answer

M1 iodine is zero order and the excess reactants hardly change, so the measured rate equals k[CH₃COCH₃][H⁺] throughout the run M1 k = 2.5 × 10⁻⁵ ÷ (1.00 × 0.500), with units (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)² A1 k = 5.0 × 10⁻⁵ mol⁻¹ dm³ s⁻¹, to two significant figures

Question 16

Explain why a reaction with rate = k[A][B] shows a constant half-life for A when B is present in a hundredfold excess, although the reaction is second order overall.

Accepted Answer

B1 with B in large excess, [B] hardly changes while all of A is used up B1 so k[B] is effectively constant and the rate equation behaves as rate = k′[A], where k′ = k[B] B1 the run is pseudo-first-order in A, so its half-life is constant, and only the order with respect to A has been found

Question 17

Explain why samples withdrawn from the propanone and iodine reaction are run into sodium hydrogencarbonate solution before they are titrated.

Accepted Answer

B1 sodium hydrogencarbonate neutralises the H⁺ catalyst, which stops the reaction in the sample B1 so the titre gives the iodine concentration at the moment the sample was taken, not some time later

Question 18

Suggest why two runs of the same initial-rates experiment, one done in the morning and one in a warm afternoon laboratory, gave different values of k, and how the method could be improved.

Accepted Answer

B1 the rate constant depends on temperature, so a warmer afternoon gives a larger k B1 carry out every run in a thermostatted water bath at the same temperature and check it with a thermometer

Question 19

Give two methods of continuous monitoring, naming the quantity measured in each.

Accepted Answer

B1 colorimetry, measuring the absorbance of a coloured reactant or product B1 gas collection in a syringe, measuring the volume of gas produced; or titration of quenched samples, or mass loss on a balance, or pH

Question 20

Suggest a continuous-monitoring method for the hydrolysis (CH₃)₃CCl(aq) + H₂O(l) → (CH₃)₃COH(aq) + H⁺(aq) + Cl⁻(aq), and why it would follow the reaction.

Accepted Answer

B1 measure the electrical conductivity, or the pH, of the mixture at regular time intervals B1 the reaction makes H⁺ and Cl⁻ ions from neutral molecules, so the conductivity rises, or the pH falls, as the reaction proceeds

Run	[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.010	0.020	3.0 × 10⁻⁵
2	0.020	0.020	1.2 × 10⁻⁴
3	0.010	0.060	9.0 × 10⁻⁵

Reactant	First t½ / s	Second t½ / s	Third t½ / s
P	25	25	25
Q	60	30	15
R	20	40	80

Volume of H₂O₂ / cm³	Volume of water / cm³	t / s
20.0	0.0	32
15.0	5.0	43
10.0	10.0	64
5.0	15.0	128

Reading	Value
Absorbance at the start	0.90
Absorbance after 240 s	0.54
Iodine concentration giving an absorbance of 0.90	0.0030 mol dm⁻³