Measuring Rates of Reaction: Mass, Gas Volume and Time

This practical, Practical Skills: Rates of Reaction on the OCR A course, measures how the rate of reaction changes when one condition is changed and everything else is kept the same.

Rate is the change in the amount or concentration of a reactant or product per unit time.

The main experiment times the reaction between sodium thiosulfate and hydrochloric acid at five temperatures: the solution slowly fills with a pale yellow precipitate of sulfur until a cross drawn under the flask can no longer be seen.

Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO₂(g) + H₂O(l)

S₂O₃²⁻(aq) + 2H⁺(aq) → S(s) + SO₂(g) + H₂O(l)

The specification asks for the techniques and procedures used to investigate reaction rates, including the measurement of mass, gas volumes and time (3.2.2(e)).

It also asks for the calculation of rate from the gradients of graphs of a physical quantity against time (3.2.2(b)).

The apparatus and techniques list (1.2.2(l)) requires rates to be measured by at least two different methods: an initial rate method such as a clock reaction, and a continuous monitoring method.

This page covers the timing method (disappearing cross and a clock reaction), gas volume in a syringe and loss of mass on a balance.

The same techniques return in the second-year rates practical activity groups, where they are used to find orders, half-lives and rate constants.

On the OCR A course this practical is examined through the written papers (practical skills are examined in every paper) and the practical endorsement.

The specification references are 3.2.2(e) (techniques to investigate rates, including the measurement of mass, gas volumes and time), 3.2.2(b) (rate from the gradients of graphs) and 1.2.2(l) (rates by at least two different methods).

Questions ask you to justify each step of the method, explain why 1/t is a measure of rate, process results into a table and graph, draw a tangent and find its gradient, explain the results with the Boltzmann distribution, and classify the errors.

The timing idea is the same one used in Practical Skills: Rates of Hydrolysis (hydrolysis of halogenoalkanes), where the time for a precipitate to appear compares three haloalkanes.

The method below is for the temperature investigation. The mixture must reach the chosen temperature before the reaction starts, and the reaction starts only when the acid goes in, so it is the thiosulfate solution that is warmed.

© Dr. Mohammed Al-Fatah – onlinelearningsystem.net

The disappearing cross method: warm the thiosulfate, add the acid and start the clock, then stop the clock when the cross can no longer be seen from above.

The table gives a complete set of results for the temperature investigation. The mixtures cooled during the run, most at the highest temperatures, so the mean temperature is the value plotted.

The times are recorded to the nearest second, the precision of the end point, and 1/t is given to three significant figures.

The rate increases with temperature, and it increases by a larger amount for each step up.

From 21.0 °C to 30.5 °C the value of 1/t rises by a factor of 0.0204 ÷ 0.0114 = 1.8, and it rises by about the same factor for each further 10 °C.

So the graph of 1/t against temperature is a curve that becomes steeper, not a straight line.

A rise of about 10 °C roughly doubles the rate of many reactions near room temperature; here the factor is 1.8.

Rate (1/t) against mean temperature for the sample data: the curve becomes steeper as temperature rises, because each 10 °C rise multiplies the rate by about 1.8.

The same reaction shows the effect of concentration. The temperature is kept constant and the thiosulfate is diluted with distilled water so that the total volume stays at 50 cm³ in every run.

A fixed total volume matters for two reasons: the concentration of thiosulfate is then proportional to the volume of thiosulfate solution used, and the depth of liquid over the cross is the same, so the same amount of sulfur still hides it.

The acid is added exactly as before, 5.0 cm³ of 2.0 mol dm⁻³, so its concentration is the same in every run.

A graph of 1/t against concentration is a straight line through the origin, so the rate is directly proportional to the concentration of thiosulfate. Halving the concentration halves the rate and doubles the time.

The explanation is collision frequency: with twice as many thiosulfate ions in the same volume there are twice as many collisions with hydrogen ions per second, and the same fraction of them have enough energy.

The concentrations in the table are those of the thiosulfate solution before the acid is added. Adding the same 5.0 cm³ of acid to every run dilutes each one by the same factor, so the proportionality is unaffected.

Rate (1/t) against thiosulfate concentration at constant temperature: a straight line through the origin, so rate is proportional to concentration.

On a graph of a quantity against time, the rate at any moment is the gradient of the tangent to the curve at that time. The tangent is a straight line that touches the curve at that point only.

Draw it long, so that the triangle used for the gradient is as large as possible and the reading errors in the rise and the run are a small percentage.

Worked Example: the initial rate. The tangent at t = 0 starts at the origin and passes through (60 s, 0.87 g).

Step 1: rise = 0.87 − 0.00 = 0.87 g; run = 60 − 0 = 60 s

Step 2: initial rate = 0.87 g ÷ 60 s = 0.0145 g s⁻¹

Step 3: in moles of CO₂, 0.0145 g s⁻¹ ÷ 44.0 g mol⁻¹ = 3.30 × 10⁻⁴ mol s⁻¹

Worked Example: the rate at 120 s. The tangent that touches the curve at 120 s passes through (40 s, 0.63 g) and (200 s, 1.11 g).

Rise = 1.11 − 0.63 = 0.48 g; run = 200 − 40 = 160 s; rate = 0.48 ÷ 160 = 0.0030 g s⁻¹.

The rate has fallen to about a fifth of its initial value because the hydrochloric acid is being used up, so acid particles collide with the chips less often.

Answer: initial rate 0.0145 g s⁻¹ (3.30 × 10⁻⁴ mol s⁻¹ of CO₂); rate at 120 s 0.0030 g s⁻¹.

The units come from the axes: g s⁻¹ for a mass graph, cm³ s⁻¹ for a gas volume graph, mol dm⁻³ s⁻¹ for a concentration graph.

Mass of carbon dioxide lost against time for the sample run, with tangents at 0 s and at 120 s: the rate falls from 0.0145 g s⁻¹ to 0.0030 g s⁻¹ as the acid is used up.

The specification asks for rates measured by at least two methods, and a clock reaction is the standard initial rate method.

In the iodine clock, hydrogen peroxide oxidises iodide ions to iodine in acid solution. A small, fixed amount of sodium thiosulfate is included in the mixture, and it turns the iodine back into iodide as fast as it forms.

When all the thiosulfate has been used up, the next iodine formed reacts with starch and the solution turns blue-black suddenly.

H₂O₂(aq) + 2I⁻(aq) + 2H⁺(aq) → I₂(aq) + 2H₂O(l)

I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq)

The colour appears when a fixed amount of iodine has been made, set by the amount of thiosulfate added, so exactly as with the disappearing cross, rate ∝ 1/t.

Only a small fraction of the reactants has been used when the colour appears, so 1/t measures the initial rate.

To investigate the effect of the iodide concentration, the volume of potassium iodide solution is changed and water is added to keep the total volume constant, while the volumes of peroxide, acid, thiosulfate and starch stay the same.

The end point is much sharper than a disappearing cross, which is one reason clock reactions give better data; the same method is used later in the course to find orders of reaction.

The same errors appear in students’ write-ups and exam answers year after year. Each one costs a mark that the corrected version earns.

• Plotting time against temperature and calling it a rate graph. Rate is 1/t: the time falls as the rate rises.
• Warming the mixture after adding the acid. The acid is added to thiosulfate that is already at the chosen temperature, and the clock starts at that moment.
• Recording only the start temperature. The mixture cools, so record the end temperature too and use the mean.
• Changing the concentration without adding water. The total volume must be constant, or the depth of liquid over the cross changes too.
• Explaining temperature only by “more collisions”. The mark is for a greater proportion of collisions with energy at least equal to the activation energy.
• Drawing the energy curve so it touches the energy axis at high energy, or making the higher-temperature curve taller.
• Joining two data points and calling it a tangent, or leaving the unit off the gradient.
• Writing “human error”. Say what is judged inconsistently (the moment the cross disappears) and call it a random error.

Say

“The same amount of sulfur hides the cross each time, so rate is proportional to 1/t.”

“At a higher temperature a greater proportion of particles have energy equal to or greater than the activation energy, so a greater frequency of collisions is successful.”

“Rate is the gradient of the tangent to the curve at that time, in g s⁻¹.”

Do not say

“The cross disappears because the reaction finishes.” (It disappears long before the reaction finishes.) “Higher temperature lowers the activation energy.” (Only a catalyst provides a lower activation energy route.) “Human error.” “The graph of 1/t against temperature is a straight line.”

Watch for

Tables that ask for 1/t to a sensible number of significant figures; a graph where you must read a time for a temperature that was not measured.

Questions that switch to a gas syringe or a balance method and ask for the rate from a tangent; the choice of a fixed total volume; and the difference between an initial rate method and continuous monitoring.