Topic 9 · page 3 of 15

Rate practicals

Two ways to follow a reaction over time, collecting a gas and measuring mass lost, and how to change one factor at a time so that the curves you draw can be compared fairly.

13 min read9.19.39.3(ii)

Read first: Rate from the gradient of a tangent

The specification suggests three practicals on rate: hydrogen peroxide decomposing, marble chips with acid, and thiosulfate with acid. The thiosulfate experiment is on page 1. This page covers the other two, and how to use all three to test a factor.

Collecting a gas: hydrogen peroxide 9.39.3(ii)

Hydrogen peroxide decomposes slowly on its own. A little solid manganese(IV) oxide speeds it up a great deal.

\ce{2H2O2(aq) -> 2H2O(l) + O2(g)}

Two ways of following a reaction. Left: a flask of hydrogen peroxide solution with manganese dioxide, stoppered, with a delivery tube leading to a gas syringe that collects oxygen. Right: a flask of marble chips and hydrochloric acid on a balance, with a cotton wool plug in the neck so that carbon dioxide can escape, and the balance reading falls. — Left: collect the gas in a syringe. Right: weigh the flask as the gas escapes.

PracticalHydrogen peroxide and manganese(IV) oxide

Put a measured volume of hydrogen peroxide solution in a conical flask and connect it, through a delivery tube, to a gas syringe.
Add a weighed amount of manganese(IV) oxide, replace the bung quickly and start the clock.
Read the volume of oxygen in the syringe every 10 s until it stops changing.
Plot volume against time. Draw a tangent at t = 0 to find the initial rate (page 2).
Repeat, changing one factor: the concentration of peroxide, the temperature, or the mass of manganese(IV) oxide.

Manganese(IV) oxide is a catalyst: it is not in the equation, and it is still there, unchanged, at the end. Illustrative readings for one run:

Time / s	Volume of O₂ / cm³
0	0
10	9
20	16
30	21
40	25
60	31
80	35
100	37
120	38

These are the points behind the graph on page 2, whose tangent at t = 0 gives an initial rate of 1.0 cm³ s⁻¹. The mean rate over the first 20 s is 16 ÷ 20 = 0.80 cm³ s⁻¹, which is lower, because the rate has already fallen by then.

Measuring mass lost: marble chips and acid 9.39.3(ii)

Marble is calcium carbonate. With hydrochloric acid it gives off carbon dioxide.

\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)}

The gas escapes, so the flask gets lighter. The mass lost is the mass of carbon dioxide that has left. A plug of cotton wool lets the gas out but stops acid spray.

PracticalMarble chips and hydrochloric acid

Put a measured volume of dilute hydrochloric acid in a conical flask. Stand it on a balance and plug the neck with cotton wool.
Add a weighed amount of marble chips. Replace the plug, zero or record the balance, and start the clock.
Record the mass every 15 or 30 s for several minutes.
Plot mass lost against time and draw tangents.
Repeat, changing one factor: acid concentration, temperature, or the size of the chips.

Worked exampleFrom mass lost to gas volume

In a run the balance reading falls by 0.44 g. How many moles and what volume of carbon dioxide is that at room temperature and pressure? ( $M_\text{r}$ of CO₂ = 44.0; molar volume 24.0 dm³ mol⁻¹)

n(CO₂) = 0.44 ÷ 44.0 = 0.0100 mol
V = 0.0100 × 24.0 = 0.240 dm³ = 240 cm³

The two methods compare like this:

Method	Good point	Weak point
Gas syringe	Reads the volume directly	A leak, or a sticking plunger, ruins the result
Mass loss	Simple; nothing to seal	Needs a balance reading to 0.01 g or better; some CO₂ dissolves in the acid

Changing one factor at a time 9.1

To show how a factor affects the rate, change that factor only and keep everything else the same. The change should alter the rate in the way collision theory predicts (page 4 onwards).

To test	Change	Keep the same	Expected result
Concentration	Acid 1.0 then 2.0 mol dm⁻³	volume of acid, mass and size of chips, temperature	steeper at the start with the stronger acid
Surface area	Chips then powder	mass of marble, acid volume and concentration, temperature	steeper at the start with powder
Temperature	Water bath at 20 °C then 40 °C	volumes, concentrations, size of chips	steeper at the start when warmer
Catalyst	With and without MnO₂	volume and concentration of peroxide, temperature	steeper with MnO₂

Reading two curves. A steeper start means a faster initial rate. Where the curve levels off depends on the amount of the limiting reagent. If only the rate changes (higher temperature, powder, a catalyst), both curves end at the same volume. If you change the amount of the limiting reagent, the final volume changes too.

Check yourself

Manganese(IV) oxide does not appear in the equation for the decomposition of hydrogen peroxide. Explain why, and why the apparatus must be sealed when you use a gas syringe.
Using the table of volumes above, calculate the mean rate over the first 20 s and explain why it is lower than the initial rate of 1.0 cm³ s⁻¹.
A flask containing marble chips and acid loses 0.44 g in 100 s. Calculate the mass of CO₂ lost per second, the amount in moles of CO₂ released, and its volume at room temperature and pressure.
State three variables to keep the same when comparing marble chips with marble powder.
Excess marble is added to 10.0 cm³ of hydrochloric acid, first with a concentration of 0.50 mol dm⁻³ and then with 1.00 mol dm⁻³. Describe the two curves of gas volume against time and calculate the final volume of CO₂ in each (molar volume 24.0 dm³ mol⁻¹).
Give one disadvantage of the mass-loss method compared with a gas syringe.

Answers cover these until you have tried the questions

It is a catalyst: it speeds up the reaction but is not used up, so it does not appear in the equation. The apparatus is sealed so that no oxygen escapes; any leak makes the measured volume too small.
Mean rate = 16 ÷ 20 = 0.80 cm³ s⁻¹. It is lower than the initial rate because the rate falls during the 20 s as the peroxide is used up, so the average over the interval is less than the rate at the start.
Mass per second = 0.44 ÷ 100 = 0.0044 g s⁻¹. n(CO₂) = 0.44 ÷ 44.0 = 0.0100 mol. V = 0.0100 × 24.0 = 0.240 dm³ = 240 cm³.
Any three of: the mass of marble, the volume of acid, the concentration of acid, the temperature.
The 1.00 mol dm⁻³ curve is steeper at the start and levels off at a higher volume. 0.50 mol dm⁻³: n(HCl) = 0.0100 × 0.50 = 0.00500 mol; CO₂ is half this, 0.00250 mol; V = 0.00250 × 24.0 = 0.0600 dm³ = 60.0 cm³. 1.00 mol dm⁻³: n(HCl) = 0.0100 mol; CO₂ 0.00500 mol; V = 0.120 dm³ = 120 cm³.
Any one of: the balance must read to a small mass (0.01 g or better) because the mass lost is small; some CO₂ dissolves in the acid, so less mass is lost than expected; it only works for a gas that escapes.