AQA's RP7 measures the rate of reaction in two ways. In the initial rate method, you time how long a clock reaction takes to reach a fixed point, repeating it with different starting concentrations. In the continuous monitoring method, you follow a single reaction as it happens. This guide uses the iodine clock for the initial rate method, and the reaction between marble chips (CaCO₃) and hydrochloric acid (HCl), followed with a gas syringe, for continuous monitoring.
Core Specification Link
This practical supports 3.1.9 Rate equations, explicitly addressing the determination of rate equations, reaction order, and rate constants (\(k\)) using graphical methods.
Reaction for the Continuous Monitoring Method
\[ \text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \]The rate of reaction is monitored by collecting and measuring the volume of carbon dioxide gas evolved over time.
Aim
To find orders of reaction from rates measured by two methods: an initial rate method, timing an iodine clock reaction at different iodide ion concentrations; and a continuous monitoring method, measuring the volume of carbon dioxide given off over time at different concentrations of hydrochloric acid.
Equipment List
Method 1: continuous monitoring (gas syringe)
- Conical flask (250 cm³)
- Gas syringe (100 cm³)
- Delivery tube fitted with a rubber bung
- Marble chips (CaCO₃ - large granules, used in excess)
- Dilute hydrochloric acid (1.0 mol dm⁻³ and 0.5 mol dm⁻³)
- Stopwatch
- Retort stand, boss, and clamps
- Electronic balance (precision to 0.01 g)
- Measuring cylinders
Method 2: initial rate (iodine clock)
- Conical flasks (100 cm³) and a white tile
- Burettes or graduated pipettes, one for each solution
- Potassium iodide solution (KI, 0.10 mol dm⁻³)
- Dilute sulfuric acid (H₂SO₄, 0.25 mol dm⁻³)
- Sodium thiosulfate solution (Na₂S₂O₃, 0.0050 mol dm⁻³)
- Freshly made starch solution
- Hydrogen peroxide solution (H₂O₂, 0.10 mol dm⁻³)
- Deionised water
- Stopwatch and thermometer
Method 1: Continuous Monitoring (Gas Syringe)
- Measure exactly 50 cm³ of 1.0 mol dm⁻³ hydrochloric acid and pour it into the 250 cm³ conical flask.
- Clamp the gas syringe horizontally in a stand and connect its inlet to the delivery tube.
- Weigh out approximately 5.00 g of large marble chips (an excess, ensuring hydrochloric acid is the limiting reactant).
- Add the marble chips to the flask, immediately insert the rubber bung tightly, and start the stopwatch.
- Record the volume of gas collected in the syringe every 10 seconds for the first 2 minutes, and then every 30 seconds until the reaction stops.
- Thoroughly clean and dry the flask . Repeat steps 1 to 5 using 50 cm³ of 0.50 mol dm⁻³ hydrochloric acid. Keep all other variables (temperature, mass, and surface area of marble chips) constant.
- Plot a graph of gas volume (y-axis) against time (x-axis) for both acid concentrations.
Safety & Risk Assessment
| Hazard | Risk | Precaution |
|---|---|---|
| 1.0 mol dm⁻³ Hydrochloric acid | Skin and eye irritation. | Wear safety goggles and lab coat. Wash spills off skin immediately. |
| Splattering of acid | Acid droplets splashing into eyes during the addition of marble chips. | Wear eye protection. Add marble chips carefully down the neck of the flask. |
| Carbon dioxide build-up | Over-pressurisation causing syringe plunger to shoot out or connections to pop. | Do not exceed the maximum volume limit of the gas syringe. Do not clamp the plunger. |
| Dilute sulfuric acid and hydrogen peroxide (iodine clock) | Can irritate the eyes and skin. | Wear eye protection. Wash any splashes off the skin with plenty of water. |
Continuous Monitoring Results & Orders
From the volume-time curves, the initial rate (rate at \(t = 0\)) is determined by drawing a tangent line at the origin and calculating its gradient:
\[ \text{Initial Rate} = \frac{\Delta \text{Volume}}{\Delta \text{Time}} \quad (\text{cm}^3\text{ s}^{-1}) \]- For 1.00 mol dm⁻³ HCl: Initial Rate = 1.80 cm³ s⁻¹
- For 0.50 mol dm⁻³ HCl: Initial Rate = 0.90 cm³ s⁻¹
Step 1: Determine the reaction order
Compare the change in concentration to the change in initial rate:
\[ \text{Concentration ratio} = \frac{1.00}{0.50} = 2.0 \quad (\text{doubled}) \] \[ \text{Rate ratio} = \frac{1.80}{0.90} = 2.0 \quad (\text{doubled}) \]Because doubling the concentration of HCl exactly doubles the initial rate of gas evolution, the reaction is first order with respect to HCl.
Step 2: Write the rate equation
\[ \text{Rate} = k[\text{HCl}]^1 \]Step 3: Calculate the rate constant (\(k\))
\[ k = \frac{\text{Rate}}{[\text{HCl}]} = \frac{1.80\text{ cm}^3\text{ s}^{-1}}{1.00\text{ mol dm}^{-3}} = 1.80\text{ cm}^3\text{ mol}^{-1}\text{ dm}^3\text{ s}^{-1} \]This is only a relative value of the rate constant, because the rate was measured as a volume of gas per second. With the rate in mol dm⁻³ s⁻¹, the rate constant of a first-order reaction has units of s⁻¹.
Method 2: Initial Rate (Iodine Clock)
In the initial rate method you make up several reaction mixtures that differ only in the concentration of one reactant, and time how long each takes to reach the same point. The iodine clock is the standard example. Hydrogen peroxide oxidises iodide ions to iodine in acidic solution:
\[ \text{H}_2\text{O}_2(\text{aq}) + 2\text{H}^+(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow \text{I}_2(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \]Each mixture also contains a small, fixed amount of sodium thiosulfate and some starch. The thiosulfate turns the iodine back into iodide as fast as it forms:
\[ \text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq}) \]Once all the thiosulfate has been used up, iodine builds up and reacts with the starch, and the mixture suddenly turns blue-black. The time, t, is recorded at that moment.
Why 1/t Measures the Initial Rate
- The colour appears when the same amount of iodine has formed in every run, because that amount is set by the fixed amount of thiosulfate.
- Rate = amount of product formed ÷ time, so the rate up to the colour change is proportional to 1/t.
- Only a small fraction of the reactants has been used by then, so this rate is close to the initial rate.
Method
- Use burettes to add 5.0 cm³ of potassium iodide solution and 20.0 cm³ of deionised water to a conical flask on a white tile.
- Add 10.0 cm³ of sulfuric acid, 5.0 cm³ of sodium thiosulfate solution and 1 cm³ of starch solution, and swirl to mix.
- Measure 10.0 cm³ of hydrogen peroxide solution into a small beaker. Pour it into the flask, swirl, and start the stopwatch at the same moment.
- Stop the stopwatch as soon as the mixture turns blue-black, and record the time.
- Repeat with 10.0, 15.0, 20.0 and 25.0 cm³ of potassium iodide solution, using less water each time so that the potassium iodide solution and water always add up to 25.0 cm³. Keep all the other volumes and the temperature the same.
- Repeat each run, calculate the mean time, and work out 1/t for each mixture.
Keeping the total volume the same means the concentration of iodide ions is proportional to the volume of potassium iodide solution, while the concentrations of all the other reactants stay the same.
Results and Analysis
| Volume of KI(aq) / cm³ | Volume of water / cm³ | Mean time, t / s | 1/t / s⁻¹ |
|---|---|---|---|
| 5.0 | 20.0 | 118 | 0.00847 |
| 10.0 | 15.0 | 60 | 0.0167 |
| 15.0 | 10.0 | 41 | 0.0244 |
| 20.0 | 5.0 | 30 | 0.0333 |
| 25.0 | 0.0 | 24 | 0.0417 |
Plot 1/t (a measure of the initial rate) against the volume of potassium iodide solution (a measure of the iodide ion concentration). The shape of the graph shows the order with respect to iodide ions:
- Zero order: a horizontal line, because 1/t does not change.
- First order: a straight line through the origin; doubling the concentration doubles 1/t.
- Second order: a curve; doubling the concentration makes 1/t four times larger.
Step 1: Compare two runs
Going from 10.0 cm³ to 20.0 cm³ of potassium iodide solution doubles the iodide ion concentration. 1/t rises from 0.0167 s⁻¹ to 0.0333 s⁻¹, so the rate also doubles.
Step 2: Check the whole data set
For every run, 1/t divided by the volume of potassium iodide solution is between 0.0016 and 0.0017 s⁻¹ cm⁻³, so 1/t is directly proportional to the iodide ion concentration. A graph of 1/t against volume is a straight line through the origin.
Step 3: Conclusion
The reaction is first order with respect to iodide ions. To find the order with respect to hydrogen peroxide, repeat the experiment varying the volume of hydrogen peroxide solution instead, again keeping the total volume the same.
Sources of Error & Improvements
| Error Source | Classification | Consequence & Mitigation |
|---|---|---|
| Gas escape during bung insertion | Systematic | Some carbon dioxide gas escapes between adding the solid marble chips and sealing the bung, reducing the recorded gas volume at early time intervals. Mitigation: Use a two-necked flask or addition funnel, allowing chips to be dropped without opening the bung. |
| Friction in the gas syringe | Random | Syringe plungers can stick due to friction, causing sudden, jumpy movements and inaccurate readings. Mitigation: Ensure the syringe is dry and clean, and rotate the plunger slightly to minimise friction. |
| Variable marble chip surface area | Random | Marble chips vary in size and shape, changing the exposed surface area and affecting reaction kinetics. Mitigation: Use uniform marble chips or ground calcium carbonate of a consistent mesh size. |
| Judging the colour change (iodine clock) | Random | Reaction time when starting and stopping the stopwatch matters most for short runs. Mitigation: Choose concentrations that give times of at least 20 s, repeat each run, or detect the colour change with a colorimeter. |
| Temperature changes between runs | Random | Rates are very sensitive to temperature, so a warmer run finishes sooner. Mitigation: Stand all the solutions in a water bath at the same temperature, and record the temperature of each mixture. |
Exam questions, marked instantly
Write your answer and the AI examiner checks it against the mark scheme, showing exactly which marks you earned and what to add.
Exam-style question
2 marksA reaction produces carbon dioxide gas. Suggest how its rate could be followed continuously without using a gas syringe.
Show Mark Scheme
- Place the reaction flask on a balance and record the mass at regular time intervals [1]
- The mass decreases as carbon dioxide escapes; the rate of mass loss shows the rate of reaction [1]
Exam-style question
2 marksDescribe how a concentration–time graph for a zero-order reactant differs from one for a first-order reactant.
Show Mark Scheme
- Zero order: a straight line with a constant (negative) gradient, because the rate does not depend on concentration [1]
- First order: a curve whose gradient decreases, with a constant half-life [1]
Exam-style question
3 marksIn an iodine clock experiment, the same small amount of sodium thiosulfate is added to every mixture. Explain why 1/t can be used as a measure of the initial rate of reaction.
Show Mark Scheme
- The blue-black colour appears when all the thiosulfate has reacted, so the same amount of iodine has formed in every run [1]
- Rate = amount of product formed ÷ time, so the rate is proportional to 1/t [1]
- Only a small fraction of the reactants has been used by the time of the colour change, so this rate is close to the initial rate [1]
CPAC Skills Assessed
- CPAC 1: Follows a written procedure to make up clock mixtures with a constant total volume.
- CPAC 2: Design, set up, and apply gas collection apparatus to measure rate variables.
- CPAC 3: Safely sets up gas collection glassware.
- CPAC 4: Records gas volume data accurately at strict time intervals.
When drawing tangents in exams to find gradients, extend your tangent line as far as possible across the graph axes. This reduces the relative error when reading the coordinates (Δy and Δx) to calculate the gradient.