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3.2Physical Chemistry

Rates of Reaction — Edexcel Linear - OL (Chemistry) Revision Notes & Topical Questions

10 min read 13 parts Lesson 5 of 9 Unit quiz

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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Lesson summary

Rates of Reaction: Measuring Rate and Effect of Concentration, Surface Area and Temperature — EDEXCEL LINEAR – OL (CHEMISTRY) notes: Plan and carry out experiments to measure rate and explain the effect of concentration, surface area and temperature using collision theory.The rat…

  • rate of a reaction
  • Worked example 1
  • Worked example 2
  • Worked example 3
  • Worked example 4
  • Rate of reaction

Section 1 of 13Defining and measuring rate

The rate of a reaction tells us how fast reactants are used up or products are formed. Rate = change in amount ÷ time. Reactions range from very slow (rusting, weathering of rocks) to very fast (explosions, precipitation of silver chloride). To measure a rate, choose a property that changes steadily during the reaction.

TechniqueMeasuredExampleLimitation
Gas collection (syringe or inverted cylinder over water)Volume of gas at timed intervalsMagnesium + hydrochloric acidGas must not be very soluble
Mass lossDecrease in mass as gas escapesMarble chips + acidLight gases such as hydrogen give small changes
Disappearing crossTime for cross to vanishSodium thiosulfate + acidSubjective endpoint; only for precipitates
Colour change / colorimeterAbsorbance of coloured solutionIodine reactionsNeeds a coloured species

Section 2 of 13Method: concentration and surface area

1
Measure 50 cm3 of sodium thiosulfate solution into a conical flask standing on a black cross on paper.
  • Use different concentrations made by diluting with water (for example 50, 40, 30, 20 and 10 cm3 of thiosulfate topped up to 50 cm3).
2
Add 10 cm3 of dilute hydrochloric acid, start the stopwatch and swirl.
3
Stop the clock when the cross can no longer be seen through the sulfur precipitate.
4
Record the time, and calculate rate as 1 ÷ time (s-1). Repeat the experiment and use a mean.
5
Plot rate against concentration; a straight line through the origin shows rate is proportional to concentration.

For surface area, react a fixed mass of calcium carbonate as large chips, small chips and powder with the same volume and concentration of acid, and measure gas volume every 10 s. Smaller pieces give a steeper curve because more particles are exposed at once. Equal masses are used so that the final volume of gas is the same.

Section 3 of 13Explaining the effects with collision theory

Particles can only react if they collide with at least the activation energy.

Explaining the effects with collision theory
Concentration or pressure: more particles per unit volume, so collisions are more frequent and the rate increases.
Surface area: more exposed particles, so more frequent collisions.
Temperature: particles gain kinetic energy, move faster, collide more often and a greater proportion of collisions have the activation energy,
  • so the rate rises steeply.

Temperature experimentwarm the thiosulfate solution to 20, 30, 40, 50 and 60 °C in a water bath, add acid and time the cross.

Explaining the effects with collision theory
A graph of 1 ÷ time against temperature curves upward: the rate roughly doubles for each 10 °C rise.
Control the volume, concentration and the same cross-and-flask setup, and note that the fastest reaction has the largest rate and the shortest time.

Section 4 of 13Fair testing and safety

Variables

  • the independent variable is the one you change;
    • the dependent variable is what you measure;
    • control variables are kept the same.

Fair testing and safety
Safety: thiosulfate with acid releases sulfur dioxide, a toxic gas,
  • so work in a well-ventilated area, use small volumes, and dispose of waste promptly.
Wear eye protection and avoid having flames when collecting hydrogen.
Industrial relevance: coal dust and flour dust explosions arise from very large surface areas,
  • and food is stored in refrigerators to slow the rate of reactions caused by bacteria.

Section 5 of 13Evaluating the experiments

Evaluation questions ask how reliable and valid the results are.

1
Reliability improves by repeating each run and ignoring anomalies when calculating the mean.
2
Validity improves by changing only one variable.
3
In the disappearing cross method, different observers may judge the end point differently,
  • so the same person should do all runs, in the same lighting, looking from above.
4
In gas collection experiments, gas may leak around a bung or dissolve in water; a gas syringe is more accurate than a measuring cylinder over water.
5
When temperature is the variable, a thermostatically controlled water bath keeps the temperature constant during the run; the reaction itself may warm the solution slightly,
  • so the starting temperature of each solution should be measured immediately before mixing.

Plotting data

  • place the independent variable on the x-axis, label axes with units, choose a scale that uses more than half the grid, plot points accurately and draw a best-fit straight line or smooth curve.
  • State the pattern in words: "as concentration doubles, rate doubles", or "as temperature increases, rate increases at an increasing gradient".

Section 6 of 13Worked examples

  1. A student times the thiosulfate cross experiment at 20 °C (80 s) and 40 °C (21 s). Calculate rate (1/time) for each and state the effect. [3]3 marks

  2. Explain why powdered zinc reacts faster with acid than zinc granules. [2]2 marks

  3. State three variables to control in a concentration experiment. [3]3 marks

  4. Describe one advantage of collecting gas in a syringe instead of measuring mass loss. [1]1 mark

Section 7 of 13Catalysts and activation energy

A catalyst increases the rate of a chemical reaction but is not used up and does not change the products.

Catalysts and activation energy
It works by providing an alternative reaction pathway with a lower activation energy,
  • so a greater proportion of collisions are successful at the same temperature.
A catalyst does not change the energy difference between reactants and products (ΔH) and in a reversible reaction it speeds up the forward and backward reactions equally,
  • so it does not alter the equilibrium yield.
ReactionCatalyst
Decomposition of hydrogen peroxide: 2H2O2 → 2H2O + O2Manganese(IV) oxide
Haber process (making ammonia)Iron
Contact process (making sulfuric acid)Vanadium(V) oxide
Hydrogenation of alkenes (margarine)Nickel
Cracking of alkanesZeolite / aluminium oxide
Catalytic convertersPlatinum, palladium, rhodium

Enzymes are biological catalysts, for example in fermentation of glucose to ethanol by yeast. Catalysts are reused, so small amounts are needed, and they allow lower temperatures, saving energy and cost. Catalysts can be poisoned by impurities, and expensive metals must be recovered.

Section 8 of 13Energy profile with and without a catalyst

On an energy profile the uncatalysed route has a high hump; the catalysed route has a lower hump but starts and ends at the same energy levels.

Energy profile with and without a catalyst
The activation energy Ea is the vertical distance from the reactants to the top of the hump.
Draw both curves on the same axes, and label Ea (uncatalysed), Ea (catalysed) and ΔH.

Section 9 of 13Reading rate graphs and calculating rates

A graph of product volume against time starts steep and flattens as reactants are used up.

Reading rate graphs and calculating rates
To find the rate at a particular time, draw a tangent to the curve and calculate its gradient (change in y ÷ change in x).
Example: a tangent at 20 s passes through (0 s, 6 cm3) and (40 s, 38 cm3), so gradient = 32 ÷ 40 = 0.80 cm3/s.
1
The average rate over an interval = volume change ÷ time change.
2
Compare curves: the steeper curve is the faster reaction.
3
If both curves finish at the same volume, equal amounts of limiting reactant were used.
4
If one curve finishes higher, it had more of the limiting reactant.

For hydrogen peroxide decomposition, use a gas syringe and compare rates with different catalysts (manganese(IV) oxide, copper(II) oxide, potato).

Reading rate graphs and calculating rates
Use the same volume and concentration of peroxide and the same mass of catalyst, and measure the volume of oxygen every 15 seconds.
After the experiment, filter, wash and dry the catalyst; its mass is unchanged, showing it was not used up.

Section 10 of 13Practical: comparing catalysts and a worked comparison

A fair comparison of catalysts for hydrogen peroxide uses 25 cm3 of 10 volume peroxide each time, 0.5 g of each catalyst, and a gas syringe to record the oxygen at 10 s intervals.

Practical: comparing catalysts and a worked comparison
Typical results: manganese(IV) oxide gives 50 cm3 in 20 s, copper(II) oxide gives 50 cm3 in 90 s and the uncatalysed peroxide produces almost nothing in the same period.
The steepest line shows the most effective catalyst.
All three reactions finish with the same volume of oxygen when the same amount of peroxide is used,
  • which shows that catalysts do not change the amount of product.

When drawing an energy profile with and without a catalyst, remember that the catalyst gives a different route through a different transition state, not a lower starting point. Industrial catalysts are often in the form of pellets, gauze or honeycomb to maximise surface area; their activity can drop if they are coated with impurities (poisoning), for example sulfur compounds poisoning the iron in the Haber process, so reactants must be purified first.

Section 11 of 13Worked examples

  1. Explain how a catalyst increases the rate of reaction. [2]2 marks

  2. A tangent to a gas volume curve goes from (0, 10) to (50, 60). Calculate the rate. [2]2 marks

  3. Describe how to show that manganese(IV) oxide is not used up in the decomposition of hydrogen peroxide. [3]3 marks

  4. Why does the final volume of gas not depend on the catalyst? [1]1 mark

Section 12 of 13Key terms

TermMeaning
Rate of reactionHow quickly reactants are used or products formed
Collision frequencyNumber of collisions per second
Surface areaExposed area of a solid
ConcentrationAmount of solute per unit volume
Control variableA factor kept constant in an experiment
PrecipitateInsoluble solid formed in a solution
TermMeaning
CatalystSubstance that speeds up a reaction and is not used up
Activation energyMinimum energy needed for a successful collision
EnzymeBiological catalyst
GradientChange in y divided by change in x
TangentStraight line touching a curve at one point
Limiting reactantReactant that runs out first

Section 13 of 13Exam technique

1
Saying a higher concentration means particles move faster — it means more particles in the same volume.
2
Forgetting to keep the total volume constant when diluting.
3
Reading the time as the rate (a longer time is a slower rate).
4
Describing the cross experiment for reactions that do not form a precipitate.
1
Saying a catalyst lowers the temperature of the reaction.
2
Drawing the catalysed pathway with lower product energy.
3
Saying catalysts are consumed.
4
Confusing the gradient with the final volume.

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