The rate and extent of chemical change

1. How fast is a reaction?

Rate means how quickly a reaction happens. Measure how much reactant is used up, or how much product is made, in a known time. A reaction can be fast at first and slower later, so a mean rate describes a whole chosen interval.

Worked example: 48 cm³ of gas forms in 40 s. Mean rate = 48 ÷ 40 = 1.2 cm³/s. If 0.80 g of a gas escapes in 2 minutes, convert 2 minutes to 120 s first: 0.80 ÷ 120 = 0.0067 g/s (2 significant figures).

Rearrange when needed: quantity changed = rate × time; time = quantity changed ÷ rate. Use the change during the specified interval, not always the total from the start.

2. Read the shape of a rate graph

On a product-versus-time graph, a steeper slope means a faster reaction. The curve often becomes less steep as reactants are used up and there are fewer successful collisions each second. A horizontal section means no more product is forming: the measured reaction has stopped, usually because a limiting reactant is used up.

Both curves finish at the same height, so both make the same total amount of gas. A reaches this amount sooner. Faster does not automatically mean more product. A graph of reactant remaining may slope down instead; use the size of the decrease to calculate the positive reaction rate.

To draw a graph, put time on the horizontal axis and measured quantity on the vertical axis. Label units, choose evenly spaced scales, plot each pair of values accurately and draw a suitable smooth curve. Do not force a straight line through a curve.

3. Rate at one moment

A tangent is a straight line that touches a curve at the chosen point and follows the curve’s direction there. Its slope shows the rate at that moment. A steeper tangent means a greater rate; a horizontal tangent means zero rate.

Higher only — calculating a tangent gradient: choose two points far apart on the tangent. Rate = change in volume ÷ change in time. For the labelled tangent, (60 − 15) ÷ (50 − 10) = 1.125 cm³/s, approximately 1.13 cm³/s. These points are on the tangent; they do not both need to be on the curve.

4. Particles must collide successfully

Reacting particles must collide to react, and the collision must have enough energy. The minimum required energy is the activation energy. More successful collisions each second mean a faster reaction.

Concentration: a more concentrated solution has more reacting particles in each cm³. They collide more frequently, so there are more successful collisions per second. Their average speed does not increase just because concentration increases.

Pressure: compressing reacting gases at the same temperature puts more particles in each cm³. Collisions become more frequent. This explanation applies to gas particles; it is not the explanation for changing a solid’s surface area.

5. Smaller pieces and higher temperatures

Surface area: breaking a solid into smaller pieces exposes more particles to the other reactant. A powder reacts faster than the same mass in large lumps because it has a larger surface area to volume ratio. Crushing does not create extra mass.

Temperature: hotter particles have more kinetic energy and move faster. They collide more often, and a greater proportion of collisions have energy equal to or above the activation energy. Both effects increase the number of successful collisions each second.

A simple model may predict that doubling concentration roughly doubles collision frequency if all other conditions are fixed. Real measured rates must still be checked; rate is not always directly proportional to every variable. Doubling temperature in °C does not mean doubling rate.

6. Catalysts offer an easier route

A catalyst increases a reaction’s rate without being used up overall. It provides a different reaction pathway with a lower activation energy. At the same temperature, more collisions now have enough energy to react.

Different reactions need suitable catalysts. Enzymes are biological catalysts. A catalyst is not a reactant that is gradually consumed; it is not included as a reactant or product in the overall balanced equation. It can be written above the reaction arrow.

A catalyst does not change the energy difference between reactants and products. In a reaction that already goes to completion, it does not make more product from the same amount of limiting reactant; it makes the product faster.

7. Required practical 11: concentration and gas volume

Aim: investigate how concentration affects rate. A testable hypothesis is: “Increasing acid concentration increases the initial rate of carbon dioxide production from a fixed mass of marble chips.”

1. Wear eye protection. Use teacher-approved dilute hydrochloric acid and small quantities of marble chips (calcium carbonate). Keep the flask stable and the gas syringe away from bench edges.
2. Measure a fixed acid volume and a fixed mass of chips. Keep chip size/surface area similar between trials. Use acid in excess over the range tested so the same chip mass limits the total gas.
3. Check that the syringe plunger moves freely and the connections are airtight. Add chips, start the timer and fit the bung promptly. Use the same starting procedure each time.
4. Read gas volume at regular intervals, for example every 10 s, until the volume stops increasing. Read the scale at eye level.
5. Repeat with several acid concentrations while keeping acid volume, chip mass and size, initial temperature and apparatus fixed. Repeat each concentration and compare mean curves or initial slopes.

Gas lost before the bung is fitted makes the collected volume too small. Leaks have the same effect. Never seal a gas-producing reaction in apparatus with no free-moving outlet. A suitable alternative is measuring loss of mass as gas escapes, using a balance; a loose cotton-wool plug may reduce spray without trapping gas.

8. Required practical 11: concentration and cloudiness

The second method measures a change in colour or turbidity (cloudiness). When dilute acid reacts with sodium thiosulfate solution, solid sulfur makes the mixture cloudy. Time how long it takes for a paper cross below the flask to disappear.

1. Make several sodium thiosulfate concentrations by dilution. Keep the total at 40 cm³ before adding acid: mix solution and water in volumes of 40 + 0, 30 + 10, 20 + 20 and 10 + 30 cm³, respectively.
2. Use the same flask on the same cross. Add a fixed volume and concentration of dilute acid, start timing, swirl in the same way, then view vertically from above.
3. Stop when the cross can no longer be seen. Keep total volume/depth, temperature, lighting, observer and cross constant.
4. Repeat, calculate a mean time and compare 1 ÷ mean time as a relative rate (s⁻¹). Shorter time means faster reaction when the same amount of sulfur is needed to hide the cross.

Safety: use small quantities and good ventilation because sulfur dioxide, an irritant gas, is formed. Do not lean over or inhale from the flask. Follow your teacher’s risk assessment and disposal instructions. A light sensor can make an appropriate colour/cloudiness endpoint more objective.

9. Process the results and test a hypothesis

For fixed-endpoint experiments, 1/time is a relative rate; it is not a measured gas volume per second. If mean time is 50 s, relative rate = 1 ÷ 50 = 0.020 s⁻¹. If it is 25 s, the relative rate is 0.040 s⁻¹, twice as large.

Use repeat measurements to calculate means and assess variation. An anomalous result is unusually far from the others. Investigate it and repeat if needed. Use a recorded method problem to justify excluding a value; do not remove results merely to improve agreement.

A gas syringe records how the rate changes throughout the reaction. A disappearing-cross experiment gives one endpoint time and involves human judgement. Both can test a concentration hypothesis if the comparisons are controlled.

10. Some reactions can go both ways

In a reversible reaction, products can react to form the original reactants. The symbol ⇌ represents two opposite directions. Changing conditions can favour one direction.

For example, hydrated copper sulfate is blue. Heating removes its water of crystallisation and produces white anhydrous copper sulfate and water: hydrated copper sulfate ⇌ anhydrous copper sulfate + water. Adding water to the anhydrous solid restores the blue colour. This is a teacher-supervised demonstration: the solid and apparatus may be hot, and copper compounds require careful handling.

Another example is ammonium chloride ⇌ ammonia + hydrogen chloride. Heating can form gases; cooling allows them to recombine. These gases are irritants, so this belongs in a properly controlled demonstration, not a home experiment.

11. Reverse reactions reverse the energy transfer

If the forward reaction is endothermic, the reverse is exothermic. For the same quantities, the sizes of the energy transfers are equal but their directions are opposite.

Heating hydrated copper sulfate takes in energy to remove water. Adding water back releases energy. If a reversible reaction transfers 75 kJ to its surroundings in the forward direction, the reverse takes in 75 kJ for the same quantities.

12. Dynamic equilibrium: busy but balanced

In a closed system, reactants and products cannot escape. A reversible reaction may reach dynamic equilibrium. The forward and reverse reactions continue at equal rates, so the amounts of reactants and products remain constant.

“Dynamic” means changes are still happening at particle level. “Equilibrium” means the two changes balance. Equal rates do not require equal amounts: the mixture may contain much more product than reactant, or the other way round.

If a gas escapes, the closed-system condition is not met. A closed system can still exchange energy with its surroundings; “closed” refers to matter not escaping.

13. Higher only: changing equilibrium conditions

Higher only. The position of equilibrium describes the relative amounts of reactants and products. Le Chatelier’s principle says that when conditions change, the system responds in a direction that opposes that change.

For A + B ⇌ C, adding A makes the forward reaction favoured until a new equilibrium forms: more C is made. Removing C also favours the forward direction, replacing some of the C removed. Adding C favours the reverse direction. The response reduces the disturbance; it does not necessarily return every concentration to its original value.

14. Higher only: temperature affects the balance

Higher only. Raising temperature favours the endothermic direction, which takes in energy. Lowering temperature favours the exothermic direction, which releases energy.

Keep rate and equilibrium separate. A higher temperature usually makes reactions faster, but can reduce the equilibrium amount of a desired product if its formation is exothermic. Use the equation and energy information supplied in the question.

15. Higher only: count gas particles

Higher only. Increasing pressure favours the side with fewer gas molecules; decreasing pressure favours the side with more. Count the numbers before gaseous formulae in the balanced equation.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), four molecules of reactant gases form two molecules of product gas. Increasing pressure favours ammonia formation. If both sides have the same number of gas molecules, pressure has no effect on equilibrium position.

A catalyst speeds up both directions and lets equilibrium be reached sooner. It does not change equilibrium position. Industrial choices may balance rate, equilibrium product amount, energy cost and the cost of high-pressure equipment; use the data supplied rather than assuming one condition is always best.

Required practical 11: compare two ways to measure rate

Calculate each mean time and 1/mean time. Decide whether these results support the hypothesis that relative rate is proportional to concentration over this range. Explain two controls for the cross method and one advantage of collecting gas at regular intervals instead.

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One final thought

Use a scientific idea and a specific example in your reflection.