Grade 12 Physical Sciences
Term 2 · Weeks 6–7

Rate and Extent of Reaction

Paper 2Chemistry · Grade 12

Chemical reactions do not all proceed at the same speed. The rate of a reaction is critical in industry, biology and everyday life. In this chapter you will define reaction rate, use collision theory to explain why reactions occur, and investigate the factors that control how fast reactions proceed.

Week 6

6.1 Reaction Rate and Collision Theory

Define reaction rate and calculate itExplain collision theory (correct energy + correct orientation)

Definition

Reaction rate

The reaction rate is the change in concentration of a reactant or product per unit time: rate = Δ[concentration]/Δtime. It is always a positive quantity — expressed as the decrease in reactant concentration (negative sign removed) or increase in product concentration per second. SI unit: mol·dm⁻³·s⁻¹. Factors affecting rate: temperature, concentration, pressure (gases), surface area, catalyst.

Definition

Collision theory

Collision theory states that for a chemical reaction to occur, the reacting particles must collide with each other with (1) sufficient energy — at least equal to the activation energy (Ea), and (2) the correct orientation so that bonds can break and reform. Not all collisions result in reaction — only effective collisions do. Increasing temperature, concentration, or surface area increases the frequency of effective collisions, thereby increasing the reaction rate.

Definition

Activation energy (Ea)

The minimum energy required for a chemical reaction to proceed.

Formula

Reaction rate

r=ΔcΔtr = \dfrac{\Delta c}{\Delta t}

r = average rate of reaction (mol·dm⁻³·s⁻¹), Δc = change in concentration (mol·dm⁻³), Δt = change in time (s)

SI unit: mol·s⁻¹

For a reaction to occur, particles must collide. But not every collision produces a reaction — only collisions with sufficient energy (≥ activation energy) AND the correct geometric orientation are successful. Increasing the number of successful collisions per second increases the reaction rate. Think of it like trying to connect two LEGO pieces: you must bring the right faces together (orientation) and press hard enough (energy).

Reaction progressPotential energyRPEaΔHEa = activation energy; ΔH < 0 = exothermic
Figure 6.1 — An energy profile diagram showing activation energy (Ea). Reactants must gain at least Ea to reach the transition state (peak). Products are lower in energy than reactants for an exothermic reaction.

Worked Example

In an experiment, 0,04 mol of gas is produced in 8 seconds. Calculate the average reaction rate.

Given

  • Δn = 0,04 mol
  • Δt = 8 s

Find

rate

Solution

  1. 1rate = Δn / Δt
  2. 2rate = 0,04 / 8
  3. 3rate = 0,005 mol·s⁻¹
Answer: rate = 0,005 mol·s⁻¹ = 5 × 10⁻³ mol·s⁻¹

Watch Out

Reaction rate is always a positive value. If you calculate using the disappearance of reactants (negative Δn), take the absolute value: rate = |Δn|/Δt.

?

Practice Question

0,12 mol of a reactant is consumed in 60 s. Calculate the average reaction rate.

(2 marks)

?

Practice Question

Explain why not every collision between reactant particles leads to a chemical reaction.

(4 marks)

Exam Tip

Exam tip: when asked to explain collision theory, always mention BOTH conditions — energy AND orientation. Mentioning only energy scores partial marks.

Week 6

6.2 Factors Affecting Reaction Rate

List and explain factors affecting reaction rate: concentration, temperature, surface area, catalystExplain how a catalyst works (alternative pathway, lower Ea)

Definition

Catalyst

A catalyst is a substance that increases the rate of a chemical reaction without itself being permanently consumed in the reaction. It works by providing an alternative reaction pathway with a lower activation energy (Ea). At the end of the reaction, the catalyst is recovered chemically unchanged. A catalyst does NOT shift the equilibrium position — it helps the system reach equilibrium faster. Example: MnO₂ catalyses decomposition of H₂O₂.

Four main factors that affect reaction rate

  • Concentration (of solutions): higher concentration → more particles per unit volume → more frequent collisions → higher rate.
  • Temperature: higher temperature → particles have more kinetic energy → more collisions have enough energy to overcome activation energy → higher rate.
  • Surface area (of solids): smaller particle size → greater surface area exposed → more surface available for collisions → higher rate.
  • Catalyst: provides an alternative reaction pathway with lower activation energy → more collisions have sufficient energy → higher rate (catalyst not consumed).
ExothermicReactantsProductsΔH < 0Heat released to surroundingsEndothermicReactantsProductsΔH > 0Heat absorbed from surroundings
Figure 6.2 — Enthalpy diagram showing the effect of a catalyst. The catalysed pathway (dashed line) has a lower activation energy Ea than the uncatalysed pathway, but the overall enthalpy change ΔH remains the same.

A catalyst works by providing a different reaction mechanism — a different series of bond-breaking and bond-forming steps. This alternative pathway has a lower energy barrier (activation energy). Because more colliding particles now have enough energy to react, the rate increases. The catalyst itself is regenerated at the end of the mechanism, so it is not consumed overall. Enzymes are biological catalysts that are highly specific — each enzyme catalyses only one reaction or class of reactions.

Effect of Increasing Each Factor on Reaction Rate

PropertyFactor IncreasedEffect and Explanation
ConcentrationRate increasesMore particles per volume → more frequent collisions
TemperatureRate increases significantlyMore particles have Ek ≥ Ea; also more frequent collisions
Surface areaRate increasesMore surface sites available for collisions
Catalyst addedRate increasesLower Ea → more collisions are successful; catalyst not consumed

Worked Example

Explain, using collision theory, why a lump of iron reacts with acid more slowly than iron powder of the same mass.

Given

  • Same mass of iron in both cases

Find

Explanation using collision theory

Solution

  1. 1Iron powder has a much smaller particle size than the lump.
  2. 2Smaller particles → much greater total surface area exposed to the acid.
  3. 3Greater surface area → more collisions between iron atoms and acid particles per second.
  4. 4More collisions per second → more successful collisions per second → faster reaction rate.
Answer: Iron powder reacts faster because its smaller particle size provides a larger surface area, increasing the frequency of collisions between reactant particles.

Watch Out

Do not say a catalyst 'provides more energy' or 'lowers the activation energy of the particles'. A catalyst lowers the activation energy OF THE REACTION (by providing an alternative pathway), not the energy of individual particles.

?

Practice Question

Explain, using collision theory, why increasing temperature increases the rate of a chemical reaction. Refer to activation energy in your answer.

(4 marks)

?

Practice Question

A student claims that adding a catalyst to a reaction changes the value of ΔH (enthalpy change). Is the student correct? Explain.

(3 marks)

Week 7

6.3 Interpreting Rate and Concentration Graphs

Interpret rate vs time and concentration vs time graphs

Two types of graphs are commonly used to represent how a reaction progresses over time. A concentration-time graph shows how the concentration of a reactant (decreasing) or product (increasing) changes. The gradient (slope) of this graph at any point gives the instantaneous rate at that time. A rate-time graph shows directly how the reaction rate changes over time — the rate is usually high initially (high reactant concentration) and decreases as reactants are used up.

Key features to identify on a concentration-time graph

  • Steep gradient early in the reaction → high rate (high concentration of reactants).
  • Graph flattens (gradient → 0) → reaction rate approaches zero (reactants nearly used up).
  • Horizontal line → reaction is complete; concentration no longer changing.
  • The steeper the initial slope, the faster the reaction.

Worked Example

On a concentration-time graph for the decomposition of hydrogen peroxide, the concentration drops from 0,50 mol·dm⁻³ at t = 0 to 0,20 mol·dm⁻³ at t = 60 s. Calculate the average rate of reaction over this period.

Given

  • c₁ = 0,50 mol·dm⁻³ at t = 0 s
  • c₂ = 0,20 mol·dm⁻³ at t = 60 s

Find

average rate

Solution

  1. 1Δc = 0,50 − 0,20 = 0,30 mol·dm⁻³
  2. 2Δt = 60 s
  3. 3rate = Δc / Δt = 0,30 / 60 = 0,005 mol·dm⁻³·s⁻¹
Answer: Average rate = 0,005 mol·dm⁻³·s⁻¹

Note

Note: when rate is expressed in terms of concentration change, the unit is mol·dm⁻³·s⁻¹. When expressed as amount change, the unit is mol·s⁻¹. Both forms appear in examinations.

?

Practice Question

Sketch the shape of a concentration-time graph for a reactant in a reaction at (a) high temperature and (b) low temperature, starting from the same initial concentration. Describe the key differences.

(5 marks)

?

Practice Question

A rate-time graph for a reaction shows a horizontal line throughout. What does this tell you about the order of the reaction with respect to all reactants?

(3 marks)

Exam Tip

Exam tip: to compare two experiments on the same graph, always note the initial rate (initial gradient) and the time to completion. A steeper initial gradient = faster initial rate.

Rate and Extent of Reaction Grade 12 Physical Sciences CAPS Notes | MathSciBuddy