Ch 1 · Rate of reaction · AS & A Level Chemistry 9701
Topic 1 of 3 in Reaction kinetics — AS & A Level Chemistry 9701 — 47 lessons.
Introduction
Reaction kinetics asks how fast a reaction proceeds, in contrast to equilibrium, which asks how far it goes. Cambridge 9701 builds every explanation of rate on collision theory: reactant particles must collide, and collide with enough energy and the right orientation, before a reaction can occur. This lesson develops the vocabulary of collision theory and the practical skill of extracting a rate from experimental data.
Key ideas — Rate of reaction and collisions
The rate of reaction is the change in concentration of a reactant or product per unit time, usually expressed in mathrmmol\ dm^-3\ s^-1. For a reaction to occur, particles must first collide; the frequency of collisions is the number of collisions occurring per unit time. Not every collision leads to a reaction: an effective collision is one in which the colliding particles have combined kinetic energy equal to or greater than the activation energy and the correct relative orientation, so that a reaction occurs; a non-effective collision is one that fails to meet either condition, so the particles simply bounce apart unchanged. Increasing the rate of reaction always means increasing the frequency of effective collisions, not simply the total number of collisions.
🧑🏫 Teacher's explanation
Imagine a busy dance floor. The 'rate of reaction' is like how many new couples are formed every minute. For a new couple to form, two people must bump into each other. However, not every bump results in a dance; they have to bump into each other with enough energy (the 'vibe') and at the right angle (facing each other) to actually start dancing. If they just brush past each other without looking, it is a 'non-effective collision'.
📖 Definitions
🔍 Worked example — Easy
A reaction produces gas at a rate of 4.5 × 10^-3 mol dm^-3 s^-1. Calculate the change in concentration of the reactant over 20 s, assuming the stoichiometry is 1:1.
🔍 Worked example — Standard
In the reaction Mg(s) + 2HCl(aq) arrow MgCl_2(aq) + H_2(g), the concentration of HCl decreases from 1.0 mol dm^-3 to 0.6 mol dm^-3 in 40 s. Calculate the average rate of decrease of HCl in mol dm^-3 s^-1.
🔍 Worked example — Exam-style
A reaction rate is proportional to the frequency of effective collisions. If the frequency of collisions is doubled by increasing pressure, but the activation energy remains the same, explain why the rate might increase by more than double if the temperature also increases slightly.
⚠️ Common errors
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🎯 Exam question
(a) Explain, in terms of collision theory, why increasing the pressure of the system increases the rate of reaction. [2] (b) A student increases the temperature of the system. Explain why this increases the rate of reaction. [2] [4 marks]
Key ideas — Effect of concentration and pressure on rate
Increasing the concentration of a reactant in solution increases the number of particles present in a given volume, so particles are closer together on average; this increases the frequency of collisions between reactant particles, and since the proportion of collisions that are effective is unchanged, the frequency of effective collisions increases, increasing the rate. For reactions involving gases, increasing the pressure has an equivalent effect to increasing concentration: it forces the same number of gas particles into a smaller volume, increasing the number of particles per unit volume, so collisions (and hence effective collisions) become more frequent and the rate increases.
🧑🏫 Teacher's explanation
Imagine a crowded dance floor. If only a few people are dancing, they rarely bump into each other. But if you pack 100 people into that same dance floor, they will bump into each other much more often. In chemistry, these 'bumps' are collisions between particles.
📖 Definitions
🔍 Worked example — Easy
A reaction produces gas. At t = 0 s, the volume of gas is 0 cm^3. At t = 40 s, the volume is 24 cm^3. Calculate the average rate of reaction in cm^3 s^-1.
🔍 Worked example — Standard
In a reaction A + B arrow C, the concentration of A decreases from 0.10 mol dm^-3 to 0.04 mol dm^-3 in 120 s. Calculate the rate of decrease of A in mol dm^-3 s^-1.
🔍 Worked example — Exam-style
Explain why increasing the pressure of a gaseous reactant increases the rate of reaction, using collision theory.
⚠️ Common errors
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🎯 Exam question
\n(a) Explain, in terms of collision theory, why the powder reacts faster than the ribbon. [3] (b) The student increases the concentration of the acid. Explain why this increases the rate of reaction. [2] [5 marks]
Key ideas — Effect of surface area
When one reactant is a solid, only particles at the exposed surface of the solid can collide with particles of the other reactant; particles inside the solid are unavailable. Breaking a solid into smaller pieces, or using a powder rather than a lump of the same total mass, increases its surface area, exposing more particles of the solid to collisions with the other reactant. This increases the frequency of effective collisions at the interface between the two reactants, increasing the rate of reaction, even though the total amount of solid, and therefore the overall extent of reaction, is unchanged.
🧑🏫 Teacher's explanation
Imagine you are trying to dissolve a large block of sugar in a glass of water versus a spoonful of fine granulated sugar. The block takes forever because the water can only touch the outside edges. The granulated sugar disappears almost instantly because there is much more 'surface' for the water to grab onto.
📖 Definitions
🔍 Worked example — Standard
A reaction involves a solid reactant. When the solid is a single cube of side length 2 cm, the rate is R. If the same mass of solid is crushed into 8 smaller cubes, each with side length 1 cm, what is the new rate R' in terms of R?
🔍 Worked example — Exam-style
Explain, using collision theory, why increasing the surface area of a solid reactant increases the rate of reaction.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Calculating rate from experimental data
The rate of reaction at any point can be found from a graph of concentration (or a measurable quantity proportional to it, such as volume of gas produced) against time, by taking the gradient of the curve at that point; the initial rate is the gradient of the tangent drawn at time zero, where the curve is steepest. Where a table of concentration and time values is given instead of a curve, an average rate over an interval can be calculated directly as the change in concentration divided by the change in time.
🧑🏫 Teacher's explanation
Imagine you are watching a movie of a race. If you want to know how fast a runner is moving at exactly the 10-minute mark, you can't just look at their start and finish times; you have to look at how much distance they cover in that tiny, split second. In chemistry, we do the same thing with reactions. We don't just care about how long the whole reaction took; we want to know how fast it was 'running' at a specific moment.
📖 Definitions
🔍 Worked example — Easy
A reaction produces gas. At t = 0 s, the volume is 0 ext cm^3. At t = 20 s, the volume is 40 ext cm^3. Calculate the average rate of gas production over this interval.
🔍 Worked example — Standard
The concentration of reactant A decreases from 1.00 mol dm^-3 to 0.60 mol dm^-3 in 40 s. Calculate the average rate of disappearance of A.
🔍 Worked example — Exam-style
A graph of concentration [C] vs time t shows a curve. At t = 50 s, a tangent is drawn to the curve. The tangent line passes through (50, 0.4) and (60, 0.7). Calculate the instantaneous rate at t = 50 s.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Worked example
Question. In an experiment, the volume of hydrogen gas produced by the reaction of excess dilute hydrochloric acid with a fixed mass of magnesium ribbon was recorded. At t = 10\ s, 24.0\ cm^3 of gas had been collected; at t = 30\ s, 60.0\ cm^3 had been collected. Calculate the average rate of reaction, in mathrmcm^3\ s^-1, between t = 10\ s and t = 30\ s. [2]
🔍 Worked example — Easy
A reaction produces 40.0 cm^3 of gas in 20 s. Calculate the average rate of reaction in cm^3 s^-1.
🔍 Worked example — Standard
In a reaction, the volume of gas increases from 15.0 cm^3 at t = 5 s to 45.0 cm^3 at t = 25 s. Calculate the average rate of reaction between t = 5 s and t = 25 s.
🔍 Worked example — Exam-style
A student measures the mass of a reactant decreasing over time. At t = 0 s, mass = 5.00 g. At t = 100 s, mass = 4.20 g. Calculate the average rate of reaction in g s^-1 (use 3 significant figures).
⚠️ Common errors
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Practice
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