Ch 1 · Simple rate equations, orders of reaction and rate constants · AS & A Level Chemistry 9701

Topic 1 of 2 in Reaction kinetics — AS & A Level Chemistry 9701 — 65 lessons.

Introduction

At AS Level you met rate of reaction only qualitatively. At A2, Cambridge examiners expect you to derive the mathematical relationship between rate and concentration from experimental data, calculate rate constants with correct units, and use that relationship to probe reaction mechanisms one collision at a time. Every quantity here — order, rate constant, half-life — must come from evidence, never from the balanced equation.

Key ideas — Rate equations, order and the rate constant

For a reaction between reactants A and B, the rate equation links rate to concentration:

🧑‍🏫 Teacher's explanation

Imagine you are running a relay race. The speed at which the whole team finishes depends on the fastest runner (the rate-determining step) and how many people are actually running (the concentration). If one runner is incredibly slow, the entire team's speed is limited by that person, no matter how fast the others are. This is exactly how reaction mechanisms work.

📖 Definitions

🔍 Worked example — Easy

For the reaction A + B arrow C, the rate equation is found to be Rate = k[A]^1[B]^1. Calculate the initial rate when [A] = 0.20 mol dm^-3, [B] = 0.10 mol dm^-3, and k = 0.50 dm^3 mol^-1 s^-1.

🔍 Worked example — Standard

A first-order reaction has a half-life (t_1/2) of 25 minutes. Calculate the rate constant k in min^-1.

🔍 Worked example — Exam-style

In an experiment, the concentration of reactant X is doubled from 0.10 mol dm^-3 to 0.20 mol dm^-3. The initial rate increases from 2.0 × 10^-3 mol dm^-3 s^-1 to 8.0 × 10^-3 mol dm^-3 s^-1. Determine the order of reaction with respect to X.

⚠️ Common errors

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🎯 Exam question

Step 1: NO_2(g) arrow NO(g) + O(g) (slow) Step 2: NO(g) + NO_2(g) arrow NO_3(g) (fast) Step 3: O(g) + NO_3(g) arrow NO_2(g) + O_2(g) (fast)

Key ideas — Deducing order from graphs, initial rates and half-lives

A concentration-time graph reveals order directly from its shape: zero order gives a straight line of constant negative gradient; first order gives an exponential decay curve with a constant half-life; second order gives a steeper decay whose half-life doubles each time the concentration halves.

🧑‍🏫 Teacher's explanation

Think of a chemical reaction like a crowd of people trying to exit a stadium through a single narrow gate. The 'rate' is how many people pass through per minute. If the gate is huge and the crowd is small, the rate doesn't depend on how many people are left (zero order). If the rate depends directly on how many people are waiting, that's first order. If the rate depends on people bumping into each other to find the exit, it's second order.

📖 Definitions

🔍 Worked example — Easy

A reaction has the rate equation rate = k[A]^1. If the initial concentration of A is 0.20 mol dm^-3 and the initial rate is 4.0 × 10^-3 mol dm^-3 s^-1, calculate the value of the rate constant k.

🔍 Worked example — Standard

For a first-order reaction, the half-life (t_1/2) is 15.0 minutes. Calculate the rate constant k.

🔍 Worked example — Exam-style

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🎯 Exam question

Key ideas — Constructing a rate equation and calculating a rate constant

Once each order is known, the rate equation is written directly, for example Rate = k[A]^2[B]. Substituting one full set of concentrations and the corresponding rate lets you calculate k by rearrangement: k = Rate/[A]^m[B]^n. The units of k depend on the overall order, since rate is always in mathrmmol\ dm^-3\ s^-1:

🧑‍🏫 Teacher's explanation

Think of a chemical reaction like a crowded hallway in a school. The 'rate' is how many students pass a certain point every second. The 'concentration' is how many students are packed into the hallway. If you double the number of students, the rate at which they pass you might double (this is 1st order) or it might quadruple because students are bumping into each other more often (this is 2nd order).

📖 Definitions

🔍 Worked example — Easy

For the reaction A arrow B, the rate equation is Rate = k[A]. If the initial concentration of A is 0.50 mol\ dm^-3 and the initial rate is 0.010\ ext mol\ dm^-3 s^-1, calculate the value of k and state its units.

🔍 Worked example — Standard

A second-order reaction follows the rate equation Rate = k[A]^2. If the initial rate is 4.0 × 10^-3 mol\ dm^-3 s^-1 and [A] = 0.20 mol\ dm^-3, calculate k.

🔍 Worked example — Exam-style

The reaction 2NO(g) + O_2(g) arrow 2NO_2(g) has a rate equation Rate = k[NO]^2[O_2]. If the rate is 1.5 × 10^-4 mol\ dm^-3 s^-1 when [NO] = 2.0 × 10^-2 mol\ dm^-3 and [O_2] = 1.0 × 10^-2 mol\ dm^-3, calculate k with units.

⚠️ Common errors

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🎯 Exam question

Key ideas — Half-life and first-order kinetics

The half-life, t_1/2, is the time for the concentration of a reactant to fall to half its value. Only for a first-order reaction is t_1/2 constant, independent of starting concentration, linked to the rate constant by

🧑‍🏫 Teacher's explanation

Imagine you are eating a bowl of popcorn. If you eat half of what is left every 5 minutes, the time it takes to finish the bowl stays the same regardless of whether you start with a huge bowl or a tiny handful. This 'constant time to halve' is the hallmark of first-order kinetics. In chemistry, first-order reactions are special because the speed of the reaction depends only on the concentration of one reactant raised to the power of 1.

📖 Definitions

🔍 Worked example — Easy

A first-order reaction has a half-life of 25.0 s. Calculate the rate constant k.

🔍 Worked example — Standard

The concentration of a reactant in a first-order reaction decreases from 0.80 mol dm^-3 to 0.40 mol dm^-3 in 15.0 s. Calculate the rate constant k.

🔍 Worked example — Exam-style

A first-order reaction has a rate constant k = 0.015 min^-1. Calculate the time required for the concentration to fall from 2.00 mol dm^-3 to 0.25 mol dm^-3.

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Key ideas — Mechanisms, intermediates and the rate-determining step

A reaction mechanism must be consistent with both the overall equation and the experimental rate equation. Species that appear in the rate equation must take part in the rate-determining step, or in a fast step before it; species that appear in the overall equation but not in the rate equation react only in steps after the rate-determining step. For example, if Rate = k[NO]^2[O_2] for 2NO(g) + O_2(g) arrow 2NO_2(g), a consistent mechanism is a fast pre-equilibrium 2NO leftharpoons N_2O_2 (an intermediate) followed by a slow rate-determining step N_2O_2 + O_2 arrow 2NO_2, since this step's molecularity matches the powers in the rate equation. Conversely, given a mechanism, the rate equation is deduced from the species present up to and including the slowest step.

🧑‍🏫 Teacher's explanation

Imagine a factory assembly line where several machines are lined up. Most machines work very quickly, but one machine in the middle is incredibly slow and can only process one item every ten minutes. No matter how fast the machines before and after it are, the entire factory's output is limited by that one slow machine. In chemistry, this 'bottleneck' is the rate-determining step (RDS).

📖 Definitions

🔍 Worked example — Easy

A reaction has the following mechanism: Step 1 (slow): A + B arrow C Step 2 (fast): C + D arrow E What is the overall equation and the rate equation?

🔍 Worked example — Standard

Given the rate equation Rate = k[NO_2]^2, suggest a possible mechanism involving an intermediate.

🔍 Worked example — Exam-style

A reaction mechanism is: Step 1 (fast): A + A leftharpoons A_2 Step 2 (slow): A_2 + B arrow C Step 3 (fast): C + A arrow D Determine the overall equation and the rate equation.

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🎯 Exam question

Step 1 (fast equilibrium): A + B leftharpoons C Step 2 (slow): C + D arrow E Step 3 (fast): E + A arrow F \n(a) Write the overall equation for the reaction. [1] (b) Deduce the rate equation for this reaction. [2] [3 marks]

Key ideas — Temperature and the rate constant: the Arrhenius equation

Increasing temperature increases k, and hence the rate, because a greater proportion of molecules then possess energy at or above the activation energy E_a. Quantitatively, k and temperature T are related by the Arrhenius equation:

🧑‍🏫 Teacher's explanation

Think of a reaction like a high-stakes obstacle course. For a reaction to happen, molecules don't just need to bump into each other; they need to bump into each other with enough 'oomph' (kinetic energy) to clear a specific hurdle called the activation energy (E_a). Even if they hit each other, if they are moving too slowly, they just bounce off without reacting.

Ch 1 · Simple rate equations, orders of reaction and rate constants · AS & A Level Chemistry 9701

All chapters in Reaction kinetics