Chemical Kinetics
How fast and why — rates, rate laws, the order of a reaction, integrated equations, half-life, and the temperature barrier set by activation energy
- Average and instantaneous rate, and rate in terms of any reactant or product.
- The rate law, the rate constant, and the order of a reaction.
- How order differs from molecularity, and how to read \(k\)'s units.
- Integrated rate laws for zero- and first-order reactions and their half-lives.
- The Arrhenius equation, activation energy, and the temperature dependence of rate.
- The idea of collision theory, pseudo-first-order reactions, and catalysis.
The Rate of a Reaction
The rate of a reaction is the change in concentration of a reactant or product per unit time. The average rate is measured over an interval; the instantaneous rate is the slope of the concentration–time curve at a single instant. Because stoichiometry links the species, the rate must be defined so it comes out the same whichever one you track.
Factors That Affect Rate
Five levers control how fast a reaction runs. Most can be understood through one idea: anything that increases the frequency of effective collisions speeds the reaction.
| Factor | Effect on rate |
|---|---|
| Concentration / pressure | more reactant ⇒ more collisions ⇒ faster |
| Temperature | rises steeply; rate roughly doubles per \(10\ ^\circ\text{C}\) |
| Catalyst | lowers activation energy ⇒ faster (both directions) |
| Surface area | finer solid ⇒ more contact ⇒ faster |
| Nature of reactants | ionic reactions are fast; covalent bond-breaking is slow |
Rate Law & Rate Constant
The rate law expresses how the rate depends on concentrations. The exponents — the orders — are found experimentally and need not match the stoichiometric coefficients. Their sum is the overall order, and \(k\) is the rate constant.
Order vs Molecularity
These two are often confused. Order is an experimental quantity for the overall reaction and can be zero, fractional, or even negative. Molecularity is the number of species colliding in a single elementary step — always a small whole number, and meaningless for an overall multi-step reaction.
| Order | Molecularity | |
|---|---|---|
| Determined by | experiment | the mechanism (an elementary step) |
| Possible values | 0, fraction, integer, negative | 1, 2, rarely 3 (whole numbers) |
| Applies to | overall reaction | a single elementary step |
Integrated Rate Laws
Integrating the rate law gives concentration as a function of time — and a straight-line plot that reveals the order. The two cases most asked at this level are zero and first order.
Half-Life
The half-life \(t_{1/2}\) is the time for the concentration to fall to half its initial value. Its dependence on starting concentration is itself a fingerprint of the order.
A first-order half-life is independent of concentration — the hallmark of radioactive decay and many reactions. A zero-order half-life shrinks as the reaction proceeds.
Temperature Dependence — the Arrhenius Equation
Rate rises sharply with temperature because more molecules carry enough energy to react. Arrhenius captured this with an exponential law relating \(k\) to the activation energy \(E_a\) and the absolute temperature.
Activation Energy & Collision Theory
For a collision to react, molecules must meet with energy at least \(E_a\) and in the right orientation. The activation energy is the barrier between reactants and products, climbed through a high-energy transition state. A catalyst opens a lower-barrier path, speeding both directions without changing \(\Delta H\).
Pseudo-First-Order Reactions
Sometimes a higher-order reaction behaves as first order because one reactant is in vast excess and its concentration barely changes. Acid hydrolysis of an ester, or the inversion of cane sugar, are classic examples — water is so abundant that its concentration is effectively constant.
A Note on Catalysis
A catalyst increases rate by providing an alternative path of lower activation energy; it is regenerated, so a small amount serves repeatedly. It speeds forward and reverse equally and does not shift the equilibrium position — only how fast equilibrium is reached. The mechanisms of surface (heterogeneous) catalysis are developed in Chapter 12, Surface Chemistry.
Putting It to Work
Problem. For \(\ce{N2 + 3H2 -> 2NH3}\), express the rate in terms of each species.
Solution. Divide each by its coefficient:
Problem. A rate constant has units \(\text{L mol}^{-1}\text{s}^{-1}\). What is the order?
Solution. Match \((\text{mol L}^{-1})^{1-n}\text{s}^{-1}\):
Problem. Doubling \([\ce{A}]\) doubles the rate; doubling \([\ce{B}]\) quadruples it. Write the rate law and overall order.
Solution. Rate \(\propto[\ce{A}]^1\) and \(\propto[\ce{B}]^2\):
Problem. A first-order reaction has \(k=2.303\times10^{-3}\ \text{s}^{-1}\). How long for \(90\%\) completion?
Solution. At \(90\%\), \([\ce{A}]_0/[\ce{A}]=10\), so \(\log=1\):
Problem. A zero-order reaction has \(k=0.02\ \text{mol L}^{-1}\text{s}^{-1}\) and \([\ce{A}]_0=0.50\ \text{M}\). Find \(t_{1/2}\).
Solution. Use \(t_{1/2}=[\ce{A}]_0/2k\):
Problem. The rate constant doubles when \(T\) rises from \(300\) to \(310\ \text{K}\). Find \(E_a\) \((R=8.314)\).
Solution. Use the two-temperature form with \(k_2/k_1=2\):
Chapter Summary
Defined per coefficient; instantaneous rate is the slope of the [ ] vs t curve.
Rate \(=k[\ce{A}]^x[\ce{B}]^y\); order \(=x+y\) is experimental; \(k\)'s units fix the order.
Order is for the overall reaction; molecularity counts colliders in one elementary step.
Zero: \([\ce{A}]=[\ce{A}]_0-kt\); first: \(k=\tfrac{2.303}{t}\log\tfrac{[\ce{A}]_0}{[\ce{A}]}\).
First order: \(t_{1/2}=0.693/k\), independent of concentration.
\(k=Ae^{-E_a/RT}\); a catalyst lowers \(E_a\) without shifting \(K\).
Problems
Identify the order first — from the rate law, the units of \(k\), or the data — then choose the matching integrated equation. Take \(R=8.314\ \text{J K}^{-1}\text{mol}^{-1}\) and \(2.303\) for the log conversion. Difficulty rises down the list.
- For \(\ce{2N2O5 -> 4NO2 + O2}\), write the rate in terms of each species.
- State the units of \(k\) for a zero-, first-, and second-order reaction.
- A reaction is \(\text{Rate}=k[\ce{A}]^2[\ce{B}]\). What is the overall order, and how does the rate change if \([\ce{A}]\) is tripled?
- Distinguish order and molecularity with one example of each that differ in value.
- A first-order reaction is \(50\%\) complete in \(20\ \text{min}\). Find \(k\) and the time for \(75\%\) completion.
- A zero-order reaction has \(k=1.0\times10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}\) and \([\ce{A}]_0=0.20\ \text{M}\). Find the time to reach \(0.05\ \text{M}\).
- The half-life of a first-order reaction is \(693\ \text{s}\). Find \(k\) and the fraction left after \(2310\ \text{s}\).
- A rate constant rises from \(1.0\times10^{-3}\) to \(2.0\times10^{-2}\ \text{s}^{-1}\) when \(T\) goes \(300\to350\ \text{K}\). Find \(E_a\).
- Explain why the acid hydrolysis of an ester is pseudo-first order.
- For a reaction, \(\log k\) vs \(1/T\) has slope \(-5000\ \text{K}\). Find \(E_a\).
- A reaction's rate doubles for every \(10\ ^\circ\text{C}\) rise. Estimate \(E_a\) near \(300\ \text{K}\), and comment on the temperature coefficient.
- For a first-order reaction, show that the time for \(99.9\%\) completion is ten times the half-life... is it? Justify your answer.