Reaction Rates and Rate Laws

What is it?

Chemical kinetics is the study of how fast reactions happen. The rate of a reaction is the change in concentration of a reactant or product per unit time:

\text{rate} = -\frac{\Delta[\text{reactant}]}{\Delta t} = \frac{\Delta[\text{product}]}{\Delta t}

The minus sign makes the rate positive, because reactant concentrations decrease. Rates are usually given in M/s (mol L⁻¹ s⁻¹).

Key idea

A rate law links the rate to the concentrations: rate $= k[\ce{A}]^m[\ce{B}]^n$ . The orders $m$ and $n$ must be found by experiment; they are not the coefficients of the balanced equation.

Why does it matter?

Industry and safety. Chemists control rates to make products efficiently and to avoid runaway reactions.
Medicine. Drug doses are spaced using the rate at which the body removes the drug (often a first-order half-life).
Mechanisms. Rate laws give clues about the steps by which a reaction actually happens.

How does it work?

1. What changes the rate (collision theory)

Particles must collide with enough energy and the right orientation to react. A reaction is faster with:

higher concentration (or pressure for gases): more frequent collisions;
higher temperature: more frequent and, above all, more energetic collisions;
larger surface area (smaller pieces of solid): more particles exposed;
a catalyst: a lower-energy pathway (see Activation Energy and Catalysts).

2. Rates and stoichiometry

For $\ce{2N2O5(g) -> 4NO2(g) + O2(g)}$ , $\ce{NO2}$ forms twice as fast as $\ce{N2O5}$ disappears. To give one rate for the reaction, divide by the coefficients:

\begin{aligned} \text{rate} &= -\frac{1}{2}\frac{\Delta[\ce{N2O5}]}{\Delta t} \\[4pt] &= \frac{1}{4}\frac{\Delta[\ce{NO2}]}{\Delta t} = \frac{\Delta[\ce{O2}]}{\Delta t} \end{aligned}

3. Rate laws and orders

In rate $= k[\ce{A}]^m[\ce{B}]^n$ :

$m$ is the order with respect to A; $n$ is the order with respect to B; $m + n$ is the overall order.
$k$ is the rate constant. It is fixed at a given temperature and increases as the temperature rises.

Order in A	Doubling [A] multiplies the rate by
0	1 (no change)
1	2
2	4

4. Units of the rate constant

Because the rate is always in M/s, the units of $k$ depend on the overall order:

Overall order	Rate law	Units of k
0	rate = k	M s⁻¹
1	rate = k[A]	s⁻¹
2	rate = k[A]² or k[A][B]	M⁻¹ s⁻¹
3	rate = k[A]²[B]	M⁻² s⁻¹

5. First-order reactions and half-life

For a first-order reaction, the concentration falls exponentially:

\begin{aligned} [\ce{A}]_t &= [\ce{A}]_0\, e^{-kt} \\[4pt] t_{1/2} &= \frac{\ln 2}{k} = \frac{0.693}{k} \end{aligned}

The half-life, $t_{1/2}$ , is the time for the concentration to halve. For a first-order reaction it does not depend on the starting concentration, so every half-life is the same length.

Think of it like this

A ticket queue at a stadium: open more gates (higher concentration of “gates”, or a catalyst) and people get through faster. A first-order half-life is like a sale where half of the remaining stock sells every hour: 100 items, then 50, then 25, then 12.5, with each halving taking the same time.

More precisely

The rate measured over a time interval is an average rate; the instantaneous rate is the slope of the tangent to the concentration–time curve. The orders in a rate law reflect the reaction mechanism (its sequence of elementary steps), which is why they can’t be read from the overall equation. Second-order reactions have $t_{1/2} = \dfrac{1}{k[\ce{A}]_0}$ , which does depend on the starting concentration.

Visualise it

Graph of concentration of A against time for a first-order reaction with initial concentration 0.100 M and k = 6.20 × 10⁻⁴ per second. The curve falls exponentially. Dashed guides mark 1 half-life at about 1120 s (0.050 M), 2 half-lives at about 2240 s (0.025 M) and 3 half-lives at about 3350 s (0.0125 M). — In a first-order reaction every half-life is the same length.

Worked example

Worked example: Average rate

Question: In $\ce{2H2O2 -> 2H2O + O2}$ , $[\ce{H2O2}]$ falls from 0.500 M to 0.400 M in 60.0 s. Find the average rate of disappearance of $\ce{H2O2}$ and the average rate of the reaction.

Rate of disappearance of $\ce{H2O2}$ :
$\begin{aligned} &-\frac{\Delta[\ce{H2O2}]}{\Delta t} \\[4pt] &= -\frac{0.400\ \text{M} - 0.500\ \text{M}}{60.0\ \text{s}} \\[4pt] &= 1.67 \times 10^{-3}\ \text{M/s} \end{aligned}$
Rate of reaction (divide by the coefficient 2): $\dfrac{1.67 \times 10^{-3}\ \text{M/s}}{2} =$ $8.33 \times 10^{-4}$ M/s

Worked example: The method of initial rates

Question: For $\ce{A + B -> products}$ :

Experiment	[A]	[B]	Initial rate
1	0.10 M	0.10 M	2.0 × 10⁻³ M/s
2	0.20 M	0.10 M	8.0 × 10⁻³ M/s
3	0.10 M	0.20 M	4.0 × 10⁻³ M/s

Find the rate law and $k$ , with units.

Experiments 1 → 2: [A] doubles, [B] constant; the rate is multiplied by $\dfrac{8.0 \times 10^{-3}\ \text{M/s}}{2.0 \times 10^{-3}\ \text{M/s}} = 4$ , so the order in A is 2.
Experiments 1 → 3: [B] doubles; the rate is multiplied by 2, so the order in B is 1.
Rate law: rate = k[A]²[B] (third order overall).
From experiment 1:
$\begin{aligned} &k = \frac{\text{rate}}{[\ce{A}]^2[\ce{B}]} \\[4pt] &= \frac{2.0 \times 10^{-3}\ \text{M s}^{-1}}{(0.10\ \text{M})^2(0.10\ \text{M})} \\[4pt] &= 2.0\ \text{M}^{-2}\ \text{s}^{-1} \end{aligned}$
The units follow from M s⁻¹ ÷ M³ = M⁻² s⁻¹.

Worked example: Half-life

Question: A first-order decomposition has $k = 6.20 \times 10^{-4}\ \text{s}^{-1}$ . Find its half-life.

\begin{aligned} t_{1/2} &= \frac{0.693}{6.20 \times 10^{-4}\ \text{s}^{-1}} \\[4pt] &= 1.12 \times 10^{3}\ \text{s} \end{aligned}

That is $1.12 \times 10^{3}\ \text{s} \times \dfrac{1\ \text{min}}{60\ \text{s}} = 18.6$ min.

Common mistake

Common mistake: Taking orders from the balanced equation

For $\ce{2NO + O2 -> 2NO2}$ the rate law happens to be rate = k[NO]²[O₂], but that is an experimental result. Many reactions have orders that differ from their coefficients, so always use data.

Common mistake: Giving every k the same units

$k$ for a first-order reaction is in s⁻¹, but for a second-order reaction it is in M⁻¹ s⁻¹. Work out the units from the rate law every time.

Common mistake: Thinking every half-life doubles the time to finish

After 2 half-lives, $\frac{1}{4}$ of the reactant remains, not none. A first-order reaction never quite reaches zero.

Notation note

M/s, M s⁻¹ and mol L⁻¹ s⁻¹ mean the same thing.
“ln” is the natural logarithm (base e); ln 2 = 0.693.
$[\ce{A}]_0$ is the initial concentration; $[\ce{A}]_t$ is the concentration at time $t$ .

Remember this

Rate = change in concentration ÷ time (M/s); divide by coefficients for one reaction rate.
Rate = k[A]^m[B]^n; orders come from experiments (method of initial rates).
Units of k: zero order M s⁻¹; first s⁻¹; second M⁻¹ s⁻¹; third M⁻² s⁻¹.
First order: $t_{1/2} = 0.693/k$ , independent of starting concentration.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/reaction-rates/