Why Doubling One Reactant Can Quadruple a Reaction's Speed

A firecracker explodes in an instant, iron rusts over months, and milk curdles in a day or two. Chemical kinetics measures how fast reactions go and finds the rules linking their speed to concentration — rules that decide the shelf life of medicines and the design of chemical plants.

This lesson covers the rate of reaction and rate laws, order and molecularity, and zero-order, first-order and pseudo-first-order reactions.

The rate of a reaction is the change in concentration of a reactant or product per unit time, and a rate law expresses that rate as a rate constant multiplied by reactant concentrations raised to powers found by experiment.

Average and instantaneous rate. For $\mathrm{R}\to \mathrm{P}$:

$$\text{Average rate}=-\frac{\Delta [\mathrm{R}]}{\Delta t}=\frac{\Delta [\mathrm{P}]}{\Delta t}\text{Instantaneous rate}=-\frac{d[\mathrm{R}]}{dt}$$

- The minus sign keeps the rate positive, since reactant concentration falls
- Units are mol L${}^{-1}$ s${}^{-1}$
- The instantaneous rate is the slope of the tangent to the concentration-time curve

Rate and stoichiometry. For $2HI\to {\mathrm{H}}_2+{\mathrm{I}}_2$, the rate is $-\frac{1}{2}\frac{\Delta [\mathrm{HI}]}{\Delta t}=\frac{\Delta [{\mathrm{H}}_2]}{\Delta t}$.

Worked example 1. In ${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2N{\mathrm{H}}_3$, ammonia forms at $4.0\times 10^{-4}$ mol L${}^{-1}$ s${}^{-1}$. Hydrogen disappears at

$$\frac{3}{2}\times 4.0\times 10^{-4}=6.0\times 10^{-4}{\text{ mol L}}^{-1}{\text{ s}}^{-1}$$

Rate law. For $a\mathrm{A}+b\mathrm{B}\to \text{products}$:

$$\text{Rate}=k[\mathrm{A}{]}^x[\mathrm{B}{]}^y$$

- k is the rate constant, which depends on temperature but not on concentration
- x and y come from experiment and need not equal a and b

An everyday example. Medicine strips carry an expiry date because the active ingredient slowly decomposes, and manufacturers use rate laws to predict how long it stays effective.

The substance. A rate law cannot be read off the balanced equation — the powers sometimes match the coefficients, but for many reactions they do not.

The order of a reaction is the sum of the powers in its experimental rate law, while molecularity is the number of species colliding in a single elementary step; order is found from data, molecularity from the mechanism.

Initial rate method. Change one reactant's concentration while keeping the others fixed: if doubling it doubles the rate, its order is 1; if the rate quadruples, the order is 2; if the rate is unchanged, the order is 0.

Worked example 2. For $\mathrm{A}+\mathrm{B}\to \text{products}$, rates are in mol L${}^{-1}$ s${}^{-1}$:

- [A] = 0.10 M, [B] = 0.10 M: rate $=2.0\times 10^{-3}$
- [A] = 0.20 M, [B] = 0.10 M: rate $=8.0\times 10^{-3}$
- [A] = 0.10 M, [B] = 0.20 M: rate $=4.0\times 10^{-3}$

Doubling [A] quadruples the rate, so the order in A is 2; doubling [B] doubles it, so the order in B is 1:

$$\text{Rate}=k[\mathrm{A}{]}^2[\mathrm{B}]k=\frac{2.0\times 10^{-3}}{(0.10{)}^2(0.10)}=2.0{\text{ L}}^2{\text{ mol}}^{-2}{\text{ s}}^{-1}$$

The overall order is 3.

Units of k. Zero order: mol L${}^{-1}$ s${}^{-1}$; first order: s${}^{-1}$; second order: L mol${}^{-1}$ s${}^{-1}$.

Order versus molecularity:

- Order can be zero, fractional or negative; molecularity is always a whole number, usually 1 or 2
- Order describes the overall reaction; molecularity describes only an elementary step
- In a complex reaction, the slowest step — the rate-determining step — controls the order

An everyday example. A single toll booth on a busy highway lets cars through only as fast as it can process them, however quickly they arrive — just as the slowest step fixes the rate of a multi-step reaction.

The substance. Molecularity has no meaning for a complex reaction as a whole — only each elementary step has one.

In a zero-order reaction the rate does not depend on concentration, in a first-order reaction it is proportional to one concentration, and a pseudo-first-order reaction is a higher-order reaction that behaves as first order because one reactant is in large excess.

Zero-order reactions:

$$\text{Rate}=k[\mathrm{R}]=[\mathrm{R}{]}_0-kt{t}_{1/2}=\frac{[\mathrm{R}{]}_0}{2k}$$

- Concentration falls in a straight line with time
- Example: decomposition of ammonia on a hot platinum surface at high pressure

First-order reactions:

$$\text{Rate}=k[\mathrm{R}]k=\frac{2.303}{t}\log \frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}{t}_{1/2}=\frac{0.693}{k}$$

- The half-life does not depend on the starting concentration
- Examples: radioactive decay and the decomposition of ${\mathrm{N}}_2{\mathrm{O}}_5$

Worked example 3. A first-order reaction has $k=1.155\times 10^{-3}$ s${}^{-1}$:

$$t_{1/2}=\frac{0.693}{1.155\times 10^{-3}}=600\text{ s}$$

For 75 per cent completion, $[\mathrm{R}]=\frac{1}{4}[\mathrm{R}{]}_0$:

$$t=\frac{2.303}{1.155\times 10^{-3}}\log 4=\frac{2.303\times 0.602}{1.155\times 10^{-3}}=1200\text{ s}$$

That is exactly two half-lives.

Pseudo-first-order reactions:

- Acid hydrolysis of ethyl acetate in a large excess of water: $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$
- The true rate is $k^{\prime }[\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5][{\mathrm{H}}_2\mathrm{O}]$, but water's concentration barely changes, so the rate becomes $k[\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5]$
- The inversion of cane sugar in acid behaves the same way

An everyday example. Radioactive iodine used in Indian hospitals to treat thyroid disorders decays by first-order kinetics, so doctors can predict how much remains after any number of half-lives.

The substance. A pseudo-first-order reaction is not truly first order — its order returns to 2 if the excess reactant is cut to a comparable concentration.