Why Milk Sours Faster on a Hot Summer Day

A firecracker explodes in an instant, milk turns sour over a day, and iron rusts over months. Chemical kinetics measures how fast reactions go and explains what controls their speed.

This part covers reaction rates, rate laws and order, molecularity and the rate-determining step, and the factors that change the rate.

**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, $-\frac{\Delta [R]}{\Delta t}$, while the instantaneous rate is the slope of the concentration–time curve at one moment, $-\frac{d[R]}{dt}$.

Signs and coefficients.** Reactant concentrations fall, so a minus sign keeps the rate positive. For $a\text{A}+b\text{B}\to c\text{C}+d\text{D}$:

$$\text{Rate}=-\frac{1}{a}\frac{d[\text{A}]}{dt}=-\frac{1}{b}\frac{d[\text{B}]}{dt}=\frac{1}{c}\frac{d[\text{C}]}{dt}=\frac{1}{d}\frac{d[\text{D}]}{dt}$$

Units are mol L${}^{-1}$ s${}^{-1}$, or atm s${}^{-1}$ for gases.

Worked example. In $2{\text{N}}_2{\text{O}}_5\to 4{\text{NO}}_2+{\text{O}}_2$, the N${}_2$O${}_5$ concentration falls from $0.50$ to $0.44$ mol L${}^{-1}$ in $30$ s:

$$-\frac{\Delta [{\text{N}}_2{\text{O}}_5]}{\Delta t}=\frac{0.06}{30}=2\times 10^{-3}{\text{ mol L}}^{-1}{\text{ s}}^{-1}$$

The rate of reaction is half of this, $1\times 10^{-3}$ mol L${}^{-1}$ s${}^{-1}$, and NO${}_2$ forms at $4\times 10^{-3}$ mol L${}^{-1}$ s${}^{-1}$.

Instantaneous rate comes from the slope of a tangent to the curve; as the interval shrinks towards zero, the average rate approaches it.

An everyday example. A bus's average speed over a whole journey and its speedometer reading at one moment differ in exactly the same way.

The substance. Rate usually slows as a reaction proceeds, because reactant concentrations fall — so an average rate depends on the interval chosen.

**The rate law, Rate $=k[\text{A}{]}^x[\text{B}{]}^y$, must be found by experiment; the order is $x+y$, each exponent comes from comparing rates when one concentration changes, and the units of $k$ are (mol L${}^{-1}$)${}^{1-n}$ s${}^{-1}$ for an order-$n$ reaction.

Initial-rate method:**

- Change one concentration while keeping the others fixed
- If doubling [A] doubles the rate, the order in A is $1$; if it quadruples the rate, $2$; if nothing changes, $0$

Worked example. For $2\text{NO}+{\text{O}}_2\to 2{\text{NO}}_2$, with concentrations and rates in mol L${}^{-1}$ and mol L${}^{-1}$ s${}^{-1}$:

- $[\text{NO}]=0.30$, $[{\text{O}}_2]=0.05$: rate $=0.096$
- $[\text{NO}]=0.60$, $[{\text{O}}_2]=0.05$: rate $=0.384$
- $[\text{NO}]=0.30$, $[{\text{O}}_2]=0.10$: rate $=0.192$

Doubling [NO] multiplies the rate by $4$, so its order is $2$; doubling [O${}_2$] doubles the rate, so its order is $1$:

$$\text{Rate}=k[\text{NO}{]}^2[{\text{O}}_2],k=\frac{0.096}{0.30^2\times 0.05}\approx 21.3{\text{ L}}^2{\text{ mol}}^{-2}{\text{ s}}^{-1}$$

**Units of $k$:

- Zero order** — mol L${}^{-1}$ s${}^{-1}$
- First order — s${}^{-1}$
- Second order — L mol${}^{-1}$ s${}^{-1}$

An everyday example. Adding more washing powder does not always clean clothes proportionally faster — only testing shows how rate depends on amount.

The substance. Order cannot be read from the balanced equation — the decomposition of N${}_2$O${}_5$ has a coefficient of $2$ but is first order.

**Order is an experimental quantity that can be zero, fractional or whole, while molecularity is the number of species colliding in a single elementary step and must be $1$, $2$ or rarely $3$; in a complex reaction, the slowest elementary step limits the overall rate.

Order versus molecularity:

- Order** — found from experiment; applies to the overall reaction; can be $0$ or a fraction
- Molecularity — theoretical; applies only to elementary steps; never $0$ or a fraction
- For an elementary reaction, the two are equal

Rate-determining step. A complex reaction proceeds through a series of elementary steps. The slowest is the bottleneck, so its rate law gives the rate of the whole reaction.

Worked example. Iodide ions catalyse the decomposition of hydrogen peroxide:

$$\text{Slow: }{\text{H}}_2{\text{O}}_2+{\text{I}}^{-}\to {\text{H}}_2\text{O}+{\text{IO}}^{-}$$

$$\text{Fast: }{\text{H}}_2{\text{O}}_2+{\text{IO}}^{-}\to {\text{H}}_2\text{O}+{\text{I}}^{-}+{\text{O}}_2$$

The slow step gives Rate $=k[{\text{H}}_2{\text{O}}_2][{\text{I}}^{-}]$ — first order in each, second order overall — even though the overall equation, $2{\text{H}}_2{\text{O}}_2\to 2{\text{H}}_2\text{O}+{\text{O}}_2$, contains no iodide.

An everyday example. A queue at a railway ticket counter moves only as fast as the clerk serves, however quickly people join it.

The substance. Molecularity has no meaning for a complex reaction as a whole — it describes individual steps only.

A reaction speeds up when reactant concentration rises, because molecules collide more often; when temperature rises, because more molecules have enough energy to react; and when a catalyst is added, because it provides a pathway with lower activation energy.

Concentration. More particles per unit volume means more collisions per second, so the rate of most reactions increases with concentration.

Temperature. For many reactions, a $10$ K rise roughly doubles the rate — the temperature coefficient is about $2$ to $3$ — because the fraction of molecules with enough energy grows sharply.

Catalyst:

- Provides an alternative pathway with lower activation energy
- Is not used up, and does not change $\Delta G$ or the equilibrium position
- Speeds up the forward and reverse reactions equally

Worked example. A reaction's rate doubles for every $10$ K rise. Warming it from $300$ K to $340$ K multiplies the rate by

$$2^{(340-300)/10}=2^4=16$$

An everyday example. Milk kept in a refrigerator stays fresh for days, but sours within hours on a hot kitchen counter — the souring reactions slow sharply at low temperature.

The substance. A catalyst cannot make an impossible reaction happen — it only speeds up reactions that are already thermodynamically feasible.