Why Some Reactions Always Take the Same Time to Halve

A rate law tells you how fast a reaction is going right now, but chemists usually need something more practical: how much reactant remains after ten minutes, or how long until half of it is gone. Integrating the rate law answers both.

This part covers zero and first order integrated rate equations, half-life, and the Arrhenius equation with collision theory.

**For a zero order reaction the rate does not depend on concentration, so $[R]=[R{]}_0-kt$; a graph of $[R]$ against $t$ is a straight line with slope $-k$ and intercept $[R{]}_0$.

Derivation:**

$$-\frac{d[R]}{dt}=k\Rightarrow {\int }_{[R{]}_0}^{[R]}d[R]=-k{\int }_0^{t}dt\Rightarrow [R]=[R{]}_0-kt$$

so $k=\frac{[R{]}_0-[R]}{t}$.

Examples: decomposition of ammonia on a hot platinum surface at high pressure, and of HI on gold — the metal surface is fully covered, so extra reactant does not speed things up.

Worked example. A zero order reaction has $k=2\times 10^{-3}$ mol L${}^{-1}$ s${}^{-1}$ and $[R{]}_0=0.10$ mol L${}^{-1}$:

$$[R{]}_{20\text{ s}}=0.10-2\times 10^{-3}\times 20=0.06{\text{ mol L}}^{-1}$$

It finishes when $[R]=0$, at $t=\frac{0.10}{2\times 10^{-3}}=50$ s.

An everyday example. A ticket counter with a single clerk serves customers at a fixed rate however long the queue grows — just like a saturated catalyst surface.

The substance. A zero order reaction runs out completely in a finite time, unlike a first order reaction.

**For a first order reaction, $k=\frac{2.303}{t}\log \frac{[R{]}_0}{[R]}$, so $\ln [R]$ against $t$ is a straight line of slope $-k$; gas-phase reactions use the same equation with pressures, and reactions with one reactant in large excess behave as pseudo first order.

Derivation:**

$$-\frac{d[R]}{dt}=k[R]\Rightarrow \ln \frac{[R]}{[R{]}_0}=-kt\Rightarrow [R]=[R{]}_0{e}^{-kt}$$

Gas phase. For $\text{A}(g)\to \text{B}(g)+\text{C}(g)$ with initial pressure $p_i$ and total pressure $p_t$, the pressure of A is $2p_i-p_t$, so

$$k=\frac{2.303}{t}\log \frac{p_i}{2p_i-p_t}$$

Pseudo first order reactions — one reactant is in such excess that its concentration barely changes:

- Acid hydrolysis of an ester such as ethyl ethanoate in excess water
- Inversion of cane sugar: ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}+{\text{H}}_2\text{O}\to {\text{C}}_6{\text{H}}_{12}{\text{O}}_6+{\text{C}}_6{\text{H}}_{12}{\text{O}}_6$

Worked example. A first order reaction falls from $0.80$ to $0.20$ mol L${}^{-1}$ in $30$ min:

$$k=\frac{2.303}{30}\log 4=\frac{2.303\times 0.602}{30}\approx 0.0462{\text{ min}}^{-1}$$

For a gas-phase reaction with $p_i=0.5$ atm and $p_t=0.6$ atm after $100$ s, $k=\frac{2.303}{100}\log \frac{0.5}{0.4}\approx 2.23\times 10^{-3}$ s${}^{-1}$.

An everyday example. Sugar syrup cooked with a little lemon juice slowly inverts its sucrose into glucose and fructose, which helps it resist crystallising.

The substance. Pseudo first order reactions are really second order — the excess reactant's constant concentration is simply folded into $k$.

**Half-life is the time for a reactant's concentration to fall to half; for a zero order reaction $t_{1/2}=\frac{[R{]}_0}{2k}$, which depends on the starting concentration, while for a first order reaction $t_{1/2}=\frac{0.693}{k}$, which does not.

Zero order.** Putting $[R]=\frac{[R{]}_0}{2}$ into $[R]=[R{]}_0-kt$:

$$t_{1/2}=\frac{[R{]}_0}{2k}$$

First order. Putting $[R]=\frac{[R{]}_0}{2}$ into the integrated equation:

$$t_{1/2}=\frac{2.303}{k}\log 2=\frac{0.693}{k}$$

Worked example 1. For the reaction above with $k=0.0462$ min${}^{-1}$:

$$t_{1/2}=\frac{0.693}{0.0462}=15\text{ min}$$

which fits: $0.80$ to $0.40$ to $0.20$ mol L${}^{-1}$ is two half-lives, $30$ min.

Worked example 2. The time for $99.9$ percent completion of a first order reaction is

$$t=\frac{2.303}{k}\log 1000=\frac{6.909}{k}\approx 10t_{1/2}$$

An everyday example. A medicine cleared from the blood by a first order process falls to half its level in the same time whatever the dose, which is why doses are spaced at regular intervals.

The substance. Equal successive half-lives are the quickest test for first order kinetics — if each halving takes the same time, the order is one.

**The Arrhenius equation, $k=A{e}^{-E_a/RT}$, shows the rate constant rising steeply with temperature as more molecules gain the activation energy; collision theory adds that molecules react only when they collide with enough energy and the proper orientation, and a catalyst lowers the activation energy.

Using two temperatures:**

$$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$$

A plot of $\ln k$ against $\frac{1}{T}$ is a straight line of slope $-\frac{E_a}{R}$.

Worked example. A rate constant doubles between $298$ K and $308$ K:

$$E_a=\frac{2.303\times 8.314\times \log 2}{\frac{1}{298}-\frac{1}{308}}=\frac{2.303\times 8.314\times 0.301\times 298\times 308}{10}\approx 52.9{\text{ kJ mol}}^{-1}$$

Collision theory:

- Molecules must collide to react: rate $=P{Z}_{AB}{e}^{-E_a/RT}$
- Only collisions with at least the threshold energy are effective
- Molecules need the proper orientation, allowed for by the probability factor $P$

Catalysts provide a path with lower $E_a$, so a far larger fraction of collisions is effective at the same temperature.

An everyday example. A car's catalytic converter uses metals such as platinum to speed up the removal of harmful exhaust gases.

The substance. **Raising temperature increases $k$ mainly through the exponential term**, not by making collisions much more frequent.