Rate equations, orders and half-lives

What you’ll learn

• Deduce reaction orders from experimental data.
• Calculate a rate constant and its units.
• Interpret concentration–time and rate–concentration graphs.

Understand the idea

A rate equation records how measured rate depends on concentration at a fixed temperature. Compare experiments in which one concentration changes while the others stay constant: doubling a concentration multiplies rate by $1$, $2$ or $4$ for zero, first or second order respectively. Orders are experimental results and cannot normally be read from the overall chemical equation.

A concentration–time graph has a gradient whose magnitude gives the rate of disappearance. A first-order reaction has a constant half-life, whereas a zero-order concentration decreases linearly. Choose a monitoring method that tracks a changing quantity: colorimetry for a coloured species, gas volume or mass for an escaping gas, or analysis of timed samples. Clock times estimate relative initial rates only when each experiment reaches the same small reaction extent.

Key formulae

$$\mathrm{rate}=k[\mathrm A]^m[\mathrm B]^n$$

$$\frac{\mathrm{rate}_2}{\mathrm{rate}_1}=\left(\frac{[\mathrm A]_2}{[\mathrm A]_1}\right)^m\quad([\mathrm B]\text{ fixed})$$

$$[\mathrm A]_t=[\mathrm A]_0\left(\frac12\right)^{t/t_{1/2}}\quad\text{(first order)}$$

$$[k]=(\mathrm{mol\,dm^{-3}})^{1-(m+n)}\mathrm{s^{-1}}$$

Worked example

Doubling $[\mathrm A]$ doubles the rate; doubling $[\mathrm B]$ at fixed $[\mathrm A]$ quadruples it. When $[\mathrm A]=0.20$ and $[\mathrm B]=0.10\,\mathrm{mol\,dm^{-3}}$, the rate is $0.0060\,\mathrm{mol\,dm^{-3}\,s^{-1}}$. Find the rate equation and $k$.

1. The reaction is first order in A and second order in B, so rate $=k[\mathrm A][\mathrm B]^2$.
2. Calculate $k=0.0060/(0.20\times0.10^2)=3.0$.
3. Overall order is three; dividing rate units by concentration cubed gives $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$.

Answer: Rate $=k[\mathrm A][\mathrm B]^2$, with $k=3.0\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}$.