Activation energy and temperature-dependent rate data

What you’ll learn

• Determine activation energy from an Arrhenius plot or two rate constants.
• Plan temperature measurements and evaluate their effect on a rate investigation.

Understand the idea

Core practical 10 investigates how a rate constant changes with temperature. The Arrhenius relationship becomes linear when $\ln k$ is plotted against $1/T$. Temperature must be in kelvin, and the natural logarithm is required. The line has gradient $-E_a/R$, so a negative gradient corresponds to a positive activation energy. Use a best-fit line through a range of data rather than joining successive points or choosing two adjacent noisy readings.

A relative rate such as $1/t$ may replace $k$ for this purpose only if it remains proportional to $k$: concentrations, clock threshold, total volume and other relevant conditions must be constant. Otherwise a temperature trend can be mixed with a concentration or endpoint trend. If both axes are calculated from measured quantities, label the transformed axes with the correct scale; a plot against $1000/T$ has a different numerical gradient from a plot against $1/T$.

The reaction mixture's temperature is the relevant measurement, not merely a bath dial setting. A supervised plan allows solutions to reach the selected temperature, checks the temperature during timing and repeats observations. Evaporation at higher temperatures can alter concentrations and is a separate source of bias. The supplied hazards determine an appropriate controlled heating method; the aim is stable measured conditions rather than simply more heating.

Key formulae

$$k=Ae^{-E_a/(RT)}$$

$$\ln k=\ln A-\frac{E_a}{R}\frac1T$$

$$E_a=-R\times\mathrm{gradient}$$

$$\ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac1{T_1}-\frac1{T_2}\right)$$

Worked example

A reaction has $k_1=0.0020\ \mathrm{s^{-1}}$ at $290\ \mathrm{K}$ and $k_2=0.0080\ \mathrm{s^{-1}}$ at $310\ \mathrm{K}$. Use $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ and the Arrhenius relationship to estimate $E_a$.

1. The rate-constant ratio is $k_2/k_1=4$, so $\ln(k_2/k_1)=\ln4=1.3863$.
2. The reciprocal-temperature difference is $1/290-1/310=2.2247\times10^{-4}\ \mathrm{K^{-1}}$.
3. Thus $E_a$ $=8.314(1.3863)/(2.2247\times10^{-4})$ $=5.18\times10^4\ \mathrm{J\,mol^{-1}}$ $=51.8\ \mathrm{kJ\,mol^{-1}}$.

Answer: Approximately $51.8\ \mathrm{kJ\,mol^{-1}}$.