Reactions speed up when heated, often a lot. Activation energy explains why: only collisions with enough energy react, and heating raises the share of such collisions sharply. The Arrhenius equation turns that into a formula, and logarithms make it easy to use.
Key idea
$k = Ae^{-E_a/RT}$. Taking logs, $\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac{1}{T}$, so a plot of $\ln k$ against $1/T$ is a line with slope $-E_a/R$. For two temperatures,
$$\ln\frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$
with $T$ in kelvins and $E_a$ in J/mol when $R = 8.314\ \text{J/(mol·K)}$.
Why it works
In a gas, the fraction of collisions with at least $E_a$ of energy is about $e^{-E_a/RT}$ (from the Maxwell–Boltzmann distribution). $A$ collects the rest: how often molecules collide and how often they meet in the right orientation. Because $T$ sits in an exponent, a small rise in temperature makes a large change in $k$. Subtract the logarithmic form at $T_1$ from the one at $T_2$ and $\ln A$ cancels, which gives the two-point form.
Worked examples
Worked example: Activation energy from two rate constants
A rate constant doubles when the temperature rises from $25\ ^\circ\text{C}$ to $35\ ^\circ\text{C}$. What is $E_a$?
- Kelvins: $T_1 = 298\ \text{K}$, $T_2 = 308\ \text{K}$. $$\dfrac{1}{T_2} - \dfrac{1}{T_1} = 0.0032468 - 0.0033557 = -1.0895 \times 10^{-4}\ \text{K}^{-1}$$
- $\ln 2 = -\dfrac{E_a}{8.314}(-1.0895 \times 10^{-4})$, so $E_a = \dfrac{(8.314)(0.693)}{1.0895 \times 10^{-4}}$ J/mol. That is $52\,900\ \text{J/mol}$, or $52.9\ \text{kJ/mol}$.
- This is the source of the old rule of thumb that a 10-degree rise doubles a rate near room temperature; it holds only for activation energies near 50 kJ/mol.
Answer: $52.9\ \text{kJ/mol}$.
Worked example: A rate constant at a new temperature
$E_a = 75.0\ \text{kJ/mol}$ and $k = 2.0 \times 10^{-3}\ \text{s}^{-1}$ at $300\ \text{K}$. Find $k$ at $320\ \text{K}$.
- Use J/mol: $E_a = 75\,000\ \text{J/mol}$. $\dfrac{1}{320} - \dfrac{1}{300} = -2.083 \times 10^{-4}\ \text{K}^{-1}$.
- $$\ln\dfrac{k_2}{k_1} = -\dfrac{75\,000}{8.314}(-2.083 \times 10^{-4}) = 1.879$$ so $\dfrac{k_2}{k_1} = e^{1.879} = 6.55$.
- $k_2 = (6.55)(2.0 \times 10^{-3}) = 1.3 \times 10^{-2}\ \text{s}^{-1}$.
Answer: $1.3 \times 10^{-2}\ \text{s}^{-1}$, about 6.5 times faster for 20 degrees.
Worked example: From an Arrhenius plot
A plot of $\ln k$ against $1/T$ has slope $-6.00 \times 10^3\ \text{K}$. Find $E_a$.
- Slope $= -E_a/R$, so $$E_a = (6.00 \times 10^3\ \text{K})(8.314\ \text{J/(mol·K)}) = 4.99 \times 10^4\ \text{J/mol}$$
- That is $49.9\ \text{kJ/mol}$.
Answer: $49.9\ \text{kJ/mol}$.
Common mistake
Using Celsius temperatures, or putting $E_a$ in kJ/mol beside $R$ in J/(mol·K). Celsius gives wrong ratios (and division by zero at $0\ ^\circ\text{C}$), and mixed units make $E_a$ off by a factor of 1000. Check: $E_a/RT$ must be a pure number, typically between 10 and 40 for reactions you can watch.
Try it on the bench: in the Mixing Lab you can warm a reaction and add a catalyst, and watch how much faster it goes.