Two-point Arrhenius and van ’t Hoff

Questions write these equations in several different-looking ways. They are all the same equation, and each comes from one line of algebra. Once you can derive it, you never need to remember which way round the fractions go.

Arrhenius: from one temperature to two

Start from the and take natural logs:

k = A\,e^{-E_\mathrm{a}/RT} \quad\Longrightarrow\quad \ln k = \ln A - \frac{E_\mathrm{a}}{RT}

Write it at two temperatures, $T_1$ and $T_2$ :

\ln k_1 = \ln A - \frac{E_\mathrm{a}}{RT_1} \qquad \ln k_2 = \ln A - \frac{E_\mathrm{a}}{RT_2}

Subtract the first from the second. $\ln A$ cancels, and $\ln k_2 - \ln k_1 = \ln(k_2/k_1)$ :

\ln\frac{k_2}{k_1} = -\frac{E_\mathrm{a}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)

That is the . The only assumption is that $A$ and $E_\mathrm{a}$ do not change between the two temperatures.

van ’t Hoff: the same algebra for K

Here the starting point is thermodynamics, not kinetics. Two expressions for $\Delta G^\circ$ (see ):

\Delta G^\circ = -RT\ln K \qquad\text{and}\qquad \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

Set them equal and divide by $-RT$ :

\ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}

This has exactly the shape of $\ln k = \ln A - E_\mathrm{a}/RT$ . Write it at $T_1$ and $T_2$ and subtract; $\Delta S^\circ/R$ cancels just as $\ln A$ did:

\ln\frac{K_2}{K_1} = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)

That is the , assuming $\Delta H^\circ$ and $\Delta S^\circ$ are constant over the range.

	Arrhenius	van ’t Hoff
Constant	rate constant $k$	equilibrium constant $K$
Energy term	$E_\mathrm{a}$ , always positive	$\Delta H^\circ$ , either sign
Cancels on subtracting	$\ln A$	$\Delta S^\circ/R$
Raising $T$	always increases $k$	increases $K$ only if endothermic

One equation, many looks

All of these say the same thing, written for $k$ (swap in $K$ and $\Delta H^\circ$ for van ’t Hoff):

Form	What changed
$\ln\dfrac{k_2}{k_1} = -\dfrac{E_\mathrm{a}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$	as derived
$\ln\dfrac{k_2}{k_1} = \dfrac{E_\mathrm{a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$	minus sign used to swap the bracket
$\ln\dfrac{k_1}{k_2} = \dfrac{E_\mathrm{a}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$	minus sign used to flip the ratio
$\ln\dfrac{k_2}{k_1} = \dfrac{E_\mathrm{a}(T_2-T_1)}{RT_1T_2}$	bracket over a common denominator
$\log_{10}\dfrac{k_2}{k_1} = \dfrac{E_\mathrm{a}}{2.303\,R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$	base-10 logs: $\ln x = 2.303\log_{10}x$

Worked example

A rate constant doubles between 298 K and 308 K. Find $E_\mathrm{a}$ .

\ln 2 = \frac{E_\mathrm{a}}{8.314}\left(\frac{1}{298}-\frac{1}{308}\right) = \frac{E_\mathrm{a}}{8.314}\times 1.090\times10^{-4}

So $E_\mathrm{a} = 0.693 \times 8.314 / (1.090\times10^{-4}) \approx 5.29\times10^{4}$ J mol⁻¹, or 52.9 kJ mol⁻¹. This is the origin of the old rule of thumb that rates double for every 10 K rise.