Temperature Dependence of Rate

Not every collision reacts. Reactant molecules collide constantly, but only a small fraction carry enough energy to cross the barrier between reactants and products. That threshold, measured above the average energy of the reactants, is the activation energy $E_a$. The Maxwell–Boltzmann distribution of molecular energies shows why temperature matters so much: raising $T$ does not shift every molecule's energy equally — it dramatically fattens the high-energy tail of the distribution, so a much larger fraction of molecules clear the barrier even for a modest rise in $T$.

The Arrhenius equation captures this quantitatively:

where: $k$ is the rate constant, $A$ is the pre-exponential factor (or frequency factor — it tracks how often molecules collide with the right orientation), $E_a$ is the activation energy, $R$ is the universal gas constant ($8.314\ \text{J mol}^{-1}\text{K}^{-1}$), and $T$ is the absolute temperature in kelvin. Because $E_a$ sits in the exponent with a negative sign, even a small increase in $T$ produces a disproportionately large jump in $k$ — the everyday rule of thumb is that rate roughly doubles for every $10\ ^\circ\text{C}$ rise near room temperature.

Linearising the equation is what makes $E_a$ measurable. Taking natural logs of both sides:

This has the form of a straight line, $y = c + mx$, if you treat $\ln k$ as $y$ and $1/T$ as $x$: the slope is $-E_a/R$ and the intercept is $\ln A$. In base-10 form, useful for calculator work, it becomes $\log k = \log A - \dfrac{E_a}{2.303\,RT}$.

The two-temperature shortcut. If $k_1$ and $k_2$ are measured at temperatures $T_1$ and $T_2$, subtracting the log-form equation at each temperature eliminates the unknown $A$ altogether:

This form is the workhorse for board-level numericals — it lets you find $E_a$ from just two rate-constant readings without ever plotting a graph.

Catalysts and the barrier. A catalyst offers the reaction an alternative pathway with a lower $E_a$, without changing $\Delta H$ of the reaction or shifting the position of equilibrium. Since $k$ depends exponentially on $-E_a$, even a small drop in $E_a$ produces a large jump in rate — this is why catalysts can be so dramatically effective while leaving the reaction's overall energetics untouched.

A quick sanity check on units. Because $E_a/RT$ must be a pure number (it sits in an exponent), $E_a$ and $RT$ always carry matching energy units — if $R$ is in joules, $E_a$ comes out in joules per mole, and it is common practice to divide by 1000 at the end to report kilojoules per mole, as the worked example below does.

The rate constant of a first order reaction is $3.0\times10^{-4}\ \text{s}^{-1}$ at $298\ \text{K}$ and $8.0\times10^{-4}\ \text{s}^{-1}$ at $308\ \text{K}$. Calculate the activation energy of the reaction. (Take $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$.)

Step 1. Write the two-temperature Arrhenius equation derived above: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303\,R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$, and identify $k_1 = 3.0\times10^{-4}$ at $T_1 = 298\ \text{K}$, $k_2 = 8.0\times10^{-4}$ at $T_2 = 308\ \text{K}$.

Step 2. Compute the left side. $\dfrac{k_2}{k_1} = \dfrac{8.0\times10^{-4}}{3.0\times10^{-4}} = 2.667$, so $\log(2.667) = 0.4260$.

Step 3. Compute the bracket on the right. $\dfrac{1}{298}-\dfrac{1}{308} = 0.0033557 - 0.0032468 = 1.089\times10^{-4}\ \text{K}^{-1}$.

Step 4. Solve for $E_a$. $0.4260 = \dfrac{E_a}{2.303\times8.314}\times(1.089\times10^{-4})$, so $E_a = \dfrac{0.4260 \times 19.147}{1.089\times10^{-4}} = \dfrac{8.157}{1.089\times10^{-4}} \approx 7.49\times10^{4}\ \text{J mol}^{-1}$. The activation energy is approximately 74.9 kJ mol⁻¹.

Competitive stretch

A plot of $\ln k$ against $1/T$ for a reaction is a straight line with slope $-9000\ \text{K}$ and y-intercept $26.0$. Find $E_a$ and $A$.

Step 1. The slope of $\ln k$ vs. $1/T$ equals $-E_a/R$, so $-E_a/R = -9000\ \text{K}$, giving $E_a = 9000 \times 8.314 = 74{,}826\ \text{J mol}^{-1} \approx 74.8\ \text{kJ mol}^{-1}$ — reassuringly close to the value found above by the two-point method.

Step 2. The y-intercept equals $\ln A$, so $\ln A = 26.0$, which gives $A = e^{26.0} \approx 1.96\times10^{11}$ (same units as $k$).

So $E_a \approx 74.8\ \text{kJ mol}^{-1}$ and $A \approx 1.96\times10^{11}$.

• Using temperature in degrees Celsius instead of kelvin — the Arrhenius equation demands absolute temperature throughout every step.
• Dropping the factor of 2.303 when switching between the natural-log and base-10 forms of the equation, which throws the final $E_a$ off by roughly that same factor.
• Believing a catalyst changes $\Delta H$ or shifts the equilibrium position — it only lowers $E_a$, speeding up the forward and reverse reactions equally so the equilibrium constant $K_{eq}$ stays unchanged.
• Reporting $E_a$ without the sign flip: the slope of $\ln k$ vs. $1/T$ is negative and equals $-E_a/R$, so $E_a$ itself must come out positive.

Ask the tutor: Stuck on rearranging the two-temperature Arrhenius equation for your own $k$ and $T$ values, or want to double-check a slope-to-$E_a$ conversion — ask the AI tutor to work through it with you.