Arrhenius equation

The Arrhenius equation links the rate constant $k$ of a reaction to the absolute temperature $T$:

where: $k$ is the rate constant at temperature $T$; $A$ is the pre-exponential factor (also called the frequency factor or Arrhenius factor), related to how often molecules collide with the correct orientation, and treated as constant for a given reaction over a moderate temperature range; $E_a$ is the activation energy in $\text{J mol}^{-1}$; $R$ is the universal gas constant, $8.314\ \text{J K}^{-1}\text{mol}^{-1}$; and $T$ is the absolute temperature in kelvin.

Reading the exponential. The term $e^{-E_a/RT}$ is always a number between $0$ and $1$, and it is the fraction of molecules whose energy meets or exceeds $E_a$ at that temperature — the same fraction pictured as the shaded tail of the Maxwell–Boltzmann distribution beyond $E_a$. Raise $T$ and $-E_a/RT$ becomes less negative, so the fraction grows and $k$ grows with it; lower $E_a$ has the identical effect. This is exactly why a modest temperature rise increases the number of effective (correctly oriented, sufficiently energetic) collisions per second — the qualitative story of collision theory, now made quantitative.

Logarithmic (graphical) form. Taking natural logs of both sides converts the exponential relation into a straight line:

or, in base-10 form, more convenient for plotting since log tables and graph slopes traditionally use base 10:

Plotting $\log k$ on the y-axis against $1/T$ on the x-axis gives a straight line with slope $=-\dfrac{E_a}{2.303R}$ (so $E_a = -2.303R\times\text{slope}$) and y-intercept $=\log A$ (so $A=10^{\text{intercept}}$). This is the standard experimental route to both quantities: measure $k$ at several temperatures, plot $\log k$ against $1/T$, and read $E_a$ off the slope and $A$ off the intercept.

Two-temperature form. Writing the logarithmic equation at two temperatures $T_1,T_2$ (with rate constants $k_1,k_2$) and subtracting, assuming $A$ is unchanged, gives:

This compact relation is the workhorse for numericals: it finds $E_a$ from just two rate-constant measurements, or predicts $k_2$ from $E_a$ and $k_1$, without ever needing the full straight-line graph.

Connecting back to collision theory. Before this chapter's Arrhenius topic, collision theory already argued qualitatively that raising temperature increases the fraction of molecules with enough kinetic energy to cross the activation-energy barrier, and that this — not simply more frequent collisions — is the dominant reason reaction rates climb with heat. The Arrhenius equation is the quantitative closure of exactly that argument: the exponential factor $e^{-E_a/RT}$ is that fraction of sufficiently energetic molecules, so every qualitative statement about "more effective collisions at higher temperature" from earlier in the unit can now be read off a number, computed from real $E_a$ and $T$ values rather than described only in words. This is also why catalysts, which do not appear explicitly in the equation, work by lowering $E_a$ rather than by changing $T$ or the collision frequency $A$: a smaller $E_a$ raises the exponential fraction at the same temperature, exactly as the equation predicts.

Question. The rate constant of a reaction at $300\ \text{K}$ is $2.0\times10^{-4}\ \text{s}^{-1}$, and at $310\ \text{K}$ it is $4.6\times10^{-4}\ \text{s}^{-1}$. Calculate the activation energy of the reaction. ($R=8.314\ \text{J K}^{-1}\text{mol}^{-1}$.)

Step 1. Identify the values: $T_1=300\ \text{K}$, $k_1=2.0\times10^{-4}$, $T_2=310\ \text{K}$, $k_2=4.6\times10^{-4}$, and use the two-temperature form derived above, $\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$.

Step 2. Compute the ratio: $\dfrac{k_2}{k_1}=\dfrac{4.6\times10^{-4}}{2.0\times10^{-4}}=2.30$, so $\log(2.30)=0.362$.

Step 3. Compute the bracket: $\dfrac{1}{300}-\dfrac{1}{310}=0.003333-0.003226=0.000108\ \text{K}^{-1}$ (i.e. $1.08\times10^{-4}\ \text{K}^{-1}$).

Step 4. Rearrange and substitute: $E_a=\dfrac{2.303\times8.314\times0.362}{1.08\times10^{-4}}=\dfrac{6.934}{1.08\times10^{-4}}\approx64{,}200\ \text{J mol}^{-1}$. The activation energy is approximately $64.2\ \text{kJ mol}^{-1}$.

Competitive stretch

For a related reaction, a plot of $\log k$ against $1/T$ over five temperatures gives slope $-5230\ \text{K}$ and y-intercept $12.5$. Using slope $=-\dfrac{E_a}{2.303R}$: $E_a=-2.303\times8.314\times(-5230)\approx100{,}139\ \text{J mol}^{-1}\approx100.1\ \text{kJ mol}^{-1}$. Using intercept $=\log A$: $A=10^{12.5}=3.16\times10^{12}\ \text{s}^{-1}$ (assuming first order, so $A$ carries the same units as $k$).

• Forgetting the factor of $2.303$ when switching between natural log (used in the raw derivation) and base-10 log (used for graph work and the two-temperature formula).
• Using Celsius instead of absolute (kelvin) temperature — $T$ must be in kelvin throughout, in both the exponential and logarithmic forms.
• Misreading the sign of the slope: the $\log k$ vs $1/T$ line has a negative slope, and $E_a=-2.303R\times\text{slope}$ must still come out positive — a positive $E_a$ requires a negative slope, not the reverse.
• Assuming $A$ always has the same units — like $k$, the units of the pre-exponential factor $A$ match the units of $k$ for that reaction's order (e.g. $\text{s}^{-1}$ for a first-order reaction).

Ask the tutor: if the switch between the two-temperature form and the slope-intercept graphical form of the Arrhenius equation feels confusing, ask your AI tutor to walk through one more numerical with you step by step.