Arrhenius Equation & Temperature Dependence

CHETemperature & Collision Theory

The Arrhenius equation quantifies how rate constants rise with temperature through the activation energy $E_a$ . ISC problems use either the two-temperature form to find $E_a$ , or the single-equation form with $A$ given; mastering the logarithmic rearrangements is essential.

Arrhenius equation

k = A\,e^{-E_a/RT}

A

frequency factor,

E_a

activation energy,

R=8.314\,J\,mol^{-1}K^{-1}

Logarithmic (single-temperature) form

\log k = \log A - \dfrac{E_a}{2.303\,RT}

slope of

\log k

1/T

-\dfrac{E_a}{2.303R}

Two-temperature form

\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303\,R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)

k_1,k_2

are rate constants at

T_1<T_2

Fraction of effective molecules

\dfrac{k}{A} = e^{-E_a/RT}

fraction of molecules with energy

\ge E_a

Arrhenius equation: $k = A\,e^{-E_a/RT}$ , where $A$ is the frequency (pre-exponential) factor and $e^{-E_a/RT}$ is the fraction of molecules with energy $\ge E_a$ .
Logarithmic form: $\log k = \log A - \dfrac{E_a}{2.303\,RT}$ ; a plot of $\log k$ vs $\dfrac{1}{T}$ is linear with slope $-\dfrac{E_a}{2.303\,R}$ .
Two-temperature form: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303\,R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ — the workhorse for finding $E_a$ or a missing $k$ .
Rate roughly doubles for every $10\,K$ rise near room temperature; this stems from the steep rise in molecules crossing $E_a$ , not a large change in collision number.
Lower $E_a$ means a larger rate constant; a catalyst raises $k$ by providing a path of lower $E_a$ , not by changing $T$ .
Keep units consistent: $R = 8.314\,J\,mol^{-1}K^{-1}$ gives $E_a$ in joules, so convert kJ to J (or use $R$ in kJ) before computing.
The frequency factor $A$ depends on how often molecules collide and on the orientation needed; the temperature dependence of $k$ enters almost entirely through $e^{-E_a/RT}$ .
From the $\log k$ vs $1/T$ graph, $E_a = -2.303\,R\times(\text{slope})$ and the intercept gives $\log A$ — a standard graphical-determination answer.
The factor $\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ is positive when $T_2>T_1$ , making $\log(k_2/k_1)$ positive — $k$ always increases with temperature for $E_a>0$ .
Always convert Celsius to Kelvin before substituting, e.g. $27^\circ C = 300\,K$ ; using $^\circ C$ directly is a fatal error.
When $A$ is given, compute $k$ directly from $k=A\,e^{-E_a/RT}$ ; evaluate the exponent $-E_a/RT$ first, then the exponential (often supplied as a hint value).
Reactions with very high $E_a$ are slow and strongly temperature-sensitive; near-zero $E_a$ reactions are fast and almost temperature-independent.

Where the marks go

Mixing units of $E_a$ and $R$ — if $E_a$ is in kJ, convert to J (multiply by $1000$ ) before dividing by $R=8.314\,J\,mol^{-1}K^{-1}$ .
Sign errors in $\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ : writing it as $\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$ flips the sign and gives a negative $E_a$ .
Reading the slope of $\log k$ vs $1/T$ as $-E_a/R$ instead of $-E_a/2.303R$ — the $2.303$ comes from using $\log_{10}$ .
Forgetting to convert temperatures from $^\circ C$ to $K$ before substituting into the two-temperature equation.
Claiming a catalyst increases $A$ or $T$ — it works by lowering $E_a$ in the exponential term, leaving $T$ unchanged.

How the board asks it

Numericaltwo-temperature form solved for the activation energy
The rate constant of a reaction is $1.5\times10^{-3}\,s^{-1}$ at $300\,K$ and $4.5\times10^{-3}\,s^{-1}$ at $320\,K$ . Calculate the activation energy $E_a$ of the reaction $(R=8.314\,J\,mol^{-1}\,K^{-1})$ .
Numericaltwo-temperature form solved for a missing rate constant
The activation energy of a reaction is $209.5\,kJ\,mol^{-1}$ and its rate constant at $581\,K$ is $1.0\times10^{-2}\,s^{-1}$ . Calculate the rate constant at $700\,K$ $(R=8.314\,J\,mol^{-1}\,K^{-1})$ .
Numericalsingle-equation form with the frequency factor given
For a first-order reaction, $A=4\times10^{13}\,s^{-1}$ and $E_a=98.6\,kJ\,mol^{-1}$ . Calculate the rate constant $k$ at $400\,K$ $(R=8.314\,J\,mol^{-1}\,K^{-1}$ ; take antilog where required $)$ .
Derive / provelogarithmic form and the $\log k$ vs $1/T$ plot
Starting from $k=A\,e^{-E_a/RT}$ , derive $\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303\,R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ and explain how $E_a$ is obtained from the slope of a graph of $\log k$ against $\dfrac{1}{T}$ .
Give reasonssteep rise of $e^{-E_a/RT}$ with temperature
Account for the fact that the rate of a reaction nearly doubles for a $10\,K$ rise in temperature near room temperature, although the number of collisions increases only slightly.
Define / statephysical meaning of the activation energy
Define activation energy $E_a$ , and state how a reaction with a high $E_a$ differs from one with a low $E_a$ in the sensitivity of its rate constant to temperature.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.