Temperature Dependence — Arrhenius Equation

Why Temperature Matters So Much

Of all the factors that govern reaction rate, temperature is the most dramatic. NCERT records a striking illustration: in the decomposition of dinitrogen pentoxide, the time taken for half the material to decompose is 12 minutes at 50 °C, 5 hours at 25 °C, and 10 days at 0 °C. A swing of 50 degrees changes the timescale by a factor of more than a thousand. The same is seen when a mixture of $\ce{KMnO4}$ and oxalic acid is decolourised — it happens far faster when warmed.

From this kind of observation chemists distilled a rough empirical rule: for a rise of 10 °C, the rate constant of a reaction is nearly doubled. This "rate doubles per 10 K" statement is a frequent NEET fact, but it is an approximation valid near room temperature, not an exact law. The exact relationship was first proposed by J. H. van't Hoff and given its physical justification by the Swedish chemist Svante Arrhenius — hence the name of the equation that follows.

The Arrhenius Equation

The temperature dependence of the rate constant is expressed by the Arrhenius equation:

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

Each symbol carries precise meaning. $k$ is the rate constant. $A$ is the Arrhenius factor, also called the frequency factor or pre-exponential factor; it is a constant specific to a given reaction and is related to the frequency of collisions and their orientation. $E_a$ is the activation energy, measured in $\text{J mol}^{-1}$. $R$ is the gas constant ($8.314\ \text{J K}^{-1}\text{mol}^{-1}$) and $T$ is the absolute temperature in kelvin.

The structure of the equation tells the whole story. Because $E_a$ sits in the exponent, even a modest change in temperature or activation energy produces a large, exponential change in $k$. Increasing $T$ makes the exponent less negative, so $e^{-E_a/RT}$ grows; lowering $E_a$ does the same. Both routes raise the rate constant — and therefore the rate.

Activation Energy and the Activated Complex

Consider the simple reaction $\ce{H2(g) + I2(g) -> 2HI(g)}$. According to Arrhenius, reaction occurs only when a hydrogen molecule and an iodine molecule collide to form an unstable intermediate. This intermediate exists for an extremely short time before breaking up to give two molecules of hydrogen iodide. The intermediate is called the activated complex (often written C), and the energy required to form it from the reactants is the activation energy, $E_a$.

Plotting potential energy against the reaction coordinate gives the energy profile below. The reactants must climb the barrier to reach the peak (the activated complex); some energy is then released as the complex decomposes into products. The net enthalpy change of the reaction depends only on the relative energies of reactants and products — not on the height of the barrier.

Fraction of Molecules with E > Ea

Not all molecules carry the same kinetic energy. Because it is impossible to track a single molecule, Maxwell and Boltzmann used statistics to describe the spread. Plotting the fraction of molecules $N_E/N_T$ against kinetic energy gives the Maxwell–Boltzmann distribution — a curve that rises to a peak at the most probable energy and tails off at higher energies. Here $N_E$ is the number of molecules with energy $E$ and $N_T$ is the total number.

Only molecules in the tail beyond $E_a$ can react. The factor $e^{-E_a/RT}$ in the Arrhenius equation is precisely the fraction of molecules with kinetic energy equal to or greater than $E_a$. When the temperature is raised, the peak of the curve shifts to higher energy and the curve broadens to the right; crucially, the area beyond $E_a$ — the reactive fraction — grows. The total area stays constant because total probability must equal one.

This is the molecular explanation of the "rate doubles per 10 K" rule: a 10 K rise near room temperature roughly doubles the fraction of molecules in the high-energy tail, doubling $e^{-E_a/RT}$ and hence $k$.

The ln k vs 1/T Plot

Taking the natural logarithm of both sides of the Arrhenius equation linearises it:

$$\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A$$

This has the form $y = mx + c$ with $y = \ln k$, $x = 1/T$, slope $m = -E_a/R$, and intercept $c = \ln A$. A plot of $\ln k$ against $1/T$ is therefore a straight line with negative slope. From it we extract both unknowns: $E_a = -(\text{slope}) \times R$, and $A = e^{\text{intercept}}$.

The Two-Temperature Form

In practice you rarely draw a full plot in an exam; instead you are given the rate constant at two temperatures and asked for $E_a$ (or vice versa). Writing the log form at $T_1$ and $T_2$ and subtracting eliminates $\ln A$:

$$\ln k_1 = -\frac{E_a}{RT_1} + \ln A \qquad \ln k_2 = -\frac{E_a}{RT_2} + \ln A$$

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

This is the single most useful working equation of the subtopic. Note the two equivalent arrangements of the bracket: $\left(\tfrac{1}{T_1}-\tfrac{1}{T_2}\right)$ and $\left(\tfrac{T_2-T_1}{T_1 T_2}\right)$. As long as the larger rate constant $k_2$ corresponds to the higher temperature $T_2$, both sides come out positive.

Worked Numericals

Effect of a Catalyst on Ea

The Arrhenius equation also explains why catalysts are so effective. A catalyst provides an alternative pathway with a lower activation energy. Because $E_a$ sits in the exponent, even a small reduction lowers the energy barrier and raises $k$ steeply. NCERT cites $\ce{2KClO3 ->[MnO2] 2KCl + 3O2}$, where $\ce{MnO2}$ accelerates the decomposition.

A catalyst does not alter the Gibbs energy $\Delta G$, the enthalpy, or the equilibrium constant of a reaction. It lowers $E_a$ for the forward and backward reactions equally, so equilibrium is reached faster but its position is unchanged. The full treatment of how collisions, orientation and barriers combine is developed alongside the steric factor in the broader study of reaction rates.