Effect of temperature on reaction rate — Arrhenius equation graphically explained

Question

How does temperature affect reaction rate? What is the Arrhenius equation, and how do we use it to calculate activation energy from experimental data?

(JEE Main, NEET, CBSE 12 — Arrhenius equation numericals are high-frequency questions in both JEE and NEET)

Solution — Step by Step

A common rule of thumb: for every 10 degC rise in temperature, the reaction rate roughly doubles. This is called the temperature coefficient.

But why? At higher temperature, molecules move faster, collide more frequently, and — more importantly — a larger fraction of molecules have energy exceeding the activation energy ( $E_a$ ). This second factor is the dominant one.

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

where:

$k$ = rate constant
$A$ = pre-exponential factor (frequency factor, related to collision frequency and orientation)
$E_a$ = activation energy (J/mol)
$R$ = gas constant (8.314 J/mol/K)
$T$ = temperature in Kelvin

Taking natural log of both sides:

\ln k = \ln A - \frac{E_a}{RT}

This is the equation of a straight line: plot $\ln k$ vs $1/T$ , and you get:

Slope = $-E_a/R$ (from which we calculate $E_a$ )
y-intercept = $\ln A$

If we know $k$ at two temperatures $T_1$ and $T_2$ :

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

Or using $\log_{10}$ :

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

Worked example: A reaction has $k_1 = 2 \times 10^{-3}$ s $^{-1}$ at 300 K and $k_2 = 8 \times 10^{-3}$ s $^{-1}$ at 320 K. Find $E_a$ .

\log\frac{8 \times 10^{-3}}{2 \times 10^{-3}} = \frac{E_a}{2.303 \times 8.314}\left(\frac{1}{300} - \frac{1}{320}\right)

\log 4 = \frac{E_a}{19.15}\left(\frac{320-300}{300 \times 320}\right)

0.602 = \frac{E_a}{19.15} \times \frac{20}{96000}

0.602 = \frac{E_a}{19.15} \times 2.083 \times 10^{-4}

E_a = \frac{0.602 \times 19.15}{2.083 \times 10^{-4}} = 55,340 \text{ J/mol} \approx \mathbf{55.3 \text{ kJ/mol}}

flowchart TD
    A["Temperature increases"] --> B["More molecules exceed Ea"]
    B --> C["Rate constant k increases exponentially"]
    C --> D["Reaction rate increases"]
    E["Arrhenius Plot"] --> F["Plot ln k vs 1/T"]
    F --> G["Straight line with negative slope"]
    G --> H["Slope = -Ea/R"]
    H --> I["Calculate Ea = -slope × R"]

Why This Works

The exponential factor $e^{-E_a/RT}$ represents the fraction of molecules with kinetic energy greater than $E_a$ (from the Maxwell-Boltzmann distribution). At low temperatures, this fraction is tiny. A small increase in $T$ dramatically increases this fraction because it appears in the exponent. This is why temperature has such a powerful effect on reaction rates — it is not linear but exponential.

The Arrhenius equation unifies two observations: reactions with high $E_a$ are slow (large exponent), and raising temperature speeds them up (smaller effective exponent).

Common Mistake

The unit trap: $E_a$ must be in J/mol (not kJ/mol) when using $R = 8.314$ J/mol/K in the Arrhenius equation. If the question gives $E_a$ in kJ/mol, multiply by 1000 before substituting. Alternatively, use $R = 8.314 \times 10^{-3}$ kJ/mol/K. Mixing units is the number one source of wrong answers in Arrhenius numericals.

For quick estimation: if the rate doubles for a 10 degC rise (temperature coefficient = 2), the activation energy is approximately 53 kJ/mol at room temperature. Most reactions tested in exams have $E_a$ between 40-100 kJ/mol. If your calculated answer falls outside this range, double-check your units.

Question

Solution — Step by Step

Why This Works

Common Mistake

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