Effect of temperature on reaction rate — Arrhenius equation graphically explained

Question

The rate constant of a reaction doubles when temperature rises from 300 K to 310 K. Calculate the activation energy. Also, explain what the Arrhenius plot ( $\ln k$ vs $1/T$ ) looks like and how to extract $E_a$ from it.

(JEE Main 2023 & NEET pattern)

Solution — Step by Step

When we have rate constants at two temperatures, we use:

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

This comes from subtracting $\ln k_1 = \ln A - E_a/(RT_1)$ from $\ln k_2 = \ln A - E_a/(RT_2)$ .

Given: $k_2/k_1 = 2$ , $T_1 = 300$ K, $T_2 = 310$ K, $R = 8.314$ J mol $^{-1}$ K $^{-1}$ .

\ln 2 = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{310}\right)

0.693 = \frac{E_a}{8.314} \times \frac{310 - 300}{300 \times 310}

0.693 = \frac{E_a}{8.314} \times \frac{10}{93000}

0.693 = \frac{E_a \times 10}{8.314 \times 93000}

E_a = \frac{0.693 \times 8.314 \times 93000}{10}

E_a = \frac{0.693 \times 773202}{10} \approx \mathbf{53.6 \text{ kJ mol}^{-1}}

The Arrhenius equation in logarithmic form:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

This is $y = c + mx$ form. Plotting $\ln k$ (y-axis) vs $1/T$ (x-axis) gives a straight line with:

Slope $= -E_a/R$ (always negative — line goes downward left to right)
y-intercept $= \ln A$ (the pre-exponential factor)

So $E_a = -\text{slope} \times R$ .

Why This Works

The Arrhenius equation $k = Ae^{-E_a/(RT)}$ captures a physical truth: molecules need a minimum energy ( $E_a$ ) to react. At higher temperatures, more molecules have kinetic energy exceeding $E_a$ (the Maxwell-Boltzmann distribution shifts right). This increases the fraction of effective collisions, raising the rate constant exponentially.

The “rate doubles every 10°C” rule is a rough approximation. The actual factor depends on $E_a$ — reactions with higher activation energy are more sensitive to temperature changes.

graph TD
    A["Temperature increases"] --> B["KE distribution shifts right"]
    B --> C["More molecules exceed Ea"]
    C --> D["More effective collisions"]
    D --> E["Rate constant k increases"]
    E --> F["Reaction rate increases"]
    G["Arrhenius Plot"] --> H["Plot ln k vs 1/T"]
    H --> I["Straight line, negative slope"]
    I --> J["Slope = -Ea/R"]
    J --> K["Ea = -slope x R"]

Alternative Method — Using log₁₀ Form

Some CBSE textbooks use the $\log_{10}$ version:

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

With $\log 2 = 0.301$ :

0.301 = \frac{E_a}{2.303 \times 8.314} \times \frac{10}{93000}

Same answer: $E_a \approx 53.6$ kJ/mol. Use whichever form matches your textbook.

For JEE numerical problems, keep $R = 8.314$ J mol $^{-1}$ K $^{-1}$ and express $E_a$ in J first, then convert to kJ by dividing by 1000. This avoids unit mismatch errors that cost marks.

Common Mistake

The biggest trap: students write $1/T_2 - 1/T_1$ instead of $1/T_1 - 1/T_2$ in the formula. This flips the sign and gives a negative $E_a$ , which is physically meaningless. Remember: $T_1$ is the lower temperature and goes first. An easy mnemonic — “small T first, big T second” — matches the formula where the positive result comes from $1/T_1 > 1/T_2$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Using log₁₀ Form

Common Mistake

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