Why rate increases with temperature. Activation energy and frequency factor.

Arrhenius Equation — k = Ae^(-Ea/RT) Explained

Question

A reaction has an activation energy of 80 kJ/mol. By what factor does the rate constant increase when the temperature is raised from 300 K to 310 K? (R = 8.314 J/mol·K)

Solution — Step by Step

We need the ratio $k_2/k_1$ . Taking the Arrhenius equation for both temperatures and dividing them gives us a clean working form:

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

This form cancels out the frequency factor $A$ entirely — we don’t need it.

\ln\frac{k_2}{k_1} = \frac{80000}{8.314}\left(\frac{1}{300} - \frac{1}{310}\right)

Calculate the bracket first: $\frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = 1.075 \times 10^{-4}$

\ln\frac{k_2}{k_1} = 9621 \times 1.075 \times 10^{-4} = 1.034

\frac{k_2}{k_1} = e^{1.034} \approx 2.81

The rate constant increases by a factor of approximately 2.81 — nearly triples with just a 10 K rise.

Why This Works

The Arrhenius equation $k = Ae^{-E_a/RT}$ captures a simple physical idea: molecules need a minimum energy ( $E_a$ ) to react. At any temperature, only a fraction of molecules have this energy — and that fraction is given by the Boltzmann factor $e^{-E_a/RT}$ .

When temperature increases, the exponent $-E_a/RT$ becomes less negative (closer to zero), so $e^{-E_a/RT}$ grows. Since $E_a$ sits in the exponent, even small temperature changes cause large changes in $k$ . This exponential sensitivity is why reactions speed up so dramatically with heat.

The frequency factor $A$ represents how often molecules collide with the correct orientation. It’s roughly constant with temperature — the exponential term does all the heavy lifting.

The 10 K rule: For reactions with $E_a \approx 50$ – $80$ kJ/mol near room temperature, rate roughly doubles or triples for every 10 K rise. This is a useful sanity check in MCQs — our answer of 2.81 fits perfectly for $E_a = 80$ kJ/mol.

Alternative Method

We can use the $\log_{10}$ form, which some students find easier to compute without a natural log table:

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

\log\frac{k_2}{k_1} = \frac{80000}{2.303 \times 8.314} \times \frac{10}{300 \times 310}

= 4176 \times \frac{10}{93000} = 4176 \times 1.075 \times 10^{-4} = 0.449

\frac{k_2}{k_1} = 10^{0.449} \approx 2.81

Same answer. The $\log_{10}$ version is actually more JEE-friendly because log tables and anti-log calculations are faster in exam conditions than working with $e$ .

Common Mistake

Putting $E_a$ in kJ without converting to J.

The gas constant $R = 8.314$ J/mol·K. If you write $E_a = 80$ instead of $80000$ in the numerator, your exponent comes out 1000× too small — you get $k_2/k_1 \approx 1.0001$ and conclude temperature has almost no effect. Always match units: $E_a$ in J/mol when using $R$ in J/mol·K. This mistake appeared as a trap option in JEE Main 2024 with distractor answers near 1.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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