Activation Energy is 80 kJ/mol — How Does Temperature Affect Rate?

Question

The activation energy of a reaction is 80 kJ/mol. Calculate the ratio of rate constants $k_2/k_1$ when the temperature is raised from 300 K to 310 K. (R = 8.314 J mol⁻¹ K⁻¹)

This is a classic Arrhenius two-temperature problem — NCERT Chapter 4, and a reliable 3-marker in CBSE 12 boards. JEE Main has repeated this exact setup with different numbers.

Solution — Step by Step

Why This Works

The Arrhenius equation tells us that only molecules with energy $\geq E_a$ can react. At any temperature, the molecular energy follows a Maxwell-Boltzmann distribution — most molecules cluster near the average energy, with a tail extending to high values. Raising the temperature shifts and broadens this distribution, so a significantly larger fraction of molecules now exceed the activation energy threshold.

For a reaction with high $E_a$ (like 80 kJ/mol here), even a small temperature rise has a dramatic effect. The exponential relationship is the key — doubling $T$ doesn’t double the rate; it can increase it by orders of magnitude. This is why storing medicines in a refrigerator is not just about “keeping them cool” — the lower temperature slows degradation reactions that have significant activation energies.

The two-temperature formula works because the pre-exponential factor $A$ (which encodes collision frequency and geometry) is approximately constant over small temperature ranges. This is a standard assumption in NCERT-level problems.

Alternative Method — Using $\log_{10}$

Some students prefer working in base 10 to avoid the $\ln \to e$ conversion at the end:

Common Mistake

A second trap: writing $\frac{1}{T_2} - \frac{1}{T_1}$ instead of $\frac{1}{T_1} - \frac{1}{T_2}$. If you get a negative $\ln(k_2/k_1)$, that means $k_2 < k_1$ — rate decreased with temperature increase, which is physically wrong for a normal reaction. That’s your signal to flip the terms.