# Why are reactions with higher activation energy more sensitive to temperature?

In the above picture it is stated clearly, with an example, that the more the Ea of a rxn the more it's sensitive to temperature, and hence even a small change in temperature in a reaction with high Ea, would have a drastic effect on the rate of rxn, as compared to a rxn with low Ea (about the probability factor and collision frequency, it's stated in the book as - "it must be understood that we are justified in doing this only when the reaction being compared are so closely related that differences in collision frequency and probability factor are comparatively insignificant" ).

My doubt is - If the energy-barrier (Ea) is high, doesn't it mean, (kind of) that, it's difficult to reach the intermediate stage, and hence a little change in temperature doesn't do good to the reaction as the barrier has still not been overcome and the reactants have a long way to go; and hence making the reaction more "temperature resistant" ? And on the other hand, if the energy-barrier is low, doesn't it mean that even a small change in temperature would be enough to overcome it and hence would increase the rate of reaction ?

After reading a bit more, I found that the statement is repeated again as - "the larger the activation energy of a reaction the larger the increase in the rate brought about by a given rise in temperature". WHY ?

My appeal to the answerers is that - Can you plz answer in what-is-actually-happening-on-the-backstage way and without ,if necessary, equations ?

Thank you very much and sorry for poor English.

• The fact will be quite obvious from the Arrhenius relationship applied at the rate constants at two different temperatures, which states that ,$$ln(\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})$$ – Soumik Das Nov 24 '18 at 14:54

The figure shows rate constants calculated with the activation energies $$E_a$$ shown (in kJ/mol). You can see how rapidly the rate constant increases with temperature when the activation energy is large, i.e. the slope is greater when the activation energy is larger, than when the activation energy is small. (The rate constant $$\displaystyle k=A_0e^{-E_a/(RT))}$$ are with $$A_0 =10,\; (E_a=10);\; A_0= 10^3,\;(E_a=20)$$ and $$A_0=10^6,\;(E_a=40)$$ with $$A_0$$ chosen only to get them on the same scale).
The gradient of the rate constant is $$\displaystyle \frac{dk}{dT}=E_a\frac{A_0e^{-E_a/(RT)}}{RT^2}$$ which shows that the slope depends on the activation energy.