Morrison and Boyd Organic chemistry

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.

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

The first thing to note is that the rate constant with a large activation barrier will be far smaller than one with a small barrier at a given temperature. If the barrier is very small (e.g. 5 kJ/mol) at, say, room temperature, almost all collisions between the reactive molecules have sufficient energy to react and the barrier is crossed and this leads to a fast reaction. This means that increasing the temperature cannot lead to much of an increase, because the reaction is going almost as fast as it can at the lower temperature.

Now consider the case with a large barrier. In this case the rate constant is far smaller than when the barrier is small, and very few collisions have enough energy to cross the barrier. Now increase the temperature and the number of filled energy levels the molecule has increased greatly, exponentially so by the Boltzmann distribution. Now very many more collisions have sufficient energy to react and so the rate constant increases greatly, although it may still be small in value.

If you plot the Arrhenius equation (given in your text) vs temperature the slope of this gives the rate of change with temperature. With larger activation energies the slope gets larger. This is the effect also described in your text.

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.


  • $\begingroup$ So as not to confuse the students, the authors have taken high temperatures (250-300 °C) and not temperatures like room temperatures, and then they have shown the variations of rates in THAT high temperature range. But you have written the first part of your answer about low range of temperatures (which kind of seems like an exception). Can you plz explain what would happen if the temp. were high enough as shown in the text ? Plz note that the authors have varied the Ea in the same range of temperatures. Thanks. $\endgroup$ – DEEKSHANT Nov 24 '18 at 18:50
  • $\begingroup$ The explanation is just the same whatever the temperature range: the physics does not change. If you re-calculate my figure and expand it in the 200-400 range you will see the same sort of figure as above. $\endgroup$ – porphyrin Nov 25 '18 at 9:42

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