According to the Arrhenius equation, the rate of reaction increases when the temperature increases. But according to Le Chatelier's principle, the reaction rate decreases when the temperature increases.

So is Arrhenius equation applicable on exothermic reactions?

  • 2
    $\begingroup$ It would then be the case that although both the forward and backward rates increase, the rate of the backward reaction increases more than the rate of the forward reaction for the exothermic reaction. Thus, both physical postulates are applicable. $\endgroup$ – Tan Yong Boon Feb 10 '18 at 8:53
  • $\begingroup$ If you keep the temperature of the reaction mixture constant, the rate of the reaction (whether endothermic or exothermic) is higher at a constant higher temperature than at a constant lower temperature. $\endgroup$ – Chet Miller Feb 10 '18 at 13:30
  • $\begingroup$ An important distinction is that Le Chatelier's principle indicates thermodynamic behavior while the Arrenius equation is a matter of kinetics. If you run a hydrogenation reaction for example, increasing the temperature has an undesireable effect on maximum yield following Le Chatelier's principle, but you are also getting that yield much quicker, so it often pays off. $\endgroup$ – Vinícius Godim Feb 12 '18 at 3:57

The Arrhenius equation is absolutely applicable to exothermic reactions and it will give results in accordance to Le Chatelier's principle.
Applying the Arrhenius equation for the forward reaction yields: $$k_\mathrm{f} = A \exp\left(\frac{-E_\mathrm{f}}{RT}\right),$$ applying the Arrhenius equation to backward reaction yields $$k_\mathrm{b} = A \exp\left(\frac{-E_\mathrm{b}}{RT}\right).$$

We know that $E_\mathrm{f} -E_\mathrm{b} = \Delta H$. Dividing $k_\mathrm{f}$ by $k_\mathrm{b}$ we obtain: $$\frac{k_\mathrm{f}}{k_b} = A \exp\left(\frac{-\Delta H}{RT}\right).$$ For an exothermic reaction $\Delta H < 0$. So $-\Delta H/(RT)$ is a positive number. As $T$ increases, the exponent decreases and thus $k_\mathrm{f}/k_\mathrm{b}$ decreases, which is nothing but the decrement of the equilibrium constant.
Thus the result is in perfect accordance with Le Chatelier's principle.

Also, you can apply the van't Hoff's equation which states: $$\ln\left(\frac{k_2}{k_1}\right) = \frac{\Delta H}{R}\cdot\left(\frac 1{T_1} - \frac 1{T_2}\right),$$ which also gives the decrease of rate of reaction with increase in temperature as $\Delta H < 0$.

  • $\begingroup$ You've assumed that the frequency factor is the same ($=A$) for both the forward and backward reactions. What is the justification for doing so? Afaik, it depicts the rate of collision of molecules with each other, and since the reactant side and the product side have different molecules, they should have different $A$ values. Thanks! $\endgroup$ – Gaurang Tandon Feb 13 '18 at 4:54

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