# If an action increases the reaction rate in a particular equilibrium mixture, does it mean that both forward and backward rates are increased?

Say for example we increase the temperature in a closed system containing gaseous reactants. The forward reaction is exothermic so the yield of the product will decrease as we increase the temperature. But increasing the temperature increases the reaction rate so does it mean that the forward and backward rates are both increased ?

## 2 Answers

The forward and reverse reaction rates both increase, but the reverse rate increases by more. By Le Chatelier's principle, we know that $$K_{eq}$$ for an exothermic reaction will decrease when the temperature is raised. We also know that $$K_{\text{eq}}=\frac{k_{\text{forward}}}{k_{\text{reverse}}}$$ and that $$k=Ae^{\frac{-E_a}{RT}}$$ (The Arrhenius equation). Since this is an exothermic reaction, $$E_a(\text{forward}), which means any increase in $$T$$ will increase $$k_{\text{forward}}$$, but it will increase $$k_{\text{reverse}}$$ by more. This lead to a decrease in $$K_{\text{eq}}$$, consistent with what was predicted by Le Chatelier's principle.

Another action you could take to speed up a reaction is to add a catalyst. In this case, if you keep the temperature the same, the proportion of the two species will remain unchanged, and both forward and reverse reactions are accelerated by the same factor.