If an action increases the reaction rate in a particular equilibrium mixture, does it mean that both forward and backward rates are increased?
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, $E_a(\text{forward})&lt;E_a(\text{reverse})$, 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.