The rate constant of a reaction is given by $k = P Z e^{-E_{a} / R T}$ under standard notation. In order to speed up the reaction,which of the following factors has to be decreased?

For a reversible reaction $A \rightleftharpoons B$,the $\Delta H_{\text{forward}} = 20 \ kJ \ mol^{-1}$. The activation energy of the uncatalysed forward reaction is $300 \ kJ \ mol^{-1}$. When the reaction is catalysed keeping the reactant concentration same,the rate of the catalysed forward reaction at $27^{\circ}C$ is found to be same as that of the uncatalysed reaction at $327^{\circ}C$. The activation energy of the catalysed backward reaction is $.... \ kJ \ mol^{-1}$.

The first order rate constant for the decomposition of ethyl iodide by the reaction $C_{2}H_{5}I_{(g)} \rightarrow C_{2}H_{4(g)} + HI_{(g)}$ at $600 \ K$ is $1.60 \times 10^{-5} \ s^{-1}$. Its energy of activation is $209 \ kJ/mol$. Calculate the rate constant of the reaction at $700 \ K$.

For an exothermic reaction $A \rightarrow B$,the activation energy of $A$ is $17 \, kJ/mol$. The heat of reaction is $40 \, kJ$. Calculate the activation energy for the reverse reaction $B \rightarrow A$.

The decomposition of formic acid on a gold surface follows first-order kinetics. If the rate constant at $300 \ K$ is $1.0 \times 10^{-3} \ s^{-1}$ and the activation energy $E_a = 11.488 \ kJ \ mol^{-1}$,the rate constant at $200 \ K$ is ............ $\times 10^{-5} \ s^{-1}$.
(Round off to the Nearest Integer).
(Given: $R = 8.314 \ J \ mol^{-1} K^{-1}$)

The activation energy for a simple chemical reaction $A \to B$ is ${E_a}$ in the forward direction. The activation energy for the reverse reaction: