$A$ rise in temperature increases the velocity of a reaction. It is because it results in

For the reaction $A_2 + B_2 \rightleftarrows 2AB$,the activation energy for the forward and backward reactions are $180 \ kJ \ mol^{-1}$ and $200 \ kJ \ mol^{-1}$,respectively. If a catalyst lowers the activation energy of both forward and backward reactions by $100 \ kJ \ mol^{-1}$,what will be the enthalpy change $(\Delta H)$ for the reaction $A_2 + B_2 \rightarrow 2AB$ in the presence of the catalyst?

The decomposition of $A$ into product has a value of $k$ as $4.5 \times 10^{3} \, s^{-1}$ at $10^{\circ} C$ and an energy of activation of $60 \, kJ \, mol^{-1}$. At what temperature would $k$ be $1.5 \times 10^{4} \, s^{-1}$?

Consider $A \xrightarrow{k_1} B$ and $C \xrightarrow{k_2} D$ are two reactions. If the rate constant $(k_1)$ of the $A \rightarrow B$ reaction can be expressed by the following equation $\log_{10} k = 14.34 - \frac{1.5 \times 10^4}{T/K}$ and activation energy of $C \rightarrow D$ reaction $(Ea_2)$ is $\frac{1}{5}$th of the $A \rightarrow B$ reaction $(Ea_1)$, then the value of $(Ea_2)$ is . . . . . . $kJ \ mol^{-1}$. (Nearest Integer)

In collision theory of chemical reaction,$Z_{AB}$ represents

The following energy profile diagram is given for the reaction $A + B \to C + D$. The enthalpy change and activation energy for the reverse reaction $C + D \to A + B$ are respectively: