In a reaction,for every $10\,^{\circ}C$ rise of temperature,the rate is doubled. If the temperature is increased from $10\,^{\circ}C$ to $100\,^{\circ}C,$ the rate of the reaction will become $.......$ times.

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)

The equilibrium constant at $27\,^{\circ}C$ and $127\,^{\circ}C$ is $30$ and $40$ respectively. What will be the ratio of activation energy $(E_{a})_{f} / (E_{a})_{b}$?

For an exothermic reaction $A \rightarrow B$,the activation energy is $15 \, K \, cal/mol$ and the heat of reaction is $5 \, K \, cal/mol$. The activation energy for the reverse reaction $B \rightarrow A$ will be ......... $K \, cal/mol$.

For a reversible chemical reaction where the forward process is exothermic, which of the following statements is correct?

On introducing a catalyst at $500 \, K,$ the rate constant of a first order reaction increases $2.718$ times. If the activation energy in the presence of a catalyst is $4.15 \, kJ \, mol^{-1},$ then what will be $E_a$ in the absence of a catalyst? (Value of $e = 2.718$ and $R = 8.314 \times 10^{-3} \, kJ \, K^{-1} \, mol^{-1}$)