For a first order reaction,the rate constant is $0.04 \, \text{min}^{-1}$ at $27 \, ^\circ\text{C}$ and $0.08 \, \text{min}^{-1}$ at $37 \, ^\circ\text{C}.$ The activation energy of reaction is .......... $\text{kcal} / \text{mol}$ $(\ln \, 2 = 0.7)$

Two reactions of the same order have equal Pre-exponential factors but their activation energies differ by $24.9 \ kJ/mol$. Calculate the ratio between the rate constants $\left( \frac{K_2}{K_1} \right)$ of these reactions at $27 \ ^\circ C$.

For reaction $A \to B$,the rate constant $k_1 = A_1 e^{-E_{a_1} / (RT)}$ and for the reaction $X \to Y$ the rate constant $k_2 = A_2 e^{-E_{a_2} / (RT)}$. If $A_1 = 10^8$,$A_2 = 10^{10}$ and $E_{a_1} = 600 \ cal/mol$,$E_{a_2} = 1800 \ cal/mol$,then the temperature at which $k_1 = k_2$ is (Given : $R = 2 \ cal/K \cdot mol$)

The energy profile diagram for a multi-step reaction,$A$ $\xrightarrow{1} B$ $\xrightarrow{2} C$ $\xrightarrow{3} D,$ is given below. The rate-determining step of the reaction is:

Catalyst $A$ reduces the activation energy for a reaction by $10 \ kJ \ mol^{-1}$ at $300 \ K$. The ratio of rate $\frac{k_{T, \text{Catalysed}}}{k_{T, \text{Uncatalysed}}}$ is $e^{x}$. Find the value of $x$ [nearest integer].
[Assume that the pre-exponential factor is same in both the cases.
Given $R = 8.31 \ J \ K^{-1} \ mol^{-1}$]

The temperature coefficient of a reaction is defined as: