At $27^{\circ} \mathrm{C}$ in presence of a catalyst, activation energy of a reaction is lowered by $10 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The logarithm of ratio of $\frac{\mathrm{k} \text { (catalysed) }}{\mathrm{k} \text { (uncatalysed) }}$ is….

(Consider that the frequency factor for both the reactions is same)

Given above is the concentration vs time plot for a dissociation reaction : $\mathrm{A} \rightarrow \mathrm{nB}$.

Based on the data of the initial phase of the reaction (initial 10 min ), the value of n is $\_\_\_\_$ .

Observe the following reactions at $\mathrm{T}(\mathrm{K})$.

I. $\mathrm{A} \rightarrow$ products.

II. $5 \mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_3{ }^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 3 \mathrm{Br}_2(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$

Both the reactions are started at 10.00 am . The rates of these reactions at 10.10 am are same. The value of $-\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}$ at 10.10 am is $2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$. The concentration of A at 10.10 am is $10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$. What is the first order rate constant (in $\mathrm{min}^{-1}$ ) of reaction $I$ ?

Correct statements regarding Arrhenius equation among the following are :

A. Factor $e^{-\mathrm{Ea} / \mathrm{RT}}$ corresponds to fraction of molecules having kinetic energy less than Ea.

B. At a given temperature, lower the Ea, faster is the reaction.

C. Increase in temperature by about $10^{\circ} \mathrm{C}$ doubles the rate of reaction.

D. Plot of $\log \mathrm{k}$ vs $\frac{1}{\mathrm{~T}}$ gives a straight line with slope $=-\frac{\mathrm{Ea}}{\mathrm{R}}$.

Choose the correct answer from the options given below :