Decomposition of a hydrocarbon follows the equation $\mathrm{k}=\left(5.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{\frac{-28000 \mathrm{~K}}{\mathrm{~T}}}$. The activation energy of reaction is $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest Integer)

Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1 / 2}=3$ hour. The percentage of sucrose remaining after 6 hours is

$\_\_\_\_$ . (Nearest integer)

(Given $: \log 2=0.3010$ and $\log 3=0.4771$ )

For a reaction $\mathrm{A} \rightarrow \mathrm{P}$ at T K , the half life $\left(\mathrm{t}_{1 / 2}\right)$ is plotted as a function of initial concentration $[\mathrm{A}]_0$ of A as given below.

The value of $x$ in the given figure is $\_\_\_\_$ s (Nearest integer)

If the half life of a first order reaction is 6.93 minutes then the time required for completion of $99 \%$ of the reaction will be $\_\_\_\_$ minutes.

(Given $: \log 2=0.3010$ )