The graph obtained between $\ln k$ ( $k=$ rate constant) on $y$-axis and $1 / T$ on $x$-axis is a straight line. The slope of it is $-4 \times 10^4 \mathrm{~K}$. The activation energy of the reaction (in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$ is $\left(R=831 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$

For a reaction, the threshold energy is $75 \mathrm{~kJ} /$ mole. If the internal energy of the reactants is 20 $\mathrm{kJ} /$ mole, the activation energy (in $\mathrm{kJ} /$ mole) is

145. Rate constants in the following reaction are Reaction 1 :

$$ A \xrightarrow{\text { Catalyst } 1} P_1, k_1=1 \mathrm{~s}^{-1} $$

Reaction 2 :

$$ A \xrightarrow{\text { Catalyst } 2} P_2, k_2=0.1 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1} $$

Reaction 3 :

$$ A \xrightarrow{\text { Catalyst } 3} P_3, k_3=0.01 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1} $$

The correct relations between the rate of the reactions at 1 M of $A$ are