For a reaction,the graph of $\ln k$ (on y-axis) and $1 / T$ (on x-axis) is a straight line with a slope $-2 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$

The influence of temperature on the rate of reaction can be found out by

Activation energy for the acid catalysed hydration of sucrose is $6.22 \ kJ \ mol^{-1}$,while the activation energy is only $2.15 \ kJ \ mol^{-1}$ when hydrolysis is catalysed by the enzyme sucrase. Explain.

All energetically effective collisions do not result in a chemical change. Explain with the help of an example.

The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

According to the Arrhenius equation,the slope of the $\log k$ vs. $\frac{1}{T}$ plot is . . . . . . .