For a first order reaction $(A \rightarrow B)$,the temperature $(T)$ dependent rate constant $(k)$ in $s^{-1}$ was found to follow the equation: $\log k = \left(-\frac{20}{T}\right)+4$. The activation energy $(E_a)$ and pre-exponential factor $(A)$ respectively,are

Slope of the straight line obtained by plotting $\log_{10} k$ against $\frac{1}{T}$ represents which term?

For a reversible reaction $A \rightleftharpoons B$,the $\Delta H_{\text{forward}} = 20 \ kJ \ mol^{-1}$. The activation energy of the uncatalysed forward reaction is $300 \ kJ \ mol^{-1}$. When the reaction is catalysed keeping the reactant concentration same,the rate of the catalysed forward reaction at $27^{\circ}C$ is found to be same as that of the uncatalysed reaction at $327^{\circ}C$. The activation energy of the catalysed backward reaction is $.... \ kJ \ mol^{-1}$.

Consider the following plots of log of rate constant $k$ $(\log k)$ vs $\frac{1}{T}$ for three different reactions. The correct order of activation energies of these reactions is

With an increase in temperature,the rate constant $k$:

The minimum energy required for molecules to participate in a chemical reaction is called $.......$.