An endothermic reaction $A \to B$ has an activation energy of $15 \ kcal/mol$ and an enthalpy change of reaction of $5 \ kcal/mol$. The activation energy of the reverse reaction $B \to A$ is ......... $kcal/mol$.

$A \rightarrow B$. The molecule $A$ changes into its isomeric form $B$ following first-order kinetics at a temperature of $1000 \ K$. If the energy barrier with respect to reactant energy for such isomeric transformation is $191.48 \ kJ \ mol^{-1}$ and the frequency factor is $10^{20} \ s^{-1}$,the time required for $50 \%$ of molecules of $A$ to become $B$ is $..............$ picoseconds (nearest integer). $[R = 8.314 \ J \ K^{-1} \ mol^{-1}]$

At $25\,^{\circ}C$,the values of rate constant,activation energy and Arrhenius constant of a reaction are $3 \times 10^{-4}\, s^{-1}$,$129\, kJ/mol$ and $2 \times 10^{15}\, s^{-1}$ respectively. The value of rate constant as $T \to \infty$ is

The half-life periods of a first order reaction at $300 \ K$ and $400 \ K$ are $50 \ s$ and $10 \ s$ respectively. The activation energy of the reaction in $kJ \ mol^{-1}$ is $(\log 5 = 0.70)$

The decomposition of a hydrocarbon follows the equation $K = (4.5 \times 10^{11} \, s^{-1}) e^{-28000 \, K/T}$. What will be the value of activation energy in $KJ \, mol^{-1}$?

Consider the following graph of the kinetic energy distribution among molecules at temperature $T.$ If the temperature is increased,how would the resulting graph differ from the one above $:-$