$A$ reaction having equal activation energies for forward and reverse reaction has:

The activation energy of a chemical reaction can be determined by:

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 $(\log 5 = 0.70)$

$A$ plot of $ \frac{1}{T} $ vs. $ \ln k $ for a reaction gives the slope $ -1 \times 10^{4} \ K $. The energy of activation for the reaction is (Given $ R = 8.314 \ J \ K^{-1} \ mol^{-1} $)

Explain the effect of an increase in temperature on the rate of reaction and the rate constant.

$A \rightarrow B$. 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). $[R = 8.314 \ J \ K^{-1} \ mol^{-1}]$