The activation energy for a reaction which doubles the rate when the temperature is raised from $298 \ K$ to $308 \ K$ is ........... $kJ \ mol^{-1}$

The time required for $10 \%$ completion of a first order reaction at $298 \,K$ is equal to that required for its $25 \%$ completion at $308 \,K$. If the value of $A$ is $4 \times 10^{10} \,s^{-1}$,calculate $k$ at $318 \,K$ and $E_a$.

For a chemical reaction at $27^{\circ} C$,the activation energy is $600 R$. The ratio of the rate constants at $327^{\circ} C$ to that of at $27^{\circ} C$ will be

The rate constant $(K')$ of one reaction is double the rate constant $(K'')$ of another reaction. What is the relationship between the corresponding activation energies of the two reactions (${E_a}'$ and ${E_a}''$)?

The plot of $\log K$ vs $\frac{1}{T}$ helps to calculate

How does the graph of the fraction of molecules versus kinetic energy change at different temperatures?