For the reaction $A \rightarrow B$,the rate constant $k$ (in $s^{-1}$) is given by $\log_{10} k = 20.35 - \frac{2.47 \times 10^{3}}{T}$. The energy of activation in $kJ \, mol^{-1}$ is ..... . (Nearest integer) [Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$]

The following figure shows a graph of $\log_{10}K$ vs $\frac{1}{T}$,where $K$ is the rate constant and $T$ is the temperature. The straight line $BC$ has a slope,$\tan \theta = -\frac{1}{2.303}$,and an intercept of $5$ on the $Y$-axis. Thus,$E_a$,the energy of activation,is ....... $cal$.

What is the activation energy $(kJ \, mol^{-1})$ for a reaction if its rate constant doubles when the temperature is raised from $300 \, K$ to $400 \, K$ ? $(R = 8.314 \, J \, mol^{-1} \, K^{-1})$

The rate constant of a reaction is $2 \times 10^{-3} \ min^{-1}$ at $300 \ K$. By increasing the temperature by $10 \ K$,its value becomes double. Calculate the energy of activation $(E_a)$ and the rate constant at $320 \ K$.

What is the slope of the graph between $\ln K$ and $\frac{1}{T}$ according to the Arrhenius equation?

The rate of a reaction quadruples when the temperature changes from $300 \, K$ to $310 \, K$. The activation energy of this reaction is ........... $kJ \, mol^{-1}$ (Assume activation energy and pre-exponential factor are independent of temperature; $\ln 2 = 0.693$; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$)