At what temperature does the rate become double that at $300 \, K$ (in $, K$)? Given $\ln k = 10 - \frac{69 \, kJ}{RT}$.

The decomposition of a hydrocarbon follows the equation $k = (4.5 \times 10^{11} \ s^{-1}) e^{-28000 \ K / T}$. Calculate the activation energy $E_a$.

The rate constant is given by the equation $k = p Z e^{-E_a/RT}$. Which factor should register a decrease for the reaction to proceed more rapidly?

For a first order reaction,a plot of $\ln k$ ($y$-axis) and $\frac{1}{T}$ ($x$-axis) gave a straight line with a slope equal to $-10^3 \ K$ and an intercept equal to $2.303$ (on $y$-axis). What is the activation energy ($E_a$ in $kJ \ mol^{-1}$) of the reaction? (Given $R = 8.314 \ J \ mol^{-1} \ K^{-1}$)

The Arrhenius plots of two reactions,$I$ and $II$ are shown graphically. The graph suggests that

In a first order reaction at $27\,^oC$ and $47\,^oC$,$50\%$ of the reaction is complete in $30\,\min$ and $10\,\min$ respectively. Calculate the energy of activation $(E_a)$.