The rate constant of a reaction increases by five times on increase in temperature from $27^{\circ} C$ to $52^{\circ} C$. The value of activation energy in $kJ \, mol^{-1}$ is $....$ . (Rounded-off to the nearest integer) $[R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

Two reactions of the same order have equal Pre-exponential factors but their activation energies differ by $24.9 \ kJ/mol$. Calculate the ratio between the rate constants $\left( \frac{K_2}{K_1} \right)$ of these reactions at $27 \ ^\circ C$.

The rate coefficient $(k)$ for a particular reaction is $1.3 \times 10^{-4} \ M^{-1} \ s^{-1}$ at $100 \ ^oC$ and $1.3 \times 10^{-3} \ M^{-1} \ s^{-1}$ at $150 \ ^oC$. What is the energy of activation $(E_a)$ (in $kJ \ mol^{-1}$) for this reaction? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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

The rate of a chemical reaction doubles with every $10^{\circ}C$ rise in temperature. If the reaction is carried out in the vicinity of $22^{\circ}C$,the activation energy of the reaction is (Given $R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$\ln 2 = 0.69$ and $\ln 3 = 1.1$)

In the plot of $\ln k$ versus $\frac{1}{T}$,what do the slope and intercept represent?