At $27^{\circ}C$ in the presence of a catalyst, the activation energy of a reaction is lowered by $10 \ kJ \ mol^{-1}$. The logarithm ratio of $\frac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is .... (Consider that the frequency factor for both the reactions is the same)

The decomposition of ethane,$\frac{d[C_2H_6]}{dt} = k[C_2H_6]$,proceeds through a complex mechanism,which includes $5$ steps. The overall rate constant $(k)$ is expressed as $k = \frac{k_1 k_2 k_3}{k_2 k_5}$,where $k_1, k_2, k_3, k_4, k_5$ are the rate constants of the $5$ steps. If the activation energies of the steps are $E_1 = 1E, E_2 = 2E, E_3 = 3E, E_4 = 4E, E_5 = 5E$,where $E = 20 \ kJ/mol$,find the overall activation energy of the decomposition.

The half-life of a first order reaction varies with temperature according to

The rate constant for a reaction can be increased by $\underline{a}$ the stability of the reactant or by $\underline{b}$ the stability of the transition state. Select the correct choice for $a$ and $b$.

For a reaction,the graph of $\ln k$ (on y-axis) and $1 / T$ (on x-axis) is a straight line with a slope $-2 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$

Write the Arrhenius equation representing the relationship between the rate constants $k_1$ and $k_2$ at two different temperatures $T_1$ and $T_2$.