The rate constant $(k)$ of a reaction is measured at different temperatures $(T),$ and the data are plotted in the given figure. The activation energy of the reaction in $kJ\, mol^{-1}$ is :
($R$ is gas constant)

If the definition of the temperature coefficient of the reaction holds good for a reaction between $27^{\circ} C$ and $37^{\circ} C$,the activation energy for the reaction in $kJ \cdot mol^{-1}$ is

The rate of the chemical reaction doubles for an increase of $10 \text{ K}$ in absolute temperature from $298 \text{ K}$. What will be the activation energy?

The activation energy of a forward reaction is $50 \, kcal$. What will be the activation energy of its reverse reaction?

The rate constant for a first order reaction is given by the following equation:
$\ln k = 33.24 - \frac{2.0 \times 10^{4} \, K}{T}$
The activation energy for the reaction is given by $... \, kJ \, mol^{-1}$. (In nearest integer)
(Given: $R = 8.3 \, J \, K^{-1} \, mol^{-1}$)

Subtract $(i)$ $\ln \, k_1 = - \frac{E_a}{R T_1} + \ln A$ and $(ii)$ $\ln \, k_2 = - \frac{E_a}{R T_2} + \ln A$ and write the resulting equation.