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

Two reactions $R_1$ and $R_2$ have identical pre-exponential factors. Activation energy of $R_1$ exceeds that of $R_2$ by $10 \, kJ \, mol^{-1}.$ If $k_1$ and $k_2$ are rate constants for reactions $R_1$ and $R_2$ respectively at $300 \, K,$ then $\ln (k_2/k_1)$ is equal to :
$(R=8.314 \, J \, mol^{-1} \, K^{-1})$

The temperature-dependent equation for the rate constant is written as:

For a first-order reaction $A \to P$,the rate constant equation is given by $\log K = -2000 \, (1/T) + 6.0$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively:

The velocity constant of a reaction at $290 \ K$ was found to be $3.2 \times 10^{-3}$. At $310 \ K$,it will be about:

For an elementary chemical reaction, the Arrhenius plot is given below. If the energy of activation is $6.64 \ kJ \ mol^{-1}$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$, the temperature at which the rate constant becomes $e^2 \ min^{-1}$, is (in $K$)