Calculate the rate constant of a first-order reaction having a pre-exponential factor $A = 1.6 \times 10^{13} \ s^{-1}$ and $\frac{E_a}{2.303 RT} = 21$.

Which of the following statements is correct for the activation energy of a reaction?

What is the activation energy for the reverse of this reaction?
$N_2O_{4(g)} \to 2NO_{2(g)}$
Data for the given reaction is:
$\Delta H = +54 \ kJ$ and $E_a = +57.2 \ kJ$
Answer in $kJ$.

For reaction $A \to B$,rate constant $K_1 = A_1 e^{-E_{a_1}/RT}$ and for the reaction $X \to Y$,rate constant $K_2 = A_2 e^{-E_{a_2}/RT}$. If $A_1 = 10^8, A_2 = 10^{10}$ and $E_{a_1} = 600 \ cal \ mol^{-1}$,$E_{a_2} = 1800 \ cal \ mol^{-1}$,then the temperature at which $K_1 = K_2$ is (given: $R = 2 \ cal \ K^{-1} \ mol^{-1}$):

For an exothermic reaction,where the enthalpy of reaction is $\Delta H$ in $kJ/mol$,the minimum value for the activation energy will be:

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: