The equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$ is followed for the decomposition of compound $A$. The activation energy for the reaction is $..... \, kJ \, mol^{-1}$. [nearest integer] (Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$)

The energies of activation for forward and reverse reactions for $A_2 + B_2 \rightleftharpoons 2AB$ are $180 \, kJ \, mol^{-1}$ and $200 \, kJ \, mol^{-1}$ respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100 \, kJ \, mol^{-1}.$ The enthalpy change of the reaction $(A_2 + B_2 \rightarrow 2AB)$ in the presence of a catalyst will be (in $kJ \, mol^{-1}$)

$A$ plot of $ \frac{1}{T} $ vs. $ \ln k $ for a reaction gives the slope $ -1 \times 10^{4} \ K $. The energy of activation for the reaction is (Given $ R = 8.314 \ J \ K^{-1} \ mol^{-1} $)

$A$ reaction rate constant is given by $k = 1.2 \times 10^{14} \, e^{-25000/RT} \, sec^{-1}$. It means

Write the logarithmic form of the Arrhenius equation $k = A e^{-\frac{E_a}{RT}}$.

Which of the following statements is correct regarding the Arrhenius equation and reaction kinetics?