For an exothermic reaction $A \rightarrow B$,the activation energy of $A$ is $17 \, kJ/mol$. The heat of reaction is $40 \, kJ$. Calculate the activation energy for the reverse reaction $B \rightarrow A$.

The relation between the fraction of molecules $\left( \frac{N_E}{N_T} \right)$ and temperature is explained by the Boltzmann and Maxwell distribution graph.

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 $(A_2 + B_2 \to 2AB)$ in the presence of catalyst will be (in $kJ \ mol^{-1}$)

The rate constant of a reaction depends on

At $527 \, ^{\circ}C$ temperature,the activation energy is $54.7 \, kJ/mol$. The value of the Arrhenius factor is $4 \times 10^{10}$. The rate constant will be:

For an exothermic reaction,the following two steps are involved:
Step $1$: $A + B \to I$ $(slow)$
Step $2$: $I \to AB$ $(fast)$
Which of the following graphs correctly represents this reaction?