The rate of gas phase chemical reactions generally increases rapidly with a rise in temperature. This is mainly because

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 $kJ \, mol^{-1}$)

Based on the provided energy profile diagram,how many steps are in the mechanism of the reaction?

$A$ graph plotted between $\log \,K$ vs $\frac{1}{T}$ for calculating activation energy is shown by:

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:

For a first order decomposition of a certain reaction,rate constant is given by the equation $\log k \left( s^{-1} \right) = 7.14 - \frac{1 \times 10^4 \ K}{T}$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$ (in $.1$)