For the reaction $X \rightleftharpoons Y$,the pre-exponential factors for the forward and backward reactions are equal. What will be the equilibrium constant of the reaction?

What is Activation Energy $(E_a)$? Explain the reaction profile graph for activation energy and discuss the probability of effective collisions.

For a reversible reaction $A \rightleftharpoons B$,the $\Delta H_{\text{forward}} = 20 \ kJ \ mol^{-1}$. The activation energy of the uncatalysed forward reaction is $300 \ kJ \ mol^{-1}$. When the reaction is catalysed keeping the reactant concentration same,the rate of the catalysed forward reaction at $27^{\circ}C$ is found to be same as that of the uncatalysed reaction at $327^{\circ}C$. The activation energy of the catalysed backward reaction is $.... \ kJ \ mol^{-1}$.

For the reaction,$A \rightleftharpoons B$,$E_a = 50 \ kJ \ mol^{-1}$ and $\Delta H = -20 \ kJ \ mol^{-1}$. When a catalyst is added,$E_a$ decreases by $10 \ kJ \ mol^{-1}$. What is the $E_a$ for the backward reaction in the presence of the catalyst?

In the Arrhenius equation,$k = A e^{-E_a/RT}$,the rate constant $k$ becomes equal to the frequency factor $A$ when:

From the given data for the reaction $H_2 + I_2 \rightarrow 2HI$,calculate the activation energy $(E_a)$:
$T_1 = 769 \ K, \ 1/T_1 = 1.3 \times 10^{-3} \ K^{-1}, \ \log_{10} K_1 = 2.9$
$T_2 = 667 \ K, \ 1/T_2 = 1.5 \times 10^{-3} \ K^{-1}, \ \log_{10} K_2 = 1.1$