The temperature-dependent equation for the rate constant is written as:

For the equilibrium $A_{(g)} \rightleftharpoons B_{(g)}$,$\Delta H$ is $-40 \ kJ/mol$. If the ratio of the activation energies of the forward $(E_f)$ and reverse $(E_b)$ reactions is $\frac{2}{3}$,then:

The minimum energy necessary to permit a reaction is

The following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

At what temperature does the rate become double that at $300 \, K$ (in $, K$)? Given $\ln k = 10 - \frac{69 \, kJ}{RT}$.

Given the activation energy for the forward direction $(E_{af})$ is $50 \ kJ$ and for the backward direction $(E_{ab})$ is $30 \ kJ$. The reaction would be: