$A$ 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} $)

How does the graph of the fraction of molecules versus kinetic energy change at different temperatures?

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 a reaction,the threshold energy is $75 \ kJ/mol$. If the internal energy of the reactants is $20 \ kJ/mol$,the activation energy (in $kJ/mol$) is

In the presence of a catalyst,the heat evolved or absorbed during the reaction . . . . . . .

What will be the effect on the fraction of molecules and the rate constant on increasing temperature?