Reactant $A$ converts to product $D$ through the given mechanism (with the net evolution of heat) :

$A \rightarrow B$	$slow ; \Delta H=+ve$
$B \rightarrow C$	$fast ; \Delta H=-ve$
$C \rightarrow D$	$fast ; \Delta H=-ve$

Which of the following represents the above reaction mechanism ?

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 $.... \ kJ \ mol^{-1}$.

According to the Arrhenius equation, which of the following statements is correct?

In a reaction, for every $10^{\circ} C$ rise of temperature, the rate is doubled. If the temperature is increased from $10^{\circ} C$ to $100^{\circ} C$, the rate of the reaction will become $:-$ (in $\times$)

For a reaction,$A \rightarrow B$,the average energies of $A$ and $B$ are $30 \ kcal/mol$ and $60 \ kcal/mol$ respectively. The energy of activation for the backward reaction is $93 \ kcal/mol$. The energy of activation for the forward reaction is:

An exothermic reaction $X \rightarrow Y$ has an activation energy of $30 \ kJ \ mol^{-1}$. If the energy change $\Delta E$ during the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction in $kJ \ mol^{-1}$ is $...$.