Reactant A converts to product D through the | vedclass

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(A) The reaction mechanism involves three steps:$1$. $A \rightarrow B$ is a slow step with $\Delta H = +ve$ (endothermic),meaning the energy of $B$ is

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(A) The reaction mechanism involves three steps:$1$. $A \rightarrow B$ is a slow step with $\Delta H = +ve$ (endothermic),meaning the energy of $B$ is higher than $A$,and it has a high activation energy $(E_{a_1})$.$2$. $B \rightarrow C$ is a fast step with $\Delta H = -ve$ (exothermic),meaning the energy of $C$ is lower than $B$,and it has a low activation energy $(E_{a_2})$.$3$. $C \rightarrow D$ is a fast step with $\Delta H = -ve$ (exothermic),meaning the energy of $D$ is lower than $C$,and it has a low activation energy $(E_{a_3})$.The overall reaction is exothermic $(\Delta H_{net} = -ve)$,so the final product $D$ must be at a lower energy level than the reactant $A$.The graph that correctly depicts these energy changes is shown in the solution image.

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 ?

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

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 ?