A rightarrow D is an endothermic reaction occ | vedclass

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(A) Given:$A ightarrow D$; $\Delta_{t} H = (+ve) = E_{D} - E_{A} 	herefore E_{D} > E_{A}$.Mechanism:$A ightarrow B$; $\Delta_{r} H = (+ve) \Right

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(A) Given:$A ightarrow D$; $\Delta_{t} H = (+ve) = E_{D} - E_{A} 	herefore E_{D} > E_{A}$.Mechanism:$A ightarrow B$; $\Delta_{r} H = (+ve) \Rightarrow E_{B} > E_{A}$$B ightarrow C$; $\Delta_{r} H = (-ve) \Rightarrow E_{C} $C ightarrow D$; $\Delta_{r} H = (-ve) \Rightarrow E_{D} Since the overall reaction is endothermic, the final energy level of $D$ must be higher than the initial energy level of $A$. However, the intermediate steps involve exothermic processes. Based on the energy relations $E_{B} > E_{A}$, $E_{C}  E_{A}$ overall.

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