Reactant A converts to product D through the given mechanism (with the net evolution of heat) :
$$A \to B$$   slow ; $$\Delta H = +ve$$
$$B \to C$$   fast ; $$\Delta H = -ve$$
$$C \to D$$   fast ; $$\Delta H = -ve$$

Which of the following represents the above reaction mechanism ?

The energy profile of a multi-step reaction is determined by both the activation energy of each step and the enthalpy change associated with it.

The speed of an elementary step depends on its activation energy $$(E_a)$$. A slow step has a large activation energy and therefore corresponds to the highest peak on the energy profile. A fast step has a comparatively smaller activation energy and is represented by a lower peak.

The sign of $$\Delta H$$ determines the relative energy levels of the reactants and products of each step:

• If $$\Delta H > 0$$ (endothermic), the product of that step lies at a higher energy than the reactant.
• If $$\Delta H < 0$$ (exothermic), the product of that step lies at a lower energy than the reactant.

For the given reaction:

• $$A \rightarrow B$$ is slow with $$\Delta H > 0$$.
  • The first transition state must therefore be the highest peak on the diagram.
  • Intermediate $$B$$ must lie above reactant $$A$$ in energy.
• $$B \rightarrow C$$ is fast with $$\Delta H < 0$$.
  • The second peak should be lower than the first.
  • Intermediate $$C$$ must lie below $$B$$.
• $$C \rightarrow D$$ is fast with $$\Delta H < 0$$.
  • The third peak should also be relatively small.
  • Product $$D$$ must lie below $$C$$.

The problem also states that there is a net evolution of heat, meaning the overall reaction is exothermic. Hence, the final product $$D$$ must be at a lower energy level than the initial reactant $$A$$.

The first graph satisfies all these conditions:

• The first peak is the highest, representing the slow rate-determining step.
• Intermediate $$B$$ is at a higher energy than $$A$$, consistent with $$\Delta H > 0$$.
• The second and third peaks are lower, corresponding to the two fast steps.
• The energy decreases successively from $$B$$ to $$C$$ and from $$C$$ to $$D$$, matching the negative values of $$\Delta H$$ for the last two steps.
• The final product $$D$$ lies below $$A$$, indicating an overall exothermic reaction.

The second graph is incorrect because its third peak is higher than the first peak, implying that the $$C \rightarrow D$$ step has the largest activation energy and is the slowest step, which contradicts the given information.

Hence, the correct energy profile is Option (A).