JEE Chemistry 2026 Question with Solution

Given: $A \rightarrow D$ is endothermic and occurs in three elementary steps:

• $A \rightarrow B$ with $\Delta H_i = +\text{ve}$
• $B \rightarrow C$ with $\Delta H_{ii} = -\text{ve}$
• $C \rightarrow D$ with $\Delta H_{iii} = -\text{ve}$

Find: The correct potential energy versus reaction coordinate graph.

For a multistep reaction, the number of elementary steps equals the number of transition states, so the graph must contain three peaks.

Since the overall reaction $A \rightarrow D$ is endothermic, the final energy of $D$ must be higher than the initial energy of $A$.

Now analyse each step:

• In step (i), $\Delta H_i = +\text{ve}$, so $B$ lies at a higher energy level than $A$.
• In step (ii), $\Delta H_{ii} = -\text{ve}$, so $C$ lies at a lower energy level than $B$.
• In step (iii), $\Delta H_{iii} = -\text{ve}$, so $D$ lies at a lower energy level than $C$.

Thus the intermediate energy pattern is:

$$A \uparrow B \downarrow C \downarrow D$$

with the additional condition that the final level of $D$ is still above that of $A$.

So the correct graph must show:

• three maxima,
• an initial rise from $A$ to $B$,
• then decreasing intermediate levels from $B$ to $C$ to $D$,
• and the final point higher than the starting point.

Among the given graphs, Graph 2 satisfies all these conditions.

Therefore, the correct option is B.

Given: A three-step endothermic reaction profile is to be identified.

Find: Which graph matches the signs of the step enthalpies and the overall enthalpy change.

Use these observations one by one:

1. Three elementary steps mean three activated complexes, so the profile must have three peaks.

2. The overall reaction is endothermic, so

$$E_D > E_A$$

where $E_A$ and $E_D$ are the energies of reactant $A$ and product $D$.

3. From

$$\Delta H_i > 0$$

we get

$$E_B > E_A$$

4. From

$$\Delta H_{ii} < 0$$

we get

$$E_C < E_B$$

5. From

$$\Delta H_{iii} < 0$$

we get

$$E_D < E_C$$

So the relative pattern must be such that the stable points follow an up, down, down trend from $A$ to $B$ to $C$ to $D$, while still keeping $D$ above $A$.

Graph 2 is the one that shows exactly this combination. Hence the correct option is B.