JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Chemical Kinetics

The overall change $\mathrm{A}\to \mathrm{D}$ is endothermic, so the product $\mathrm{D}$ must lie above the reactant $\mathrm{A}$. Step by step, $\mathrm{A}\to \mathrm{B}$ with $\Delta H>0$ raises the energy to the highest intermediate $\mathrm{B}$; $\mathrm{B}\to \mathrm{C}$ with $\Delta H<0$ lowers it to $\mathrm{C}$; and $\mathrm{C}\to \mathrm{D}$ with $\Delta H<0$ lowers it further to $\mathrm{D}$. The correct profile shows three maxima, the intermediate minima decreasing in the order $\mathrm{B}>\mathrm{C}>\mathrm{D}$, with $\mathrm{A}$ the lowest point and $\mathrm{D}$ still above $\mathrm{A}$. The graph that reproduces this ordering is the required reaction profile.