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JEE Mains Chemistry · Chemical Kinetics

Mechanisms, Energy Profiles and Catalysts

The slow step of a mechanism writes the rate law once its intermediates are removed; an energy profile shows each step's barrier, the intermediates and ΔH, and a catalyst lowers the barriers without moving the start or the end.

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Why this matters

Fifteen PYQs, eight of them multiple choice, and one from 2026. Six take a mechanism apart — the order from a slow step after a fast equilibrium, the steady state, or the overall activation energy from a composite rate constant; nine read energy profiles for barriers, intermediates, the rate-determining step, ΔH and what a catalyst changes. Two ideas cover the page.

Concept 1 of 2: Rate law and activation energy from a mechanism

A reaction can go no faster than its slowest step, so that step writes the rate law. It often contains an intermediate, which cannot appear in the final answer. Replace it using the fast equilibrium before the slow step, or by setting its net rate of formation to zero (the steady state). When the overall kk is a product or ratio of step constants, the activation energies add and subtract the same way.

Definition

  • Rate law = the slow step's rate law, with its molecularity as the exponents.
  • Fast pre-equilibrium A2⇌2A\mathrm{A_2 \rightleftharpoons 2A}: K=[A]2[A2]K = \dfrac{[A]^2}{[A_2]}, so [A]=(K[A2])1/2[A] = (K[A_2])^{1/2}.
  • Steady state for A→k1B→k2CA \xrightarrow{k_1} B \xrightarrow{k_2} C: d[B]dt=k1[A]−k2[B]=0\dfrac{d[B]}{dt} = k_1[A] - k_2[B] = 0, so [B]=k1k2[A][B] = \dfrac{k_1}{k_2}[A].
  • Composite constant k=k1k2k3k = \dfrac{k_1k_2}{k_3}: Ea=Ea1+Ea2−Ea3E_a = E_{a1} + E_{a2} - E_{a3}.
  • A power carries over: k=k1k = \sqrt{k_1} gives Ea=12Ea1E_a = \tfrac12E_{a1}.

Composite activation energy

k=k1k2k3 ⇒ Ea=Ea1+Ea2−Ea3k = \frac{k_1k_2}{k_3}\ \Rightarrow\ E_a = E_{a1} + E_{a2} - E_{a3}

Worked example

The reaction 2NO+O2→2NO2\mathrm{2NO + O_2 \rightarrow 2NO_2} follows the mechanism 2NO⇌N2O2\mathrm{2NO \rightleftharpoons N_2O_2} (fast), N2O2+O2→2NO2\mathrm{N_2O_2 + O_2 \rightarrow 2NO_2} (slow). Find the rate law and the overall order.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q39Moderate

Example 1 · Chemical Kinetics · Mechanisms, Energy Profiles and Catalysts

The reaction A2+B2→2ABA_{2}+B_{2}\rightarrow 2AB follows the mechanism
A2⇌k1k−1 A+A( fast )A+B2→K2AB+B (slow) A+B→AB (fast) \begin{matrix} & A_{2}\underset{k_{- 1}}{\overset{k_{1}}{\rightleftharpoons}}\text{ }A + A(\text{~fast~}) \\ & A +B_{2}\overset{K_{2}}{\rightarrow}AB + B\text{~(slow)~} \\ & A + B \rightarrow AB\text{~(fast)~} \end{matrix}
The overall order of the reaction is :

An intermediate left in the rate law

The final rate law contains only species in the overall equation (and any catalyst). Replace an intermediate using the fast equilibrium or the steady state before counting the order.

Rate law written from the overall equation

The exponents come from the SLOW step, not from the overall stoichiometry. 2NO+Br2→2NOBr\mathrm{2NO + Br_2 \rightarrow 2NOBr} is third order because of its mechanism, not because three molecules appear on the left.
Drill 5 more on rate law and activation energy from a mechanism

Concept 2 of 2: Reading energy profiles and the effect of a catalyst

An energy profile follows the energy along the reaction path. Each hump is an activated complex; each valley between humps is an intermediate. The climb from a step's start to its hump is that step's barrier. The start and the end fix ΔH\Delta H, so a catalyst — which only offers a lower path — cannot change it.

Definition

  • Humps = activated complexes (one per step); valleys between humps = intermediates.
  • The step with the largest barrier, measured from its own starting level, is the slow, rate-determining step.
  • ΔH=Ea,f−Ea,b\Delta H = E_{a,f} - E_{a,b}. Products above reactants: endothermic, products less stable. Products below: exothermic.
  • A catalyst gives a new path with a lower hump. It lowers Ea,fE_{a,f} and Ea,bE_{a,b} by the SAME amount, and leaves ΔH\Delta H, ΔG\Delta G and KK unchanged; the catalysed curve starts and ends at the same levels.
  • A catalyst cannot make a non-spontaneous reaction happen; it only speeds up both directions.

Barriers and enthalpy

ΔH=Ea,f−Ea,b\Delta H = E_{a,f} - E_{a,b}

Worked example

An exothermic reaction has ΔH=−35\Delta H = -35 kJ mol−1^{-1} and Ea,f=60E_{a,f} = 60 kJ mol−1^{-1}. Find Ea,bE_{a,b}. A catalyst then lowers Ea,fE_{a,f} to 42 kJ mol−1^{-1}; find the new Ea,bE_{a,b}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q104Moderate

Example 2 · Chemical Kinetics · Mechanisms, Energy Profiles and Catalysts

Consider the given figure and choose the correct option :

A catalyst that changes ΔH

A catalyst lowers both barriers by the same amount, so ΔH=Ea,f−Ea,b\Delta H = E_{a,f} - E_{a,b} does not change; nor do ΔG\Delta G and KK. It cannot make a non-spontaneous reaction occur.

The sign of ΔH flipped

ΔH=Ea,f−Ea,b\Delta H = E_{a,f} - E_{a,b}, forward minus backward. With Ea,f=120E_{a,f} = 120 and Ea,b=150E_{a,b} = 150 kJ mol−1^{-1}, ΔH=−30\Delta H = -30 kJ mol−1^{-1}: exothermic, not +30+30.

Valleys counted as activated complexes

Peaks are activated complexes; the valleys between them are intermediates. The start and end levels are the reactants and products and are neither.
Drill 8 more on reading energy profiles and the effect of a catalyst

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Watch out for (5)

Test yourself on Chemical Kinetics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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