Chemical Kinetics JEE 2024: Why the Overall Order Is 3

What is the answer to the Chemical Kinetics JEE 2024 mechanism question?

The overall order in this chemical kinetics JEE 2024 question is 3, not 2: eliminate the intermediate before adding the concentration powers. The source is JEE Advanced 2024, Paper 1, Chemistry. It is a numerical-answer question, not an MCQ.

The task is to determine the overall order of this reaction using the supplied mechanism:

2 H_{2} (g) + 2 N O (g) \to N_{2} (g) + 2 H_{2} O (g)

The three elementary steps are:

Fast equilibrium, with the forward and reverse rate constants shown:

2 N O (g) \xrightleftharpoons [k_{- 1}] k_{1} N_{2} O_{2} (g)

Slow reaction:

N_{2} O_{2} (g) + H_{2} (g) \overset{k_{2}}{\to} N_{2} O (g) + H_{2} O (g)

Fast reaction:

N_{2} O (g) + H_{2} (g) \overset{k_{3}}{\to} N_{2} (g) + H_{2} O (g)

The question bank tags this problem as medium, with an expected solving time of 90 seconds. These are question-bank tags, not official exam classifications or a guaranteed solving time. The working below follows the official solution’s fast-equilibrium method.

How do you write the rate law for the slow step?

For the supplied mechanism, the second elementary step is the rate-determining step. Start with its rate law, as in the official solution:

N_{2} O_{2} (g) + H_{2} (g) \overset{k_{2}}{\to} N_{2} O (g) + H_{2} O (g)

r = k_{2} [N_{2} O_{2}] [H_{2}]

Each reacting species has concentration power one because this is an elementary step involving one molecule of each. These powers do not come from the coefficients in the overall balanced equation.

The nitrogen oxide dimer is an intermediate: it is formed in the first step and consumed in the second. The derivation is unfinished because the final reactant-based rate law must eliminate its concentration.

How does the fast equilibrium eliminate the intermediate?

The first step expresses the intermediate concentration in terms of nitric oxide concentration. Under the supplied fast-equilibrium treatment, equate its forward and reverse rates, then isolate the intermediate.

Return to the first step, labelled fast equilibrium:

2 N O \xrightleftharpoons [k_{- 1}] k_{1} N_{2} O_{2}

The forward elementary step involves two nitric oxide molecules, while the reverse step involves one dimer molecule. Their rates are:

r_{f o r w a r d} = k_{1} [N O]^{2}

r_{r e v e r s e} = k_{- 1} [N_{2} O_{2}]

Equating these rates gives:

k_{1} [N O]^{2} = k_{- 1} [N_{2} O_{2}]

Divide by the reverse rate constant:

[N_{2} O_{2}] = \frac{k_{1}}{k_{- 1}} [N O]^{2}

The squared nitric oxide dependence comes from the forward elementary equilibrium step, not from the overall balanced equation. At fixed temperature, doubling nitric oxide concentration therefore makes the equilibrium intermediate concentration four times as large.

The equilibrium ratio is not the effective overall rate constant. It must still multiply the slow-step rate constant when we substitute into the rate law.

How do you substitute and get the numerical answer 3?

Substitution makes the rate second order in nitric oxide and first order in hydrogen, giving overall order 3. Start again from the slow-step law:

r = k_{2} [N_{2} O_{2}] [H_{2}]

Replace the intermediate concentration explicitly:

r = k_{2} (\frac{k_{1}}{k_{- 1}} [N O]^{2}) [H_{2}]

Collect the constants:

r = \frac{k_{1} k_{2}}{k_{- 1}} [N O]^{2} [H_{2}] = k_{e f f} [N O]^{2} [H_{2}]

where:

k_{e f f} = \frac{k_{1} k_{2}}{k_{- 1}}

Read the concentration exponents:

Order in N O = 2, order in H_{2} = 1

$Overall order = 2 + 1 = 3$ Enter the numerical answer:

3

The final fast step’s rate constant does not enter this expression because that step is not rate-controlling under the supplied mechanism. It completes the conversion to products.

Why is answering 2 an unfinished solution?

Answering 2 counts the powers in the intermediate-containing expression, not the final rate law in terms of starting reactants. Here, 2 is an illustrative incorrect result, not an official distractor: this numerical-answer question has no options.

The incorrect route is:

r = k_{2} [N_{2} O_{2}] [H_{2}] ⟹ 1 + 1 = 2

The slow-step expression is valid, but reporting that sum as the overall order stops the method too early. The intermediate concentration itself depends on nitric oxide concentration.

Slow-step molecularity is 2: two reacting molecules participate in that elementary event.
Overall reaction order is 3: the final reactant-based rate law contains nitric oxide squared and hydrogen to the first power.

The repair is specific: replace the intermediate using the fast equilibrium before adding exponents. Molecularity describes one elementary event; overall order comes from the completed rate law.

Which three related questions check the same method?

Use these original related practice questions, not additional verified PYQs, to check concentration changes, rate-constant units and intermediate elimination. Each includes its working and answer.

What happens if nitric oxide is doubled and hydrogen is halved?

At fixed temperature, the rate doubles. Apply the changes to the derived rate law:

r = k_{e f f} [N O]^{2} [H_{2}]

\frac{r_{n e w}}{r_{o l d}} = 2^{2} \times \frac{1}{2} = 2

Answer: the new rate is twice the original rate.

What are the units of the effective rate constant?

The units are litres squared per mole squared per second. For rate measured in moles per litre per second and concentration in moles per litre, divide rate units by concentration cubed:

[k_{e f f}] = \frac{m o l L^{- 1} s^{- 1}}{(m o l L^{- 1})^{3}} = L^{2} m o l^{- 2} s^{- 1}

Answer:

L^{2} m o l^{- 2} s^{- 1}

What is the rate law for a different fast-equilibrium mechanism?

The hypothetical mechanism below gives first order in the first reactant and second order in the second, hence total order three. Derive its rate law using the labelled rate constants:

A + B \xrightleftharpoons [k_{- a}] k_{a} I fast equilibrium

I + B \overset{k_{b}}{\to} P slow elementary step

Equate forward and reverse equilibrium rates, then isolate the intermediate:

k_{a} [A] [B] = k_{- a} [I]

[I] = \frac{k_{a}}{k_{- a}} [A] [B]

Substitute into the slow-step rate law:

r = k_{b} [I] [B] = \frac{k_{a} k_{b}}{k_{- a}} [A] [B]^{2}

Answer:

Order in A = 1, order in B = 2, total order = 3

Cover the working and reproduce the intermediate-elimination step before attempting another mechanism question.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

Why is the order 3 and not 2 in the JEE Advanced 2024 kinetics question?

The slow-step rate law contains the intermediate N₂O₂, so adding its concentration powers gives an unfinished result. Eliminating N₂O₂ using the fast equilibrium gives r = k_eff[NO]²[H₂]. The overall order is therefore 2 + 1 = 3; the slow step's molecularity is 2.

How do you eliminate the intermediate using fast equilibrium?

For the fast equilibrium 2NO ⇌ N₂O₂, equate the forward and reverse rates: k₁[NO]² = k₋₁[N₂O₂]. Rearranging gives [N₂O₂] = (k₁/k₋₁)[NO]². Substitute this into r = k₂[N₂O₂][H₂] to obtain the final reactant-based rate law.

What are the units of the effective rate constant in this question?

The effective rate constant has units L² mol⁻² s⁻¹ when rate is measured in mol L⁻¹ s⁻¹ and concentration in mol L⁻¹. This follows by dividing the rate units by concentration cubed, since the overall reaction order is 3.

What happens to the rate if NO is doubled and H₂ is halved?

At fixed temperature, the rate doubles. From r = k_eff[NO]²[H₂], the rate multiplier is 2² × 1/2 = 2.