JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
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This MCQ module is based on: Rate Law Order
TOPIC 10 OF 23
Rate Law Order
🎓 Class 12
Chemistry
CBSE
Theory
Ch 3 – Chemical Kinetics
⏱ ~14 min
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Rate Law Order
3.4 Rate Expression and Rate Constant
For a general reaction
\[ aA + bB \;\longrightarrow\; cC + dD \]
experiment shows that the rate depends on reactant concentrations as
\[ \text{Rate} \;\propto\; [A]^x\,[B]^y \]
which on introducing a proportionality constant becomes the rate law:
\[ \boxed{\; \text{Rate} \;=\; k\,[A]^x\,[B]^y \;} \]
The exponents x and y may or may not be equal to the stoichiometric coefficients a and b; they are determined experimentally.
Rate Law (Differential rate equation): An expression that relates the rate of a reaction to the concentrations of the reactants, each raised to the experimentally determined power.
The constant of proportionality k is called the rate constant (or specific reaction rate). It is independent of the concentrations of the reactants but is a strong function of temperature.
Determining the Rate Law from Experiment
For \(2\,NO(g) + O_2(g) \to 2\,NO_2(g)\) the following initial-rate data were obtained:
| Run | [NO]/mol L⁻¹ | [O₂]/mol L⁻¹ | Initial rate / mol L⁻¹ s⁻¹ |
|---|---|---|---|
| 1 | 0.30 | 0.30 | 0.096 |
| 2 | 0.60 | 0.30 | 0.384 |
| 3 | 0.30 | 0.60 | 0.192 |
| 4 | 0.60 | 0.60 | 0.768 |
Comparing runs 1 and 2: [O₂] is unchanged, [NO] doubles, rate becomes 4× → order in NO is 2.
Comparing runs 1 and 3: [NO] is unchanged, [O₂] doubles, rate doubles → order in O₂ is 1.
Therefore Rate \(= k\,[NO]^2[O_2]\), and from run 1: \(k = \dfrac{0.096}{(0.30)^2(0.30)} = 3.56\) L² mol⁻² s⁻¹.
3.4.1 Order of a Reaction
The order of a reaction is the sum of the exponents of the concentration terms in the rate law. For Rate = \(k[A]^x[B]^y\), the overall order is \(x+y\).
| Reaction | Experimental Rate Law | Order |
|---|---|---|
| \(2NO + O_2 \to 2NO_2\) | \(k[NO]^2[O_2]\) | 3 |
| \(CHCl_3 + Cl_2 \to CCl_4 + HCl\) | \(k[CHCl_3][Cl_2]^{1/2}\) | 1.5 (fractional) |
| \(CH_3COOC_2H_5 + H_2O \to CH_3COOH + C_2H_5OH\) | \(k[CH_3COOC_2H_5]\) | 1 (pseudo) |
| \(2NH_3 \xrightarrow{Pt} N_2 + 3H_2\) | \(k\,[NH_3]^0 = k\) | 0 (zero) |
Units of the Rate Constant
From the rate law: Rate = \(k\,[\text{conc}]^n\). So
\[ k \;=\; \frac{\text{Rate}}{[\text{conc}]^n} \;=\; \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^n} \;=\; (\text{mol L}^{-1})^{1-n}\,\text{s}^{-1} \]
| Order (n) | Units of k |
|---|---|
| 0 (zero) | mol L⁻¹ s⁻¹ |
| 1 (first) | s⁻¹ |
| 2 (second) | L mol⁻¹ s⁻¹ |
| 3 (third) | L² mol⁻² s⁻¹ |
Worked Example 3.4 L3 Apply
Calculate the overall order of a reaction with the rate expression:
(a) Rate = \(k\,[A]^{1/2}[B]^{3/2}\) (b) Rate = \(k\,[A]^{3/2}[B]^{-1}\).
(a) Order = ½ + 3/2 = 2.
(b) Order = 3/2 + (−1) = ½.
Worked Example 3.5 L3 Apply
Identify the reaction order from the units of the rate constant given below:
(a) k = 3.0 × 10⁻⁴ s⁻¹ (b) k = 5.0 × 10⁻³ L mol⁻¹ s⁻¹ (c) k = 1.5 × 10⁻² mol L⁻¹ s⁻¹.
Use units (mol L⁻¹)¹⁻ⁿ s⁻¹.
(a) s⁻¹ → 1 − n = 0 → first order.
(b) L mol⁻¹ s⁻¹ = (mol L⁻¹)⁻¹ s⁻¹ → 1 − n = −1 → second order.
(c) mol L⁻¹ s⁻¹ → 1 − n = 1 → zero order.
3.4.2 Molecularity of a Reaction
The molecularity of an elementary reaction is the number of reacting species that must collide simultaneously to bring about the chemical change.
| Molecularity | Type | Example |
|---|---|---|
| 1 | Unimolecular | \(NH_4NO_2 \to N_2 + 2H_2O\) |
| 2 | Bimolecular | \(2HI \to H_2 + I_2\) |
| 3 | Termolecular (rare) | \(2NO + O_2 \to 2NO_2\) |
Order ≠ Molecularity in general!
- Order is experimental; molecularity is theoretical (and applies only to elementary steps).
- Order can be 0, fractional, or even negative; molecularity is always a small positive integer.
- For an elementary reaction, order = molecularity. For a complex reaction, the order is decided by the slowest (rate-determining) step, while the overall stoichiometry is irrelevant.
Rate-Determining Step
For a sequence of steps
e.g. \( 2NO_2 + F_2 \to 2NO_2F \) is believed to occur as
- Step 1 (slow): \(NO_2 + F_2 \to NO_2F + F\)
- Step 2 (fast): \(NO_2 + F \to NO_2F\)
The slow step (1) is bimolecular, so the experimental rate law is Rate \(= k[NO_2][F_2]\) — first order in each, even though the balanced equation has \(2\,NO_2\).
3.5 Integrated Rate Equations
The differential rate equation \(-d[R]/dt = k[R]^n\) is hard to test directly because it involves derivatives. Integration converts it into a form linking measurable quantities — concentration and time.
3.5.1 Zero Order Reaction
The rate is independent of reactant concentration:
\[ R \to P, \quad \text{Rate} = -\frac{d[R]}{dt} = k\,[R]^0 = k \]
Separating variables and integrating between \(t=0\) (\([R]=[R]_0\)) and time \(t\) (\([R]=[R]\)):
\[ d[R] = -k\,dt \;\Rightarrow\; [R] = -kt + I \]
At \(t=0\), \(I=[R]_0\). Therefore:
\[ \boxed{\; [R] \;=\; [R]_0 - kt \;} \quad\text{or}\quad k = \frac{[R]_0 - [R]}{t} \]
A plot of [R] vs t is a straight line of slope −k and intercept [R]₀.
Examples: photochemical decomposition of HI on a gold surface, decomposition of N₂O on a hot platinum filament, enzyme reactions at saturating substrate.
3.5.2 First Order Reaction
The rate is proportional to the first power of reactant concentration:
\[ -\frac{d[R]}{dt} = k\,[R] \;\Rightarrow\; \frac{d[R]}{[R]} = -k\,dt \]
Integrating between \(t=0,\,[R]=[R]_0\) and time \(t,\,[R]=[R]\):
\[ \ln[R] = -kt + \ln[R]_0 \]
\[ \boxed{\; \ln\frac{[R]_0}{[R]} = kt \;\Leftrightarrow\; k = \frac{2.303}{t}\,\log\frac{[R]_0}{[R]} \;} \]
A plot of \(\ln[R]\) vs t is a straight line of slope −k; equivalently, a plot of \(\log\bigl([R]_0/[R]\bigr)\) vs t has slope \(k/2.303\).
Examples of first-order reactions:
- All radioactive decays: \(^{226}_{88}Ra \to ^{4}_{2}He + ^{222}_{86}Rn\)
- Decomposition of N₂O₅, of H₂O₂ in aqueous solution
- Hydrogenation of ethene: \(C_2H_4 + H_2 \xrightarrow{Ni} C_2H_6\) (Rate = \(k[C_2H_4]\))
Worked Example 3.6 L3 Apply
The initial concentration of \(N_2O_5\) in the first order reaction \(N_2O_5 \to 2NO_2 + ½\,O_2\) was 1.24 × 10⁻² mol L⁻¹ at 318 K. After 60 minutes, it became 0.20 × 10⁻² mol L⁻¹. Calculate the rate constant.
For first order: \(k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}\)
\(k = \dfrac{2.303}{60\text{ min}}\log\dfrac{1.24 \times 10^{-2}}{0.20 \times 10^{-2}} = \dfrac{2.303}{60}\log(6.2)\)
\(= \dfrac{2.303}{60}(0.7924) = 0.0304\) min⁻¹.
Worked Example 3.7 L3 Apply
The decomposition of NH₃ on platinum is zero order with k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹. What are the rates of production of N₂ and H₂?
The reaction: \(2NH_3(g) \xrightarrow{Pt} N_2(g) + 3H_2(g)\). For zero order, rate = k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹.
Rate \(= -\dfrac{1}{2}\dfrac{d[NH_3]}{dt} = +\dfrac{d[N_2]}{dt} = +\dfrac{1}{3}\dfrac{d[H_2]}{dt}\).
\(\Rightarrow \dfrac{d[N_2]}{dt} = 2.5 \times 10^{-4}\) mol L⁻¹ s⁻¹
\(\Rightarrow \dfrac{d[H_2]}{dt} = 3 \times 2.5 \times 10^{-4} = 7.5 \times 10^{-4}\) mol L⁻¹ s⁻¹.
Worked Example 3.8 L3 Apply
The first order rate constant for decomposition of an organic compound A is 2.5 × 10⁻³ s⁻¹. Calculate the time required for 25% of A to decompose.
If 25% has reacted, [R]/[R]₀ = 0.75. Use \(t = \dfrac{2.303}{k}\log\dfrac{[R]_0}{[R]}\):
\(t = \dfrac{2.303}{2.5\times 10^{-3}}\log\dfrac{1}{0.75} = \dfrac{2.303}{2.5\times 10^{-3}}\times 0.1249\)
\( = 115.0 \) s.
Interactive: First Order Decay Predictor
Given a rate constant and initial concentration, predict the concentration of the reactant after time t.
Predicted [R] at time t = — mol L⁻¹
% reacted = — %
Activity 3.2 — Detecting the Order Graphically
Setup: You collect [R] vs t data for a reaction. You plot three graphs: (i) [R] vs t, (ii) ln[R] vs t, (iii) 1/[R] vs t.
Predict: Which plot must be linear if the reaction is first order? Zero order?
Zero order: Plot (i) [R] vs t is linear (slope −k).
First order: Plot (ii) ln[R] vs t is linear (slope −k).
Second order: Plot (iii) 1/[R] vs t is linear (slope +k). (You will study this in higher classes.)
This trial-and-error method of plotting is the standard experimental way to discover the order of an unknown reaction.
Competency-Based Questions
Q1. The rate law for the reaction \(A + B \to P\) was found to be Rate = \(k[A][B]^2\). The order with respect to A and B and the overall order respectively are: L2
(a) First in A, second in B, overall third.
Q2. The unit of rate constant of a reaction is L mol⁻¹ s⁻¹. The order is: L3
(c) Second. Units of k = (mol L⁻¹)⁽¹⁻ⁿ⁾ s⁻¹ = (mol L⁻¹)⁻¹ s⁻¹ implies 1−n = −1, so n = 2.
Q3. (Short answer) Distinguish between order and molecularity giving one example of each. L4
Order is the experimentally determined sum of exponents in the rate law (can be 0, fractional, integer); molecularity is the number of species colliding in an elementary step (always a small positive integer). Example: hydrolysis of ester is pseudo-first-order (experimental); decomposition of HI: \(2HI \to H_2 + I_2\) is bimolecular (theoretical).
Q4. (Fill in the blank) The slope of a graph of ln[R] versus t for a first order reaction is _________. L1
−k (negative of the rate constant).
Q5. (Long answer) The decomposition of dimethyl ether on a hot Pt surface, \( (CH_3)_2 O \to CH_4 + H_2 + CO\), proceeds with a rate that is independent of dimethyl ether pressure. Identify the order, propose a reason, and write the integrated rate law. L5
It is zero order. The Pt surface saturates with adsorbed dimethyl ether at any reasonable pressure, so additional gas does not increase the surface coverage and hence does not increase the rate. Integrated form: \([R] = [R]_0 - kt\) (linear decrease).
Assertion–Reason Questions
(A) Both true & R explains A. (B) Both true but R does not explain A. (C) A true, R false. (D) A false, R true.
Assertion: Order of an elementary reaction equals its molecularity.
Reason: An elementary reaction proceeds in a single step and the rate law follows directly from the stoichiometry.
(A) Both true; R explains A. For elementary steps the powers in the rate law equal the coefficients.
Assertion: The rate constant of a reaction depends on temperature.
Reason: Increasing temperature increases the concentration of reactants.
(C) Assertion is true (Arrhenius law). Reason is false — temperature changes the fraction of molecules with sufficient energy, not concentration.
Assertion: A reaction can have fractional order.
Reason: Order is determined experimentally and reflects a complex underlying mechanism.
(A) Both true; R correctly explains A.
Frequently Asked Questions - Rate Law Order
What is the main concept covered in Rate Law Order?
In NCERT Class 12 Chemistry Chapter 3 (Chemical Kinetics), "Rate Law Order" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Rate Law Order useful in real-life or applied chemistry?
Real-life applications of "Rate Law Order" from NCERT Class 12 Chemistry Chapter 3 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Rate Law Order?
Key reactions in "Rate Law Order" (NCERT Class 12 Chemistry Chapter 3 Chemical Kinetics) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 3?
NCERT Class 12 Chemistry Chapter 3 (Chemical Kinetics) is structured so each part builds chemical understanding sequentially. "Rate Law Order" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Rate Law Order?
CBSE board questions from "Rate Law Order" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Rate Law Order" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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