where k is a proportionality constant called rate constant. The equation like (3.4), which relates the rate of a reaction to concentration of reactants is called rate law or rate expression.
-- NCERT Class 12 Chemistry, Ch. 3, p. 67Rate Law Rate Constant
8 MCQs9-step worked example
Source: NCERT Chemical KineticsPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 7 Oct 2026
Try this first
How is the rate law of a reaction established?
- A.By adding the stoichiometric coefficients of the reactants in the balanced equation
- B.By counting the molecules that collide in the slowest step of the mechanism
- C.By experiment, measuring how the rate changes with the concentrations
- D.By taking the coefficients of the products as exponents
Tap to see the answer
Answer: C. The rate law must be determined experimentally; it cannot be predicted from the balanced equation alone (NCERT Class 12 Chemistry, Chapter 3, page 68).
A is wrong: A is wrong because the coefficients of the balanced equation do not fix the exponents; NCERT says the rate law must be determined experimentally (trap: reading the rate law from stoichiometry).
B is wrong: B is wrong because collisions in an elementary step define molecularity, a different quantity; the rate law itself comes from rate measurements (trap: mixing order with molecularity).
D is wrong: D is wrong because exponents in a rate law belong to reactant concentrations and are found by experiment, not read from product coefficients (trap: reading the rate law from stoichiometry).
Rate Law Rate Constant, explained for NEET
The trap is writing the rate law straight from the balanced equation. For 2A + B → products the reflex answer is rate = k[A]²[B], and it is only right if experiment says so. NCERT is blunt about it: the rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation but must be determined experimentally (NCERT Class 12 Chemistry, Chapter 3, page 68).
For a reaction aA + bB → cC + dD, the rate is written as Rate = k[A]^x[B]^y. The exponents x and y may or may not equal the stoichiometric coefficients, and k is the proportionality constant called the rate constant. The equation that relates rate to the concentrations of reactants is the rate law, also called the rate expression (NCERT Class 12 Chemistry, Chapter 3, page 67).
NCERT shows both cases on page 68. For CHCl₃ + Cl₂ → CCl₄ + HCl the experimental expression is Rate = k[CHCl₃][Cl₂]^½, so the exponents differ from the coefficients. In the ester hydrolysis the water has exponent 0.
The sum x + y is the overall order, and x and y are the orders with respect to A and B. Order can be 0, 1, 2, 3 or a fraction, and zero order means the rate is independent of reactant concentration (NCERT Class 12 Chemistry, Chapter 3, page 68).
The unit of k follows from the order: k = rate / [conc]^n. A second-order constant has the unit L mol⁻¹ s⁻¹ and a first-order constant has s⁻¹ (NCERT Class 12 Chemistry, Chapter 3, page 69). So the unit of k alone identifies the order.
Watch out: when a table of initial rates is given, change one concentration at a time. If doubling a concentration quadruples the rate, that exponent is 2. The coefficient in the equation never enters.
How do you solve a Rate Law Rate Constant question? A worked example
- 1
Given
For A + B → C, three initial-rate experiments give: (1) [A] = 0.10 mol L⁻¹, [B] = 0.10 mol L⁻¹, rate = 2.0 × 10⁻³ mol L⁻¹ s⁻¹; (2) [A] = 0.20 mol L⁻¹, [B] = 0.10 mol L⁻¹, rate = 8.0 × 10⁻³ mol L⁻¹ s⁻¹; (3) [A] = 0.10 mol L⁻¹, [B] = 0.30 mol L⁻¹, rate = 6.0 × 10⁻³ mol L⁻¹ s⁻¹.
- 2
Required
The rate law, the overall order, and the rate constant k with its unit.
- 3
Concept
The exponents are found only from the data, one concentration at a time, as NCERT does for 2NO + O₂ → 2NO₂ (NCERT Class 12 Chemistry, Chapter 3, page 67). The equation A + B → C has coefficients of 1 and says nothing about the exponents.
- 4
Formula
Rate = k[A]^x[B]^y, and k = rate / ([A]^x[B]^y).
- 5
Substitution
Experiments 1 and 2 change only [A], which doubles, and the rate goes from 2.0 × 10⁻³ to 8.0 × 10⁻³, a factor of 4, so 2^x = 4. Experiments 1 and 3 change only [B], which triples, and the rate goes from 2.0 × 10⁻³ to 6.0 × 10⁻³, a factor of 3, so 3^y = 3.
- 6
Calculation
x = 2 and y = 1, so Rate = k[A]²[B] and the overall order is 2 + 1 = 3. From experiment 1, k = 2.0 × 10⁻³ / (0.10² × 0.10) = 2.0 × 10⁻³ / 1.0 × 10⁻³ = 2.0. The counting numbers 2 and 3 in the ratios are exact and do not limit the significant figures of k.
- 7
Final answer
Rate = k[A]²[B], third order overall, k = 2.0 L² mol⁻² s⁻¹.
- 8
Common trap
Reading the rate law from the equation A + B → C would give Rate = k[A][B], a second-order law. The data show [A] enters squared.
- 9
Similar NEET-style question
For X + Y → products, doubling [X] at constant [Y] doubles the rate, and doubling [Y] at constant [X] raises the rate by a factor of 4. For [X] = 0.20 mol L⁻¹, [Y] = 0.20 mol L⁻¹ the rate is 1.6 × 10⁻² mol L⁻¹ s⁻¹. Find the rate law and k. (Answer: 2^x = 2 gives x = 1 and 2^y = 4 gives y = 2, so Rate = k[X][Y]², third order; k = 1.6 × 10⁻² / (0.20 × 0.20²) = 1.6 × 10⁻² / 8.0 × 10⁻³ = 2.0 L² mol⁻² s⁻¹.)
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Can you answer these Rate Law Rate Constant MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
How is the rate law of a reaction established?
Show answer and why every option is right or wrong
Answer: C. The rate law must be determined experimentally; it cannot be predicted from the balanced equation alone (NCERT Class 12 Chemistry, Chapter 3, page 68).
Why A is wrong: A is wrong because the coefficients of the balanced equation do not fix the exponents; NCERT says the rate law must be determined experimentally (trap: reading the rate law from stoichiometry).
Why B is wrong: B is wrong because collisions in an elementary step define molecularity, a different quantity; the rate law itself comes from rate measurements (trap: mixing order with molecularity).
Why D is wrong: D is wrong because exponents in a rate law belong to reactant concentrations and are found by experiment, not read from product coefficients (trap: reading the rate law from stoichiometry).
MCQ 2Easy RecallPractice
In the rate law Rate = k[A]^x[B]^y, what is the sum x + y called?
Show answer and why every option is right or wrong
Answer: A. The sum of the powers of the concentration terms in the rate law is the order of the reaction; x and y are the orders with respect to A and B (NCERT Class 12 Chemistry, Chapter 3, page 68).
Why B is wrong: B is wrong because molecularity is the number of species colliding in an elementary reaction, not the sum of rate-law exponents (trap: mixing order with molecularity).
Why C is wrong: C is wrong because the rate constant k is the proportionality constant in the rate law, not the sum of the exponents (NCERT Class 12 Chemistry, Chapter 3, page 67).
Why D is wrong: D is wrong because the sum of the coefficients comes from the balanced equation, while order comes from the experimental rate law (trap: reading the rate law from stoichiometry).
MCQ 3Easy RecallPractice
What is the unit of the rate constant of a first-order reaction when concentration is in mol L⁻¹ and time in seconds?
Show answer and why every option is right or wrong
Answer: D. For a first-order reaction k = rate / concentration, which gives s⁻¹ (NCERT Class 12 Chemistry, Chapter 3, page 69).
Why A is wrong: A is wrong because mol L⁻¹ s⁻¹ is the unit of the rate of reaction and of k for a zero-order reaction (trap: mixing the zero-order unit with the first-order unit).
Why B is wrong: B is wrong because L mol⁻¹ s⁻¹ is the unit of a second-order rate constant (NCERT Class 12 Chemistry, Chapter 3, page 69).
Why C is wrong: C is wrong because L² mol⁻² s⁻¹ belongs to a third-order reaction, which divides the rate by concentration cubed (trap: mixing the order with the unit).
MCQ 4Direct ApplicationPractice
For CHCl₃ + Cl₂ → CCl₄ + HCl, the experimental rate expression is Rate = k[CHCl₃][Cl₂]^½. What is the overall order of the reaction?
Show answer and why every option is right or wrong
Answer: B. Overall order is the sum of the exponents: 1 + ½ = 3/2. NCERT lists this rate expression, with exponents unlike the coefficients, on page 68 (NCERT Class 12 Chemistry, Chapter 3, page 68).
Why A is wrong: A is wrong because ½ is only the exponent of Cl₂, not the sum over all reactants (trap: counting only one reactant).
Why C is wrong: C is wrong because 2 is the sum of the stoichiometric coefficients of the reactants, which is not the order (trap: reading the rate law from stoichiometry).
Why D is wrong: D is wrong because 1 is only the exponent of CHCl₃; the exponent ½ of Cl₂ must be added (trap: counting only one reactant).
MCQ 5Direct ApplicationPractice
A reaction has a rate constant k = 4.0 × 10⁻³ mol L⁻¹ s⁻¹. What is its order?
Show answer and why every option is right or wrong
Answer: A. The unit mol L⁻¹ s⁻¹ is the unit of k only when rate = k, with no concentration term, so the reaction is zero order. NCERT matches units of k to orders in Table 3.3 and Example 3.4 (NCERT Class 12 Chemistry, Chapter 3, page 69).
Why B is wrong: B is wrong because a first-order constant has the unit s⁻¹, with no concentration in it (NCERT Class 12 Chemistry, Chapter 3, page 69).
Why C is wrong: C is wrong because a second-order constant has the unit L mol⁻¹ s⁻¹ (trap: mixing the order with the unit).
Why D is wrong: D is wrong because a third-order constant has the unit L² mol⁻² s⁻¹ (trap: mixing the order with the unit).
MCQ 6Direct ApplicationPractice
In NCERT's data for 2NO + O₂ → 2NO₂, keeping [NO] at 0.30 mol L⁻¹ and doubling [O₂] from 0.30 to 0.60 mol L⁻¹ changes the initial rate from 0.096 to 0.192 mol L⁻¹ s⁻¹. What is the order with respect to O₂?
Show answer and why every option is right or wrong
Answer: D. The rate doubles when [O₂] doubles, so the rate depends on [O₂] to the first power (NCERT Class 12 Chemistry, Chapter 3, page 67).
Why A is wrong: A is wrong because a zero-order species would leave the rate unchanged when its concentration doubles.
Why B is wrong: B is wrong because a second-order species would quadruple the rate; here it only doubles (trap: confusing the coefficient 1 of O₂ with a squared term).
Why C is wrong: C is wrong because a half-order species would raise the rate by a factor of about 1.4, not 2.
MCQ 7CalculationPractice
For a reaction with Rate = k[A]²[B], the concentration of A is tripled and the concentration of B is halved. By what factor does the rate change?
Show answer and why every option is right or wrong
Answer: C. Tripling [A] multiplies the rate by 3² = 9, and halving [B] multiplies it by ½ (first order in B), so the rate changes by 9 × ½ = 4.5. This is the same proportionality NCERT uses when doubling [NO] quadruples the rate (NCERT Class 12 Chemistry, Chapter 3, page 67).
Why A is wrong: A is wrong because 3 × ½ = 1.5 treats A as first order; the exponent of A is 2 (trap: ignoring the exponent).
Why B is wrong: B is wrong because 3.0 applies the exponent 1 to A and ignores the change in B.
Why D is wrong: D is wrong because 9 accounts for A only; the halving of [B] still has to be applied (trap: changing one species and forgetting the other).
MCQ 8CalculationPractice
In NCERT's data for 2NO + O₂ → 2NO₂ with Rate = k[NO]²[O₂], experiment 2 has [NO] = 0.60 mol L⁻¹, [O₂] = 0.30 mol L⁻¹ and an initial rate of 0.384 mol L⁻¹ s⁻¹. What is k, with its unit?
Show answer and why every option is right or wrong
Answer: B. k = rate / ([NO]²[O₂]) = 0.384 / (0.36 × 0.30) = 0.384 / 0.108 = 3.6. The reaction is third order overall, so the unit is L² mol⁻² s⁻¹ (NCERT Class 12 Chemistry, Chapter 3, pages 67 and 69).
Why A is wrong: A is wrong because 0.384 / (0.60 × 0.30) = 2.1 leaves [NO] unsquared (trap: dropping the exponent).
Why C is wrong: C is wrong because the value 3.6 is right but L mol⁻¹ s⁻¹ is the unit of a second-order constant; this reaction is third order (trap: mixing the order with the unit).
Why D is wrong: D is wrong because 0.384 / 0.30 = 1.3 divides by [O₂] alone and skips the [NO]² term.
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What to remember before solving Rate Law Rate Constant questions
7 NCERT lines
Key Fact
Exponents may differ from stoichiometric coefficients (CHCl3 + Cl2 and ester hydrolysis examples)
In these reactions, the exponents of the concentration terms are not the same as their stoichiometric coefficients.
-- NCERT Class 12 Chemistry, Ch. 3, p. 68Key Fact
Rate law cannot be predicted from the balanced equation; it must be determined experimentally
Rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation, i.e., theoretically but must be determined experimentally.
-- NCERT Class 12 Chemistry, Ch. 3, p. 68Sum of these exponents, i.e., x + y in (3.4) gives the overall order of a reaction whereas x and y represent the order with respect to the reactants A and B respectively. Hence, the sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.
-- NCERT Class 12 Chemistry, Ch. 3, p. 68Order of a reaction can be 0, 1, 2, 3 and even a fraction. A zero order reaction means that the rate of reaction is independent of the concentration of reactants.
-- NCERT Class 12 Chemistry, Ch. 3, p. 68The unit of second order rate constant is L mol–1 s–1, therefore k = 2.3 × 10–5 L mol–1 s–1 represents a second order reaction. (ii) The unit of a first order rate constant is s–1 therefore k = 3 × 10–4 s–1 represents a first order reaction.
-- NCERT Class 12 Chemistry, Ch. 3, p. 69Rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which need not equal the stoichiometric coefficients. For 2NO(g)+O2(g)->2NO2(g): doubling [NO] at constant [O2] increases the initial rate by a factor of four (order 2 in NO); doubling [O2] at constant [NO] doubles the rate (order 1 in O2), giving Rate = k[NO]^2[O2].
-- NCERT Class 12 Chemistry, Ch. 3, p. 67Where do students lose marks on Rate Law Rate Constant?
1 trap or mistake
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Root cause: concept gap
Correction
Rate law from EXPERIMENT, not stoichiometry. Order = sum of exponents in rate = k[A]^x[B]^y. Molecularity follows mechanism.
Rate Law Rate Constant: NEET previous year questions (PYQs) with answers
1 question from NEET 2023, answers verified against NTA official keys
NEET 2023
1Increase by a factor of six
2Increase by a factor of nine
3Increase by a factor of three
4Decrease by a factor of nine
NTA Answer: Option 2(final)
Why the other options are wrong
- Option 1: Six would be 2 × 3. In Rate = k [A]x [B]y the order in A is x = 2, so tripling [A] multiplies the rate by 3² = 9.
- Option 3: A threefold increase would need first order in A. Here the power of [A] is 2, so the rate rises 3² = 9 times.
- Option 4: The rate cannot fall when [A] rises with a positive exponent (2). It increases 3² = 9 times.
More in Chemical Kinetics: 4 exam traps and mistakes · 4 formulas · 2 question patterns from its other lessons.
Sources
NCERT refs: Class 12 Chemistry Chapter 3, p.68 | Class 12 Chemistry Chapter 3, p.67 | Class 12 Chemistry Chapter 3, p.69
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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