CUET UG Categorised PYQ Chemistry Unit 3 -

📚

CUET UG Categorised PYQ Chemistry Unit 3

Chemistry

⏱️ 30 Minutes

❓ 19 Questions

📅 2022, 2023& 2024

✍️ Long Answers

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Categorised PYQ Chemistry Unit 3

1 / 19

QUESTION 1 OF 19

What is the overall order of the reaction?
Rate = k[A]¹ᐟ²[B]³ᐟ² (PYQ 2024)

Answer Review

1 What is the overall order of the reaction?
Rate = k[A]¹ᐟ²[B]³ᐟ² (PYQ 2024)

A 2

B 0

C 1

D 0.5

💡 Short Explanation

Overall order is the sum of powers in the rate law. Power of [A] = 1/2. Power of [B] = 3/2. Overall order = 1/2 + 3/2 = 2.

✅ Correct Answer Explanation

(Detailed) → The order of a reaction with respect to a reactant is the power to which its concentration is raised in the rate equation. → In the given rate law: → Rate = k[A]¹ᐟ²[B]³ᐟ² → Order with respect to A = 1/2. → Order with respect to B = 3/2. → Overall order = 1/2 + 3/2 = 4/2 = 2. → Hence, the reaction is second order overall.

❓ Why Other Options Are Incorrect

B) 0 → This would mean the rate is independent of concentration, which is not the case here.
C) 1 → This is incorrect because the powers add up to 2, not 1.
D) 0.5 → This is only the order with respect to A, not the overall order.

🎯 Strategy

Used

Option A → Selected because 1/2 + 3/2 = 2.
Option B → Rejected because the rate depends on concentrations.
Option C → Rejected because it is an incorrect addition of powers.
Option D → Rejected because it considers only [A].

Final Answer → Overall order = sum of powers in rate law.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Order of Reaction

2 Which factor in Arrhenius equation corresponds to fraction of molecules having kinetic energy greater than activation energy? (PYQ 2024)

A ln k

B ln A

C RT

D e^(-Ea/RT)

💡 Short Explanation

Arrhenius equation is k = A e^(-Ea/RT). The exponential term represents the fraction of molecules having enough energy. This term is also called the Boltzmann factor.

✅ Correct Answer Explanation

(Detailed) → The Arrhenius equation is: → k = A e^(-Ea/RT) → Here, A is the frequency factor. → Ea is the activation energy. → R is the gas constant. → T is the temperature. → The term e^(-Ea/RT) represents the fraction of molecules having kinetic energy greater than or equal to activation energy. → As temperature increases, this fraction increases, and the reaction rate also increases. → Hence, e^(-Ea/RT) is the correct answer.

❓ Why Other Options Are Incorrect

A) ln k → It is the natural logarithm of the rate constant, not the energy fraction term.
B) ln A → It is the natural logarithm of the frequency factor, not the fraction of energetic molecules.
C) RT → It represents thermal energy term, but not the probability fraction by itself.

🎯 Strategy

Used

Option A → Reject because ln k is not the Boltzmann factor.
Option B → Reject because ln A is related to frequency factor.
Option C → Reject because RT alone is not a probability factor.
Option D → Select because the exponential term represents the required fraction.

Final Answer → Fraction of molecules with sufficient energy is represented by e^(-Ea/RT).

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Arrhenius Equation

3 What is the another term used for probability factor (P) in collision theory? (PYQ 2024)

A Temperature factor

B Compressibility factor

C Steric factor

D Concentration factor

💡 Short Explanation

Collision theory explains reaction rate using effective collisions. Probability factor accounts for proper orientation of colliding molecules. It is also called steric factor.

✅ Correct Answer Explanation

(Detailed) → According to collision theory, a reaction occurs only when reactant molecules collide effectively. → For a collision to be effective, molecules must have sufficient energy and proper orientation. → The probability factor P represents the fraction of collisions having correct orientation for product formation. → This orientation factor is known as the steric factor. → Hence, Steric factor is the correct answer.

❓ Why Other Options Are Incorrect

A) Temperature factor → Temperature affects kinetic energy, not directly the orientation probability.
B) Compressibility factor → This is used in real gas behaviour and is unrelated to collision orientation.
D) Concentration factor → Concentration affects number of collisions, not the orientation probability.

🎯 Strategy

Used

Option A → Reject because temperature affects energy.
Option B → Reject because compressibility belongs to real gases.
Option C → Select because probability factor means steric factor.
Option D → Reject because concentration affects collision frequency.

Final Answer → Probability factor in collision theory = Steric factor.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Collision Theory

4 A first order reaction has a half-life of 693 sec. What will be its rate constant? (PYQ 2024)

A 0.01 sec⁻¹

B 1 sec⁻¹

C 0.001 sec⁻¹

D 0.1 sec⁻¹

💡 Short Explanation

For a first order reaction, t₁/₂ = 0.693/k. Given half-life = 693 sec. Substituting gives k = 0.001 sec⁻¹.

✅ Correct Answer Explanation

(Detailed) → For a first order reaction: → t₁/₂ = 0.693/k → Therefore: → k = 0.693/t₁/₂ → Given, t₁/₂ = 693 sec → k = 0.693/693 → k = 0.001 sec⁻¹ → Hence, 0.001 sec⁻¹ is the correct answer.

❓ Why Other Options Are Incorrect

A) 0.01 sec⁻¹ → If k = 0.01 sec⁻¹, then t₁/₂ = 69.3 sec, not 693 sec.
B) 1 sec⁻¹ → If k = 1 sec⁻¹, then t₁/₂ = 0.693 sec, not 693 sec.
D) 0.1 sec⁻¹ → If k = 0.1 sec⁻¹, then t₁/₂ = 6.93 sec, not 693 sec.

🎯 Strategy

Used

Option A → Reject because it gives half-life 69.3 sec.
Option B → Reject because it gives half-life 0.693 sec.
Option C → Select because 0.693/693 = 0.001 sec⁻¹.
Option D → Reject because it gives half-life 6.93 sec.

Final Answer → Use first order half-life formula.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: First Order Reaction

5 Arrange the following rate constant units in increasing order of their order of reaction: (PYQ 2024)
Statements:
(A) sec⁻¹
(B) mol L⁻¹ sec⁻¹
(C) mol⁻¹ L sec⁻¹
(D) mol⁻² L² sec⁻¹

A (C) < (A) < (B) < (D)

B (C) < (B) < (A) < (D)

C (B) < (A) < (C) < (D)

D (A) < (B) < (C) < (D)

💡 Short Explanation

Units of rate constant depend on reaction order. Zero order: mol L⁻¹ sec⁻¹. First order: sec⁻¹. Second order: mol⁻¹ L sec⁻¹. Third order: mol⁻² L² sec⁻¹.

✅ Correct Answer Explanation

(Detailed) → The unit of rate constant changes with the order of reaction. → For zero order reaction, unit of k = mol L⁻¹ sec⁻¹. → Therefore, statement (B) represents zero order. → For first order reaction, unit of k = sec⁻¹. → Therefore, statement (A) represents first order. → For second order reaction, unit of k = L mol⁻¹ sec⁻¹ or mol⁻¹ L sec⁻¹. → Therefore, statement (C) represents second order. → For third order reaction, unit of k = L² mol⁻² sec⁻¹ or mol⁻² L² sec⁻¹. → Therefore, statement (D) represents third order. → Increasing order of reaction order is: → Zero order < First order < Second order < Third order → Hence, (B) < (A) < (C) < (D) is the correct sequence.

❓ Why Other Options Are Incorrect

A) (C) < (A) < (B) < (D) → This incorrectly places second order before first and zero order.
B) (C) < (B) < (A) < (D) → This incorrectly places second order before zero order.
D) (A) < (B) < (C) < (D) → This incorrectly places first order before zero order.

🎯 Strategy

Used

Option A → Reject because second order cannot come before zero and first order.
Option B → Reject because second order is wrongly placed first.
Option C → Select because it follows zero, first, second and third order.
Option D → Reject because zero order should come before first order.

Final Answer → Match each unit with its reaction order first.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Units of Rate Constant

6 The decomposition of NH₃ on a platinum surface is a zero-order reaction. If k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹, the rate of production of H₂ is: (PYQ 2022)

A 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹

B 5.0 × 10⁻⁴ mol L⁻¹ s⁻¹

C 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹

D 1.25 × 10⁻⁴ mol L⁻¹ s⁻¹

💡 Short Explanation

For zero-order reaction, rate = k. Reaction: 2NH₃ → N₂ + 3H₂. Rate of production of H₂ = 3 × rate of reaction.

✅ Correct Answer Explanation

(Detailed) → For a zero-order reaction, rate of reaction is equal to rate constant k. → Given: → k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹ → The balanced reaction is: → 2NH₃ → N₂ + 3H₂ → From the equation, 3 moles of H₂ are produced for one reaction unit. → Therefore, rate of production of H₂ = 3 × k → Rate of production of H₂ = 3 × 2.5 × 10⁻⁴ → Rate of production of H₂ = 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹ → Hence, 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹ is the correct answer.

❓ Why Other Options Are Incorrect

A) 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹ → This is only the rate constant k, not the rate of production of H₂.
B) 5.0 × 10⁻⁴ mol L⁻¹ s⁻¹ → This corresponds to 2 × k, not 3 × k.
D) 1.25 × 10⁻⁴ mol L⁻¹ s⁻¹ → This incorrectly divides the value instead of multiplying.

🎯 Strategy

Used

Option A → Reject because it ignores coefficient of H₂.
Option B → Reject because it uses coefficient 2 instead of 3.
Option C → Select because H₂ coefficient is 3.
Option D → Reject because division is not required here.

Final Answer → Use stoichiometric coefficient of H₂.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Zero Order Reaction

7 The molecularity of the elementary reaction NH₄NO₂ → N₂ + 2H₂O is: (PYQ 2022)

A Zero

B One

C Two

D Three

💡 Short Explanation

Molecularity is defined for elementary reactions. It is the number of reacting species involved in a step. Here, only one molecule of NH₄NO₂ decomposes.

✅ Correct Answer Explanation

(Detailed) → Molecularity is the number of reacting species taking part in an elementary reaction. → In the given reaction: → NH₄NO₂ → N₂ + 2H₂O → Only one reactant molecule, NH₄NO₂, is involved on the reactant side. → Therefore, the reaction is unimolecular. → Hence, molecularity is one.

❓ Why Other Options Are Incorrect

A) Zero → Molecularity can never be zero.
C) Two → This would mean two reacting molecules are involved, but only one molecule is present on the reactant side.
D) Three → This would mean three reacting molecules are involved, which is not true here.

🎯 Strategy

Used

Option A → Reject because molecularity cannot be zero.
Option B → Select because only one reactant molecule is involved.
Option C → Reject because there are not two reacting molecules.
Option D → Reject because there are not three reacting molecules.

Final Answer → Count reactant molecules in the elementary step.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Molecularity of Reaction

8 A reaction proceeds through four elementary steps, and the overall rate constant follows the relation:
k = (k₁ × k₂ × √k₄) / k₃
The activation energies of the steps Ea₁, Ea₂, Ea₃ and Ea₄ are 10 kJ mol⁻¹, 30 kJ mol⁻¹, 40 kJ mol⁻¹ and 20 kJ mol⁻¹, respectively. Determine the overall activation energy for the reaction. (PYQ 2023)

A 0 kJ mol⁻¹

B 100 kJ mol⁻¹

C 20 kJ mol⁻¹

D 10 kJ mol⁻¹

💡 Short Explanation

Use the relation between rate constants. Activation energies in numerator are added. Activation energy in denominator is subtracted. Powers act as multipliers.

✅ Correct Answer Explanation

(Detailed) → Given relation: → k = (k₁ × k₂ × √k₄) / k₃ → Therefore, the overall activation energy is: → Ea = Ea₁ + Ea₂ − Ea₃ + 1/2 Ea₄ → Substitute the values: → Ea = 10 + 30 − 40 + 1/2(20) → Ea = 40 − 40 + 10 → Ea = 10 kJ mol⁻¹ → Hence, 10 kJ mol⁻¹ is the correct answer.

❓ Why Other Options Are Incorrect

A) 0 kJ mol⁻¹ → This may be obtained if the √k₄ contribution is ignored.
B) 100 kJ mol⁻¹ → This is obtained by simply adding all activation energies without considering the relation.
C) 20 kJ mol⁻¹ → This may result from incorrect subtraction or power treatment.

🎯 Strategy

Used

Option A → Reject because √k₄ contribution must be included.
Option B → Reject because all terms cannot be simply added.
Option C → Reject because the denominator and square-root effect are not handled correctly.
Option D → Select because correct substitution gives 10 kJ mol⁻¹.

Final Answer → Add numerator terms, subtract denominator terms and multiply by powers.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Temperature Dependence of Rate Constant

9 A chemical reaction is second order with respect to A and first order with respect to B. If the concentration of A is reduced to half and the concentration of B becomes one-third of its original value, what will be the new reaction rate compared to the initial rate? (PYQ 2023)

A 12 times

B 1/12 times

C 6 times

D 1/6 times

💡 Short Explanation

Rate law is Rate = k[A]²[B]. A becomes half, so its effect is squared. B becomes one-third. New rate = (1/2)² × (1/3) = 1/12.

✅ Correct Answer Explanation

(Detailed) → Since the reaction is second order with respect to A and first order with respect to B: → Rate = k[A]²[B] → New concentration of A = 1/2[A] → New concentration of B = 1/3[B] → New rate: → R′ = k(1/2[A])²(1/3[B]) → R′ = k × 1/4[A]² × 1/3[B] → R′ = 1/12 k[A]²[B] → R′ = 1/12 R → Therefore, the new rate is 1/12 times the initial rate.

❓ Why Other Options Are Incorrect

A) 12 times → This would apply if concentration increased instead of decreased.
C) 6 times → This ignores the second-order effect of A.
D) 1/6 times → This treats the concentration changes incorrectly.

🎯 Strategy

Used

Option A → Reject because concentration is reduced, so rate cannot increase.
Option B → Select because (1/2)² × (1/3) = 1/12.
Option C → Reject because the reaction rate decreases, not increases.
Option D → Reject because it does not square the change in A correctly.

Final Answer → Apply concentration changes using powers from the rate law.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Rate Law and Order of Reaction

10 For the reaction:
CHCl₃ + Cl₂ → CCl₄ + HCl
Rate law:
Rate = k[CHCl₃][Cl₂]¹ᐟ²
Determine the order and molecularity of the reaction. (PYQ 2023)

A 3/2 and 0

B 3/2 and 2

C 3/2 and 3/2

D 2 and 3/2

💡 Short Explanation

Order is calculated from the rate law. Molecularity is counted from reacting molecules. Order = 1 + 1/2 = 3/2. Molecularity = 2.

✅ Correct Answer Explanation

(Detailed) → Given rate law is: → Rate = k[CHCl₃][Cl₂]¹ᐟ² → Order with respect to CHCl₃ = 1. → Order with respect to Cl₂ = 1/2. → Overall order = 1 + 1/2 = 3/2. → Molecularity is the number of reacting molecules involved in the reaction step. → In the reaction: → CHCl₃ + Cl₂ → CCl₄ + HCl → One molecule of CHCl₃ and one molecule of Cl₂ are involved. → Therefore, molecularity = 2. → Hence, order and molecularity are 3/2 and 2 respectively.

❓ Why Other Options Are Incorrect

A) 3/2 and 0 → Molecularity cannot be zero.
C) 3/2 and 3/2 → Molecularity cannot be fractional.
D) 2 and 3/2 → Order is not 2 and molecularity cannot be fractional.

🎯 Strategy

Used

Option A → Reject because molecularity cannot be zero.
Option B → Select because order = 3/2 and molecularity = 2.
Option C → Reject because molecularity must be a whole number.
Option D → Reject because both order and molecularity are incorrectly stated.

Final Answer → Use rate law for order and reaction equation for molecularity.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Order and Molecularity

11 For the reaction:
aX → bY
log₁₀(dx/dt) = log₁₀(dy/dt) + 0.4771
Given: log 3 = 0.4771
X and Y respectively are: (PYQ 2023)

A C₂H₂ and C₆H₆

B Pentane and 2,2-dimethylpropane

C O₂ and O₃

D N₂O₄ and NO₂

💡 Short Explanation

The logarithmic relation shows a rate ratio of 3 : 1. This means 3 molecules of X form 1 molecule of Y. The reaction matching this is 3C₂H₂ → C₆H₆.

✅ Correct Answer Explanation

(Detailed) → Given: → log₁₀(dx/dt) = log₁₀(dy/dt) + 0.4771 → Since log 3 = 0.4771, → log₁₀(dx/dt) = log₁₀(dy/dt) + log 3 → Therefore: → log₁₀(dx/dt) = log₁₀[3(dy/dt)] → So, → dx/dt = 3dy/dt → This indicates the stoichiometric relation: → 3X → Y → The reaction that matches this is: → 3C₂H₂ → C₆H₆ → This is the trimerisation of acetylene to benzene. → Hence, C₂H₂ and C₆H₆ is the correct answer.

❓ Why Other Options Are Incorrect

B) Pentane and 2,2-dimethylpropane → This represents isomerisation with 1 : 1 ratio, not 3 : 1.
C) O₂ and O₃ → The relation is 3O₂ → 2O₃, giving 3 : 2 ratio, not 3 : 1.
D) N₂O₄ and NO₂ → The relation is N₂O₄ → 2NO₂, giving 1 : 2 ratio, not 3 : 1.

🎯 Strategy

Used

Option A → Matches 3X → Y stoichiometry.
Option B → Reject because it is 1 : 1 conversion.
Option C → Reject because it gives 3 : 2 ratio.
Option D → Reject because it gives 1 : 2 ratio.

Final Answer → log 3 indicates 3 : 1 stoichiometry.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Rate of Reaction and Stoichiometric Relationship

12 Which of the following is not the graphical representation of first-order kinetics? (PYQ 2023)

💡 Short Explanation

For a first-order reaction, half-life is independent of initial concentration. t₁/₂ = 0.693/k. Therefore, t₁/₂ versus [A]₀ should be a horizontal line, not a vertical line.

✅ Correct Answer Explanation

(Detailed) → For a first-order reaction: → t₁/₂ = 0.693/k → This formula shows that half-life depends only on the rate constant k. → It does not depend on the initial concentration [A]₀. → Therefore, a graph of t₁/₂ against [A]₀ should be a horizontal straight line. → A vertical line for t₁/₂ versus [A]₀ is not a correct graphical representation of first-order kinetics. → Hence, option D is not the correct graphical representation of first-order kinetics.

❓ Why Other Options Are Incorrect

A) [Graph option A] → This is correct because log[A]ₜ versus time gives a straight line with negative slope for a first-order reaction.
B) [Graph option B] → This is correct because concentration decreases exponentially with time in a first-order reaction.
C) [Graph option C] → This is correct because ln([A]₀/[A]ₜ) = kt gives a straight line passing through the origin.

🎯 Strategy

Used

Option A → Reject because it represents valid first-order kinetics.
Option B → Reject because exponential decrease is valid for first-order reaction.
Option C → Reject because ln([A]₀/[A]ₜ) versus time is valid.
Option D → Select because first-order half-life cannot depend on initial concentration.

Final Answer → First-order half-life remains constant.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: First Order Reaction and Integrated Rate Law

13 Which of the following could not be a rate determining step? (PYQ 2023)

A NO₂ + NO₂ → NO₃ + NO

B O₃ ⇌ O₂ + O

C H₂O₂ + I⁻ → H₂O + IO⁻

D NO₂ + F₂ → NO₂F + F

💡 Short Explanation

Rate-determining step is usually the slowest step. It is generally written as an irreversible elementary step. A reversible equilibrium step is usually not considered the rate-determining step.

✅ Correct Answer Explanation

(Detailed) → A rate-determining step controls the overall rate of a reaction mechanism. → It is generally the slowest elementary step in the mechanism. → Such a step is usually written as a one-way reaction. → In option B: → O₃ ⇌ O₂ + O → This is a reversible equilibrium step. → Since equilibrium is established through both forward and backward reactions, it is generally not considered the slow rate-controlling step. → Hence, O₃ ⇌ O₂ + O could not be the rate determining step.

❓ Why Other Options Are Incorrect

A) NO₂ + NO₂ → NO₃ + NO → This is an elementary bimolecular step and can be slow, so it may act as a rate-determining step.
C) H₂O₂ + I⁻ → H₂O + IO⁻ → This can act as a slow step in the iodide-catalysed decomposition of hydrogen peroxide.
D) NO₂ + F₂ → NO₂F + F → This is an elementary step and can be slow, so it may act as a rate-determining step.

🎯 Strategy

Used

Option A → Reject because it can be a slow elementary step.
Option B → Select because it is a reversible equilibrium step.
Option C → Reject because it can be a slow reaction step.
Option D → Reject because it can be a slow elementary step.

Final Answer → Rate-determining step is usually a one-way slow step, not an equilibrium step.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Reaction Mechanism and Rate Determining Step

14 For the following reaction:
2A₂(g) + 1/4 X(g) → 2A₂X(g)
Volume is increased to double its value by decreasing the pressure on it. If the reaction is first order with respect to X and second order with respect to A₂, the rate of reaction will: (PYQ 2024)

A Decrease by eight times of its initial value

B Increase by eight times of its initial value

C Increase by four times of its initial value

D Remain unchanged

💡 Short Explanation

Rate = k[A₂]²[X]. Doubling volume halves the concentration of gases. New rate = (1/2)² × (1/2) = 1/8 of the original rate.

✅ Correct Answer Explanation

(Detailed) → The reaction is second order with respect to A₂ and first order with respect to X. → Therefore, the rate law is: → Rate = k[A₂]²[X] → When volume is doubled, the concentration of each gaseous reactant becomes half. → New concentration of A₂ = [A₂]/2 → New concentration of X = [X]/2 → New rate = k([A₂]/2)²([X]/2) → New rate = k[A₂]²[X] × 1/4 × 1/2 → New rate = original rate × 1/8 → Therefore, the rate decreases to 1/8 of its initial value. → Hence, it decreases by eight times of its initial value.

❓ Why Other Options Are Incorrect

B) Increase by eight times of its initial value → This is incorrect because concentration decreases when volume increases.
C) Increase by four times of its initial value → This is incorrect because the rate does not increase.
D) Remain unchanged → This is incorrect because rate depends on concentration of reactants.

🎯 Strategy

Used

Option A → Select because new rate becomes 1/8 of original rate.
Option B → Reject because direction of change is wrong.
Option C → Reject because rate decreases, not increases.
Option D → Reject because concentration affects rate.

Final Answer → Double volume → half concentration → apply powers in rate law.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Rate Law and Concentration Dependence

15 Identify the order of reaction if its rate constant is k = 2 × 10⁻² s⁻¹. (PYQ 2024)

A Zero order

B First order

C Second order

D Half order

💡 Short Explanation

Unit of rate constant helps identify order. For first order reaction, unit of k is s⁻¹. Given k = 2 × 10⁻² s⁻¹.

✅ Correct Answer Explanation

(Detailed) → The unit of rate constant depends on the order of reaction. → For a first order reaction, the unit of rate constant is time⁻¹. → In the given question: → k = 2 × 10⁻² s⁻¹ → The unit s⁻¹ clearly matches a first order reaction. → Therefore, the reaction is first order. → Hence, First order is the correct answer.

❓ Why Other Options Are Incorrect

A) Zero order → For zero order reaction, unit of k is mol L⁻¹ s⁻¹, not s⁻¹.
C) Second order → For second order reaction, unit of k is L mol⁻¹ s⁻¹.
D) Half order → Half order reactions have different units and do not match s⁻¹.

🎯 Strategy

Used

Option A → Reject because zero order does not have unit s⁻¹.
Option B → Select because first order has unit s⁻¹.
Option C → Reject because second order has concentration⁻¹ time⁻¹ unit.
Option D → Reject because half order unit does not match the given unit.

Final Answer → Unit s⁻¹ directly indicates first order.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Units of Rate Constant

16 For a complex reaction, the order of reaction is equal to: (PYQ 2024)

A Sum of stoichiometric coefficients in balanced chemical reaction

B The molecularity of overall reaction

C Order of fastest step of the reaction

D The molecularity of slowest step of reaction

💡 Short Explanation

Complex reactions occur in multiple steps. The slowest step controls the rate. The slowest step is called the rate-determining step.

✅ Correct Answer Explanation

(Detailed) → Complex reactions do not usually occur in a single step. → They proceed through a series of elementary steps. → The slowest step in the mechanism determines the overall rate of reaction. → This slowest step is known as the rate-determining step. → For such reactions, the order is related to the molecularity of the slowest step. → Hence, the molecularity of slowest step of reaction is the correct answer.

❓ Why Other Options Are Incorrect

A) Sum of stoichiometric coefficients in balanced chemical reaction → Order cannot generally be obtained from the balanced chemical equation for complex reactions.
B) The molecularity of overall reaction → Molecularity is defined for elementary steps, not for the overall complex reaction.
C) Order of fastest step of the reaction → The fastest step does not determine the rate of the reaction.

🎯 Strategy

Used

Option A → Reject because stoichiometric coefficients do not give order for complex reactions.
Option B → Reject because overall molecularity is not used for complex reactions.
Option C → Reject because fastest step does not control rate.
Option D → Select because slowest step is rate-determining.

Final Answer → Slowest step decides the rate and hence the order.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Complex Reactions and Rate Determining Step

17 If 75% of a first order reaction gets completed in 32 minutes, time taken for 50% completion of this reaction is: (PYQ 2024)

A 16 minutes

B 78 minutes

C 8 minutes

D 4 minutes

💡 Short Explanation

50% completion means one half-life. 75% completion means two half-lives. 32 minutes = 2 half-lives, so one half-life = 16 minutes.

✅ Correct Answer Explanation

(Detailed) → For a first order reaction, half-life is independent of initial concentration. → 50% completion means half of the reactant has reacted. → This time is one half-life, t₁/₂. → 75% completion means only 25% of the reactant remains. → This means the reaction has completed two half-lives. → Given time for 75% completion = 32 minutes. → Therefore: → 2t₁/₂ = 32 minutes → t₁/₂ = 16 minutes → Hence, time taken for 50% completion is 16 minutes.

❓ Why Other Options Are Incorrect

B) 78 minutes → This is much higher and does not follow first-order half-life logic.
C) 8 minutes → This is half of the correct half-life value.
D) 4 minutes → This is too low and not related to the given 75% completion time.

🎯 Strategy

Used

Option A → Select because 75% completion = 2 half-lives = 32 minutes, so 1 half-life = 16 minutes.
Option B → Reject because it overestimates the time.
Option C → Reject because it divides 32 incorrectly.
Option D → Reject because it is not supported by first-order kinetics.

Final Answer → 50% completion time = one half-life = 16 minutes.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Half-Life of First Order Reaction

18 What is the overall order of reaction which has the following rate expression?
Rate = k[A]²[B]¹ᐟ² (PYQ 2024)

A One

B Two

C Zero

D Three

💡 Short Explanation

Overall order is the sum of powers in rate law. Power of [A] = 2. Power of [B] = 1/2. The calculated order is 2.5, but the provided answer key marks Three.

✅ Correct Answer Explanation

(Detailed) → The order of reaction is calculated from the powers of concentration terms in the rate law. → Given rate expression is: → Rate = k[A]²[B]¹ᐟ² → Order with respect to A = 2. → Order with respect to B = 1/2. → Overall order = 2 + 1/2 = 2.5. → However, the given options do not contain 2.5. → According to the provided answer key, option D) Three is selected. → Hence, Three is given as the correct answer from the provided source.

❓ Why Other Options Are Incorrect

A) One → This is too low and ignores the powers in the rate law.
B) Two → This counts only [A]² and ignores [B]¹ᐟ².
C) Zero → This is incorrect because the rate depends on reactant concentrations.

🎯 Strategy

Used

Option A → Reject because total order is not 1.
Option B → Reject because [B]¹ᐟ² is not counted.
Option C → Reject because the reaction is not zero order.
Option D → Selected according to the provided answer key.

Final Answer → Sum the powers in rate expression, then match with given answer key.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Order of Reaction

19 Inverse of rate constant is: (PYQ 2024)

A Zero order reaction

B Pseudo first order reaction

C Second order reaction

D Third order reaction

💡 Short Explanation

Unit of rate constant depends on reaction order. For zero order reaction, k has unit concentration/time. Inverse of k gives time/concentration.

✅ Correct Answer Explanation

(Detailed) → The unit of rate constant changes according to the order of reaction. → For zero order reaction: → Rate = k → Therefore, k has the same unit as rate. → Unit of k = concentration per unit time. → The inverse of rate constant is therefore associated with time/concentration. → Among the given options, zero order reaction is the correct answer. → Hence, Zero order reaction is the correct answer.

❓ Why Other Options Are Incorrect

B) Pseudo first order reaction → It behaves like a first order reaction and its rate constant has unit time⁻¹.
C) Second order reaction → Its rate constant has unit concentration⁻¹ time⁻¹.
D) Third order reaction → Its rate constant has unit concentration⁻² time⁻¹.

🎯 Strategy

Used

Option A → Select because zero order has rate constant directly related to rate unit.
Option B → Reject because pseudo first order behaves like first order.
Option C → Reject because second order has a different rate constant unit.
Option D → Reject because third order has a different dimensional form.

Final Answer → Identify order using unit/dimension of rate constant.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Chemical Kinetics, Topic: Units of Rate Constant

CUET UG Categorised PYQ Chemistry Unit 3

Test Complete!

Answer Review