NCERT Class 12 Chemistry Chapter 4 Exercise Solutions - Chemical Kinetics

NCERT Class 12 Chemistry Chapter 4 - Exercise Solutions

Chemical Kinetics

Welcome to comprehensive exercise solutions for NCERT Class 12 Chemistry Chapter 4: Chemical Kinetics. These solutions provide detailed explanations for all textbook exercises, helping you master reaction rates and mechanisms.

📚 Exercise 4.1 - Multiple Choice Questions

Question 1

The rate of a chemical reaction tells us about:

(a) the concentration of reactants and products
(b) the direction in which the reaction proceeds
(c) how fast the reactants are converted into products
(d) the energy change during the reaction

Solution: Answer: (c) how fast the reactants are converted into products

Explanation: Chemical kinetics is the study of reaction rates. The rate of a chemical reaction tells us:

What reaction rate measures:

Speed of reaction: How quickly reactants are consumed
Formation speed: How quickly products are formed
Time dependency: Concentration changes with time

Analysis of options:

(a) Concentration: Chemical equilibrium deals with concentrations ✗
(b) Direction: Thermodynamics deals with reaction direction ✗
(c) Speed: Chemical kinetics deals with reaction speed ✓
(d) Energy change: Thermochemistry deals with energy changes ✗

Key Concept: Reaction rate = Change in concentration/Change in time

Question 2

In a reaction, 2A → Products, the concentration of A decreases from 0.5 M to 0.4 M in 10 minutes. The rate of appearance of product is:

(a) 0.01 M min⁻¹
(b) 0.005 M min⁻¹
(c) 0.001 M min⁻¹
(d) 0.002 M min⁻¹

Solution: Answer: (b) 0.005 M min⁻¹

Explanation:

Given:

Reaction: 2A → Products
Initial concentration of A, [A]₀ = 0.5 M
Final concentration of A, [A] = 0.4 M
Time interval, Δt = 10 minutes

Step 1: Calculate change in concentration of A Δ[A] = [A]final - [A]initial = 0.4 - 0.5 = -0.1 M

Step 2: Calculate rate of disappearance of A Rate of disappearance of A = -(1/2) × (Δ[A]/Δt) Rate = -(1/2) × (-0.1/10) = (1/2) × 0.01 = 0.005 M min⁻¹

Step 3: Apply rate relationship For the reaction: 2A → Products Rate of disappearance of A = Rate of appearance of Products

Therefore, Rate of appearance of Products = 0.005 M min⁻¹

Important Note: The stoichiometric coefficient (2) must be considered when calculating rates.

Question 3

The rate constant of a reaction depends on:

(a) temperature
(b) concentration of reactants
(c) time
(d) pressure

Solution: Answer: (a) temperature

Explanation:

Rate Constant (k):

Definition: Proportionality constant in rate law
Nature: Constant for a given reaction at a specific temperature
Dependence: Varies with temperature, not with concentration

Analysis of options:

(a) Temperature: Rate constant increases with temperature (Arrhenius equation) ✓
(b) Concentration: Rate constant is independent of concentration ✗
(c) Time: Rate constant is independent of time ✗
(d) Pressure: Rate constant is independent of pressure (except for gas-phase reactions at different pressures) ✗

Key Concept: Rate constant depends only on temperature and nature of the reaction.

Arrhenius Equation: k = A × e^(-Ea/RT) where k varies with temperature T.

📚 Exercise 4.2 - Very Short Answer Questions

Question 1

What is the effect of temperature on the rate constant of a reaction?

Solution:

Effect of Temperature on Rate Constant:

General Rule: Rate constant increases with temperature.

Quantitative Relationship:

Arrhenius Equation: k = A × e^(-Ea/RT)
Where:
- k = rate constant
- A = frequency factor (pre-exponential factor)
- Ea = activation energy
- R = gas constant
- T = absolute temperature

Key Observations:

Exponential Increase: Small temperature increase causes large increase in k
Activation Energy: Higher Ea means stronger temperature dependence
Rule of Thumb: Rate approximately doubles for every 10°C increase in temperature

Example: If rate constant at 300K is k₁, then at 310K: k₂ ≈ 2k₁ (approximately)

Physical Reason:

Higher temperature → More molecules have energy ≥ Ea
More effective collisions → Increased reaction rate
Therefore, rate constant increases

Question 2

What do you understand by the rate of a reaction?

Solution:

Rate of Reaction:

Definition: The rate of reaction is the change in concentration of reactants or products per unit time.

Mathematical Expression: For a general reaction: aA + bB → cC + dD

Rate of Reaction = -1/a × d[A]/dt = -1/b × d[B]/dt = 1/c × d[C]/dt = 1/d × d[D]/dt

Key Points:

Units: Usually mol L⁻¹ s⁻¹ or M s⁻¹
Negative Sign: Indicates decrease in reactant concentration
Positive Sign: Indicates increase in product concentration
Stoichiometric Coefficients: Must be considered in rate calculations

Example: For 2A → B

If [A] decreases from 0.2 M to 0.1 M in 5 minutes
Rate = -(1/2) × (0.1-0.2)/5 = 0.01 M min⁻¹

Physical Meaning: Rate tells us how fast the reaction proceeds - how quickly reactants are consumed and products are formed.

📚 Exercise 4.3 - Short Answer Questions

Question 1

For a reaction A + B → Products, the rate law is given by: Rate = k[A]²[B]. What is the overall order of the reaction?

Solution:

Given Rate Law: Rate = k[A]²[B]

Step 1: Identify the order with respect to each reactant

Order with respect to A: Power of [A] = 2 (second order)
Order with respect to B: Power of [B] = 1 (first order)

Step 2: Calculate overall order Overall Order = Order with respect to A + Order with respect to B Overall Order = 2 + 1 = 3

Answer: The overall order of the reaction is 3.

Additional Information:

Nature of Reaction: Third order reaction
Units of Rate Constant: M⁻² time⁻¹
Implications: Rate is highly dependent on concentrations, especially [A]

Question 2

The rate constant of a reaction is 2.5 × 10⁻³ s⁻¹. If the initial concentration of the reactant is 0.1 M, what will be its concentration after 2 minutes?

Solution:

Given:

Rate constant, k = 2.5 × 10⁻³ s⁻¹
Initial concentration, [R]₀ = 0.1 M
Time, t = 2 minutes = 120 seconds

Step 1: Identify the order of reaction The units of rate constant (s⁻¹) indicate a first-order reaction.

Step 2: Use the first-order integrated rate law For first-order reaction: [R] = [R]₀ × e^(-kt)

Step 3: Substitute the values [R] = 0.1 × e^(-2.5 × 10⁻³ × 120) [R] = 0.1 × e^(-0.3)

Step 4: Calculate e^(-0.3) e^(-0.3) ≈ 0.7408

Step 5: Find final concentration [R] = 0.1 × 0.7408 = 0.07408 M

Answer: The concentration after 2 minutes will be approximately 0.074 M.

Alternative Method Using Logarithms: ln([R]/[R]₀) = -kt ln([R]/0.1) = -2.5 × 10⁻³ × 120 = -0.3 [R]/0.1 = e^(-0.3) = 0.7408 [R] = 0.07408 M

📚 Exercise 4.4 - Long Answer Questions

Question 1

Derive the integrated rate law for a zero-order reaction and discuss its characteristics.

Solution:

Zero-Order Reaction: A reaction where the rate is independent of the concentration of reactants.

Rate Law: Rate = k (constant)

Derivation:

Step 1: Write the differential rate equation For reaction: R → Products Rate = -d[R]/dt = k

Step 2: Separate variables and integrate -d[R] = k dt ∫d[R] = -k∫dt

Step 3: Apply limits Let [R]₀ be initial concentration at t = 0 Let [R] be concentration at time t

∫[R]₀^[R] d[R] = -k∫₀^t dt [R] - [R]₀ = -kt

Step 4: Rearrange to get the integrated rate law [R] = [R]₀ - kt

Integrated Rate Law for Zero-Order Reaction: [R] = [R]₀ - kt

Characteristics of Zero-Order Reactions:

1. Rate Expression:

Rate = k (constant)
Independent of reactant concentration

2. Integrated Rate Law:

[R] = [R]₀ - kt
Linear relationship between concentration and time

3. Graphical Representation:

Plot of [R] vs t is a straight line
Slope = -k
Y-intercept = [R]₀

4. Half-Life:

Half-life (t₁/₂) = [R]₀/2k
Depends on initial concentration
Not constant for different initial concentrations

5. Units of Rate Constant:

k = [R]/t
Units = concentration/time = M s⁻¹

6. Examples:

Photochemical reactions: Constant light intensity
Catalytic reactions: Saturated catalyst surface
Surface reactions: Constant surface area

7. Conditions for Zero-Order Kinetics:

Heterogeneous catalysis: Reactant adsorption sites saturated
Photochemical reactions: Constant light intensity
Enzyme kinetics: At very high substrate concentrations

Graphical Analysis:

Concentration vs Time Plot:

Straight line with negative slope
Extends to [R] = 0 at t = [R]₀/k
Cannot be negative

Rate vs Concentration Plot:

Horizontal line (constant rate)
Independent of concentration changes

Significance:

Important in industrial processes where conditions are controlled
Helps in understanding catalyst saturation
Useful in photochemical reaction design

Question 2

Explain the collision theory of chemical reactions. What are the limitations of this theory?

Solution:

Collision Theory of Chemical Reactions:

Basic Principle: Chemical reactions occur when reactant molecules collide with sufficient energy and proper orientation.

Key Postulates: Collision Theory

1. Molecular Collisions:

Reactant molecules must collide for a reaction to occur
More collisions → Higher reaction rate
Collision frequency ∝ [A] × [B] for bimolecular reactions

2. Activation Energy:

Molecules must possess minimum kinetic energy (activation energy, Ea)
Only molecules with energy ≥ Ea can overcome the energy barrier
Fraction of molecules with sufficient energy follows Maxwell-Boltzmann distribution

3. Proper Orientation:

Molecules must collide in the correct orientation
Steric factor (P) accounts for orientation requirements
P = Number of effective collisions/Total collisions

Mathematical Expression:

Rate of Reaction = P × Z × e^(-Ea/RT)

Where:

P = Steric factor (orientation factor)
Z = Collision frequency
e^(-Ea/RT) = Fraction of molecules with energy ≥ Ea

Energy Profile Diagram:

Reactants start at certain energy level
Peak represents transition state (activation energy)
Products at lower or higher energy level
Energy difference = ΔH (enthalpy change)

Factors Affecting Reaction Rate (According to Collision Theory):

1. Temperature:

Increases kinetic energy of molecules
More molecules have energy ≥ Ea
Increases collision frequency
Effect: Exponential increase in rate

2. Concentration:

Higher concentration → More molecules → More collisions
Direct relationship between concentration and rate
Effect: Linear increase in rate

3. Nature of Reactants:

Different molecules have different activation energies
Steric factors vary with molecular structure
Size and complexity affect collision efficiency

4. Catalyst:

Lowers activation energy
Provides alternative reaction pathway
Increases rate without being consumed

Limitations of Collision Theory:

1. Complexity of Real Reactions:

Theory works well for simple gas-phase reactions
Fails to explain complex reactions involving multiple steps
Doesn’t account for reaction mechanisms

2. Quantitative Predictions:

Cannot accurately predict rate constants
Steric factor is often determined experimentally
Doesn’t explain temperature dependence quantitatively

3. Transition State Theory:

Collision theory doesn’t explain the nature of activated complex
Doesn’t account for molecular rearrangements
Cannot explain reactions with negative activation energy

4. Solution-Phase Reactions:

Theory developed for gas-phase reactions
Limited applicability to reactions in solutions
Doesn’t account for solvent effects

5. Catalysis:

Cannot explain catalytic action completely
Doesn’t account for surface catalysis mechanisms
Cannot explain enzyme catalysis

6. Energy Distribution:

Assumes Maxwell-Boltzmann distribution
Doesn’t account for energy redistribution during collision
Cannot explain reactions involving energy transfer

7. Molecular Complexity:

Simple theory doesn’t account for molecular structure
Cannot explain isotopic effects
Doesn’t consider internal degrees of freedom

Modifications and Improvements:

1. Transition State Theory (Activated Complex Theory):

Considers formation of activated complex
Better explains temperature dependence
Accounts for molecular rearrangements

2. Rice-Ramsperger-Kassel-Marcus (RRKM) Theory:

Explains unimolecular reactions
Accounts for energy redistribution
Better for complex molecules

3. Molecular Dynamics:

Computer simulations of molecular collisions
Accounts for molecular orientations
Predicts reaction outcomes

Significance Despite Limitations:

Provides basic understanding of reaction mechanisms
Explains temperature and concentration effects
Foundation for more sophisticated theories
Useful for qualitative predictions

🎯 Key Takeaways

Important Formulas:

Rate Law: Rate = k[A]ᵐ[B]ⁿ
Arrhenius Equation: k = A × e^(-Ea/RT)
Integrated Rate Laws:
- Zero order: [R] = [R]₀ - kt
- First order: ln([R]/[R]₀) = -kt
- Second order: 1/[R] - 1/[R]₀ = kt

Problem-Solving Strategy:

Identify reaction order from rate law or units
Choose appropriate integrated rate law
Pay attention to units and conversions
Verify if answer is physically reasonable

Common Mistakes to Avoid:

Order vs Molecularity: Order is experimental, molecularity is theoretical
Rate Constant: Depends on temperature, not concentration
Half-life: Different expressions for different orders
Units: Ensure consistency in units throughout calculations

Remember: Chemical kinetics is about understanding how fast reactions occur and why. Practice with different types of problems to build confidence.

Happy Learning! ⚗️