NCERT Solutions Class 12 Chemistry Chapter 4 Chemical Kinetics

Mastering NCERT Solutions Class 12 Chemistry Chapter 4 Chemical Kinetics is crucial for students aiming to achieve top scores in CBSE Class 12 board examinations, as well as competitive entrance tests like JEE and NEET. This branch of physical chemistry deals with the rates of chemical reactions, the factors influencing these speeds, and the microscopic mechanisms through which chemical transformations take place. In this detailed guide, you will find comprehensive NCERT Chemistry Class 12 solutions covering every concept, derivation, intext problem, and board question with step-by-step working.

Whether you are revising for upcoming pre-boards or seeking NCERT Solutions step by step to solidify your conceptual foundation, working through structured solutions ensures complete clarity. Accessing quality NCERT Solutions Chemistry chapter wise enables you to tackle both theoretical derivations and tricky numerical problems with absolute confidence.

Chapter Overview: Chemical Kinetics

In physical chemistry, thermodynamics informs us about the feasibility of a reaction (whether ΔG is negative), but it provides no information regarding the time taken for the process to complete. Chemical Kinetics bridges this gap by investigating reaction velocities, reaction pathways, and transition states. In the CBSE Class 12 curriculum, this chapter carries substantial weightage (typically 7 marks) within the physical chemistry section.

The core syllabus of Chemical Kinetics encompasses the following major topics:

• Rate of a Chemical Reaction: Average rate vs. instantaneous rate, mathematical expression in terms of stoichiometric coefficients, and graphical representation.
• Factors Influencing Reaction Rates: Concentration of reactants, temperature, presence of a catalyst, and surface area.
• Rate Law and Specific Rate Constant: Differential rate equations, definition of rate constant (k), and its characteristic units across various reaction orders.
• Order vs. Molecularity: Distinguishing between empirical reaction order and theoretical molecularity for elementary and complex reactions.
• Integrated Rate Equations: Mathematical derivations of concentration-time relationships and half-life expressions (t_1/2) for zero-order and first-order reactions.
• Pseudo First-Order Reactions: Reactions that follow first-order kinetics due to one reactant being present in large excess (e.g., acid-catalyzed ester hydrolysis).
• Temperature Dependence and Arrhenius Theory: Activation energy (E_a), concept of activated complex, Maxwell-Boltzmann energy distribution curves, and the logarithmic Arrhenius equation.
• Collision Theory of Chemical Reactions: Effective collisions, threshold energy, steric/probability factor (P), and collision frequency (Z).

Key Comparison: Order of Reaction vs. Molecularity

Feature	Order of Reaction	Molecularity of Reaction
Definition	The sum of powers of the concentration terms of reactants in the experimentally determined rate law.	The number of reacting species (atoms, ions, or molecules) colliding simultaneously in an elementary step.
Nature	Experimental quantity; cannot be deduced purely from a balanced equation.	Theoretical concept derived from the reaction mechanism.
Values	Can be zero, whole number, fractional, or negative.	Always a positive non-zero integer (1, 2, or 3). Cannot be zero or fractional.
Applicability	Applicable to both elementary and complex reactions.	Meaningful only for elementary reactions; has no significance for complex multi-step reactions.

Fundamental Derivations and Mathematical Formulations

1. Integrated Rate Equation for a Zero-Order Reaction

For a zero-order reaction: R → P

The differential rate equation is given by:

Rate = -d[R]/dt = k[R]0 = k

Rearranging the variables gives:

d[R] = -k dt

Integrating both sides:

∫ d[R] = -k ∫ dt ⇒ [R] = -kt + I

At initial time t = 0, [R] = [R]₀ (initial concentration). Substituting these boundary conditions yields I = [R]0.

Therefore, the integrated rate equation is:

[R] = -kt + [R]₀  or  k = ([R]₀ - [R]) / t

Half-Life Period (t_1/2): At t = t_1/2, [R] = [R]₀ / 2.

k = ([R]0 - [R]0/2) / t1/2 = [R]0 / (2 t1/2)

t_1/2 = [R]₀ / (2k) (Directly proportional to initial concentration).

2. Integrated Rate Equation for a First-Order Reaction

For a first-order reaction: R → P

The differential rate expression is:

-d[R]/dt = k[R] ⇒ d[R]/[R] = -k dt

Integrating both sides:

ln[R] = -kt + I

At t = 0, [R] = [R]₀ ⇒ I = ln[R]0.

Substituting I into the expression gives:

ln[R] = -kt + ln[R]0 ⇒ ln([R]0 / [R]) = kt

Converting natural logarithm (ln) to common base-10 logarithm (log₁₀):

k = (2.303 / t) × log₁₀([R]₀ / [R])

Half-Life Period (t_1/2): When t = t_1/2, [R] = [R]₀ / 2.

k = (2.303 / t1/2) × log10([R]0 / ([R]0 / 2)) = (2.303 / t1/2) × log10(2)

Since log₁₀(2) ≈ 0.3010:

t_1/2 = (2.303 × 0.3010) / k = 0.693 / k

Note: The half-life of a first-order reaction is completely independent of the initial concentration of reactants.

3. Temperature Dependence: The Arrhenius Equation

Svante Arrhenius proposed an empirical relationship quantifying the influence of temperature on the rate constant:

k = A × e^{-E_a / (RT)}

Where A is the Arrhenius pre-exponential frequency factor, E_a is the activation energy in J·mol^-1, R is the universal gas constant (8.314 J·K^-1·mol^-1), and T is absolute temperature in Kelvin.

Taking the natural logarithm on both sides:

ln k = ln A - (Ea / RT)

Converting to base-10 logarithms:

log₁₀ k = log₁₀ A - E_a / (2.303 RT)

If the rate constants are k₁ and k₂ at temperatures T₁ and T₂ respectively, the formula becomes:

log₁₀(k₂ / k₁) = (E_a / 2.303 R) × [(T₂ - T₁) / (T₁ × T₂)]

NCERT Intext and Textbook Exercise Solutions

Below are step-by-step whiteboard-style solutions to representative numerical problems from the NCERT Solutions Class 12 Chemistry Chapter exercises.

Problem 1: Expressing Rates of Reaction

Question: For the reaction 2A + B → 3C + D, the concentration of A decreases from 0.5 mol·L^-1 to 0.3 mol·L^-1 in 10 minutes. Calculate the average rate of reaction and the rate of production of C.

1. Given: Initial concentration [A]₁ = 0.5 mol·L^-1, Final concentration [A]₂ = 0.3 mol·L^-1, Time interval Δt = 10 min.
2. To Find: Average rate of reaction (r_avg) and rate of formation of C (d[C]/dt).
3. Formula:
   • Rate = -(1/2) × (Δ[A] / Δt)
   • Rate = +(1/3) × (Δ[C] / Δt) ⇒ Δ[C]/Δt = 3 × Rate
4. Substitution and Calculation:
   • Δ[A] = [A]₂ - [A]₁ = 0.3 - 0.5 = -0.2 mol·L^-1
   • Rate = -(1/2) × (-0.2 mol·L^-1 / 10 min) = 0.01 mol·L^-1·min^-1
   • Rate of production of C = 3 × 0.01 = 0.03 mol·L^-1·min^-1
5. Final Answer: Average Rate = 1.0 × 10^-2 mol·L^-1·min^-1; Rate of formation of C = 3.0 × 10^-2 mol·L^-1·min^-1

Problem 2: Time Required for Fraction of First-Order Reaction

Question: Show that in a first-order reaction, the time required for 99.9% completion is 10 times the time required for 50% completion (half-life) of the reaction.

1. Given: First-order reaction; Reaction completions = 99.9% and 50%.
2. To Find: Ratio t_99.9% / t_50%.
3. Formula:
   • t = (2.303 / k) × log₁₀([R]₀ / [R])
   • t_50% = 0.693 / k = (2.303 / k) × log₁₀(2) ≈ (2.303 / k) × 0.3010
4. Substitution and Calculation:
   • For 99.9% completion: [R] = [R]₀ - 0.999[R]₀ = 0.001[R]₀ = 10^-3[R]₀
   • t_99.9% = (2.303 / k) × log₁₀([R]₀ / 10^-3[R]₀) = (2.303 / k) × log₁₀(10³) = (2.303 / k) × 3
   • Taking the ratio: t_99.9% / t_50% = [(2.303 / k) × 3] / [(2.303 / k) × 0.3010] = 3 / 0.3010 ≈ 9.967 ≈ 10
5. Final Answer: Hence proved: t_99.9% ≈ 10 × t_50%

Problem 3: Calculating Activation Energy Using Arrhenius Equation

Question: The rate constant of a reaction doubles when the temperature increases from 298 K to 308 K. Calculate the activation energy (E_a) of the reaction. (Given: R = 8.314 J·K^-1·mol^-1, log 2 = 0.3010).

1. Given: T₁ = 298 K, T₂ = 308 K, k₂ / k₁ = 2, R = 8.314 J·K^-1·mol^-1.
2. To Find: Activation energy (E_a).
3. Formula: log10(k2 / k1) = [Ea / (2.303 × R)] × [(T2 - T1) / (T1 × T2)]
4. Substitution:
   • log₁₀(2) = [E_a / (2.303 × 8.314)] × [(308 - 298) / (298 × 308)]
   • 0.3010 = [E_a / 19.147] × [10 / 91784]
   • E_a = (0.3010 × 19.147 × 91784) / 10
   • E_a = 52897.8 J·mol^-1 = 52.90 kJ·mol^-1
5. Final Answer: Activation Energy (E_a) = 52.9 kJ·mol^-1

Important CBSE Board Questions and Detailed Solutions

The following selected questions from the official CBSE question bank represent key high-yield topics frequently tested in Class 12 Chemistry board examinations. Study these fully worked solutions carefully.

CBSE Question 1: Raoult's Law for Volatile Liquids

Question: State Raoult's Law for volatile liquids.

1. Statement: Raoult's Law states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction present in the solution.
2. Mathematical Expression: For a binary solution of components 1 and 2:
   • p1 = p1° × x1
   • p2 = p2° × x2
   • Total Vapour Pressure: ptotal = p1 + p2 = p1°x1 + p2°x2 = p1°(1 - x2) + p2°x2
   Where p₁° and p₂° are the vapour pressures of pure components 1 and 2, and x₁, x₂ are their respective mole fractions in the solution.
3. Final Answer: p₁ = p₁° × x₁ (Partial vapour pressure = Pure vapour pressure × Mole fraction)

CBSE Question 2: Molarity vs. Molality

Question: Define molarity and molality. Which one is temperature independent and why?

1. Definitions:
   • Molarity (M): Defined as the number of moles of solute dissolved in one litre (1 L or 1 dm³) of solution. Formula: M = (Moles of solute) / (Volume of solution in Litres).
   • Molality (m): Defined as the number of moles of solute present per kilogram (1 kg) of solvent. Formula: m = (Moles of solute) / (Mass of solvent in kg).
2. Temperature Dependence Analysis:
   • Molarity depends on the total volume of the solution. Since volume expands or contracts with changes in temperature (volume is temperature-dependent), molarity changes with temperature.
   • Molality depends solely on the masses of solute and solvent. Since mass is an invariant quantity that does not alter with temperature changes, molality is strictly temperature independent.
3. Final Answer: Molality is temperature independent because mass does not change with temperature.

CBSE Question 3: Ideal Solutions and Characteristics

Question: What is an ideal solution? Give two characteristics.

1. Definition: An ideal solution is a homogeneous mixture in which the intermolecular attractive forces between solute-solvent (A-B) molecules are identical to those between solute-solute (A-A) and solvent-solvent (B-B) molecules, obeying Raoult's Law across the entire range of concentrations and temperatures.
2. Key Characteristics:
   • Enthalpy of Mixing is zero (ΔH_mix = 0): No heat is absorbed or evolved during the formation of the solution.
   • Volume of Mixing is zero (ΔV_mix = 0): The total volume of the solution equals the sum of the volumes of the individual components before mixing.
3. Example: Benzene + Toluene, n-hexane + n-heptane, or Bromoethane + Chloroethane.
4. Final Answer: An ideal solution obeys Raoult's Law at all concentrations with ΔH_mix = 0 and ΔV_mix = 0.

CBSE Question 4: Molarity Calculation for 5% (w/v) NaOH

Question: Calculate the molarity of 5% (w/v) NaOH solution.

1. Given: Concentration = 5% (w/v) NaOH (meaning 5.0 g of NaOH dissolved in 100 mL of solution). Molar mass of NaOH (M_B) = 23 + 16 + 1 = 40.0 g·mol^-1.
2. To Find: Molarity (M) of the solution.
3. Formula: M = (Mass of solute in g × 1000) / (Molar mass of solute × Volume of solution in mL)
4. Substitution:
   • Mass of NaOH (w_B) = 5.0 g
   • Volume of solution (V) = 100 mL
   • Moles of NaOH = 5.0 / 40.0 = 0.125 mol
   • Volume in Litres = 100 / 1000 = 0.10 L
   • Molarity M = 0.125 mol / 0.10 L = 1.25 M
5. Final Answer: Molarity of 5% (w/v) NaOH = 1.25 M (or 1.25 mol·L^-1)

CBSE Question 5: Molar Mass Determination from Relative Lowering of Vapour Pressure

Question: The vapour pressure of pure benzene is 640 mm Hg. A non-volatile solute of mass 2.175 g is dissolved in 39.0 g of benzene. The vapour pressure of the solution becomes 600 mm Hg. Calculate the molar mass of the solute. (Molar mass of benzene C₆H₆ = 78 g·mol^-1).

1. Given: Pure vapour pressure (p°) = 640 mm Hg, Solution vapour pressure (p) = 600 mm Hg, Mass of non-volatile solute (w₂) = 2.175 g, Mass of benzene solvent (w₁) = 39.0 g, Molar mass of benzene (M₁) = 78.0 g·mol^-1.
2. To Find: Molar mass of non-volatile solute (M₂).
3. Formula: Using Raoult's Law for dilute solutions:

   (p° - p) / p° = n2 / n1 = (w2 / M2) / (w1 / M1) = (w2 × M1) / (M2 × w1)

4. Step-by-Step Substitution:
   • Relative lowering of vapour pressure: (640 - 600) / 640 = 40 / 640 = 1 / 16 = 0.0625
   • Moles of solvent n₁ = w₁ / M₁ = 39.0 / 78.0 = 0.50 mol
   • 0.0625 = (2.175 / M₂) / 0.50
   • 0.0625 × 0.50 = 2.175 / M₂
   • 0.03125 = 2.175 / M₂
   • M₂ = 2.175 / 0.03125 = 69.6 g·mol^-1
5. Final Answer: Molar mass of the solute = 69.6 g·mol^-1

CBSE Question 6: Osmotic Pressure and Biomolecules

Question: Explain osmotic pressure and its application in determining the molecular mass of biomolecules.

1. Definition: Osmotic pressure (Π) is the excess hydrostatic pressure that must be applied to the solution side to prevent the inward osmosis of pure solvent across a semipermeable membrane (SPM).
2. Mathematical Formulation: According to the van 't Hoff equation:

   Π = CRT = (nB / V)RT = (wB × R × T) / (MB × V)

   Rearranging for molar mass: MB = (wB × R × T) / (Π × V)
3. Why Osmotic Pressure is Preferred for Biomolecules:
   • Biomolecules (proteins, enzymes, nucleic acids, and polymers) are unstable and decompose at elevated temperatures; osmotic pressure is measured at ambient room temperature (unlike elevation in boiling point).
   • Biomolecules have very high molecular masses, leading to negligible depression in freezing point or elevation in boiling point. However, osmotic pressure values remain appreciably large and accurately measurable even for highly dilute solutions.
   • It uses molarity (mol·L^-1) rather than molality, making preparation and measurement convenient in biological buffers.
4. Final Answer: Osmotic pressure (Π = CRT) allows accurate molar mass determination at room temperature without denaturing delicate biomolecules.

CBSE Question 7: Molal Depression Constant (K_f)

Question: What is the freezing point depression constant (K_f)? Derive the expression relating it to the enthalpy of fusion.

1. Definition: The molal depression constant (K_f), also known as the cryoscopic constant, is defined as the depression in freezing point produced when 1 mole of a non-volatile, non-electrolyte solute is dissolved in 1 kilogram (1000 g) of a solvent (i.e., for a 1 molal solution).
2. Equation: ΔTf = Kf × m (where m is molality).
3. Thermodynamic Relationship: From thermodynamic principles, K_f depends exclusively on the intrinsic properties of the solvent:

   Kf = (R × M1 × Tf°2) / (1000 × ΔHfus)

   Where:
   • R = Universal gas constant (8.314 J·K^-1·mol^-1)
   • M₁ = Molar mass of the solvent (in g·mol^-1)
   • T_f° = Normal freezing point of pure solvent (in K)
   • ΔH_fus = Molar enthalpy of fusion of the solvent (in J·mol^-1)
   For water, K_f = 1.86 K·kg·mol^-1.
4. Final Answer: K_f represents freezing point depression for a 1 m solution, given by K_f = (R × M₁ × T_f°²) / (1000 × ΔH_fus).

CBSE Question 8: Calculating Osmotic Pressure of Glucose Solution

Question: Calculate the osmotic pressure of a solution containing 1.0 g of glucose (C₆H₁₂O₆) dissolved in 100 mL of solution at 300 K. (R = 0.0821 L·atm·K^-1·mol^-1).

1. Given: Mass of glucose (w_B) = 1.0 g, Molar mass of glucose (M_B) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g·mol^-1, Volume of solution (V) = 100 mL = 0.10 L, Temperature (T) = 300 K, Gas constant (R) = 0.0821 L·atm·K^-1·mol^-1.
2. To Find: Osmotic pressure (Π).
3. Formula: Π = (wB × R × T) / (MB × V)
4. Substitution:
   • Moles of glucose = 1.0 / 180 = 0.005556 mol
   • Molar concentration (C) = 0.005556 mol / 0.10 L = 0.05556 mol·L^-1
   • Π = 0.05556 × 0.0821 × 300
   • Π = 1.368 ≈ 1.37 atm
   • In SI units (Pascals): 1.37 × 101325 Pa = 1.39 × 10⁵ Pa
5. Final Answer: Osmotic Pressure (Π) = 1.37 atm (or 1.39 × 10⁵ Pa)

Important Formulas and Key Equations

Here is a quick-revision summary of all crucial mathematical relationships in Class 12 Chemical Kinetics and Physical Chemistry solutions for quick board exam recall:

Concept / Parameter	Formula	Key Units / Notes
General Rate of Reaction (aA + bB → cC + dD)	Rate = -(1/a)(d[A]/dt) = +(1/c)(d[C]/dt)	mol·L-1·s-1 (or bar·s-1 for gases)
Units of Rate Constant (k) (Reaction of order n)	Units = (mol·L-1)1-n s-1	n=0: mol·L-1·s-1 n=1: s-1 n=2: L·mol-1·s-1
Zero-Order Integrated Rate Law	k = ([R]0 - [R]) / t	Linear plot: [R] vs t (Slope = -k)
Zero-Order Half-Life	t1/2 = [R]0 / (2k)	Proportional to initial concentration [R]0
First-Order Integrated Rate Law	k = (2.303 / t) × log10([R]0 / [R])	Linear plot: log[R] vs t (Slope = -k/2.303)
First-Order Half-Life	t1/2 = 0.693 / k	Independent of initial concentration [R]0
Arrhenius Two-Temperature Form	log(k2/k1) = [Ea / (2.303R)] × [(T2 - T1)/(T1T2)]	Plot: log k vs 1/T (Slope = -Ea / 2.303R)
Colligative: Osmotic Pressure	Π = CRT = (wB R T) / (MB V)	R = 0.0821 L·atm·K-1·mol-1 or 8.314 J·K-1·mol-1
Colligative: Freezing Point Depression	ΔTf = Kf × m = (1000 × Kf × wB) / (MB × wA)	Kf in K·kg·mol-1

Common Mistakes and How to Avoid Them

When solving Chemical Kinetics and Solution problems in board exams, students often lose marks due to subtle procedural oversights. Keep the following pointers in mind:

• Stoichiometric Coefficients in Rate Expressions: For a reaction like 2N2O5 → 4NO2 + O2, remember that Rate of Disappearance of N₂O₅ is -d[N2O5]/dt, while the Rate of Reaction is -(1/2)d[N2O5]/dt. Do not equate the overall rate to the disappearance rate without dividing by its coefficient.
• Units of the Rate Constant: Never write generic units for k. For a first-order reaction, the unit is s-1 or min-1; for a zero-order reaction, it is mol·L-1·s-1; and for second-order, it is L·mol-1·s-1. Always double check using (mol/L)1-n s-1.
• Temperature in Kelvin: In all Arrhenius equation and osmotic pressure (Π = CRT) calculations, temperatures must strictly be converted to Kelvin (K = °C + 273.15). Leaving temperature in Celsius leads to incorrect exponential values.
• Gas Constant (R) Consistency: When calculating activation energy (E_a) in J·mol^-1, use R = 8.314 J·K-1·mol-1. When computing osmotic pressure with pressure in atmospheres and volume in litres, use R = 0.0821 L·atm·K-1·mol-1.
• Concentration Remaining vs. Reacted: In first-order problems where "a reaction is 75% complete," the concentration remaining is [R] = [R]0 - 0.75[R]0 = 0.25[R]0. Do not substitute 0.75[R]₀ into the denominator of the logarithmic formula.

CBSE Board Exam Weightage and Strategy

According to the latest CBSE Class 12 Chemistry examination blueprint, Physical Chemistry chapters (Solutions, Electrochemistry, and Chemical Kinetics) collectively carry 23 out of 70 marks in the theory paper. Chemical Kinetics typically accounts for 7 marks.

Expected question typology for Chemical Kinetics includes:

1. Multiple Choice Questions (1 Mark): Determining reaction order from rate constants, units of k, and effects of positive catalysts on activation energy barriers.
2. Assertion-Reason Questions (1 Mark): Conceptual tests on elementary reactions, zero-order conditions, or molecularity limits (molecularity > 3 is rare due to unlikelihood of simultaneous collisions).
3. Short Answer Questions (2 or 3 Marks): Derivations of integrated rate equations, half-life problems, or table-based initial rate method numericals.
4. Long Answer / Case-Based Questions (4 or 5 Marks): Multi-part problems combining Arrhenius graphs, temperature coefficients, and reaction mechanism step-determination.

More NCERT Solutions and Practice

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