Chapter 12: Chemical Kinetics

Introduction

Chemical Kinetics Overview

• Chemical kinetics studies reaction rates and mechanisms.
• It helps determine how fast reactions occur and how conditions affect rates.

Solved Examples

• Ex 1: Identify the branch of chemistry studying reaction speed.
  Answer: Chemical kinetics.

Practice Set

• Q1: What does chemical kinetics study?
  A) Reaction equilibrium
  B) Reaction rates and mechanisms
  C) Atomic structure
  D) Molecular bonding
• Q2: Which factor is NOT studied in chemical kinetics?
  A) Reaction speed
  B) Reaction mechanism
  C) Reaction equilibrium constant
  D) Effect of temperature on rate
• Q3: Chemical kinetics helps to:
  A) Predict product stability
  B) Determine reaction speed and rate changes
  C) Calculate molecular mass
  D) Identify element types

Answer Key

• Q1: B - Chemical kinetics focuses on reaction rates and mechanisms.
• Q2: C - Equilibrium constant is studied in chemical equilibrium, not kinetics.
• Q3: B - Kinetics determines speed and how rates change with conditions.

Memory Aids

• "Kinetics = Speed + Path" (rate and mechanism)
• Focus on how fast and how reactions proceed.

Rate of a Chemical Reaction

Definition and Units

• Rate = change in concentration of reactant or product per unit time.
• For reaction R → P, rate = -Δ[R]/Δt or +Δ[P]/Δt.
• Units: mol L^-1 s^-1 or atm s^-1 for gases.

Average and Instantaneous Rate

• Average rate: change over a large time interval, r_av = -Δ[R]/Δt or +Δ[P]/Δt.
• Instantaneous rate: rate at a specific time, r_inst = -d[R]/dt or +d[P]/dt.
• Δ indicates larger change (average), d indicates small change (instantaneous).

Solved Examples

• Ex 1: Calculate average rate if [R] changes from 0.5 to 0.2 mol/L in 10 s.
  Rate = -(0.2 - 0.5)/10 = 0.03 mol L^-1 s^-1.
• Ex 2: Find instantaneous rate at t = 5 s from slope of tangent on concentration-time graph.

Practice Set

• Q1: Rate of reaction is defined as:
  A) Change in temperature per time
  B) Change in concentration per time
  C) Change in pressure per time
  D) Change in volume per time
• Q2: Units of reaction rate for gases are:
  A) mol L^-1 s^-1
  B) atm s^-1
  C) Both A and B
  D) mol s^-1
• Q3: Instantaneous rate differs from average rate because:
  A) It uses larger time intervals
  B) It is rate at a specific instant
  C) It ignores concentration changes
  D) It is always zero

Answer Key

• Q1: B - Rate measures concentration change per unit time.
• Q2: C - Both mol L^-1 s^-1 and atm s^-1 are used.
• Q3: B - Instantaneous rate is at a specific moment, unlike average rate.

Memory Aids

• Average rate uses Δ (big change), instantaneous uses d (small change).
• Rate = concentration change/time interval.

Definition and Units

• Rate = change in concentration of reactant or product per unit time.
• For reaction R → P, rate = -Δ[R]/Δt or +Δ[P]/Δt.
• Units: mol L^-1 s^-1 or atm s^-1 for gases.

Average and Instantaneous Rate

• Average rate: change over a large time interval, r_av = -Δ[R]/Δt or +Δ[P]/Δt.
• Instantaneous rate: rate at a specific time, r_inst = -d[R]/dt or +d[P]/dt.
• Δ indicates larger change (average), d indicates small change (instantaneous).

Solved Examples

• Ex 1: Calculate average rate if [R] changes from 0.5 to 0.2 mol/L in 10 s.
  Rate = -(0.2 - 0.5)/10 = 0.03 mol L^-1 s^-1.
• Ex 2: Find instantaneous rate at t = 5 s from slope of tangent on concentration-time graph.

Practice Set

• Q1: Rate of reaction is defined as:
  A) Change in temperature per time
  B) Change in concentration per time
  C) Change in pressure per time
  D) Change in volume per time
• Q2: Units of reaction rate for gases are:
  A) mol L^-1 s^-1
  B) atm s^-1
  C) Both A and B
  D) mol s^-1
• Q3: Instantaneous rate differs from average rate because:
  A) It uses larger time intervals
  B) It is rate at a specific instant
  C) It ignores concentration changes
  D) It is always zero

Answer Key

• Q1: B - Rate measures concentration change per unit time.
• Q2: C - Both mol L^-1 s^-1 and atm s^-1 are used.
• Q3: B - Instantaneous rate is at a specific moment, unlike average rate.

Memory Aids

• Average rate uses Δ (big change), instantaneous uses d (small change).
• Rate = concentration change/time interval.

Integrated Rate Equations

Zero Order Reaction

• Rate is independent of reactant concentration.
• Rate remains constant: rate = k.
• Reaction: R → Products, rate = -d[R]/dt = k[R]⁰ = k.
• Integrated rate equation: k = ([R]₀ - [R]_t)/t.
• Concentration-time relation: [R]_t = -kt + [R]₀ (linear with slope -k).
• Half-life: t_1/2 = [R]₀/(2k), directly proportional to initial concentration.

Solved Examples

• Ex 1: Calculate rate constant k if [R]₀ = 0.5 M, [R]_t = 0.2 M after 10 s.
  Solution: k = (1/10)(0.5 - 0.2) = 0.03 M/s.

Practice Set

• Q1: For a zero order reaction, rate is:
  A. Proportional to [R]
  B. Independent of [R]
  C. Proportional to [R]²
  D. Inversely proportional to [R]
• Q2: The half-life of a zero order reaction is:
  A. Constant
  B. Proportional to [R]₀
  C. Inversely proportional to [R]₀
  D. Independent of [R]₀
• Q3: If k = 0.04 M/s and [R]₀ = 0.8 M, time to reduce concentration to 0.4 M is:
  A. 10 s
  B. 20 s
  C. 5 s
  D. 15 s

Answer Key

• Q1: B - Rate is independent of concentration in zero order.
• Q2: B - Half-life is directly proportional to initial concentration.
• Q3: B - t = ([R]₀ - [R]_t)/k = (0.8 - 0.4)/0.04 = 10 s.

Memory Aids

• Zero order: Rate constant, straight line [R] vs. t.
• Half-life ∝ initial concentration.
• Rate independent of [R].

First Order Reaction

• Rate depends on one reactant concentration: rate = k[R].
• Integrated rate equation: k = (2.303/t) log ([R]₀/[R]_t).
• Half-life: t_1/2 = 0.693/k, independent of concentration.
• Time for n^th fraction completion: t_1/n = (2.303/k) log (n/(n-1)).
• Time to reduce to n^th fraction: t = (2.303/k) log n.

Solved Examples

• Ex 1: Calculate k if [R]₀ = 0.5 M, [R]_t = 0.25 M after 20 s.
  Solution: k = (2.303/20) log(0.5/0.25) = 0.115 s^-1.
• Ex 2: Find half-life if k = 0.1 s^-1.
  Solution: t_1/2 = 0.693/0.1 = 6.93 s.

Practice Set

• Q1: Half-life of first order reaction is:
  A. Proportional to [R]₀
  B. Independent of [R]₀
  C. Proportional to 1/[R]₀
  D. Proportional to [R]₀²
• Q2: For first order, integrated rate law is:
  A. [R]_t = -kt + [R]₀
  B. k = (2.303/t) log([R]₀/[R]_t)
  C. k = (1/t)([R]₀ - [R]_t)
  D. k = (1/t) log([R]₀/[R]_t)
• Q3: Time to reduce to half concentration for k = 0.2 s^-1 is:
  A. 3.465 s
  B. 0.693 s
  C. 6.93 s
  D. 1.386 s

Answer Key

• Q1: B - Half-life is independent of initial concentration.
• Q2: B - Correct integrated rate law for first order.
• Q3: A - t_1/2 = 0.693/0.2 = 3.465 s.

Memory Aids

• First order: Half-life constant, independent of concentration.
• Plot log[R] vs. time for straight line.
• Use t_1/2 = 0.693/k for calculations.

General Integrated Rate Equations and Graphs

• Zero order: Rate = k, [R] = -kt + [R]₀.
• First order: Rate = k[R], log[R] = -kt/2.303 + log[R]₀.
• Second order: Rate = k[R]², 1/[R] = kt + 1/[R]₀.
• Third order: Rate = k[R]³, 1/[R]² = 2kt + 1/[R]₀².
• Half-life varies with order:
  • Zero order: t_1/2 ∝ [R]₀
  • First order: t_1/2 = constant
  • Second order: t_1/2 ∝ 1/[R]₀
  • Third order: t_1/2 ∝ 1/[R]₀²

Time Relationships in First Order Reactions

• Time for different completion fractions related to half-life:
• 75% completion ≈ 2 t_1/2, 87.5% ≈ 3 t_1/2, 93.75% ≈ 4 t_1/2, etc.
• Amount left after n half-lives: [R]₀/2ⁿ.
• Pseudo first order reactions behave like first order under certain conditions.

Examples of Pseudo First Order Reactions

• Acid hydrolysis of ethyl acetate: Rate = k[ethyl acetate].
• Acid-catalyzed inversion of cane sugar: Rate = k[cane sugar].

Zero Order Reaction

• Rate is independent of reactant concentration.
• Rate remains constant: rate = k.
• Reaction: R → Products, rate = -d[R]/dt = k[R]⁰ = k.
• Integrated rate equation: k = \(\frac{1}{t}([R]_0 - [R]_t)\).
• Concentration-time relation: [R]_t = -kt + [R]₀ (linear with slope -k).
• Half-life: t_1/2 = \(\frac{[R]_0}{2k}\), directly proportional to initial concentration.

Solved Examples

• Ex 1: Calculate rate constant k if [R]₀ = 0.5 M, [R]_t = 0.2 M after 10 s.
  Solution: k = (1/10)(0.5 - 0.2) = 0.03 M/s.

Practice Set

• Q1: For a zero order reaction, rate is:
  A. Proportional to [R]
  B. Independent of [R]
  C. Proportional to [R]²
  D. Inversely proportional to [R]
• Q2: The half-life of a zero order reaction is:
  A. Constant
  B. Proportional to [R]₀
  C. Inversely proportional to [R]₀
  D. Independent of [R]₀
• Q3: If k = 0.04 M/s and [R]₀ = 0.8 M, time to reduce concentration to 0.4 M is:
  A. 10 s
  B. 20 s
  C. 5 s
  D. 15 s

Answer Key

• Q1: B - Rate is independent of concentration in zero order.
• Q2: B - Half-life is directly proportional to initial concentration.
• Q3: B - t = ([R]₀ - [R]_t)/k = (0.8 - 0.4)/0.04 = 10 s.

Memory Aids

• Zero order: Rate constant, straight line [R] vs. t.
• Half-life ∝ initial concentration.
• Rate independent of [R].

First Order Reaction

• Rate depends on one reactant concentration: rate = k[R].
• Integrated rate equation: k = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]_t}\).
• Half-life: t_1/2 = 0.693/k, independent of concentration.
• Time for n^th fraction completion: t_1/n = \(\frac{2.303}{k} \log \frac{n}{n-1}\).
• Time to reduce to n^th fraction: t = \(\frac{2.303}{k} \log n\).

Solved Examples

• Ex 1: Calculate k if [R]₀ = 0.5 M, [R]_t = 0.25 M after 20 s.
  Solution: k = (2.303/20) log(0.5/0.25) = 0.115 s^-1.
• Ex 2: Find half-life if k = 0.1 s^-1.
  Solution: t_1/2 = 0.693/0.1 = 6.93 s.

Practice Set

• Q1: Half-life of first order reaction is:
  A. Proportional to [R]₀
  B. Independent of [R]₀
  C. Proportional to 1/[R]₀
  D. Proportional to [R]₀²
• Q2: For first order, integrated rate law is:
  A. [R]_t = -kt + [R]₀
  B. k = (2.303/t) log([R]₀/[R]_t)
  C. k = (1/t)([R]₀ - [R]_t)
  D. k = (1/t) log([R]₀/[R]_t)
• Q3: Time to reduce to half concentration for k = 0.2 s^-1 is:
  A. 3.465 s
  B. 0.693 s
  C. 6.93 s
  D. 1.386 s

Answer Key

• Q1: B - Half-life is independent of initial concentration.
• Q2: B - Correct integrated rate law for first order.
• Q3: A - t_1/2 = 0.693/0.2 = 3.465 s.

Memory Aids

• First order: Half-life constant, independent of concentration.
• Plot log[R] vs. time for straight line.
• Use t_1/2 = 0.693/k for calculations.

General Integrated Rate Equations and Graphs

• Zero order: Rate = k, [R] = -kt + [R]₀.
• First order: Rate = k[R], log[R] = -kt/2.303 + log[R]₀.
• Second order: Rate = k[R]², 1/[R] = kt + 1/[R]₀.
• Third order: Rate = k[R]³, 1/[R]² = 2kt + 1/[R]₀².
• Half-life varies with order:
  • Zero order: t_1/2 ∝ [R]₀
  • First order: t_1/2 = constant
  • Second order: t_1/2 ∝ 1/[R]₀
  • Third order: t_1/2 ∝ 1/[R]₀²

Time Relationships in First Order Reactions

• Time for different completion fractions related to half-life:
• 75% completion ≈ 2 t_1/2, 87.5% ≈ 3 t_1/2, 93.75% ≈ 4 t_1/2, etc.
• Amount left after n half-lives: [R]₀/2ⁿ.
• Pseudo first order reactions behave like first order under certain conditions.

Examples of Pseudo First Order Reactions

• Acid hydrolysis of ethyl acetate: Rate = k[ethyl acetate].
• Acid-catalyzed inversion of cane sugar: Rate = k[cane sugar].

Collision Theory

Explanation

• Reaction occurs due to collision between reactant molecules (assumed as hard spheres).
• Not all collisions lead to reaction; only effective collisions do.
• Effective collisions require sufficient kinetic energy (threshold energy = activation energy + energy of reactants) and proper molecular orientation.
• Rate equation: Rate = PZ_ABe^-E_a/RT, where P = steric factor, Z_AB = collision frequency.
• Activation energy and orientation determine reaction rate.
• Drawback: Ignores molecular structure, treats molecules as hard spheres.

Solved Examples

• Ex 1: Calculate rate if P = 0.5, Z_AB = 10¹⁰ collisions/s, E_a = 50 kJ/mol, T = 300 K, R = 8.314 J/mol·K.
  Solution: Rate = 0.5 × 10¹⁰ × e^{-(50000)/(8.314×300)} = 0.5 × 10¹⁰ × e^-20.04 ≈ negligible (very slow reaction).

Practice Set

• Q1: What is required for an effective collision?
  • A. Any collision
  • B. Collision with sufficient energy and proper orientation
  • C. Collision with low energy
  • D. Collision without orientation
• Q2: Which factor does NOT affect collision frequency?
  • A. Concentration of reactants
  • B. Temperature
  • C. Molecular orientation
  • D. Volume of reaction mixture
• Q3: Rate equation in collision theory includes:
  • A. Activation energy and steric factor
  • B. Only activation energy
  • C. Only collision frequency
  • D. None of these

Answer Key

• Q1: B - Effective collisions need sufficient energy and proper orientation.
• Q2: C - Molecular orientation affects effectiveness, not frequency.
• Q3: A - Rate = PZ_ABe^-E_a/RT includes activation energy and steric factor.

Memory Aids

• "Collision + Energy + Orientation = Reaction"
• Only collisions with energy ≥ activation energy and correct orientation produce products.
• Hard sphere model ignores molecular shape.

Explanation

• Reaction occurs due to collision between reactant molecules (assumed as hard spheres).
• Not all collisions lead to reaction; only effective collisions do.
• Effective collisions require sufficient kinetic energy (threshold energy = activation energy + energy of reactants) and proper molecular orientation.
• Rate equation: Rate = PZ_ABe^\u2212E_a/RT, where P = steric factor, Z_AB = collision frequency.
• Activation energy and orientation determine reaction rate.
• Drawback: Ignores molecular structure, treats molecules as hard spheres.

Solved Examples

• Ex 1: Calculate rate if P = 0.5, Z_AB = 10¹⁰ collisions/s, E_a = 50 kJ/mol, T = 300 K, R = 8.314 J/mol·K.
  Solution: Rate = 0.5 × 10¹⁰ × e^{\u2212(50000)/(8.314×300)} = 0.5 × 10¹⁰ × e^\u221220.04 ≈ negligible (very slow reaction).

Practice Set

• Q1: What is required for an effective collision?
  • A. Any collision
  • B. Collision with sufficient energy and proper orientation
  • C. Collision with low energy
  • D. Collision without orientation
• Q2: Which factor does NOT affect collision frequency?
  • A. Concentration of reactants
  • B. Temperature
  • C. Molecular orientation
  • D. Volume of reaction mixture
• Q3: Rate equation in collision theory includes:
  • A. Activation energy and steric factor
  • B. Only activation energy
  • C. Only collision frequency
  • D. None of these

Answer Key

• Q1: B - Effective collisions need sufficient energy and proper orientation.
• Q2: C - Molecular orientation affects effectiveness, not frequency.
• Q3: A - Rate = PZ_ABe^\u2212E_a/RT includes activation energy and steric factor.

Memory Aids

• "Collision + Energy + Orientation = Reaction"
• Only collisions with energy ≥ activation energy and correct orientation produce products.
• Hard sphere model ignores molecular shape.

Reaction Order and Rate Constants

Explanation

• Zero order: k = ([A]₀ - [A]) / t; half-life t_1/2 = [A]₀ / 2k.
• First order: k = (2.303 / t) log ([A]₀ / [A]) = (2.303 / t) log (a / (a - x)).
• In terms of pressure: k = (2.303 / t) log (P_i / (2P_i - P_t)).
• Half-life for first order: t_1/2 = 0.693 / k.
• Other times: t_60% = 0.9166 / k, t_75% = 1.386 / k, t_90% = 2.303 / k, t_99.9% = 6.93 / k.
• Time to reduce to n^th fraction: t = (2.303 / k) log (a / (a / n)) = (2.303 / k) log n.
• Concentration after n half-lives: [A] = [A]₀ / 2ⁿ.
• Order determination by half-life method: n = 1 + (log t_1/21 - log t_1/22) / (log a₂ - log a₁).

Solved Examples

• Ex 1: For a zero order reaction with [A]₀ = 0.1 M, k = 0.02 M/s, find t_1/2.
  Solution: t_1/2 = [A]₀ / 2k = 0.1 / (2 × 0.02) = 2.5 s.
• Ex 2: For a first order reaction, if k = 0.1 s^-1, find t_1/2.
  Solution: t_1/2 = 0.693 / k = 0.693 / 0.1 = 6.93 s.

Practice Set

• Q1: For a zero order reaction, half-life is:
  • A. Independent of initial concentration
  • B. Directly proportional to initial concentration
  • C. Inversely proportional to initial concentration
  • D. Equal to 0.693/k
• Q2: For a first order reaction, rate constant k is given by:
  • A. k = (2.303 / t) log (a / (a - x))
  • B. k = (a - x) / t
  • C. k = (a + x) / t
  • D. k = (2.303 / t) log (a + x)
• Q3: The half-life of a first order reaction is 10 min. What is the rate constant?
  • A. 0.0693 min^-1
  • B. 0.693 min^-1
  • C. 6.93 min^-1
  • D. 0.00693 min^-1

Answer Key

• Q1: B - Half-life for zero order is proportional to initial concentration.
• Q2: A - k = (2.303 / t) log (a / (a - x)) for first order.
• Q3: A - k = 0.693 / t_1/2 = 0.693 / 10 = 0.0693 min^-1.

Memory Aids

• Zero order: t_1/2 [A]₀, First order: t_1/2 constant.
• First order k from log formula: k = (2.303 / t) log (a / (a - x)).
• Half-life formula: t_1/2 = 0.693 / k for first order.

Explanation

• Zero order: k = ([A]₀ - [A]) / t; half-life t_1/2 = [A]₀ / 2k.
• First order: k = (2.303 / t) log ([A]₀ / [A]) = (2.303 / t) log (a / (a - x)).
• In terms of pressure: k = (2.303 / t) log (P_i / (2P_i - P_t)).
• Half-life for first order: t_1/2 = 0.693 / k.
• Other times: t_60% = 0.9166 / k, t_75% = 1.386 / k, t_90% = 2.303 / k, t_99.9% = 6.93 / k.
• Time to reduce to n^th fraction: t = (2.303 / k) log (a / (a / n)) = (2.303 / k) log n.
• Concentration after n half-lives: [A] = [A]₀ / 2ⁿ.
• Order determination by half-life method: n = 1 + (log t_1/21 - log t_1/22) / (log a₂ - log a₁).

Solved Examples

• Ex 1: For a zero order reaction with [A]₀ = 0.1 M, k = 0.02 M/s, find t_1/2.
  Solution: t_1/2 = [A]₀ / 2k = 0.1 / (2 × 0.02) = 2.5 s.
• Ex 2: For a first order reaction, if k = 0.1 s⁻¹, find t_1/2.
  Solution: t_1/2 = 0.693 / k = 0.693 / 0.1 = 6.93 s.

Practice Set

• Q1: For a zero order reaction, half-life is:
  • A. Independent of initial concentration
  • B. Directly proportional to initial concentration
  • C. Inversely proportional to initial concentration
  • D. Equal to 0.693/k
• Q2: For a first order reaction, rate constant k is given by:
  • A. k = (2.303 / t) log (a / (a - x))
  • B. k = (a - x) / t
  • C. k = (a + x) / t
  • D. k = (2.303 / t) log (a + x)
• Q3: The half-life of a first order reaction is 10 min. What is the rate constant?
  • A. 0.0693 min⁻¹
  • B. 0.693 min⁻¹
  • C. 6.93 min⁻¹
  • D. 0.00693 min⁻¹

Answer Key

• Q1: B - Half-life for zero order is proportional to initial concentration.
• Q2: A - k = (2.303 / t) log (a / (a - x)) for first order.
• Q3: A - k = 0.693 / t_1/2 = 0.693 / 10 = 0.0693 min⁻¹.

Memory Aids

• Zero order: t_1/2  [A]₀, First order: t_1/2 constant.
• First order k from log formula: k = (2.303 / t) log (a / (a - x)).
• Half-life formula: t_1/2 = 0.693 / k for first order.