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.

• Key focus: speed of reaction and factors altering it.

Solved Examples

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

Practice Set

• Q1: What does chemical kinetics study?
  A. Reaction equilibrium
  B. Reaction rates and mechanisms
  C. Reaction products
  D. Reaction catalysts

• Q2: Which factor is NOT studied in chemical kinetics?
  A. Reaction speed
  B. Reaction mechanism
  C. Reaction yield
  D. Effect of temperature on rate

• Q3: Chemical kinetics helps to:
  A. Determine reaction speed only
  B. Describe conditions altering reaction rates
  C. Both A and B
  D. None of the above

Answer Key

• Q1: B - Chemical kinetics focuses on reaction rates and mechanisms.

• Q2: C - Reaction yield is not a primary focus of kinetics.

• Q3: C - It determines speed and conditions affecting rates.

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 = -∆[R]/∆t = +∆[P]/∆t.

• Units: mol L-1 s-1 or atm s-1 (gases).

Average and Instantaneous Rates

• Average rate: over a time interval, rav = -∆[R]/∆t or +∆[P]/∆t.

• Instantaneous rate: at a specific time, rinst = -d[R]/dt or +d[P]/dt.

• ∆ denotes finite change; d denotes infinitesimal change.

Solved Examples

• Ex 1: Calculate average rate if [R] changes from 0.5 to 0.3 mol/L in 10 s.
  Solution: Rate = -∆[R]/∆t = -(0.3 - 0.5)/10 = 0.02 mol L-1 s-1.

Practice Set

• Q1: Rate of reaction is expressed as:
  A. Change in volume/time
  B. Change in concentration/time
  C. Change in mass/time
  D. Change in temperature/time

• Q2: For R → P, if [P] increases by 0.4 mol/L in 20 s, average rate is:
  A. 0.02 mol L-1 s-1
  B. 0.4 mol L-1 s-1
  C. 8 mol L-1 s-1
  D. 0.5 mol L-1 s-1

• Q3: Instantaneous rate differs from average rate because:
  A. It uses finite changes
  B. It is rate at a specific instant
  C. It ignores concentration changes
  D. It is always zero

Answer Key

• Q1: B - Rate is change in concentration per unit time.

• Q2: A - Rate = 0.4/20 = 0.02 mol L-1 s-1.

• Q3: B - Instantaneous rate is at a specific time (differential).

Memory Aids

• Average rate uses ∆ (big change), instantaneous uses d (small change).

• Rate units always concentration/time.

Factors Influencing Rate of a Reaction

Key Factors

• Concentration: Higher reactant concentration increases collision frequency, speeding up the reaction.

• Physical state: Gaseous reactants react faster than liquids or solids due to better mixing and mobility.

• Temperature: Increasing temperature raises kinetic energy, roughly doubling rate every 106C.

• Catalyst: Lowers activation energy, increasing reaction speed without being consumed.

• Surface area: More surface area of solids or catalysts exposes more reactive sites, accelerating reaction.

• Light presence: Photochemical reactions require light to proceed.

• Activation energy: Lower activation energy means faster reaction.

Solved Examples

Ex 1: How does increasing temperature from 306C to 406C affect reaction rate?

• Rate approximately doubles due to 106C rise.

Practice Set

• Q1: Which factor does NOT increase reaction rate?
  A) Increasing concentration
  B) Increasing surface area
  C) Decreasing temperature
  D) Adding catalyst

• Q2: Why do gaseous reactions proceed faster than solids?
  A) Higher activation energy
  B) Better mixing and mobility
  C) Lower surface area
  D) Presence of light

• Q3: What effect does a catalyst have?
  A) Increases activation energy
  B) Decreases reaction rate
  C) Lowers activation energy
  D) Changes reactant concentration

Answer Key

• Q1: C - Decreasing temperature lowers kinetic energy, slowing reaction.

• Q2: B - Gases mix and move freely, increasing collision frequency.

• Q3: C - Catalysts lower activation energy, speeding up reaction.

Memory Aids

• Rate increases with concentration, temperature, surface area, catalyst presence.

• Gases react faster than solids/liquids.

• Light needed only for photochemical reactions.

Rate Expression and Rate Constant

Key Concepts

• For reaction: aA + bB 12121212121212121212121212121212121212121212121212121212121212 cC + dD

• Rate 27 [A]x [B]y, where x,y may differ from stoichiometric coefficients a,b.

• Rate law: Rate = k [A]x [B]y, k = rate constant.

• Order of reaction: Sum of powers x + y.

• Order can be 0, 1, 2, 3 or fractional.

Solved Examples

Ex 1: For Rate = k[A]2[B], find order.

• Order = 2 + 1 = 3 (third order reaction).

Practice Set

• Q1: Rate = k[A]0 means order is?
  A) 0
  B) 1
  C) 2
  D) Fractional

• Q2: If Rate = k[A][B], overall order is?
  A) 1
  B) 2
  C) 3
  D) 0

• Q3: Rate = k[A]1.5, order is?
  A) 1
  B) 1.5
  C) 2
  D) 0

Answer Key

• Q1: A - Zero order means rate independent of [A].

• Q2: B - Sum of powers 1 + 1 = 2.

• Q3: B - Order equals 1.5 (fractional order).

Memory Aids

• Order = sum of exponents in rate law.

• Exponents ≠ stoichiometric coefficients unless elementary reaction.

• Fractional orders possible.

Elementary and Complex Reactions

Key Points

• Elementary reactions: Single-step, order equals stoichiometric coefficients.

• Complex reactions: Multi-step, overall rate determined by slowest step (rate-determining step).

• Rate law derived from mechanism, not overall balanced equation.

Solved Examples

Ex 1: For elementary reaction A + B 12121212121212121212121212121212121212121212121212121212121212 C, order is?

• Order = 1 + 1 = 2 (second order).

Practice Set

• Q1: In complex reactions, rate is controlled by?
  A) Fastest step
  B) Slowest step
  C) Overall equation
  D) Catalyst only

• Q2: For elementary reaction 2A 121212121212121212121212121212121212121212121212 B, order is?
  A) 1
  B) 2
  C) 3
  D) 0

• Q3: Rate law for complex reaction depends on?
  A) All steps equally
  B) Slowest step only
  C) Fastest step only
  D) Reactant concentration only

Answer Key

• Q1: B - Slowest step limits overall rate.

• Q2: B - Order equals stoichiometric coefficient sum (2).

• Q3: B - Rate law depends on rate-determining (slowest) step.

Memory Aids

• Elementary: order = stoichiometric coefficients.

• Complex: rate = slowest step rate.

• Rate law from mechanism, not overall equation.

Units of Rate Constant

Key Points

• Rate constant units depend on reaction order.

• General formula: k = rate / [concentration]order.

• Concentration in mol L-1, time in seconds (s).

Solved Examples

Ex 1: Units of k for first order reaction?

• Units = s-1.

Practice Set

• Q1: Units of k for zero order reaction?
  A) mol L-1 s-1
  B) s-1
  C) mol-1 L s-1
  D) mol L-1

• Q2: Units of k for second order reaction?
  A) s-1
  B) mol L-1 s-1
  C) mol-1 L s-1
  D) mol2 L-2 s-1

• Q3: Units of k for third order reaction?
  A) mol-2 L2 s-1
  B) mol-1 L s-1
  C) s-1
  D) mol L-1 s-1

Answer Key

• Q1: A - Zero order k units are mol L-1 s-1.

• Q2: C - Second order k units are mol-1 L s-1.

• Q3: A - Third order k units are mol-2 L2 s-1.

Memory Aids

• Order 0: mol L-1 s-1

• Order 1: s-1

• Order 2: mol-1 L s-1

• Order 3: mol-2 L2 s-1

Molecularity of a Reaction

Key Concepts

• Number of reacting species colliding simultaneously in an elementary reaction.

• Types: Unimolecular, Bimolecular, Trimolecular.

• Trimolecular reactions are rare due to low probability of three-body collisions.

Examples

• Unimolecular: NH4NO2 121212121212121212 N2 + 2H2O

• Bimolecular: 2HI 121212121212121212 H2 + I2

• Trimolecular: 2NO + O2 1212121212121212 2NO2

Solved Examples

Ex 1: Molecularity of 2HI 1212121212121212 H2 + I2?

• Bimolecular (2 molecules collide).

Practice Set

• Q1: Molecularity of NH4NO2 decomposition?
  A) 1
  B) 2
  C) 3
  D) 0

• Q2: Which molecularity is rare?
  A) Unimolecular
  B) Bimolecular
  C) Trimolecular
  D) Zero molecular

• Q3: Molecularity applies to?
  A) Complex reactions
  B) Elementary reactions
  C) Both
  D) None

Answer Key

• Q1: A - Unimolecular (single molecule decomposes).

• Q2: C - Trimolecular collisions are rare.

• Q3: B - Molecularity only applies to elementary reactions.

Memory Aids

• Molecularity = number of molecules colliding simultaneously.

• Unimolecular = 1, Bimolecular = 2, Trimolecular = 3.

• Trimolecular rare due to collision probability.

Order vs Molecularity

Comparison Points

• Order: Determined experimentally, can be zero, fractional, or integer.

• Molecularity: Number of molecules in elementary step, always whole number.

• Order applies to all reactions; molecularity only to elementary reactions.

• Order can change with conditions; molecularity is fixed.

• For complex reactions, order depends on slowest step; molecularity not defined.

Solved Examples

Ex 1: Can order be fractional?

• Yes, order can be fractional; molecularity cannot.

Practice Set

• Q1: Molecularity can be?
  A) Fractional
  B) Zero
  C) Whole number
  D) Negative

• Q2: Order is determined by?
  A) Experiment
  B) Stoichiometry
  C) Molecularity
  D) Catalyst

• Q3: Which changes with temperature?
  A) Molecularity
  B) Order
  C) Both
  D) None

Answer Key

• Q1: C - Molecularity is always a whole number.

• Q2: A - Order is found experimentally.

• Q3: B - Order can change with conditions; molecularity does not.

Memory Aids

• Order = experimental, can be fractional.

• Molecularity = theoretical, whole number.

• Order varies; molecularity fixed per elementary step.

Integrated Rate Equations

Overview

• Describe how reactant concentration changes with time.

• Used to determine reaction order and rate constant.

• Different orders have distinct integrated rate laws.

Solved Examples

• Ex 1: Calculate rate constant for zero order reaction with [R]0=0.5 M, [R]t=0.2 M after 10 s.
  Solution: k = (1/t)([R]0 - [R]t) = (1/10)(0.5 - 0.2) = 0.03 M/s

Practice Set

• Q1: Which integrated rate law corresponds to a first order reaction?
  A. [R]t = -kt + [R]0
  B. ln[R]t = -kt + ln[R]0
  C. 1/[R]t = kt + 1/[R]0
  D. [R]t2 = kt + [R]02

• Q2: For a zero order reaction, the plot of [R] vs time is:
  A. Exponential decay
  B. Straight line with negative slope
  C. Curved upward
  D. Horizontal line

• Q3: The half-life of a first order reaction is:
  A. Dependent on initial concentration
  B. Independent of initial concentration
  C. Directly proportional to concentration squared
  D. Zero

Answer Key

• Q1: B - First order integrated rate law is ln[R]t = -kt + ln[R]0.

• Q2: B - Zero order reaction shows linear decrease of [R] with time.

• Q3: B - Half-life of first order reaction is constant, independent of concentration.

Memory Aids

• Zero order: Rate constant, linear [R] vs t plot.

• First order: ln[R] linear with time, half-life constant.

• Half-life formulas: Zero order t1/2 = [R]0/2k; First order t1/2 = 0.693/k.

Zero Order Reaction

Key Points

• Rate independent of reactant concentration: rate = k.

• Rate law: rate = k [R]0 = k.

• Integrated rate law: [R]t = -kt + [R]0.

• Graph of rate vs concentration is a horizontal line.

• Half-life: t1/2 = [R]0 / 2k (directly proportional to initial concentration).

Solved Examples

• Ex 1: Calculate half-life for zero order reaction with [R]0 = 0.4 M and k = 0.02 M/s.
  Solution: t1/2 = [R]0 / 2k = 0.4 / (2 × 0.02) = 10 s.

Practice Set

• Q1: For zero order reaction, rate is:
  A. Proportional to [R]
  B. Independent of [R]
  C. Proportional to [R]2
  D. Zero

• Q2: The graph of [R] vs time for zero order reaction is:
  A. Exponential decay
  B. Straight line with positive slope
  C. Straight line with negative slope
  D. Horizontal line

• Q3: Half-life of zero order reaction:
  A. Decreases with increasing [R]0
  B. Is constant
  C. Increases with increasing [R]0
  D. Depends on temperature only

Answer Key

• Q1: B - Rate does not depend on concentration.

• Q2: C - [R] decreases linearly with time.

• Q3: C - Half-life is directly proportional to initial concentration.

Memory Aids

• Zero order: Rate constant, linear decrease of [R].

• Half-life increases with initial concentration.

First Order Reaction

Key Points

• Rate depends on one reactant concentration: rate = k [R].

• Integrated rate law: ln[R]t = -kt + ln[R]0.

• Graph of ln[R] vs time is a straight line with slope -k.

• Half-life: t1/2 = 0.693 / k (independent of concentration).

• Time for nth fraction completion: t1/n = (2.303 / k) log(n / (n-1)).

• Pseudo first order reactions behave like first order under certain conditions.

Solved Examples

• Ex 1: Calculate k if [R]0 = 0.8 M, [R]t = 0.2 M after 20 s.
  Solution: k = (2.303 / t) log([R]0 / [R]t) = (2.303 / 20) log(0.8 / 0.2) = 0.0693 s-1.

• Ex 2: Find half-life for k = 0.1 s-1.
  Solution: t1/2 = 0.693 / k = 0.693 / 0.1 = 6.93 s.

Practice Set

• Q1: The half-life of a first order reaction:
  A. Depends on initial concentration
  B. Is constant
  C. Increases with concentration
  D. Decreases with temperature

• Q2: For first order reaction, plot of:
  A. [R] vs t is linear
  B. ln[R] vs t is linear
  C. 1/[R] vs t is linear
  D. [R]2 vs t is linear

• Q3: Time taken for 3/4th completion of first order reaction is:
  A. t1/2
  B. 2 t1/2
  C. 3 t1/2
  D. 4 t1/2

Answer Key

• Q1: B - Half-life is independent of concentration.

• Q2: B - ln[R] vs time is a straight line.

• Q3: C - 3/4th completion corresponds to 2 half-lives, so time = 2 t1/2.

Memory Aids

• First order: ln[R] linear with time.

• Half-life constant, t1/2 = 0.693/k.

• Pseudo first order: apparent first order under excess of one reactant.

Rate Laws and Graphs

Key Points

• Zero order: rate constant, rate vs [R] is flat line.

• First order: rate ∝ [R], rate vs [R] is straight line.

• Second order: rate ∝ [R]2, rate vs [R]2 is linear.

• Third order: rate ∝ [R]3, rate vs [R]3 is linear.

• Integrated rate plots help determine reaction order.

• Half-life varies with order: zero (∝ [R]0), first (constant), second (∝ 1/[R]0), third (∝ 1/[R]02).

Solved Examples

• Ex 1: Identify reaction order if plot of 1/[R] vs time is linear.
  Solution: Second order reaction.

• Ex 2: Half-life decreases with increasing concentration. Reaction order is:
  Solution: Second or third order.

Practice Set

• Q1: Which plot is linear for a second order reaction?
  A. [R] vs t
  B. ln[R] vs t
  C. 1/[R] vs t
  D. 1/([R])2 vs t

• Q2: Half-life of a third order reaction varies as:
  A. Constant
  B. Directly proportional to [R]0
  C. Inversely proportional to [R]0
  D. Inversely proportional to [R]02

• Q3: Rate law for a reaction with rate ∝ [A]3 is:
  A. Zero order
  B. First order
  C. Second order
  D. Third order

Answer Key

• Q1: C - 1/[R] vs time linear indicates second order.

• Q2: D - Half-life ∝ 1/[R]02 for third order.

• Q3: D - Rate proportional to cube of concentration is third order.

Memory Aids

• Order from plots: [R] linear zero order, ln[R] first order, 1/[R] second order.

• Half-life trends: zero (∝ [R]0), first (constant), second (∝ 1/[R]0), third (∝ 1/[R]02).

Half-life and Fraction Times

Key Points

• Half-life (t1/2): time for half reactant to react.

• Amount left after n half-lives: [R]0 / 2n.

• Time for nth fraction completion for first order: t1/n = (2.303 / k) log(n / (n-1)).

• Time for reactant to reduce to nth fraction: t = (2.303 / k) log n.

• Fraction completion times are multiples of half-life.

Solved Examples

• Ex 1: Amount left after 3 half-lives if initial concentration is 0.8 M.
  Solution: 0.8 / 23 = 0.1 M.

• Ex 2: Time to complete 75% reaction for first order with t1/2 = 10 min.
  Solution: 75% completion ≈ 2 half-lives = 20 min.

Practice Set

• Q1: After 4 half-lives, amount left is:
  A. [R]0/4
  B. [R]0/8
  C. [R]0/16
  D. Zero

• Q2: Time to complete 87.5% reaction corresponds to:
  A. 2 half-lives
  B. 3 half-lives
  C. 4 half-lives
  D. 5 half-lives

• Q3: For first order reaction, time to reduce to 1/5th is:
  A. (2.303/k) log 5
  B. (2.303/k) log 1/5
  C. (k/2.303) log 5
  D. (k/2.303) log 1/5

Answer Key

• Q1: C - Amount left = [R]0/24 = [R]0/16.

• Q2: B - 87.5% completion ≈ 3 half-lives.

• Q3: A - Time = (2.303/k) log n, here n=5.

Memory Aids

• Amount left halves every half-life.

• Fraction completion times are multiples of half-life.

• Use log relations for fractional times in first order.

Pseudo First Order Reactions

Key Points

• Reactions not truly first order but behave as such under certain conditions.

• One reactant in large excess makes its concentration effectively constant.

• Rate law simplifies to first order with respect to limiting reactant.

• Examples: Acid hydrolysis of ethyl acetate, acid-catalyzed inversion of cane sugar.

• Rate = k [compound], where k includes constant concentration of other reactant.

Solved Examples

• Ex 1: In acid hydrolysis of ethyl acetate, if [H+] is constant, rate law is:
  Solution: Rate = k' [ethyl acetate], pseudo first order.

Practice Set

• Q1: Pseudo first order reactions occur when:
  A. All reactants are in equal concentration
  B. One reactant is in large excess
  C. Reaction is zero order
  D. Reaction is second order

• Q2: Acid hydrolysis of ethyl acetate is:
  A. Zero order
  B. First order
  C. Pseudo first order
  D. Second order

• Q3: Rate law for pseudo first order reaction is:
  A. Rate = k [A]0
  B. Rate = k [A]1
  C. Rate = k [A]2
  D. Rate = k [A][B]

Answer Key

• Q1: B - Large excess of one reactant makes reaction pseudo first order.

• Q2: C - Acid hydrolysis is pseudo first order.

• Q3: B - Rate depends linearly on limiting reactant concentration.

Memory Aids

• Pseudo first order: one reactant constant, rate depends on other.

• Common in acid/base catalyzed reactions.

Temperature Dependence of Rate

Key Points

• Increasing temperature by 10 °C nearly doubles the rate constant (k) of a reaction.

• Arrhenius equation: k = A e^{-E_a/RT}, where A is frequency factor, Ea is activation energy.

• Only collisions with energy ≥ Ea lead to reaction.

• Reactant molecules collide to form an unstable activated complex, which then forms products.

• Example: H2(g) + I2(g) → 2HI(g)

Solved Examples

• Ex 1: Calculate the effect on rate constant if temperature increases from 300 K to 310 K, given Ea = 50 kJ/mol.

  • Use Arrhenius equation ratio form:
    log(k2/k1) = (Ea/2.303R)(1/T1 - 1/T2)

  • Substitute values and solve for k2/k1.

Practice Set

• Q1: What happens to the rate constant when temperature increases by 10 °C?

  • A) Halves

  • B) Doubles

  • C) Triples

  • D) No change

• Q2: In Arrhenius equation, what does the term e^{-E_a/RT} represent?

  • A) Frequency of collisions

  • B) Fraction of molecules with energy ≥ Ea

  • C) Activation energy

  • D) Rate constant

• Q3: For a reaction with Ea = 60 kJ/mol, calculate log(k2/k1) when temperature changes from 300 K to 310 K. (R = 8.314 J/mol·K)

  • A) 0.1

  • B) 0.2

  • C) 0.3

  • D) 0.4

Answer Key

• Q1: B - Rate constant nearly doubles with 10 °C rise.

• Q2: B - e^{-E_a/RT} is the Boltzmann factor, fraction of molecules with energy ≥ Ea.

• Q3: B - Using formula, log(k2/k1) ≈ 0.2.

Memory Aids

• "10 °C rise doubles rate" - quick rule for temperature effect.

• Arrhenius factor = frequency × Boltzmann factor.

• Activation energy = energy barrier to reaction.

Activation Energy & Energy Profile

Key Points

• Activation energy (Ea) = Energy of activated complex − Energy of reactants.

• Energy is released when activated complex forms products.

• Energy profile shows reactants → activated complex → products.

• Exothermic: Products have lower energy than reactants (ΔH < 0).

• Endothermic: Products have higher energy than reactants (ΔH > 0).

Solved Examples

• Ex 1: Identify if reaction is exothermic or endothermic from energy profile where products are lower in energy.

• Answer: Exothermic, as products have less energy than reactants.

Practice Set

• Q1: Activation energy is the energy required to:

  • A) Break all bonds in reactants

  • B) Form products

  • C) Form activated complex

  • D) Cool the system

• Q2: In an exothermic reaction, the energy of products is:

  • A) Higher than reactants

  • B) Equal to reactants

  • C) Lower than reactants

  • D) Zero

• Q3: Which of the following represents activation energy?

  • A) Energy difference between reactants and products

  • B) Energy difference between activated complex and reactants

  • C) Energy difference between products and activated complex

  • D) Total energy of reactants

Answer Key

• Q1: C - Activation energy is energy to form activated complex.

• Q2: C - Exothermic reactions have products lower in energy.

• Q3: B - Activation energy is difference between activated complex and reactants.

Memory Aids

• Activation energy = "hill" reactants must climb.

• Exothermic = energy released, products lower.

• Endothermic = energy absorbed, products higher.

Boltzmann Factor & Kinetic Energy Distribution

Key Points

• Not all molecules have same kinetic energy; distribution follows Boltzmann statistics.

• Fraction of molecules with energy ≥ E: N_E/N_T = e^{-E/RT} (Boltzmann factor).

• Graph of fraction vs kinetic energy peaks at most probable kinetic energy.

• Increasing temperature shifts peak right and broadens curve, increasing molecules with energy ≥ Ea.

Solved Examples

• Ex 1: Explain why reaction rate increases with temperature using kinetic energy distribution.

  • At higher T, more molecules have energy ≥ Ea, increasing effective collisions and rate.

Practice Set

• Q1: Boltzmann factor represents:

  • A) Total molecules

  • B) Fraction with energy ≥ E

  • C) Activation energy

  • D) Rate constant

• Q2: Increasing temperature causes kinetic energy distribution curve to:

  • A) Shift left

  • B) Shift right and broaden

  • C) Narrow

  • D) Remain same

• Q3: Which factor increases reaction rate with temperature?

  • A) More collisions with energy < Ea

  • B) More molecules with energy ≥ Ea

  • C) Decreased frequency factor

  • D) Lower activation energy

Answer Key

• Q1: B - Boltzmann factor is fraction with energy ≥ E.

• Q2: B - Curve shifts right and broadens with temperature.

• Q3: B - More molecules exceed activation energy at higher T.

Memory Aids

• Boltzmann factor = fraction of "energetic" molecules.

• Higher T = more molecules can react.

• Distribution curve peak = most probable kinetic energy.

Effect of Catalyst on Reaction Rate

Key Points

• Catalyst increases reaction rate without permanent chemical change.

• Provides alternate pathway with lower activation energy.

• Does not change Gibbs energy (ΔG) or equilibrium constant.

• Helps system reach equilibrium faster.

Solved Examples

• Ex 1: Explain how catalyst affects activation energy and rate.

  • Catalyst lowers activation energy peak, increasing rate by allowing easier reaction path.

Practice Set

• Q1: A catalyst:

  • A) Changes equilibrium constant

  • B) Increases activation energy

  • C) Lowers activation energy

  • D) Is consumed permanently

• Q2: Catalyst affects reaction by:

  • A) Changing ΔG

  • B) Providing alternate pathway

  • C) Increasing reactant energy

  • D) Changing products

• Q3: Which statement is true?

  • A) Catalyst speeds up non-spontaneous reactions

  • B) Catalyst changes equilibrium position

  • C) Catalyst lowers potential energy barrier

  • D) Catalyst is consumed in reaction

Answer Key

• Q1: C - Catalyst lowers activation energy.

• Q2: B - Catalyst provides alternate pathway.

• Q3: C - Catalyst lowers energy barrier but does not change equilibrium.

Memory Aids

• "Catalyst lowers the hill" to speed reaction.

• Does not change equilibrium, only rate.

• Not consumed, acts repeatedly.

Negative Temperature Coefficient

Key Points

• Some reactions show rate decrease with temperature rise (negative temperature coefficient).

• Example: NO + ½ O2 → NO2

Solved Examples

• Ex 1: Identify reaction with negative temperature coefficient.

  • Given NO + ½ O2 → NO2 shows rate decrease with temperature increase.

Practice Set

• Q1: Negative temperature coefficient means rate:

  • A) Increases with temperature

  • B) Decreases with temperature

  • C) Remains constant

  • D) Doubles every 10 °C

• Q2: Which reaction shows negative temperature coefficient?

  • A) H2 + I2 → 2HI

  • B) NO + ½ O2 → NO2

  • C) Combustion of methane

  • D) Acid-base neutralization

• Q3: Negative temperature coefficient reactions are:

  • A) Common in all reactions

  • B) Due to complex mechanisms

  • C) Due to catalysts

  • D) Due to high activation energy

Answer Key

• Q1: B - Rate decreases with temperature.

• Q2: B - NO + ½ O2 → NO2 shows this behavior.

• Q3: B - Negative coefficient often due to complex reaction steps.

Memory Aids

• Negative temperature coefficient = "rate slows as heat rises".

• Rare, mostly in complex reactions.

Collision Theory of Chemical Reactions

Fundamentals

• Reactant molecules are treated as hard spheres.

• Reaction occurs due to collision between molecules.

• Not all collisions lead to reaction; only effective collisions do.

• Effective collisions require:

  • Sufficient kinetic energy (threshold energy = activation energy + energy of reactants).

  • Proper molecular orientation.

• Activated complex (transition state) forms temporarily during collision before products form.

• Higher concentration increases collision frequency, speeding up reaction.

Rate Expression

• Rate = P ZAB e\(-\frac{E_a}{RT}\)

• P = steric factor (probability of proper orientation).

• ZAB = collision frequency (collisions per second per unit volume).

• Activation energy and orientation determine effective collisions and reaction rate.

Limitations

• Assumes molecules as hard spheres, ignoring molecular structure.

• Does not account for complex molecular interactions.

Solved Examples

• Ex 1: Calculate rate if P = 0.5, ZAB = 1010 s−1, Ea = 50 kJ/mol, T = 300 K, R = 8.314 J/mol·K.

  • Convert Ea to J: 50,000 J/mol.

  • Calculate exponent: e−Ea/RT = e−50000/(8.314×300) ≈ e−20.04 ≈ 2.06×10−9.

  • Rate = 0.5 × 1010 × 2.06×10−9 = 10.3 s−1.

Practice Set

• Q1: Which factor does NOT affect effective collision?

  • A. Activation energy

  • B. Molecular orientation

  • C. Collision frequency

  • D. Molecular color

• Q2: Increasing concentration of reactants affects reaction rate by:

  • A. Decreasing collision frequency

  • B. Increasing collision frequency

  • C. Increasing activation energy

  • D. Changing molecular orientation

• Q3: For the rate expression Rate = P ZAB e−Ea/RT, what does P represent?

  • A. Activation energy

  • B. Collision frequency

  • C. Steric factor

  • D. Temperature

Answer Key

• Q1: D. Molecular color — It does not influence collision effectiveness.

• Q2: B. Increasing collision frequency — More particles increase collision chances.

• Q3: C. Steric factor — Probability of proper molecular orientation during collision.

Memory Aids

• "Collision + Orientation + Energy = Reaction"

• Effective collision = Threshold energy + Proper orientation

• Higher concentration → More collisions → Faster reaction

Kinetics Equations

Zero Order Reaction

• Rate constant: k = ([A]0 − [A]) / t

• Half-life: t1/2 = [A]0 / 2k

First Order Reaction

• Rate constant: k = (2.303 / t) log ([A]0 / [A]) = (2.303 / t) log (a / (a − x))

• In terms of pressure: k = (2.303 / t) log (Pi / (2Pi − Pt))

• Half-life: t1/2 = 0.693 / k

• Other times: t60% = 0.9166 / k, t75% = 1.386 / k, t90% = 2.303 / k, t99.9% = 6.93 / k

General Relations

• Time to reduce to nth 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 t1/2₁ − log t1/2₂) / (log a₂ − log a₁)

Solved Examples

• Ex 1: Calculate t1/2 for a first order reaction with k = 0.1 s−1.

  • t1/2 = 0.693 / 0.1 = 6.93 s

• Ex 2: For zero order reaction, if [A]0 = 0.5 M, [A] = 0.3 M after 10 s, find k.

  • k = (0.5 − 0.3) / 10 = 0.02 M/s

Practice Set

• Q1: For a zero order reaction, the 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: The rate constant k for a first order reaction is 0.2 s−1. What is the half-life?

  • A. 3.465 s

  • B. 0.693 s

  • C. 5 s

  • D. 2 s

• Q3: For a first order reaction, if initial concentration is 0.8 M, what is concentration after 3 half-lives?

  • A. 0.1 M

  • B. 0.2 M

  • C. 0.4 M

  • D. 0.05 M

Answer Key

• Q1: B. Directly proportional to initial concentration — t1/2 = [A]0 / 2k for zero order.

• Q2: A. 3.465 s — t1/2 = 0.693 / k = 0.693 / 0.2 = 3.465 s.

• Q3: B. 0.2 M — After 3 half-lives, concentration = 0.8 / 23 = 0.1 M (corrected: 0.1 M, option A).

Memory Aids

• Zero order half-life ∝ initial concentration

• First order half-life = constant = 0.693/k

• Concentration after n half-lives = initial / 2n