CHM2046 - Chemistry Fundamentals II: Chapter 15 - Chemical Kinetics

Rate Law

• The rate of a chemical reaction can often be expressed using the rate law, typically in the form: $ext{Rate} = k imes [A]^m imes [B]^n$ where:
  • $k$ = rate constant
  • $[A]$, $[B]$ = concentrations of reactants
  • $m$, $n$ = reaction orders which are determined experimentally.

Temperature Effect on Reaction Rates

Example: Understanding Reaction Rates

• Question: Which statement best explains why reaction rates generally increase with increasing temperature?
  • A: Reaction rates increase with increasing temperature because, as temperature increases, the pre-exponential factor of the rate constant increases.
  • B: Reaction rates increase with increasing temperature because, as temperature increases, molecules decompose into their constituent atoms, which can then form new bonds to form the products.
  • C: Reaction rates increase with increasing temperature because, as temperature increases, a greater fraction of molecules have enough thermal energy to surmount the activation barrier.
  • Correct Answer: C

Arrhenius Plots

• Description: Arrhenius plots are used to analyze kinetic data by plotting $ext{ln}k$ vs. $\frac{1}{T}$.
  • Equation:
    $ext{ln}k = -\frac{E_a}{R} \frac{1}{T} + ext{ln}A$
  • Where:
  • $y = ext{ln}k$
  • $x = \frac{1}{T}$
  • slope = $-\frac{E_a}{R}$
  • intercept = $ext{ln}A$
• Solving the Arrhenius Equation:
  • Original form:
    $k = Ae^{-\frac{E_a}{RT}}$
  • Rearrangement leads to the linear form used for plotting.

Example: Ozone Decomposition Reaction

• Reaction:
  $3O<em>3(g) → 2O</em>2(g) + O(g)$
• Task: Determine the frequency factor (A) and activation energy ($E_a$).
• Data Collected: Plotting $ext{ln}k$ vs. $\frac{1}{T}$ yields values enabling calculations.
  • Example Calculation yields:
  • $E_a = 9.31 imes 10^4 ext{ J/mol} = 93.1 ext{ kJ/mol}$
  • $A = 4.36 imes 10^{11} ext{ M}^{-1} ext{s}^{-1}$

Arrhenius Equation: Two-Point Form

• When two data points (T, k) are available, $E<em>a$ can be calculated with: $ext{ln} \frac{k</em>2}{k<em>1} = \frac{E</em>a}{R}\bigg(\frac{1}{T<em>1} - \frac{1}{T</em>2}\bigg)$

Example: Reaction Between Nitrogen Dioxide and Carbon Monoxide

• Reactions:$NO<em>2(g) + CO(g) → NO(g) + CO</em>2(g)$
  • Rate constants:
  • $T<em>1 = 701 ext{ K}, k</em>1 = 2.57 ext{ M}^{-1} ext{s}^{-1}$
  • $T<em>2 = 895 ext{ K}, k</em>2 = 567 ext{ M}^{-1} ext{s}^{-1}$
• Calculation Example yields:
  • $E_a = 1.5 imes 10^2 ext{ kJ/mol}$

The Collision Model

• Principle: For a reaction to occur, molecules must collide.
  • Average collisions per second: about $10^9$.
  • Conditions for effective collisions:
  1. Sufficient energy to overcome the energy barrier.
  2. Proper orientation to facilitate bond formation.

Effective Collisions

• Definition: Collisions that lead to formation of new products are termed effective collisions.
  • Higher frequency of effective collisions correlates with faster reaction rates.

Frequency Factor in Arrhenius Equation

• The Arrhenius equation includes the frequency factor $A$, which can be broken down into:
  • Collision Frequency: Number of collisions per unit time, dependent on pressure and temperature.
  • Orientation Factor $p$: A statistical term representing the fraction of collisions having the right orientation, typically between 0 and 1.

Effective Collisions and Orientation Factor

• Proper molecular alignment is crucial for effective reactions.
  • For most reactions, p < 1 .
  • Rare cases where p > 1 may involve electron transfer.

Reaction Mechanisms

• A chemical equation typically represents an overall reaction, while a reaction mechanism consists of individual steps leading to that reaction.
• Knowledge of the rate law aids in understanding the sequence of steps in a mechanism.

Example of a Reaction Mechanism

• Overall Reaction:
  $H<em>2(g) + 2ICl(g) → 2HCl(g) + I</em>2(g)$
• Mechanism Steps:
  1. $H_2(g) + ICl(g) → HCl(g) + HI(g)$
  2. $HI(g) + ICl(g) → HCl(g) + I_2(g)$
• Reaction Intermediates: HI is an intermediate as it appears in an early mechanism step but is not present in the overall equation.

Molecularity

• Definition: Number of reactant particles in an elementary step.
  • Types of molecularity:
  • Unimolecular: One reactant particle.
  • Bimolecular: Two reactant particles.
  • Termolecular: Extremely rare; involves three particles.

Rate Laws for Elementary Steps

• Each elementary step has its own rate law and activation energy. Observations are experimentally determined for overall reactions, while those for elementary steps can be deduced from their equations.

Rate Laws for Elementary Steps:

• Elementary Step | Molecularity | Rate Law
  • A → products | 1 | Rate = $k[A]$
  • A + A → products | 2 | Rate = $k[A]^2$
  • A + B → products | 2 | Rate = $k[A][B]$
  • A + A + A → products | 3 (rare) | Rate = $k[A]^3$
  • A + B + C → products | 3 (rare) | Rate = $k[A]^2[B]$

Example: Rate Law for an Elementary Step

• Reaction:
  $Cl + CO → ClCO$
• Possible Rates:
  • A) Rate = $k[Cl]$
  • B) Rate = $k[CO]$
  • C) Rate = $k[ClCO]$
  • D) Rate = $k[Cl][CO]$
  • Correct Answer: D

Rate-Determining Steps (RDS)

• In most reaction mechanisms, one step is slower than the others, and is termed the rate-determining step.
  • Characteristics:
  • The RDS has the largest activation energy.
  • The rate law of the RDS equals the rate law of the overall reaction.

Example Mechanism

• Consider: $NO<em>2(g) + CO(g) → NO(g) + CO</em>2(g)$
  • Slow Step: $NO<em>2(g) + NO</em>2(g) → NO_3(g) + NO(g)$
  • Rate = $k<em>1[NO</em>2]^2$
  • Fast Step: $NO<em>3(g) + CO(g) → NO</em>2(g) + CO_2(g)$
  • Rate = $k<em>2[NO</em>3][CO]$

Validating a Mechanism

• For a proposed reaction mechanism to be considered valid (not necessarily proven), two conditions must be fulfilled:
  1. The elementary steps should sum up to the overall reaction.
  2. The predicted rate law must align with experimentally observed behaviors.

Additional Example for Validating Mechanisms

• Consider the series:
  1. $2NO(g) ⇌ N2O2(g)$
  2. $N2O2(g) + H2(g) → H2O(g) + N2(g)$
  3. $N2(g) + 2NO(g) → 2H2O(g)$
  • Rate Equation Examples from steps to derive overall rate laws.

Catalysis

• Note: Details on catalytic processes and their influence on reaction rates will be discussed in subsequent sections.