Chemical Kinetics and Reaction Mechanisms

Pseudo First-Order Kinetics 15.4

• Consider a rate law with more than one component:
  $Rate = -\frac{d[A]}{dt} = k [A] [B]^2 [C]^3$
• To simplify, we can work under pseudo-first order conditions.
• If [B] >> [A] and [C] >> [A], then the kinetics will appear as:
  $Rate = k' [A]$
• Where $k' = k [B]^2 [C]^3$
• Key Points:
  • $[B]$ and $[C]$ stay relatively constant compared to $[A]$.
  • $[A]$ is the limiting factor.
  • The simplified rate constant is: $k = \frac{k'}{[B]^2 [C]^3}$

Reaction Mechanisms

• $CH<em>3Cl + Br^- \rightarrow CH</em>3Br + Cl^-$
• Mechanism 1:
  $CH<em>3Cl + Br^- \rightarrow CH</em>3Br + Cl^-$
  $Rate = -\frac{d[CH<em>3Cl]}{dt} = k [CH</em>3Cl] [Br^-]$
• Mechanism 2:
  1. $CH<em>3Cl \rightarrow CH</em>3^+ + Cl^-$ (slow)
  2. $CH<em>3^+ + Br^- \rightarrow CH</em>3Br$ (fast)
     $Rate = -\frac{d[CH<em>3Cl]}{dt} = k [CH</em>3Cl]$
• The rate law is based on the rate-determining step (RDS).

Pseudo First-Order Kinetics Example Problem

• Reaction: $BrO<em>3^- + 5Br^- + 6H^+ \rightarrow 3Br</em>2 + 3H_2O$
• Rate law: $Rate = -\frac{d[BrO<em>3^-]}{dt} = k[BrO</em>3^-][Br^-][H^+]^2$
• Conditions:
  • $[BrO<em>3^-]</em>0 = 0.0010 M$
  • $[Br^-] = 2.0 M$
  • $[H^+] = 1.5 M$
• Assumption:
  • Assume pseudo-first order kinetics: $Rate = k'[BrO_3^-]$
• Given slope from plot $ln[BrO_3^-]$ vs time is $-0.087 s^{-1}$, therefore $k' = 0.087 s^{-1}$.
• $k' = k [Br^-] [H^+]^2$
• $k = \frac{k'}{[Br^-][H^+]^2} = \frac{0.087 s^{-1}}{(2.0 M)(1.5 M)^2} = 0.019 M^{-2}s^{-1}$

Reaction Mechanisms 15.6

• A reaction mechanism is the step-by-step process by which reactants evolve to products.
• Very rarely is the reaction mechanism reflected in the overall balanced equation.
• Intermediates are formed and consumed during the reaction.
• Example: $NO<em>2 + CO \rightarrow NO + CO</em>2$ Mechanism:
  1. $NO<em>2 + NO</em>2 \rightarrow NO_3 + NO$
  2. $NO<em>3 + CO \rightarrow NO</em>2 + CO_2$

Rate Determining Step

• The rate law is determined by the rate-determining step (RDS) of the reaction mechanism.
• The RDS is the slowest step and therefore is the observable step.
• Rate Law stoichiometry of RDS.
• Example:
  $2NO<em>2 + F</em>2 \rightarrow 2NO<em>2F$$Rate = -\frac{d[NO</em>2]}{dt} = k [NO<em>2] [F</em>2]$
• Mechanism:
  1. $NO<em>2 + F</em>2 \rightarrow NO_2F + F$ (slow)
  2. $F + NO<em>2 \rightarrow NO</em>2F$ (fast)
• The sum of the elementary steps gives the overall balanced equation confirming it is a good mechanism
• The first step is the rate-limiting step, determining the overall reaction rate.

Mechanisms Example

• $CH<em>3Cl + Br^- \rightarrow CH</em>3Br + Cl^-$
• Mechanism 1: $CH<em>3Cl + Br^- \rightarrow CH</em>3Br + Cl^-$$Rate = -\frac{d[CH<em>3Cl]}{dt} = k [CH</em>3Cl] [Br^-]$
  • Pseudo 1st order
• Mechanism 2:
  1. $CH<em>3Cl \rightarrow CH</em>3^+ + Cl^-$ (slow)
  2. $CH<em>3^+ + Br^- \rightarrow CH</em>3Br$ (fast)
     $Rate = -\frac{d[CH<em>3Cl]}{dt} = k [CH</em>3Cl]$
  • Reflected in data
• Based on the graphs:
  • Mechanism 1 is 1st order in $CH_3Cl$ and 0th order in $Br^-$.
  • Mechanism 2 is 1st order in $CH_3Cl$.

Determining Rate Law and Mechanism

• Reaction: $H<em>2(g) + 2BrCl(g) \rightarrow 2HCl(g) + Br</em>2(g)$
• Experimental Rate Law: $Rate = k [H_2] [BrCl]$
• Proposed Mechanism:
  1. $H_2 + BrCl \rightarrow HCl + HBr$ (slow)
  2. $HBr + BrCl \rightarrow HCl + Br_2$ (fast)
• The intermediate in this mechanism is HBr.
• The sum of the elementary steps gives the overall balanced equation verifying the mechanism. The first step is the rate-determining step. (✓ Mechanism)