Kinetics Review Problem Set

Validating a Mechanism

• Two Steps to Validate a Mechanism

1. Elementary steps must add up to the overall equation.
2. The derived rate law from the mechanism must equal the experimentally determined rate law.

• Example 1:
  • Overall reaction: $H<em>2 + I</em>2 \rightarrow 2HI$
  • Elementary steps:
  1. $I_2 \rightleftharpoons 2I$ (fast, equilibrium)
  2. $I + H<em>2 \rightleftharpoons H</em>2I$ (fast, equilibrium)

1. $H_2I + I \rightarrow 2HI$ (slow)

• Rate Law Derivation

  • Rate law for the slow step: $rate = k[H_2I][I]$
  • Eliminate reaction intermediates using fast equilibrium steps.
  • $k<em>{f1}[I</em>2] = k_{r1}[I]^2$
  • $k<em>{f2}[I][H</em>2] = k<em>{r2}[H</em>2I]$
  • Solving for intermediates:
  • $[H<em>2I] = \frac{k</em>{f2}}{k<em>{r2}}[I][H</em>2]$
  • $[I]^2 = \frac{k<em>{f1}}{k</em>{r1}}[I_2]$
  • Substitute into the rate law:
  • $rate = k \frac{k<em>{f2}}{k</em>{r2}}[I][H<em>2][I] = k \frac{k</em>{f2}}{k<em>{r2}}[I]^2[H</em>2]$
  • $rate = k \frac{k<em>{f2}}{k</em>{r2}} \frac{k<em>{f1}}{k</em>{r1}}[I<em>2][H</em>2]$
  • $rate = k'[I<em>2][H</em>2]$ where $k' = k \frac{k<em>{f2}k</em>{f1}}{k<em>{r2}k</em>{r1}}$

• A mechanism can be validated but it doesn't mean that the mechanism is the true mechanism. Further experimentation is required.

Invalid Mechanism Example

• Reaction: $2NO<em>2 + O</em>3 \rightarrow N<em>2O</em>5 + O_2$
• Experimental rate law: first order with respect to $NO<em>2$ and $O</em>3$
• Proposed mechanism:

1. $NO<em>2 + NO</em>2 \rightleftharpoons N<em>2O</em>4$ (fast, equilibrium)
2. $N<em>2O</em>4 + O<em>3 \rightarrow N</em>2O<em>5 + O</em>2$ (slow)

• Rate law for the slow step: $rate = k[N<em>2O</em>4][O_3]$
• Eliminate the reaction intermediate, $N<em>2O</em>4$
  • $k<em>{f1}[NO</em>2]^2 = k<em>{r1}[N</em>2O_4]$
  • $[N<em>2O</em>4] = \frac{k<em>{f1}}{k</em>{r1}}[NO_2]^2$
• Substitute into the rate law:
  • $rate = k \frac{k<em>{f1}}{k</em>{r1}}[NO<em>2]^2[O</em>3]$
  • The derived rate law expression indicates that $NO<em>2$ is 2nd order and $O</em>3$ is 1st order, which does not match the experimental rate law.

Arrhenius Equation and Activation Energy

• Arrhenius equation: $k = Ae^{-\frac{E_a}{RT}}$
• To calculate $E<em>a$ using rate constants at two different temperatures: $ln(\frac{k</em>1}{k<em>2}) = \frac{E</em>a}{R} (\frac{1}{T<em>2} - \frac{1}{T</em>1})$
• Example:
  • $k<em>1 = 2.1 \times 10^{-10} s^{-1}$ at $T</em>1 = 330 ^\circ C = 603.15 K$
  • $k<em>2 = 2.6 \times 10^{-11} s^{-1}$ at $T</em>2 = 300 ^\circ C = 573.15 K$
    $ln(\frac{2.1 \times 10^{-10}}{2.6 \times 10^{-11}}) = \frac{E<em>a}{8.314 J/mol\cdot K} (\frac{1}{573.15 K} - \frac{1}{603.15 K})$$E</em>a = 200.1 kJ/mol$

Arrhenius Equation in Linear Form

• Linear form:$ln(k) = -\frac{E_a}{R} (\frac{1}{T}) + ln(A)$
• The slope of the plot of $ln(k)$ vs. $\frac{1}{T}$ is $-\frac{E_a}{R}$
• To find $E<em>a$ from the slope: $E</em>a = |slope| \times R$
• Example: slope = -32713 K
  $E_a = |-32713 K| \times 8.314 J/mol\cdot K = 272.0 kJ/mol$

First-Order Kinetics

• Integrated rate law: $ln(\frac{[A]<em>t}{[A]</em>0}) = -kt$
• Example: A first-order reaction with $k = 3.58 \times 10^{-5} s^{-1}$
• Initial concentration: $[A]_0 = 5 M$
• Time: t = 2.5 hours = 9000 seconds
  $ln(\frac{[A]<em>t}{5.0 M}) = -(3.58 \times 10^{-5} s^{-1})(9000 s) = -0.3222$$[A]</em>t = 5.0 M \times e^{-0.3222} = 3.62 M$

Method of Initial Rates

• Generic rate law: $rate = k[A]^x[B]^y$
• Use changes in initial concentrations to determine reaction orders
  $rate = k [A]^1[B]^2$

Determining Reaction Order from Plots

• 0th order: plot of [A] vs time is linear
• 1st order: plot of ln[A] vs time is linear
• 2nd order: plot of 1/[A] vs time is linear