\( \newcommand{\xrightleftharpoons}[2]{\overset{#1}{\underset{#2}{\rightleftharpoons}}} \) \( \newcommand{\conc}[1]{\left[\mathrm{#1}\right]} \) \( \newcommand{\chem}[1]{\mathrm{#1}} \)

Chapter 13
Chemical Equilibrium

Shaun Williams, PhD

Reaction Rate

The Decomposition of Nitrogen Dioxide
\( \chem{2NO_2(g) \rightarrow 2NO(g) + O_2(g)} \) (at \( 300^\circ \textrm{C}\))

Time
(sec)
\( [\mathrm{NO}_2] \)
(mol/L)
\( [\mathrm{NO}] \)
(mol/L)
\( [\mathrm{O}_2] \)
(mol/L)
0 0.0100 0 0
50 0.0079 0.0021 0.0011
100 0.0065 0.0035 0.0018
150 0.0055 0.0045 0.0023
200 0.0048 0.0052 0.0026
250 0.0043 0.0057 0.0029
350 0.0034 0.0066 0.0033
400 0.0031 0.0069 0.0035

The Decomposition of Nitrogen Dioxide

Starting with a flask of nitrogen dioxide at 300oC, the concentration of nitrogen dioxide falls exponentially over time while the concentrations of nitrogen monoxide and oxygen gas rise exponentially over the same time.

Instantaneous Rate

Rate Law

Types of Rate Laws

Rate Laws: A Summary

What does it mean to determine the form of the rate law?

Method of Initial Rates

Overall Reaction Order

First-Order

Plot of \(\ln\conc{N_2O_5}\) versus Time

\(\ln\conc{N_2O_5}\) Time(s)
-2.303 0
-2.649 50
-2.996 100
-3.689 200
-4.382 300
-5.075 400
Because there data here shows a linear trend, this is a first-irder reaction.

Half-Life in First-Order Reactions

Example Problem

A first order reaction is 35% complete at teh end of 55 minutes. What is the value of \(k\)?

\(k=7.8 \times 10^{-3}\,\mathrm{min}^{-1}\)

Second-Order Reactions

Plot of \(\ln\conc{C_4H_6}\) versus Time and Plot of \(\bfrac{1}{\conc{C_4H_6}}\) versus Time

Comparison of the natural log of the concentration of C4H6 versus time and one over the concentration of C4H6 versus time. Because the one over plot is linear whereas the natural log plot is not linear, we can identify this reaction as second-order.

Half-Life in Second-Order Reactions

Practice Problem

For a reaction \(\chem{aA\rightarrow Products}\), \(\conc{A}_0=5.0\,\mathrm{M}\), and the first two half-lives are 25 and 50 minutes, respectively.

  1. Write the rate law for this reaction

    \(\mathrm{rate}=k\conc{A}^2\)

  2. Calculate \(k\).

    \(k=8.0 \times 10^{-3}\,\mathrm{M}^{-1}\,\mathrm{min}^{-1}\)

  3. Calculate \(\conc{A}\) at \(t=525\,\mathrm{minutes}\).

    \(\conc{A}=0.23\,\mathrm{M}\)

Zero-Order Reactions

Plot of \(\conc{A}\) vs Time

A plot of [A] vs time for a zero-order reaction shows a linear trend. The slope of the linear trend is equal to the negative of the rate constant.

Half-Life in a Zero-Order Reaction

Reaction Mechanism

A Molecular Representation of the Elementary Steps in the Reaction of \(\chem{NO_2}\) and \(\chem{CO}\)

$$ \chem{NO_2(g)+CO(g)\rightarrow NO(g)+CO_2(g)} $$ In the first step, two nitrogen dioxide molecules react to form nitrogen trioxide and nitrogen monoxide. In the second step, a nitrogen trioxide molecule reacts with carbon monoxide to form nitrogen dioxide and carbon dioxide.

Elementary Steps (Molecularity)

Rate-Determining Step

Reaction Mechanism Requirements

Decomposition of \(\chem{N_2O_5}\)

$$ \chem{2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)} $$

$$ \begin{align} \text{Step 1:} & \; \chem{2N_2O_5 \rightleftharpoons 2NO_2+2NO_3} \; & \text{(fast)} \\ \text{Step 2:} & \; \chem{NO_2+NO_3 \rightarrow NO+O_2+NO_2} \; & \text{(slow)} \\ \text{Step 3:} & \; \chem{NO_3+NO \rightarrow 2NO_2} \; & \text{(fast)} \end{align} $$

Collision Model

Change in Potential Energy

The potential energy of a reacting system increases as we move from the reactants. The potential energy reaches its maximum value at the transition state before falling to the product's potential energy.

For Reactants to Form Products

Arrhenius Equation

$$ k=Ae^{-\bfrac{E_a}{RT}} $$

Linear form: $$\ln(k)=-\frac{E_a}{R}\left(\frac{1}{T}\right)+\ln(A)$$

Change in Potential Energy

A plot of the natural log of the rate constant versus one over the temperature in Kelvin will yeild a straight line.

Catalyst

Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction

The catalyzed pathway has a substantially lower potential energy peak, energy of the transition state, than the uncatalyzed pathway.

Effect of a Catalyst on the Number of Reaction-Producing Collisions

Based on a natural kinetic energy distribution of molecules, a considerably greater number of molecules have an energy greater than the catalyzed activation energy than the uncatalyzed activation energy.

Heterogeneous Catalysis

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