Chapter 16: Kinetics: Reaction Rates, Mechanisms, and Catalysis Principles

Chapter 16: Kinetics: Rates and Mechanisms of Chemical Reactions

Focusing on Reaction Rate

• Chemical Kinetics: The study of reaction rates and the factors that affect them.

• Reaction Rate: Defined as the change in the concentrations of reactants or products as a function of time.

• Factors Influencing Reaction Rate:

  • Collision Principle: Particles must collide in order to react.

  • Concentration: The higher the concentration of reactants, the greater the reaction rate. A higher concentration of reactant particles allows for a greater number of collisions.

  • Physical State: The physical state of the reactants influences the reaction rate because substances must mix in order for particles to collide.

  • Surface Area: Decreasing the size of solid particles increases total surface area. For example, a hot steel nail glows feebly when placed in $O_2$, but the same mass of steel wool bursts into flame due to increased surface area.

  • Temperature: The higher the temperature, the greater the reaction rate. At higher temperatures, particles possess more energy; therefore, they collide more often and more effectively.

  • Activation Energy: A reaction can occur only if the collision energy is high enough. No reaction can occur if collision energy is too low.

Expressing the Reaction Rate

• Measurement: Reaction rate is measured in terms of the changes in concentrations of reactants or products per unit time.

• Mathematical Expression: For a general reaction $A \rightarrow B$, the rate is measured by the concentration of $A$ at $t_1$ and $t_2$:

  • $Rate = -\frac{[A]_2 - [A]_1}{t_2 - t_1} = -\frac{\Delta [A]}{\Delta t}$

  • Square brackets $[\,]$ indicate concentration in moles per liter ($mol/dm^3$).

  • The negative sign is applied to reactant concentrations because they decrease over time; this ensures the calculated rate is a positive value.

  • Product concentrations increase over time, so their rates are expressed positively: $Rate = \frac{\Delta [B]}{\Delta t}$.

• Types of Reaction Rates:

  • Average Rate: Changes in concentration over a finite period of time.

  • Instantaneous Rate: The rate at a specific point in time, determined by the slope of a tangent line to the concentration-time curve.

  • Initial Rate: The instantaneous rate at $t = 0$.

• General Form of Reaction Rates: For the general balanced equation $aA + bB \rightarrow cC + dD$, the rate is expressed as:

  • $Rate = -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = \frac{1}{c} \frac{\Delta [C]}{\Delta t} = \frac{1}{d} \frac{\Delta [D]}{\Delta t}$

Sample Problem 16.1: Expressing Rate in Terms of Changes in Concentration

• Problem: Hydrogen gas combustion: $2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$.

  • (a) Express the rate in terms of changes in $[H_2]$, $[O_2]$, and $[H_2O]$ with time.

  • (b) When $[O_2]$ is decreasing at $0.23\,mol/L \cdot s$, at what rate is $[H_2O]$ increasing?

• Solution (a):

  • $Rate = -\frac{1}{2} \frac{\Delta [H_2]}{\Delta t} = -\frac{\Delta [O_2]}{\Delta t} = \frac{1}{2} \frac{\Delta [H_2O]}{\Delta t}$

• Solution (b):

  • $\frac{\Delta [H_2O]}{\Delta t} = -2 \frac{\Delta [O_2]}{\Delta t}$

  • $\frac{\Delta [H_2O]}{\Delta t} = -2(-0.23\,mol/L \cdot s) = 0.46\,mol/L \cdot s$

The Rate Law and Its Components

• General Rate Law: For any reaction occurring at a fixed temperature:

  • $Rate = k[A]^m[B]^n$

• Rate Constant ($k$): Specific for a given reaction at a given temperature; it does not change as the reaction proceeds.

• Reaction Orders ($m$ and $n$): Determined by experiment, not necessarily related to the stoichiometric coefficients ($a$ and $b$).

  • First Order (with respect to A): If the rate doubles when $[A]$ doubles ($Rate = k[A]^1$).

  • Second Order (with respect to A): If the rate quadruples when $[A]$ doubles ($Rate = k[A]^2$).

  • Zero Order (with respect to A): If the rate remains unchanged when $[A]$ doubles ($Rate = k[A]^0 = k$).

• Overall Reaction Order: The sum of the individual reactant orders ($m + n$).

Sample Problem 16.2: Determining Reaction Orders from Rate Laws

• Problem (a): Reaction $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$, Rate Law: $Rate = k[NO]^2[O_2]$.

  • Order in NO: 2 (Second order).

  • Order in $O_2$: 1 (First order).

  • Overall order: 3 (Third order).

  • Rate change if $[NO]$ is tripled and $[O_2]$ is doubled: Rate increases by a factor of $3^2 \times 2 = 18$.

• Problem (b): $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$, Rate Law: $Rate = k[CH_3CHO]^{3/2}$.

  • Order in $CH_3CHO$: $3/2$.

  • Overall order: $3/2$.

• Problem (c): $H_2O_2(aq) + 3I^-(aq) + 2H^+(aq) \rightarrow I_3^-(aq) + 2H_2O(l)$, Rate Law: $Rate = k[H_2O_2][I^-]$.

  • Order in $H_2O_2$: 1.

  • Order in $I^-$: 1.

  • Order in $H^+$: 0.

  • Overall order: 2.

Determining Reaction Orders and Rate Constants

• Experimental Series: To find $m$ and $n$, a series of experiments is conducted where one concentration is varied while the other is kept constant.

• Calculating Order with respect to $O_2$:

  • $\frac{Rate_2}{Rate_1} = \frac{k[O_2]^m_2[NO]^n_2}{k[O_2]^m_1[NO]^n_1}$

  • If concentration doubles and rate doubles, $2^m = 2$, so $m = 1$.

• Using Logarithms: $\log\left(\frac{Rate_2}{Rate_1}\right) = m \log\left(\frac{[A]_2}{[A]_1}\right)$.

• Units of Rate Constant $k$ (Table 16.2):

  • Overall Order 0: Units = $mol/L \cdot s$ (or $mol \cdot L^{-1} \cdot s^{-1}$)

  • Overall Order 1: Units = $1/s$ (or $s^{-1}$)

  • Overall Order 2: Units = $L/mol \cdot s$ (or $L \cdot mol^{-1} \cdot s^{-1}$)

  • Overall Order 3: Units = $L^2/mol^2 \cdot s$ (or $L^2 \cdot mol^{-2} \cdot s^{-1}$)

Sample Problem 16.3: Determining Orders from Rate Data

• Data:

  • Expt 1: Rate 0.0050, $[NO_2]$ 0.10, $[CO]$ 0.10

  • Expt 2: Rate 0.080, $[NO_2]$ 0.40, $[CO]$ 0.10

  • Expt 3: Rate 0.0050, $[NO_2]$ 0.10, $[CO]$ 0.20

• Solution:

  • Order of $NO_2$ ($m$): Compare Expt 1 and 2. Rate increases 16x when concentration increases 4x: $4^m = 16 \rightarrow m = 2$.

  • Order of $CO$ ($n$): Compare Expt 1 and 3. Rate does not change. $n = 0$.

  • Rate Law: $Rate = k[NO_2]^2$.

  • Calculate $k$: $0.0050 = k(0.10)^2 \rightarrow k = 0.50\,L/mol \cdot s$.

Integrated Rate Laws: Concentration Changes Over Time

• First-Order Reactions:

  • Integrated Law: $\ln\left(\frac{[A]_0}{[A]_t}\right) = kt$ or $\ln[A]_t = -kt + \ln[A]_0$.

  • Half-life ($t_{1/2}$): The time taken for concentration to drop to half its initial value. For first-order: $t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k}$.

  • Key Characteristic: $t_{1/2}$ is constant and does not depend on $[A]_0$.

• Second-Order Reactions (One Reactant):

  • Integrated Law: $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$.

  • Half-life: $t_{1/2} = \frac{1}{k[A]_0}$.

  • Key Characteristic: $t_{1/2}$ is inversely proportional to initial concentration.

• Zero-Order Reactions:

  • Integrated Law: $[A]_t = -kt + [A]_0$.

  • Half-life: $t_{1/2} = \frac{[A]_0}{2k}$.

  • Key Characteristic: $t_{1/2}$ is directly proportional to initial concentration.

Sample Problem 16.5: First-Order Reaction Concentration

• Problem: Cyclobutane ($C_4H_8$) decomposes to ethylene. $k = 87\,s^{-1}$.

  • (a) Initial concentration is $2.00\,M$. Concentration after $0.010\,s$?

    • $\ln\left(\frac{2.00}{[A]_t}\right) = (87\,s^{-1})(0.010\,s) = 0.87$

    • $\frac{2.00}{[A]_t} = e^{0.87} = 2.38 \rightarrow [A]_t = 0.84\,M$

  • (b) Time for 70% decomposition (30% remains)?

    • $\ln\left(\frac{100}{30}\right) = (87\,s^{-1})t \rightarrow t = 0.014\,s$

Theories of Chemical Kinetics

• Collision Theory: Particles must collide with sufficient energy ($E_a$) and the correct orientation to react.

  • The Orientation Probability Factor ($p$): The ratio of effectively oriented collisions to total collisions. More complex structures have smaller $p$ values.

• Transition State Theory: An effective collision forms a transition state (activated complex), an unstable species with partial bonds existing at the maximum potential energy point.

• Reaction Energy Diagrams:

  • $E_{a(fwd)}$: Activation energy for the forward reaction.

  • $E_{a(rev)}$: Activation energy for the reverse reaction.

  • $\Delta H_{rxn} = E_{a(fwd)} - E_{a(rev)}$.

• Arrhenius Equation:

  • $k = A e^{-E_a/RT}$

  • $A$: Frequency factor ($A = pZ$, where $Z$ is collision frequency).

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

  • Two-temperature form: $\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$.

  • $R = 8.314\,J/mol \cdot K$.

Sample Problem 16.10: Determining Activation Energy

• Problem: HI decomposition. $k = 3.52 \times 10^{-7}\,L/mol \cdot s$ at $500.\,K$ and $k = 3.95 \times 10^{-2}\,L/mol \cdot s$ at $700.\,K$.

• Solution:

  • $\ln\left(\frac{3.95 \times 10^{-2}}{3.52 \times 10^{-7}}\right) = -\frac{E_a}{8.314} \left(\frac{1}{700} - \frac{1}{500}\right)$

  • $11.63 = -\frac{E_a}{8.314} (-0.000571)$

  • $E_a = 1.69 \times 10^5\,J/mol = 169\,kJ/mol$

Reaction Mechanisms

• Elementary Steps: Single molecular events that sum to the overall balanced equation.

• Molecularity: The number of reactant particles in an elementary step.

  • Unimolecular: One particle.

  • Bimolecular: Two particles (Rate Law: $Rate = k[A]^2$ or $k[A][B]$).

  • Termolecular: Three particles (rare).

• Rate Law for Elementary Steps: Can be deduced directly from stoichiometry. Reaction order equals molecularity.

• Reaction Intermediate: A substance formed in one step and consumed in a later step (not in the overall equation).

• Rate-Determining Step (RDS): The slowest step in the mechanism determines the overall rate law.

• Criteria for a Valid Mechanism:

  1. Elementary steps sum to the overall balanced equation.

  2. Steps are molecularly reasonable (usually 1 or 2 particles).

  3. Mechanism correlates with the observed (experimental) rate law.

Catalysis: Speeding Up a Reaction

• Catalyst: Increases the reaction rate without being consumed. It provides an alternative pathway with a lower total activation energy ($E_a$).

• Properties of Catalysts:

  • Affects both forward and reverse reaction rates.

  • Does NOT change the enthalpy change ($\Delta H$) or the overall yield.

• Homogeneous Catalysis: Catalyst is in the same phase as reactants (e.g., $Br^-(aq)$ acting as a catalyst for $H_2O_2$ decomposition).

• Heterogeneous Catalysis: Catalyst is in a different phase (e.g., solid metal surfaces in hydrogenation reactions).

• Biological Catalysts (Enzymes):

  • Active Site: Region on the enzyme where the substrate binds.

  • Substrate: The reactant.

  • Lock-and-Key Model: Active site has a fixed shape matching the substrate.

  • Induced-Fit Model: Active site changes shape to bind substrate more effectively.

Case Study: Ozone Depletion

• Uncatalyzed Reaction: $O_3 + O \rightarrow 2O_2$, $E_a = 17\,kJ$, $\Delta H = -392\,kJ$.

• Catalyzed Reaction (by Cl atoms):

  • Step 1: $O_3 + Cl \rightarrow ClO + O_2$, $E_{a1} = 2\,kJ$.

  • Step 2: $ClO + O \rightarrow Cl + O_2$, $E_{a2} = 0.4\,kJ$.

  • Overall: $O_3 + O \rightarrow 2O_2$.

  • $Cl$ is the catalyst; $ClO$ is the intermediate.