Chemical Kinetics

1. Rate Laws and Reaction Order

1.1 Rate of Reaction

For the reaction $aA + bB \to cC + dD$, the rate of reaction is:

$v = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$

1.2 The Rate Law

Definition 1 (Rate Law): For many reactions, the rate is proportional to the concentrations of reactants raised to powers:

$v = k[A]^m[B]^n$

where $k$ is the rate constant, $m$ is the order with respect to $A$, $n$ is the order with respect to $B$, and the overall order is $m + n$. The orders $m$ and $n$ are experimentally determined — they need not equal the stoichiometric coefficients.

1.3 Elementary Reactions and molecularity

For an elementary reaction (single molecular event), the order equals the molecularity:

• Unimolecular: $A \to$ products, rate $= k[A]$ (first order)
• Bimolecular: $A + B \to$ products, rate $= k[A][B]$ (second order)
• Termolecular: $A + B + C \to$ products, rate $= k[A][B][C]$ (third order, rare)

2. Integrated Rate Laws

2.1 Zeroth-Order Reactions

$\frac{d[A]}{dt} = -k$

$[A] = [A]_0 - kt$

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

2.2 First-Order Reactions

$\frac{d[A]}{dt} = -k[A]$

$\ln[A] = \ln[A]_0 - kt \quad \text{or} \quad [A] = [A]_0 e^{-kt}$

Half-life: $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$

The half-life is independent of initial concentration.

Example 1: Radioactive decay of ${}^{14}\text{C}$ has $t_{1/2} = 5730$ years. What fraction remains after 10000 years?

$k = \frac{0.693}{5730} = 1.21 \times 10^{-4} \text{ yr}^{-1}$

$\frac{[A]}{[A]_0} = e^{-kt} = e^{-1.21 \times 10^{-4} \times 10000} = e^{-1.21} = 0.298$

About 29.8% remains.

$\blacksquare$

2.3 Second-Order Reactions

Type I: $A + A \to$ products, rate $= k[A]^2$:

$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$

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

Type II: $A + B \to$ products with $[A]_0 = [B]_0$:

$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$

2.4 Pseudo-First-Order Reactions

When one reactant is in large excess ($[B]_0 \gg [A]_0$):

$v = k[A][B] \approx k"[A]$

where $k' = k[B]_0$ is the pseudo-first-order rate constant.

3. Methods for Determining Reaction Order

3.1 Differential Method (Initial Rates)

Measure initial rates at different initial concentrations:

$v_0 = k[A]_0^m \implies \log v_0 = \log k + m\log[A]_0$

A plot of $\log v_0$ vs $\log[A]_0$ has slope $m$.

3.2 Integral Method

Assume a reaction order, plot the corresponding linearized form:

• Zeroth order: $[A]$ vs $t$ (linear)
• First order: $\ln[A]$ vs $t$ (linear)
• Second order: $1/[A]$ vs $t$ (linear)

3.3 Half-Life Method

• If $t_{1/2}$ is constant: first order.
• If $t_{1/2}$ doubles when $[A]_0$ halves: second order.
• If $t_{1/2} \propto 1/[A]_0$: second order.

4. The Arrhenius Equation

4.1 Temperature Dependence of Rate Constants

Theorem 1 (Arrhenius Equation):

$k = A\,e^{-E_a/RT}$

where $A$ is the pre-exponential (frequency) factor and $E_a$ is the activation energy.

Logarithmic form:

$\ln k = \ln A - \frac{E_a}{RT}$

A plot of $\ln k$ vs $1/T$ gives a straight line with slope $-E_a/R$.

4.2 Two-Point Form

$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Example 2: A reaction has $k_1 = 3.46 \times 10^{-5}$ s$^{-1}$ at 298 K and $k_2 = 1.35 \times 10^{-3}$ s$^{-1}$ at 350 K. Find $E_a$.

$E_a = R\frac{\ln(k_2/k_1)}{1/T_1 - 1/T_2} = 8.314 \times \frac{\ln(1.35 \times 10^{-3}/3.46 \times 10^{-5})}{1/298 - 1/350}$

$= 8.314 \times \frac{3.66}{5.0 \times 10^{-4}} = 60.9 \text{ kJ/mol}$

$\blacksquare$

4.3 Modified Arrhenius Equation

For more accurate descriptions over wide temperature ranges:

$k = A\,T^n\,e^{-E_a/RT}$

5. Collision Theory

5.1 Basic Collision Theory

Theorem 2 (Collision Theory Rate Constant):

$k = \sigma\,N_A\,\langle v_r \rangle\,e^{-E_a/RT}$

where $\sigma = \pi(d_A + d_B)^2$ is the collision cross-section and $\langle v_r \rangle$ is the relative mean speed.

The mean relative speed from kinetic theory:

$\langle v_r \rangle = \sqrt{\frac{8k_BT}{\pi\mu}}$

where $\mu = \frac{m_A m_B}{m_A + m_B}$ is the reduced mass.

5.2 Steric Factor

Definition 2 (Steric Factor): Not every collision leads to reaction. The steric factor $P$ accounts for orientation requirements:

$k = P\,\sigma\,N_A\,\langle v_r \rangle\,e^{-E_a/RT}$

For simple collisions, $P \approx 1$; for complex molecules, $P \ll 1$.

6. Transition State Theory

6.1 Activated Complex

Definition 3 (Transition State): The transition state (activated complex) is the highest energy configuration along the reaction coordinate. The energy difference between reactants and the transition state is the activation energy.

6.2 The Eyring Equation

Theorem 3 (Eyring Equation):

$k = \frac{k_B T}{h}\,e^{-\Delta^{\ddagger} G^\circ/RT} = \frac{k_B T}{h}\,e^{\Delta^{\ddagger} S^\circ/R}\,e^{-\Delta^{\ddagger} H^\circ/RT}$

where $k_B$ is Boltzmann’s constant, $h$ is Planck’s constant, $\Delta^{\ddagger} G^\circ$, $\Delta^{\ddagger} H^\circ$, and $\Delta^{\ddagger} S^\circ$ are the standard Gibbs energy, enthalpy, and entropy of activation.

6.3 Relation to Arrhenius Parameters

At moderate temperatures:

$E_a = \Delta^{\ddagger} H^\circ + RT$

$A = e\,\frac{k_B T}{h}\,e^{\Delta^{\ddagger} S^\circ/R}$

A large positive $\Delta^{\ddagger} S^\circ$ means a loose, disordered transition state (typical for unimolecular reactions). A negative $\Delta^{\ddagger} S^\circ$ means a rigid, ordered transition state (typical for bimolecular reactions).

7. Reaction Mechanisms

7.1 Elementary Steps and Mechanisms

Definition 4 (Mechanism): A reaction mechanism is a sequence of elementary steps that accounts for the overall stoichiometry and the observed rate law.

Theorem 4 (Rate-Determining Step): If one elementary step is much slower than all others, the overall rate is approximately equal to the rate of that step.

7.2 Steady-State Approximation

Definition 5 (Steady-State Approximation): For reactive intermediates, assume $d[\text{intermediate}]/dt \approx 0$ after a short induction period.

Example 3: The decomposition of $\text{N}_2\text{O}_5$: $2\text{N}_2\text{O}_5 \to 4\text{NO}_2 + \text{O}_2$.

Proposed mechanism:

1. $\text{N}_2\text{O}_5 \xrightarrow{k_1} \text{NO}_2 + \text{NO}_3$ (slow)
2. $\text{NO}_2 + \text{NO}_3 \xrightarrow{k_{-1}} \text{N}_2\text{O}_5$ (fast)
3. $\text{NO}_2 + \text{NO}_3 \xrightarrow{k_2} \text{NO} + \text{O}_2 + \text{NO}_2$ (slow)
4. $\text{NO} + \text{NO}_3 \xrightarrow{k_3} 2\text{NO}_2$ (fast)

Steady-state for $\text{NO}_3$:

$\frac{d[\text{NO}_3]}{dt} = k_1[\text{N}_2\text{O}_5] - k_{-1}[\text{NO}_2][\text{NO}_3] - (k_2 + k_3)[\text{NO}_2][\text{NO}_3] = 0$

$[\text{NO}_3] = \frac{k_1[\text{N}_2\text{O}_5]}{(k_{-1} + k_2 + k_3)[\text{NO}_2]}$

The rate of formation of $\text{O}_2$ (from step 3): $v = k_2[\text{NO}_2][\text{NO}_3]$.

Substituting: $v = k_{\text{eff}}[\text{N}_2\text{O}_5]$ where $k_{\text{eff}} = \frac{k_1 k_2}{k_{-1} + k_2 + k_3}$.

$\blacksquare$

7.3 Pre-Equilibrium Approximation

When a rapid equilibrium precedes the rate-determining step:

$K = \frac{k_1}{k_{-1}} = \frac{[\text{intermediate}]}{[\text{reactant}]}$

The rate is determined by the slow step with the intermediate concentration expressed through $K$.

8. Chain Reactions

8.1 Chain Reaction Mechanism

1. Initiation: Formation of reactive intermediates (radicals).
2. Propagation: Intermediate reacts with reactant to form product and regenerate the intermediate.
3. Termination: Intermediates combine to form stable products.

Example 4: $\text{H}_2 + \text{Br}_2 \to 2\text{HBr}$ (Bodenstein mechanism).

$v = \frac{k[\text{H}_2][\text{Br}_2]^{1/2}}{1 + k'[\text{HBr}/\text{Br}_2]}$

The term $[\text{Br}_2]^{1/2}$ arises from the chain initiation/termination steps.

$\blacksquare$

8.2 Chain Length

Definition 6 (Chain Length): The number of product molecules formed per initiation event:

$\nu = \frac{\text{rate of propagation}}{\text{rate of initiation}}$

8.3 Explosions

Chain-branching reactions can lead to explosions (e.g., $\text{H}_2 + \text{O}_2$):

• Thermal explosion: Exothermic reaction heats the system, increasing the rate exponentially.
• Chain-branching explosion: Each propagation step produces more radicals than it consumes.

9. Enzyme Kinetics

9.1 Michaelis-Menten Mechanism

$E + S \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} ES \xrightarrow{k_2} E + P$

9.2 Michaelis-Menten Equation

Theorem 5 (Michaelis-Menten Equation): Under steady-state approximation for $[ES]$:

$v = \frac{V_{\max}[S]}{K_M + [S]}$

where $V_{\max} = k_2[E]_0$ is the maximum velocity and $K_M = (k_{-1} + k_2)/k_1$ is the Michaelis constant.

9.3 Lineweaver-Burk Plot

Taking reciprocals:

$\frac{1}{v} = \frac{K_M}{V_{\max}}\frac{1}{[S]} + \frac{1}{V_{\max}}$

A plot of $1/v$ vs $1/[S]$ gives slope $K_M/V_{\max}$ and intercept $1/V_{\max}$.

9.4 Catalytic Efficiency

Definition 7 (Catalytic Efficiency): For $[S] \ll K_M$:

$\frac{v}{[E][S]} = \frac{k_2}{K_M}$

The quantity $k_2/K_M$ is the catalytic efficiency. The diffusion-controlled limit is $\sim 10^8$–$10^9$ M$^{-1}$s$^{-1}$.

9.5 Enzyme Inhibition

For competitive inhibition:

$v = \frac{V_{\max}[S]}{K_M(1 + [I]/K_I) + [S]}$

10. Catalysis

10.1 Types of Catalysis

• Homogeneous catalysis: Catalyst and reactants in the same phase.
• Heterogeneous catalysis: Catalyst in a different phase (in most cases solid catalyst, gaseous/liquid reactants). Involves adsorption, surface reaction, and desorption.
• Autocatalysis: Product catalyzes its own formation (S-shaped kinetics).

10.2 Langmuir-Hinshelwood Mechanism

For heterogeneous catalysis on a surface:

1. Adsorption of reactants onto the surface.
2. Surface reaction between adsorbed species.
3. Desorption of products.

Rate depends on surface coverage $\theta$, described by the Langmuir isotherm:

$\theta = \frac{KP}{1 + KP}$

11. Complex Reaction Mechanisms

11.1 Parallel Reactions

$A \xrightarrow{k_1} B$ $A \xrightarrow{k_2} C$

$\frac{[B]}{[C]} = \frac{k_1}{k_2}$

The ratio of products is constant and determined by the ratio of rate constants.

11.2 Consecutive Reactions

$A \xrightarrow{k_1} B \xrightarrow{k_2} C$

$[B] = \frac{k_1[A]_0}{k_2 - k_1}\left(e^{-k_1 t} - e^{-k_2 t}\right)$

Maximum concentration of $B$ occurs at $t_{\max} = \frac{\ln(k_2/k_1)}{k_2 - k_1}$.

11.3 Reversible Reactions

$A \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} B$

$\frac{[B]_{\text{eq}}}{[A]_{\text{eq}}} = \frac{k_1}{k_{-1}} = K$

$[A] = [A]_0\frac{k_{-1} + k_1 e^{-(k_1 + k_{-1})t}}{k_1 + k_{-1}}$

12. Photochemistry

12.1 Beer-Lambert Law

Theorem 6 (Beer-Lambert Law):

$A = \varepsilon\,c\,l = \log_{10}\frac{I_0}{I}$

where $A$ is absorbance, $\varepsilon$ is the molar absorptivity, $c$ is concentration, $l$ is path length, $I_0$ is incident intensity, and $I$ is transmitted intensity.

12.2 Quantum Yield

Definition 8 (Quantum Yield):

$\Phi = \frac{\text{number of reaction events}}{\text{number of photons absorbed}}$

For a chain reaction, $\Phi \gg 1$; for fluorescence, $\Phi \leq 1$.

12.3 Stern-Volmer Equation

For fluorescence quenching:

$\frac{I_0}{I} = 1 + k_q\,\tau_0\,[Q] = 1 + K_{SV}[Q]$

where $[Q]$ is the quencher concentration, $\tau_0$ is the fluorescence lifetime without quencher, and $K_{SV}$ is the Stern-Volmer constant.

13. Fast Reactions

13.1 Relaxation Methods

For a reaction perturbed from equilibrium by a rapid temperature jump ($T$-jump):

Theorem 7 (Relaxation Time): For a single-step reaction $A \rightleftharpoons B$:

$\frac{1}{\tau} = k_1 + k_{-1}$

For $A + B \rightleftharpoons C + D$:

$\frac{1}{\tau} = k_1([A]_{\text{eq}} + [B]_{\text{eq}}) + k_{-1}([C]_{\text{eq}} + [D]_{\text{eq}})$

13.2 Flash Photolysis

A short laser pulse initiates the reaction; time-resolved spectroscopy monitors the decay of intermediates. Can measure rate constants up to $\sim 10^{12}$ s$^{-1}$.

Common Pitfalls

1. Confusing molecularity with reaction order. Molecularity applies only to elementary steps; the overall reaction order is determined experimentally. Fix: Never assign orders from the balanced equation unless the reaction is known to be elementary.
2. Using integrated rate laws for non-elementary reactions. The integrated forms assume a single step of that order. Fix: First determine the rate law experimentally, then check which integrated form is consistent.
3. Ignoring the steady-state approximation validity. The approximation requires the intermediate to be consumed as fast as it is formed. Fix: Check that $k_2 \gg k_1$ or verify the result by numerical integration.
4. Wrong activation energy units in the Arrhenius equation. $E_a$ must be in J/mol (not kJ/mol) when using $R = 8.314$ J/(mol·K). Fix: Always convert to consistent units before substituting.
5. Confusing $K_M$ with $K_d$ (dissociation constant). $K_M = (k_{-1} + k_2)/k_1$, not $k_{-1}/k_1$. Fix: $K_M = K_d$ only when $k_2 \ll k_{-1}$.
6. Misapplying Michaelis-Menten. The equation assumes steady-state $[ES]$, not equilibrium, and $[E]_0 \ll [S]$. Fix: When $[S]$ is comparable to $[E]_0$, use the full quadratic solution.
7. Forgetting that the Eyring equation uses $\Delta^{\ddagger} H^\circ$, not $E_a$. $E_a = \Delta^{\ddagger} H^\circ + RT$. Fix: For reactions in solution, $E_a \approx \Delta^{\ddagger} H^\circ$, but in the gas phase the $RT$ term matters at high temperatures.

Summary

• Rate law: $v = k[A]^m[B]^n$; order determined experimentally.
• Integrated rate laws: Zeroth ($[A]$ vs $t$), first ($\ln[A]$ vs $t$), second ($1/[A]$ vs $t$).
• Arrhenius equation: $k = A e^{-E_a/RT}$; activation energy from slope of $\ln k$ vs $1/T$.
• Collision theory: $k = P\sigma N_A\langle v_r\rangle e^{-E_a/RT}$.
• Eyring equation: $k = (k_B T/h)e^{-\Delta^{\ddagger}G^\circ/RT}$; connects kinetics to thermodynamics.
• Steady-state approximation: $d[\text{intermediate}]/dt \approx 0$; simplifies complex mechanisms.
• Michaelis-Menten: $v = V_{\max}[S]/(K_M + [S])$; Lineweaver-Burk plot for parameter extraction.
• Chain reactions: Initiation, propagation, termination; chain length $\nu$.
• Enzyme inhibition: Competitive, uncompetitive, noncompetitive effects on $K_M$ and $V_{\max}$.

Worked Examples

Example 1: Determining Reaction Order from Initial Rate Data

Problem: For the reaction A + 2B -> C, experiments yield: [A]=0.1, [B]=0.1, Rate=0.002; [A]=0.2, [B]=0.1, Rate=0.004; [A]=0.1, [B]=0.2, Rate=0.002. Determine the rate law. Solution: Doubling [A] (experiments 1 to 2) doubles the rate: order in A = 1. Doubling [B] (experiments 1 to 3) does not change the rate: order in B = 0. Rate law: v = k[A]. Rate constant k = 0.002/0.1 = 0.02 mol^-1 L s^-1.

Example 2: Arrhenius Plot Analysis

Problem: The rate constant doubles when temperature increases from 300 K to 310 K. Calculate the activation energy (R = 8.314 J mol^-1 K^-1). Solution: ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). ln(2) = (Ea/8.314)(1/300 - 1/310) = (Ea/8.314)(10/93000). Ea = 0.693 x 8.314 x 93000/10 = 53,570 J/mol = 53.6 kJ/mol.

Cross-References