Chemical Kinetics | Chemistry - Wyatt's Notes

sources:

• text: “Atkins, P., & de Paula, J. (2014). Atkins’ Physical Chemistry (10th ed.). Oxford University Press.”
• text: “Chang, R., & Goldsby, K. A. (2016). Chemistry (12th ed.). McGraw-Hill.” import Citations from ‘@components/Citations.astro’

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

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

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.

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)

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

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

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

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

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

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.

$\blacksquare$

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 $[A]_0 = [B]_0$:

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

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.

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 $m$.

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)

• 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.

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 $-E_a/R$.

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

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 $E_a$.

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

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

$\blacksquare$

For more accurate descriptions over wide temperature ranges:

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

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.

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, $P \ll 1$.

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.

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.

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).

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.

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 $k_{\text{eff}} = \frac{k_1 k_2}{k_{-1} + k_2 + k_3}$.

$\blacksquare$

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 $K$.

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.

$\blacksquare$

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

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

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.

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

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.

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 $1/V_{\max}$.

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$^{-1}$.

For competitive inhibition:

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

• 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).

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:

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

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

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

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

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

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

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

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

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

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

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.

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, $\Phi \leq 1$.

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.

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 $A + B \rightleftharpoons C + D$:

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

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$^{-1}$.

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.

flowchart TD
    A[Chemical Kinetics] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

• 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 $V_{\max}$.

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.

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.

Problem: An enzyme-catalyzed reaction has V_max = 0.50 micromol/min and K_M = 25 microM. What is the reaction velocity when [S] = 50 microM? At what substrate concentration is v = 0.25 micromol/min?

Solution: Using the Michaelis-Menten equation v = V_max[S]/(K_M + [S]):

For [S] = 50 microM: v = 0.50 x 50/(25 + 50) = 25/75 = 0.33 micromol/min.

For v = 0.25 micromol/min: 0.25 = 0.50[S]/(25 + [S]). Rearranging: 0.25(25 + [S]) = 0.50[S] => 6.25 + 0.25[S] = 0.50[S] => 6.25 = 0.25[S] => [S] = 25 microM.

Note that when [S] = K_M, v = V_max/2 = 0.25 micromol/min, which is always true by definition of K_M.

Common mistake: Confusing K_M with the dissociation constant K_d. K_M = (k_{-1} + k_2)/k_1, which equals K_d only when k_2 << k_{-1}. Never assume K_M equals the substrate concentration at half-maximal velocity unless the steady-state assumption holds.

$\blacksquare$

Chemical kinetics is the study of how fast reactions happen and what controls their speed. Think of activation energy as a hill that molecules must climb before they can react. Temperature gives molecules more kinetic energy, helping more of them clear the hill, which is why reactions speed up when heated. The Arrhenius equation quantifies this: higher temperature means more molecules have enough energy to overcome the barrier. Catalysts work by building a tunnel through the hill rather than going over it, lowering the activation energy without being consumed. Enzymes are nature’s catalysts, binding substrates in exactly the right orientation to react. The rate law tells us which ingredients matter most: doubling the concentration of a reactant that appears squared in the rate law quadruples the speed, while a reactant with zero order has no effect at all. Reaction mechanisms reveal the step-by-step choreography behind the overall transformation, with the slowest step acting as the bottleneck that determines the overall pace.

import { Citation } from “@components/Citations.astro”

<Citations sources={[ {title=“Physical Chemistry”, author=“Atkins and de Paula”, year=“2014”, type=“book”}, {title=“Chemical Kinetics and Dynamics”, author=“Steinfeld, Francisco and Hase”, year=“1998”, type=“book”}, ]} />

• Thermodynamics
• Calculus
• Quantum Mechanics
• Linear Algebra