Chemical Kinetics

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.

• 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.