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Chapter 10 • Theory & Derivations

Unit 10: Molecular Reaction Dynamics, Potential Energy Surfaces & Transition State Theory

Microscopic molecular dynamics of chemical transformations: hard-sphere collision theory, steric orientation factors, the Smoluchowski-Debye diffusion-controlled limits in solution, statistical thermodynamic partition function derivation of the Eyring-Polanyi Transition State Theory (TST), thermodynamic activation parameters, primary and secondary kinetic salt effects, crossed molecular beam reactive scattering, and London-Eyring-Polanyi-Sato (LEPS) potential energy surfaces.

§10.1 Hard-Sphere Collision Theory of Gas Reactions & Steric Factor P

Simple collision theory (Trautz and Lewis, 1916-1918) provided the earliest microscopic physical model for bimolecular reaction rate constants.

Fundamental Formulation

For an elementary bimolecular reaction $A + B \longrightarrow P$:

$$r = Z_{AB} \cdot f_{\text{Boltzmann}} \cdot P$$

where:

  1. $Z_{AB}$ is the collision density between species $A$ and $B$:
$$Z_{AB} = \sigma_{AB} \bar{v}_{\text{rel}} \mathcal{N}_A \mathcal{N}_B = \pi d_{AB}^2 \sqrt{\frac{8 k_B T}{\pi \mu}} \mathcal{N}_A \mathcal{N}_B$$

where $d_{AB} = \frac{d_A + d_B}{2}$ is the collision diameter, and $\mu = \frac{m_A m_B}{m_A + m_B}$ is the reduced mass.

  1. $f_{\text{Boltzmann}} = \exp(-E_0 / R T)$ is the fraction of collisions possessing relative kinetic energy along the line of centers exceeding threshold $E_0$.
  2. $P$ is the steric factor ($0 < P \le 1$).

The Bimolecular Rate Constant

Expressing in molar units ($k = r / [A][B]$):

$$k(T) = P \cdot N_A \pi d_{AB}^2 \sqrt{\frac{8 k_B T}{\pi \mu}} \exp\left( -\frac{E_0}{R T} \right)$$

Comparing with the Arrhenius equation $k = A \exp(-E_a / R T)$:

$$A = P \cdot N_A \sigma_{AB} \sqrt{\frac{8 k_B T}{\pi \mu}} \propto P T^{1/2}$$

Physical Meaning of the Steric Factor $P$

For simple spherical atom-atom reactions (e.g., $K + Br_2$), $P \approx 1\text{--}4$ (the "harpoon mechanism" gives $P > 1$ due to long-range electron transfer). For polyatomic molecules with complex geometry (e.g., cyclization of hexatriene), $P$ drops to $10^{-4}\text{--}10^{-7}$, indicating that less than one in a million energetic collisions occurs with proper mutual orientation.

Polanyi Rules & Potential Energy Surface Topology Matrix

| Reaction Class | Canonical Chemical Reaction | $\Delta H_r^\circ$ ($\text{kJ/mol}$) | Barrier Location | Polanyi Classification | Optimum Energy for Reaction | Product Energy Disposal | |---|---|---|---|---|---|---| | Exothermic | $F + H_2 \longrightarrow HF + H$ | $-134.0$ | Entrance Valley | Early Barrier (Attractive PES) | Translational Energy ($E_{\text{trans}}$) | High Product Vibration ($v' = 2, 3$) | | Exothermic | $H + Cl_2 \longrightarrow HCl + Cl$ | $-188.0$ | Entrance Valley | Early Barrier (Attractive PES) | Translational Energy ($E_{\text{trans}}$) | High Product Vibration ($v' = 3 - 6$) | | Thermoneutral| $H + H_2 \longrightarrow H_2 + H$ | $0.0$ | Symmetric Saddle Point | Central Barrier | Both Translation & Vibration | Moderate Translation & Vibration | | Endothermic | $H + HF \longrightarrow H_2 + F$ | $+134.0$ | Exit Valley | Late Barrier (Repulsive PES) | Reactant Vibration ($E_{\text{vib}}$) | High Product Translation ($E_{\text{trans}}'$) | | Endothermic | $Cl + HCl \longrightarrow Cl_2 + H$| $+188.0$ | Exit Valley | Late Barrier (Repulsive PES) | Reactant Vibration ($E_{\text{vib}}$) | High Product Translation ($E_{\text{trans}}'$) | | Harpoon Reaction| $K + Br_2 \longrightarrow KBr + Br$ | $-175.0$ | Long-Range Curve Crossing | Harpoon Electron Jump ($R_c \approx 8\text{ Å}$)| Low Thermal Energy Suffices | Forward Stripping Rebound ($KBr$) |

§10.2 Diffusion-Controlled Reactions in Solution: Smoluchowski & Debye Limits

In liquid solution, molecules are enclosed within solvent "cages", undergoing dozens of rapid repeated collisions (an encounter) before diffusing apart.

The Smoluchowski Equation (1917)

For an activationless reaction between neutral species where every encounter leads immediately to reaction ($E_a \approx 0$), the rate is strictly limited by the rate at which reactants diffuse toward one another. Solving Fick's second law for spherical diffusion to an absorbing sphere of radius $R_{AB} = r_A + r_B$:

$$k_{\text{diff}} = 4 \pi N_A (D_A + D_B) R_{AB}$$

where $D_A, D_B$ are diffusion coefficients.

Substituting the Stokes-Einstein relation $D_i = \frac{k_B T}{6 \pi \eta r_i}$: Assuming equal hydrodynamic radii $r_A = r_B = r$:

$$D_A + D_B = \frac{2 k_B T}{6 \pi \eta r} = \frac{k_B T}{3 \pi \eta r}$$
$$R_{AB} = 2 r$$
$$k_{\text{diff}} = 4 \pi N_A \left( \frac{k_B T}{3 \pi \eta r} \right) (2 r) = \frac{8 N_A k_B T}{3 \eta} = \frac{8 R T}{3 \eta}$$

Remarkable Feature: The diffusion-controlled rate constant is independent of solute size and depends solely on temperature and solvent dynamic viscosity $\eta$! For water at $298.15\text{ K}$ ($\eta = 0.890\text{ cP}$):

$$k_{\text{diff}} = \frac{8 \times 8.314 \times 298.15}{3 \times (0.890 \times 10^{-3})} \approx 7.4 \times 10^9\text{ M}^{-1}\text{s}^{-1}$$

The Debye Modification for Ionic Encounters (1942)

For charged ions ($z_A e$ and $z_B e$), Coulombic attraction or repulsion alters encounter flux:

$$k_{\text{diff, ions}} = k_{\text{diff}} \cdot \left[ \frac{\Phi}{e^\Phi - 1} \right], \quad \text{where } \Phi = \frac{z_A z_B e^2}{4 \pi \varepsilon_r \varepsilon_0 R_{AB} k_B T}$$

Oppositely charged ions ($z_A z_B < 0$) accelerate diffusion ($k > k_{\text{diff}}$), while like-charged ions repel ($k < k_{\text{diff}}$).

§10.3 Transition State Theory (TST): Eyring Equation via Statistical Partition Functions

Henry Eyring, Michael Polanyi, and Eugene Wigner (1935) established Transition State Theory (also known as Activated Complex Theory), which calculates absolute reaction rates from first-principles statistical mechanics without empirical parameters.

Postulates of Classical TST

  1. Reactants are in quasi-thermal equilibrium with the activated complex:
$$A + B \rightleftharpoons [AB]^\ddagger \\xrightarrow{k^\ddagger} P$$
  1. The activated complex $[AB]^\ddagger$ is treated as an ordinary molecule, except that one vibrational mode along the reaction coordinate has transformed into a loose translational motion across the saddle point.
  2. Every activated complex crossing the barrier in the forward direction proceeds irreversibly to product (transmission coefficient $\kappa_{\text{trans}} \approx 1$).

Partition Function Derivation

The pseudo-equilibrium constant is:

$$K^\ddagger = \frac{[AB]^\ddagger}{[A][B]} = \frac{q^\ddagger}{q_A q_B} \exp\left( -\frac{\Delta E_0^\ddagger}{k_B T} \right)$$

where $q$ are molecular partition functions per unit volume.

Factoring out the critical reaction-coordinate vibration:

$$q^\ddagger = q_{\text{rc}} \cdot q_{\text{int}}^\ddagger$$

In the classical limit ($h \nu \ll k_B T$):

$$q_{\text{rc}} = \frac{k_B T}{h \nu^\ddagger}$$

The rate of barrier crossing is the vibrational frequency $\nu^\ddagger$:

$$k^\ddagger = \nu^\ddagger$$

The reaction rate is:

$$r = k^\ddagger [AB]^\ddagger = \nu^\ddagger \left( \frac{k_B T}{h \nu^\ddagger} \right) \frac{q_{\text{int}}^\ddagger}{q_A q_B} \exp\left( -\frac{\Delta E_0^\ddagger}{k_B T} \right) [A][B]$$

The unknown imaginary frequency $\nu^\ddagger$ cancels identically, yielding the foundational Eyring Equation:

$$k(T) = \kappa_{\text{trans}} \frac{k_B T}{h} \frac{q^\ddagger}{q_A q_B} \exp\left( -\frac{\Delta E_0^\ddagger}{k_B T} \right)$$

The universal prefactor $\frac{k_B T}{h} \approx 6.21 \times 10^{12}\text{ s}^{-1}$ at $298.15\text{ K}$ represents the fundamental attempt frequency of chemical transformation.

§10.4 Thermodynamic Formulation of TST: Activation Enthalpy, Entropy & Free Energy

Eyring recast Transition State Theory into macroscopic thermodynamic language by expressing the quasi-equilibrium constant in terms of standard activation free energy:

$$K^\ddagger = \exp\left( -\frac{\Delta G^{\ddagger\circ}}{R T} \right) = \exp\left( \frac{\Delta S^{\ddagger\circ}}{R} \right) \exp\left( -\frac{\Delta H^{\ddagger\circ}}{R T} \right)$$

The Thermodynamic Eyring Equation

$$k(T) = \frac{k_B T}{h} (c^\circ)^{1 - m} \exp\left( \frac{\Delta S^{\ddagger\circ}}{R} \right) \exp\left( -\frac{\Delta H^{\ddagger\circ}}{R T} \right)$$

where $m$ is molecularity and $c^\circ = 1\text{ M}$ is standard state concentration.

Relation Between Arrhenius and Eyring Parameters

From the Arrhenius definition $E_a \equiv R T^2 \frac{d \ln k}{dT}$:

$$\ln k = \ln\left( \frac{k_B}{h} \right) + \ln T - \frac{\Delta H^{\ddagger\circ}}{R T} + \frac{\Delta S^{\ddagger\circ}}{R}$$
$$\frac{d \ln k}{dT} = \frac{1}{T} + \frac{\Delta H^{\ddagger\circ}}{R T^2}$$
$$E_a = R T^2 \left( \frac{1}{T} + \frac{\Delta H^{\ddagger\circ}}{R T^2} \right) = \Delta H^{\ddagger\circ} + R T \quad (\text{liquid-phase reactions})$$

For ideal gas reactions of molecularity $m$:

$$E_a = \Delta H^{\ddagger\circ} + m R T$$
  • For unimolecular gas reactions ($m = 1$): $E_a = \Delta H^{\ddagger\circ} + R T$.
  • For bimolecular gas reactions ($m = 2$): $E_a = \Delta H^{\ddagger\circ} + 2 R T$.

Physical Meaning of Activation Entropy ($\Delta S^{\ddagger\circ}$)

  • $\Delta S^{\ddagger\circ} < 0$ (associative transition state): Two independent molecules combine to form a rigid, highly ordered transition state, losing translational and rotational degrees of freedom (typical bimolecular additions).
  • $\Delta S^{\ddagger\circ} > 0$ (dissociative transition state): A molecule loosens bonds, releasing fragments or solvent molecules, increasing disorder.

§10.5 Kinetic Salt Effects in Solution: The Brønsted-Bjerrum Equation

The rate of ionic reactions in solution depends strongly on the ionic strength $I$ of the medium.

Brønsted-Bjerrum Formulation (1922-1924)

Consider an elementary reaction between ions $A^{z_A}$ and $B^{z_B}$ forming an activated complex $[AB]^{\ddagger (z_A + z_B)}$:

$$A^{z_A} + B^{z_B} \rightleftharpoons [AB]^\ddagger \\xrightarrow{k_0} P$$

Thermodynamic equilibrium requires activities:

$$K^\ddagger = \frac{a_\ddagger}{a_A a_B} = \frac{[AB]^\ddagger \gamma_\ddagger}{[A]\gamma_A [B]\gamma_B} \implies [AB]^\ddagger = K^\ddagger [A][B] \left( \frac{\gamma_A \gamma_B}{\gamma_\ddagger} \right)$$

The observed rate is:

$$r = k_0 [AB]^\ddagger = k_0 K^\ddagger [A][B] \left( \frac{\gamma_A \gamma_B}{\gamma_\ddagger} \right) = k [A][B]$$

Thus, the rate constant $k$ is:

$$k = k_0 \left( \frac{\gamma_A \gamma_B}{\gamma_\ddagger} \right) \iff \log_{10} k = \log_{10} k_0 + \log_{10} \gamma_A + \log_{10} \gamma_B - \log_{10} \gamma_\ddagger$$

where $k_0$ is the rate constant at infinite dilution ($I \to 0$).

Incorporation of Debye-Hückel Limiting Law

At low ionic strength ($I < 0.05\text{ M}$), the activity coefficient is given by $\log_{10} \gamma_i = -A z_i^2 \sqrt{I}$:

$$\log_{10}\left( \frac{\gamma_A \gamma_B}{\gamma_\ddagger} \right) = -A \left[ z_A^2 + z_B^2 - (z_A + z_B)^2 \right] \sqrt{I} = -A [ -2 z_A z_B ] \sqrt{I} = 2 A z_A z_B \sqrt{I}$$

For water at $298.15\text{ K}$, $A = 0.509\;(\text{mol/kg})^{-1/2}$:

$$\log_{10}\left( \frac{k}{k_0} \right) = 2 (0.509) z_A z_B \sqrt{I} = 1.018 \, z_A z_B \sqrt{I}$$

Diagnostic Regimes of Primary Salt Effect

1. Like Charges ($z_A z_B > 0$): Slope $> 0$. Adding inert electrolyte shields repulsion, stabilizing the $[AB]^\ddagger$ complex and accelerating the reaction.

2. Opposite Charges ($z_A z_B < 0$): Slope $< 0$. Adding inert electrolyte stabilizes separated reactants more than the complex, decelerating the reaction.

3. Neutral Reactant ($z_A z_B = 0$): Slope $= 0$. Rate is virtually independent of ionic strength.

University Honors Research Monograph: Quantum Resonances & Feshbach Bound States in the F + H2 Reaction

The benchmark elementary bimolecular reaction $F(^2P) + H_2 \longrightarrow HF(v', j') + H$ is the gold standard of modern quantum reaction dynamics:

  • Reactive Scattering Resonances:

In crossed molecular beam experiments with Velocity Map Imaging, Yuan T. Lee, K. Liu, and X. Yang detected sharp step-function peaks in the backward-scattered differential cross-section for producing $HF(v'=3)$ at precise collision energies ($E_{\text{coll}} \approx 0.040 - 0.052\text{ eV}$).

  • The Feshbach Transition State Bound State:

Full quantum 3D wavepacket calculations on highly accurate ab initio potential energy surfaces (such as the FXZ and CBS surfaces) revealed that these resonance spikes arise from quasi-bound quantum states trapped inside a dynamic potential well located in the transition state region:

$$F\cdots H-H \rightleftharpoons [F\cdots H\cdots H]^\ddagger \longrightarrow HF(v'=3) + H$$

Because the entrance channel is vibrationally adiabatic, the colliding system is temporarily trapped in a metastable quantum state for several vibrational periods ($\sim 30 - 50\text{ fs}$) before tunneling out into the exit channel, proving that chemical reactions exhibit discrete quantum bound states at the transition state barrier.

§10.6 Molecular Reaction Dynamics: Crossed Beams & State-to-State Scattering

While classical kinetics measures macroscopic ensemble thermal averages $k(T)$, molecular reaction dynamics probes single-collision events with defined quantum states, velocity vectors, and scattering angles.

Crossed Molecular Beam Experiments (Herschbach and Lee, Nobel Prize 1986)

Two collimated supersonic molecular beams collide at right angles in an ultra-high vacuum chamber ($P < 10^{-10}\text{ Torr}$):

1. Velocity Selection: Choppers select initial relative kinetic energy $E_{\text{coll}}$.

2. Rotatable Mass Spectrometer: Measures angular distribution $d\sigma / d\Omega$ and time-of-flight velocity distributions of scattered reaction products.

Dynamics Classification: Rebound vs. Stripping

1. Rebound Mechanism ($K + CH_3I \longrightarrow KI + CH_3$):

  • Small impact parameter ($b < d$).
  • Direct backward scattering ($\theta \approx 180^\circ$).
  • Low product vibrational excitation; high translational energy release.

2. Stripping Mechanism ($K + Br_2 \longrightarrow KBr + Br$):

  • Large impact parameter ($b > d$).
  • Forward scattering ($\theta \approx 0^\circ$).
  • Harpoon Model: At long distance ($R_c \approx 4\text{--}6\text{ Å}$), an electron jumps from potassium to bromine ($K + Br_2 \to K^+ + Br_2^-$), followed by strong Coulombic attraction that strips the halogen atom into high vibrational excitation.

§10.7 Potential Energy Surfaces (PES): LEPS Formulation & Saddle Points

The Potential Energy Surface (PES) is the hyper-dimensional function $V(\vec{R})$ expressing the electronic energy of a reacting system as a function of nuclear geometry under the Born-Oppenheimer approximation.

Collinear Triatomic Reactions ($A + B-C \longrightarrow A-B + C$)

For a collinear arrangement, the potential energy depends on only two internuclear distances: $R_{AB}$ and $R_{BC}$.

The London-Eyring-Polanyi-Sato (LEPS) Surface

Constructed from semi-empirical valence bond theory by summing Coulombic ($Q$) and exchange ($J$) integrals:

$$V(R_{AB}, R_{BC}, R_{AC}) = Q_1 + Q_2 + Q_3 - \sqrt{\frac{1}{2} \left[ (J_1 - J_2)^2 + (J_2 - J_3)^2 + (J_3 - J_1)^2 \right]}$$

calibrated using experimental diatomic Morse potentials.

Topography of the PES

1. Reactant Valley: Large $R_{AB}$, equilibrium $R_{BC}$.

2. Product Valley: Equilibrium $R_{AB}$, large $R_{BC}$.

3. Transition State (Saddle Point $\ddagger$): A first-order saddle point where the gradient vanishes ($\nabla V = 0$), corresponding to a maximum along the reaction coordinate and a minimum along all perpendicular vibrational coordinates.

Polanyi's Rules for Barrier Location

John Polanyi (Nobel Prize 1986) related barrier location to energy partitioning:

1. Early Barrier (Attractive Surface, reactant valley):

Translational energy $E_{\text{trans}}$ is highly effective in promoting reaction; product energy appears primarily as vibration ($E_{\text{vib}}'$). Exothermic reactions generally possess early barriers.

2. Late Barrier (Repulsive Surface, product valley):

Vibrational energy $E_{\text{vib}}$ in the reactant bond is far more effective than translational energy in crossing the barrier. Endothermic reactions generally possess late barriers.

§10.8 Velocity Map Imaging (VMI) & Femtosecond Transition State Spectroscopy

Observing chemical reactions at the most fundamental quantum mechanical level requires measuring 3D product velocity vectors using ion imaging and observing transient Transition States in real time using femtosecond pump-probe laser pulses.

1. Velocity Map Imaging (VMI) in Crossed Molecular Beams

Velocity Map Imaging, introduced by David Parker and André Eppink (1997), revolutionized experimental reaction dynamics by measuring full 3D differential cross-sections $\frac{d^2\sigma}{d\Omega dE}$ with 100% collection efficiency.

Principle of Electrostatic Immersion Lens Focusing

In a crossed molecular beam experiment:

  1. Two collimated supersonic molecular beams (e.g., $F + H_2$ or $O(^1D) + CH_4$) intersect at $90^\circ$ inside an ultra-high vacuum scattering chamber ($P < 10^{-10}\text{ Torr}$).
  2. Products formed in specific quantum states are selectively ionized by Resonance-Enhanced Multi-Photon Ionization (REMPI).
  3. The resulting ion cloud expands according to product recoil velocities. An open electrostatic immersion lens system (repeller, extractor, and ground electrodes with smooth curved potentials) accelerates ions toward a 2D position-sensitive microchannel plate (MCP) detector backed by a phosphor screen and CCD camera.
  4. Crucially, the inhomogeneous electric field configuration functions as an ion telescope: All ions having identical initial velocity vectors $(v_x, v_y, v_z)$ are mapped onto the exact same spatial coordinate $(X, Y)$ on the detector, regardless of where they were formed within the finite spatial laser ionization volume.
Image Reconstruction via Abel Inversion

The 2D CCD image represents a cylindrical projection of the true 3D cylindrically symmetric scattering distribution around the relative velocity vector $\vec{v}_{\text{rel}}$. Applying the mathematical Inverse Abel Transform:

$$F(r, \theta) = -\frac{1}{\pi} \int_r^\infty \frac{d P(x, y)/dx}{\sqrt{x^2 - r^2}} dx$$

reconstructs the true 3D velocity slice, revealing product scattering angle $\theta_{\text{c.m.}}$ (forward vs. backward rebound) and kinetic energy disposal $E_{\text{trans}}'$, proving the microscopic collision mechanism.

2. Femtosecond Transition State Spectroscopy (Ahmed Zewail)

Prior to femtosecond laser spectroscopy (Nobel Prize in Chemistry 1999 to Ahmed Zewail), the Transition State was regarded as an unobservable theoretical abstraction lasting only $\tau \approx 10 - 100\text{ fs}$ ($10^{-14} - 10^{-13}\text{ s}$).

The Pump-Probe Ultrafast Paradigm

1. Pump Pulse ($t = 0$): An ultra-short laser pulse ($\tau_{\text{pulse}} \approx 20 - 50\text{ fs}$) promotes a stable reactant molecule ($ICN$ or $NaI$) from its ground potential energy surface $V_0(R)$ onto an excited repulsive surface $V_1(R)$, launching a coherent nuclear wavepacket:

$$ICN + h\nu_{\text{pump}} \longrightarrow [I\cdots CN]^{\ddagger *}$$

2. Dynamic Evolution: The wavepacket slides down the repulsive potential curve as the bond stretches ($R = R_0 \to R^\ddagger \to \infty$).

3. Probe Pulse ($t = \Delta t$): After an adjustable optical delay time $\Delta t$ (varied by moving a computer-controlled translation stage by $\Delta x = c \Delta t$, where $1\;\mu\text{m} \leftrightarrow 6.67\text{ fs}$):

  • Tuning the probe laser to frequency $\lambda_{\text{free}}$ monitors the appearance of free dissociated products ($CN^\bullet$).
  • Tuning the probe laser to a shifted frequency $\lambda_{\text{TS}}$ excites the transient complex $[I\cdots CN]^{\ddagger *}$ while the atoms are at intermediate separation $R^\ddagger$, emitting characteristic fluorescence.

By recording fluorescence as a function of optical delay $\Delta t$, the birth of a chemical molecule is watched in real time, clocking the transition state lifetime at precisely $200\text{ femtoseconds}$.

Easy Example 10.1: Bimolecular Collision Theory Pre-Exponential Factor for Gas Reactions

For the gas-phase reaction $H + O_2 \longrightarrow OH + O$ at $T = 1000.0\text{ K}$, atomic and molecular diameters are $d_H = 0.100\text{ nm}$ and $d_{O_2} = 0.360\text{ nm}$. Molar masses are $M_H = 1.008\text{ g/mol}$ and $M_{O_2} = 31.999\text{ g/mol}$. Assuming a steric factor of $P = 0.400$: (a) calculate the collision cross-section $\sigma_{AB}$, (b) calculate the reduced mass $\mu$, and (c) calculate the theoretical collision-theory pre-exponential factor $A$ in $\text{M}^{-1}\text{s}^{-1}$.

Step 1: Calculate collision cross-section $\sigma_{AB}$

$$d_{AB} = \frac{d_H + d_{O_2}}{2} = \frac{0.100 + 0.360}{2} = 0.230\text{ nm} = 2.30 \times 10^{-10}\text{ m}$$
$$\sigma_{AB} = \pi d_{AB}^2 = \pi (2.30 \times 10^{-10}\text{ m})^2 = 1.6619 \times 10^{-19}\text{ m}^2$$

Step 2: Calculate reduced mass $\mu$

$$\mu = \frac{m_H m_{O_2}}{m_H + m_{O_2}} = \frac{1.008 \times 31.999}{(1.008 + 31.999) \times (6.02214 \times 10^{26}\text{ kg/mol})} = \frac{32.255}{33.007 \times 6.02214 \times 10^{26}} = 1.6225 \times 10^{-27}\text{ kg}$$

Step 3: Average relative speed $\bar{v}_{\text{rel}}$

$$\bar{v}_{\text{rel}} = \sqrt{\frac{8 k_B T}{\pi \mu}} = \sqrt{\frac{8 \times (1.380649 \times 10^{-23}) \times 1000.0}{\pi \times (1.6225 \times 10^{-27})}} = \sqrt{\frac{1.10452 \times 10^{-19}}{5.0973 \times 10^{-27}}} = \sqrt{2.16687 \times 10^7} = 4654.96\text{ m/s}$$

Step 4: Compute pre-exponential factor $A$

$$A = P \cdot N_A \sigma_{AB} \bar{v}_{\text{rel}}$$
$$A = 0.400 \times (6.02214 \times 10^{23}\text{ mol}^{-1}) \times (1.6619 \times 10^{-19}\text{ m}^2) \times (4654.96\text{ m/s})$$
$$A = 0.400 \times 1.00082 \times 10^5 \times 4654.96 = 1.8635 \times 10^8\text{ m}^3/(\text{mol}\cdot\text{s})$$

Converting to $\text{M}^{-1}\text{s}^{-1}$ ($1\text{ m}^3 = 1000\text{ L}$):

$$A = (1.8635 \times 10^8) \times 1000 = 1.86 \times 10^{11}\text{ M}^{-1}\text{s}^{-1}$$
Medium Example 10.2: Thermodynamic Activation Parameters ($\Delta H^\ddagger, \Delta S^\ddagger, \Delta G^\ddagger$) from Eyring Plot

A second-order liquid-phase nucleophilic substitution reaction has measured rate constants of $k_1 = 3.20 \times 10^{-4}\text{ M}^{-1}\text{s}^{-1}$ at $T_1 = 298.15\text{ K}$ and $k_2 = 2.85 \times 10^{-3}\text{ M}^{-1}\text{s}^{-1}$ at $T_2 = 318.15\text{ K}$. Calculate: (a) the activation enthalpy $\Delta H^{\ddagger\circ}$ in $\text{kJ/mol}$, (b) the activation entropy $\Delta S^{\ddagger\circ}$ in $\text{J}/(\text{mol}\cdot\text{K})$, and (c) the Gibbs activation free energy $\Delta G^{\ddagger\circ}$ at $298.15\text{ K}$.

Step 1: Eyring two-point equation

$$\ln\left( \frac{k_2 / T_2}{k_1 / T_1} \right) = -\frac{\Delta H^{\ddagger\circ}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) = \frac{\Delta H^{\ddagger\circ}}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$$

Evaluate ratios:

$$\frac{k_2 / T_2}{k_1 / T_1} = \frac{2.85 \times 10^{-3} / 318.15}{3.20 \times 10^{-4} / 298.15} = \frac{8.95804 \times 10^{-6}}{1.07329 \times 10^{-6}} = 8.3463$$
$$\ln(8.3463) = 2.1218$$
$$\frac{T_2 - T_1}{T_1 T_2} = \frac{20.0}{298.15 \times 318.15} = \frac{20.0}{9.4856 \times 10^4} = 2.10846 \times 10^{-4}\text{ K}^{-1}$$
$$\Delta H^{\ddagger\circ} = \frac{R \ln(8.3463)}{2.10846 \times 10^{-4}} = \frac{8.314462 \times 2.1218}{2.10846 \times 10^{-4}} = \frac{17.6416}{2.10846 \times 10^{-4}} = 8.3670 \times 10^4\text{ J/mol} = 83.67\text{ kJ/mol}$$

Step 2: Activation entropy $\Delta S^{\ddagger\circ}$ Using data at $T_1 = 298.15\text{ K}$:

$$k = \frac{k_B T}{h} \exp\left( \frac{\Delta S^{\ddagger\circ}}{R} \right) \exp\left( -\frac{\Delta H^{\ddagger\circ}}{R T} \right)$$
$$\frac{k_B T_1}{h} = \frac{(1.380649 \times 10^{-23}) \times (298.15)}{6.62607 \times 10^{-34}} = 6.2124 \times 10^{12}\text{ s}^{-1}$$
$$\frac{k_1}{k_B T_1 / h} = \frac{3.20 \times 10^{-4}}{6.2124 \times 10^{12}} = 5.1510 \times 10^{-17}$$
$$\ln(5.1510 \times 10^{-17}) = -37.505$$
$$\frac{\Delta S^{\ddagger\circ}}{R} - \frac{\Delta H^{\ddagger\circ}}{R T_1} = -37.505$$
$$\frac{\Delta H^{\ddagger\circ}}{R T_1} = \frac{83670}{8.314462 \times 298.15} = \frac{83670}{2478.96} = 33.752$$
$$\frac{\Delta S^{\ddagger\circ}}{R} = -37.505 + 33.752 = -3.753$$
$$\Delta S^{\ddagger\circ} = -3.753 \times 8.314462 = -31.20\text{ J}/(\text{mol}\cdot\text{K})$$

The negative activation entropy signifies an associative bimolecular transition state.

Step 3: Activation free energy $\Delta G^{\ddagger\circ}$ at $298.15\text{ K}$

$$\Delta G^{\ddagger\circ} = \Delta H^{\ddagger\circ} - T_1 \Delta S^{\ddagger\circ} = 83.67\text{ kJ/mol} - (298.15\text{ K}) \times (-0.03120\text{ kJ}/(\text{mol}\cdot\text{K})) = 83.67 + 9.30 = 92.97\text{ kJ/mol}$$
Medium Example 10.3: Primary Kinetic Salt Effect on Persulfate-Iodide Redox Reaction

The reaction between persulfate ion and iodide ion ($S_2O_8^{2-} + 2 I^- \longrightarrow 2 SO_4^{2-} + I_2$) involves an initial rate-determining step between $S_2O_8^{2-}$ ($z_A = -2$) and $I^-$ ($z_B = -1$). At $T = 25.0^\circ\text{C}$ in water, the rate constant at infinite dilution is $k_0 = 1.05 \times 10^{-3}\text{ M}^{-1}\text{s}^{-1}$. Using the Brønsted-Bjerrum equation $\log_{10}(k / k_0) = 1.018 \, z_A z_B \sqrt{I}$: (a) calculate the rate constant $k$ in an electrolyte solution of ionic strength $I = 0.0100\text{ M}$, and (b) at $I = 0.0400\text{ M}$.

Step 1: Calculate charge product

$$z_A = -2, \quad z_B = -1 \implies z_A z_B = (-2) \times (-1) = +2$$

Because both reactants carry negative charges, the charge product is positive ($+2$), predicting a positive salt effect.

Step 2: Rate constant at $I = 0.0100\text{ M}$

$$\sqrt{I} = \sqrt{0.0100} = 0.100\text{ M}^{1/2}$$
$$\log_{10}\left( \frac{k}{k_0} \right) = 1.018 \times (+2) \times 0.100 = 0.2036$$
$$\frac{k}{k_0} = 10^{0.2036} = 1.598$$
$$k = 1.598 \times (1.05 \times 10^{-3}\text{ M}^{-1}\text{s}^{-1}) = 1.68 \times 10^{-3}\text{ M}^{-1}\text{s}^{-1}$$

Step 3: Rate constant at $I = 0.0400\text{ M}$

$$\sqrt{I} = \sqrt{0.0400} = 0.200\text{ M}^{1/2}$$
$$\log_{10}\left( \frac{k}{k_0} \right) = 1.018 \times (+2) \times 0.200 = 0.4072$$
$$\frac{k}{k_0} = 10^{0.4072} = 2.554$$
$$k = 2.554 \times (1.05 \times 10^{-3}\text{ M}^{-1}\text{s}^{-1}) = 2.68 \times 10^{-3}\text{ M}^{-1}\text{s}^{-1}$$

Adding inert salt increases ionic strength from $0$ to $0.04\text{ M}$, accelerating the reaction rate by $155\%$.

Hard Example 10.4: Debye Diffusion-Controlled Limit for Oppositely Charged Ions

The neutralization reaction between hydronium and hydroxide ions ($H_3O^+ + OH^- \longrightarrow 2 H_2O$) is diffusion-controlled in water at $T = 298.15\text{ K}$ ($\varepsilon_r = 78.36$). The encounter distance is $R_{AB} = 0.450\text{ nm}$, and diffusion coefficients are $D(H_3O^+) = 9.31 \times 10^{-9}\text{ m}^2/\text{s}$ and $D(OH^-) = 5.30 \times 10^{-9}\text{ m}^2/\text{s}$. (a) Calculate the neutral Smoluchowski rate constant $k_{\text{diff}}$. (b) Calculate the Debye Coulombic electrostatic factor $f_{\text{Debye}} = \frac{\Phi}{e^\Phi - 1}$ where $\Phi = \frac{z_A z_B e^2}{4 \pi \varepsilon_r \varepsilon_0 R_{AB} k_B T}$. (c) Calculate the true ionic diffusion-controlled rate constant $k_{\text{diff, ions}}$.

Step 1: Calculate neutral Smoluchowski rate constant

$$D_A + D_B = (9.31 + 5.30) \times 10^{-9} = 1.461 \times 10^{-8}\text{ m}^2/\text{s}$$
$$R_{AB} = 4.50 \times 10^{-10}\text{ m}$$
$$k_{\text{diff}} = 4 \pi N_A (D_A + D_B) R_{AB}$$
$$k_{\text{diff}} = 4 \pi \times (6.02214 \times 10^{23}\text{ mol}^{-1}) \times (1.461 \times 10^{-8}\text{ m}^2/\text{s}) \times (4.50 \times 10^{-10}\text{ m})$$
$$k_{\text{diff}} = 4.9669 \times 10^7\text{ m}^3/(\text{mol}\cdot\text{s}) = 4.97 \times 10^{10}\text{ M}^{-1}\text{s}^{-1}$$

Step 2: Calculate Debye factor $\Phi$

$$z_A = +1, \quad z_B = -1 \implies z_A z_B = -1$$
$$e^2 = (1.6021766 \times 10^{-19})^2 = 2.56697 \times 10^{-38}\text{ C}^2$$
$$4 \pi \varepsilon_r \varepsilon_0 = 4 \pi \times 78.36 \times (8.85419 \times 10^{-12}) = 8.7188 \times 10^{-10}\text{ F/m}$$
$$k_B T = (1.380649 \times 10^{-23}) \times (298.15) = 4.1164 \times 10^{-21}\text{ J}$$

Denominator:

$$\text{Denom} = (8.7188 \times 10^{-10}) \times (4.50 \times 10^{-10}) \times (4.1164 \times 10^{-21}) = 1.6151 \times 10^{-39}$$
$$\Phi = \frac{-2.56697 \times 10^{-38}}{1.6151 \times 10^{-39}} = -1.5893$$

Debye enhancement factor:

$$e^\Phi = e^{-1.5893} = 0.20407$$
$$f_{\text{Debye}} = \frac{\Phi}{e^\Phi - 1} = \frac{-1.5893}{0.20407 - 1} = \frac{-1.5893}{-0.79593} = 1.9968 \approx 2.00$$

Step 3: Calculate true ionic rate constant

$$k_{\text{diff, ions}} = k_{\text{diff}} \cdot f_{\text{Debye}} = (4.9669 \times 10^{10}\text{ M}^{-1}\text{s}^{-1}) \times 1.9968 = 9.92 \times 10^{10}\text{ M}^{-1}\text{s}^{-1} \approx 1.0 \times 10^{11}\text{ M}^{-1}\text{s}^{-1}$$

Electrostatic Coulombic attraction doubles the rate of diffusional encounter, matching the experimental neutralization rate ($1.3 \times 10^{11}\text{ M}^{-1}\text{s}^{-1}$).

Medium Example 10.5: Harpoon Mechanism Electron Jump Radius and Cross-Section

In the crossed molecular beam reaction $K + Br_2 \longrightarrow KBr + Br$, electron transfer occurs via the harpoon mechanism when the covalent and ionic potential energy curves cross at radius $R_c$:

$$I_P(K) - E_A(Br_2) = \frac{e^2}{4 \pi \varepsilon_0 R_c}$$

Given the ionization potential of potassium $I_P(K) = 4.34\text{ eV}$ and the electron affinity of bromine $E_A(Br_2) = 2.50\text{ eV}$: (a) calculate the electron jump radius $R_c$ in Ångströms, and (b) calculate the reaction cross-section $\sigma = \pi R_c^2$ and compare it with the hard-sphere gas kinetic cross-section ($\sigma_{\text{hs}} \approx 35\text{ Å}^2$).

Step 1: Calculate energy difference

$$\Delta E = I_P(K) - E_A(Br_2) = 4.34\text{ eV} - 2.50\text{ eV} = 1.84\text{ eV}$$

Convert to Joules:

$$\Delta E = 1.84 \times (1.6021766 \times 10^{-19}\text{ J}) = 2.9480 \times 10^{-19}\text{ J}$$

Step 2: Solve for crossing radius $R_c$

$$R_c = \frac{e^2}{4 \pi \varepsilon_0 \Delta E} = \frac{(8.98755 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2) \times (1.6021766 \times 10^{-19}\text{ C})^2}{2.9480 \times 10^{-19}\text{ J}}$$
$$e^2 / (4 \pi \varepsilon_0) = 2.30708 \times 10^{-28}\text{ J}\cdot\text{m}$$
$$R_c = \frac{2.30708 \times 10^{-28}\text{ J}\cdot\text{m}}{2.9480 \times 10^{-19}\text{ J}} = 7.8259 \times 10^{-10}\text{ m} = 7.83\text{ Å}$$

Step 3: Calculate reaction cross-section $\sigma$

$$\sigma = \pi R_c^2 = \pi (7.8259\text{ Å})^2 = 192.4\text{ Å}^2$$

Ratio to hard-sphere cross-section:

$$\frac{\sigma}{\sigma_{\text{hs}}} = \frac{192.4\text{ Å}^2}{35.0\text{ Å}^2} = 5.50$$

The harpoon electron jump occurs at a distance of almost $8\text{ Å}$, yielding a reaction cross-section over 5.5 times larger than the physical hard-sphere collision size.

Easy Example 10.6: Universal Eyring Attempt Frequency Calculation

The fundamental prefactor in Transition State Theory is the universal attempt frequency $\nu_0 = \frac{k_B T}{h}$. Calculate $\nu_0$ and the corresponding period $\tau_0 = 1/\nu_0$ at: (a) cryogenic temperature $T = 77.0\text{ K}$ (liquid nitrogen), (b) room temperature $T = 298.15\text{ K}$, and (c) flame temperature $T = 2000.0\text{ K}$.

Step 1: Formula for attempt frequency

$$\nu_0 = \frac{k_B T}{h} = \frac{1.380649 \times 10^{-23}\text{ J/K}}{6.62607 \times 10^{-34}\text{ J}\cdot\text{s}} \times T = (2.08366 \times 10^{10}\text{ s}^{-1}\text{K}^{-1}) \times T$$

Step 2: At $T = 77.0\text{ K}$

$$\nu_0(77\text{ K}) = (2.08366 \times 10^{10}) \times 77.0 = 1.6044 \times 10^{12}\text{ s}^{-1} = 1.60\text{ THz}$$
$$\tau_0(77\text{ K}) = \frac{1}{1.6044 \times 10^{12}\text{ s}^{-1}} = 6.233 \times 10^{-13}\text{ s} = 623.3\text{ fs}$$

Step 3: At $T = 298.15\text{ K}$

$$\nu_0(298.15\text{ K}) = (2.08366 \times 10^{10}) \times 298.15 = 6.2124 \times 10^{12}\text{ s}^{-1} = 6.21\text{ THz}$$
$$\tau_0(298.15\text{ K}) = \frac{1}{6.2124 \times 10^{12}\text{ s}^{-1}} = 1.610 \times 10^{-13}\text{ s} = 161.0\text{ fs}$$

Step 4: At $T = 2000.0\text{ K}$

$$\nu_0(2000\text{ K}) = (2.08366 \times 10^{10}) \times 2000.0 = 4.1673 \times 10^{13}\text{ s}^{-1} = 41.67\text{ THz}$$
$$\tau_0(2000\text{ K}) = \frac{1}{4.1673 \times 10^{13}\text{ s}^{-1}} = 2.400 \times 10^{-14}\text{ s} = 24.0\text{ fs}$$

Across all temperatures, the transit time across the transition state barrier is on the femtosecond ($10^{-14}\text{--}10^{-13}\text{ s}$) timescale.

Hard Example 10.7: Polanyi Parameter Classification of Early vs. Late Barrier Dynamics

For the collinear atom-diatom reaction $A + BC \longrightarrow AB + C$, the barrier location parameter is defined as $\mathcal{L} = \frac{R_{AB}^\ddagger - R_{AB, e}}{R_{BC}^\ddagger - R_{BC, e}}$. Consider two reactions:

  1. Reaction 1 ($F + H_2 \longrightarrow HF + H$): $\Delta H_r^\circ = -134\text{ kJ/mol}$, $R_{FH}^\ddagger = 1.54\text{ Å}$ ($R_{FH, e} = 0.92\text{ Å}$), $R_{HH}^\ddagger = 0.76\text{ Å}$ ($R_{HH, e} = 0.74\text{ Å}$).
  2. Reaction 2 ($H + HF \longrightarrow H_2 + F$): $\Delta H_r^\circ = +134\text{ kJ/mol}$, $R_{HH}^\ddagger = 0.76\text{ Å}$ ($R_{HH, e} = 0.74\text{ Å}$), $R_{FH}^\ddagger = 1.54\text{ Å}$ ($R_{FH, e} = 0.92\text{ Å}$).

(a) Classify each reaction as having an early or late barrier according to the Hammond-Polanyi postulate. (b) For each reaction, state whether translational kinetic energy or vibrational reactant excitation is more effective at driving the reaction across the barrier.

Step 1: Analyze Reaction 1 ($F + H_2 \longrightarrow HF + H$)

  • $\Delta H_r^\circ = -134\text{ kJ/mol}$ (strongly exothermic).
  • At the transition state:
$$\Delta R_{HH} = R_{HH}^\ddagger - R_{HH, e} = 0.76 - 0.74 = 0.02\text{ Å} \quad (\text{bond barely stretched by } 2.7\%)$$
$$\Delta R_{FH} = R_{FH}^\ddagger - R_{FH, e} = 1.54 - 0.92 = 0.62\text{ Å} \quad (\text{forming bond is still very distant})$$

Because the reactant $H-H$ bond is virtually unstretched at the saddle point, the transition state resembles the reactants: This is an Early Barrier (Attractive PES), located in the entrance valley.

  • Dynamic consequence: According to Polanyi's rules, relative translational kinetic energy ($E_{\text{trans}}$) is far more effective than reactant vibrational energy in crossing an early barrier. Excess energy in products appears as vibrational excitation of $HF$ (the basis of the $HF$ chemical laser).

Step 2: Analyze Reaction 2 ($H + HF \longrightarrow H_2 + F$)

  • $\Delta H_r^\circ = +134\text{ kJ/mol}$ (strongly endothermic, the microscopic reverse of Reaction 1).
  • At the transition state:
$$\Delta R_{FH} = R_{FH}^\ddagger - R_{FH, e} = 1.54 - 0.92 = 0.62\text{ Å} \quad (\text{reactant bond is stretched by } 67\%!)$$
$$\Delta R_{HH} = R_{HH}^\ddagger - R_{HH, e} = 0.76 - 0.74 = 0.02\text{ Å} \quad (\text{product bond is almost formed})$$

Because the reactant $F-H$ bond must be stretched extensively to reach the saddle point, the transition state resembles the products: This is a Late Barrier (Repulsive PES), located in the exit valley.

  • Dynamic consequence: According to Polanyi's rules, vibrational excitation of the reactant $HF$ bond ($v \ge 1$) is overwhelmingly more effective than translational energy in promoting reaction across a late barrier.
Hard Example 10.8: Velocity Map Imaging Newton Sphere Radius and Center-of-Mass Product Recoil

In a Velocity Map Imaging (VMI) crossed molecular beam experiment, the elementary bimolecular reaction:

$$O(^1D) + CH_4 \longrightarrow OH(v=0, j) + CH_3$$

is studied at collision energy $E_{\text{coll}} = 32.5\text{ kJ/mol}$ ($0.3368\text{ eV}$). The reaction exothermicity is $\Delta_r H^\circ = -181.5\text{ kJ/mol}$ ($1.881\text{ eV}$). Total available energy is $E_{\text{avail}} = E_{\text{coll}} - \Delta_r H^\circ = 214.0\text{ kJ/mol}$ ($2.218\text{ eV}$). A flight tube of length $D = 0.650\text{ m}$ guides photoions to a position-sensitive detector with calibration constant $\mathcal{N} = 42.50\text{ m}/(\text{s}\cdot\text{mm})$ ($1\text{ mm}$ on detector corresponds to $42.50\text{ m/s}$ in laboratory velocity). Molar masses: $M_O = 15.999\text{ g/mol}$, $M_{CH_4} = 16.043\text{ g/mol}$, $M_{OH} = 17.007\text{ g/mol}$, $M_{CH_3} = 15.035\text{ g/mol}$. Total mass $M_{\text{tot}} = 32.042\text{ g/mol}$.

(a) Calculate total available energy $E_{\text{avail}}$ in Joules per molecule. (b) If all available energy is converted into center-of-mass product translation ($E_{\text{trans}}' = E_{\text{avail}}$), calculate the maximum center-of-mass recoil velocity $u_{OH, \max}$ of the $OH$ fragment in $\text{m/s}$. (c) Calculate the maximum outer radius $R_{\max}$ of the $OH$ Newton sphere on the 2D VMI phosphor detector in millimeters. (d) If the experimental VMI image exhibits peak intensity at radius $R_{\text{obs}} = 38.2\text{ mm}$, calculate the actual translational energy disposal $E_{\text{trans}}'$ and the internal vibrational/rotational excitation energy of the $CH_3$ and $OH$ fragments $E_{\text{int}}'$.

Step 1: Calculate total available energy per molecule Total available energy:

$$E_{\text{avail}} = 214.0\text{ kJ/mol} = 2.140 \times 10^5\text{ J/mol}$$

Per molecule:

$$E_{\text{avail}} = \frac{2.140 \times 10^5\text{ J/mol}}{6.02214 \times 10^{23}\text{ mol}^{-1}} = 3.5535 \times 10^{-19}\text{ J}$$

Step 2: Center-of-mass momentum conservation and product recoil velocity The reduced mass of the products ($OH + CH_3$):

$$\mu' = \frac{M_{OH} \cdot M_{CH_3}}{M_{OH} + M_{CH_3}} = \frac{17.007 \times 15.035}{17.007 + 15.035} = \frac{255.700}{32.042} = 7.9801\text{ g/mol} = 1.3251 \times 10^{-26}\text{ kg}$$

In the center-of-mass frame:

$$E_{\text{trans}}' = \frac{1}{2} \mu' v_{\text{rel}}'^2 \implies v_{\text{rel}}' = \sqrt{\frac{2 E_{\text{trans}}'}{\mu'}}$$

From center-of-mass velocity partition:

$$u_{OH} = \frac{M_{CH_3}}{M_{\text{tot}}} v_{\text{rel}}' = \frac{15.035}{32.042} \sqrt{\frac{2 E_{\text{trans}}'}{\mu'}}$$

For maximum translation ($E_{\text{trans}}' = E_{\text{avail}} = 3.5535 \times 10^{-19}\text{ J}$):

$$v_{\text{rel}, \max}' = \sqrt{\frac{2 \times 3.5535 \times 10^{-19}\text{ J}}{1.3251 \times 10^{-26}\text{ kg}}} = \sqrt{5.3634 \times 10^7} = 7323.5\text{ m/s}$$
$$u_{OH, \max} = \frac{15.035}{32.042} \times 7323.5\text{ m/s} = 0.46923 \times 7323.5 = 3436.4\text{ m/s}$$

Step 3: Maximum Newton sphere radius on detector Using instrument velocity scaling $\mathcal{N} = 42.50\text{ m}/(\text{s}\cdot\text{mm})$:

$$R_{\max} = \frac{u_{OH, \max}}{\mathcal{N}} = \frac{3436.4\text{ m/s}}{42.50\text{ m}/(\text{s}\cdot\text{mm})} = 80.86\text{ mm} \approx 80.9\text{ mm}$$

Step 4: Energy disposal from observed radius $R_{\text{obs}} = 38.2\text{ mm}$ Observed center-of-mass velocity:

$$u_{OH, \text{obs}} = R_{\text{obs}} \cdot \mathcal{N} = (38.2\text{ mm}) \times (42.50\text{ m}/(\text{s}\cdot\text{mm})) = 1623.5\text{ m/s}$$

Observed relative velocity:

$$v_{\text{rel}, \text{obs}}' = \frac{u_{OH, \text{obs}}}{0.46923} = \frac{1623.5}{0.46923} = 3459.9\text{ m/s}$$

Actual translational energy disposal:

$$E_{\text{trans}}' = \frac{1}{2} \mu' (v_{\text{rel}, \text{obs}}')^2 = \frac{1}{2} (1.3251 \times 10^{-26}\text{ kg}) \times (3459.9\text{ m/s})^2$$
$$(3459.9)^2 = 1.1971 \times 10^7\text{ m}^2/\text{s}^2$$
$$E_{\text{trans}}' = 0.5 \times (1.3251 \times 10^{-26}) \times (1.1971 \times 10^7) = 7.931 \times 10^{-20}\text{ J}$$

In $\text{kJ/mol}$:

$$E_{\text{trans}}' = (7.931 \times 10^{-20}\text{ J}) \times (6.02214 \times 10^{23}) = 47.76\text{ kJ/mol}$$

Fraction of energy in translation:

$$f_{\text{trans}} = \frac{E_{\text{trans}}'}{E_{\text{avail}}} = \frac{47.76\text{ kJ/mol}}{214.0\text{ kJ/mol}} = 0.223 = 22.3\%$$

Internal excitation of products:

$$E_{\text{int}}' = E_{\text{avail}} - E_{\text{trans}}' = 214.0 - 47.76 = 166.24\text{ kJ/mol} = 77.7\%$$

Over $77\%$ of the available energy flows into internal umbrella vibration and rotation of $CH_3$ and $OH$, revealing a direct insertion dynamics mechanism.

Hard Example 10.9: Quasiclassical Trajectory (QCT) State-to-State Opacity & Differential Cross-Section

A Quasiclassical Trajectory (QCT) numerical simulation is executed on the London-Eyring-Polanyi-Sato (LEPS) potential energy surface for the collinear atom-diatom collision:

$$H + D_2(v=0, j=0) \longrightarrow HD(v', j') + D$$

at fixed collision energy $E_{\text{coll}} = 0.500\text{ eV}$ ($48.24\text{ kJ/mol}$, initial relative velocity $v_{\text{rel}} = 7580\text{ m/s}$). A Monte Carlo ensemble of $N_{\text{tot}} = 10,000$ trajectories was integrated by sampling impact parameters $b$ uniformly distributed from $b = 0$ to maximum cutoff $b_{\max} = 1.40\text{ Å}$ ($1.40 \times 10^{-10}\text{ m}$). The numerical opacity function (reaction probability as a function of impact parameter) fits the linear triangular profile:

$$P(b) = P_0 \left( 1 - \frac{b}{b_{\max}} \right) \quad \text{for } b \le b_{\max}$$

with zero-impact-parameter head-on probability $P_0 = P(0) = 0.720$.

(a) Show by integration that the total integral reactive cross-section $\sigma_r$ is given by:

$$\sigma_r = 2 \pi \int_0^{b_{\max}} P(b) b db = \frac{\pi b_{\max}^2 P_0}{3}$$

(b) Calculate the total reaction cross-section $\sigma_r$ in Ångströms squared ($\text{Å}^2$) and in $\text{m}^2$. (c) Calculate the bimolecular rate constant $k(E_{\text{coll}}) = v_{\text{rel}} \cdot \sigma_r$ at this monoenergetic collision velocity in $\text{m}^3/\text{s}$ and $\text{M}^{-1}\text{s}^{-1}$. (d) Explain dynamically why head-on collisions ($b \to 0$) have maximum reaction probability while glancing collisions ($b \to b_{\max}$) fail to react.

Step 1: Integration of total reaction cross-section $\sigma_r$ The classical definition of integral reaction cross-section is:

$$\sigma_r = 2 \pi \int_0^{b_{\max}} P(b) b db$$

Substitute $P(b) = P_0 \left( 1 - \frac{b}{b_{\max}} \right)$:

$$\sigma_r = 2 \pi P_0 \int_0^{b_{\max}} \left( b - \frac{b^2}{b_{\max}} \right) db$$

Evaluate the definite integral:

$$\int_0^{b_{\max}} b db = \left[ \frac{b^2}{2} \right]_0^{b_{\max}} = \frac{b_{\max}^2}{2}$$
$$\int_0^{b_{\max}} \frac{b^2}{b_{\max}} db = \frac{1}{b_{\max}} \left[ \frac{b^3}{3} \right]_0^{b_{\max}} = \frac{b_{\max}^2}{3}$$

Difference:

$$\frac{b_{\max}^2}{2} - \frac{b_{\max}^2}{3} = \frac{b_{\max}^2}{6}$$

Multiply by $2 \pi P_0$:

$$\sigma_r = 2 \pi P_0 \left( \frac{b_{\max}^2}{6} \right) = \frac{\pi b_{\max}^2 P_0}{3}$$

Step 2: Calculate $\sigma_r$ numerically Given $b_{\max} = 1.40\text{ Å}$ and $P_0 = 0.720$:

$$b_{\max}^2 = (1.40\text{ Å})^2 = 1.960\text{ Å}^2$$
$$\pi b_{\max}^2 = \pi \times 1.960 = 6.1575\text{ Å}^2$$
$$\sigma_r = \frac{6.1575 \times 0.720}{3} = 1.4778\text{ Å}^2 \approx 1.48\text{ Å}^2$$

In $\text{m}^2$:

$$\sigma_r = 1.4778 \times 10^{-20}\text{ m}^2$$

Step 3: Calculate monoenergetic rate constant $k(E_{\text{coll}})$

$$k(E_{\text{coll}}) = v_{\text{rel}} \cdot \sigma_r = (7580\text{ m/s}) \times (1.4778 \times 10^{-20}\text{ m}^2) = 1.1202 \times 10^{-16}\text{ m}^3/\text{s}$$

Convert to chemical molar units ($\text{M}^{-1}\text{s}^{-1} = \text{L}/(\text{mol}\cdot\text{s})$):

$$k = (1.1202 \times 10^{-16}\text{ m}^3/\text{s}) \times (10^3\text{ L/m}^3) \times (6.02214 \times 10^{23}\text{ molecules/mol})$$
$$k = 1.1202 \times 10^{-13} \times 6.02214 \times 10^{23} = 6.746 \times 10^{10}\text{ L}/(\text{mol}\cdot\text{s}) = 6.75 \times 10^{10}\text{ M}^{-1}\text{s}^{-1}$$

Step 4: Dynamic interpretation of the opacity function

  • In head-on collisions ($b \to 0$), orbital angular momentum $L = \mu v_{\text{rel}} b \to 0$. Nearly $100\%$ of the initial relative translational kinetic energy is directed along the line-of-centers directly into the collinear $H\cdots D-D$ reaction coordinate, efficiently conquering the saddle point barrier ($V^\ddagger \approx 0.42\text{ eV}$).
  • In glancing collisions ($b \to b_{\max}$), large orbital angular momentum creates an effective centrifugal potential barrier $V_{\text{cent}}(R) = \frac{L^2}{2 \mu R^2}$. The colliding partners are deflected before reaching the transition state saddle point, causing $P(b)$ to decay linearly to zero.

Solved Honors Problems & Derivations

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