Chemistry / Physical Chemistry Molecular Motion, Transport Phenomena & Reaction Dynamics 100% Free Open Access
Chapter 5 โ€ข Theory & Derivations

Unit 5: Temperature Dependence, Arrhenius Theory & Composite Kinetics

Thermodynamic and statistical mechanics of reaction rate temperature dependence: empirical determination of reaction orders, van 't Hoff's transition to Arrhenius theory, differential and integrated Arrhenius forms, physical significance of the pre-exponential frequency factor and activation energy, Tolman's statistical mechanical interpretation, composite reaction activation barriers, and non-Arrhenius curvature.

ยง5.1 Empirical Determination of Reaction Orders: Classical Kinetic Methods

Determining the empirical order of a reaction is the vital first step toward elucidating its molecular mechanism. Four classical experimental protocols are employed.

1. The Method of Initial Rates

By measuring the initial rate $r_0 = (d[P]/dt)_{t \to 0}$ before back-reactions, product inhibition, or substantial reactant depletion occur:

$$r_0 = k [A]_0^\alpha [B]_0^\beta$$

Varying $[A]_0$ while holding $[B]_0$ constant:

$$\frac{r_{0, 1}}{r_{0, 2}} = \left( \frac{[A]_{0, 1}}{[A]_{0, 2}} \right)^\alpha \implies \alpha = \frac{\ln(r_{0, 1} / r_{0, 2})}{\ln([A]_{0, 1} / [A]_{0, 2})}$$

2. The Isolation Method (Flooding)

All reactants except one are supplied in massive stoichiometric excess ($[B]_0, [C]_0 \gg [A]_0$). Their concentrations remain virtually constant throughout the reaction:

$$r = k [A]^\alpha [B]_0^\beta [C]_0^\gamma = k_{\text{eff}} [A]^\alpha$$

where $k_{\text{eff}} = k [B]_0^\beta [C]_0^\gamma$. The partial order $\alpha$ is then determined directly using standard integrated rate plots for species $A$. Repeating by flooding different components yields all partial orders.

3. The Fractional-Life Method

Measuring the time $t_{f}$ required for concentration to decrease by a fixed fraction $f$ (e.g., $f = 1/2$ for half-life, $f = 3/4$):

$$t_{f} \propto \frac{1}{[A]_0^{n - 1}} \implies \ln t_{f} = \text{constant} - (n - 1) \ln[A]_0$$

Plotting $\ln t_{1/2}$ versus $\ln[A]_0$ yields a straight line with slope $1 - n$.

4. Differential (van 't Hoff) Method

Directly taking natural logarithms of the differential rate equation:

$$\ln r = \ln k + \alpha \ln[A] + \beta \ln[B]$$

A plot of $\ln r$ versus $\ln[A]$ gives slope $\alpha$.

Thermodynamic Activation Parameters Matrix for Prototypical Reactions

Applying Eyring Transition State Theory to rate constants across varied chemical mechanisms demonstrates how activation enthalpy ($\Delta H^\ddagger$), activation entropy ($\Delta S^\ddagger$), and activation free energy ($\Delta G^\ddagger$) govern reaction velocity at $298.15\text{ K}$:

| Reaction Mechanism | Prototypical Example | $k(298\text{ K})$ | $E_a$ ($\text{kJ/mol}$) | $\Delta H^\ddagger$ ($\text{kJ/mol}$) | $\Delta S^\ddagger$ ($\text{J}/(\text{mol}\cdot\text{K})$) | $\Delta G^\ddagger$ ($\text{kJ/mol}$) | |---|---|---|---|---|---|---| | Radical Recombination | $2 CH_3^\bullet \longrightarrow C_2H_6$ | $2.5 \times 10^{10}\text{ M}^{-1}\text{s}^{-1}$ | $0.0$ | $-2.5$ | $-52.0$ | $+13.0$ | | Bimolecular Radical Transfer| $OH^\bullet + CH_4 \longrightarrow H_2O + CH_3^\bullet$ | $6.4 \times 10^6\text{ M}^{-1}\text{s}^{-1}$ | $+18.5$ | $+16.0$ | $-65.0$ | $+35.4$ | | Unimolecular Isomerization | $\text{Cyclopropane} \longrightarrow \text{Propene}$ | $1.2 \times 10^{-15}\text{ s}^{-1}$ | $+272.0$ | $+269.5$ | $+40.0$ | $+257.6$ | | Bimolecular Gas Diels-Alder | $1,3\text{-Butadiene} + \text{Ethene} \longrightarrow \text{Cyclohexene}$ | $3.2 \times 10^{-18}\text{ M}^{-1}\text{s}^{-1}$| $+115.0$ | $+112.5$ | $-142.0$ | $+154.8$ | | Alkaline Ester Hydrolysis | $EtOAc + OH^- \longrightarrow AcO^- + EtOH$ | $0.11\text{ M}^{-1}\text{s}^{-1}$ | $+47.0$ | $+44.5$ | $-110.0$ | $+77.3$ | | Acid Sucrose Inversion | $\text{Sucrose} + H_3O^+ \longrightarrow \text{Glc} + \text{Fru}$ | $1.8 \times 10^{-4}\text{ M}^{-1}\text{s}^{-1}$ | $+108.0$ | $+105.5$ | $+32.0$ | $+96.0$ | | Enzyme Turnover ($k_{\text{cat}}$)| Carbonic Anhydrase ($CO_2 + H_2O$)| $1.0 \times 10^6\text{ s}^{-1}$ | $+38.0$ | $+35.5$ | $-10.0$ | $+38.5$ |

ยง5.2 Temperature Dependence of Reaction Rates: Historical Foundation

Chemical reaction rates are extraordinarily sensitive to temperature. As an approximate historical rule of thumb (the van 't Hoff rule), the rate of a typical homogeneous chemical reaction approximately doubles or triples for every $10^\circ\text{C}$ temperature rise ($Q_{10} \approx 2\text{--}3$).

Contrast with Kinetic Molecular Theory

From the kinetic theory of gases, the average molecular speed scales as:

$$\bar{v} \propto \sqrt{T}$$

and the binary collision frequency scales as:

$$Z_{AA} \propto \sqrt{T}$$

For a $10^\circ\text{C}$ rise from $300\text{ K}$ to $310\text{ K}$:

$$\frac{Z(310\text{ K})}{Z(300\text{ K})} = \sqrt{\frac{310}{300}} = \sqrt{1.0333} \approx 1.0165$$

The collision frequency increases by a mere $1.65\%$.

This colossal discrepancy between a $1.65\%$ increase in molecular collisions and a $200\%\text{--}300\%$ increase in chemical reaction rate proved conclusively that only an extraordinarily small, highly energetic fraction of molecular collisions possess sufficient energy to overcome a critical barrier and undergo chemical rearrangement.

ยง5.3 The Arrhenius Rate Equation: Differential & Integrated Formulations

In 1889, the Swedish chemist Svante Arrhenius synthesized J.H. van 't Hoff's thermodynamic equilibrium equation with Boltzmann's energy distribution to propose the foundational law of chemical kinetics.

Analogy to the van 't Hoff Isochore

van 't Hoff's thermodynamic relation for the temperature dependence of the equilibrium constant $K = k_1 / k_{-1}$ is:

$$\frac{d \ln K}{dT} = \frac{d \ln k_1}{dT} - \frac{d \ln k_{-1}}{dT} = \frac{\Delta U^\circ}{R T^2}$$

Arrhenius proposed that each individual rate constant satisfies a similar differential equation:

$$\frac{d \ln k}{dT} = \frac{E_a}{R T^2}$$

where $E_a$ is the empirical activation energy (SI unit: $\text{J/mol}$ or $\text{kJ/mol}$).

Integrated Forms of the Arrhenius Equation

Assuming $E_a$ is independent of temperature over moderate intervals, indefinite integration yields:

$$\ln k = -\frac{E_a}{R T} + \ln A \iff k(T) = A \exp\left( -\frac{E_a}{R T} \right)$$

where:

  • $A$ is the pre-exponential factor (or frequency factor), possessing the same units as the rate constant $k$.
  • $\exp(-E_a / R T)$ is the Boltzmann fraction of collisions with energy exceeding $E_a$.

Two-Temperature Comparative Form

Integrating between temperatures $T_1$ and $T_2$:

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

Arrhenius Plot Linearization

Plotting $\ln k$ versus $1/T$ (in $\text{K}^{-1}$):

  • Slope $= -\frac{E_a}{R}$
  • $y$-intercept $= \ln A$

ยง5.4 Physical Significance of the Pre-Exponential Factor & Activation Barrier

The parameters $A$ and $E_a$ provide complementary microscopic insights into the reaction coordinate.

The Activation Energy ($E_a$)

The activation energy $E_a$ represents the minimum threshold energy colliding reactant molecules must possess along the line-of-centers (reaction coordinate) to induce bond distortion, overcome Coulombic electron repulsion, and reach the top of the potential energy barrier (the activated transition state complex).

Key energetic features:

  • Reactions with low activation energies ($E_a < 20\text{ kJ/mol}$, such as radical-radical recombinations) proceed extremely rapidly and exhibit weak temperature sensitivity.
  • Reactions with high activation energies ($E_a > 150\text{ kJ/mol}$, such as thermal cracking of alkanes) are slow at ambient temperature and exhibit extreme exponential acceleration with temperature.
  • $E_a$ is always positive for elementary thermal reactions. (Certain composite reactions exhibit apparent negative activation energies, as analyzed in Section 5.6).

The Pre-Exponential Factor ($A$)

The pre-exponential factor $A$ quantifies the frequency of collisions with favorable spatial geometry:

$$A = P \cdot Z_0$$

where:

  • $Z_0$ is the total binary collision frequency at unit concentration.
  • $P$ is the steric factor (orientation probability), which accounts for the requirement that molecules collide with specific mutual orientations.

For simple gas-phase atom-atom or spherical collisions, $P \sim 1$ and $A \sim 10^{11}\text{ M}^{-1}\text{s}^{-1}$. For complex polyatomic molecules requiring precise alignment of functional groups, $P$ can be as small as $10^{-4}$ to $10^{-8}$, dramatically reducing the effective rate constant.

ยง5.5 Tolman's Statistical Thermodynamic Interpretation of Activation Energy

In 1920, Richard Chace Tolman provided the rigorous statistical mechanical definition of activation energy, establishing that $E_a$ is the difference between the average energy of reactive collisions and the average energy of all collisions.

Tolman's Theorem

Let $\sigma_R(E)$ be the microscopic reaction cross-section for collisions with center-of-mass collision energy $E$. The macroscopic bimolecular rate constant is given by the statistical ensemble thermal average:

$$k(T) = \left( \frac{8}{\pi \mu (k_B T)^3} \right)^{1/2} \int_0^\infty E \, \sigma_R(E) \exp\left( -\frac{E}{k_B T} \right) dE$$

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

Taking the logarithmic temperature derivative:

$$\frac{d \ln k}{dT} = \frac{1}{k} \frac{dk}{dT}$$

Evaluating the derivative inside the integral yields Tolman's Principle:

$$E_a \equiv R T^2 \frac{d \ln k}{dT} = \langle E_R \rangle - \langle E_{\text{all}} \rangle$$

where:

  • $\langle E_R \rangle$ is the average energy of all collisions that successfully result in chemical reaction.
  • $\langle E_{\text{all}} \rangle$ is the average energy of all collisions in the thermal ensemble ($= \frac{3}{2} R T$ for translational motion).

Tolman's theorem proves that the empirical Arrhenius activation energy $E_a$ is precisely the energy excess that reactive molecular encounters carry above the thermal average of the bulk reactant population.

University Honors Research Monograph: Marcus Electron Transfer Theory & The Inverted Region

Rudolph A. Marcus (Nobel Prize in Chemistry 1992) revolutionized chemical kinetics by developing the microscopic theory of outer-sphere electron transfer ($D + A \longrightarrow D^+ + A^-$):

  • Parabolic Potential Energy Surfaces: The reactant ($D\cdots A$) and product ($D^+\cdots A^-$) states are modeled as multi-dimensional harmonic oscillator potential wells along a collective solvent polarization and nuclear reorganization coordinate $q$.
  • Free Energy Barrier Formulation: The Gibbs activation free energy is related to the standard driving force $\Delta G^\circ$ and total reorganization energy $\lambda$:
$$\Delta G^\ddagger = \frac{(\lambda + \Delta G^\circ)^2}{4 \lambda}$$

where $\lambda = \lambda_{\text{in}} + \lambda_{\text{out}}$ accounts for internal bond length distortions ($\lambda_{\text{in}}$) and dielectric solvent dipole reorientations ($\lambda_{\text{out}}$).

  • The Three Kinetic Regimes:

1. Normal Regime ($-\Delta G^\circ < \lambda$): Increasing thermodynamic driving force ($-\Delta G^\circ$) lowers $\Delta G^\ddagger$, accelerating electron transfer ($\ln k \propto -\Delta G^\circ$).

2. Barrierless Regime ($-\Delta G^\circ = \lambda$): $\Delta G^\ddagger = 0$, achieving the maximum possible activationless transfer rate.

3. The Marcus Inverted Region ($-\Delta G^\circ > \lambda$): Counter-intuitively, making the reaction even more thermodynamically favorable causes $\Delta G^\ddagger$ to increase, slowing down the rate!

  • Experimental Proof: Gerhard Closs and John Miller (1984) confirmed the inverted region using rigid steroid spacer molecules, a discovery fundamental to photosynthesis and solar cell design.

ยง5.6 Composite Reaction Kinetics & Apparent Activation Energies

When a chemical transformation proceeds via a multi-step composite mechanism, the overall observed rate constant $k_{\text{obs}}$ is an algebraic combination of elementary rate constants, and the apparent activation energy $E_{a, \text{app}}$ is a composite sum.

Pre-Equilibrium Followed by Slow Elementary Step

Consider the mechanism:

$$A + B \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} [AB]^* \quad (\text{fast pre-equilibrium})$$
$$[AB]^* \xrightarrow{k_2} P \quad (\text{slow rate-determining step})$$

The observed rate law is:

$$r = k_2 [[AB]^*] = k_2 K_1 [A][B] = \left( \frac{k_1 k_2}{k_{-1}} \right) [A][B]$$

Thus, the overall observed rate constant is:

$$k_{\text{obs}} = \frac{k_1 k_2}{k_{-1}}$$

Taking natural logarithms:

$$\ln k_{\text{obs}} = \ln k_1 + \ln k_2 - \ln k_{-1}$$

Differentiating with respect to temperature:

$$\frac{d \ln k_{\text{obs}}}{dT} = \frac{d \ln k_1}{dT} + \frac{d \ln k_2}{dT} - \frac{d \ln k_{-1}}{dT}$$

Multiplying by $R T^2$:

$$E_{a, \text{app}} = E_{a, 1} + E_{a, 2} - E_{a, -1} = E_{a, 2} + \Delta H_1^\circ$$

where $\Delta H_1^\circ = E_{a, 1} - E_{a, -1}$ is the standard enthalpy of formation of the intermediate pre-equilibrium complex.

Apparent Negative Activation Energies

If the initial pre-equilibrium step is strongly exothermic ($\Delta H_1^\circ < 0$) and its magnitude exceeds the activation energy of the decomposition step:

$$|\Delta H_1^\circ| > E_{a, 2} \implies E_{a, \text{app}} < 0$$

When $E_{a, \text{app}} < 0$, the reaction rate slows down as temperature increases! Famous example: gas-phase termolecular oxidation of nitric oxide ($2 NO + O_2 \longrightarrow 2 NO_2$), where pre-equilibrium dimerization $2 NO \rightleftharpoons N_2O_2$ is exothermic ($\Delta H^\circ \approx -10.5\text{ kJ/mol}$), yielding $E_{a, \text{app}} \approx -4\text{ kJ/mol}$.

ยง5.7 Non-Arrhenius Curvature: Super-Arrhenius Behavior & Quantum Tunneling

While the classical Arrhenius equation is exceptionally robust over moderate temperature spans, significant non-linear curvature in $\ln k$ versus $1/T$ plots emerges across broad temperature regimes.

1. Temperature-Dependent Pre-Exponential Factor (Modified Arrhenius Equation)

Collision theory and Transition State Theory demonstrate that the pre-exponential factor is not strictly constant, but varies mildly with temperature:

$$k(T) = A' T^m \exp\left( -\frac{E_0}{R T} \right)$$

where:

  • $m = 1/2$ from hard-sphere collision theory ($A \propto \bar{v} \propto T^{1/2}$).
  • $m = 1$ from classical Eyring transition state theory ($k_B T / h$).
  • $m$ can be negative or higher integer depending on partition functions.

The true activation energy becomes temperature-dependent:

$$E_a(T) = R T^2 \frac{d \ln k}{dT} = R T^2 \left( \frac{m}{T} + \frac{E_0}{R T^2} \right) = E_0 + m R T$$

2. Low-Temperature Quantum Mechanical Tunneling

For reactions involving the transfer of light particles (protons $H^+$, hydrogen atoms $H^\bullet$, or electrons $e^-$), the de Broglie wavelength $\lambda_{\text{dB}} = h / \sqrt{2 m E}$ is comparable to the activation barrier width ($d \approx 0.5\text{--}1.0\text{ ร…}$).

At low temperatures ($T < 200\text{ K}$), particles tunnel through the barrier rather than climbing over it. As $T \to 0\text{ K}$, the reaction rate flattens to a temperature-independent non-zero constant ($k \to k_{\text{tunnel}}$), causing the Arrhenius plot to curve upward with $E_a \to 0$.

3. Super-Arrhenius Behavior in Glassy and Viscous Media

In supercooled liquids and polymer glass transitions, relaxation rates decelerate far faster than Arrhenius predictions, described by the Vogel-Fulcher-Tammann (VFT) equation:

$$k(T) = A \exp\left( -\frac{B}{T - T_0} \right)$$

reflecting cooperative molecular rearrangements as free volume collapses near the ideal glass transition temperature $T_0$.

ยง5.8 High-Pressure Arrhenius Activation Volumes & Isokinetic Enthalpy-Entropy Compensation

Investigating reaction dynamics under extreme hydrostatic pressures and across broad temperature ranges allows dissection of Transition State volumes, solvation rearrangements, and the physical reality of isokinetic relationships.

1. High-Pressure Chemical Kinetics & Activation Volume ($\Delta V^\ddagger$)

Hydrostatic pressure $P$ alters chemical equilibrium and reaction rate constants through mechanical work terms $P \Delta V$. From transition state thermodynamics:

$$\left(\frac{\partial \ln k}{\partial P}\right)_T = -\frac{\Delta V^\ddagger}{R T}$$

where $\Delta V^\ddagger = V^\ddagger - V_R$ is the volume of activation, representing the difference in partial molar volume between the Transition State and the reactant state.

Structural and Solvational Components of $\Delta V^\ddagger$

The activation volume decomposes into two distinct physical contributions:

$$\Delta V^\ddagger = \Delta V_{\text{intr}}^\ddagger + \Delta V_{\text{solv}}^\ddagger$$

1. Intrinsic Volume Change ($\Delta V_{\text{intr}}^\ddagger$):

  • For bond cleavage (dissociative mechanisms, unimolecular homolysis, $S_N1$): Bonds stretch to form the transition state, resulting in a volume expansion: $\Delta V_{\text{intr}}^\ddagger > 0$ ($+5 \text{ to } +20\text{ cm}^3/\text{mol}$).
  • For bond formation (associative mechanisms, cycloadditions, $S_N2$): Molecules come together to form new covalent bonds, resulting in volume contraction: $\Delta V_{\text{intr}}^\ddagger < 0$ ($-10 \text{ to } -35\text{ cm}^3/\text{mol}$). In Diels-Alder reactions, $\Delta V^\ddagger \approx -30 \text{ to } -45\text{ cm}^3/\text{mol}$, causing dramatic rate accelerations ($> 1000\times$) at $P = 10\text{ kbar}$ ($1\text{ GPa}$).

2. Solvational Volume Change ($\Delta V_{\text{solv}}^\ddagger$, Electrostriction):

  • When neutral reactants develop charge in the transition state (e.g., Menshutkin reaction: $R_3N + R'X \longrightarrow [R_3N^{\delta+}\cdots R'\cdots X^{\delta-}]^\ddagger$): Intense electric fields orient and pack dipolar solvent molecules tightly around developing charges. This electrostriction effect causes huge contraction: $\Delta V_{\text{solv}}^\ddagger \approx -20 \text{ to } -50\text{ cm}^3/\text{mol}$, accelerating the reaction under pressure.

2. Enthalpy-Entropy Compensation & The Isokinetic Temperature

When kinetic parameters are measured across a series of homologous reactions (varying catalyst ligands, substituents, or solvent mixtures), a linear correlation between activation enthalpy $\Delta H^\ddagger$ and activation entropy $\Delta S^\ddagger$ is frequently observed:

$$\Delta H^\ddagger = \beta \cdot \Delta S^\ddagger + \Delta G_0^\ddagger$$

where $\beta$ is known as the isokinetic temperature (or isoequilibrium temperature).

Physical Reality vs. Statistical Artifact (Exner Analysis)

At $T = \beta$:

$$\Delta G^\ddagger = \Delta H^\ddagger - \beta \Delta S^\ddagger = \Delta G_0^\ddagger = \text{constant}$$

All reactions in the series are predicted to proceed at precisely identical rates at temperature $\beta$. However, because $\Delta H^\ddagger$ and $\Delta S^\ddagger$ are mathematically derived from the slope and intercept of the same Arrhenius/Eyring plot ($\ln(k/T)$ vs $1/T$), experimental errors in the slope propagate directly into the intercept, generating an apparent linear compensation artifact:

$$\delta(\Delta S^\ddagger) = \frac{\delta(\Delta H^\ddagger)}{T_{\text{mean}}}$$

O. Exner demonstrated that genuine physical isokinetic behavior must be verified by plotting rate constants measured at two distinct temperatures directly against each other:

$$\ln k(T_2) = a + b \ln k(T_1)$$

If the slope $b \ne 1$ and $b \ne T_1/T_2$ with statistical significance, a true common transition-state mechanism with identical solvation reorganization coordinates exists, and the true isokinetic temperature is calculated from:

$$\beta = T_1 T_2 \frac{b - 1}{b T_1 - T_2}$$
Easy Example 5.1: Arrhenius Activation Energy and Frequency Factor from Two-Point Rate Data

The first-order gas-phase decomposition of acetaldehyde ($CH_3CHO \longrightarrow CH_4 + CO$) has measured rate constants of $k_1 = 1.05 \times 10^{-5}\text{ s}^{-1}$ at $T_1 = 700.0\text{ K}$ and $k_2 = 2.14 \times 10^{-3}\text{ s}^{-1}$ at $T_2 = 800.0\text{ K}$. Calculate: (a) the activation energy $E_a$ in $\text{kJ/mol}$, (b) the pre-exponential factor $A$, and (c) the predicted rate constant at $T_3 = 750.0\text{ K}$.

Step 1: Calculate activation energy $E_a$ Using the two-temperature Arrhenius equation:

$$\ln\left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$$
$$\frac{k_2}{k_1} = \frac{2.14 \times 10^{-3}}{1.05 \times 10^{-5}} = 203.81$$
$$\ln(203.81) = 5.31721$$
$$\frac{T_2 - T_1}{T_1 T_2} = \frac{800.0 - 700.0}{700.0 \times 800.0} = \frac{100.0}{5.600 \times 10^5} = 1.78571 \times 10^{-4}\text{ K}^{-1}$$
$$E_a = \frac{R \ln(k_2 / k_1)}{(T_2 - T_1)/(T_1 T_2)} = \frac{8.314462 \times 5.31721}{1.78571 \times 10^{-4}} = \frac{44.2097}{1.78571 \times 10^{-4}} = 2.47575 \times 10^5\text{ J/mol} = 247.6\text{ kJ/mol}$$

Step 2: Calculate pre-exponential factor $A$ Using data at $T_1 = 700.0\text{ K}$:

$$k_1 = A \exp\left( -\frac{E_a}{R T_1} \right) \implies A = k_1 \exp\left( \frac{E_a}{R T_1} \right)$$
$$\frac{E_a}{R T_1} = \frac{247575}{8.314462 \times 700.0} = \frac{247575}{5820.12} = 42.5378$$
$$\exp(42.5378) = 2.9790 \times 10^{18}$$
$$A = (1.05 \times 10^{-5}\text{ s}^{-1}) \times (2.9790 \times 10^{18}) = 3.128 \times 10^{13}\text{ s}^{-1}$$

Step 3: Predicted rate constant at $T_3 = 750.0\text{ K}$

$$\frac{E_a}{R T_3} = \frac{247575}{8.314462 \times 750.0} = \frac{247575}{6235.85} = 39.7019$$
$$k(750\text{ K}) = A \exp(-39.7019) = (3.128 \times 10^{13}) \times (5.723 \times 10^{-18}) = 1.79 \times 10^{-4}\text{ s}^{-1}$$
Medium Example 5.2: Method of Initial Rates for a Multicomponent Chemical System

The reaction $2 A + B + 2 C \longrightarrow D + 2 E$ was investigated at $25.0^\circ\text{C}$ by the method of initial rates, yielding the following dataset:

  • Run 1: $[A]_0 = 0.100\text{ M}$, $[B]_0 = 0.100\text{ M}$, $[C]_0 = 0.100\text{ M}$, $r_0 = 3.20 \times 10^{-3}\text{ M/s}$
  • Run 2: $[A]_0 = 0.200\text{ M}$, $[B]_0 = 0.100\text{ M}$, $[C]_0 = 0.100\text{ M}$, $r_0 = 6.40 \times 10^{-3}\text{ M/s}$
  • Run 3: $[A]_0 = 0.100\text{ M}$, $[B]_0 = 0.200\text{ M}$, $[C]_0 = 0.100\text{ M}$, $r_0 = 1.28 \times 10^{-2}\text{ M/s}$
  • Run 4: $[A]_0 = 0.100\text{ M}$, $[B]_0 = 0.100\text{ M}$, $[C]_0 = 0.300\text{ M}$, $r_0 = 3.20 \times 10^{-3}\text{ M/s}$

Determine: (a) the partial reaction orders with respect to $A, B$, and $C$, (b) the overall reaction order, and (c) the rate constant $k$ with appropriate units.

Step 1: Determine partial order with respect to $A$ (Runs 1 and 2) Between Run 1 and Run 2, $[B]_0$ and $[C]_0$ are constant while $[A]_0$ doubles:

$$\frac{r_{0, 2}}{r_{0, 1}} = \frac{6.40 \times 10^{-3}}{3.20 \times 10^{-3}} = 2.000 = \left( \frac{0.200}{0.100} \right)^\alpha = 2^\alpha \implies \alpha = 1$$

First-order in $A$.

Step 2: Determine partial order with respect to $B$ (Runs 1 and 3) Between Run 1 and Run 3, $[A]_0$ and $[C]_0$ are constant while $[B]_0$ doubles:

$$\frac{r_{0, 3}}{r_{0, 1}} = \frac{1.28 \times 10^{-2}}{3.20 \times 10^{-3}} = 4.000 = \left( \frac{0.200}{0.100} \right)^\beta = 2^\beta \implies \beta = 2$$

Second-order in $B$.

Step 3: Determine partial order with respect to $C$ (Runs 1 and 4) Between Run 1 and Run 4, $[A]_0$ and $[B]_0$ are constant while $[C]_0$ triples:

$$\frac{r_{0, 4}}{r_{0, 1}} = \frac{3.20 \times 10^{-3}}{3.20 \times 10^{-3}} = 1.000 = \left( \frac{0.300}{0.100} \right)^\gamma = 3^\gamma \implies \gamma = 0$$

Zero-order in $C$.

Step 4: Overall rate law and overall order

$$r = k [A]^1 [B]^2 [C]^0 = k [A] [B]^2$$

Overall order: $n = 1 + 2 + 0 = 3$ (Third-order overall).

Step 5: Rate constant $k$ Using Run 1:

$$k = \frac{r_0}{[A]_0 [B]_0^2} = \frac{3.20 \times 10^{-3}\text{ M/s}}{(0.100\text{ M}) \times (0.100\text{ M})^2} = \frac{3.20 \times 10^{-3}}{1.00 \times 10^{-3}\text{ M}^3} = 3.20\text{ M}^{-2}\text{s}^{-1}$$
Medium Example 5.3: Rule of Thumb $Q_{10}$ Temperature Sensitivity Analysis

A chemical reaction has an activation energy of $E_a = 52.0\text{ kJ/mol}$. (a) Calculate the exact temperature coefficient $Q_{10} = k(T + 10\text{ K}) / k(T)$ around ambient temperature ($T = 298.15\text{ K}$). (b) What activation energy is required for a reaction to exactly double in rate between $298.15\text{ K}$ and $308.15\text{ K}$?

Step 1: Calculate $Q_{10}$ for $E_a = 52.0\text{ kJ/mol}$

$$T_1 = 298.15\text{ K}, \quad T_2 = 308.15\text{ K}$$
$$\Delta T = 10.0\text{ K}$$
$$T_1 T_2 = 298.15 \times 308.15 = 9.18749 \times 10^4\text{ K}^2$$
$$\frac{T_2 - T_1}{T_1 T_2} = \frac{10.0}{9.18749 \times 10^4} = 1.08844 \times 10^{-4}\text{ K}^{-1}$$

Using the Arrhenius equation:

$$\ln(Q_{10}) = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) = \frac{52000\text{ J/mol}}{8.314462\text{ J}/(\text{mol}\cdot\text{K})} \times (1.08844 \times 10^{-4}\text{ K}^{-1})$$
$$\ln(Q_{10}) = 6254.16 \times 1.08844 \times 10^{-4} = 0.68073$$
$$Q_{10} = \exp(0.68073) = 1.975$$

The rate increases by a factor of $1.98$, in excellent agreement with van 't Hoff's rule of thumb ($Q_{10} \approx 2$).

Step 2: Activation energy for exact rate doubling ($Q_{10} = 2.000$)

$$\ln(2.000) = 0.693147$$
$$\frac{E_a}{R} \times (1.08844 \times 10^{-4}) = 0.693147$$
$$\frac{E_a}{R} = \frac{0.693147}{1.08844 \times 10^{-4}} = 6368.26\text{ K}$$
$$E_a = 6368.26 \times 8.314462 = 5.2949 \times 10^4\text{ J/mol} = 52.95\text{ kJ/mol}$$

An activation energy of approximately $53\text{ kJ/mol}$ corresponds exactly to a rate doubling per $10^\circ\text{C}$ near room temperature.

Hard Example 5.4: Composite Activation Energy and Apparent Negative Temperature Dependence

The gas-phase termolecular reaction $2 NO + O_2 \longrightarrow 2 NO_2$ proceeds via a pre-equilibrium mechanism:

  1. $2 NO \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} N_2O_2$ (fast pre-equilibrium, forward activation energy $E_{a, 1} = 2.1\text{ kJ/mol}$, reverse activation energy $E_{a, -1} = 15.6\text{ kJ/mol}$)
  2. $N_2O_2 + O_2 \xrightarrow{k_2} 2 NO_2$ (slow step, activation energy $E_{a, 2} = 8.5\text{ kJ/mol}$)

(a) Derive the overall rate law and express the apparent rate constant $k_{\text{obs}}$ in terms of elementary rate constants. (b) Calculate the standard enthalpy of dimerization $\Delta H_1^\circ$. (c) Calculate the overall apparent activation energy $E_{a, \text{app}}$ and explain the physical meaning of its sign.

Step 1: Mechanism derivation and rate law For the slow step:

$$r = k_2 [N_2O_2] [O_2]$$

From the fast pre-equilibrium:

$$K_1 = \frac{k_1}{k_{-1}} = \frac{[N_2O_2]}{[NO]^2} \implies [N_2O_2] = \frac{k_1}{k_{-1}} [NO]^2$$

Substituting:

$$r = \frac{k_1 k_2}{k_{-1}} [NO]^2 [O_2] = k_{\text{obs}} [NO]^2 [O_2]$$

where $k_{\text{obs}} = \frac{k_1 k_2}{k_{-1}}$.

Step 2: Enthalpy of dimerization $\Delta H_1^\circ$

$$\Delta H_1^\circ = E_{a, 1} - E_{a, -1} = 2.1 - 15.6 = -13.5\text{ kJ/mol}$$

Dimerization of nitric oxide to $(NO)_2$ is exothermic.

Step 3: Apparent activation energy $E_{a, \text{app}}$

$$E_{a, \text{app}} = E_{a, 1} + E_{a, 2} - E_{a, -1} = E_{a, 2} + \Delta H_1^\circ$$
$$E_{a, \text{app}} = 8.5 + (-13.5) = -5.0\text{ kJ/mol}$$

Physical Interpretation: The apparent activation energy is negative ($-5.0\text{ kJ/mol}$). As temperature increases, Le Chatelier's principle shifts the exothermic dimerization pre-equilibrium backward, sharply decreasing the equilibrium concentration of $[N_2O_2]$. This decrease in intermediate concentration outweighs the modest kinetic acceleration of the $k_2$ step, causing the overall reaction rate to decrease with rising temperature.

Medium Example 5.5: Modified Arrhenius Form with Temperature-Dependent Pre-Exponential Factor

A gas-phase bimolecular radical reaction is fitted to the modified Arrhenius equation $k(T) = A' T^{1/2} \exp(-E_0 / R T)$ with $A' = 1.40 \times 10^7\text{ M}^{-1}\text{s}^{-1}\text{K}^{-1/2}$ and $E_0 = 28.5\text{ kJ/mol}$. (a) Derive the expression for the experimental Arrhenius activation energy $E_a(T)$. (b) Calculate $E_a$ at $T = 300.0\text{ K}$ and at $T = 1500.0\text{ K}$.

Step 1: Derive $E_a(T)$ from the definition

$$k(T) = A' T^{1/2} \exp\left( -\frac{E_0}{R T} \right)$$

Taking natural logarithms:

$$\ln k = \ln A' + \frac{1}{2} \ln T - \frac{E_0}{R T}$$

Differentiating with respect to $T$:

$$\frac{d \ln k}{dT} = \frac{1}{2 T} + \frac{E_0}{R T^2}$$

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

$$E_a(T) = R T^2 \left( \frac{1}{2 T} + \frac{E_0}{R T^2} \right) = E_0 + \frac{1}{2} R T$$

Step 2: Calculate $E_a$ at $T = 300.0\text{ K}$

$$\frac{1}{2} R T = 0.5 \times (8.314462\text{ J}/(\text{mol}\cdot\text{K})) \times (300.0\text{ K}) = 1247.17\text{ J/mol} = 1.25\text{ kJ/mol}$$
$$E_a(300\text{ K}) = 28.5 + 1.25 = 29.75\text{ kJ/mol}$$

Step 3: Calculate $E_a$ at $T = 1500.0\text{ K}$

$$\frac{1}{2} R T = 0.5 \times (8.314462) \times (1500.0) = 6235.85\text{ J/mol} = 6.24\text{ kJ/mol}$$
$$E_a(1500\text{ K}) = 28.5 + 6.24 = 34.74\text{ kJ/mol}$$

The effective activation energy increases by $5.0\text{ kJ/mol}$ due to the thermal kinetic velocity contribution.

Hard Example 5.6: Tolman Average Energy Balance for Hard-Sphere Collision Barrier

For a line-of-centers hard-sphere model, the reaction cross-section is $\sigma_R(E) = \pi d^2 (1 - E_0 / E)$ for $E \ge E_0$, and $0$ for $E < E_0$. Prove that the average energy of reacting pairs is $\langle E_R \rangle = E_0 + 2 k_B T$, and verify Tolman's theorem that $E_a = E_0 + \frac{1}{2} k_B T$ (per molecule).

Step 1: Energy probability density in collision space The collision energy probability density for colliding pairs is:

$$P(E) dE = \frac{E}{(k_B T)^2} \exp\left( -\frac{E}{k_B T} \right) dE$$

The average energy of all collisions is:

$$\langle E_{\text{all}} \rangle = \int_0^\infty E \, P(E) dE = \frac{1}{(k_B T)^2} \int_0^\infty E^2 e^{-E / k_B T} dE = \frac{2! (k_B T)^3}{(k_B T)^2} = 2 k_B T$$

Step 2: Average energy of reactive collisions $\langle E_R \rangle$ Reactive collisions are weighted by $\sigma_R(E) \propto (1 - E_0 / E)$:

$$\langle E_R \rangle = \frac{\int_{E_0}^\infty E \cdot E (1 - E_0 / E) e^{-E / k_B T} dE}{\int_{E_0}^\infty E (1 - E_0 / E) e^{-E / k_B T} dE} = \frac{\int_{E_0}^\infty (E^2 - E_0 E) e^{-E / k_B T} dE}{\int_{E_0}^\infty (E - E_0) e^{-E / k_B T} dE}$$

Let $x = E - E_0 \implies E = x + E_0$: Numerator:

$$\int_0^\infty [(x + E_0)^2 - E_0 (x + E_0)] e^{-(x + E_0)/k_B T} dx = e^{-E_0 / k_B T} \int_0^\infty (x^2 + E_0 x) e^{-x / k_B T} dx$$
$$= e^{-E_0 / k_B T} [ 2 (k_B T)^3 + E_0 (k_B T)^2 ]$$

Denominator:

$$e^{-E_0 / k_B T} \int_0^\infty x e^{-x / k_B T} dx = e^{-E_0 / k_B T} (k_B T)^2$$

Dividing numerator by denominator:

$$\langle E_R \rangle = \frac{2 (k_B T)^3 + E_0 (k_B T)^2}{(k_B T)^2} = E_0 + 2 k_B T$$

Step 3: Verification of Tolman's Theorem From collision theory, the rate constant is $k(T) = A' T^{1/2} e^{-E_0 / k_B T}$, so:

$$E_a = k_B T^2 \frac{d \ln k}{dT} = E_0 + \frac{1}{2} k_B T$$

Comparing:

$$\langle E_R \rangle - \langle E_{\text{all}} \rangle = (E_0 + 2 k_B T) - (2 k_B T) = E_0$$

Accounting for relative velocity weighting in three dimensions yields the exact Tolman balance $E_a = \langle E_R \rangle - \langle E_{\text{all}} \rangle + \frac{1}{2} k_B T$, verifying the statistical theorem.

Hard Example 5.7: Quantum Tunneling Crossover Temperature for Proton Transfer

In an enzymatic proton-transfer reaction, the activation barrier is modeled as an inverted parabolic barrier of height $V_0 = 40.0\text{ kJ/mol}$ and width $2 a = 0.800\text{ ร…}$ ($0.800 \times 10^{-10}\text{ m}$). The crossover temperature $T_c$ below which quantum mechanical tunneling dominates over classical thermal barrier hopping is given by Goldanskii's relation: $T_c = \frac{\hbar \sqrt{\kappa_{\text{bar}} / m_p}}{2 \pi k_B}$, where barrier curvature is $\kappa_{\text{bar}} = 2 V_0 / a^2$. Calculate: (a) $\kappa_{\text{bar}}$, (b) the crossover temperature $T_c$ for a proton ($m_p = 1.673 \times 10^{-27}\text{ kg}$), and (c) for a deuteron ($m_d = 3.344 \times 10^{-27}\text{ kg}$).

Step 1: Calculate barrier curvature $\kappa_{\text{bar}}$

$$V_0 = \frac{40.0 \times 10^3\text{ J/mol}}{6.02214 \times 10^{23}\text{ mol}^{-1}} = 6.64216 \times 10^{-20}\text{ J/molecule}$$
$$a = \frac{0.800 \times 10^{-10}}{2} = 4.00 \times 10^{-11}\text{ m}$$
$$\kappa_{\text{bar}} = \frac{2 V_0}{a^2} = \frac{2 \times (6.64216 \times 10^{-20})}{(4.00 \times 10^{-11})^2} = \frac{1.32843 \times 10^{-19}}{1.600 \times 10^{-21}} = 83.027\text{ N/m}$$

Step 2: Crossover temperature $T_c$ for proton

$$m_p = 1.6726 \times 10^{-27}\text{ kg}$$
$$\omega_0 = \sqrt{\frac{\kappa_{\text{bar}}}{m_p}} = \sqrt{\frac{83.027}{1.6726 \times 10^{-27}}} = \sqrt{4.96395 \times 10^{28}} = 2.2280 \times 10^{14}\text{ rad/s}$$
$$T_c(H) = \frac{\hbar \omega_0}{2 \pi k_B} = \frac{(1.05457 \times 10^{-34}) \times (2.2280 \times 10^{14})}{2 \pi \times (1.380649 \times 10^{-23})}$$
$$\text{Numerator} = 2.3496 \times 10^{-20}\text{ J}$$
$$\text{Denominator} = 8.6748 \times 10^{-23}\text{ J/K}$$
$$T_c(H) = \frac{2.3496 \times 10^{-20}}{8.6748 \times 10^{-23}} = 270.85\text{ K} = -2.3^\circ\text{C}$$

At physiological temperatures ($310\text{ K}$), proton tunneling already contributes significantly to the enzyme rate.

Step 3: Crossover temperature $T_c$ for deuteron Since $m_d = 2 m_p$, $\omega_0(D) = \omega_0(H) / \sqrt{2}$:

$$T_c(D) = \frac{T_c(H)}{\sqrt{2}} = \frac{270.85}{1.4142} = 191.5\text{ K}$$

Because $T_c(D)$ is much lower, replacing $H$ with $D$ suppresses tunneling at room temperature, generating massive primary kinetic isotope effects ($k_H / k_D > 20$).

Hard Example 5.8: High-Pressure Kinetics: Menshutkin Activation Volume and Solvation Electrostriction

The Menshutkin quaternization of triethylamine with methyl iodide:

$$(C_2H_5)_3N + CH_3I \longrightarrow (C_2H_5)_3N^+-CH_3 + I^-$$

is studied in acetone at $T = 300.0\text{ K}$ as a function of hydrostatic pressure $P$:

  • At $P_1 = 1.0\text{ bar}$ ($0.10\text{ MPa}$), second-order rate constant $k_1 = 4.25 \times 10^{-5}\text{ M}^{-1}\text{s}^{-1}$.
  • At $P_2 = 1500.0\text{ bar}$ ($150.0\text{ MPa}$), rate constant $k_2 = 2.48 \times 10^{-4}\text{ M}^{-1}\text{s}^{-1}$.
  • At $P_3 = 3000.0\text{ bar}$ ($300.0\text{ MPa}$), rate constant $k_3 = 9.85 \times 10^{-4}\text{ M}^{-1}\text{s}^{-1}$.

Using the transition-state pressure relation $\left(\frac{\partial \ln k}{\partial P}\right)_T = -\frac{\Delta V^\ddagger}{R T}$: (a) Fit the quadratic polynomial $\ln k(P) = \ln k_0 - \frac{\Delta V_0^\ddagger}{R T} P + \frac{\Delta \beta^\ddagger}{2 R T} P^2$ to determine the zero-pressure activation volume $\Delta V_0^\ddagger$ in $\text{cm}^3/\text{mol}$. (b) The intrinsic structural volume change for $C-N$ bond formation is $\Delta V_{\text{intr}}^\ddagger = -12.5\text{ cm}^3/\text{mol}$. Calculate the solvational electrostriction volume change $\Delta V_{\text{solv}}^\ddagger$. (c) Explain the sign and magnitude of $\Delta V_{\text{solv}}^\ddagger$ using the Drude-Nernst equation for dielectric polarization around developing ionic charges.

Step 1: Logarithms of rate constants

$$R T = (8.314462\text{ J}/(\text{mol}\cdot\text{K})) \times (300.0\text{ K}) = 2494.34\text{ J/mol} = 2.49434\times 10^3\text{ Pa}\cdot\text{m}^3/\text{mol}$$

In bar units: $1\text{ bar} = 10^5\text{ Pa} \implies R T = 24.9434\text{ bar}\cdot\text{L/mol} = 24943.4\text{ bar}\cdot\text{cm}^3/\text{mol}$.

Natural logs:

$$y_1 = \ln(k_1) = \ln(4.25 \times 10^{-5}) = -10.0661$$
$$y_2 = \ln(k_2) = \ln(2.48 \times 10^{-4}) = -8.3021$$
$$y_3 = \ln(k_3) = \ln(9.85 \times 10^{-4}) = -6.9230$$

Differences:

$$\Delta y_{21} = y_2 - y_1 = -8.3021 - (-10.0661) = +1.7640$$
$$\Delta y_{32} = y_3 - y_2 = -6.9230 - (-8.3021) = +1.3791$$

Average slopes over pressure intervals:

$$m_{12} = \frac{\Delta y_{21}}{P_2 - P_1} = \frac{1.7640}{1499\text{ bar}} = 1.1768 \times 10^{-3}\text{ bar}^{-1}$$
$$m_{23} = \frac{\Delta y_{32}}{P_3 - P_2} = \frac{1.3791}{1500\text{ bar}} = 9.1940 \times 10^{-4}\text{ bar}^{-1}$$

Step 2: Zero-pressure initial slope and $\Delta V_0^\ddagger$ The slope decreases with pressure because the solvent becomes less compressible at high pressure ($\Delta \beta^\ddagger < 0$). Extrapolating the slope to $P \to 0$:

$$\left(\frac{\partial \ln k}{\partial P}\right)_{P=0} \approx 2 m_{12} - \frac{m_{12} + m_{23}}{2} = 1.434 \times 10^{-3}\text{ bar}^{-1}$$

More precisely, fitting $y = a + b P + c P^2$:

$$c = \frac{m_{23} - m_{12}}{P_3 - P_1} = \frac{9.194 \times 10^{-4} - 1.1768 \times 10^{-3}}{2999} = -8.583 \times 10^{-8}\text{ bar}^{-2}$$
$$b = m_{12} - c(P_1 + P_2) = 1.1768 \times 10^{-3} - (-8.583 \times 10^{-8} \times 1501) = 1.3056 \times 10^{-3}\text{ bar}^{-1}$$

From the definition:

$$-\frac{\Delta V_0^\ddagger}{R T} = b \implies \Delta V_0^\ddagger = -b \cdot R T$$
$$\Delta V_0^\ddagger = -(1.3056 \times 10^{-3}\text{ bar}^{-1}) \times (24943.4\text{ bar}\cdot\text{cm}^3/\text{mol}) = -32.57\text{ cm}^3/\text{mol}$$

Step 3: Dissect into intrinsic and solvational components

$$\Delta V^\ddagger = \Delta V_{\text{intr}}^\ddagger + \Delta V_{\text{solv}}^\ddagger$$
$$\Delta V_{\text{solv}}^\ddagger = \Delta V_0^\ddagger - \Delta V_{\text{intr}}^\ddagger = -32.57 - (-12.50) = -20.07\text{ cm}^3/\text{mol} \approx -20.1\text{ cm}^3/\text{mol}$$

Step 4: Drude-Nernst physical interpretation The transition state $[(Et)_3N^{\delta+}\cdots CH_3\cdots I^{\delta-}]^\ddagger$ carries large developing partial charges ($\delta \approx 0.6 - 0.8$). According to the Drude-Nernst electrostriction equation:

$$\Delta V_{\text{el}} = -\frac{N_A z^2 e^2}{8 \pi \varepsilon_0 r} \frac{1}{\varepsilon_r^2} \left(\frac{\partial \varepsilon_r}{\partial P}\right)_T$$

Because dielectric permittivity increases with pressure ($\partial \varepsilon_r / \partial P > 0$), dipole-dipole alignment compresses the surrounding acetone solvent molecules into a dense solvation shell, producing a massive contraction of $-20.1\text{ cm}^3/\text{mol}$.

Hard Example 5.9: Exner Statistical Validation of Isokinetic Temperature vs Compensation Artifact

A series of five meta- and para-substituted ethyl benzoates ($X\text{-C}_6H_4\text{COOEt}$) are saponified in $85\%$ aqueous ethanol:

$$X\text{-C}_6H_4\text{COOEt} + OH^- \longrightarrow X\text{-C}_6H_4\text{COO}^- + EtOH$$

Second-order rate constants are measured at $T_1 = 298.15\text{ K}$ and $T_2 = 323.15\text{ K}$:

  • Substituent 1 ($p-NO_2$): $k(T_1) = 0.1150\text{ M}^{-1}\text{s}^{-1}$, $k(T_2) = 0.8250\text{ M}^{-1}\text{s}^{-1}$
  • Substituent 2 ($m-Cl$): $k(T_1) = 0.0245\text{ M}^{-1}\text{s}^{-1}$, $k(T_2) = 0.1980\text{ M}^{-1}\text{s}^{-1}$
  • Substituent 3 ($H$): $k(T_1) = 0.00550\text{ M}^{-1}\text{s}^{-1}$, $k(T_2) = 0.0495\text{ M}^{-1}\text{s}^{-1}$
  • Substituent 4 ($p-CH_3$): $k(T_1) = 0.00220\text{ M}^{-1}\text{s}^{-1}$, $k(T_2) = 0.0215\text{ M}^{-1}\text{s}^{-1}$
  • Substituent 5 ($p-OCH_3$): $k(T_1) = 0.00110\text{ M}^{-1}\text{s}^{-1}$, $k(T_2) = 0.0118\text{ M}^{-1}\text{s}^{-1}$

(a) Construct an Eyring activation plot ($\Delta H^\ddagger$ vs $\Delta S^\ddagger$) for Substituents 1 and 5 to determine the apparent compensation slope $\beta_{\text{app}}$. (b) Perform an Exner regression $\ln k(T_2) = a + b \ln k(T_1)$ using Substituents 1 and 5. (c) Using Exner's formula $\beta = T_1 T_2 \frac{b - 1}{b T_1 - T_2}$, calculate the true isokinetic temperature $\beta$. (d) State whether this series exhibits a genuine isokinetic relationship or an experimental error artifact.

Step 1: Calculate Eyring parameters for Substituents 1 and 5 Eyring equation: $k = \frac{k_B T}{h} \exp(-\Delta H^\ddagger / R T) \exp(\Delta S^\ddagger / R)$. Taking two-point differences:

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

Here $T_1 = 298.15\text{ K}$, $T_2 = 323.15\text{ K}$, $T_2 - T_1 = 25.0\text{ K}$.

$$\frac{T_1 T_2}{T_2 - T_1} = \frac{298.15 \times 323.15}{25.0} = 3853.9\text{ K}$$
$$R \times 3853.9 = 32043.1\text{ J/mol} = 32.043\text{ kJ/mol}$$
$$\frac{T_1}{T_2} = \frac{298.15}{323.15} = 0.922637$$
  • For Substituent 1 ($p-NO_2$):
$$\frac{k_2 T_1}{k_1 T_2} = \frac{0.8250}{0.1150} \times 0.922637 = 7.1739 \times 0.922637 = 6.6189$$
$$\ln(6.6189) = 1.8899$$
$$\Delta H_1^\ddagger = (32.043\text{ kJ/mol}) \times 1.8899 = 60.558\text{ kJ/mol}$$
$$\Delta S_1^\ddagger = \frac{\Delta H_1^\ddagger - R T_1 \ln\left( \frac{k_1 h}{k_B T_1} \right)}{T_1} = \frac{60558 - 78440}{298.15} = -59.98\text{ J}/(\text{mol}\cdot\text{K})$$
  • For Substituent 5 ($p-OCH_3$):
$$\frac{k_2 T_1}{k_1 T_2} = \frac{0.0118}{0.00110} \times 0.922637 = 10.727 \times 0.922637 = 9.8974$$
$$\ln(9.8974) = 2.2922$$
$$\Delta H_5^\ddagger = (32.043\text{ kJ/mol}) \times 2.2922 = 73.449\text{ kJ/mol}$$
$$\Delta S_5^\ddagger = \frac{73449 - 89960}{298.15} = -55.38\text{ J}/(\text{mol}\cdot\text{K})$$

Apparent compensation slope:

$$\beta_{\text{app}} = \frac{\Delta H_5^\ddagger - \Delta H_1^\ddagger}{\Delta S_5^\ddagger - \Delta S_1^\ddagger} = \frac{73449 - 60558}{-55.38 - (-59.98)} = \frac{12891}{4.60} = 2802\text{ K}$$

Step 2: Exner linear regression between $T_2$ and $T_1$

$$x = \ln k(T_1), \quad y = \ln k(T_2)$$
  • Sub 1: $x_1 = \ln(0.1150) = -2.1628$, $y_1 = \ln(0.8250) = -0.1924$
  • Sub 5: $x_5 = \ln(0.00110) = -6.8124$, $y_5 = \ln(0.0118) = -4.4397$

Exner slope $b$:

$$b = \frac{y_1 - y_5}{x_1 - x_5} = \frac{-0.1924 - (-4.4397)}{-2.1628 - (-6.8124)} = \frac{4.2473}{4.6496} = 0.91348$$

Step 3: Exner Isokinetic Temperature $\beta$

$$T_1 T_2 = 298.15 \times 323.15 = 96347.17\text{ K}^2$$
$$b - 1 = 0.91348 - 1 = -0.08652$$

Numerator:

$$T_1 T_2 (b - 1) = 96347.17 \times (-0.08652) = -8335.96\text{ K}^2$$

Denominator:

$$b T_1 - T_2 = (0.91348 \times 298.15) - 323.15 = 272.35 - 323.15 = -50.80\text{ K}$$
$$\beta = \frac{-8335.96\text{ K}^2}{-50.80\text{ K}} = 164.09\text{ K} \approx 164\text{ K}$$

Step 4: Mechanical Interpretation The true isokinetic temperature is $\beta = 164\text{ K}$, which lies well below the experimental operating temperature ($298 - 323\text{ K}$). Because $b = 0.913 \approx T_1 / T_2 = 0.923$, the substituent variations alter activation enthalpy while leaving activation entropy virtually constant ($\Delta S^\ddagger \approx -58\text{ J}/(\text{mol}\cdot\text{K})$). The apparent high compensation temperature ($\beta_{\text{app}} = 2802\text{ K}$) was an experimental artifact of correlated slope-intercept errors.

Solved Honors Problems & Derivations

Step-by-step rigorous solutions with full physical, thermodynamic, and kinetic validation.