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Chapter 4 โ€ข Theory & Derivations

Unit 4: Empirical Chemical Kinetics & Integrated Rate Laws

Mathematical and empirical formulation of chemical reaction rates: extent of reaction, differential rate laws, reaction order versus molecularity, analytical integration of zero, first, second, and third-order rate equations, pseudo-order approximations, fractional-life analysis, and the kinetic approach of reversible reactions toward dynamic chemical equilibrium.

ยง4.1 Reaction Rates, Extent of Reaction & Differential Rate Formulations

Chemical kinetics investigates the time evolution of reacting systems and the microscopic pathways through which reactants transform into products.

Definition of Reaction Rate and Extent of Reaction

Consider a general closed homogeneous chemical reaction with stoichiometric coefficients $\nu_i$:

$$\sum_i \nu_i A_i = 0$$

where $\nu_i < 0$ for reactants and $\nu_i > 0$ for products.

The extent of reaction $\xi$ (SI unit: moles) is defined by de Donder as:

$$d\xi = \frac{dn_i}{\nu_i} \implies n_i(t) = n_i(0) + \nu_i \xi(t)$$

The intensive rate of reaction $r$ (or $v$) per unit volume $V$ is defined as:

$$r = \frac{1}{V} \frac{d\xi}{dt} = \frac{1}{\nu_i V} \frac{dn_i}{dt}$$

For a constant-volume system, introducing concentration $c_i = [A_i] = n_i / V$:

$$r = \frac{1}{\nu_i} \frac{d[A_i]}{dt}$$

For a specific reaction such as $a A + b B \longrightarrow c C + d D$:

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

This definition guarantees that the reaction rate $r$ is a uniquely defined, positive quantity independent of which participant species is monitored.

Master Classification Table: Analytical Solutions to Empirical Rate Laws

| Reaction Order | Differential Rate Law | Integrated Rate Law | Linear Coordinate Plot | Slope ($m$) & Intercept ($b$) | Rate Constant Units | Half-Life Expression ($t_{1/2}$) | Three-Quarter Life ($t_{3/4}$) | |---|---|---|---|---|---|---|---| | Zero Order ($n=0$) | $-\frac{d[A]}{dt} = k$ | $[A]_t = [A]_0 - k t$ | $[A]_t \text{ vs } t$ | $m = -k$, $b = [A]_0$ | $\text{M}\cdot\text{s}^{-1}$ | $t_{1/2} = \frac{[A]_0}{2 k}$ | $t_{3/4} = \frac{3 [A]_0}{4 k}$ | | First Order ($n=1$) | $-\frac{d[A]}{dt} = k [A]$ | $\ln[A]_t = \ln[A]_0 - k t$ | $\ln[A]_t \text{ vs } t$ | $m = -k$, $b = \ln[A]_0$ | $\text{s}^{-1}$ | $t_{1/2} = \frac{\ln 2}{k}$ | $t_{3/4} = \frac{\ln 4}{k} = 2 t_{1/2}$ | | Second Order (symmetric) | $-\frac{d[A]}{dt} = k [A]^2$ | $\frac{1}{[A]_t} = \frac{1}{[A]_0} + k t$ | $\frac{1}{[A]_t} \text{ vs } t$ | $m = +k$, $b = \frac{1}{[A]_0}$ | $\text{M}^{-1}\cdot\text{s}^{-1}$ | $t_{1/2} = \frac{1}{k [A]_0}$ | $t_{3/4} = \frac{3}{k [A]_0} = 3 t_{1/2}$ | | Second Order (asymmetric)| $-\frac{d[A]}{dt} = k [A][B]$ | $\ln\left(\frac{[B]_t [A]_0}{[A]_t [B]_0}\right) = ([B]_0 - [A]_0) k t$ | $\ln\left(\frac{[B]_t}{[A]_t}\right) \text{ vs } t$ | $m = ([B]_0 - [A]_0) k$ | $\text{M}^{-1}\cdot\text{s}^{-1}$ | Dependent on $[B]_0 / [A]_0$ | Dependent on $[B]_0 / [A]_0$ | | Third Order (symmetric) | $-\frac{d[A]}{dt} = k [A]^3$ | $\frac{1}{[A]_t^2} = \frac{1}{[A]_0^2} + 2 k t$ | $\frac{1}{[A]_t^2} \text{ vs } t$ | $m = +2k$, $b = \frac{1}{[A]_0^2}$ | $\text{M}^{-2}\cdot\text{s}^{-1}$ | $t_{1/2} = \frac{3}{2 k [A]_0^2}$ | $t_{3/4} = \frac{15}{2 k [A]_0^2} = 5 t_{1/2}$ | | $n$-th Order ($n \ne 1$) | $-\frac{d[A]}{dt} = k [A]^n$ | $\frac{1}{[A]_t^{n-1}} = \frac{1}{[A]_0^{n-1}} + (n-1) k t$ | $\frac{1}{[A]_t^{n-1}} \text{ vs } t$ | $m = (n-1)k$ | $\text{M}^{1-n}\cdot\text{s}^{-1}$ | $t_{1/2} = \frac{2^{n-1} - 1}{(n-1) k [A]_0^{n-1}}$ | $t_{3/4} = \frac{4^{n-1} - 1}{(n-1) k [A]_0^{n-1}}$ | | Reversible First-Order | $-\frac{d[A]}{dt} = k_1 [A] - k_{-1} [B]$ | $\ln\left(\frac{[A]_0 - [A]_{\text{eq}}}{[A]_t - [A]_{\text{eq}}}\right) = (k_1 + k_{-1}) t$| $\ln([A]_t - [A]_{\text{eq}}) \text{ vs } t$| $m = -(k_1 + k_{-1})$ | $\text{s}^{-1}$ | $t_{1/2} = \frac{\ln 2}{k_1 + k_{-1}}$ | Relaxation time $\tau = \frac{1}{k_1 + k_{-1}}$ |

ยง4.2 Reaction Order vs. Molecularity: Empirical Rate Laws

A critical distinction in chemical kinetics lies between the purely empirical concept of reaction order and the theoretical concept of molecularity.

Empirical Differential Rate Law

For many reactions far from equilibrium, the reaction rate depends on instantaneous reactant concentrations via an empirical power law:

$$r = k [A]^\alpha [B]^\beta [C]^\gamma \cdots$$

where:

  • $k$ is the rate constant (or specific reaction rate), independent of concentration but a strong function of temperature and ionic strength.
  • $\alpha, \beta, \gamma$ are the partial orders of reaction with respect to species $A, B, C$.
  • The overall order of reaction is $n = \alpha + \beta + \gamma + \cdots$.

Reaction Order:

  • An experimentally determined, empirical quantity.
  • Can be an integer ($0, 1, 2, 3$), a fraction ($1/2, 3/2$), or even negative.
  • Does NOT necessarily bear any relation to stoichiometric coefficients $a, b, c$ unless the reaction is elementary.

Molecularity:

  • A theoretical concept applied strictly to an elementary reaction step.
  • Represents the exact number of reactant particles that must collide simultaneously to form the transition state.
  • Strictly a positive integer: unimolecular ($1$), bimolecular ($2$), or termolecular ($3$). Quadrimolecular elementary steps have zero probability of occurrence.

ยง4.3 Integrated Rate Laws: Zero-Order Reactions & Surface Heterogeneity

In a zero-order reaction, the rate of reaction is entirely independent of the reactant concentration.

Differential and Integrated Equations

$$-\frac{d[A]}{dt} = k_0$$

where $k_0$ has SI units of $\text{mol}/(\text{m}^3\cdot\text{s})$ or $\text{M}\cdot\text{s}^{-1}$.

Separating variables and integrating from $t = 0$ ($[A] = [A]_0$) to time $t$:

$$\int_{[A]_0}^{[A]} d[A] = -k_0 \int_0^t dt$$
$$[A](t) = [A]_0 - k_0 t$$

A plot of $[A]$ versus $t$ is a straight line with slope $-k_0$ and intercept $[A]_0$.

Half-Life ($t_{1/2}$)

The half-life $t_{1/2}$ is the time required for concentration to decrease to half its initial value ($[A] = [A]_0 / 2$):

$$\frac{[A]_0}{2} = [A]_0 - k_0 t_{1/2} \implies t_{1/2} = \frac{[A]_0}{2 k_0}$$

In zero-order kinetics, the half-life is directly proportional to initial concentration $[A]_0$.

Total Reaction Lifetime ($t_{\text{end}}$)

The reaction terminates completely when $[A] = 0$:

$$t_{\text{end}} = \frac{[A]_0}{k_0} = 2 t_{1/2}$$

Physical Occurrence in Heterogeneous Surface Catalysis

Zero-order kinetics commonly occur when a reaction takes place on a saturated solid catalyst surface or enzyme active site (e.g., decomposition of ammonia on hot tungsten, $2 NH_3 \xrightarrow{W} N_2 + 3 H_2$). When all catalytic active sites are fully covered by adsorbed molecules (Langmuir coverage $\theta \approx 1$), increasing gas concentration cannot increase the reaction rate.

ยง4.4 Integrated Rate Laws: First-Order Reactions & Radioactive Decay Analogy

First-order kinetics govern processes where the rate is directly proportional to the concentration of a single reactant.

Differential and Integrated Equations

$$-\frac{d[A]}{dt} = k_1 [A]$$

where $k_1$ has SI units of $\text{s}^{-1}$ (time$^{-1}$).

Separating variables:

$$\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = -k_1 \int_0^t dt$$
$$\ln\left( \frac{[A]}{[A]_0} \right) = -k_1 t \iff [A](t) = [A]_0 \exp(-k_1 t)$$

Linearized form:

$$\ln[A] = \ln[A]_0 - k_1 t$$

A plot of $\ln[A]$ versus $t$ yields a straight line with slope $-k_1$.

Concentration of Product ($P$)

For $A \longrightarrow P$, with $[P]_0 = 0$:

$$[P](t) = [A]_0 - [A](t) = [A]_0 \left( 1 - e^{-k_1 t} \right)$$

Half-Life ($t_{1/2}$)

At $t = t_{1/2}$, $[A] = [A]_0 / 2$:

$$\ln\left( \frac{1}{2} \right) = -k_1 t_{1/2} \implies t_{1/2} = \frac{\ln 2}{k_1} = \frac{0.69315}{k_1}$$

Cardinal Feature: The half-life of a first-order process is strictly independent of the initial concentration $[A]_0$. Successive half-lives remain constant: after $n$ half-lives, $[A] = [A]_0 / 2^n$.

Mean Lifetime ($\tau$)

The average lifetime of a reacting particle is the reciprocal of the first-order rate constant:

$$\tau = \langle t \rangle = \frac{\int_0^\infty t \, e^{-k_1 t} dt}{\int_0^\infty e^{-k_1 t} dt} = \frac{1}{k_1} = \frac{t_{1/2}}{\ln 2} \approx 1.443 \, t_{1/2}$$

ยง4.5 Integrated Rate Laws: Second-Order Reactions (Equal & Unequal Reactants)

Second-order reactions involve the collision of two molecules, classified into two distinct cases.

Case 1: Single Reactant or Equal Initial Concentrations ($2 A \to P$ or $A + B \to P$ with $[A]_0 = [B]_0$)

$$-\frac{d[A]}{dt} = k_2 [A]^2$$

where $k_2$ has SI units of $\text{m}^3/(\text{mol}\cdot\text{s})$ or $\text{M}^{-1}\text{s}^{-1}$.

Separating variables:

$$\int_{[A]_0}^{[A]} \frac{d[A]}{[A]^2} = -k_2 \int_0^t dt \implies -\left[ \frac{1}{[A]} - \frac{1}{[A]_0} \right] = -k_2 t$$
$$\frac{1}{[A](t)} = \frac{1}{[A]_0} + k_2 t$$

A plot of $1/[A]$ versus $t$ is linear with slope $+k_2$ and intercept $1/[A]_0$.

Half-Life:

$$\frac{1}{[A]_0/2} - \frac{1}{[A]_0} = k_2 t_{1/2} \implies t_{1/2} = \frac{1}{k_2 [A]_0}$$

In second-order kinetics, the half-life is inversely proportional to initial concentration. Each successive half-life doubles ($t_{1/2}, 2 t_{1/2}, 4 t_{1/2}$).

Case 2: Unequal Initial Concentrations ($A + B \to P$ with $[A]_0 \neq [B]_0$)

Let $x$ be the extent of concentration reacted at time $t$: $[A] = [A]_0 - x$, $[B] = [B]_0 - x$.

$$\frac{dx}{dt} = k_2 ([A]_0 - x)([B]_0 - x)$$

Using partial fractions:

$$\frac{1}{([A]_0 - x)([B]_0 - x)} = \frac{1}{[B]_0 - [A]_0} \left( \frac{1}{[A]_0 - x} - \frac{1}{[B]_0 - x} \right)$$

Integrating from $x = 0$ at $t = 0$:

$$\frac{1}{[B]_0 - [A]_0} \left[ \ln\left( \frac{[A]_0}{[A]_0 - x} \right) - \ln\left( \frac{[B]_0}{[B]_0 - x} \right) \right] = k_2 t$$
$$\frac{1}{[B]_0 - [A]_0} \ln\left( \frac{[A]_0 [B]}{[B]_0 [A]} \right) = k_2 t \iff \ln\left( \frac{[B]}{[A]} \right) = \ln\left( \frac{[B]_0}{[A]_0} \right) + ([B]_0 - [A]_0) k_2 t$$

Plotting $\ln([B]/[A])$ versus $t$ yields a straight line with slope $([B]_0 - [A]_0) k_2$.

University Honors Research Monograph: Microfluidic Flow Kinetics & Automated Machine-Learning Profiling

Continuous-flow microchemical reactors have transformed empirical reaction kinetics from laborious manual batch aliquot sampling into continuous, autonomous real-time multidimensional mapping:

  • Peclet and Reynolds Number Hydrodynamics: In microchannels of hydraulic diameter $D_h \approx 200 - 500\;\mu\text{m}$, fluid flow is strictly laminar:
$$\text{Re} = \frac{\rho u D_h}{\eta} < 100$$

Turbulent back-mixing is eliminated, while molecular diffusion distances are miniature ($< 100\;\mu\text{m}$), ensuring complete radial mixing in milliseconds.

  • Plug-Flow Reaction Profiling: Under plug-flow conditions, axial distance $z$ downstream from the mixing junction translates linearly to reaction residence time:
$$\tau_{\text{res}} = \frac{z}{u} = \frac{V_{\text{channel}}}{Q_{\text{total}}}$$

Varying total volumetric pumping rate $Q_{\text{total}}$ rapidly sweeps reaction times from $10\text{ ms}$ to $10\text{ minutes}$.

  • Closed-Loop Bayesian Optimization: Coupling microfluidic flow reactors to in-line high-pressure NMR, HPLC-MS, and ATR-FTIR flow cells enables machine learning algorithms (Bayesian optimization with Gaussian process regression) to automatically map multi-variable kinetic response surfaces ($T$, $P$, catalyst loading, residence time), resolving full multi-step reaction mechanisms and rate constants in under 48 hours.

ยง4.6 Third-Order Kinetics, Fractional Orders & Pseudo-Molecular Regimes

Higher-order and non-integer kinetics emerge in complex termolecular gas reactions and catalytic systems.

Third-Order Reactions ($3 A \to P$)

$$-\frac{d[A]}{dt} = k_3 [A]^3$$

where $k_3$ has SI units of $\text{M}^{-2}\text{s}^{-1}$.

Integrating:

$$\int_{[A]_0}^{[A]} \frac{d[A]}{[A]^3} = -k_3 t \implies -\frac{1}{2} \left( \frac{1}{[A]^2} - \frac{1}{[A]_0^2} \right) = -k_3 t$$
$$\frac{1}{[A]^2} = \frac{1}{[A]_0^2} + 2 k_3 t$$

Half-life for third-order kinetics:

$$t_{1/2} = \frac{3}{2 k_3 [A]_0^2}$$

True gas-phase termolecular elementary reactions are rare due to the vanishing probability of simultaneous three-body collisions; prominent examples include gas-phase oxidation of nitric oxide ($2 NO + O_2 \longrightarrow 2 NO_2$) and radical recombination ($H + H + M \longrightarrow H_2 + M$).

Fractional-Life Method for General Order $n$

For an $n$-th order reaction ($-\frac{d[A]}{dt} = k [A]^n$, $n \neq 1$):

$$\frac{1}{[A]^{n-1}} = \frac{1}{[A]_0^{n-1}} + (n - 1) k t$$

The half-life scaling is:

$$t_{1/2} = \frac{2^{n-1} - 1}{(n - 1) k [A]_0^{n-1}} \implies t_{1/2} \propto [A]_0^{1 - n}$$

Pseudo-Unimolecular Reactions (Isolation Method)

When a bimolecular reaction $A + B \to P$ is conducted with species $B$ in overwhelming stoichiometric excess ($[B]_0 \gg [A]_0$): $[B](t) \approx [B]_0 = \text{constant}$ throughout the course of reaction.

$$-\frac{d[A]}{dt} = k_2 [A] [B] \approx (k_2 [B]_0) [A] = k_{\text{obs}} [A]$$

where $k_{\text{obs}} = k_2 [B]_0$ is the pseudo-first-order rate constant (units: $\text{s}^{-1}$). This reduces second-order mathematics to simple first-order exponential decays, forming the basis of the isolation method.

ยง4.7 Reversible Reactions Approaching Equilibrium: Microscopic Reversibility

When the forward and reverse reactions proceed at comparable rates, the system approaches a dynamic chemical equilibrium.

First-Order Opposing Reaction ($A \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} B$)

Let initial concentrations be $[A]_0$ and $[B]_0 = 0$. At time $t$, $[A] = [A]_0 - x$ and $[B] = x$.

The net rate of consumption of $A$ is:

$$\frac{dx}{dt} = k_1 [A] - k_{-1} [B] = k_1 ([A]_0 - x) - k_{-1} x = k_1 [A]_0 - (k_1 + k_{-1}) x$$

At dynamic equilibrium ($t \to \infty$), net rate vanishes ($dx/dt = 0$):

$$k_1 [A]_{\text{eq}} = k_{-1} [B]_{\text{eq}} \implies \frac{[B]_{\text{eq}}}{[A]_{\text{eq}}} = \frac{x_{\text{eq}}}{[A]_0 - x_{\text{eq}}} = \frac{k_1}{k_{-1}} = K_c$$

where $K_c$ is the thermodynamic equilibrium constant. This satisfies the Principle of Microscopic Reversibility.

Expressing $k_1 [A]_0$ in terms of equilibrium extent:

$$k_1 [A]_0 = (k_1 + k_{-1}) x_{\text{eq}}$$

Substituting back into the rate differential equation:

$$\frac{dx}{dt} = (k_1 + k_{-1}) (x_{\text{eq}} - x)$$

Separating variables and integrating:

$$\int_0^x \frac{dx}{x_{\text{eq}} - x} = (k_1 + k_{-1}) \int_0^t dt \implies \ln\left( \frac{x_{\text{eq}}}{x_{\text{eq}} - x} \right) = (k_1 + k_{-1}) t$$
$$x(t) = x_{\text{eq}} \left[ 1 - e^{-(k_1 + k_{-1}) t} \right]$$

In terms of reactant concentration:

$$[A](t) - [A]_{\text{eq}} = ([A]_0 - [A]_{\text{eq}}) \exp\left( -(k_1 + k_{-1}) t \right)$$

Key Insight: The approach to equilibrium is a first-order exponential relaxation whose apparent rate constant is the sum of forward and backward rate constants ($k_{\text{obs}} = k_1 + k_{-1}$). Measuring $k_{\text{obs}}$ along with the equilibrium ratio $K_c = k_1 / k_{-1}$ allows exact evaluation of both individual rate constants.

ยง4.8 Operando Reaction Calorimetry & Online Flow-Spectroscopy for Real-Time Rate Tracking

Determining multi-step kinetic rate laws in modern organic synthesis and industrial process development relies on continuous, in situ physical property tracking using operando reaction calorimetry and continuous-flow spectroscopy.

1. In Situ Heat-Flow Reaction Calorimetry

Because virtually all chemical transformations exhibit a non-zero enthalpy of reaction ($\Delta_r H \ne 0$), the instantaneous rate of heat generation $q_r(t)$ in a closed or flow reactor is directly proportional to the instantaneous chemical reaction rate $r(t) = -d[A]/dt$:

$$q_r(t) = V_{\text{rxn}} \cdot (-\Delta_r H) \cdot r(t)$$
Dynamic Heat Balance in a Controlled Reactor

In an automated heat-flow reaction calorimeter (such as an RC1 or Omnical system), a thermostat jacket maintains reactor temperature $T_r$ under isothermal control. The heat balance equation is:

$$q_{\text{flow}}(t) = U A (T_r(t) - T_j(t)) = q_r(t) + q_{\text{stir}} + q_{\text{loss}} - C_p \frac{dT_r}{dt}$$

where:

  • $U$ is overall heat transfer coefficient ($\text{W}/(\text{m}^2\cdot\text{K})$).
  • $A$ is wetted heat exchange area ($\text{m}^2$).
  • $T_j$ is cooling/heating jacket temperature.
  • $q_{\text{stir}}$ is mechanical agitator power dissipation.
  • $C_p$ is total heat capacity of the reactor assembly and liquid contents.

Under true isothermal steady conditions ($dT_r/dt = 0$), calibration pulses determine the instantaneous heat transfer product $U A$. Subtracting baseline power yields direct, millisecond-resolved chemical heat release rates $q_r(t)$ without withdrawing samples or perturbing chemical equilibria.

Reaction Progress Kinetic Analysis (RPKA) via Calorimetry

Integrating heat release from time $t = 0$ to completion ($t = \infty$) yields total chemical enthalpy release:

$$Q_{\text{tot}} = \int_0^\infty q_r(t) dt = V_{\text{rxn}} \cdot (-\Delta_r H) \cdot [A]_0$$

The instantaneous fractional conversion $\chi(t)$ and reactant concentration $[A](t)$ are given by:

$$\chi(t) = \frac{\int_0^t q_r(t') dt'}{Q_{\text{tot}}} \implies [A](t) = [A]_0 (1 - \chi(t))$$

Plotting instantaneous reaction rate $r(t) = q_r(t) / (V (-\Delta_r H))$ directly against concentration $[A](t)$ generates "graphical rate equations" that reveal reaction orders, catalyst deactivation, and product inhibition in a single experiment.

2. Online UV-Vis / FTIR Attenuated Total Reflectance (ATR) Spectroscopy

In situ ATR-FTIR utilizes a zinc selenide ($\text{ZnSe}$) or diamond internal reflection element placed inside the reaction medium. An infrared beam undergoes multiple total internal reflections at the crystal-solution interface, generating an evanescent wave that penetrates $1 - 3\;\mu\text{m}$ into the solution.

By Beer-Lambert's law for evanescent absorption:

$$A_i(\nu, t) = \epsilon_i(\nu) \cdot d_p(\nu) \cdot N_{\text{refl}} \cdot [C_i](t)$$

where $d_p = \frac{\lambda}{2 \pi \sqrt{n_{\text{crystal}}^2 \sin^2 \theta - n_{\text{sol}}^2}}$ is penetration depth. Fourier transform deconvolution generates complete concentration profiles $[C_i](t)$ for reactants, transient reactive intermediates, and products with sub-second temporal resolution.

Easy Example 4.1: First-Order Hydrolysis Kinetics and Half-Life Evaluation

The thermal decomposition of dinitrogen pentoxide ($2 N_2O_5(g) \longrightarrow 4 NO_2(g) + O_2(g)$) in carbon tetrachloride is first-order with a rate constant $k_1 = 6.20 \times 10^{-4}\text{ s}^{-1}$ at $45.0^\circ\text{C}$. Calculate: (a) the half-life $t_{1/2}$, (b) the time required for $90.0\%$ of the $N_2O_5$ to decompose, and (c) the fraction remaining after $t = 30.0\text{ minutes}$.

Step 1: Calculate half-life

$$t_{1/2} = \frac{\ln 2}{k_1} = \frac{0.693147}{6.20 \times 10^{-4}\text{ s}^{-1}} = 1117.98\text{ s} = 18.63\text{ minutes}$$

Step 2: Time for $90.0\%$ decomposition When $90.0\%$ has decomposed, the fraction remaining is:

$$\frac{[A]}{[A]_0} = 1.00 - 0.900 = 0.100$$

Using the integrated first-order rate law:

$$\ln\left( \frac{[A]}{[A]_0} \right) = -k_1 t$$
$$t = -\frac{\ln(0.100)}{k_1} = \frac{2.302585}{6.20 \times 10^{-4}\text{ s}^{-1}} = 3713.8\text{ s} = 61.90\text{ minutes}$$

Step 3: Fraction remaining after $30.0\text{ minutes}$

$$t = 30.0 \times 60 = 1800\text{ s}$$
$$k_1 t = (6.20 \times 10^{-4}\text{ s}^{-1}) \times (1800\text{ s}) = 1.116$$
$$\frac{[A]}{[A]_0} = \exp(-1.116) = 0.3276 = 32.76\%$$

After 30 minutes, $32.8\%$ of the original $N_2O_5$ remains unreacted.

Medium Example 4.2: Second-Order Saponification Rate Constant Determination

The alkaline saponification of ethyl acetate ($CH_3COOC_2H_5 + NaOH \longrightarrow CH_3COONa + C_2H_5OH$) is second-order overall. In an experiment at $25.0^\circ\text{C}$, initial concentrations of both ester and sodium hydroxide were equal at $[A]_0 = [B]_0 = 0.0500\text{ M}$. After $t = 15.0\text{ minutes}$, the remaining concentration of $NaOH$ was titrated to be $0.0180\text{ M}$. Calculate: (a) the second-order rate constant $k_2$ in $\text{M}^{-1}\text{s}^{-1}$, and (b) the half-life $t_{1/2}$.

Step 1: Integrated second-order equation with equal initial concentrations

$$\frac{1}{[A]} - \frac{1}{[A]_0} = k_2 t$$

where $[A]_0 = 0.0500\text{ M}$ and $[A] = 0.0180\text{ M}$.

$$t = 15.0 \times 60 = 900.0\text{ s}$$
$$\frac{1}{0.0180} - \frac{1}{0.0500} = 55.5556 - 20.0000 = 35.5556\text{ M}^{-1}$$
$$k_2 = \frac{35.5556\text{ M}^{-1}}{900.0\text{ s}} = 0.039506\text{ M}^{-1}\text{s}^{-1} = 0.0395\text{ L}/(\text{mol}\cdot\text{s})$$

Step 2: Half-life calculation

$$t_{1/2} = \frac{1}{k_2 [A]_0} = \frac{1}{0.039506 \times 0.0500} = \frac{1}{1.9753 \times 10^{-3}} = 506.26\text{ s} = 8.44\text{ minutes}$$
Hard Example 4.3: Second-Order Reaction with Unequal Initial Reactant Concentrations

The reaction $A + B \longrightarrow P$ is first order with respect to $A$ and first order with respect to $B$ ($r = k_2 [A][B]$). The initial concentrations are $[A]_0 = 0.0400\text{ M}$ and $[B]_0 = 0.0800\text{ M}$. The second-order rate constant is $k_2 = 0.0750\text{ M}^{-1}\text{s}^{-1}$. Calculate: (a) the concentrations of $A$ and $B$ after $t = 300.0\text{ s}$, and (b) the time required for $80.0\%$ of reactant $A$ to be consumed.

Step 1: Evaluate integrated form for $[A]_0 \neq [B]_0$

$$\ln\left( \frac{[B]}{[A]} \right) = \ln\left( \frac{[B]_0}{[A]_0} \right) + ([B]_0 - [A]_0) k_2 t$$
$$[B]_0 - [A]_0 = 0.0800 - 0.0400 = 0.0400\text{ M}$$
$$\frac{[B]_0}{[A]_0} = \frac{0.0800}{0.0400} = 2.000 \implies \ln(2.000) = 0.69315$$

At $t = 300.0\text{ s}$:

$$([B]_0 - [A]_0) k_2 t = (0.0400\text{ M}) \times (0.0750\text{ M}^{-1}\text{s}^{-1}) \times (300.0\text{ s}) = 0.9000$$
$$\ln\left( \frac{[B]}{[A]} \right) = 0.69315 + 0.9000 = 1.59315$$
$$\frac{[B]}{[A]} = \exp(1.59315) = 4.9192$$

Step 2: Relate $[B]$ and $[A]$ by stoichiometry Let $x$ be the reacted concentration: $[A] = 0.0400 - x$, $[B] = 0.0800 - x = [A] + 0.0400$.

$$\frac{[A] + 0.0400}{[A]} = 4.9192 \implies 1 + \frac{0.0400}{[A]} = 4.9192$$
$$\frac{0.0400}{[A]} = 3.9192 \implies [A] = \frac{0.0400}{3.9192} = 0.01021\text{ M}$$
$$[B] = [A] + 0.0400 = 0.05021\text{ M}$$

Step 3: Time for $80.0\%$ conversion of $A$ When $80.0\%$ of $A$ is consumed:

$$[A] = (1 - 0.800) \times 0.0400 = 0.00800\text{ M}$$
$$x = 0.03200\text{ M}$$
$$[B] = 0.0800 - 0.0320 = 0.04800\text{ M}$$
$$\frac{[B]}{[A]} = \frac{0.04800}{0.00800} = 6.000$$

Using the integrated rate law:

$$\ln(6.000) = \ln(2.000) + (0.0400 \times 0.0750) t$$
$$1.79176 = 0.69315 + 0.00300 \, t$$
$$0.00300 \, t = 1.09861 \implies t = \frac{1.09861}{0.00300} = 366.2\text{ s} = 6.10\text{ minutes}$$
Medium Example 4.4: Reaction Order Determination via Fractional-Life Method

In a kinetic study of the decomposition of an organic peroxide, the measured half-life $t_{1/2}$ varied with initial concentration $[A]_0$ as follows: at $[A]_0 = 0.0200\text{ M}$, $t_{1/2} = 245.0\text{ s}$; at $[A]_0 = 0.0800\text{ M}$, $t_{1/2} = 61.25\text{ s}$. Determine: (a) the overall reaction order $n$, and (b) the rate constant $k$.

Step 1: Fractional-life scaling relation For reaction order $n$:

$$t_{1/2} \propto [A]_0^{1 - n} \implies \frac{t_{1/2, 1}}{t_{1/2, 2}} = \left( \frac{[A]_{0, 1}}{[A]_{0, 2}} \right)^{1 - n}$$

Substituting experimental values:

$$\frac{245.0}{61.25} = 4.000$$
$$\frac{[A]_{0, 1}}{[A]_{0, 2}} = \frac{0.0200}{0.0800} = 0.250 = \frac{1}{4}$$

Thus:

$$4.000 = \left( \frac{1}{4} \right)^{1 - n} = 4^{n - 1}$$

Taking logarithms:

$$\ln(4.000) = (n - 1) \ln(4) \implies n - 1 = 1 \implies n = 2$$

The reaction is strictly second-order ($n = 2$).

Step 2: Evaluate the second-order rate constant $k_2$ For $n = 2$:

$$t_{1/2} = \frac{1}{k_2 [A]_0} \implies k_2 = \frac{1}{t_{1/2} [A]_0}$$

Using condition 1:

$$k_2 = \frac{1}{245.0\text{ s} \times 0.0200\text{ M}} = \frac{1}{4.900} = 0.2041\text{ M}^{-1}\text{s}^{-1}$$

Using condition 2 to verify:

$$k_2 = \frac{1}{61.25\text{ s} \times 0.0800\text{ M}} = \frac{1}{4.900} = 0.2041\text{ M}^{-1}\text{s}^{-1}$$
Easy Example 4.5: Pseudo-First-Order Saponification by Reactant Isolation

The ester hydrolysis reaction $RCOOR' + H_2O \xrightarrow{H^+} RCOOH + R'OH$ has a true second-order rate constant of $k_2 = 1.85 \times 10^{-4}\text{ M}^{-1}\text{s}^{-1}$. In dilute aqueous solution, water is the solvent with concentration $[H_2O] = 55.5\text{ M}$. (a) Calculate the pseudo-first-order rate constant $k_{\text{obs}}$, and (b) calculate the time required for $50.0\%$ and $99.0\%$ of the ester to hydrolyze.

Step 1: Calculate pseudo-first-order rate constant Because $[H_2O] = 55.5\text{ M} \gg [\text{ester}]_0 \sim 0.01\text{ M}$, $[H_2O]$ remains virtually constant:

$$k_{\text{obs}} = k_2 [H_2O] = (1.85 \times 10^{-4}\text{ M}^{-1}\text{s}^{-1}) \times (55.5\text{ M}) = 1.02675 \times 10^{-2}\text{ s}^{-1}$$

Step 2: Half-life ($50.0\%$ conversion)

$$t_{1/2} = \frac{\ln 2}{k_{\text{obs}}} = \frac{0.693147}{1.02675 \times 10^{-2}\text{ s}^{-1}} = 67.51\text{ s}$$

Step 3: Time for $99.0\%$ conversion

$$\frac{[A]}{[A]_0} = 1 - 0.990 = 0.0100$$
$$t = -\frac{\ln(0.0100)}{k_{\text{obs}}} = \frac{4.60517}{1.02675 \times 10^{-2}\text{ s}^{-1}} = 448.52\text{ s} = 7.48\text{ minutes}$$
Hard Example 4.6: First-Order Reversible Kinetics Approaching Dynamic Equilibrium

The isomerization of cis-stilbene ($A$) to trans-stilbene ($B$) ($A \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} B$) is reversible and first-order in both directions. Starting with pure cis-stilbene at $[A]_0 = 0.150\text{ M}$, the equilibrium concentration of cis-isomer is $[A]_{\text{eq}} = 0.0300\text{ M}$. The time required for $[A]$ to reach $0.0900\text{ M}$ is $t = 120.0\text{ s}$. Calculate: (a) the equilibrium constant $K_c$, (b) the relaxation sum $(k_1 + k_{-1})$, and (c) the individual rate constants $k_1$ and $k_{-1}$.

Step 1: Determine equilibrium constant $K_c$ At equilibrium:

$$[B]_{\text{eq}} = [A]_0 - [A]_{\text{eq}} = 0.150 - 0.0300 = 0.120\text{ M}$$
$$K_c = \frac{[B]_{\text{eq}}}{[A]_{\text{eq}}} = \frac{0.120\text{ M}}{0.0300\text{ M}} = 4.000$$

Therefore:

$$k_1 = 4.000 \, k_{-1}$$

Step 2: Integrated reversible rate law

$$[A](t) - [A]_{\text{eq}} = ([A]_0 - [A]_{\text{eq}}) \exp\left( -(k_1 + k_{-1}) t \right)$$
$$\frac{[A](t) - [A]_{\text{eq}}}{[A]_0 - [A]_{\text{eq}}} = \exp\left( -(k_1 + k_{-1}) t \right)$$

Substitute given concentrations at $t = 120.0\text{ s}$:

$$[A](t) - [A]_{\text{eq}} = 0.0900 - 0.0300 = 0.0600\text{ M}$$
$$[A]_0 - [A]_{\text{eq}} = 0.150 - 0.0300 = 0.120\text{ M}$$
$$\frac{0.0600}{0.120} = 0.5000 = \exp\left( -(k_1 + k_{-1}) \times 120.0 \right)$$

Taking logarithms:

$$-(k_1 + k_{-1}) \times 120.0 = \ln(0.5000) = -0.69315$$
$$k_1 + k_{-1} = \frac{0.69315}{120.0} = 5.77625 \times 10^{-3}\text{ s}^{-1}$$

Step 3: Solve for individual rate constants Substitute $k_1 = 4.000 \, k_{-1}$:

$$4.000 \, k_{-1} + k_{-1} = 5.000 \, k_{-1} = 5.77625 \times 10^{-3}\text{ s}^{-1}$$
$$k_{-1} = \frac{5.77625 \times 10^{-3}}{5.000} = 1.155 \times 10^{-3}\text{ s}^{-1}$$
$$k_1 = 4.000 \times (1.155 \times 10^{-3}) = 4.621 \times 10^{-3}\text{ s}^{-1}$$
Hard Example 4.7: Gas-Phase Termolecular Nitric Oxide Oxidation Kinetics

The homogeneous gas-phase oxidation of nitric oxide ($2 NO + O_2 \longrightarrow 2 NO_2$) follows the rate law $r = k_3 [NO]^2 [O_2]$. At $T = 300.0\text{ K}$, initial partial pressures are $P_{NO} = 20.0\text{ Torr}$ and $P_{O_2} = 10.0\text{ Torr}$. The rate constant in pressure units is $k_p = 3.50 \times 10^{-3}\text{ Torr}^{-2}\text{s}^{-1}$. (a) Calculate the initial rate of pressure decrease in $\text{Torr/s}$, and (b) convert $k_p$ into concentration units $k_c$ in $\text{M}^{-2}\text{s}^{-1}$.

Step 1: Initial rate of reaction in pressure units The reaction rate is:

$$r_p = -\frac{d P_{O_2}}{dt} = -\frac{1}{2} \frac{d P_{NO}}{dt} = k_p P_{NO}^2 P_{O_2}$$
$$r_p = (3.50 \times 10^{-3}\text{ Torr}^{-2}\text{s}^{-1}) \times (20.0\text{ Torr})^2 \times (10.0\text{ Torr})$$
$$r_p = (3.50 \times 10^{-3}) \times (400.0) \times (10.0) = 14.00\text{ Torr/s}$$

Total pressure rate of change:

$$P_{\text{tot}} = P_{NO} + P_{O_2} + P_{NO_2}$$
$$\Delta P_{\text{tot}} = \Delta n = 2 - (2 + 1) = -1\text{ mole per extent}$$
$$\frac{d P_{\text{tot}}}{dt} = -r_p = -14.00\text{ Torr/s}$$

Step 2: Conversion of $k_p$ to concentration units $k_c$ Using ideal gas law $P_i = c_i R T$:

$$r = -\frac{d c_{O_2}}{dt} = -\frac{1}{R T} \frac{d P_{O_2}}{dt} = \frac{1}{R T} k_p P_{NO}^2 P_{O_2} = \frac{k_p}{R T} (c_{NO} R T)^2 (c_{O_2} R T)$$
$$r = k_p (R T)^2 [NO]^2 [O_2] = k_c [NO]^2 [O_2]$$

Therefore:

$$k_c = k_p (R T)^2$$

In SI / standard laboratory units:

$$R = 0.0820574\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}) = 62.3637\text{ L}\cdot\text{Torr}/(\text{mol}\cdot\text{K})$$
$$R T = 62.3637 \times 300.0 = 1.87091 \times 10^4\text{ L}\cdot\text{Torr/mol}$$
$$(R T)^2 = (1.87091 \times 10^4)^2 = 3.5003 \times 10^8\text{ L}^2\cdot\text{Torr}^2\text{/mol}^2$$
$$k_c = (3.50 \times 10^{-3}\text{ Torr}^{-2}\text{s}^{-1}) \times (3.5003 \times 10^8\text{ L}^2\cdot\text{Torr}^2\text{/mol}^2) = 1.225 \times 10^6\text{ L}^2/(\text{mol}^2\cdot\text{s}) = 1.225 \times 10^6\text{ M}^{-2}\text{s}^{-1}$$
Hard Example 4.8: RPKA Heat-Flow Reaction Calorimetry for Catalytic Hydroformylation

The rhodium-catalyzed hydroformylation of 1-octene ($C_8H_{16} + CO + H_2 \longrightarrow C_9H_{18}O$) is monitored by in situ heat-flow reaction calorimetry in an isothermal batch reactor at $T = 80.0^\circ\text{C}$ ($353.15\text{ K}$). Reaction parameters:

  • Reactor liquid volume: $V_{\text{rxn}} = 0.500\text{ L}$
  • Initial 1-octene concentration: $[A]_0 = 1.200\text{ M}$
  • Enthalpy of hydroformylation: $\Delta_r H^\circ = -138.0\text{ kJ/mol}$
  • Total heat exchange coefficient $\times$ area: $U A = 18.50\text{ W/K}$

Calorimetric observations:

  • At time $t_1 = 600.0\text{ s}$, the temperature difference across the reactor cooling jacket is $\Delta T_1 = T_r - T_{j, 1} = 2.450\text{ K}$.
  • At time $t_2 = 2400.0\text{ s}$, the temperature difference drops to $\Delta T_2 = T_r - T_{j, 2} = 0.6125\text{ K}$.
  • Numerical integration of the thermal power from $t = 0$ to $t_1$ yields total accumulated heat $Q(t_1) = 24.84\text{ kJ}$.

(a) Calculate the instantaneous heat release rates $q_r(t_1)$ and $q_r(t_2)$ in Watts ($\text{J/s}$). (b) Calculate the instantaneous reaction rates $r(t_1)$ and $r(t_2)$ in $\text{mol}/(\text{L}\cdot\text{s})$. (c) Using $Q(t_1)$ and total theoretical heat $Q_{\text{tot}}$, calculate the remaining 1-octene concentration $[A](t_1)$ at $600\text{ s}$. (d) If the reaction is first-order with respect to 1-octene under constant syngas pressure, calculate the observed rate constant $k_{\text{obs}}$ in $\text{s}^{-1}$.

Step 1: Calculate instantaneous heat release rates $q_r$ Under steady isothermal control ($dT_r/dt = 0$, baseline mechanical power negligible):

$$q_r(t) = U A \Delta T(t)$$
  • At $t_1 = 600\text{ s}$:
$$q_r(t_1) = (18.50\text{ W/K}) \times (2.450\text{ K}) = 45.325\text{ W} = 45.33\text{ J/s}$$
  • At $t_2 = 2400\text{ s}$:
$$q_r(t_2) = (18.50\text{ W/K}) \times (0.6125\text{ K}) = 11.331\text{ W} = 11.33\text{ J/s}$$

Step 2: Calculate instantaneous chemical reaction rates $r(t)$ The relationship between heat flow and volumetric rate is:

$$q_r(t) = V_{\text{rxn}} \cdot (-\Delta_r H) \cdot r(t) \implies r(t) = \frac{q_r(t)}{V_{\text{rxn}} \cdot (-\Delta_r H)}$$

Conversion factor:

$$V_{\text{rxn}} \cdot (-\Delta_r H) = (0.500\text{ L}) \times (138.0 \times 10^3\text{ J/mol}) = 69000\text{ J}\cdot\text{L/mol}$$
  • At $t_1$:
$$r(t_1) = \frac{45.325\text{ J/s}}{69000\text{ J}\cdot\text{L/mol}} = 6.5688 \times 10^{-4}\text{ mol}/(\text{L}\cdot\text{s}) = 6.57 \times 10^{-4}\text{ M/s}$$
  • At $t_2$:
$$r(t_2) = \frac{11.331\text{ J/s}}{69000\text{ J}\cdot\text{L/mol}} = 1.6422 \times 10^{-4}\text{ mol}/(\text{L}\cdot\text{s}) = 1.64 \times 10^{-4}\text{ M/s}$$

Step 3: Calculate remaining concentration $[A](t_1)$ Total initial moles of 1-octene:

$$n_{A, 0} = [A]_0 \cdot V_{\text{rxn}} = (1.200\text{ mol/L}) \times (0.500\text{ L}) = 0.600\text{ mol}$$

Total theoretical reaction heat at complete conversion:

$$Q_{\text{tot}} = n_{A, 0} \cdot (-\Delta_r H) = (0.600\text{ mol}) \times (138.0\text{ kJ/mol}) = 82.80\text{ kJ}$$

Fractional conversion at $t_1 = 600\text{ s}$:

$$\chi(t_1) = \frac{Q(t_1)}{Q_{\text{tot}}} = \frac{24.84\text{ kJ}}{82.80\text{ kJ}} = 0.3000 = 30.0\%$$

Remaining concentration:

$$[A](t_1) = [A]_0 (1 - \chi(t_1)) = 1.200\text{ M} \times (1 - 0.300) = 0.8400\text{ M}$$

Step 4: Calculate observed rate constant $k_{\text{obs}}$ Assuming first-order kinetics $r(t_1) = k_{\text{obs}} [A](t_1)$:

$$k_{\text{obs}} = \frac{r(t_1)}{[A](t_1)} = \frac{6.5688 \times 10^{-4}\text{ M/s}}{0.8400\text{ M}} = 7.820 \times 10^{-4}\text{ s}^{-1}$$

Using integrated first-order law for conversion verification:

$$\chi(t_1) = 1 - e^{-k_{\text{obs}} t_1} \implies 1 - e^{-(7.820 \times 10^{-4} \times 600)} = 1 - e^{-0.4692} = 1 - 0.6255 = 0.374$$

Taking the exact differential rate constant:

$$k_{\text{obs}} = 7.82 \times 10^{-4}\text{ s}^{-1}$$

Calorimetry allows immediate extraction of the instantaneous rate constant without sampling!

Hard Example 4.9: Gas-Phase Nitric Oxide Oxidation: Ter-Molecular vs Pre-Equilibrium Dimer Kinetics

The homogeneous gas-phase oxidation of nitric oxide:

$$2 NO(g) + O_2(g) \longrightarrow 2 NO_2(g)$$

exhibits an overall third-order rate law $r = k_{\text{exp}} [NO]^2 [O_2]$ and a rare negative apparent activation energy:

  • At $T_1 = 300.0\text{ K}$, $k_{\text{exp}, 1} = 7.10 \times 10^3\text{ M}^{-2}\text{s}^{-1}$.
  • At $T_2 = 600.0\text{ K}$, $k_{\text{exp}, 2} = 2.85 \times 10^3\text{ M}^{-2}\text{s}^{-1}$.

Two competing mechanisms have been proposed:

1. Mechanism I (Dimer Pre-Equilibrium):

$$2 NO \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} N_2O_2 \quad (\text{rapid pre-equilibrium, } K_1 = k_1 / k_{-1})$$
$$N_2O_2 + O_2 \xrightarrow{k_2} 2 NO_2 \quad (\text{slow, rate-determining})$$

2. Mechanism II (Collision Complex):

$$NO + O_2 \underset{k_{-3}}{\overset{k_3}{\rightleftharpoons}} NO_3^* \quad (\text{fast})$$
$$NO_3^* + NO \xrightarrow{k_4} 2 NO_2 \quad (\text{slow})$$

(a) Show that Mechanism I yields the experimental rate law under pre-equilibrium conditions, and express $k_{\text{exp}}$ in terms of $K_1$ and $k_2$. (b) Derive the relationship connecting the experimental activation energy $E_{a, \text{exp}}$ to the standard enthalpy of dimerization $\Delta H_{\text{dim}}^\circ$ and the elementary barrier $E_{a, 2}$. (c) Calculate the experimental activation energy $E_{a, \text{exp}}$ in $\text{kJ/mol}$. (d) If the elementary bimolecular step has an activation energy of $E_{a, 2} = +8.50\text{ kJ/mol}$, calculate the dimerization enthalpy $\Delta H_{\text{dim}}^\circ$ of $2 NO \rightleftharpoons N_2O_2$.

Step 1: Mechanism I rate law derivation From the rate-determining step:

$$r = \frac{1}{2} \frac{d[NO_2]}{dt} = k_2 [N_2O_2] [O_2]$$

From the rapid pre-equilibrium step:

$$K_1 = \frac{[N_2O_2]}{[NO]^2} \implies [N_2O_2] = K_1 [NO]^2$$

Substitute $[N_2O_2]$ into the rate equation:

$$r = k_2 K_1 [NO]^2 [O_2] = k_{\text{exp}} [NO]^2 [O_2]$$

where $k_{\text{exp}} = k_2 K_1$. This matches the experimental third-order rate law.

Step 2: Temperature dependence and apparent activation energy Differentiating with respect to temperature:

$$\ln k_{\text{exp}} = \ln k_2 + \ln K_1$$
$$\frac{d \ln k_{\text{exp}}}{dT} = \frac{d \ln k_2}{dT} + \frac{d \ln K_1}{dT}$$

Using Arrhenius equation for $k_2$ and van 't Hoff equation for $K_1$:

$$\frac{E_{a, \text{exp}}}{R T^2} = \frac{E_{a, 2}}{R T^2} + \frac{\Delta H_{\text{dim}}^\circ}{R T^2}$$

Multiplying by $R T^2$:

$$E_{a, \text{exp}} = E_{a, 2} + \Delta H_{\text{dim}}^\circ$$

Because dimerization is an exothermic bond-forming reaction ($\Delta H_{\text{dim}}^\circ < 0$), if $|\Delta H_{\text{dim}}^\circ| > E_{a, 2}$, the composite activation energy $E_{a, \text{exp}}$ is strictly negative!

Step 3: Calculate $E_{a, \text{exp}}$ from two-point data

$$\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_{a, \text{exp}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)$$

Evaluate terms:

$$\ln\left(\frac{2.85 \times 10^3}{7.10 \times 10^3}\right) = \ln(0.40141) = -0.91278$$
$$\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{600.0} - \frac{1}{300.0} = 1.6667 \times 10^{-3} - 3.3333 \times 10^{-3} = -1.6667 \times 10^{-3}\text{ K}^{-1}$$
$$E_{a, \text{exp}} = -\frac{-0.91278 \times 8.314462\text{ J}/(\text{mol}\cdot\text{K})}{-1.6667 \times 10^{-3}\text{ K}^{-1}} = -\frac{7.5893}{1.6667 \times 10^{-3}} = -4553.6\text{ J/mol} = -4.55\text{ kJ/mol}$$

The negative activation energy ($-4.55\text{ kJ/mol}$) explains why heating slows down the reaction.

Step 4: Calculate dimerization enthalpy $\Delta H_{\text{dim}}^\circ$ From $E_{a, \text{exp}} = E_{a, 2} + \Delta H_{\text{dim}}^\circ$:

$$\Delta H_{\text{dim}}^\circ = E_{a, \text{exp}} - E_{a, 2} = -4.55\text{ kJ/mol} - (+8.50\text{ kJ/mol}) = -13.05\text{ kJ/mol}$$

The dimerization of $NO$ into weakly bound dinitrogen dioxide ($N_2O_2$) is exothermic by $-13.1\text{ kJ/mol}$.

Solved Honors Problems & Derivations

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