13.3 Collision Theory, Activation Energy & the Arrhenius Equation

13.3 Collision Theory, Activation Energy & the Arrhenius Equation

Quick Summary: Collision theory explains reaction rates through three essential criteria: particles must collide, possess kinetic energy at or above the activation energy ($E_a$), and strike with correct steric orientation. The Maxwell-Boltzmann distribution shows that the fraction of molecules with $E \ge E_a$ is $f = e^{-E_a / RT}$. Reaction energy profiles map potential energy changes from reactants through the activated complex (transition state) to products, where $\Delta H = E_{a,\text{fwd}} - E_{a,\text{rev}}$. The Arrhenius equation ($k = A e^{-E_a / RT}$) and its linear plot ($\ln k$ vs $1/T$, $\text{slope} = -E_a/R$) quantify the exponential temperature dependence of reaction rate constants.

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1. Postulates of Collision Theory

Collision theory provides a microscopic mechanical explanation for macroscopic reaction rates, particularly in gas and solution phases. The theory establishes three mandatory criteria for an effective chemical transformation:

1. Molecular Contact: Reacting particles must physically collide. In a typical gas at standard conditions, molecules undergo approximately $10^{30}\text{ collisions}/(\text{L}\cdot\text{s})$. If every collision yielded products, all chemical reactions would reach completion in fractions of a microsecond. In reality, only a minute fraction of collisions results in a reaction.
2. Activation Energy Threshold ($E \ge E_a$): Colliding species must possess a minimum quantity of kinetic energy called the activation energy ($E_a$). This kinetic energy is converted upon impact into potential energy to overcome mutual electron cloud repulsions and stretch, distort, and break existing chemical bonds.
3. Proper Steric Orientation: Molecules are not point particles; they possess distinct geometric shapes and localized electron densities. The colliding molecules must strike each other with a spatial orientation that brings the reacting atoms into direct contact with appropriate orbital overlap.

The Steric Factor ($p$)

The frequency factor $A$ in kinetics incorporates both the collision frequency ($z$) and a dimensionless steric factor ($p$): $A = p \times z$ For simple spherical atoms or ions, $p \approx 1$. For complex polyatomic molecules, $p$ can be $10^{-2}$ to $10^{-6}$, because only a narrow envelope of collision angles permits productive bond formation. For example, in the reaction $\text{NO}(g) + \text{O}_3(g) \longrightarrow \text{NO}_2(g) + \text{O}_2(g)$, an effective reaction requires the nitrogen atom of NO to strike an oxygen atom of ozone. Collisions between the oxygen atom of NO and ozone produce rebound without reaction.

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2. Maxwell-Boltzmann Distribution and Temperature Effects

The kinetic energies of molecules in a gas or liquid sample follow a Maxwell-Boltzmann distribution curve. At any given absolute temperature ($T$), molecules exhibit a wide range of velocities and kinetic energies.

The Fraction of Reactive Collisions ($f$)

According to statistical mechanics, the fraction ($f$) of molecular collisions possessing kinetic energy equal to or greater than the activation energy threshold ($E_a$) is given by the exponential factor: $f = e^{-E_a / RT}$ where:

• $E_a$ is the activation energy in $\text{J/mol}$.
• $R = 8.314\text{ J}/(\text{mol}\cdot\text{K})$ is the universal gas constant.
• $T$ is absolute temperature in Kelvin ($\text{K}$).

Why Temperature Accelerates Reaction Rates Dramatically

When temperature rises from $T_1$ to $T_2$:

1. The average molecular speed increases by only a modest factor (proportional to $\sqrt{T}$), increasing collision frequency by roughly $1\%$ to $2\%$ for a 10 °C rise.
2. Crucially, the Maxwell-Boltzmann distribution curve broadens and shifts toward higher energies. This causes an exponential increase in the area under the curve past $E_a$ in the high-energy tail.
3. Consequently, the fraction of molecules with $E \ge E_a$ ($f = e^{-E_a / RT}$) expands dramatically. For reactions with typical activation energies around $50\text{ kJ/mol}$, a 10 °C temperature increase near room temperature doubles the reaction rate constant.

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3. Reaction Energy Profiles & Transition State Theory

A reaction energy profile (or reaction coordinate diagram) plots potential energy along the progress of the reaction.

Structure of the Energy Profile

• Reactants: Initial potential energy level of the starting materials.
• Transition State (Activated Complex): The transient, high-energy arrangement of atoms located at the absolute apex of the potential energy barrier. In the transition state, original bonds are partially broken while new bonds are partially formed. The transition state cannot be isolated because it represents an unstable energy maximum with a lifetime on the order of molecular vibrations (~10^{-13} s).
• Forward Activation Energy ($E_{a,\text{fwd}}$): The potential energy difference between the transition state and the reactants: $E_{a,\text{fwd}} = E_{\text{transition state}} - E_{\text{reactants}}$
• Reverse Activation Energy ($E_{a,\text{rev}}$): The potential energy difference between the transition state and the products: $E_{a,\text{rev}} = E_{\text{transition state}} - E_{\text{products}}$
• Reaction Enthalpy ($\Delta H_{\text{rxn}}$): The net thermodynamic energy change of the reaction: $\Delta H_{\text{rxn}} = E_{a,\text{fwd}} - E_{a,\text{rev}} = E_{\text{products}} - E_{\text{reactants}}$

Exothermic vs. Endothermic Profiles

• Exothermic Reaction ($\Delta H < 0$): Products lie lower in potential energy than reactants ($E_{\text{products}} < E_{\text{reactants}}$). Consequently, $E_{a,\text{fwd}} < E_{a,\text{rev}}$.
• Endothermic Reaction ($\Delta H > 0$): Products lie higher in potential energy than reactants ($E_{\text{products}} > E_{\text{reactants}}$). Consequently, $E_{a,\text{fwd}} > E_{a,\text{rev}}$.

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4. The Arrhenius Equation: Linear and Two-Point Forms

In 1889, Svante Arrhenius synthesized collision theory and thermal distributions into the fundamental equation of chemical kinetics: $k = A e^{-E_a / RT}$

Linearization and Graphical Determination of $E_a$

Taking the natural logarithm of both sides: $\ln k = -\frac{E_a}{RT} + \ln A \implies \ln k = \left(-\frac{E_a}{R}\right)\left(\frac{1}{T}\right) + \ln A$ Comparing this relationship to $y = mx + b$ demonstrates:

• x-axis variable: Reciprocal temperature, $\frac{1}{T}$ (with $T$ in Kelvin, units $\text{K}^{-1}$).
• y-axis variable: Natural logarithm of the rate constant, $\ln k$.
• Slope: $\text{Slope} = -\frac{E_a}{R} \implies E_a = -\text{Slope} \times R$.
• y-Intercept: $\text{Intercept} = \ln A$.

Because $E_a$ and $R$ are positive values, the slope of an Arrhenius plot is always negative. A steeper downward slope corresponds to a larger activation energy, meaning the reaction is more sensitive to temperature fluctuations.

The Two-Point Arrhenius Equation

When evaluating rate constants $k_1$ and $k_2$ at two absolute temperatures $T_1$ and $T_2$: $\ln k_2 - \ln k_1 = \left(-\frac{E_a}{R}\frac{1}{T_2} + \ln A\right) - \left(-\frac{E_a}{R}\frac{1}{T_1} + \ln A\right)$ $\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

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5. Worked Quantitative Example: Two-Point Arrhenius Calculation

Problem: In an illustrative data set, a second-order gas-phase decomposition has a rate constant of $k_1 = 0.544\text{ M}^{-1}\cdot\text{s}^{-1}$ at $592\text{ K}$ and $k_2 = 2.44\text{ M}^{-1}\cdot\text{s}^{-1}$ at $627\text{ K}$. Calculate the activation energy ($E_a$) of the reaction in $\text{kJ/mol}$.

Step 1: Compute the temperature reciprocals and difference $\frac{1}{T_1} = \frac{1}{592\text{ K}} = 1.6892 \times 10^{-3}\text{ K}^{-1}$ $\frac{1}{T_2} = \frac{1}{627\text{ K}} = 1.5949 \times 10^{-3}\text{ K}^{-1}$ $\left(\frac{1}{T_2} - \frac{1}{T_1}\right) = 1.5949 \times 10^{-3}\text{ K}^{-1} - 1.6892 \times 10^{-3}\text{ K}^{-1} = -9.43 \times 10^{-5}\text{ K}^{-1}$

Step 2: Calculate the natural logarithm of the rate constant ratio $\ln\left(\frac{k_2}{k_1}\right) = \ln\left(\frac{2.44}{0.544}\right) = \ln(4.4853) = 1.5008$

Step 3: Solve for $E_a$ using $R = 8.314\text{ J}/(\text{mol}\cdot\text{K})$ $1.5008 = -\frac{E_a}{8.314\text{ J}/(\text{mol}\cdot\text{K})}\left(-9.43 \times 10^{-5}\text{ K}^{-1}\right)$ $1.5008 = E_a \times \left(1.1342 \times 10^{-5}\text{ mol/J}\right)$ $E_a = \frac{1.5008}{1.1342 \times 10^{-5}\text{ mol/J}} = 132320\text{ J/mol} = 132\text{ kJ/mol}$ The activation energy for this data set is $132\text{ kJ/mol}$. Notice that a rate constant rising about 4.5-fold over only 35 K signals a large $E_a$.