Activation Energy

Overview

Activation energy ( $E_a$ ) is the minimum energy required for a chemical reaction to occur. It represents the energy barrier that reactants must overcome to form products.

Energy Diagrams

Exothermic Reaction

Energy
  ↑
  |     ╱╲ ← Transition State
  |    ╱  ╲
  | Ea↑    ╲
  |  ╱      ╲
  |─╱        ╲
  |           ╲────  ← Products
  | Reactants      ↓ ΔH < 0
  |________________→ Reaction Progress

Endothermic Reaction

Energy
  ↑
  |           ╱╲ ← Transition State
  |          ╱  ╲
  |         ╱    ╲
  |        ╱      ╲──── ← Products
  |    Ea ↑        ↑ ΔH > 0
  |      ╱
  |─────╱ Reactants
  |________________→ Reaction Progress

Key Concepts

Transition State (Activated Complex)

Highest energy point along reaction path
Unstable, short-lived species
Bonds are partially broken/formed

Relationship

E_a(\text{forward}) - E_a(\text{reverse}) = \Delta H

The Arrhenius Equation

Standard Form

k = Ae^{-E_a/RT}

Where:

$k$ = rate constant
$A$ = frequency factor (pre-exponential factor)
$E_a$ = activation energy (J/mol)
$R$ = gas constant (8.314 J/mol·K)
$T$ = temperature (K)

Logarithmic Form

\ln k = \ln A - \frac{E_a}{RT}

Or:

\log k = \log A - \frac{E_a}{2.303RT}

Determining Activation Energy

Method 1: Two-Temperature Form

\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Method 2: Graphical (Arrhenius Plot)

Plot $\ln k$ vs $1/T$ :

Slope = $-E_a/R$
Intercept = $\ln A$

Examples

Example 1: Finding $E_a$ from Two Temperatures

At 25°C (298 K), $k = 2.5 \times 10^{-3}$ s⁻¹ At 50°C (323 K), $k = 3.2 \times 10^{-2}$ s⁻¹

\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

\ln\left(\frac{3.2 \times 10^{-2}}{2.5 \times 10^{-3}}\right) = \frac{E_a}{8.314}\left(\frac{1}{298} - \frac{1}{323}\right)

\ln(12.8) = E_a \times (3.13 \times 10^{-5})

2.55 = E_a \times 3.13 \times 10^{-5}

E_a = 81,500 \text{ J/mol} = 81.5 \text{ kJ/mol}

Example 2: Finding k at New Temperature

Given: $k = 0.015$ s⁻¹ at 300 K, $E_a = 50$ kJ/mol Find k at 350 K.

\ln\left(\frac{k_2}{0.015}\right) = \frac{50,000}{8.314}\left(\frac{1}{300} - \frac{1}{350}\right)

\ln\left(\frac{k_2}{0.015}\right) = 6015 \times (4.76 \times 10^{-4}) = 2.86

\frac{k_2}{0.015} = e^{2.86} = 17.5

k_2 = 0.26 \text{ s}^{-1}

Effect of Temperature

Why Rate Increases with Temperature

Molecules move faster → more collisions
More molecules have energy ≥ $E_a$
The fraction with $E \geq E_a$ follows Boltzmann distribution

Boltzmann Factor

\text{Fraction with } E \geq E_a = e^{-E_a/RT}

Catalysis and Activation Energy

Catalyst Effect

Catalysts lower activation energy by providing an alternative reaction pathway.

Energy
  ↑
  |     ╱╲ Uncatalyzed (high Ea)
  |    ╱  ╲
  |   ╱    ╲
  |  ╱  ╱╲  ╲ Catalyzed (lower Ea)
  | ╱  ╱  ╲  ╲
  |╱__╱    ╲__╲___
  |________________→ Reaction Progress

Catalyst Properties

Lowers $E_a$ for both forward and reverse reactions
Does not change $\Delta H$
Is not consumed in the reaction
Increases rate without changing equilibrium

Types of Catalysts

Homogeneous Catalysts

Same phase as reactants Example: Acid catalysis in solution

Heterogeneous Catalysts

Different phase from reactants Example: Metal surface catalysts (Pt, Ni, Pd)

Biological Catalysts (Enzymes)

Highly specific protein catalysts Lower $E_a$ dramatically ( $10^6 - 10^{12}$ rate increase)

Frequency Factor (A)

A = pZ