For reaction to occur, particles must:

1. Collide with proper orientation
2. Have sufficient energy to break bonds

Not all collisions lead to reaction:

• Most collisions lack enough energy
• Wrong orientation → no reaction

Activation Energy (Ea)

Activation energy: Minimum energy needed for reaction

Energy diagram:

• Reactants at lower energy
• Transition state (activated complex) at peak
• Products at final energy
• Ea = barrier height from reactants

Key points:

• Higher Ea → slower reaction
• Lower Ea → faster reaction
• Catalysts lower Ea

Arrhenius Equation

Temperature dependence of k:

$k = Ae^{-E_a/RT}$

Where:

• k = rate constant
• A = frequency factor (collision frequency × orientation factor)
• Ea = activation energy (J/mol)
• R = 8.314 J/(mol·K)
• T = temperature (Kelvin)

Linear form:

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

Plot ln(k) vs 1/T: Straight line, slope = -Ea/R

Two-Point Form

Compare k at two temperatures:

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

Useful for:

• Finding Ea from two k values at different T
• Predicting k at new temperature

Temperature Effects

General rule: Rate roughly doubles for every 10°C increase

Why higher T increases rate:

1. More collisions (faster molecules)
2. More energetic collisions (more exceed Ea)
3. Effect 2 dominates

Exponential dependence: Small T change → large k change

Energy Diagrams

Exothermic reaction:

• Products lower than reactants
• Releases energy
• ΔH < 0

Endothermic reaction:

• Products higher than reactants
• Absorbs energy
• ΔH > 0

Both have Ea barrier:

• Ea,forward for reactants → products
• Ea,reverse for products → reactants
• Ea,forward + ΔH = Ea,reverse (endothermic)
• Ea,forward - ΔH = Ea,reverse (exothermic)

Catalysts

Catalyst: Substance that increases rate without being consumed

How catalysts work:

• Provide alternative pathway with lower Ea
• Does NOT change ΔH
• Does NOT change equilibrium position
• Increases both forward and reverse rates equally

Types:

• Homogeneous: Same phase as reactants
• Heterogeneous: Different phase (often solid surface)
• Enzymes: Biological catalysts

Example: Decomposition of H₂O₂

• Uncatalyzed: slow
• With MnO₂: fast
• With catalase (enzyme): very fast

Calculate ln(k₂/k₁):

$\frac{k_2}{k_1} = \frac{1.5 \times 10^{-2}}{3.2 \times 10^{-4}} = 46.9$

$\ln(46.9) = 3.85$

Calculate 1/T₁ - 1/T₂:

$\frac{1}{500} - \frac{1}{600} = 0.00200 - 0.00167 = 0.000333 \text{ K}^{-1}$

Solve for Ea:

$3.85 = \frac{E_a}{8.314} \times 0.000333$

$E_a = \frac{3.85 \times 8.314}{0.000333} = 96,100 \text{ J/mol}$

$E_a = 96 \text{ kJ/mol}$

Answer (a): Ea = 96 kJ/mol

(b) Calculate A

Use Arrhenius equation at either temperature (use T₁ = 500 K):

$k_1 = Ae^{-E_a/RT_1}$

$A = k_1 e^{E_a/RT_1}$

Calculate Ea/(RT₁):

$\frac{E_a}{RT_1} = \frac{96100}{8.314 \times 500} = \frac{96100}{4157} = 23.1$

Calculate A:

$A = (3.2 \times 10^{-4}) \times e^{23.1}$

$e^{23.1} = 1.1 \times 10^{10}$

$A = 3.5 \times 10^6 \text{ s}^{-1}$

Answer (b): A = 3.5 × 10⁶ s⁻¹

Check with T₂: k₂ = (3.5 × 10⁶)e^(-96100/(8.314×600)) = (3.5 × 10⁶)e^(-19.25) = (3.5 × 10⁶)(4.3 × 10⁻⁹) = 1.5 × 10⁻² ✓

Calculate Ea(reverse):

Relationship for exothermic:

$E_{a,\text{reverse}} = E_{a,\text{forward}} - \Delta H$

But ΔH = -50 kJ (negative), so:

$E_{a,\text{reverse}} = 80 - (-50) = 80 + 50 = 130 \text{ kJ/mol}$

Or think of it as:

• Forward: reactants → transition state = +80 kJ
• Total drop: reactants → products = -50 kJ
• Reverse: products → transition state = +130 kJ

Answer: Ea(reverse) = 130 kJ/mol

General relationships:

Exothermic (ΔH < 0):

• Ea,reverse > Ea,forward
• Ea,reverse = Ea,forward + |ΔH|

Endothermic (ΔH > 0):

• Ea,forward > Ea,reverse
• Ea,forward = Ea,reverse + ΔH

Always: $E_{a,\text{forward}} - E_{a,\text{reverse}} = \Delta H$

If catalyst added:

• Both Ea,forward and Ea,reverse decrease by same amount
• ΔH unchanged (path doesn't affect thermodynamics)
• Transition state lower, but products/reactants same