Activation Energy and Temperature Effects

Understand activation energy, collision theory, the Arrhenius equation, and how temperature affects reaction rates.

Activation Energy and Temperature Effects

Collision Theory

For reaction to occur, particles must:

Collide with proper orientation
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:

More collisions (faster molecules)
More energetic collisions (more exceed Ea)
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

📚 Practice Problems

1Problem 1easy

❓ Question:

A reaction has Ea = 75 kJ/mol. At 300 K, k = 2.0 × 10⁻³ s⁻¹. What is k at 350 K?

💡 Show Solution

Given:

Ea = 75 kJ/mol = 75,000 J/mol
T₁ = 300 K, k₁ = 2.0 × 10⁻³ s⁻¹
T₂ = 350 K, k₂ = ?
R = 8.314 J/(mol·K)

Use two-point Arrhenius:

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

Calculate 1/T₁ - 1/T₂:

$\frac{1}{300} - \frac{1}{350} = 0.003333 - 0.002857 = 0.000476 \text{ K}^{-1}$

Calculate Ea/R:

$\frac{E_a}{R} = \frac{75000}{8.314} = 9020 \text{ K}$

Calculate ln(k₂/k₁):

$\ln\left(\frac{k_2}{k_1}\right) = 9020 \times 0.000476 = 4.29$

Solve for k₂:

$\frac{k_2}{k_1} = e^{4.29} = 73.0$

$k_2 = 73.0 \times k_1 = 73.0 \times (2.0 \times 10^{-3})$

$k_2 = 0.146 \text{ s}^{-1} = 1.5 \times 10^{-1} \text{ s}^{-1}$

Answer: k₂ = 0.15 s⁻¹

Interpretation: 50°C increase → rate constant increased 73-fold!

2Problem 2medium

❓ Question:

For a reaction, k = 3.2 × 10⁻⁴ s⁻¹ at 500 K and k = 1.5 × 10⁻² s⁻¹ at 600 K. Calculate (a) Ea and (b) the frequency factor A.

💡 Show Solution

Given:

T₁ = 500 K, k₁ = 3.2 × 10⁻⁴ s⁻¹
T₂ = 600 K, k₂ = 1.5 × 10⁻² s⁻¹
R = 8.314 J/(mol·K)

(a) Calculate Ea

Use two-point form:

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

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 ✓

3Problem 3hard

❓ Question:

Draw and label an energy diagram for an exothermic reaction with Ea = 80 kJ/mol and ΔH = -50 kJ/mol. What is the activation energy for the reverse reaction?

💡 Show Solution

Given:

Exothermic: ΔH = -50 kJ/mol
Ea(forward) = 80 kJ/mol

Energy Diagram:

Energy
  ↑
  |    Transition State
  |         ↗ ↘
  |      ↗     ↘  Ea(reverse) = 130 kJ
  |   ↗           ↘
  |  ↗  Ea(fwd)     ↘
  | ↗   80 kJ        ↘
Reactants              Products
  |                    ↓ ΔH = -50 kJ
  |___________________
  |__________________|
  
  Reaction Progress →

Key points:

Reactants start at reference level
Transition state is 80 kJ above reactants
Products are 50 kJ below reactants (exothermic)

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

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