Activation Energy and Temperature Effects

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

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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: