Activation Energy and Temperature Effects

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

Written and reviewed by the Study Mondo Education TeamLast updated February 14, 2026

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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:

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

4Problem 4easy

❓ Question:

From an Arrhenius plot, ln(k) = −9500(1/T) + 18.2. Find (a) Ea and (b) A.

💡 Show Solution

Slope m = −9500 ⇒ Ea = −mR = 9500×8.314 = 7.90×10^4 J/mol = 79.0 kJ/mol. Intercept b = 18.2 ⇒ A = e^{18.2} ≈ 9.91×10^7 (units of s⁻¹ for first-order).

5Problem 5medium

❓ Question:

Two reactions: (I) A₁ = 2.0×10^12 s⁻¹, Ea₁ = 85 kJ/mol; (II) A₂ = 4.0×10^10 s⁻¹, Ea₂ = 65 kJ/mol. At 298 K, which is faster and by what factor?

💡 Show Solution

Compute k = A e^{−Ea/RT}. Use R = 8.314 J/(mol·K). Convert Ea to J/mol. For I: exponent = −85000/(8.314×298) = −34.3 ⇒ k₁ ≈ 2.0×10^12 × e^{−34.3} = 2.0×10^12 × 1.2×10^{−15} ≈ 2.4×10^{−3} s⁻¹. For II: exponent = −65000/(8.314×298) = −26.2 ⇒ k₂ ≈ 4.0×10^10 × e^{−26.2} = 4.0×10^10 × 4.3×10^{−12} ≈ 0.172 s⁻¹. Reaction II is faster by factor ≈ k₂/k₁ ≈ 0.172 / 2.4×10^{−3} ≈ 71.

6Problem 6medium

❓ Question:

By what temperature increase ΔT (°C) will a reaction with Ea = 95 kJ/mol approximately double its rate near 300 K?

💡 Show Solution

Use two-point form with k₂/k₁ = 2 and T₁ = 300 K, T₂ = 300 + ΔT. ln(2) = (Ea/R)(1/T₁ − 1/T₂). Solve approximately with T₂ ≈ T₁ + ΔT and small ΔT: 1/T₁ − 1/T₂ ≈ ΔT/T₁^2. Thus ln(2) ≈ (Ea/R)(ΔT/T₁^2) ⇒ ΔT ≈ ln(2)·R·T₁^2 / Ea. Plug in: ΔT ≈ 0.693×8.314×(300)^2 / 95000 ≈ (0.693×8.314×90000)/95000 ≈ (5187)/95000 ≈ 0.0546×10^2 ≈ 5.46 °C. So roughly a 5–6 °C increase doubles the rate (depends on Ea and starting T).

7Problem 7hard

❓ Question:

A catalyst lowers Ea from 120 kJ/mol to 80 kJ/mol. At 350 K, by what factor does the rate constant increase (assume A unchanged)?

💡 Show Solution

Factor = k_cat/k_uncat = e^{−Ea_cat/RT} / e^{−Ea_uncat/RT} = e^{(Ea_uncat − Ea_cat)/(RT)}. ΔEa = 40 kJ/mol = 4.0×10^4 J/mol; RT = 8.314×350 ≈ 2910 J/mol. Factor ≈ e^{40000/2910} = e^{13.7} ≈ 9.0×10^5. So k increases by ~9×10^5.

Explain using:

📋 AP Chemistry — Exam Format Guide

⏱ 3 hours 15 minutes📝 67 questions📊 3 sections

Section	Format	Questions	Time	Weight	Calculator
Multiple Choice	MCQ	60	90 min	50%	✅
Free Response (Long)	FRQ	3	69 min	30%	✅
Free Response (Short)	FRQ	4	36 min	20%	✅

📊 Scoring: 1-5

Extremely Qualified

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Well Qualified

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Qualified

~24%

Possibly Qualified

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No Recommendation

~24%

💡 Key Test-Day Tips

✓Memorize common polyatomic ions
✓Practice dimensional analysis
✓Know your gas laws

⚠️ Common Mistakes: Activation Energy and Temperature Effects

Avoid these 3 frequent errors

🌍 Real-World Applications: Activation Energy and Temperature Effects

See how this math is used in the real world

📝 Worked Example: Stoichiometry — Limiting Reagent

Problem:

$2$ mol of $H_2$ reacts with $1$ mol of $O_2$ . How many grams of water are produced? Which is the limiting reagent? ( $2H_2 + O_2 \to 2H_2O$ )

2Determine the limiting reagent

3Calculate moles of product

4Convert moles to grams

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

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Activation Energy and Temperature Effects

Try the Interactive Version!

Activation Energy and Temperature Effects

Collision Theory

Activation Energy (Ea)

Arrhenius Equation

Two-Point Form

Temperature Effects

Energy Diagrams

Catalysts

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2medium

❓ Question:

3Problem 3hard

❓ Question:

4Problem 4easy

❓ Question:

5Problem 5medium

❓ Question:

6Problem 6medium

❓ Question:

7Problem 7hard

❓ Question:

📋 AP Chemistry — Exam Format Guide

📊 Scoring: 1-5

💡 Key Test-Day Tips

⚠️ Common Mistakes: Activation Energy and Temperature Effects

🌍 Real-World Applications: Activation Energy and Temperature Effects

📝 Worked Example: Stoichiometry — Limiting Reagent

Practice Lab

Practice with Flashcards

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📌 Related Topics in Kinetics

❓ Frequently Asked Questions