5.2.5Additional lessons
Activation Energy — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 100 of 172
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 5What Activation Energy Is
Every reaction needs an energy input to get started, even a strongly exothermic one such as burning a fuel.
Definition
- Definition Activation energy, Eₐ, is the minimum energy that colliding particles must have for a reaction to occur.
Energy barrier
- Energy barrier It acts as a barrier between reactants and products, because some bonds must begin to break before new bonds can form.
Units
- Units It is quoted in kJ mol⁻¹, which means the energy needed by one mole of reacting particles.
Size and speed
- Size and speed A reaction with a high Eₐ is slow at a given temperature, because only a tiny fraction of collisions are energetic enough.
Section 2 of 5Eₐ in Practice
Eₐ in Practice
Fuel in air
- Fuel in air Methane does not ignite at room temperature because the collisions lack energy;
- a spark supplies the activation energy to start the reaction.
Temperature
- Temperature Heating shifts more particles above Eₐ, which is why raising the temperature speeds up almost every reaction.
Catalysts
- Catalysts A catalyst gives a pathway with lower Eₐ, so a larger fraction of particles can react without any extra heating.
Exothermic reactions
- Exothermic reactions The overall energy released may be large, but the reaction still needs Eₐ first to get over the barrier.
Section 3 of 5The Arrhenius Equation (HL)
1
The Arrhenius equation links the rate constant to temperature: k = A e^(−Eₐ/RT), where A is the frequency factor and R is 8.31 J K⁻¹ mol⁻¹.
2
Taking natural logarithms gives ln k = −Eₐ/R × (1/T) + ln A, which has the form y = mx + c.
3
A graph of ln k against 1/T is therefore a straight line with gradient −Eₐ/R and intercept ln A.
4
The activation energy is found from the gradient: Eₐ = −gradient × R, and the temperature must always be in kelvin.
5
Two rate constants at two temperatures also give Eₐ through ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂).
Section 4 of 5Worked Example
Question
- The rate constant of a reaction doubles when the temperature rises from 298 K to 308 K.
- Calculate Eₐ (R = 8.31 J K⁻¹ mol⁻¹).
- ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂), with k₂/k₁ = 2, so ln 2 = 0.693.
- 1/298 − 1/308 = 3.356 × 10⁻³ − 3.247 × 10⁻³ = 1.09 × 10⁻⁴ K⁻¹.
- Eₐ = 0.693 × 8.31 / 1.09 × 10⁻⁴ ≈ 5.29 × 10⁴ J mol⁻¹.
Answer Eₐ ≈ 53 kJ mol⁻¹
Section 5 of 5Exam Checklist & Tips
5.2.5 – Key Points for the Exam
- Activation energy The minimum energy that colliding particles need in order to react, in kJ mol⁻¹.
- Frequency factor (A) A constant in the Arrhenius equation that takes account of collision frequency and orientation.
- Arrhenius equation k = A e^(−Eₐ/RT), relating the rate constant to temperature and activation energy.
- Gradient of ln k against 1/T Equal to −Eₐ/R.
Common Mistakes
- Using temperature in °C in the Arrhenius equation instead of kelvin.
- Forgetting to convert the gradient result from J mol⁻¹ to kJ mol⁻¹ by dividing by 1000.
- Believing that exothermic reactions have no activation energy.
- Thinking that Eₐ is the enthalpy change of the reaction;
- ΔH is the difference between reactants and products, not the barrier.
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