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5.2.13Additional lessons

The Arrhenius Equation (HL) — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 107 of 172

📚 IB Chemistry HL Hub / Topical Revision
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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.
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Section 1 of 5The Equation

The factor e^(−Ea/RT) is the fraction of collisions with energy of at least Ea.

Exponential form
  • Exponential form k = A e^(−Ea/RT), where k is the rate constant and T is the absolute temperature.
Pre-exponential factor
  • Pre-exponential factor A, also called the Arrhenius factor, relates to collision frequency and the orientation requirement.
Activation energy
  • Activation energy Ea is in J mol⁻¹ in the equation, though values are usually quoted in kJ mol⁻¹.
Gas constant
  • Gas constant R = 8.31 J K⁻¹ mol⁻¹, which fixes the units of the exponent as dimensionless.

Section 2 of 5Logarithmic Forms

1
Take natural logs of both sides to get ln k = −Ea/(RT) + ln A.
2
This resembles y = mx + c, so plotting ln k against 1/T gives a straight line.
3
For two temperatures, subtract the equations to obtain ln(k₁/k₂) = (Ea/R)(1/T₂ − 1/T₁).
4
Use this form to find the ratio of rate constants, or to calculate Ea if both k values are known.
5
Always use temperatures in kelvin and Ea in J mol⁻¹ when R is 8.31 J K⁻¹ mol⁻¹.

Section 3 of 5Predictions from the Equation

Predictions from the Equation
Higher T
  • Higher T Increases e^(−Ea/RT), so k and the rate rise.
Larger Ea
  • Larger Ea Reduces the exponential factor, so a reaction with high Ea is slower and more sensitive to temperature.
Catalyst
  • Catalyst Lowers Ea, so k increases without any change in temperature.
Rule of thumb
  • Rule of thumb Many reactions near room temperature roughly double in rate for a 10 K rise.

Section 4 of 5Worked Example

Question

  • A reaction has Ea = 50.0 kJ mol⁻¹.
  • By what factor does k increase when T rises from 298 K to 308 K?
  • (R = 8.31 J K⁻¹ mol⁻¹)

  1. ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) = (50 000 ÷ 8.31)(1/298 − 1/308).
  2. 1/298 − 1/308 = 1.0894 × 10⁻⁴ K⁻¹, giving ln(k₂/k₁) = 6017 × 1.0894 × 10⁻⁴ = 0.656.
  3. k₂/k₁ = e0.656 = 1.93.

Answer k increases by a factor of about 1.9

Section 5 of 5Exam Checklist & Tips

5.2.13 – Key Points for the Exam

  • Arrhenius equation k = A e^(−Ea/RT), linking rate constant, temperature and activation energy.
  • Arrhenius factor The constant A that takes account of collision frequency and orientation.
  • Activation energy Minimum energy needed for a successful collision.
  • Absolute temperature Temperature in kelvin, required in all exponential rate calculations.

Common Mistakes

  • Using Ea in kJ mol⁻¹ with R in J K⁻¹ mol⁻¹ without converting.
  • Using Celsius temperatures in the Arrhenius equation.
  • Swapping T₁ and T₂ in the two-temperature equation and obtaining the wrong sign.
  • Saying that a catalyst changes the value of A rather than lowering Ea.

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