5.2.4Kinetics
The Arrhenius Equation — AP CHEMISTRY Revision Notes & Topical Questions
5 min read 5 parts Lesson 16 of 16 Unit quiz
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.
Part 1 of 5Why Temperature Matters
A small rise in temperature can multiply a rate constant many times because it changes the fraction of energetic collisions.
Why Temperature Matters
Energy distribution
- Energy distribution At higher temperature the distribution of molecular energies spreads out and a larger tail lies above the activation energy.
Collision frequency
- Collision frequency The frequency of collisions rises only slightly with temperature, so it is not the main cause of the increase.
Exponential effect
- Exponential effect The fraction of molecules with enough energy changes exponentially, which is why rates are so sensitive to temperature.
Rate constant
- Rate constant Temperature changes k, not the concentrations, so it changes the rate law constant itself.
Part 2 of 5The Equation and Its Forms
The relationship can be written in exponential, logarithmic and two-point forms.
Exponential form
- Exponential form k = Ae^(−Eₐ/RT), where A is the frequency factor, Eₐ the activation energy, R = 8.314 J mol⁻¹ K⁻¹ and T in kelvin.
Logarithmic form
- Logarithmic form ln k = −(Eₐ/R)(1/T) + ln A, which has the form of a straight line y = mx + c.
Graph
- Graph Plotting ln k against 1/T gives a gradient of −Eₐ/R and an intercept of ln A.
Two-point form
- Two-point form ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂) lets you compare rate constants at two temperatures.
Units
- Units Use Eₐ in J mol⁻¹ with R in J mol⁻¹ K⁻¹; convert kJ to J before substituting.
| Quantity | Effect on k |
|---|---|
| Higher temperature | k increases |
| Higher activation energy | k decreases |
| Catalyst (lower Eₐ) | k increases |
Part 3 of 5Reading the Graph
Experimental values of k at several temperatures give the activation energy graphically.
1
Measure the rate constant k at several known temperatures, keeping all other conditions the same.
2
Convert each temperature to kelvin, then calculate 1/T and ln k for every data point.
3
Plot ln k on the vertical axis against 1/T on the horizontal axis and draw the best-fit straight line.
4
The line slopes downwards, so the gradient is negative and equal to −Eₐ/R.
5
Multiply the magnitude of the gradient by 8.314 to obtain Eₐ in J mol⁻¹.
Part 4 of 5Worked Example
Question
- A reaction's rate constant doubles when the temperature rises from 300 K to 310 K.
- Estimate the activation energy.
- Use ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂) with k₂/k₁ = 2, so ln 2 = 0.6931.
- 1/300 − 1/310 = 3.333 × 10⁻³ − 3.226 × 10⁻³ = 1.075 × 10⁻⁴ K⁻¹.
- Eₐ = R ln2 ÷ (1/T₁ − 1/T₂) = 8.314 × 0.6931 ÷ 1.075 × 10⁻⁴ ≈ 5.36 × 10⁴ J mol⁻¹.
Answer Eₐ ≈ 53.6 kJ mol⁻¹
Part 5 of 5Exam Checklist & Tips
5.2.4 – Key Points for the Exam
- Activation energy The minimum energy needed for a collision to lead to reaction.
- Frequency factor, A A constant related to collision frequency and the orientation requirement.
- Arrhenius plot A graph of ln k against 1/T whose gradient is −Eₐ/R.
- Rate constant The proportionality constant in the rate law, which depends on temperature.
- Gas constant, R 8.314 J mol⁻¹ K⁻¹ when working in joules.
Common Mistakes
- Using temperature in °C instead of kelvin in the equation.
- Substituting Eₐ in kJ mol⁻¹ while R is in J mol⁻¹ K⁻¹.
- Reading the gradient of the plot as Eₐ itself, instead of −Eₐ/R.
- Thinking that a higher temperature changes the activation energy of the reaction.
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