4.1.10Unit 4. Rates, Equilibria and Further Organic Chemistry
Activation Energy and Arrhenius — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 10 of 56 Unit quiz
Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution
R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O
Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء
Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.
Section 1 of 6The Arrhenius Equation
The equation links the rate constant to temperature and to the energy barrier.
Form
- Form k = A e^(−Eₐ/RT), where A is a constant, Eₐ is the activation energy, R is the gas constant and T is the temperature in kelvin.
Exponential term
- Exponential term e^(−Eₐ/RT) represents the fraction of molecules with energy at least equal to Eₐ.
Effect of T
- Effect of T Raising T makes the exponent less negative, so k increases rapidly with temperature.
Effect of Eₐ
- Effect of Eₐ A larger activation energy gives a smaller k at the same temperature, so the reaction is slower.
Section 2 of 6Logarithmic Form
Logarithmic Form
Equation
- Equation Taking natural logs gives ln k = ln A − Eₐ/(RT).
Straight line
- Straight line This has the form y = mx + c, where y is ln k and x is 1/T.
Gradient
- Gradient The gradient of the line is −Eₐ/R, which is negative.
Intercept
- Intercept The intercept on the y-axis is ln A.
| Quantity | Value or unit |
|---|---|
| R | 8.31 J K⁻¹ mol⁻¹ |
| T | Kelvin, K |
| Eₐ | J mol⁻¹, then converted to kJ mol⁻¹ |
| Plot | ln k against 1/T |
Section 3 of 6Finding Eₐ Experimentally
1
Measure the rate constant k at several different temperatures with the other conditions kept the same.
2
Convert each temperature to kelvin and calculate 1/T, and also ln k.
3
Plot ln k on the y-axis against 1/T on the x-axis and draw the best-fit line.
4
Calculate the gradient of the line using two points far apart.
5
Eₐ = −gradient × R, converting from J mol⁻¹ to kJ mol⁻¹ by dividing by 1000.
Section 4 of 6Interpreting Temperature Effects
Rule of thumb
- Rule of thumb The rate roughly doubles for a 10 K rise in many reactions near room temperature, but this depends on Eₐ.
Comparison
- Comparison The effect of temperature is larger for reactions with a higher activation energy.
Catalysts
- Catalysts A catalyst lowers Eₐ, so the exponential term and so k increase at the same temperature.
Units
- Units If Eₐ is in kJ mol⁻¹ it must be converted to J mol⁻¹ before using R in J K⁻¹ mol⁻¹.
Section 5 of 6Worked Example
Question
- A graph of ln k against 1/T passes through (0.00320 K⁻¹, −6.30) and (0.00300 K⁻¹, −4.50).
- Calculate Eₐ.
- Gradient = (−4.50 − (−6.30)) ÷ (0.00300 − 0.00320) = 1.80 ÷ (−0.00020) = −9000 K.
- Eₐ = −gradient × R = 9000 × 8.31 = 74 790 J mol⁻¹.
- Eₐ = 74.8 kJ mol⁻¹ to three significant figures.
Answer Eₐ = 74.8 kJ mol⁻¹
Section 6 of 6Exam Checklist & Tips
4.1.10 – Key Points for the Exam
- Arrhenius equation An equation relating the rate constant to temperature and activation energy.
- Activation energy The minimum energy that colliding particles need to react.
- Gas constant R = 8.31 J K⁻¹ mol⁻¹.
- Natural logarithm A logarithm to base e, written ln.
- Pre-exponential factor The constant A, related to collision frequency and orientation.
Common Mistakes
- Using temperatures in °C instead of kelvin.
- Forgetting to convert Eₐ from J mol⁻¹ to kJ mol⁻¹ at the end.
- Losing the minus signthe gradient is −Eₐ/R, and so is negative.
- Plotting ln k against T instead of against 1/T.
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