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5.1.7Physical Chemistry and Transition Elements

The Arrhenius Equation

5 min read 5 parts Lesson 8 of 40

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Section 1 of 5The Equation and Its Terms

5.1.7 The Arrhenius Equation

The Arrhenius equation is k = Ae−Eₐ/RT.

k
  • The rate constant, whose units depend on the overall order of the reaction.
A
  • The pre-exponential factor, which reflects how often collisions occur with a suitable orientation and has the same units as k.
Eₐ
  • The activation energy in J mol⁻¹, so a value given in kJ mol⁻¹ must be multiplied by 1000 before use.
R and T
  • R is the gas constant, 8.314 J K⁻¹ mol⁻¹, and T is the temperature in kelvin, never in degrees Celsius.
e−Eₐ/RT
  • This exponential term is the fraction of molecules with energy at least Eₐ, and it increases rapidly as T rises.

Section 2 of 5The Logarithmic Form and the Graph

Taking natural logarithms turns the equation into a straight line.

1
Take natural logs of both sides to give ln k = −Eₐ/(RT) + ln A.
2
Compare this with y = mx + c, so plotting ln k on the vertical axis against 1/T on the horizontal axis gives a straight line.
3
The gradient is −Eₐ/R, so the line slopes downwards and Eₐ = −gradient × R.
4
The y-intercept, found where 1/T = 0, equals ln A, so A = eintercept.
5
Using the gradient needs 1/T in K⁻¹, and values are often multiplied by a power of ten, which must be corrected in the final answer.

Section 3 of 5Using the Equation

Using the Equation
Effect of Eₐ
  • A larger activation energy gives a smaller fraction of successful collisions, so k is lower and more sensitive to temperature.
Effect of temperature
  • Raising T makes the exponent less negative, so k rises exponentially rather than in proportion to T.
Two-point method
  • Subtracting ln k₂ − ln k₁ gives ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂), which avoids drawing a graph.
Catalysts
  • A catalyst lowers Eₐ, which increases the exponential term and so raises k at the same temperature.

Section 4 of 5Worked Example

Question

  • A plot of ln k against 1/T has a gradient of −6.0 × 10³ K and a y-intercept of 25.0.
  • Calculate Eₐ and A, and the factor by which k increases from 300 K to 310 K.

  1. Eₐ = −gradient × R = 6.0 × 10³ × 8.314 = 49 884 J mol⁻¹ ≈ 49.9 kJ mol⁻¹.
  2. ln A = 25.0, so A = e²⁵ = 7.2 × 10¹⁰ (same units as k).
  3. ln(k₂/k₁) = 6000 × (1/300 − 1/310) = 6000 × 1.075 × 10⁻⁴ = 0.645.
  4. k₂/k₁ = e0.645 = 1.9.

Answer Eₐ ≈ 49.9 kJ mol⁻¹, A ≈ 7.2 × 10¹⁰, and k increases by a factor of about 1.9.

Section 5 of 5Exam Checklist & Tips

5.1.7 – Key Points for the Exam

  • Arrhenius equation k = Ae−Eₐ/RT, which relates the rate constant to temperature and activation energy.
  • Pre-exponential factor The constant A, related to collision frequency and orientation, with the same units as k.
  • Gas constant R = 8.314 J K⁻¹ mol⁻¹.
  • Natural logarithm ln, the logarithm to base e, used to linearise the Arrhenius equation.

Common Mistakes

  • Using temperature in °C instead of kelvin when evaluating 1/T or the exponent.
  • Leaving Eₐ in kJ mol⁻¹ when R is in J K⁻¹ mol⁻¹, which makes the answer wrong by a factor of 1000.
  • Forgetting the negative sign of the gradient and so quoting a negative activation energy.
  • Plotting ln k against T rather than against 1/T, which does not give a straight line.
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