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1.11.2Physical Chemistry

The Arrhenius Equation

5 min read 6 parts Lesson 59 of 63

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Section 1 of 6Meaning of the Equation

1.11.2 The Arrhenius Equation

The equation links the rate constant to temperature and activation energy: k = Ae^(−Ea/RT).

k
  • The rate constant, which changes with temperature and with the use of a catalyst but not with concentration.
A
  • The Arrhenius (pre-exponential) constant, linked to how often collisions occur with a suitable orientation;
    • it is treated as constant for a given reaction.
Ea
  • The activation energy in J mol⁻¹, the minimum energy per mole that colliding particles need for a reaction to occur.
R
  • The gas constant, 8.314 J K⁻¹ mol⁻¹, so energy must be in joules, not kilojoules, when substituting.
T
  • The absolute temperature in kelvin; converting from °C by adding 273 is essential before substituting.
e^(−Ea/RT)
  • The fraction of collisions that have at least the activation energy, which grows rapidly as T increases.

Section 2 of 6Taking Logarithms

The logarithmic form turns the exponential into a straight-line graph that is easy to use.

1
Start with k = Ae^(−Ea/RT), an exponential form that is awkward to rearrange directly.
2
Take natural logarithms of both sides to obtain ln k = ln A − Ea/RT.
3
Rewrite this as ln k = −(Ea/R)(1/T) + ln A, which matches the straight-line form y = mx + c.
4
Plotting ln k on the y-axis against 1/T on the x-axis gives a straight line with a negative gradient.
5
The gradient equals −Ea/R, so Ea = −gradient × R, and the intercept on the y-axis equals ln A.

Section 3 of 6Effects on k

Effects on k
Raising T
  • Higher temperature makes Ea/RT smaller, so e^(−Ea/RT) and therefore k both increase.
Raising Ea
  • A larger activation energy means a smaller fraction of collisions succeed, so k falls and the reaction is slower.
Catalyst
  • A catalyst provides a route with lower Ea, which increases k at the same temperature.
Sensitivity
  • Reactions with a high Ea show a bigger percentage increase in k for the same temperature rise.

Section 4 of 6Using the Arrhenius Plot

Straight line
  • ln k against 1/T is linear because ln A and Ea/R are constants over the temperature range studied.
Gradient
  • Read from two points far apart on the line, then Ea = −gradient × 8.314, giving joules per mole.
Intercept
  • The value of ln k when 1/T is zero equals ln A, so A = e^(intercept); this is found by extrapolation.
Units
  • Using 1/T in K⁻¹ means the gradient has units of kelvin, which converts to J mol⁻¹ after multiplying by R.

Section 5 of 6Worked Example

Question

  • The rate constant of a reaction is 2.0 × 10⁻⁴ s⁻¹ at 300 K and 1.5 × 10⁻³ s⁻¹ at 320 K.
  • Calculate Ea, using R = 8.314 J K⁻¹ mol⁻¹.

  1. Write the difference of two logarithmic equations: ln k₂ − ln k₁ = (Ea/R)(1/T₁ − 1/T₂).
  2. ln(1.5 × 10⁻³ / 2.0 × 10⁻⁴) = ln 7.5 = 2.015.
  3. 1/300 − 1/320 = 3.333 × 10⁻³ − 3.125 × 10⁻³ = 2.083 × 10⁻⁵ K⁻¹.
  4. Ea = 2.015 × 8.314 / 2.083 × 10⁻⁵ = 8.04 × 10⁴ J mol⁻¹.

Answer Ea ≈ 80 kJ mol⁻¹ (80.4 kJ mol⁻¹).

Section 6 of 6Exam Checklist & Tips

1.11.2 – Key Points for the Exam

  • Arrhenius equation k = Ae^(−Ea/RT), relating the rate constant to temperature and activation energy.
  • Activation energy The minimum energy that particles must have when they collide for a reaction to take place.
  • Rate constant
    • The proportionality constant in the rate equation;
      • it depends on temperature and catalyst.
  • Arrhenius plot A graph of ln k against 1/T whose gradient is −Ea/R.

Common Mistakes

  • Using Ea in kJ mol⁻¹ with R in J K⁻¹ mol⁻¹ without converting to joules first.
  • Using temperature in °C instead of kelvin in the equation or when finding 1/T.
  • Forgetting the minus sign, so Ea comes out negative when found from the gradient.
  • Saying the rate constant depends on concentration; it does not.
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