Arrhenius EquationCourse level: College Extension

The Arrhenius Equation and Temperature Dependence

What you'll be able to do: Use the Arrhenius equation in exponential, linear and two-point forms to relate the rate constant, activation energy and temperature.

Best after: Collision Theory and Activation Energy, Rate Laws and Reaction Order from Initial Rates

Introduction

Collision theory says temperature matters enormously. The Arrhenius equation makes that quantitative, and its linear form turns a set of rate constants measured at different temperatures into an activation energy.

These are recommended, not required. You can start this lesson at any time.

Learning objectives

  • Interpret each term of the Arrhenius equation
  • Determine Ea from the slope of an ln k versus 1/T plot
  • Apply the two-point Arrhenius form to find Ea or a new rate constant
  • Analyse errors caused by unit and temperature-scale mistakes

The equation and what each part means

A is the frequency factor, combining how often correctly oriented collisions occur. The exponential term e-Ea/RT is the Arrhenius energy-weighting factor for crossing the activation barrier. A large Ea or a low temperature makes that fraction tiny, and k is small.

k = A e-Ea/RT

The linear form

Taking natural logarithms produces a straight-line relationship in the variables ln k and 1/T. The slope is -Ea/R, so Ea = -R times slope, and the intercept is ln A. Note that 1/T on the horizontal axis means higher temperatures lie to the left.

Read the horizontal axis label before using its slope. For x = 1/T in K⁻¹, slope = −Ea/R. If x = 1000/T, then ln k = [−Ea/(1000R)]x + ln A, so Ea = −1000R × slope. A graph scaling mistake changes the activation energy by a factor of 1000 even if the line was fitted perfectly. In the usual positive-Ea case the slope is negative because larger 1/T means a lower temperature and a smaller k.

ln k = -(Ea/R)(1/T) + ln A

Read temperature sensitivity from the slope. Illustrative models and stated conditions apply; use the surrounding explanation for the calculation and its limits.
A steeper downward slope means a larger activation energy and therefore a more temperature-sensitive reaction.

The two-point form

Subtracting the linear equation at two temperatures eliminates ln A entirely, leaving a relationship between the two rate constants, the two temperatures and Ea. Any one unknown can be found from the other four.

To predict k₂, rearrange before calculating: k₂ = k₁ exp[(Ea/R)(1/T₁ − 1/T₂)]. With Ea = 50.0 kJ/mol, T₁ = 300 K and T₂ = 320 K, the exponent is (50000/8.314)(1/300 − 1/320) = 1.253. Thus k₂/k₁ ≈ 3.50. The positive exponent agrees with heating increasing k in this positive-Ea model. The ratio is unitless only if both rate constants use the same time and concentration units. A Celsius difference may equal a kelvin difference, but reciprocal Celsius temperatures cannot replace reciprocal kelvin temperatures.

ln(k₂/k₁) = -(Ea/R)(1/T₂ - 1/T₁)

Units, carefully

Temperature must be in kelvin, never Celsius, because the equation depends on absolute temperature. With R = 8.314 J/(mol K), Ea comes out in J/mol and usually needs dividing by 1000 to report in kJ/mol.

Mixing R = 8.314 with an Ea already in kJ/mol is the single most common numerical error in this topic.

Convert Celsius to kelvin first and decide up front whether Ea is in J/mol or kJ/mol.

Error analysis and comparison

Two reactions can share the same rate at one temperature yet respond very differently to heating: the one with the larger Ea has the steeper ln k against 1/T line and gains far more from a temperature rise. When an experimental plot curves, suspect a change of mechanism over the temperature range rather than a failure of the equation.

A curved Arrhenius plot can signal changing mechanism, changing apparent parameters or other experimental effects; it is a reason to investigate rather than fit one line blindly. Extrapolation far beyond the measured range is uncertain. Comparing temperature sensitivity at equal initial k assumes comparable, approximately constant Ea and A over that range. A larger Ea alone does not guarantee a larger k at one temperature because A also matters.

Interpret the pre-exponential factor

In k = A e-Ea/RT, A sets the multiplicative scale of the rate constant for a given Arrhenius model. If two reactions have the same Ea but different A values, their k ratio equals their A ratio at every common temperature. A embodies collision frequency and orientation within the model; it should not be mistaken for activation energy or concentration.

A has the same units as k, and represents more than a bare count of collisions in many reactions. A catalyst changes the mechanism and can change both apparent Ea and A. For the simple positive-barrier model, lowering Ea with A fixed raises k exponentially; this comparison is useful but is not a rule that every catalyst leaves A untouched. Complex mechanisms can even show an apparent negative Ea; introductory questions normally specify ordinary positive-Ea Arrhenius behavior.

At the same Ea and T, k₁/k₂ = A1/A2.

Key ideas

Definition
Frequency factor A

A constant capturing collision frequency and the fraction of collisions with acceptable orientation.

Rule
Slope gives Ea

On a plot of ln k against 1/T, Ea = -R times the slope.

Key concept
Temperature sensitivity

Reactions with large activation energies are the most sensitive to a change in temperature.

Assumption
Constant mechanism

Arrhenius behaviour assumes Ea and A do not change over the temperature range studied.

Equation
Arrhenius equation

k = A e-Ea/RT

  • A = frequency factor, same units as k
  • Ea = activation energy in J/mol when R = 8.314
  • R = 8.314 J/(mol K)
  • T = absolute temperature in K
Equation
Linear form

ln k = -(Ea/R)(1/T) + ln A

  • slope = -Ea/R
  • intercept = ln A
Equation
Two-point form

ln(k₂/k₁) = -(Ea/R)(1/T₂ - 1/T₁)

  • k₁, k₂ = rate constants at T₁ and T₂

Worked examples

Worked example 1

A reaction has k = 2.0 × 10⁻³ s⁻¹ at 300. K and k = 8.0 × 10⁻³ s⁻¹ at 320. K. Find Ea in kJ/mol.

Try it first: Write the two-point form and check that both temperatures are already in kelvin.

    0 of 4 steps revealed.

    Worked example 2

    A plot of ln k against 1/T has slope -6.50 × 10³ K. Find Ea.

      0 of 3 steps revealed.

      Worked example 3

      Reaction P has Ea = 30 kJ/mol and reaction Q has Ea = 90 kJ/mol. Both have the same k at 300 K. Which is faster at 350 K, and why?

        0 of 3 steps revealed.

        Worked example 4

        Reactions X and Y share Ea = 50 kJ/mol, but AX = 10 AY. Compare their k values at the same temperature and after both are heated.

        Try it first: The exponential factor is identical for both at each common temperature.

          0 of 3 steps revealed.

          Common mistakes

          Common mistakes and better checks
          MistakeWhy it's wrongCheck instead
          Using Celsius in the Arrhenius equation.The exponent needs absolute temperature; Celsius can even be negative, which is meaningless here.Add 273.15 to every temperature before substituting.
          Mixing R = 8.314 J/(mol K) with Ea in kJ/mol.The factor of 1000 makes the answer wrong by three orders of magnitude.Work in J/mol with R = 8.314 and convert only at the end.
          Reporting a negative activation energy from the slope.The slope is -Ea/R, so a negative slope gives a positive Ea.Multiply the slope by -R and expect a positive value.
          Assuming a catalyst can change only one Arrhenius parameter.A new catalytic mechanism can change both the apparent activation energy and the pre-exponential factor.Use the stated catalytic model; do not assume A must remain fixed.

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          What you should now know

          The Arrhenius equation k = A e-Ea/RT splits the rate constant into a frequency-and-orientation factor A and an exponential energy-weighting term associated with crossing the barrier. Taking logarithms gives ln k = -(Ea/R)(1/T) + ln A, so a plot of ln k against 1/T is a straight line of slope -Ea/R. Two rate constants at two temperatures are enough to find Ea through the two-point form, provided the temperatures are in kelvin and R is 8.314 J/(mol K), which returns Ea in joules per mole.

          • k = A e-Ea/RT separates collision frequency from the energy requirement
          • ln k versus 1/T is linear with slope -Ea/R and intercept ln A
          • The two-point form finds Ea from two rate constants
          • Always use kelvin, and keep R and Ea in matching energy units
          • Larger Ea means greater temperature sensitivity

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          Sources and further reading

          • Chemistry 2e, Section 12.5: Collision Theory
            Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 12.5
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          • AP Chemistry Course and Exam Description
            College Board · Unit 5: Kinetics
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          • General Chemistry bookshelf (LibreTexts)
            LibreTexts, University of California, Davis
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