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.
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Chemical KineticsArrhenius EquationContent level: Challenge 26 min
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
Lesson
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 fraction of collisions carrying at least the activation energy. 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.
ln k = -(Ea/R)(1/T) + ln A
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.
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.
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 2
A plot of ln k against 1/T has slope -6.50 × 10³ K. Find Ea.
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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?
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Common mistakes
Using Celsius in the Arrhenius equation.
Why it's wrong: The exponent needs absolute temperature; Celsius can even be negative, which is meaningless here.
Check instead: Add 273.15 to every temperature before substituting.
Mixing R = 8.314 J/(mol K) with Ea in kJ/mol.
Why it's wrong: The factor of 1000 makes the answer wrong by three orders of magnitude.
Check instead: Work in J/mol with R = 8.314 and convert only at the end.
Reporting a negative activation energy from the slope.
Why it's wrong: The slope is -Ea/R, so a negative slope gives a positive Ea.
Check instead: Multiply the slope by -R and expect a positive value.
Believing a catalyst changes A rather than Ea.
Why it's wrong: The dominant catalytic effect is a lower activation energy through a new pathway.
Check instead: Attribute the increased k mainly to a smaller Ea in the exponential term.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
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 term giving the fraction of collisions that clear 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
Sources and further reading
- Chemistry 2e, Section 12.5: Collision TheoryPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 12.5View source
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License