KINETICS 6.1 The Arrhenius equation - Its implications and use to calculate activation energies and the relate the effect of temperature on the value of the rate constant

[Author ©  Dr Phil Brown PhD: Doc Brown's exam revision notes suitable for students of advanced pre-university/college level theoretical-physical chemistry courses: rates of reaction factors: KINETICS - using the Arrhenius equation [kinetics page updated Mar 13th 2026 *]

6.1 Deriving the activation energy E_a from kinetics–rates data

• CAN YOU DETERMINE THE VALUE OF THE ACTIVATION ENERGY FOR A GIVEN REACTION?

• IF SO, WHAT DATA DO WE NEED? AND HOW DO WE DO THE CALCULATION?

• (a) The Arrhenius equation

  • The Arrhenius equation quantitatively describes the relationship between the rate constant k, temperature and the activation energy.

  • The rate constant value increases with increase in temperature and nothing else varies it!

  • k = A exp^(–Ea/RT) 

    • k = rate constant (from the rate expression) e.g. rate = k [A]^a [B]^b etc.

    • A = a constant for a given reaction, sometimes called the 'frequency' or 'pre–exponential' factor and it seems to be linked to stereochemical factors i.e. the spatial aspects of a reactant particle collision.

    • E_a = activation energy in Jmol^–1 

    • R = ideal gas constant = 8.314 Jmol^–1K^–1 

    • T = temperature in K (Kelvin = ^oC + 273)

• Rewriting the Arrhenius equation in logarithmic form gives:

  • ln(k) = ln(A) – E_a/(RT)

  • or log₁₀(k) = log₁₀(A) – E_a/(2.303RT)

• (b) What data do you need to calculate the activation energy from the Arrhenius equation?

• From accurate rate data at different temperatures (e.g. 5 or 10^o intervals and a minimum of four results) you can calculate the values of k OR more simply, for a fixed 'recipe', a set of 'rate' results.

• You then plot  the value of ln(k or relative rate) versus the reciprocal temperature in Kelvin. 

• Table of some rate data for the reaction between hydrogen and iodine

• Temp./K	1/T	k	ln(k)
  556	0.001799	0.0000445	-10.02
  575	0.001739	0.0001370	-8.896
  629	0.001589	0.0025200	-5.983
  666	0.001502	0.0014100	-4.262
  700	0.001429	0.0643000	-2.744

• The negative gradient of the graph is equal to –E_a/R,

• so, E_a = –R x gradient (in J, and /1000 => kJ).

  • In terms of y = mx + c, m = gradient = dy/dx = E_a/R, c = a constant = ln(A),

  • so, in terms of the Arrhenius equation, the algebra equates to

  • ln(k) = – E_a/(RT) + ln(A)

(c) Example calculation for 6.1 to determine the activation energy of a chemical reaction

• Some accurate rate constant data for various temperatures are tabulated above for the reaction between hydrogen and iodine to form hydrogen iodide.

• H_2(g) + I_2(g) ==> 2HI_(g)

• rate = k₂ [H_2(g)] [I_2(g)]

• The Excel Arrhenius plot of ln(k) versus 1/T ...

• The Excel plot routine conveniently provided the best line fit.

• y = –19867 + 25.651

• therefore gradient = –19867 = –E_a/R

• so activation energy E_a = 19867 x 8.314 = 165174 J mol^–1

• activation energy, E_a = 165 kJ mol^–1 (3 sf, 165.2 4sf)

• The constant A can be calculated as follows:

• The constant 25.651 = ln(A), therefore A = e^25.651 , so A = 1.325 x 10¹¹,

• so the full Arrhenius expression for the hydrogen iodine reaction is

• k = Ae^(Ea/RT) = 1.325 x 10¹¹ x e^{(165174/8.314T)}