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Advanced level chemistry kinetics notes: Deducing activation energy from rate data

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Advanced A Level Chemistry

Kinetics-Rates Part 6

6.1 The Arrhenius equation

Its implications and use to calculate activation energies

and the effect of temperature on the value of the rate constant


Advanced A Level Kinetics Index

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6.1 Deriving the activation energy Ea 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?

  • 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.

      • Ea = 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) – Ea/(RT)

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

  • From accurate rate data at different temperatures (e.g. 5 or 10o 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

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

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

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

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

    • ln(k) = – Ea/(RT) + ln(A)

  • 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.

    • H2(g) + I2(g) ==> 2HI(g)

    • rate = k2 [H2(g)] [I2(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 = –Ea/R

    • so, Ea = 19867 x 8.314 = 165174 J mol–1,

    • activation energy, Ea = 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 = e25.651 , 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)


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