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Arrhenius equation and rate of change

  1. Apr 15, 2014 #1
    I wanted to see how the rate of change of rate constant with temperature (dk/dT) changes with activation energy. I tried to do this with differentials: k=A*e-EA/RT so

    [tex]\frac{dk}{dT} = \frac{A E_A \cdot e^{-\frac{E_A}{RT}} } {RT^2}[/tex]

    and then

    [tex]\frac{d(\frac{dk}{dT})}{dE_A} = A \cdot e^{-\frac{E_A}{RT}} \cdot (\frac{1}{RT^2} - \frac{1}{R^2T^3})[/tex]

    So long as EA>RT, it appears that rate of change of rate constant with temperature would decrease for increasing activation energy (i.e. if you increase activation energy, then the rate constant is less susceptible to increasing when you change the temperature). EA>RT would be almost universally the case I imagine - at 298 K that's less than 2.5 kJ/mol.

    But this doesn't tally up with what we know, which is that for a given temperature increase, a greater activation energy leads to a greater increase in rate constant. e.g. if we write well-known

    k2/k1 = exp( -EA/R * (1/T22 - 1/T12) )

    and try some values we get this conclusion. So what went wrong or what is wrong with my idea to do it by differentiation?
  2. jcsd
  3. Apr 18, 2014 #2
    Oops I made a stupid mistake here. Sorry if this was confusing people. It should be

    k2/k1 = exp( EA/R * (1/T1 - 1/T2) )

    No squared terms.
  4. Apr 18, 2014 #3
    You differentiated with respect to EA incorrectly. Try again. The two terms in parenthesis do not have consistent units.
  5. Apr 20, 2014 #4
    Thanks, I had indeed made this mistake - the actual differential should be

    [tex]\frac{d(\frac{dk}{dT})}{d(E_A)} = \frac{A}{RT^2} \cdot e^{-\frac{E_A}{RT}} \cdot (1 - \frac{E_A}{RT})[/tex]

    But it seems that my original conclusion still holds true - that EA < RT (incredibly small activation energy) is necessary for rate of change of rate constant with temperature to increase at a higher activation energy, which however is the trend which the expression for k2/k1 indicates is true for all EA (it seems to me)?
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