Posted by: Mark Foreman | November 25, 2016

How to get the energy of activation.

In the past, yesteryear we used to use graph paper to estimate the activation energy. We would plot graphs of ln(k) against 1/T (Kelvin -1).

Now a modern student wanted to do it without drawing a graph. Given data obtained at two different temperatures lets see how we can do it.

K1 = A exp (-Ea/RT1)


K2 = A exp (-Ea/RT2)


K1 / K2 = exp (-Ea/RT1) / exp (-Ea/RT2)


ln (K1 / K2) = ln (exp (-Ea/RT1)) – ln ( exp (-Ea/RT2))

is it a false start ?

No we can now write

ln (K1 / K2) =  (-Ea/RT1) –  (-Ea/RT2)

Which can be rearranged to

ln (K1 / K2) =  -Ea {(1/RT1) –  (1/RT2)}

Then to

ln (K1 / K2) =  -Ea/R {(1/T1) –  (1/T2)}

This gives us an equation which is very similar to the Clausius–Clapeyron equation.

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