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.

K_{1} = A exp (-E_{a}/RT_{1})

And

K_{2} = A exp (-E_{a}/RT_{2})

So

K_{1} / K_{2} = exp (-E_{a}/RT_{1}) / exp (-E_{a}/RT_{2})

Next

ln (K_{1} / K_{2}) = ln (exp (-E_{a}/RT_{1})) – ln ( exp (-E_{a}/RT_{2}))

*is it a false start ?*

No we can now write

ln (K_{1} / K_{2}) = (-E_{a}/RT_{1}) – (-E_{a}/RT_{2})

Which can be rearranged to

ln (K_{1} / K_{2}) = -E_{a }{(1/RT_{1}) – (1/RT_{2})}

Then to

ln (K_{1} / K_{2}) = -E_{a}/R {(1/T_{1}) – (1/T_{2})}

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

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