Activation energy can be determined graphically from the Ink vs 1/T graph. From the graph,
In k = ln(A\(e^{-E_a/RT}\))
In k = In A + in \(e^{-E_a/RT}\)
In k = In A + - \(E_a/RT\)
This is in the form of y = mx + c
When a graph is plotted between 1/T and ln
k, a straight line is obtained.
Slope of the line = -Ea/R
Therefore, Ea = slope × R
