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Activation Energy and graphs

If I plot ln k against 1/T
the gradient Δlnk / Δ1/T = -Ea/R

This is the same as saying Δlnk*ΔT
could somone explain how this can be -Ea/R

Reply 1

charco

Study Forum Helper

Original post by jsmith6131

If I plot ln k against 1/T
the gradient Δlnk / Δ1/T = -Ea/R

This is the same as saying Δlnk*ΔT
could somone explain how this can be -Ea/R

It's not "change in T" it's T

Arrhenius:

k = Ae^-Ea/RT

take natural logs

lnk = lnA -Ea/RT

plot lnk against 1/RT

gradient = -Ea

EDIT: Corrected as I'd missed out the lnA ...

(edited 12 years ago)

Reply 2

username459260

meh, ok

thanks

if I ploted ln k against RT
would the gradient be 1/Ea

Reply 3

charco

Study Forum Helper

Original post by jsmith6131

meh, ok

thanks

if I ploted ln k against RT
would the gradient be 1/Ea

oh why?

Why don't you plot ln(k) against 1/T. The gradient equals -Ea/R.

Reply 6

username459260

I was just wondering what the gradeint WOULD be if I plotted ln k against R or RT

Reply 7

charco

Study Forum Helper

Original post by jsmith6131

I was just wondering what the gradeint WOULD be if I plotted ln k against R or RT

The whole idea is to get an expression which can be represented by the equation of a straight line, y = mx + c, where x and y are the axes (variables) and m is the gradient.

The log form of Arrhenius conforms to this equation:

lnk = -Ea/RT + lnA

so you need to put lnk on the y axis and a variable from -Ea/RT on the x axis leaving the remainder of the -Ea/RT as the gradient.

You have not got T you have got 1/T in the function. In the same way as 2 is not the same as 1/2.

A plot involving T will not now give a straight line as to get T on the top of this function you have to multiply through by T² through the entire equation.

lnk*T² = -EaT/R + lnAT²

Now it does not conform to requirements because lnAT² is not constant.

You cannot just choose to turn one part of an equation upside down.

To go back to the original straight line, y = mx + c

If you plot y against x you get a straight line gradient m

If you plot y against 1/x do you get a straight line gradient 1/m?