How do you find A in Arrhenius equation?

1 Answer
Mar 21, 2016

Answer:

Plot a graph of #1/T# versus #ln(k)#. The #y#-intercept is #ln(A)#.

Explanation:

The Arrhenius equation is

#k = A e^{-E_"a"/(RT)}#,

where #A# and #E_"a"# are assumed to be independent on the temperature.

This can be rewritten as

#ln(k) = ln(A) - E_"a"/R * 1/T#

If you measured the rate constant, #k#, at 2 different temperatures, #T_1# and #T_2#, you can write

#ln(k_1) = ln(A) - E_"a"/R * 1/T_1#
#ln(k_2) = ln(A) - E_"a"/R * 1/T_2#

Solve them simultaneously and you will get

#E_"a" = frac{Rln(k_2/k_1)}{1/T_1 - 1/T_2}#

#A = e^{frac{T_2ln(k_2) - T_1ln(k_1)}{T_2 - T_1}}#

If you measured the rate constant at many different temperatures, you can plot a graph of #1/T# versus #ln(k)#. If the Arrhenius relation is true, then the graph would look linear.

Extrapolate and get the #y#-intercept of the graph. That would be #ln(A)#. You can get #A# using #A = e^{ln(A)}#.