How do you solve the Arrhenius equation for activation energy?

1 Answer
Apr 1, 2016

Generally, it can be done by graphing. The Arrhenius equation is:

#\mathbf(k = Ae^(-E_a"/"RT))#

where:

  • #k# is the rate constant, in units that depend on the rate law. For instance, if #r(t) = k[A]^2#, then #k# has units of #"M"/"s" * 1/"M"^2 = 1/("M"cdot"s")#.
  • #A# is the "pre-exponential factor", which is merely an experimentally-determined constant correlating with the frequency of properly oriented collisions.
  • #E_a# is the activation energy in units of, say, #"J"#.
  • #R# is the universal gas constant. Using #"J"#, #R = "8.314472 J/mol"cdot"K"#.
  • #T# is temperature in #"K"#.

METHOD A

To "solve for it", just divide by #A# and take the natural log.

#k/A = e^(-E_a"/"RT)#

#ln(k/A) = -(E_a)/(RT)#

#lnk - lnA = -(E_a)/(RT)#

#color(blue)(stackrel(y)overbrace(lnk) = stackrel(m)overbrace(-(E_a)/R) stackrel(x)overbrace(1/T) + stackrel(b)overbrace(lnA))#

So now, if you grab a bunch of rate constants for the same reaction at different temperatures, graphing #lnk# vs. #1/T# would give you a straight line with a negative slope.

The slope is #m = -(E_a)/R#, so now you can solve for #E_a#. You can also easily get #A# from the y-intercept.

METHOD B

Or, if you meant literally solve for it, you would get:

#ln(k/A) = -(E_a)/(RT)#

#color(blue)(E_a = -RTln(k/A))#

So knowing the temperature, rate constant, and #A#, you can solve for #E_a#.

However, since #A# is experimentally determined, you shouldn't anticipate knowing #A# ahead of time (unless the reaction has been done before), so the first method is more foolproof. Furthermore, using #k# and #T# for one trial is not very good science. It's better to do multiple trials and be more sure.