If the rate constant is #"109 s"^(-1)# at #"352 K"# and #"185 s"^(-1)# at #"426 K"#, what is the activation energy for this reaction?

#E_a = "8.91 kJ/mol"#

Well, having two rate constants at two corresponding temperatures should immediately remind you of the Arrhenius equation.

#k_1 = Ae^(-E_a//RT_1)#

#k_2 = Ae^(-E_a//RT_2)#

where #A#, the frequency factor, is assumed the same at the two temperatures #T#.

As before, take the ratio and #ln#, and rearrange to obtain:

#ln(k_1/k_2) = E_a/R[1/T_2 - 1/T_1]# #" "" "bb((1))#

Alternatively, take one state and take the #ln# of that instead to obtain:

#lnk = -E_a/R 1/T + lnA# #" "" "bb((2))#

The first equation could be algebraically solved, or the second could be graphed using the above data.

METHOD 1: GRAPH

Here's the graph:

And from the slope we can easily obtain the activation energy:

#color(blue)(E_a) = -Rcdot"slope"#

#= -"0.008314472 kJ/mol"cdotcancel"K" cdot -1072 cancel"K"#

#=# #color(blue)ul"8.91 kJ/mol"#

The benefit of this first method is you cannot get the frequency factor #A# from the second method. What is the value of #A# in #"s"^(-1)#?

METHOD 2: ALGEBRA

Solve for #E_a#:

#color(blue)(E_a) = Rln(k_1/k_2) cdot [1/T_2 - 1/T_1]^(-1)#

#= "0.008314472 kJ/mol" cdot ln(109/185) cdot [1/"426 K" - 1/"352 K"]^(-1)#

#=# #color(blue)ul"8.91 kJ/mol"#