# How do you solve the Arrhenius equation for activation energy?

##### 1 Answer

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

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

#k# is therate 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 theactivation energyin units of, say,#"J"# .#R# is theuniversal gas constant. Using#"J"# ,#R = "8.314472 J/mol"cdot"K"# .#T# istemperaturein#"K"# .

**METHOD A**

To "solve for it", just divide by

#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

The slope is

**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

However, since