What is the activation energy for a reaction with a frequency factor of #1.2 xx 10^13 "s"^(-1)# and a rate constant of #5.9 xx 10^(-2) "s"^(-1)# at #31.00^@ "C"#?

1 Answer
Truong-Son N.
Oct 6, 2017

#E_a = "82.30 kJ/mol"#

This is a usual reaction. Nothing really fast, but not that slow either.

The relation between the rate constant and temperature is something you should be getting to know really well at this point:

#k = Ae^(-E_a//RT)#

(what is the equation called?)

  • #k# is the rate constant.
  • #A# is the frequency factor, which has the same units as the rate constant.
  • #E_a# is the activation energy, usually in #"kJ/mol"#.
  • #R = "0.008314472 kJ/mol"cdot"K"# is the universal gas constant that ensures the units cancel in the exponential.
  • #T# is the temperature in #"K"#.

And so, given:

  • #A = 1.2 xx 10^13 "s"^(-1)# is the frequency factor for this first-order process (how did I know it was first order?)
  • #k = 5.9 xx 10^(-2) "s"^(-1)# is the rate constant for this first-order process at #31.00 + 273.15 = "304.15 K"#.

That's enough info to solve for the activation energy. Take the #ln# of both sides:

#lnk = lnA - E_a/(RT)#

Then solve for #E_a#.

#E_a/(RT) = lnA - lnk#

Thus, the activation energy is:

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

#= "0.008314472 kJ/mol"cdotcancel"K" cdot 304.15 cancel"K" cdot ln((1.2 xx 10^13 cancel("s"^(-1)))/(5.9 xx 10^(-2) cancel("s"^(-1))))#

#=# #"2.5284 kJ/mol" cdot 32.946#

#=# #ulcolor(blue)("82.30 kJ/mol")#

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