The rate constant of a certain reaction is given by #log_(10)k = 5.4 - 212/T#. Calculate the activation energy at #127^@ "C"#?

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Sep 21, 2017

I got #"4.059 kJ/mol"#, so this is an extremely fast reaction... Ordinary activation energies are in the range of #25# to about #"150 kJ/mol"#.


You're saying...

#log k = 5.4 - 212/T#

is the expression for some rate constant #k# as a function of temperature #T#. Well, recall the Arrhenius equation:

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

where #A# is the frequency factor, #R# is the universal gas constant, and #E_a# is the activation energy.

You can see that we do not know #A#. One can obtain #A# as follows. Take the #ln# of both sides of the equation:

#lnk = lnA - E_a/(RT)#

Since #(lnx)/(logx) ~~ 2.303#, we rewrite this as:

#2.303logk = -E_a/(RT) + lnA#

And thus,

#color(green)(barul(|stackrel(" ")(" "overbrace(logk)^(y) = overbrace(-E_a/R)^(m) overbrace(1/(2.303T))^(x) + overbrace(logA)^(b)" ")|))#

So, now we can plot #log k# vs. #1/(2.303T)# in a small enough temperature range to have a slope #-E_a/R# and a y-intercept #logA#.

(A large temperature range would show the curved, inverse relationship of #k# with #T#, which would only inconvenience us.)

This gives us a general way to determine the activation energy given a particular temperature in #"K"#. For now, the y-intercept gives the frequency factor #A#:

#log A = 5.4#

#=> A = 10^(5.4) = ul(2.512 xx 10^5 "s"^(-1))#

Now, we can solve for the activation energy in two ways... either by the slope in the graph, or by solving the original Arrhenius equation.

METHOD 1

The slope is #-"488.14804 K"^(-1)#, so:

#"slope" = -"488.14804 K"^(-1) = -E_a/R#

#=> color(blue)(E_a) = -R(-"488.14804 K"^(-1))#

#= -"0.008314472 kJ/mol"cdotcancel"K" cdot -488.14804 cancel("K"^(-1))#

#=# #ulcolor(blue)("4.059 kJ/mol")#

METHOD 2

Solve for the activation energy algebraically, plugging in #k = 10^(logk) = 10^(5.4 - 212/T)#:

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

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

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

#= -("0.008314472 kJ/mol"cdot"K")("400.15 K"")ln(10^(5.4 - 212/(400.15))/(2.512 xx 10^5))#

#= "3.327 kJ/mol" cdot ln(0.2952)#

#=# #ulcolor(blue)("4.059 kJ/mol")#

Either way, we got the same thing.

This illustrates that the activation energy is HARDLY a function of temperature, especially because we considered a small enough temperature range that the dependence of #k# on #T# was linear.

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