Question

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

Answer 1

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"`.

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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.