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

Sep 21, 2017

I got $\text{4.059 kJ/mol}$, so this is an extremely fast reaction... Ordinary activation energies are in the range of $25$ to about $\text{150 kJ/mol}$.

You're saying...

$\log k = 5.4 - \frac{212}{T}$

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

$k = A {e}^{- {E}_{a} / R T}$

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:

$\ln k = \ln A - {E}_{a} / \left(R T\right)$

Since $\frac{\ln x}{\log x} \approx 2.303$, we rewrite this as:

$2.303 \log k = - {E}_{a} / \left(R T\right) + \ln A$

And thus,

$\textcolor{g r e e n}{\overline{\underline{| \stackrel{\text{ ")(" "overbrace(logk)^(y) = overbrace(-E_a/R)^(m) overbrace(1/(2.303T))^(x) + overbrace(logA)^(b)" }}{|}}}}$

So, now we can plot $\log k$ vs. $\frac{1}{2.303 T}$ in a small enough temperature range to have a slope $- {E}_{a} / R$ and a y-intercept $\log A$.

(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 $\text{K}$. For now, the y-intercept gives the frequency factor $A$:

$\log A = 5.4$

$\implies A = {10}^{5.4} = \underline{2.512 \times {10}^{5} {\text{s}}^{- 1}}$

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 $- {\text{488.14804 K}}^{- 1}$, so:

${\text{slope" = -"488.14804 K}}^{- 1} = - {E}_{a} / R$

$\implies \textcolor{b l u e}{{E}_{a}} = - R \left(- {\text{488.14804 K}}^{- 1}\right)$

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

$=$ $\underline{\textcolor{b l u e}{\text{4.059 kJ/mol}}}$

METHOD 2

Solve for the activation energy algebraically, plugging in $k = {10}^{\log k} = {10}^{5.4 - \frac{212}{T}}$:

$\frac{k}{A} = {e}^{- {E}_{a} / R T}$

$\ln \left(\frac{k}{A}\right) = - {E}_{a} / \left(R T\right)$

$\implies \textcolor{b l u e}{{E}_{a}} = - R T \ln \left(\frac{k}{A}\right)$

$= - \left(\text{0.008314472 kJ/mol"cdot"K")("400.15 K}\right) \ln \left({10}^{5.4 - \frac{212}{400.15}} / \left(2.512 \times {10}^{5}\right)\right)$

$= \text{3.327 kJ/mol} \cdot \ln \left(0.2952\right)$

$=$ $\underline{\textcolor{b l u e}{\text{4.059 kJ/mol}}}$

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.