# Can you experimentally determine activation energy if the rate expression and rate constant are unknown?

When drawing a graph to find the activation energy of a reaction, is it possible to use ln(1/time taken to reach certain point) instead of ln(k), as k is proportional to 1/time?

For example, if an experiment I'm doing has a specific end point, can I use ln(1/time taken to reach that endpoint) instead of ln(k), and do this for a range of temperatures in order to work out the activation energy? Or would I have to use the rate equation and determine the rate constant for each temperature, as I am not sure of the rate equation for the reaction.

Thank you:)

• Well, of course you may plug random combinations of anything into the equations instead of anything else, just don't expect the result to be very useful. Do you know how the concentration changes with time in a (say) 2nd order rate law? Try treating that expression as you suggest. Will the resulting graph turn into a straight line? See where this gets you. – Ivan Neretin Aug 25 '17 at 19:57

What you need for calculating $E_a$ or $\Delta H^\ddagger$ and $\Delta S^\ddagger$ are rate constants at different temperatures. So the real question here is how to get the rate constant from a, what you call "specific end point". I am assuming here that you are talking about reactant/product concentrations at a specific time point and you also know the the starting concentrations.

This is in principle, at least for simple lower order reactions, possible, but only if you know the order of the reaction. How to do it depends on the order of reaction and the setup.

For example for zero order:

$A \rightarrow B$

$[A]_t=[A]_0-kt$

If we got 10% left after 90 seconds and we started with a concentration of 1 M this means:

$0.1=1-90 k$

which means:

$k=0.01 M/s$

Or for a second order reaction following $2A \rightarrow B$:

$\frac{1}{[A]_t}-\frac{1}{[A]_0}=kt$

Let's say again that after 90 seconds 10% of A was left and we started with 1M:

$\frac{1}{0.1}-\frac{1}{1}=90k$

$k=0.1M^{-1}s^{s-1}$

Please note that this is very close related to half-lifes of reactions. You might want to look into how to calculate those, the reverse process is how you can calculate the rate constant knowing the conversion and the starting concentration.

This is valid only provided that the reaction is strictly first-order and provided that the “specific end-point” is a fixed concentration ratio. Note that the slope of $\ln(t_\text{end})$ vs. $\frac{1}{T}$ is $\frac{E_a}{R}$ but the intercept is not $\ln(A)$.

If the rection is first order ( $X {\xrightarrow{k}} Y$) then the product concentration at time t is:

$[Y]_t = [X]_0 .e^{k. t}$

ie:

$\ln\left( \frac{[Y]_t}{[X]_0}\right) = k. t$

If the end point is some fixed concentration ratio ($\frac{[Y]_{end}}{[X]_0}$), the time to reach the endpoint is:

$t_{end} = \frac{\ln\left( \frac{[Y]_{end}}{[X]_0}\right)}{k}$

If we substitute this into the Arrhenius equation ($\ln(k) = \ln(A) - \frac{E_{a}}{R.T}$), we get:

$\ln(t_{end}) = \ln\left( \ln\left( \frac{[Y]_{end}}{[X]_0}\right)\right) - \ln A + \frac{E_{a}}{R.T}$

so a plot of $ln(t_{end})$ vs. $\frac{1}{T}$ will have a slope of $\frac{E_{a}}{R}$

• This seems more like a comment than an answer. Could you expand, please? – Jan Aug 26 '17 at 5:02