I conducted a chemical experiment (degree in chemical engineering) to understand the isotopic effect in chemical kinetics. In the experiment, I reacted isopropanol and D-isopropanol (D stands for deuterium) with chromic acid (H2CrO4).

In order to measure the rate of the reaction - we used a spectrophotometer, to measure the absorption of the Cr(+6) ion as a function of time. As expected - the absorption of the reaction solutions diminished over time, as more Cr(+6) reacted and turned into Cr(+3).

The graphs in the experiment are attached here: enter image description here

enter image description here

Now, I need to somehow find the connection between the rate equations (written in the pictures) and the Arrhenius equation.

If we write the Arrhenius equation for the reactions (with isopropanol and D-isopropanol) we get:

K(iso) = A*exp(-Ea_iso/RT)

K(D-iso) = A*exp(-Ea_Diso/RT)


K(iso) / K(D-iso) = exp(-dEa/RT)


dEa = Ea_iso - Ea_Diso

I need to find the ratio K(iso) / K(D-iso), given the rate equations in the graphs above. How can I do that?

Thanks in advance

  • 1
    $\begingroup$ There would be A(iso) and A(D-iso), the latter lower. // Do you consider isotope exchange before oxidation? like IPrOD + H+ <=> IPrOH + D+? $\endgroup$
    – Poutnik
    Jun 6, 2023 at 0:57
  • $\begingroup$ I'm not sure I understand your question. Can you explain it again? $\endgroup$ Jun 6, 2023 at 1:15
  • $\begingroup$ Do you ask whether the coefficient A is different for the reactions? In the analysis we assume that A is equal for both reactions. $\endgroup$ Jun 6, 2023 at 1:17
  • 1
    $\begingroup$ The first part is the statement, not a question. IMHO, the isotopic effect on frequency factor is more significant than on activation energy. $\endgroup$
    – Poutnik
    Jun 6, 2023 at 1:55
  • 1
    $\begingroup$ You have the two rate constants as shown in your data so you can calculate what you want. Normally to get at activation energies measurements at different temperature are made. I'm assuming you use D8 propanol. $\endgroup$
    – porphyrin
    Jun 6, 2023 at 7:27

1 Answer 1


To Include or not to Include

The constants in your fitted equation make your kinetics different than first order kinetics. Which is why, if we consider the constatnts, this problem cannot be solved.

Explanation: Because the reactions are following different order kinetics, the data provided is not enough to determine the activation energies of the two reactions. More data needs to be retrieved to determine the activation energies.

If we represent $Ab$ as absorption, then:

$$ \ln{Ab_\text{iso}} = \ln{(0.5682\times \mathrm{e}^{-0.04958\times t}+0.007084)}\\ \ln{Ab_\text{D-iso}} = \ln{(0.6474\times \mathrm{e}^{-0.02649\times t}-0.005161)}\\ $$

graph of logarithm of absorbance with time

Now, $Ab_\text{D-iso}$ seems to be following first order kinetics, because the graph is linear. However, $Ab_\text{iso}$ doesn't seem to follow first order kinetics. This makes our analysis somewhat trickier.

However, if you do want to include the constants, I have exactly outlined the steps needed to solve the question, which is attached at the end.

Not Including the Constants

If we consider the constants as an error term, your equations become:

$$ Ab_\text{iso} = 0.5682\times \mathrm{e}^{-0.04958\times t}\\ Ab_\text{D-iso} = 0.6474\times \mathrm{e}^{-0.02649\times t}\\ $$

These equations follow first order kinetics exactly:

$$ \text{first order rate law}\\ \ce{[A]} = \ce{[A]}_\circ\mathrm{e}^{-kt} $$

Since $\ce{[A]} \propto Ab$:

$$ Ab = Ab_\circ\mathrm{e}^{-kT} $$

By comparing the equations, we obtain:

$$ Ab_\text{iso} = (Ab_\text{iso})_\circ\mathrm{e}^{-k_\text{iso}t}\\ k_\text{iso} = 0.04958 \\ Ab_\text{D-iso} = (Ab_\text{D-iso})_\circ\mathrm{e}^{-k_\text{D-iso}t}\\ k_\text{D-iso} = 0.02649 \\ $$

Using the same method outlined by you, at room temperature:

$$ \ln{\dfrac{k_\text{D-iso}}{k_\text{iso}}}\approx -0.62682\\ \boxed{\Delta E_\mathrm{a} \approx \pu{1.527 kJ mol^{-1}}} $$

Including the Constants

Let us assume the order of reaction with D-isopropanol ($\ce{R1}$) is one and that with isopropanol $\ce{R_2}$ is $n$.

$$ \ce{R1 -> P1}; \text{first order}\\ \ce{R2 -> P2}; n^\text{th}\text{ order} $$

The rate equations are given by:

$$ \ce{[R1]} = \ce{[R1]}_\circ \mathrm{e}^{-k_1t}\\ \ce{[R2]} = \dfrac{1}{\left( k_2t - \dfrac{1}{\ce{[R2]}_\circ^{n+1}}\right)^{n+1}} $$

Assuming $Ab_{\ce{[X]}} = c\ce{[X]}$

$$ Ab_\text{D-iso} = (Ab_\text{D-iso})_\circ \mathrm{e}^{-k_1t}\\ cAb_\text{iso} = \dfrac{1}{\left( k_2t - \dfrac{1}{c^{n+1}(Ab_\text{iso})_\circ^{n+1}}\right)^{n+1}} $$

We have arrived at two equations with six unknown parameters: $(Ab_\text{D-iso})_\circ$, $(Ab_\text{iso})_\circ$, $k_1$, $k_2$ ,$c$, and $n$. This is unsolvable. You will need to make a few adjustments and require more data:

Steps to Determine Activation Energies

  1. Ensure initial concentrations of all reactants are equal and known, reducing two unknown parameters. These can also be approximated as the initial values of your data:

$$ Ab_\text{D-iso} \approx 0.642 \mathrm{e}^{-k_1t}\\ cAb_\text{iso} \approx \dfrac{1}{\left( k_2t - \dfrac{1}{c^{n+1}0.575^{n+1}}\right)^{n+1}} $$

We still need to reduce two more unknown parameters, which are $n$ and $c$.

  1. Find the value of $c$ by preparing samples of different concentration and measuring the absorbance, reducing one unknown parameter.
  2. Determine the value of $n$ using the differential rate law, reducing one unknown parameter.

That should reduce the variables to two and the equations will be solvable for $k_1$ and $k_2$. Then, you can use the Arrhenius approximation to calculate activation energies.

  • 2
    $\begingroup$ This is nonsense. The equations that you are treating as his data are best fit first-order curves, so they are by definition first-order. The reason you get non-linearity is because you're including the non-zero y-intercept from the fitted curve. And the $k$ values that you already have from his non-linear least squares fit are likely to be much more accurate than those that you extracted from a linear fit to the transformed points. $\endgroup$
    – Andrew
    Jun 7, 2023 at 11:00
  • 1
    $\begingroup$ I'm not sure why you do not expect a reply, as I have replied to your previous questions in other answers. To your question 'What if the fitted equation is actually correct and kinetics do change under the isotope effect, as seen in the logarithm graph." If the equation is correct, the reaction is first-order. That's the point. The equation is a fit that assumes first order kinetics. If you removed the y-intercept which your linearization did not account for, you would necessarily have a perfect straight line, since you are "fitting" a first order curve to a first-order equation. $\endgroup$
    – Andrew
    Jun 7, 2023 at 13:37
  • 3
    $\begingroup$ yes, I'm using "y-intercept" sloppily here, since "asymptotic as t goes to infinity" is wordier. A non-zero asymptote can arise experimentally in many ways, for example if the instrument is not zeroed properly if the product has a small absorbance at the wavelength of interest. It is up to the experimenter to determine if it is better to fit with a constant or without. If you want to use a linear transformation that does not account for that constant, you can simply remove it, but keeping it and not accounting for it in your transformation is incorrect. $\endgroup$
    – Andrew
    Jun 7, 2023 at 14:04
  • 2
    $\begingroup$ Furthermore, I do not understand the point of transforming his first-order fit equation in order to extract rate constants that are already available in his equation. Since you are working with the equation rather than the raw data, there is no possibility that you will somehow get more accurate values. And more generally fitting first-order data to a linear equation is less accurate than fitting directly to a curve (though it is possible that the original fit was done via linearization.) $\endgroup$
    – Andrew
    Jun 7, 2023 at 14:08
  • 1
    $\begingroup$ "Probably should be OP's call whether to ignore the constants or not." Right. And they chose to include them. So we don't "have analysis with. . .neglecting the constants.", because that can't be done without the original data. Your "analysis" is just incorrect math. $\endgroup$
    – Andrew
    Jun 7, 2023 at 14:26

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