0
$\begingroup$

I need to find by how much the activation energy decreases when the rate constant increases by a '$x$' factor when a catalyst is used in the reaction.

I'm just super confused about how to solve this. Using Arrhenius relation I have

$$\frac{k_2}{k_1} = \frac{A\mathrm e^{-E_\mathrm{a2}/(RT)}}{A\mathrm e^{-E_\mathrm{a1}/(RT)}}$$

$$x = \mathrm e^{(E_\mathrm{a1} - E_\mathrm{a2})/(RT)}$$

I took natural log on both sides to get rid of the exponent and then ended up finally with an equation I can't seem to get any relation from

$$RT\ln x = E_\mathrm{a1} - E_\mathrm{a2}$$

Now I need a ratio of the two activation energies, but I don't know how I get that from here as I have just a addition/subtraction relation between the two. Would be grateful if someone has a clue about how I could proceed.

The problem is I have come across a question in Atkins' Physical Chemistry [1, p. 841] and the answer is provided in percent, but I don't see how you could arrive at a solution like that with where I am:

Self-test 20D.4 Consider the decomposition of hydrogen peroxide, which can be catalysed in solution by iodide ion. By how much is the activation energy of the reaction reduced if the rate constant of reaction increases by a factor of 2000 at 298 K upon addition of the catalyst?

Answer: 25 per cent

References

  1. Atkins, P. W.; De Paula, J. Physical Chemistry: Thermodynamics, Structure, and Change, 10th ed.; W.H. Freeman: New York, 2014. ISBN 978-1-4641-2452-5.
$\endgroup$
  • 1
    $\begingroup$ Usually these questions ask you to find the difference between the two energies, not the ratio, because exactly as you said, you can only find the difference of those. $\endgroup$ – Ezze Oct 30 '19 at 8:48
  • 1
    $\begingroup$ As Ezze has commented, you already answered your question (or rather, there is not much to add to your answer). You can express the required change in the activation energy as an absolute change Ea2−Ea1 , as a fraction change (Ea2−Ea1)/Ea1 or as percent change 100(Ea2−Ea1)/Ea1 $\endgroup$ – Buck Thorn Oct 30 '19 at 17:17
1
$\begingroup$

It appears that you are correct in your thought process. Determining by how much the activation energy of the reaction is reduced requires you to basically find

$$Δ(\%) = \frac{E_\mathrm{a1} - E_\mathrm{a2}}{E_\mathrm{a1}} × 100\% = \left(1 - \frac{E_\mathrm{a2}}{E_\mathrm{a1}}\right) × 100\%$$

There is a system of three independent equations

$$ \begin{cases} k_1 = A\mathrm e^{E_\mathrm{a1}/(RT)}\\ k_2 = A\mathrm e^{E_\mathrm{a2}/(RT)}\\ \frac{k_2}{k_1} = x \end{cases} $$

with five unknowns $k_1$, $k_2$, $E_\mathrm{a1}$, $E_\mathrm{a2}$, $A$, which requires either one more equation, or to know one more parameter of the system in order to find $E_\mathrm{a2}/E_\mathrm{a1}.$ The textbook provides you with activation energy in the previous illustration right next to your problem [1, p. 841], so I assume you are supposed to use $E_\mathrm{a1} = \pu{76 kJ mol−1}:$

Screenshot

At this point the problem can indeed be solved to a certain numerical value:

$$E_\mathrm{a1} - E_\mathrm{a2} = RT\ln x = \pu{8.3145 J mol-1 K-1} × \pu{298 K} × \ln{2000} \approx \pu{18.8 kJ mol-1}$$

$$Δ(\%) = \frac{\pu{76 kJ mol-1} - \pu{18.8 kJ mol-1}}{\pu{76 kJ mol-1}} × 100\% \approx 25\%$$

The question whether it's a textbook flaw, or authors assumed the reader would read the material more carefully still remains open.

References

  1. Atkins, P. W.; De Paula, J. Physical Chemistry: Thermodynamics, Structure, and Change, 10th ed.; W.H. Freeman: New York, 2014. ISBN 978-1-4641-2452-5.
| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.