Finding change in activation energy from change in rate constant in catalysis 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

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.
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 $E_\mathrm{a1} = \pu{76 kJ mol−1}:$


  


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


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.