# Solving Activation Energy Equation in AP Chemistry

• Sir_Viking
In summary: Your expert summarizerIn summary, an AP Chemistry student in Texas is seeking help with their homework on calculating activation energy. They have correctly solved for the activation energy of a given reaction, but are unsure about their accuracy. They are asking for someone to check their work and have been informed of the correct units for activation energy. The expert summarizer encourages them to continue their diligent studies and offers assistance if needed.
Sir_Viking
First off, an introduction.
I am a AP Chemistry student (though not a very good one, to be honest) in Texas.
My homework has an equation for activation energy that I have to solve, and I'm having a bit of trouble with it.

CH3I + C2H5ONa --> CH3OC2H5+ NaI
the following data has been obtained
Temperature (K)...K (M-1 s-1)

273...... 5.60 x 10-5
279...... 11.8 x 10-5
285...... 24.5 x 10-5
291...... 48.8 x 10-5
297...... 100. x 10-5

That is the information I have been given.

ln((5.6 x 10-5)/(11.8 x 10-5)) = Ea/8.31 (1/279 - 1/273)

-.74533 = Ea/8.31 (-7.877 x 10-5)
-6.1937 = Ea(-7.877 x 10-5)

Ea = 78.6 KJ

That's what I got, but I'm doubtful about how well I did it.

Anyway, I guess I'm just asking someone to check my work.
[EDIT] Whoops, I must've put the numbers in the calculator wrong, it was 78.6 KJ, not 7.86 of anything

Last edited:
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Dear AP Chemistry student,

Thank you for reaching out for help with your homework. It's great to see that you are taking your studies seriously and trying to understand the material.

I have checked your work and it looks like you have correctly solved for the activation energy of the reaction. Your calculation and reasoning seem to be sound.

However, just to clarify, the units for activation energy are typically given in kJ/mol, not just kJ. So the correct answer would be 78.6 kJ/mol.

Also, make sure to double-check your calculations to avoid any small errors, as you mentioned in your edit. It's always a good practice to go back and review your work to ensure accuracy.

Overall, it seems like you have a good understanding of activation energy and its calculation. Keep up the good work and don't hesitate to reach out for help if you have any further questions.

Best of luck with your studies!

Hello there, as a fellow scientist, I would be happy to provide some feedback on your work. First of all, it's great that you are trying to solve the activation energy equation in AP Chemistry. It shows that you are dedicated to learning and improving in your studies.

Your approach seems to be correct. You have used the Arrhenius equation, ln(k2/k1) = (Ea/R)(1/T1 - 1/T2), where k1 and T1 are the initial rate and temperature, and k2 and T2 are the final rate and temperature.

However, there are a few things that I would like to point out. Firstly, when solving for Ea, you need to use the natural log of the ratio of the rates, not the rates themselves. So it would be ln((11.8 x 10^-5)/(5.6 x 10^-5)) = Ea/8.31 (1/279 - 1/273). This will give you a slightly different value for Ea.

Secondly, when solving for Ea, it is important to use consistent units. In this case, the temperature should be in Kelvin (K) and the rate constant should be in inverse seconds (s^-1). So when you calculate the natural log of the ratio of rates, you should have units of s^-1/s^-1, which will cancel out and leave you with just a number.

Finally, when you plug in the numbers, make sure to use all the significant figures given in the data. This will give you a more accurate answer.

Overall, your approach and solution seem to be correct. Just pay attention to the units and significant figures, and you should be able to solve similar problems in the future with ease. Keep up the good work!

## 1. What is activation energy?

Activation energy is the minimum amount of energy required for a chemical reaction to occur. It is often described as the energy barrier that must be overcome for a reaction to take place.

## 2. How do you calculate activation energy?

The activation energy can be calculated using the Arrhenius equation, which takes into account the reaction rate constant, temperature, and the frequency factor. The equation is: Ea = -RT ln(k/T), where Ea is the activation energy, R is the gas constant, T is the temperature in Kelvin, and k is the reaction rate constant.

## 3. What is the significance of activation energy in chemical reactions?

Activation energy plays a crucial role in determining the rate of a chemical reaction. A higher activation energy means that the reaction will occur at a slower rate, while a lower activation energy indicates a faster reaction. It also determines the feasibility of a chemical reaction, as reactions with high activation energy may be too difficult to occur under certain conditions.

## 4. How does temperature affect the activation energy?

An increase in temperature typically results in a decrease in activation energy. This is because a higher temperature provides more energy to the reactant molecules, making them more likely to overcome the energy barrier and react. In general, a 10°C increase in temperature can double the reaction rate.

## 5. How can activation energy be lowered?

Activation energy can be lowered by adding a catalyst to the reaction. Catalysts provide an alternative pathway for the reaction, reducing the amount of energy required for the reaction to occur. This allows the reaction to happen at a faster rate and at a lower temperature. Another way to lower activation energy is by increasing the concentration of reactants, which increases the likelihood of successful collisions between molecules.

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