Sapling #15 Wk.9-10

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IsabelMurillo3K
Posts: 100
Joined: Wed Sep 30, 2020 9:51 pm

Sapling #15 Wk.9-10

Postby IsabelMurillo3K » Sun Mar 07, 2021 11:28 pm

Calculate the activation energy, Ea , in kilojoules per mole for a reaction at 69.0 ∘C that has a rate constant of 0.298 s−1 and a frequency factor of 8.07×1011 s−1 .
Hey guys! I was wondering if anyone could explain what formula is necessary to solve this problem?
Thank you!

Grace_Remphrey_2J
Posts: 100
Joined: Wed Sep 30, 2020 9:54 pm

Re: Sapling #15 Wk.9-10

Postby Grace_Remphrey_2J » Sun Mar 07, 2021 11:54 pm

For this problem, I used the Arrhenius equation to calculate the activation energy but rearranged it to solve for Ea. Hope this helps!
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Olivia Yang 3J
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Re: Sapling #15 Wk.9-10

Postby Olivia Yang 3J » Mon Mar 08, 2021 6:52 am

For this problem use the arrhenius equation k = Ae^−EaRT where A is the frequency factor 8.07×1011. You should convert the temperature into Kelvin from Celsius, rearrange the equation to kA = e^−EaRT and take the natural log of both sides so that your equation becomes Ea =−RTln(k/A). You can then input the given values for R, T, A, and k and solve for the activation energy. Lastly, you should convert your answer from joules to kilojoules

Gicelle Rubin 1E
Posts: 101
Joined: Fri Oct 02, 2020 12:16 am

Re: Sapling #15 Wk.9-10

Postby Gicelle Rubin 1E » Fri Mar 12, 2021 12:38 am

Olivia Yang 3J wrote:For this problem use the arrhenius equation k = Ae^−EaRT where A is the frequency factor 8.07×1011. You should convert the temperature into Kelvin from Celsius, rearrange the equation to kA = e^−EaRT and take the natural log of both sides so that your equation becomes Ea =−RTln(k/A). You can then input the given values for R, T, A, and k and solve for the activation energy. Lastly, you should convert your answer from joules to kilojoules


Thank you!!! I was stuck on this question :,)

Ally M
Posts: 51
Joined: Wed Feb 17, 2021 12:22 am

Re: Sapling #15 Wk.9-10

Postby Ally M » Sat Mar 12, 2022 1:40 pm

For this question, I first took the natural log of both sides, then the common log, and then just substituted the rest of the values into the Arrhenius equation!

RalphGuevarra
Posts: 35
Joined: Mon Jan 09, 2023 10:14 am

Re: Sapling #15 Wk.9-10

Postby RalphGuevarra » Wed Mar 08, 2023 10:48 am

Olivia Yang 3J wrote:For this problem use the arrhenius equation k = Ae^−EaRT where A is the frequency factor 8.07×1011. You should convert the temperature into Kelvin from Celsius, rearrange the equation to kA = e^−EaRT and take the natural log of both sides so that your equation becomes Ea =−RTln(k/A). You can then input the given values for R, T, A, and k and solve for the activation energy. Lastly, you should convert your answer from joules to kilojoules

This is so helpful, thank you!

405718358
Posts: 30
Joined: Mon Jan 09, 2023 9:07 am

Re: Sapling #15 Wk.9-10

Postby 405718358 » Mon Mar 20, 2023 10:37 am

Olivia Yang 3J wrote:For this problem use the arrhenius equation k = Ae^−EaRT where A is the frequency factor 8.07×1011. You should convert the temperature into Kelvin from Celsius, rearrange the equation to kA = e^−EaRT and take the natural log of both sides so that your equation becomes Ea =−RTln(k/A). You can then input the given values for R, T, A, and k and solve for the activation energy. Lastly, you should convert your answer from joules to kilojoules



This is a really helpful explanation, i was stuck on this questions for a really long time thank you for the help !


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