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Topic: Activation Energy  (Read 1948 times)

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Offline Seoirse14

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Activation Energy
« on: April 01, 2017, 03:17:44 PM »
Hi,
I was wondering if you could help me with this 1st year physical chemistry exam question.
Here's the question:
The rate constant exactly doubles when the temperature is raised from 280K to 290K. Estimate the activation energy for the reaction.

I'm not sure how to approach it, any help would be much appreciated!

Offline Babcock_Hall

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Re: Activation Energy
« Reply #1 on: April 01, 2017, 03:25:25 PM »
What are your thoughts?  It is a forum rule (red link above) that you must show your attempt or give your thinking before we can help you.

Offline Seoirse14

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Re: Activation Energy
« Reply #2 on: April 01, 2017, 04:22:59 PM »
I think I need to use the Arrhenius equation : k = Ae-Eact/RT  and calculate lnk and 1/T to get data points that I can then graph.  I should be able to determine the slope from that graph and multiply it by the gas constant R to get Ea. I haven't been given a value for k so I wonder if I should use the Boltzmann constant 1.3807x 10-23 J/K .



Offline Borek

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Re: Activation Energy
« Reply #3 on: April 01, 2017, 05:44:49 PM »
The only thing that Boltzman constant has in common with the Arrhenius equation is the accidental use of the same letter as a symbol.

You don't need value of k - what is important is that the reaction gets twice as fast when the temperature changes. Ratio of speeds doesn't depend on the initial k.
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Offline Enthalpy

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Re: Activation Energy
« Reply #4 on: April 01, 2017, 06:00:36 PM »
Here k is a rate, like mol/(mol*s). It could be different units, since many processes follow such a law, for instance leakage currents in semiconductor components. But a "rate" can't be Boltzmann's constant which converts a temperature into an energy.

Apparently you don't obtain help from the units. You should, they're your allies. Try hard to memorize what is each quantity, and double-check every time if your grub makes sense. This is one single thing that every professional scientist does.

You don't need any fixed value of k to find the activation energy. "Doubles" is enough. You have to compare both rates.

Drawing a graph isn't necessary here. This is algebra, accessible by a few transformations.

Graphs can be useful with Arrhenius law, especially when you have made a set of measurements and want to check if they follow an Arrhenius law and if there is one single activation energy over the data set, or where the data switches from one activation energy to an other.

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