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Topic: Transition state theory : Eyring equation  (Read 5314 times)

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Offline dipesh747

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Transition state theory : Eyring equation
« on: April 06, 2012, 01:05:38 PM »
Hello,

I'm working through a question where I am asked to calculate the rate constant for a reaction. The units are supposed to be in dm3 mol-1 s-1

I'm working through the Eyring equation and I'm stuck on the translational contribution part. I've got the answers, which I have uploaded.



What I'm confused about is how answer has been calculated in units of m3. When I go through the units for the mass part it is Kg/ (Kg x Kg) and for the second part the units are (JS)^2 / J K-1 Mol-1 mol K

So when I work through that I get units off JS^2 / Kg

I'm not sure if I've missed out something because it does say qtrans/ volume but I don't really understand why the volume part is there and what units that should be in.

Very confused so any help would be appreciated!

Offline juanrga

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Re: Transition state theory : Eyring equation
« Reply #1 on: April 07, 2012, 07:27:45 AM »
Hello,

I'm working through a question where I am asked to calculate the rate constant for a reaction. The units are supposed to be in dm3 mol-1 s-1

I'm working through the Eyring equation and I'm stuck on the translational contribution part. I've got the answers, which I have uploaded.



What I'm confused about is how answer has been calculated in units of m3. When I go through the units for the mass part it is Kg/ (Kg x Kg) and for the second part the units are (JS)^2 / J K-1 Mol-1 mol K

So when I work through that I get units off JS^2 / Kg

I'm not sure if I've missed out something because it does say qtrans/ volume but I don't really understand why the volume part is there and what units that should be in.

Very confused so any help would be appreciated!

q has dimension 1 as you can check in the "Electronic contribution".

In the "Translational contribution" you have
$$\frac{[q/V]}{[q/V] · [q/V]} = [V]$$
therefore the right hand side is correctly given in units of m3.

You got the correct units J s2 / kg, when taking into account the power 3/2, we obtain

( J s2 / kg )3/2

Using the definition of Joule

J s = m2 kg / s

and substituting back we obtain the correct m3 units.

( m2 )3/2

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Offline dipesh747

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Re: Transition state theory : Eyring equation
« Reply #2 on: April 07, 2012, 10:42:08 AM »


and substituting back we obtain the correct m3 units.

( m2 )3/2



(m^2)^(3/2) doesn't equal 3 though :s 2^3 = 8. 8^0.5 isn't 3.....or have I missed something?

Offline juanrga

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Re: Transition state theory : Eyring equation
« Reply #3 on: April 07, 2012, 11:53:20 AM »


and substituting back we obtain the correct m3 units.

( m2 )3/2

(m^2)^(3/2) doesn't equal 3 though :s 2^3 = 8. 8^0.5 isn't 3.....or have I missed something?

It equals 3, just apply the algebra of exponents (this is explained in math textbooks)

( m2 )3/2 = m(2 · 3/2) = m3

Online ref: http://www.mathsisfun.com/algebra/exponent-laws.html
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Offline dipesh747

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Re: Transition state theory : Eyring equation
« Reply #4 on: April 07, 2012, 11:56:39 AM »
Oh yeah so it does.....bit rusty on my maths nowadays lol cheers

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