Eyring Equation Derivation

[shaneo]

Below is my shot at explaining how to get to the second to last step of the eyring equation from Gibbs laws. However I don't understand two mathematical parts as highlighted and described in the image below.


Could someone please explain how they come about?


Thanks




note:
eq 4 should equal eq 2 (just another way of writing it) (not sure hence the highlighted T in eq 3)
to get to eq 5. i have cancelled the "T" in the equation
to get to eq 6. reversed the minus
to get to eq. 3 is the part that i don't understand.



update: Please see comments below
(not sure how to get from bttom eq in picture to the Eyring equation

Edited January 1, 2015 by shaneo

[studiot]

You need an equation for K^** (which is dimensionless or in moles) in terms of K, which is the rate constant in moles/sec.

 

that is you need to multiply by sec^-1

 

The equation you require is

 

K = K^**(kT/h)

 

Note also that equation1 is a statement of the difference between the free energy of the activated complex and the reactants

 

[math]\Delta {G^{**}} = \Delta {H^{**}} - T\Delta {S^{**}} = - RT\ln ({K^{**}})[/math]

If you take the exponential of each side you end up with the product of two exponentials

 

[math]K = \frac{{kT}}{h}{e^{ - \frac{{\Delta {G^{**}}}}{{RT}}}}[/math]

[math] = \frac{{kT}}{h}{e^{\frac{{\Delta {S^{**}}}}{R}}}{e^{ - \frac{{\Delta {H^{**}}}}{{RT}}}}[/math]

Edited January 1, 2015 by studiot

[shaneo]

thanks for your response.. however what i require is an in detail step by step approach to the derivation :/ the mathematics doesn't quite add up in my mind. However i do now know that we take the exponential of the above metioned equation in order to reverse the natural logs in a revised form of the eyring equation.

The equation that i am mentioning is on the bottom of the description

Edited January 1, 2015 by shaneo

[studiot]

OK, my post showed how to get your first highlighted (in yellow) questions ie the negative delta H double double star and T in the denominator.

 

It's to late to assemble the rest of the derivation tonight, I will see what i can post tomorrow, if no one else does it first.

 

It would be very helpful if you would label your notation, since everyone has a different notation for this.

Particularly with the rate and equilibrium constants.

 

I assume you have realised I can't do double daggers and have used double stars instead.

 

Edit, Do you realise you have incorrectly multiplied equation 1 through by -1 to form equation 2 incorrectly?

 

[math]\Delta H - T\Delta S = - RT\ln (K)[/math]

[math] - \Delta H + T\Delta S = + RT\ln (K)[/math]

Edited January 2, 2015 by studiot

[shaneo]

ah that last part makes much more sense now! no i didn't realise that i hadn't rearranged eq 1 correctly. My lecturer put the equations down without much explanation. I think i have it up the the point of the last equation in my second image on the description now.

I will post my workings out either tonight or tomorrow.
I think they are correct!


Thanks again,

Shaneo

[studiot]

Yes you were on the right track apart from then -1.