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Chemical Reaction Procession what pKA determines whether a rxn proceeds Rate Topic: -----

#1 10 January 2012 - 04:47 AM blazinfury 


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Does a chemical reaction proceed from low pKa to high pKa or high pKa to low pKa?

A large Ka implies how readily a reaction goes toward the products. Now since pKa=-log(Ka), wouldn't that imply that a chemcial reaction goes from high pKa to low pKa?
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#2 10 January 2012 - 11:02 PM User is online  mississippichem 


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View Postblazinfury, on 10 January 2012 - 04:47 AM, said:

Does a chemical reaction proceed from low pKa to high pKa or high pKa to low pKa?

A large Ka implies how readily a reaction goes toward the products. Now since pKa=-log(Ka), wouldn't that imply that a chemcial reaction goes from high pKa to low pKa?


Remember that we can write a pKa for a reaction but not for a compound per se. Often times it looks like a pKa is written for a compound but it is really just the reaction:

HA \rightarrow H^{+}+A^{-}

So obviously here the pKa is:

pKa = -\log \left ( \frac{[H^{+}][A^{-}]}{[HA]}\cdot \frac{\gamma_{H} \gamma_{A}}{\gamma_{\mathrm{HA}}} \right )

Where the all the activity coefficients, \gamma are very close to unity for an ideal solution.

So the answer to your question is neither. An acid/base equilibrium reaction with a low pKa will be product favored when written in the form:

HA \rightarrow H^{+}+A^{-}

Or will be reactant favored if written as:

H^{+} + A^{-} \rightarrow HA

*I don't know how to LaTeX equilibrium arrows so forgive my notational travesty please :o .

Remember that the equilibrium constant says nothing about the kinetics of a reaction. You can use equilibrium to obtain thermodynamic state functions like Gibbs energy, a measure of spontaneity, but this says nothing about the reaction "proceeding". A reaction can be both thermodynamically and equilibrium favored but proceed at a rate that is slow enough to be practically unobservable.

This post has been edited by mississippichem: 10 January 2012 - 11:04 PM

You've come a long way. Remember back when we defined what a velocity meant? Now we are talking about an antisymmetric tensor of second rank in four dimensions.

-Feynman Lectures on Physics II
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#3 10 January 2012 - 11:47 PM hypervalent_iodine 


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http://www.sciencefo...latex-tutorial/

There is no excuse!! :P

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Bidirectional Equations

We sometimes need an equation that is not mono-directional but can flow in two ways and we thus require another arrow form:

6CO2 + 6H2O <=> C6H12O6 + 6O2

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#4 10 January 2012 - 11:52 PM User is online  mississippichem 


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View Posthypervalent_iodine, on 10 January 2012 - 11:47 PM, said:

http://www.sciencefo...latex-tutorial/

There is no excuse!! :P


Bah.

\mathrm{nice} \ \mathrm{person} <=>\mathrm{bad} \ \mathrm{person}

You are being mostly product favored right now. :P, K_{c} >>> 1 .
You've come a long way. Remember back when we defined what a velocity meant? Now we are talking about an antisymmetric tensor of second rank in four dimensions.

-Feynman Lectures on Physics II
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#5 11 January 2012 - 12:03 AM hypervalent_iodine 


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Maybe I'll go to a neo-nazi convention and try to shift the equilibrium to the left.
I have a blog!
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