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?

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:

 

[ce] HA \rightarrow H^{+}+A^{-}[/ce]

 

So obviously here the pKa is:

 

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

 

Where the all the activity coefficients, [math] \gamma [/math] 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:

 

[ce] HA \rightarrow H^{+}+A^{-}[/ce]

 

Or will be reactant favored if written as:

 

[ce] H^{+} + A^{-} \rightarrow HA [/ce]

 

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

 

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.

Edited January 10, 201213 yr by mississippichem

http://www.scienceforums.net/topic/17898-the-chemistry-latex-tutorial/

 

There is no excuse!!

 

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:

 

[ce]6CO2 + 6H2O <=> C6H12O6 + 6O2[/ce]

http://www.scienceforums.net/topic/17898-the-chemistry-latex-tutorial/

 

There is no excuse!!

 

Bah.

 

[ce]\mathrm{nice} \ \mathrm{person} <=>\mathrm{bad} \ \mathrm{person} [/ce]

 

You are being mostly product favored right now. , [math] K_{c} >>> 1 [/math].

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