[Miffymycat]

We are taught that adding a catalyst to an equilibrium system does not alter the position, just the rate of equilibrium attainment, because it increases the forward and backward rate by the same amount.  Why and how can that be?  The activation energies will be different for the forward and backward direction.  Does a change in forward/backward pre-exponential factors somehow exactly cancel the difference in Ea?  Seems hard to believe.  Or is it some outcome from the Boltzmann distribution curve?  Or some other reason??!!

[Dan]

The activation energies will be different for the forward and backward direction. 

Not true, on what do you base this statement?

See for example: http://www.chemguide.co.uk/physical/basicrates/energyprofiles.html

[Miffymycat]

Sorry Dan - which statement?!!

[curiouscat]

The equilibrium constant is given by:



The ΔG is a difference between products and reactants. No matter what you do to the activation energies ΔG remains the same.

So why would Keq change? Why would it matter that "activation energies will be different for the forward and backward direction."?

[Borek]

So why would Keq change?

Would it?

[curiouscat]

So why would Keq change?

Would it?

You think it would? I'm confused what you meant.   

[Borek]

Presence of catalyst doesn't change equilibrium position. K is a constant (for a given temperature).

[fledarmus]

The acitivation energy of a reaction is the difference in energy between the starting material and the highest energy transition state the the system has to go through on the path between starting material and product. The activation energy will obviously be different between the forward reaction and the reverse reaction if they are going through the same transition state. If the starting material is higher energy than the product, then the highest energy transition state will be closer in energy to the starting material than to the product, and the activation energy of the forward reaction will be less than the activation energy of the reverse reaction. If the activation energy for the reaction to go one direction is so much higher than the other that there is insufficient ambient energy available for the reaction to occur, then the reaction is said to be irreversible, and no equilibirum will be established.

HOWEVER, if you are establishing an equilibrium between starting material and products, one of the inherent assumptions is that there is sufficient free energy available for both the forward and reverse reactions to occur. UNDER THOSE CONDITIONS, the concentration of the starting material and the products (the position of the equilibrium) can be measured and defined as an equilibrium constant. The ratio of starting materials to products depends only on their relative energies, not of the energies of any of the transition states between them.

Catalysts act by stabilizing the highest energy transition state, which lowers the energy of that state. Again, if both the forward and reverse reactions are going through the same high-energy transition state, lowering the energy of that transition state would lower the activation energy of both forward and reverse reactions, making both reaction rates faster. This does not change the relative energy of the starting materials and products, however, so the final concentrations of these compounds, once you have achieved equilibrium, will be the same as it was with no catalyst. However, since the reaction rates both directions have been increased, you will achieve equilibrium more rapidly.