Is there always symmetry in enzyme effect of Fwd/Reverse reaction rate?

[nomdeplume1234]

For example, if an enzyme speeds up the rate of the forward reaction by 10 does it necessarily speed up the reverse ration by 10 as well?

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[mcatjelly]

Yep. Enzymes affect only the rate/activation energy of a reaction--not its equilibrium. You know that when an enzyme decreases a reaction's activation energy, it increases the rate of reaction. However, if it only affected the reaction in the forward direction, then it would change the equilibrium constant as there would be many more products than reactants. So by affecting both the forward and reverse reactions equally, enzymes don't affect equilibrium.

 

[nomdeplume1234]

Yep. Enzymes affect only the rate/activation energy of a reaction--not its equilibrium. You know that when an enzyme decreases a reaction's activation energy, it increases the rate of reaction. However, if it only affected the reaction in the forward direction, then it would change the equilibrium constant as there would be many more products than reactants. So by affecting both the forward and reverse reactions equally, enzymes don't affect equilibrium.

Thanks for the explanation. It makes sense logically but is not very intuitive to me yet.

 

[mcatjelly]

Thanks for the explanation. It makes sense logically but is not very intuitive to me yet.

It's not intuitive to me either. :/

 

[ElectricNoogie]

Think about it this way. If an enzyme "lowers the hump" of the reaction by lowering activation energy, then it should be easier now to get over the hump going in either direction, forwards or reverse. This explains why an enzyme will not change Keq, but can allow a reaction to reach Keq faster.

 

[intrepid19]

http://academic.pgcc.edu/~kroberts/Lecture/Chapter 5/05-05_CatalystGraph_L.jpg

If you look at this picture, you see that the enzyme lowers the activation energy, just as ElectricNoogle was saying. So it's now faster to go from reactants to products because there's a lower hump to get over. If you look at going from products to reactants, it's still more difficult than going from reactants to products, because you're going "uphill" more. However, the hump is STILL lower than it was without the enzyme, so the reverse reaction will be sped up, as well.

 

[nomdeplume1234]

http://academic.pgcc.edu/~kroberts/Lecture/Chapter 5/05-05_CatalystGraph_L.jpg

If you look at this picture, you see that the enzyme lowers the activation energy, just as ElectricNoogle was saying. So it's now faster to go from reactants to products because there's a lower hump to get over. If you look at going from products to reactants, it's still more difficult than going from reactants to products, because you're going "uphill" more. However, the hump is STILL lower than it was without the enzyme, so the reverse reaction will be sped up, as well.

Nice picture. Alluding to what you mention, based on the pic it looks like the forward reaction would be faster than the reverse but it seems that since the rate is a constant that the forward rate = backward rate.