Okay so if i have two enzyme catalyzed reactions:
1) delta g of the ES = -22kJ/mol
2) delta g od the ES = 42 kJ/mol
Is this first reaction going to have a higher reaction rate because the of the negative delta g, because doesnt a negative delta g give you a spontaneous reaction? But then in the back of my mind im thinking just because it is spontaneous doesnt mean that it can be fast; it could take a million years to have the reaction go to completion right? But since it is a catalyzed reaction does that mean that the reaction rates are increased and the first reaction would have a faster reaction rate?
If anyone could please help me understand this concept I would really appreciate it. Thank you!
3 answers
You have it right on both accounts. The exothermic will go faster (generally). In general, the catalyzed reactions go "faster" (generally). In both cases, the first reaction will go faster.
Thank you!
Cindy,
I saw this post earlier and wanted to answer it, but I had to take care of some things. You are about 89% correct: ∆G only tells you if the reaction is spontaneous or not; I can not stress this enough, but it DOES NOT tell you anything about the rate of a reaction. The only thing that ∆G tells you, with regards to an enzymatic reaction, is that the reaction will occur spontaneously (is favored) or that it will not occur unless you put in energy. Look at the following equation:
A + B ---> C +D
∆G= Delta Gº +RTlog[C*D/A*B]
Looking at the following equation, you can see that it tells you nothing about the rate and is mainly dependent on the concentrations. The reason why one ES complex has a lower ∆G is because of the concentrations of the substrates/reactants (look at the equation). The ∆G for one is lower then the other, because of the amount of substrates (plug in the numbers to see how it will change for yourself). When comparing ∆G values between ES complexes, the one that has a lower ∆G is the one that is more favored, not which one will have a faster reaction rate.
As far as reaction rates go, their are three things that can lower the reaction rate: catalysts/enzymes, heat, and reactants. The rate of a reaction for ES complexes are dependent on a ∆G, but not this ∆G (change in free energy). The rate of a reaction is dependent upon ∆G‡, which is equal to the difference in free energy between the transition state and the substrate. Without geting to bogged down in equations, just know that enzymes accelerate the reaction rate by binding to the transition state and lowering the energy of activation ∆G‡.
I saw this post earlier and wanted to answer it, but I had to take care of some things. You are about 89% correct: ∆G only tells you if the reaction is spontaneous or not; I can not stress this enough, but it DOES NOT tell you anything about the rate of a reaction. The only thing that ∆G tells you, with regards to an enzymatic reaction, is that the reaction will occur spontaneously (is favored) or that it will not occur unless you put in energy. Look at the following equation:
A + B ---> C +D
∆G= Delta Gº +RTlog[C*D/A*B]
Looking at the following equation, you can see that it tells you nothing about the rate and is mainly dependent on the concentrations. The reason why one ES complex has a lower ∆G is because of the concentrations of the substrates/reactants (look at the equation). The ∆G for one is lower then the other, because of the amount of substrates (plug in the numbers to see how it will change for yourself). When comparing ∆G values between ES complexes, the one that has a lower ∆G is the one that is more favored, not which one will have a faster reaction rate.
As far as reaction rates go, their are three things that can lower the reaction rate: catalysts/enzymes, heat, and reactants. The rate of a reaction for ES complexes are dependent on a ∆G, but not this ∆G (change in free energy). The rate of a reaction is dependent upon ∆G‡, which is equal to the difference in free energy between the transition state and the substrate. Without geting to bogged down in equations, just know that enzymes accelerate the reaction rate by binding to the transition state and lowering the energy of activation ∆G‡.
1 answer