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!
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‡.
Great question! Understanding the relationship between delta G and reaction rates can be a bit tricky, but I'll do my best to explain it.
First, let's clarify what delta G represents. Delta G, or Gibbs free energy, is a measure of the energy change that occurs during a chemical reaction. A negative delta G indicates that the reaction is spontaneous, meaning it can proceed without an input of energy. Conversely, a positive delta G would indicate a non-spontaneous reaction that requires an input of energy to proceed.
Now, let's relate this to reaction rates. The reaction rate refers to how quickly a reaction occurs. A reaction with a higher reaction rate means that it completes more quickly.
While a negative delta G suggests a spontaneous reaction, it does not directly indicate the reaction rate. As you correctly mentioned, a spontaneous reaction can still proceed slowly and take a long time to complete. However, when it comes to enzyme-catalyzed reactions, things work a bit differently.
Enzymes are biological catalysts that speed up the rate of a chemical reaction by lowering the activation energy required for the reaction to occur. Activation energy is the energy barrier that reactants must overcome for a reaction to proceed.
Enzymes achieve this by binding to the reactant molecules (substrates) and altering their structure, making the reaction more favorable. The enzyme-substrate complex (ES) forms during the reaction and is stabilized by certain interactions.
In your first question, where the delta G of the ES complex is -22 kJ/mol, this suggests that the formation of the ES complex is energetically favorable. So, in this case, the enzyme would lower the activation energy and speed up the reaction rate, making it faster compared to a reaction without an enzyme.
In your second question, where the delta G of the ES complex is 42 kJ/mol, this indicates that the formation of the ES complex is energetically unfavorable. This means that the reaction without an enzyme would proceed slowly or not at all. However, an enzyme can still potentially catalyze this reaction, but it may not increase the reaction rate as much as in the first question.
So, to summarize, while a negative delta G does suggest a spontaneous reaction, the reaction rate is influenced by both the presence of an enzyme and the delta G of the ES complex. Enzymes can lower the activation energy, making the reaction faster. However, the magnitude of the effect depends on the specific reaction and the energetic favorability of the ES complex.
I hope this explanation helps! Let me know if you have any further questions.