enzymes and rate constant

[destroythemcat]

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This question is from Kaplan

The rate of an enzyme catalyzed reaction can be influenced by all of the following EXCEPT

A. Substrate concentration
B. Temperature
C. Enzyme concentration
D. Rate Constant
E. pH

The answer provided is D. Rate Constant. The explanation states that: the rate constant (k) which appears in the rate law is a temperature dependent constant. For a given temperature nothing will change k, so it can have no influence on an enzyme catalyzed reaction.

I know that the addition of an enzyme causes the reaction to have a new rate constant (is that correct?) but this new rate constant has an effect on the rxn rate? Can anyone help?

Does addition of an enzyme influence the rate constant K or give the rate a new rate constant?

 

[el_duderino]

This is a dumb question. The rate constant doesn't change, which is why it's called a constant. All the others are variables.

 

[destroythemcat]

so if rate=K [A] --> how does an enzyme increase the rate? (i know that it lower the activation energy needed) but do it also substitute the rate constant K for a new one?

 

[Alucard6]

More enzyme = more places for substrate to bind which means more product.

 

[MedHopeful20134]

I think the answer is A here. Once the enzyme is saturated, substrate concentration would have no effect. k is actually affected by activation energy.

 

[destroythemcat]

the answer is D though…..?

 

[hemaoncol]

This is a dumb question. The rate constant doesn't change, which is why it's called a constant. All the others are variables.

Such an unnecessary reaction. If you have nothing nice to say, don't say anything at all...

I think the answer is A here. Once the enzyme is saturated, substrate concentration would have no effect. k is actually affected by activation energy.

The answer is not A. Be careful when making assumptions. You are assuming that the substrate concentration is at saturation. Well, what if it is not? The substrate would then have an effect.

the answer is D though…..?

Rate constant and the rate of reaction are two different things. In the equation: rate of rxn = k [A]^a [C]^c; the only things that vary are the rate of reaction, compound A and B. When the rate of reaction is changed, the concentrations of A and B will change accordingly but the rate constant will remain the same, a constant. So that equation is to be looked at differently. Think of this way. In the equation above, if concentration [A] is increased, we should expect the rate of reaction to go faster. Now looking at it in the reverse way, when you make a reaction go faster, you would expect the amount of product to go up. However, only those would go up; k would not be affected.

 

[lumpyduster]

so if rate=K [A] --> how does an enzyme increase the rate? (i know that it lower the activation energy needed) but do it also substitute the rate constant K for a new one?

Yep. There's an equation called the Arrhenius equation. k = Ae^(-Ea/[RT]), where A is a constant, Ea is activation energy, R is a constant, and T is temperature. If you lower Ea (which is what enzymes do) you increase k which in turn increases rate. Similarly, if you increase temperature you also increase k/rate.

 

[hemaoncol]

Yep. There's an equation called the Arrhenius equation. k = Ae^(-Ea/[RT]), where A is a constant, Ea is activation energy, R is a constant, and T is temperature. If you lower Ea (which is what enzymes do) you increase k which in turn increases rate. Similarly, if you increase temperature you also increase k/rate.

Please note that it is not as if the more enzyme you add, the more of the activation Energy you reduce and therefore the larger the k constant. That's actually not how it works. You need to look at the enzyme as a reactant that is a regenerated at the end of the reaction, after all that is what an enzyme is. k is a constant. In that equation you posted, the only variable is T. If you have two reactants, if you increase the concentration of any of them, you would favor the production of more product. However, you would also agree that you did not suddenly reduce the activation energy.

 

[lumpyduster]

Please note that it is not as if the more enzyme you add, the more of the activation Energy you reduce and therefore the larger the k constant. That's actually not how it works. You need to look at the enzyme as a reactant that is a regenerated at the end of the reaction, after all that is what an enzyme is. k is a constant. In that equation you posted, the only variable is T. If you have two reactants, if you increase the concentration of any of them, you would favor the production of more product. However, you would also agree that you did not suddenly reduce the activation energy.

I'm not sure where I said/implied any of that, but I agree with you.

 

[el_duderino]

It's a dumb question. You wouldn't see that question on the MCAT.

 

[hemaoncol]

I'm not sure where I said/implied any of that, but I agree with you.

Oh no, I am not saying you did. I am just clarifying things.

It's a dumb question. You wouldn't see that question on the MCAT.

You are adding zero substance to this conversation. Are you just trolling?

 

[MedHopeful20134]

I think I misread the question.
The key word is "enzyme catalyzed reaction". The reaction is already catalyzed, so Ea would stay constant once catalyzed.
But in uncatalyzed reactions, k is affected by activation energy.