Free Energy

[ucla2134]

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Does free energy relate to how fast (more rapidly, less rapidly, more slowly..) the reaction goes?

In kaplan subject test, there is this answer states that if the reaction X to Z is exergonic(the final energy is smaller than the initial energy), it will occur MORE RAPIDLY than the endogernic reaction (the final energy is larger than the initial energy).

I don't understand why SINCE FREE ENERGY is thermodynamic, it will tell us only whether the reaction is spontaneous or non-spontaneous. It doesn't relate to the kinetic energy, right?
Am i right or wrong or am i just missing something here
THANKS

 

[ucla2134]

Anyone???🙁

 

[drteeth]

Gibbs free E only tells us about how spontaneous a rxn. might be. For example, if you have a very negative deltaG this would mean that the rxn. is highly spontaneous and the majority of the rxn. would go to completion. For a rxn. to be spontaneous >50% must go to product. If the deltaG is highly positive the rxn. will be nonspontaneous and very little product will be formed. I am almost positive that in either of these cases deltaG does not tell us about the speed or the kinetics of the rxn....

Hope this helps

 

[sacjumpman]

Does free energy relate to how fast (more rapidly, less rapidly, more slowly..) the reaction goes?

In kaplan subject test, there is this answer states that if the reaction X to Z is exergonic(the final energy is smaller than the initial energy), it will occur MORE RAPIDLY than the endogernic reaction (the final energy is larger than the initial energy).

I don't understand why SINCE FREE ENERGY is thermodynamic, it will tell us only whether the reaction is spontaneous or non-spontaneous. It doesn't relate to the kinetic energy, right?
Am i right or wrong or am i just missing something here
THANKS

which kappy subject test. and which number?

I'll look.

 

[ucla2134]

which kappy subject test. and which number?

I'll look.

Kaplan subject test number 2. If you dont have it, i can pm you

 

[151AND8TH]

Dr Teeth Explained it right. deltalG does NOT tell the kenetic of a rx.. it only deal with spontaneity..
dG<0 sponteneous dG>0 non-spontaneous dG=0 equilibrium

 

[ucla2134]

That is what i though that the free energy has nothing related to the kinetic. So the KAPLAN is WRONG👎


Ok, here is their exact answer and explanation:
Kaplan's Answer: " Although in the forward direction the reaction X to Z will occur more rapidly than from U to X, in the
reverse reaction Z to X will occur more slowly than X to U."
Note: they give us the diagram, reaction coordinate versus free energy. Reaction X to Z has free energy smaller than 0 whereas reaction U to X has free energy bigger than 0.

Kaplan's explanation: "Choice C is true. In the
forward direction, X &#8594; Z has a smaller energy of activation (and is therefore a faster reaction) than does U &#8594; X. Meanwhile, the free energy of activation for Z &#8594; X, that is, the energy required to go from Z to Y, is larger than the
energy required to move from X to W. Therefore, in the reverse direction, the reaction Z &#8594; X will be slower than the reaction X &#8594; U"


How on earth they can relate activation energy to Kinetic???

 

[sacjumpman]

That is what i though that the free energy has nothing related to the kinetic. So the KAPLAN is WRONG👎


Ok, here is their exact answer and explanation:
Kaplan's Answer: " Although in the forward direction the reaction X to Z will occur more rapidly than from U to X, in the
reverse reaction Z to X will occur more slowly than X to U."
Note: they give us the diagram, reaction coordinate versus free energy. Reaction X to Z has free energy smaller than 0 whereas reaction U to X has free energy bigger than 0.

Kaplan's explanation: "Choice C is true. In the
forward direction, X &#8594; Z has a smaller energy of activation (and is therefore a faster reaction) than does U &#8594; X. Meanwhile, the free energy of activation for Z &#8594; X, that is, the energy required to go from Z to Y, is larger than the
energy required to move from X to W. Therefore, in the reverse direction, the reaction Z &#8594; X will be slower than the reaction X &#8594; U"


How on earth they can relate activation energy to Kinetic???

You have to remember that activation energy IS directly related to kinetics and how fast a reaction will go.

So while the y-axis on the graph is Gibbs Free energy, they can determine how fast it will go (relatively to the other reactions on the graph) based on the activation energies.

I don't think we have to know this but the way they are related is through the Arrhenius equation:

k=Ae^-(AE/RT)

k=rate constant
AE=Activation Energy
R= gas constant
T= temp in Kelvins

So you are on the right track, but you can make some determination of how fast a reaction will go: the smaller the activation energy, the faster the reaction.

 

[ucla2134]

You have to remember that activation energy IS directly related to kinetics and how fast a reaction will go.

So while the y-axis on the graph is Gibbs Free energy, they can determine how fast it will go (relatively to the other reactions on the graph) based on the activation energies.

I don't think we have to know this but the way they are related is through the Arrhenius equation:

k=Ae^-(AE/RT)

k=rate constant
AE=Activation Energy
R= gas constant
T= temp in Kelvins

So you are on the right track, but you can make some determination of how fast a reaction will go: the smaller the activation energy, the faster the reaction.

I got it now. Thanks sac. Another equation has to memorize.

 

[sacjumpman]

I got it now. Thanks sac. Another equation has to memorize.

But remember, more so than just memorizing the equation, understand the concept.

The idea that activation energy and kinetics are very connected and how. And how this ties in with Gibbs free energy.

The idea is probably more important than being able to utilize that equation.