December 14, 2019, 11:41:14 PM
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### Topic: Gibbs Free Energy Change and Activation Energy  (Read 436 times)

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#### byli2223

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##### Gibbs Free Energy Change and Activation Energy
« on: July 31, 2019, 03:00:36 PM »
Isn't when the Gibbs free energy is negative, the reaction is spontaneous? which means the reaction doesn't need other energy input? How is activation energy related to this question then? The answer was B. Thanks

#### Corribus

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##### Re: Gibbs Free Energy Change and Activation Energy
« Reply #1 on: July 31, 2019, 03:53:57 PM »
Isn't when the Gibbs free energy is negative, the reaction is spontaneous? which means the reaction doesn't need other energy input?
No, that's not what a spontaneous reaction means. Spontaneity, and ΔG, refers to the free energy difference (chemical potential energy, if you will) for the reactants vs the products. A spontaneous reaction is one in which the energy of the products is lower than the reactants, so that at equilibrium products are favored over reactants. This says nothing about how long it takes to reach equilibrium or how much thermal energy is required to create the rearrangements that have to take place to get the products.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### sunkal

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##### Re: Gibbs Free Energy Change and Activation Energy
« Reply #2 on: August 02, 2019, 11:36:34 AM »
The standard Gibbs energy change given allows us to calculate the equilibrium constant K through the equation ΔGo = - RT ln K. Given that ΔGo = - 2.90 kJ, K = 3.22 at 298 K. To predict the spontaneity of the reaction you need to know the value of ΔG under the given conditions. If ΔG is negative, then the reaction is spontaneous up to equilibrium. The two should not be mixed up, although it is a common mistake made.

Now, if you know the transition state theory in chemical kinetics, the rate of a reaction is controlled by its activation Gibbs energy ΔG#. If the value of ΔS# is zero or negligible, then activation energy controls the rate. It so happens that for the reaction given the activation energy is very large so that the reaction is extremely slow at 298 K, so slow that diamond doesn't appear at all to change to graphite.

#### Corribus

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##### Re: Gibbs Free Energy Change and Activation Energy
« Reply #3 on: August 02, 2019, 12:28:10 PM »
btw, this thermodynamics of diamond to graphite transition was discussed here not too long ago:

https://www.chemicalforums.com/index.php?topic=95325.0
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### sunkal

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##### Re: Gibbs Free Energy Change and Activation Energy
« Reply #4 on: August 02, 2019, 05:15:57 PM »
A spontaneous reaction is one in which the energy of the products is lower than the reactants, so that at equilibrium products are favored over reactants.

That is not correct. The correct statement is: A spontaneous reaction at constant temperature and pressure is one in which the Gibbs energy (not just energy meaning internal energy) of the products is lower than that of the reactants, so that at equilibrium products are favored over reactants.

#### Corribus

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##### Re: Gibbs Free Energy Change and Activation Energy
« Reply #5 on: August 03, 2019, 08:47:49 AM »
That is not correct. The correct statement is: A spontaneous reaction at constant temperature and pressure is one in which the Gibbs energy (not just energy meaning internal energy) of the products is lower than that of the reactants, so that at equilibrium products are favored over reactants.
You shouldn't quote people out of context. I think it was pretty clear I was referring to Gibbs free energy.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### sunkal

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##### Re: Gibbs Free Energy Change and Activation Energy
« Reply #6 on: August 03, 2019, 09:39:24 AM »
You shouldn't quote people out of context. I think it was pretty clear I was referring to Gibbs free energy.
Both terms, energy and Gibbs energy, with very different meanings, are in use. So, using the term energy to imply Gibbs energy is not a good idea. Students may not be able to realize the implication. It is better to be more careful. Even I didn't understand that Gibbs energy was implied. Sorry if this caused a concern.