October 21, 2021, 05:42:15 PM
Forum Rules: Read This Before Posting


Topic: gibbs free energy  (Read 543 times)

0 Members and 1 Guest are viewing this topic.

Offline carlepy

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
  • Gender: Female
  • Hi my name is Carla. I am a nuclear scientist
gibbs free energy
« on: May 21, 2021, 01:41:15 AM »
Fellow chems (hi from a physicist)

If I have two different reactions with ΔG1= -10 and ΔG2 = -100 can I say that ΔG2 is more favorable than ΔG1? That reaction 2 would ocurr easier that reaction 1?

Thank you

Offline Meter

  • Full Member
  • ****
  • Posts: 232
  • Mole Snacks: +14/-3
  • Take what I say with a grain of salt
Re: gibbs free energy
« Reply #1 on: May 21, 2021, 03:15:55 AM »
AFAIK Gibbs free energy can be used as a measure for how easy it is to make a reaction happen under the given conditions. Even reactions with positive Gibbs energy (under standard conditions) can also be made favorable through for example heating, but then the Gibbs energy of the heated system would change.

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 1955
  • Mole Snacks: +284/-12
Re: gibbs free energy
« Reply #2 on: May 21, 2021, 06:32:19 AM »
Reaction 2 is thermodynamically more favourable than reaction 1. Whether it occurs more easily is a question of kinetics, and depends on activation energy rather than ΔG.

Imagine you have to cross a hill to get to a valley on the other side. You might be more eager to get there the lower the valley is (warmer climate perhaps), but how hard it is to get there depends on the height of the hill you have to cross.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3186
  • Mole Snacks: +485/-22
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: gibbs free energy
« Reply #3 on: May 21, 2021, 09:40:32 AM »
Honestly, I've never liked speaking in terms of "more favorable" or "less favorable", particularly when describing two completely different reactions. I think you have to be a little careful when using colloquial words like "favorable" to describe the implications of scientific concepts - it is easily misleading. What does favorable mean, exactly? Based on the common meaning of "favorable" in English, a student might be inclined to think it speaks to rate. (Or, in the OP's only language, "easiness".) But it doesn't. To latch on to mjc's analogy, the deepness of the valley on the other side says nothing of how high of a mountain you have to cross to get there, and therefore how long the journey will take. The favorability, as indicated by Gibbs energy, refers to how far away from equilibrium a system is and in which direction the system will move/change under the current conditions. I've always thought "spontaneity" is a better word to use than "favorability", and I also think it's better to imagine it generally being a binary concept - either a reaction/process is spontaneous in a forward direction or it isn't under a certain set of conditions or a certain point in time, rather than it's more or less so. Considering (as an example) that the conversion of diamond to graphite is a spontaneous/favorable process that basically doesn't happen over any realistic timescale, I think teachers and textbooks do a disservice to students when they speak of Gibbs energy in terms that could be mistaken to relate to how fast the process is likely to happen.

Just some food for thought.
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

Offline carlepy

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
  • Gender: Female
  • Hi my name is Carla. I am a nuclear scientist
Re: gibbs free energy
« Reply #4 on: May 24, 2021, 12:56:49 AM »
thank you guys!!!!
yall are fantastic :)

Offline Meter

  • Full Member
  • ****
  • Posts: 232
  • Mole Snacks: +14/-3
  • Take what I say with a grain of salt
Re: gibbs free energy
« Reply #5 on: May 24, 2021, 03:17:05 AM »
Another note:

In the equation ΔGo = -RT ln(K) => K = e-ΔGo/RT. K depends on ΔGo and the equation shows that large ΔGo implies small K and very negative ΔGo implies large K.

The size of ΔGo tells us how favorable the reactants/products of an equilibrium is!

Sponsored Links