October 15, 2019, 07:34:55 AM
Forum Rules: Read This Before Posting


Topic: Delta G Naught and spontaneity  (Read 5095 times)

0 Members and 1 Guest are viewing this topic.

Offline galpinj

  • Full Member
  • ****
  • Posts: 102
  • Mole Snacks: +2/-5
Delta G Naught and spontaneity
« on: January 09, 2016, 11:54:05 PM »
Hey Everyone,

     I've been having some trouble with Delta Go, and couldn't find the answer online/in the forum.

1. I read in a few places that Delta Go is at STP and that all substances be 1M. This doesn't make sense to me because, given that  Delta Go = -RT ln Keq, the concentrations of Keq will often be different, not just [1]/[1].

2. I've had some trouble understanding Delta Go on a conceptual level. If Delta Go is itself negative, one can infer that the Keq > 1 and that the process will be spontaneous. My question is this, given that Delta Go is negative at equilibrium, won't the products continue to be favored, ultimately throwing the reaction out of equilibrium.

Thank you for the assistance

Offline thetada

  • Rhyming Chemist
  • Full Member
  • ****
  • Posts: 182
  • Mole Snacks: +18/-0
    • Rhyming Chemist
Re: Delta G Naught and spontaneity
« Reply #1 on: January 10, 2016, 03:31:08 AM »
K can't be less than 0 but it can be 1, more than 1 or greater than zero but less than 1. In the latter instance ln k is negative.  Since R and T (in kelvin) are always positive, G is negative for values of k greater than 1. This shows what dG really means, it's a gauge of the extent to which a reaction will go to completion. At 0oC, (and atmospheric pressure) water has a dG value of zero and the mixture should be half water and half ice. As T raises, dG becomes positive, and the proportion of ice decreases rapidly. The reverse is true as T decreases.

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 1626
  • Mole Snacks: +230/-11
Re: Delta G Naught and spontaneity
« Reply #2 on: January 11, 2016, 08:35:38 AM »
(note: there are Greek letters, symbols etc. just above the typing window as you type a post. There is no need to write stuff like "Delta Go" etc., which can be ambiguous.)

ΔG° refers to a defined standard state (which may be 1M at STP, but needn't be). Keq refers to the equilibrium situation, which is usually different from standard states. The difference of the equilibrium concentrations from 1M is related to the difference of ΔG° from 0.
Have you studied the derivation of the equation ΔG° = -RTlnKeq, or just been given it as something to memorise?
What quantity is equal to 0 at equilibrium? How is it related to ΔG°?

Quote
At 0oC, (and atmospheric pressure) water has a dG value of zero
That's ambiguous, what do you mean?
Quote
and the mixture should be half water and half ice.
No, it is water and ice, relative amounts irrelevant.
Quote
As T raises, dG becomes positive, and the proportion of ice decreases rapidly
There is no ice in equilibrium with water at any temp above 0°C at AP. All the ice will melt, given time.

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 3840
  • Mole Snacks: +243/-16
Re: Delta G Naught and spontaneity
« Reply #3 on: January 11, 2016, 09:38:00 AM »
Speaking only for myself, I like to reserve the use of the word "spontaneous" for cases in which ΔG is negative.  When ΔG = 0, we are at chemical equilibrium.  When ΔG° is negative, I use the word "favorable," meaning that Keq is greater than one, as you pointed out.

Offline thetada

  • Rhyming Chemist
  • Full Member
  • ****
  • Posts: 182
  • Mole Snacks: +18/-0
    • Rhyming Chemist
Re: Delta G Naught and spontaneity
« Reply #4 on: January 13, 2016, 04:50:39 AM »
This is very interesting. I'd appreciate it if anyone could explain what's wrong with my reasoning.

ΔG = ΔH - TΔS

I have the following data:
ΔH = -6010J mol-1
ΔS = -22.0
(Why Chemical Reactions happen, Keeler and Wothers, 2003)

I was a bit off, ΔG = 0 when T = 273.18K (Not exactly 0oC)

If ΔG = 0, (and T ≠ 0) then ln K must = 0 meaning that K = 1
That means that the ratio of reactants and products = 1 and in this case, since there is only one of each, their molar quantities must be equal.

If we go half a degree above water's freezing point (T = 273.65K),
ΔG = 10.3

Which gives the following value of K: 1.004537233

This is only a bit more than one, suggesting that products are only slightly more numerous than reactants, ie that some ice should remain.

What's the flaw in my reasoning? Is it that the expression for k is affected by the difference in phase between ice and water (in a similar vein, for example, to the way that solids / liquids are omitted from K expressions for gas-phase equilibria)?

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 1626
  • Mole Snacks: +230/-11
Re: Delta G Naught and spontaneity
« Reply #5 on: January 13, 2016, 08:28:10 AM »
This illustrates the dangers of applying equilibrium theory to heterogeneous (multi-phase) systems.
Quote
That means that the ratio of reactants and products = 1 and in this case, since there is only one of each, their molar quantities must be equal.
Ratio of what? Not amounts, but concentrations (strictly, activities, which are idealised concentrations). What are the concentrations of ice and water? They are not solutes in a solution. By convention, all pure substances (solid or liquid) are assigned an activity of 1. This does not change when you change the amounts of substances (as long as they are separate phases).
Thus at the equilibrium condition K = 1 = [water]/[ice] =1/1. This is irrespective of how much water or ice is present.
If you raise the temperature slightly K = 1.0045. But if you melt a little ice, [water] is still 1 and [ice] is still 1. You will never reestablish equilibrium, however much ice you melt. That is why pure phases have sharp melting points.
Do you see the difference between this situation and an equilibrium A  ::equil:: B in solution (or gas), where if at T1 K = 1 and [A] = [B ], and at T2 K > 1, some A reacts to B, [A] decreases and [B ] increases?

Offline thetada

  • Rhyming Chemist
  • Full Member
  • ****
  • Posts: 182
  • Mole Snacks: +18/-0
    • Rhyming Chemist
Re: Delta G Naught and spontaneity
« Reply #6 on: January 13, 2016, 08:33:20 AM »
I do. Thanks for your help.

Offline galpinj

  • Full Member
  • ****
  • Posts: 102
  • Mole Snacks: +2/-5
Re: Delta G Naught and spontaneity
« Reply #7 on: January 15, 2016, 09:32:43 PM »
Speaking only for myself, I like to reserve the use of the word "spontaneous" for cases in which ΔG is negative.  When ΔG = 0, we are at chemical equilibrium.  When ΔG° is negative, I use the word "favorable," meaning that Keq is greater than one, as you pointed out.

Thank you for the help everyone. I will try to be clearer when creating questions in the future, although I think most people grasped my dilemma. I think Babcock's response helps me to resolve the problem I was having with ΔG°. In essence, ΔG° doesn't need to have reactants and products in 1M concentrations. Moreover, a negative ΔG° is not so much that the process will continue spontaneously, but rather that the [products] is higher than the [reactants].

Sponsored Links