How would you define the concept of equilibrium in terms of free energy and entropy?

2 Answers
Feb 4, 2016

Answer:

At equilibrium: #DeltaG=G_B-G_A=0#.

#DeltaS=(DeltaH)/T#

Explanation:

Consider the following equilibrium:

#A(g)rightleftharpoonsB(g)#

Initially we have #1.0 mol# of #A# and none of #B#.

The total free energy of the system #G# is defined by:

#G=G_A+G_B#

As #A# changes to #B#, #G_A# will decrease and #G_B# will increase. The reaction will proceed as long as #G_B# is less than #G_A#. When the system reaches equilibrium , #G_A# and #G_B# will have the same value #DeltaG=G_B-G_A=0#. The system has reached the minimum free energy.

Regarding entropy #DeltaS#. Since the change of free energy is given by:

#DeltaG=DeltaH-TDeltaS=> " At equilibrium "DeltaS=(DeltaH)/T#

where #DeltaH# is the change on the enthalpy of the system at equilibrium, and #T# is the Kelvin temperature.

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For much more details on this topic, watch this video on:
Thermodynamics | Free Energy, Pressure & Equilibrium.

Feb 4, 2016

At a general equilibrium, #DeltaG = 0#, and it is likely that #DeltaH# and #DeltaS# are nonzero.


An example of an equilibrium is a phase transition, such as vaporization, freezing, etc. Let's say we looked at vaporization, the liquid-vapor dynamic equilibrium at a boiling point.

That would be looking at the enthalpy change (e.g. #DeltaH_"vap"#) of a substance with respect to pressure (because vapor pressure has to reach atmospheric pressure) but a constant temperature---the temperature stays the same until the substance fully vaporizes.

ENTHALPY VS. TEMPERATURE, ENTROPY, PRESSURE, AND VOLUME

From this equation we can derive the relationship between enthalpy change, temperature, entropy, pressure, and volume (Physical Chemistry: A Molecular Approach, McQuarrie):

#H = U + PV#

#\mathbf(dH = dU + d(PV))#

#= delq_"rev" + delw_"rev" + PdV + VdP#

Since #dS = (delq_"rev")/(T)#:

#= TdS - cancel(PdV + PdV) + VdP#

#color(blue)(dH = TdS + VdP)#

or

#color(blue)(DeltaH = TDeltaS + VDeltaP)#

ACHIEVING "THE" THERMODYNAMICS RELATIONSHIP

Now, it would be nice to know how to rewrite #V#, because I want to achieve #DeltaG = DeltaH - TDeltaS#, which relates the three state functions that we utilize probably the most in thermochemistry. That means I should introduce #DeltaG# somehow.

Therefore, an expression for Gibbs' free energy is a way to do it. We could start at the Maxwell relation, which is:

#\mathbf(dG = -SdT + VdP)#

At a constant temperature, the partial derivative of the Gibbs' free energy with respect to pressure is the volume (#DeltaT = 0, :. -SDeltaT = 0#):

#((delG)/(delP))_T = V#

Therefore, we can substitute to get:

#dH = TdS + ((delG)/(delP))_TdP#

Phase changes where you watch something while it boils is at a constant atmospheric and vapor pressure (stuff is vaporizing while the vapor pressure is equal to the atmospheric pressure) as well as a constant temperature.

Since the Gibbs' free energy is supposed to change with respect to pressure, when pressure doesn't change, at least for an ideal substance, #DeltaG = 0# and #DeltaP = 0#. So, #cancel(((delG_"vap")/(delP))_T)^(0)cancel(dP)^(0) = 0#.

Therefore, for something boiling:

#dG_"vap" = 0 -> color(blue)(dH_"vap" = T_"vap"dS_"vap")#

or

#DeltaG_"vap" = 0 -> color(blue)(DeltaH_"vap" = T_"vap"DeltaS_"vap")#

So when something is boiling, if you know the enthalpy and the temperature, you should able to calculate the entropy if the pressure is also constant.