# Is delta G zero at boiling point?

##### 1 Answer

Yes, it is. At the boiling point,

For a proof of this, look below.

Take a look at this diagram:

where

#mu = barG = G/n# ;

#mu = # Chemical Potential (like high and low energy states)#G = # Gibbs' Free Energy#barG = # Molar Gibbs' Free Energy#S = # Entropy#barS = # Molar Entropy#n = # moles

This graph basically says, within each particular phase, the negative change in Gibbs' Free Energy per mole of substance as the temperature changes at a constant pressure is equal to the entropy per mole of substance. This can be written as:

#\mathbf(-((delmu)/(delT))_P = -((delbarG)/(delT))_P = barS)#

or

#-n((delmu)/(delT))_P = -((delG)/(delT))_P = S#

On this graph, outside of phase changes, the pressure is assumed to be *constant* within a particular phase (that's what the subscript P means) but the temperature is *not* (

At the boiling point, you're at *constant* but the temperature is *also* constant.

Let us write out the following **Maxwell relation**:

#DeltaH = TDeltaS + VDeltaP#

or the differential form

#\mathbf(dH = TdS + VdP)# where

#H# is enthalpy,#V# is volume, and#P# is pressure, we can see thatenthalpy depends on temperature, pressure, and entropy(entropy is not constant in a phase change since the moving particles are starting to change speed due to the change in volume).

When the temperature is *constant* but the pressure is *not* constant, this relationship holds true as-is. Since pressure *is* constant, we can eliminate

#dH = TdS + cancel(VdP)#

Thus, integrating with respect to entropy at a constant temperature, we get:

#DeltaH_(vap) = T_(vap) int_(S_1)^(S_2) dS = T_(vap)DeltaS_(vap)#

But wait, that means...

#DeltaH_(vap) - T_(vap)DeltaS_(vap) = 0#

#DeltaG_(vap) = cancel(DeltaH_(vap) - T_(vap)DeltaS_(vap))#

#color(blue)(DeltaG_(vap) = 0)#

**So yes,** **0 at the boiling point (and also at the freezing point).**