4th Set of Notes
How can one measure the entropy?
Relies on the idea that S = ∫ dQrev / T or S > ∫ dQirrev / T
Also based on the Third Law of Thermodynamics:
“The entropy of any pure, perfect crystal is zero at 0 Kelvin"
Thus unlike the enthalpy and the internal energy cases, for which we defined the
____________ ______________ as the reference point, and declared the
___________ of ______________ as being equal to zero for the _________
elemental substance in its most ______________ state at _____ ____________ of
pressure, here there is a more formal definition based on the disorder of the substance.
Thus define the standard molar entropy So, which is like the Hof for the enthalpy,
BUT, since there is a third law reference based on disorder, then So(elemental substance)
is NOT equal to zero at 1atm pressure, the normal std state and 298K.
Application to Chemical Rxns.
For a chemical reaction then: define Sorxn as being equal to:
Sorsn = prod nprod So (prod) - react nreact So(react)
Examples:
The third law definition bases entropies on zero disorder at 0K, so values of So in the
tables derived from knowing how entropy changes with Temp.
Temp Dependence of the Entropy
If we pick the heating to be done in a reversible way, then we can say that
S(T2) - S(T1) = ∫ dQrev/T
At Constant Pressure
S =
At Constant Volume:
dQ = C dT
so
S =
So to calculate the entropy at some other temp., assume a reversible change done at
constant Pressure or constant Volume
What if in raising the temperature, there is a phase transition?
In general, heating a compound from a low temp to high temp may involve several phase
changes.
If the phase transition is in EQUILIBRIUM, then it is a REVERSIBLE phase transition,
and:
For a phase transition at constant P and T,
Qp = Qrev = Htrans
So:
Strans = trans / Ttrans
So the ideal is to get So for the substance at any temp and pressure by increasing the
temperature of the substance to the equilibrium phase transition temp, use the enthalpy
and temp of transition, then increase the temp to the next transition, and so on and so on.
Example:
Obtain the entropy of a compound that is a gas at room temp and pressure.
So(25oC, 1atm) = So(0K) +
0
∫ T melt .......................
.................... Tmelt ∫ T boil ..................
+
+
Is there a pressure dependence?
Pressure Dependence of the Entropy
Liquids and Solids, Effect is very small, or negligible under normal circumstances.
For Gases this is different:
Consider a reversible change due to a small change dP in the pressure.
dS = dQrev / T
E = Q + W
so
dS =
For an ideal gas changing pressure form P1 to P2 at const. T,
dE =
So dS = - dWrev/T = P dV/ T
But PV = nRT = constant (for constant temperature)
So d(PV) = P ......... + ........V ................ = 0
PdV = -VdP,..........(at const. T)
So dS = PdV/ T ..................= - V dP/ T .......... = -nRdP / P
Integrating both sides we find:
So: S = S(P2) - S(P1) = -nR ln(P2/P1)
Application to the mixing of two ideal gases:
Initial Condition: Two gases A and B at pressures P
Final Condition: Two gases at mole fraction XA and XB
SA =
SB =
Smix = SA + SB =
What if the gases were identical?
ENTROPY From a Molecular Point of View
Microscopically, entropy is a measure of disorder. More disorder
_____________________ entropy.
Universe is becoming more disordered.
Feynman: disorder is the number of ways that the insides of a system can be arranged so
that from the outside the system looks the same.
Consider molar entropies of ice, water, and steam
Suggests that the water molecules in ice are more highly ordered.
May be qualitatively related to hardness: consider graphite vs diamond
Sometimes can rationalize it in certain reactions:
CO32- (aq) + H+(aq) -> HCO3-(aq) .................................S0 = + 148.1 J/(Kmol)
NH4+(aq) + CO32-(aq) ->
CH4 (aq) -> CH4(CCl4)
Quantitative interpretation obtained by Boltzmann.
S = k ln W
k = Boltzmann’s constant (1.318 x 10-23 J K-1)
R/NA
W = number of microscopic states of the system
REMEMBER ONE OF OUR CURRENT STATEMENTS OF THE SECOND LAW is
The sum of the entropy changes of the system and surroundings is positive.
Want a Thermodynamic variable that allows us to look only at the system!
Gibbs Free Energy
Want predictor of spontaneity based on common conditions used for chemical reactions.
Often Temp and Pressure are held constant!
G = Gibbs Free Energy = H-TS
Is G a state variable ? ...................................extensive or intensive
Definition chosen because it allows prediction of spontaneity for a constant temperature
and constant pressure process.
Can we show this?
Based on this statement of 2nd Law of Thermodynamics is: The Total Entropy of the
System and Its Surroundings Increases for an Irreversible Process. (Applies to both, not
to one independently)
dSsys + dSsurr > or = 0 ......................... OR dSsys > or = -dSsurr
Consider a system in thermal but not mechanical contact with the surroundings, both at
the same temperature. If dQ is the heat supplied to the system, then -dQ is the heat of the
surroundings.
Thus dSsys
Impose further constraint (Const. Volume)
PV work? dUsys = dQv; dSsys - dU/T > 0................... All terms for the system.
............................OR .......0 > dU - TdS ........................................... (A)
Further constraint of constant entropy:
................................... 0 > dUsys ............ OR (dUsys)S,V < 0 (closed, PV work)
OR with a further constraint of constant internal energy:
.......................................... T dSsys > 0
OR
The last criterion of a spontaneous process only based on the system properties for a
_________
system.
If instead the constraint of constant system pressure:
...........................dHsys = dQp so dSsys
................OR .................0 > dH - TdS ........................................... (B)
and the further constraints of Constant Entropy or Constant Enthalpy yield:
...........................(dH)P,S < 0 ................ OR ............. (dS)P,H > 0 ............(Closed System)
Again, these are criterion for spontaneity base on SYSTEM properties.
Consider Eqns. A and B (closed, no non-PV work)
dUsys -TdSsys < 0 (const V) ............................. dHsys - TdSsys < 0
New Thermodynamic Functions:
HELMHOLTZ free energy: A = U - TS
GIBBS free energy: G = H - TS
Expressing the differentials (these should be memorized)
dA =
dG =
At constant temp.
dA =
dG =
Are “A” and “G” State Functions?
Holding V and T constant for a closed system with only PV work, a criterion for
spontaneity is:
Holding P and T constant for a closed system with only PV work a criterion for
spontaneity is:
Note that dAsys tells about spontaneity for .....
And dGsys tells about spontaneity for .......
At const. T, V then, systems move spontaneously towards states of lower ________, the
equilibrium is reached when neither the forward or reverse process can occur.
So ........... (dA)T,Vsys = ____
(lower system internal energy, raise system entropy, ?????
For chemists, Gsys is more often used to predict the spontaneity, or the equilibrium
situation
Why?
(Lower system enthalpy, raise system entropy; minimize _______