From WolfWikis

LECTURE 11

Enthalpy H versus pressure

In Lecture 6 we were forced to leave some unfinished business: how does pressure affect enthalpy H?.

As we showed above we have the following relations of first and second order for G

We also know that by definition:

G = H - TS

Consider an isothermal change in pressure. We get:

As we can see for an ideal gas there is no dependence at all.

Gibbs free energy G versus pressure

The first order partial on G versus P (see above) is the volume V; this allows us to find the dependence of G on P by simply integrating over the volume V from one pressure to the other.

Ideal gas

For a mole of ideal gas we can use the gas law to integrate volume over pressure and we get (see Eq 22.57/58)

ΔG_molar = RT ln [P₂/P₁]

It is customary to identify one of the pressures (P₁} with the standard state of 1 bar and use the plimsoll to indicate the fact that we are referring to a standard state by writing:

G_molar(P) = G^o_molar + RT ln[P/1]=G^o_molar + RT ln[P]

The fact that we are making the function intensive (per mole) is usually indicated by putting a bar over the G symbol, although this is often omitted for G^o_molar

Solids

For solids the volume does not change very much with pressure ( the isothermal compressibility κ is small), so can assume it more or less constant:

G(P_final)=G(P_initial)+ ∫ VdP (from init to final) ≈ G(P_initial)+ V.∫ dP (from init to final)=G(P_initial)+ VΔP

The Gibbs-Helmholtz expression

For step 3 we made use of the relationship between C_p and H and C_p and S

We said before that S is a first order derivative of G. As you can see from this derivation the enthalpy H is also a first order derivative, albeit not of G itself but of G/T.

The last step in the derivation simply takes the step before twice -say for the G and H at the begin and end of a process- and subtracts the two identical equations leading to a Δ symbol. In this differential form the Gibbs-Helmholtz equation can be applied to any process.

Gibbs free energy at function of T.

If heat capacities are know from 0K we could -as we have seen- determine both enthalpy and entropy by integration:

As we have seen we must be careful at phase transitions such as melting or vaporization. At these points the curves are discontinuous and the derivative C_p is undefined.

We also discussed the fact that the third law allows us to define S(0) as zero in most cases. For the enthalpy we cannot do that so that our curve is with respect to an undefined zero point. We really should plot H(T) - H(0) and leave H(0) undefined.

Because the Gibbs free energy G= H-TS we can also construct a curve for G as a function of temperature, simply by combining the H and the S curve:

Interestingly, if we do so, the discontinuties at the phase transition points will drop out for G because at these points Δ_trsH = T_trsΔ_trsS.

The H(0) problem does not disappear so that once again our curve is subject to an arbitrary offset in the y-direction. The best thing we can do is plot the quantity G(T) - H(0) and leave the offset H(0) undefined.

We have seen above that the derivative of G with temperature is -S. As entropy is always positive, this means that the G-curve is always descending with temperature. It also means that although the curve is continuous even at the phase transitions, the slope of the G curve is not, because the derivative -S makes a jump there. Fig. 22.7 in the book shows an example of such a curve for benzene. Note the kinks in the curve at the mp and the bp.

Tabulation of H,S and G. Frozen entropy

There are tables for H^o(T)-H^o(0), S^o(T) and G^o(T)-H^o(0) as a function of temperature for numerous substances. As we discussed before the plimsoll defines are standard state in terms of pressure (1 bar) and of concentration reference states -if applicable-, but temperature is the one of interest.

For most substances the Third Law assumption that lim[S^o(T)] for T->0 = 0 is a reasonable one but there are notable exceptions. Carbon monoxide is a good example. In the solid form the molecules should ideally be fully ordered at absolute zero, but because the size of the carbon atom and the oxigen atom is not very different and the dipole of the molecule very small, it is quite possible to put in a molecule 'upside down', i.e. with the oxygen on a carbon site and vice versa. At higher temperatures -at which the crystal is formed- this lowers the G function, because it increases entropy. In fact we could say that if we could put each molecule into the lattice in two different ways, the number of ways W_disorder we can put N molecules in into the lattice is 2^N.

This leads to an additional contribution to the entropy of

S_disorder = k.lnW_disorder= N.k.ln2 = Rln2 = 5.7 J/Kmole.

Although at lower temparatures the entropy term in G = H-TS becomes less and less significant and the ordering of the crystal to a state of lower entropy should become a spontaneous process, in reality the kinetics are so slow that the ordering process does not happen and solid CO therefore has a non-zero entropy when approaching 0K.

In principle all crystalline materials have this effect to some extent but CO is unusual because the concentration of 'wrongly aligned' entities is of the order of 50% rather than say 1 in 10¹³ (a typical defect concentration in say single crystal silicon).

Transitions of order one and two

Note that although the G curve is continuous, its first order derivatives (such as S and H) are discontinuous at melting. This is why this transition it called a first order transition. We could say that:

G is continuous but has a kink

The first order derivatives (H,S,..) are discontinuous (have a jump)

The second order derivatives (C_p, ..) have a singularity (go to ∞)

More subtle transitions where G is continuous, H and S are also continuous but have a kink and the discontinuity is only found in the second order derivatives (such as C_p) also exist. They are called second order transitions. In such a case:

G is continuous and has no kink

The first order derivatives (H,S,..) are continuous (but have a kink)

The second order derivatives (C_p, ..) are discontinuous (have a jump)

This classification goes back to Ehrenfest. Obviously it based on the question: what order derivative is the first to go discontinuous? Of course we could extend this principle and define third order transitions but there are reasons to be doubtful that such things exist.

Another problem is that it is assumed that the order must be integer: 1,2, etc. Is it possible to have a transition of intermediate non-integer order, say 1.3? Although derivatives of fractional order are beyond the scope of the chemistry curriculum the mathematics does exist (Liouville).

Schematic comparison of G,S and C_p for 1st and 2nd order transitions