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Home , Gibbs free energy

Chem 4501 Introduction to , 3 Credits Kinetics, and

Module Number 9—Active Learning Answers

1. Imagine you have an elementally pure substance at a constant of 273 K that behaves as an right up to the point of condensation at e4 bar; the molar of the liquid is 0.1 L. Plot the molar Gibbs free of this substance from 1 to 100 bar, labeling the axes of your plot including values, assuming no change from e4–100 bar.

Recall that the molar free energy of an ideal gas is

o ! P $ G = G + RT ln# & " Po %

so, at 1 bar, G = Go , since Po = 1 bar. And, since it’s an elementally pure substance, we can take the standard-state molar free energy as the free energy of formation, which will be 0. We also know (from ) that free energy varies with according to

"∂G % $ ' = V # ∂P &T

so at constant temperature, we may compute molar G(P) according to

P G P = Go + V dP! ( ) ∫1 or, more generally,

P G P = G P + 2 V dP! ( 2 ) ( 1) ∫ P 1

In the case of an ideal gas, we may use the ideal gas to compute the molar volume (RT/P). In the case of a liquid, we may consider the substance to be incompressible (i.e., the molar volume is a constant). That means

G P Go RT ln P G P G P V P P ( )gas = + ( )liq = ( cond ) + ( − cond ) where Pcond is the pressure at which the vapor condenses to a liquid.

So, a plot of molar free energy over the pressure range will begin at Go at P = 1 bar, and increase as RT ln P up to e4 bar (at which point the free energy will have increased by 4RT, or 9.0 kJ mol–1). The molar volume of the ideal gas at that pressure will be RT / P = 0.42 L. This value is greater than the molar volume of the liquid, which is consistent with condensation taking place.

Afterwards, the free energy will go up linearly with a slope of V (with volume expressed in the mildly unusual units of kJ mol–1 bar–1). As e4 ≈ 54.6, that means the endpoint will be

o G(P =100 bar) = G + 4RT + 0.1 L•(100 − 54.6 bar)

Making all the appropriate unit substitutions/conversions, that is equal to 9.5 kJ mol–1 at 100 bar.

G, kJ/mol vs P, bar 10 9 8 7 6 5 4 3 2 1 0 0 20 40 60 80 100 120

If the solid were to be included, it too would be incompressible, and of smaller volume (if it were not of smaller volume, the loss of on going from liquid to solid would not be overcome by a reduction in PV , and solidification would not take place), so another non-differentiable point on the curve would appear, and the free energy curve would then increase with a still smaller slope than that over the liquid range.

o –1 2. ΔrG / J mol = 1895 +3.363T for the phase change of carbon from coal () to diamond. At 298 K, how hard does Superman have to press to turn coal into diamonds? FYI, the densities of coal and diamond are 2.25 and 3.51 g cm–3, respectively.

As we learned in the last problem, for the two solids, we expect the free to be

G P Go V P 1 ( )s = s + s ( − )

When both solids are present at equilibrium (i.e., at the pressure at which coal turns into diamond), the two solids will have the same molar free energy, or

o o Ggraphite +Vgraphite (Peq −1) = Gdiamond +Vdiamond (Peq −1)

Let’s assume the equilibrium pressure is very high, for a moment. Then we can solve for Peq as

o o o o Gdiamond −Ggraphite Gdiamond −Ggraphite P = ( ) = ( ) eq " % (Vgraphite −Vdiamond ) 1 1 MWC $ − ' # ρ graphite ρ diamond & 1895+ 3.363•298 J mol-1 = ( ) " 1 1 % 12.0115 g mol-1 ( )$ 3 −1 − 3 −1 ' # 2.25×10 g L 3.51×10 g L &

Again, reconciling units, we have that Peq = 15,133 bar (justifying our approximation that Peq >> 1.

So, that Superman needs to press pretty darn hard to turn those charcoal briquets into diamonds…

3. For a pure-substance pressure/temperature (P/T) phase diagram, which of the below statements is/are always true when a given P/T point is on one of the lines on the diagram?

(a) The molar of all (d) Solid and liquid are in phases present are equal equilibrium (b) The chemical potentials of all (e) (a) and (d) phases present are equal

(c) ΔStrs = 0 (f) (b) and (d)