Gibbs Free Energy

Gibbs free energy is a measure of chemical energy

All chemical systems tend naturally toward states of minimum Gibbs free energy G = H - TS Where: G = Gibbs Free Energy H = Enthalpy (heat content) T = Temperature in Kelvins S = Entropy (can think of as randomness) Gibbs Free Energy

• Products and reactants are in equilibrium when their Gibbs free energies are equal

• A chemical reaction will proceed in the direction of lower Gibbs free energy (i.e., ΔGr < 0)

…so the reaction won’t proceed if the reaction produces an increase in Gibbs free energy Gibbs Free Energy

ΔG°r = ΣnG°f (products) - ΣnG°f (reactants)

ΔG°r > 0, backwards reaction with deficient energy ΔG°r < 0, forwards reaction with excess energy ΔG°r = 0, reaction is in equilibrium ΔG°r is a measure of the driving force

o ΔG f = free energy of formation Thermodynamics

For a phase we can determine V, T, P, etc., but not G or H

We can only determine changes in G or H as we change some other parameters of the system

Example: measure ΔH for a reaction by calorimetry - the heat given off or absorbed as a reaction proceeds

Arbitrary reference state and assign an equally arbitrary value of H to it: Choose 298.15 K/25°C and 0.1 MPa/1 atm/1 bar (lab conditions) ...and assign H = 0 for pure elements (in their natural state - gas, liquid, solid) at that reference Thermodynamics

In our calorimeter we can then determine ΔH for the reaction:

Si (metal) + O2 (gas) = SiO2 ΔH = -910,648 J/mol

= molar enthalpy of formation of quartz (at 25°C, 1 atm) It serves quite well for a standard value of H for the phase

Entropy has a more universal reference state: entropy of every substance = 0 at 0K, so we use that (and adjust for temperature)

Then we can use G = H - TS to determine G for quartz = -856,288 J/mol Thermodynamics

o ΔGR = −RT ln K

K=equilibrium constant at standard T T in kelvin 298.18K R=gas constant=1.987 cal/molo

o ΔGR = −1.364log K

o −ΔGR K =10 1.364 o Example: What is the ΔG R of calcite dissociation? 2+ 2- CaCO3 Ca + CO3

o Use data in appendix B for ΔG f

o ΔG R = [(-132.3)+(-126.17)] - [(-269.9)] = +11.43 kcal

(+) means that the reaction goes from right to left so K must be small

What is the value of K?

o −ΔGR K =10 1.364

K = 10(-11.43/1.364) = 10-8.3798 = 4.171 x 10-9 What if T ≠ 25oC? Use the Van’t Hoff Equation

o ΔGR = −RT ln K and ΔG°r = ΔH°r-TΔS°r We can derive: lnKT - lnKT° = (-ΔH°r/R)(1/T-1/T°) o o ΔH R  1 1  log KT = log KT −  −  2.3025R  T 298.15 

o ΔH R Enthalpy of reaction R=1.987 cal/mol° T in Kelvin Example: What is KT of calcite dissociation at T=38°C?

o o ΔH R  1 1  log KT = log KT −  −  2.3025R  T 298.15  o = [(-129.74)+(-161.8)] - [(-288.46)] = -3.08 ΔH R

−9 − 3.08  1 1  log KT = log(4.71x10 ) −  −  = −9.0532 2.3025(1.987)  311 298.15  −10 -9 KT = 8.85x10 (KT° = 4.171 x 10 )

When T increases, K decreases Thermodynamics Summary thus far: – G is a measure of relative chemical stability for a phase – We can determine G for any phase by measuring H and S for the reaction creating the phase from the elements – We can then determine G at any T and P mathematically • Most accurate if know how V and S vary with P and T – dV/dP is the coefficient of isothermal compressibility – dS/dT is the heat capacity (Cp)

If we know G for various phases, we can determine which is most stable • Why is melt more stable than solids at high T? • Is diamond or graphite stable at 150 km depth? • What will be the effect of increased P on melting?