Exploring Equilibrium and Free Energy in Chemical Reactions

1. Free Energy and Equilibrium

2. When components of a reaction are mixed, they will proceed, rapidly or slowly (depending on kinetics) to the equilibriumposition. • Equilibrium position: defined as the point at which forward and reverse reaction rates are equal (Chapter 13). • Thermodynamics point of view: equilibrium point occurs at the lowest value of free energy available to the reaction system.

3. Consider the simple reaction below: • A(g) ⇌ B(g) • where 1.0 mole of gaseous A is placed in a reaction vessel at a pressure of 2.0 atm. • Total free energy of system: • GTotal = GA + GB • Figure (a) illustrates the initial free energies of A and B.

4. As A changes to B, GA will decrease because PA is decreasing (b). • GB will increase because PB is increasing. • Reaction will proceed to right (products) as long as GTotal decreases (GB is less than GA). • At some point the pressures of A and B reach the values such as GA is equal to GB.

5. The system is at equilibrium and has reached minimum free energy (c). • No longer driving force to change A to B or B to A, so system remains at this position and pressures of A and B remain constant.

6. A(g) ⇌ B(g) 1.0 mol A(g) at 2.0 atm • Plot of free energy versus the mole fraction of A is shown in (a) below. • Minimum free energy is reached when 75% of A has been changed to B. • At this point, the pressure of A is 0.25 times the original pressure or • (0.25)(2.0 atm) = 0.50 atm • Pressure of B is • (0.75)(2.0 atm) = 1.5 atm

7. K (equilibrium constant) can be calculated using equilibrium pressures: • Overall free energy curve is shown in (c). • This demonstrates any mixture of A(g) and B(g) containing 1.0 mol (A plus B) at a total pressure of 2.0 atm will react until it reaches the minimum in the curve.

8. In summary, • Reactions proceed to minimum free energy (equilibrium) • Corresponds to the point where • Gproducts = Greactants • or • ΔG = Gproducts – Greactants = 0

9. Remember, • ΔG = ΔGo + RT ln(Q) • At equilibrium, ΔG equals 0 and Q = K. • So ΔG = 0 = ΔGo + RT ln(K) • Or ΔGo = -RT ln(K)

10. ΔGo = -RT ln(K) • Case 1: ΔGo = 0. • When ΔGo = 0 the free energies of reactants and products are equal when all components are in standard states (1 atm for gases). • System is at equilibrium when the pressures of all reactants and products are 1 atm, which means K = 1.

11. ΔGo = -RT ln(K) • Case 2:ΔGo < 0. • ΔGo (Goproducts – Goreactants) is negative • Goproducts < Goreactants • System is not at equilibrium and system will adjust to the right to reach equilibrium. • K will be greater than 1 (pressures of products at equilibrium are greater than pressures of reactants).

12. ΔGo = -RT ln(K) • Case 3:ΔGo > 0. • ΔGo (Goproducts – Goreactants) is positive • Goreactants< Goproducts • System will not be at equilibrium and system will adjust to the left to reach equilibrium. • K will be less than 1(pressure of the reactants will be greater than 1 atm and the pressure of the products will be less than 1 atm).

13. Cases 1, 2 and 3 are summarized in Table 16.6. • The value of K for a specific reaction can be calculated from the equation • ΔGo = -RT ln(K).

14. Free Energy and Work • The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy. • “Free energy” is energy “free” to do work. • wmax = ΔG • The amount of work obtained is always less than the maximum.

15. Henry Bent’s First Two Laws of Thermodynamics • First law: You can’t win, you can only break even • Second law: You can’t break even • Read this section in the book!!!!!!!!!!!!