Chapter 19 Chemical Thermodynamics (Entropy, Free Energy and Equilibrium)

Why do some reactions happen spontaneously?

Spontaneous Change • Changes that occur without (continuous) outside influence. • Many spontaneous changes are exothermic (DH < 0) C3H8(g) + 5 O2 (g)  3 CO2(g) + 4 H2O(g) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

• Others are not (DH > 0) H2O

NaCl(s)  Na+(aq) + Cl-(aq)

• A process that is spontaneous in one direction is not spontaneous in the opposite direction. – Spontaneous processes are not reversible • What happens when you drop an egg to the floor?

• A reversible process is non-spontaneous – An equilibrium reaction for example

Spontaneous Change • There is natural tendency towards disorder. • When Temp > 0oC, ice spontaneously melts.

• Spontaneity depends on temperature.

Spontaneous Change • The tendency towards disorder is affected by: – Volume – Temperature – Gases produced in chemical reaction CaCO3(s) + HCl(aq)  H2O(l) + CO2(g) + CaCl2(aq)

– Reactions resulting in less complex products. N2O4(g)  2 NO2(g)

Entropy • Entropy, S: The thermodynamic property which is a measure of the randomness of a system. • It is a state function. DS = Sfinal - Sinitial DS = Sproducts - Sreactants • DS > 0, entropy (randomness) increases • DS < 0, entropy (randomness) decreases • An increase in entropy favors spontaneity.

Entropy & Spontaneity; The Second Law of Thermodynamics • Whenever a spontaneous process takes place, the entropy of the universe increases. DSuniverse > 0 DSuniverse = DSsystem + DSsurroundings • Entropy is not conserved • For a reversible process, DSuniv = 0 • For a spontaneous process, DSuniv > 0

Note: DSsys can actually decrease in a spontaneous process as long as DSsurr increases to offset it.

Entropy & Molecular Motion • Modes of motion: – translation – vibration – rotation

• The more energy stored in molecular motions, the higher the entropy.

The Third Law of Thermodynamics: The entropy of a perfect crystal at 0 K is zero.

T = 0 K, S = 0

Entropy Calculations • Standard Entropy, So, is reported in the Appendix. • In a “Hess’s Law type calculation” you can determine the entropy change for a reaction. o DS rxn   nS o (products)   mS o (reactants )

• What is the entropy change for the following reaction at 25oC and 1 atm? N2O4(g)  2 NO2(g)

Entropy, Enthalpy, & Gibb’s Free Energy • J. Willard Gibbs determined a relationship between enthalpy and entropy that describes the maximum useful energy obtainable (to do work) from a process at constant T & P. G = H - TS • G = Gibbs free energy • G is another thermodynamic state function, so: DG = Gfinal - Ginitial DG = DH - TDS • This relationship also allows us to determine the spontaneity of a system. DG > 0, reaction is non-spontaneous DG = 0, system is at equilibrium DG < 0, reaction is spontaneous

Thermodynamics • Enthalpy – DH > 0, Endothermic – DH < 0, Exothermic

• Entropy – DS > 0, more Random – DS < 0, more Ordered

• Gibbs free energy – DG < 0, Spontaneous – DG > 0, Non-spontaneous

Entropy, Enthalpy, & Gibb’s Free Energy DH

DS

-

-

-

+

+

-

+

+

DG = DH - TDS

Thermodynamics • At standard conditions we can use the thermodynamic tables (Appendix) to calculate changes in enthalpy, entropy, or Gibbs free energy. – 25oC and 1 atm (also 1 M for aqueous solutions) o DH rxn   nDH of (products)   mDH of (reactants ) o DS rxn   nS o (products)   mS o (reactants )

o DGrxn   nDG of (products)   mDG of (reactants )

Note: In the Thermodynamic Table (1) values for DGfo and DHfo are zero for elements in their standard state. (2) values for So are not zero, (the third law specifies that S = 0 only at 0 K.)

Spontaneous Change & DG • Standard Gibbs free energy, DGo, is the specific for 298K. DGo = DHo – TDSo DGo = DHo – (298K)DSo • However, at temperatures other than 298 K, we will assume that the DHo and DSo do not change significantly. – (DSo may change significantly if there is a phase change.)

• This allows us to estimate DG at other than standard temperature, yet based on the tabulated DHo and So. DGT ≈ DHo – TDSo

Spontaneous Change & DG for N2O4 Will N2O4(g) decompose spontaneously at 25oC? N2O4(g)  2 NO2(g) • At standard conditions we calculated DSo = 0.177 kJ/mol-K

Spontaneous Change & DG for N2O4 • Will the reaction proceed spontaneously at 100 oC? • At 100 oC (373 K) we will assume DSo = 0.177 kJ/mol-K DHo = 58.02 kJ/mol

Spontaneous Change & DG for N2O4 At what temperature will N2O4(g) begin to decompose spontaneously? N2O4(g)  2 NO2(g)

DG & Equilibrium Temperature • When DG = 0, the system is at equilibrium  G Products = G Reactants

DG  0 DH o  TDS o  0 DH o  TDS o DH o T DS o Assuming there is not significant change for DSo or DHo with temperature, we can estimate the Temperature for a system at equilibrium.

Calculate the temperature at which liquid water will be in equilibrium with vapor water. H2O(l) D H2O(g)

DG & Equilibrium • We know that equilibrium can be approached from either side of the reaction. Ca(OH)2(s) D Ca2+(aq) + 2 OH-(aq) • Starting with Ca(OH)2(s) – Q K – Reaction goes left to reach equilibrium

Q = [Ca2+][OH-]2

DG & Equilibrium • The Gibbs free energy change for a reaction is related to R = 8.314 J/K-mol equilibrium by the equation T = Kelvin Temp o DGrxn = DG rxn + RTlnQ Q = reaction quotient Ca(OH)2(s) D Ca2+(aq) + 2 OH-(aq) Q = [Ca2+][OH-]2 Spontaneous changes will occur if there is a decrease in free energy.

Starting with only Ca(OH)2(s) the reaction proceeds in the forward direction, but not far.

Ca(OH)2(s)

Ca2+(aq) + 2OH-(aq)

DG & Equilibrium • For the reverse reaction Ca2+(aq) + 2 OH-(aq) D Ca(OH)2(s)

1 Q [Ca 2 ][OH - ]2 Spontaneous changes will occur if there is a decrease in free energy.

Starting with only Ca2+ and OH- the reaction proceeds in the forward direction, to a large extent.

Ca2+(aq) + 2OH-(aq)

Ca(OH)2(s)

DG & Equilibrium • Since DGrxn is related to Q by; DGrxn = DGorxn + RTlnQ • and at equilibrium, DGrxn = 0, and Q = K 0 = DGorxn + RTlnK • we can derive an expression that allows determination of the thermodynamic equilibrium constant.

K e

 DG o / RT

K can be any of the equilibrium constants that we have discussed. Ksp, Kc, Ka, Kb, Kform, Kinst

DG & Equilibrium • Using the thermodynamic tables, determine a value for the solubility product constant of Ca(OH)2 under standard conditions. o DGrxn   nDG of (products)   mDG of (reactants )

o DGrxn  [DG of, Ca 2 ( aq)  2DG of, OH- ( aq) ] - [DG of, Ca(OH)2 ( s ) ] o DGrxn  [553.0  2(-157.3) ] - [-898.5] o DGrxn  30.9 kJ/mol