Chapter 19: Thermodynamics First Law of Thermodynamics: energy cannot be created or destroyed -total energy of the universe cannot change -you can transfer energy *∆E universe = 0 = ∆E system + ∆E surroundings

-for an exo. reaction, “lost” heat from the system goes into the surroundings -two ways energy “lost” from a system *converted to heat, q *used to do work, w *∆E = q + w *∆E is a state function

*-thermodynamics predicts whether a process will proceed under the given conditions spontaneous process- proceeds on its own without outside assistance regardless of speed (irreversible) ex- iron rusting, water in a freezer turns to ice -occur in one direction only ex: H 2 O forming from H 2 + O 2 is spon, but the reverse is not -reverse process is nonspontaneous  requires energy input -if there is less free energy after the reaction than before, the reaction is thermodynamically favorable

-a reversible process will proceed back and forth between the two end conditions -at equilibrium -results in no change in free energy -temp is important when determining if a reaction is spon. ex: ice melting above 0°C is spon, but the reverse process is not

Factors that determine favorability: 1. Enthalpy (∆H) -comparison of the bond energy of the reactants to the products bond energy: energy needed to break a bond -enthalpy is favorable for exo reactions and unfavorable for endo reactions, b/c products are more stable than reactants

2. Entropy (∆S) -relates to the randomness of a system -entropy change is favorable when the result is a more random system *∆S is + Factors that inc the entropy: 1. solid < liquid < gas 2.reactions with more moles gaseous product than reactant 3.inc in temp *equal # mol of gases have = entropy unless one is an atom and one a molecule (molec higher b/c they can rotate and vibrate and atoms cannot)

2 nd Law of Thermodynamics -entropy of the universe must be + to be spontaneous ∆S univ = ∆ S sys + ∆ S surr *for reversible  ∆ S univ = 0 *for irreversible/spontaneous  ∆ S univ > 0

Predict if ∆S is + or - (at constant temp) 1.H 2 O(ℓ)  H 2 O(g) * + b/c gas  more freedom to move 2.Ag + (aq) + Cℓ - (aq)  AgCℓ(s) * - b/c solid is less free to move 3.4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) * - b/c solid is less free to move 4.N 2 (g) + O 2 (g)  2NO(g) * close to zero b/c mol of gas are the same -page 799

3 rd Law of Thermodynamics -entropy of a pure crystalline substance at absolute zero (0K) is zero -entropy will inc as temp inc

standard molar entropies (S°)- entropies for substances in their standard states (1atm) -page 801 table 19.1 *gases > liquids > solids *inc with inc molar mass *inc with inc # atoms

Entropy Change in a Chem Reaction ∆ S° = ∑nS°(products) - ∑nS°(reactants) (-) = decrease in entropy (+) = increase in entropy **equations must be balanced **coefficient from balanced equation goes in front of S°

Gibbs Free Energy (G) ∆G = ∆H - T∆S -using standard conditions  ∆G° = ∆H° - T∆S° -if ∆G < 0, the reaction is spon. in the forward direction -if ∆G = 0, the reaction is at equilibrium -if ∆G > 0, the reaction in the forward direction is nonspon., but reverse is spon.

Standard Free Energies of Formation (∆G° f ) -change of free energy when a substance is formed from its elements in their standard state -pg 806 table page 1059 for values ∆G° = ∑nG f °(products) - ∑nG f °(reactants) -page 806

Free Energy and Temperature -may want to examine reactions not at 25°C -think about: ∆G = ∆H - T∆S -rewrite as: ∆G = ∆H + (-T∆S) -copy table 19.3 page 809

Free Energy and the Equilibrium Constant Nonstandard Conditions -most reactions occur under nonstandard conditions ∆G = ∆G° + RT ln Q R= 8.31J/Kmol T= temp Q= reaction quotient (partial pressure in atm for gases and solutes conc in M) *under std. conditions, Q=1(ln Q=0) and the equation becomes ∆G = ∆G°

Relationship Between ∆G° and K -at equilibrium, ∆G = 0 and Q = K ∆G = ∆G° + RT ln Q -from above we get: 0 = ∆G° + RT ln K ∆G°= - RT ln K -can find K if we know ∆G° ln K = ∆G°/-RT K = e -∆G°/RT

-if ∆G° is -, ln K must be + and K>1 -the more - ∆G° is the larger the K is -if ∆G° is +, ln K must be - and K<1