Chapter 19: Thermodynamics

1. Chapter 19: Thermodynamics First Law of Thermodynamics: energy cannot be created or destroyed, energy is conserved *total energy of the universe cannot change *you can transfer energy *∆Euniverse = ∆Esystem+ ∆Esurroundings= 0

2. -for an exo. reaction, “lost” energy 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

4. *-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 -occur in one direction only ex: H2O forming from H2 + O2 is spon, but the reverse is not -reverse process is nonspontaneous requires energy input

5. -temp is important when determining if a reaction is spon. ex: ice melting above 0°C is spon, but the reverse process is not

6. -a reversible process will proceed back and forth between the two end conditions -at equilibrium -results in no change in free energy

7. 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 incthe entropy: 1. solid <liquid < gas • reactions with more moles gaseous product than reactant *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) 3. inc in temp

8. 2nd Law of Thermodynamics -entropy of the universe must be + to be spontaneous ∆Suniv= ∆ Ssys+ ∆ Ssurr *for reversible∆ Suniv= 0 *for irreversible/spontaneous∆ Suniv> 0

9. Predict if ∆S is + or - (at constant temp) • H2O(ℓ)  H2O(g) * + b/c gas more freedom to move • Ag+(aq) + Cℓ-(aq)  AgCℓ(s) * - b/c solid is less free to move • 4Fe(s) + 3O2(g)  2Fe2O3(s) * - b/c solid is less free to move • N2(g) + O2(g)  2NO(g) * close to zero b/c mol of gas are the same -page 799

10. 3rd Law of Thermodynamics -entropy of a pure crystalline substance at absolute zero (0K) is zero -entropy will inc as temp inc

11. 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

12. 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°

13. 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.

14. 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 19.2 -page 1059 for values ∆G°= ∑nGf°(products) - ∑nGf°(reactants) -page 806

15. 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

16. 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°

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

18. -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