AP Chemistry Chapter 17

1. AP Chemistry Chapter 17 Spontaneity of Reaction

2. Spontaneous reactions • What does that mean? • Some occur without any “help” • Others require some “help” • No help – ice cube melting • Help – wood burning

3. If a reaction is spontaneous under a certain set of conditions, the reverse reaction must be nonspontaneous • In any spontaneous change, the amount of free energy available decreases toward zero as the process proceeds towards equilibrium.

4. Reactions tend to be spontaneous when: • it leads to lower energy = -∆H • But not always!!!! • Also tend to be spontaneous if the reaction results in an increase in randomness • Entropy S • Greater entropy – more random the system is. +∆S increase in entropy ∆S>0 • - ∆S decrease in entropy ∆S<0

5. Page 448 Example 17.1 Predict sign of ∆S

6. In general, nature tends to move spontaneously from more ordered to more random state • Entropy increases in the order: • s <l < g • Increasing temperature of a substance increased its entropy

7. Third Law of Thermodynamics • A completely ordered pure crystalline solid has an entropy of zero at 0K

8. ∆S for reactions • Pg. 450 Table of Standard Entropies • Used to calculate the standard entropy change, ∆So, for reactions. • ∆So = ∑ So products – ∑So reactants • Must remember to multiply by the number of moles from balanced equation

9. Note that So is a positive quantity for both compounds and elements; can be negative for ions in solutions

10. Reactions which So is positive tend to be spontaneous, at least at high temperatures. • H2O(s) H2O(l) ( ∆S > 0) • H2O(l) H2O(g) (∆S > 0) • Fe2O3(s) + 3H2(g)  2Fe(s) + 3H2O(g) (∆S > 0) • All of these reactions are endothermic (∆H>0) • They become spontaneous at high temperatures

11. A reaction that results in an increase in the number of moles of gas is accompanied by an increase in entropy. • If the number of moles of gas decreases, ∆S is a negative quantity

12. Elements have nonzero standard entropies • Standard molar entropies of pure substances are always positive quantities • Aqueous ions may have negative So values

13. Among substances of similar structure and physical state, entropy usually increases with molar mass • Molecule becomes more complex, more ways for the atoms to move about with respect to one another (higher entropy)

14. Pg. 451 sample • Example 17.2

15. Second Law of Thermodynamics • In a spontaneous process, there is a net increase in entropy, taking into account both system and surroundings. • ∆Suniverse = ∆Ssystem + ∆Ssurroundings > 0 • spontaneous

16. Gibbs Free Energy G • Two quantities affect reaction spontaneity; • enthalpy, H and entropy, S • Put them together in a way that the signs will give us a clue • G = H – TS • T = kelvin temp

17. ∆G – for a reaction at constant temp and pressure, represents that portion of the total energy change that is available to do useful work – is a state property • Depends only on the nature of products and reactants and the conditions (temp/pressure/concentration), not on the path by which the reaction is carried out

18. - ∆G = spontaneous • + ∆G = not spontaneous (reverse is spontaneous ∆G = 0 system is at equilibrium (no tendency for reaction to occur in either direction)

19. ∆G measure of the driving force of a reaction • Reaction, at constant pressure and temperature, go in such a direction as to decrease the free energy of the system • Products have lower free energy, reaction will go in that direction • Reactants have lower free energy, reaction will go in that direction (means the reverse rxn spontaneous)

20. Gibbs-Helmholtz equation • ∆G = ∆H - T∆S • To make ∆G negative; • Negative value for ∆H (exothermic) • Positive value for ∆S (less ordered)

21. Gibbs-Helmholtz equation • Valid under all conditions but we will apply it only under “standard conditions” • Meaning: gases are at one atmosphere partial pressure • Ions or molecules in solution are at one molar concentration • ∆G = standard free energy change

22. ∆Go = ∆Ho - T∆So • now we can use the tables in the book • If ∆Go is negative = spontaneous at standard conditions • If ∆Go is positive = nonspontaneous at standard conditions • ∆G = 0 system is at equilibrium at standard conditions

23. Calculation of ∆G at 25oCFree Energies of Formation!! • Make sure units are correct • Use ∆H is kJ, convert ∆S for J/K to kJ/K • Pg. 455 Example 17.3 • Pg. 456 Example 17.4, 17.5

24. Chart on pg. 458