1. ENTROPY Measure of: Level of disorder in a system or Number of Microscopic Energy Levels Available to a Molecule (i.e. microstates)

2. SPONTANEOUS CHEMICAL PROCESSES Spontaneity = Tendency for process to occur naturally e.g. Iron tends to rust, diamond tends to turn to graphite, (fortunately, over a very long time period - relax, ladies & DeBeers!) dead plant material tends to decay, ice tends to melt at room temperature, brain cells tend to decay with time!

3. Spontaneity  Are all spontaneous changes exothermic?

4.  No, but most exothermic processes are spontaneous.  Some reactions may be spontaneous under one set of conditions, but non-spontaneous under other conditions.  e.g. Formation of Lime from Chalk: CaCO 3 (s) CaO (s) + CO 2 (g) H = +178.3 kJ Reaction becomes spontaneous > 800 0 C, even though it is endothermic! Spontaneity

5. Absolute Entropy (S)  With increasing temperature, the entropy of a molecule increases (more microscopic energy levels beome available, e.g. vibrations, rotations etc.)  As the temperature of a molecule approaches absolute zero, microscopic energy levels also approach zero, so its entropy also approaches zero.

6. S = k lnW W = No. of microscopic energy levels In perfectly ordered solid, at 0K, W = 1 S = k ln 1 = 0 Absolute Entropy (S)

7. this leads to:

8. the 3rd Law of Thermodynamics

9. 3 rd Law of Thermodynamics  "the Entropy of a perfect crystalline substance approaches zero as the absolute temperature appraches zero"

10. Entropy Changes ( S)  Reaction Entropy ( S r )  S r = S products - S reactants

11. Standard Entropy of Reaction ( S 0 r )  ∆ S 0 r = (Entropy of products in standard states at temp. T) - (Entropy of reactants in standard states at temp. T) for : aA + bB cC + dD  ∆ S 0 r = cS 0 (C) + dS 0 (D) - aS 0 (A) - bS 0 (B)

12. Standard Entropy Change Calculation  Calculate S 0 r at 25 0 C for: N 2 H 4 (l) + 3 O 2 (g) 2NO 2 (g) + 2 H 2 O (l) S 0 r = 2 x S 0 (H 2 O (l) )+ 2S 0 (NO 2 (g) )- S 0 ( N 2 H 4 (l) ) - 3S 0 (O 2 (g) ) = 2 mol (69.91 JK -1 mol -1 ) + 2 mol (239.95 JK -1 mol -1 ) - 1 mol (121.21 JK -1 mol -1 ) - 3 mol (205.03 JK -1 mol -1 ) = -116.58 JK -1

13. So what happened to the 2nd Law of Thermodynamics?

14. 2 nd Law of Thermodynamics  In any spontaneous process, the Entropy of the Universe (i.e. system + surroundings) must increase: so  S sys + S surr >0