Introduction to Entropy by Mike Roller

Entropy (S) = a measure of randomness or disorder MATTER IS ENERGY. ENERGY IS INFORMATION. EVERYTHING IS INFORMATION. PHYSICS SAYS THAT STRUCTURES... BUILDINGS, SOCIETIES, IDEOLOGIES... WILL SEEK THEIR POINT OF LEAST ENERGY. THIS MEANS THAT THINGS FALL. THEY FALL FROM HEIGHTS OF ENERGY AND STRUCTURED INFORMATION INTO MEANINGLESS, POWERLESS DISORDER. THIS IS CALLED ENTROPY.

Entropy: Time’s Arrow

In any spontaneous process, the entropy of the universe increases. ΔS universe > 0 Another version of the 2 nd Law: Energy spontaneously spreads out if it has no outside resistance Entropy measures the spontaneous dispersal of energy as a function of temperature How much energy is spread out How widely spread out it becomes Entropy change = “energy dispersed”/T Second Law of Thermodynamics occurs without outside intervention 

Entropy of the Universe ΔS universe = ΔS system + ΔS surroundings Positional disorderEnergetic disorder ΔS universe > 0  spontaneous process Both ΔS sys and ΔS surr positive Both ΔS sys and ΔS surr negative ΔS sys negative, ΔS surr positive ΔS sys positive, ΔS surr negative spontaneous process. nonspontaneous process. depends depends

Entropy of the Surroundings (Energetic Disorder) System Heat Entropy Surroundings System Heat Entropy Surroundings Low T  large entropy change (surroundings) High T  small entropy change (surroundings) ΔH sys < 0 ΔH sys > 0 ΔS surr > 0 ΔS surr < 0

Positional Disorder and Probability Probability of 1particle in left bulb= ½ " 2particles both in left bulb= (½)(½) = ¼ " 3particlesall in left bulb= (½)(½)(½) = 1 / 8 " 4"all " = (½)(½)(½)(½) = 1 / 16 "10"all " = (½) 10 = 1 / 1024 " 20"all " = (½) 20 = 1 / " a mole of"all " = (½) 6.02  The arrangement with the greatest entropy is the one with the highest probability (most “spread out”).

S solid < S liquid << S gas Entropy of the System: Positional Disorder Ludwig Boltzmann Ordered states Disordered states Low probability (few ways) High probability (many ways) Low S High S S system  Positional disorder S increases with increasing # of possible positions Ludwig Boltzmann

The Third Law: The entropy of a perfect crystal at 0 K is zero. Everything in its place No molecular motion The Third Law of Thermodynamics

Entropy Curve SolidGasLiquid S (q rev /T) (J/K) Temperature (K) 0 0  fusion  vaporization S° (absolute entropy) can be calculated for any substance

Entropy Increases with... Melting (fusion)S liquid > S solid ΔH fusion /T fusion = ΔS fusion VaporizationS gas > S liquid ΔH vaporization /T vaporization = ΔS vaporization Increasing n gas in a reaction HeatingS T2 > S T1 if T 2 > T 1 Dissolving (usually)S solution > (S solvent + S solute ) Molecular complexitymore bonds, more entropy Atomic complexitymore e -, protons, neutrons

Recap: Characteristics of Entropy S is a state function S is extensive (more stuff, more entropy) At 0 K, S = 0 (we can know absolute entropy) S > 0 for elements and compounds in their standard states ΔS° rxn =  nS° products -  nS° reactants Raise T  increase S Increase n gas  increase S More complex systems  larger S

Entropy and Gibbs Free Energy by Mike Roller

Entropy (S) Review ΔS universe > 0 for spontaneous processes ΔS universe = ΔS system + ΔS surroundings  positional  energetic We can know the absolute entropy value for a substance S° values for elements & compounds in their standard states are tabulated (Appendix C, p. 1019) For any chemical reaction, we can calculate ΔS° rxn : ΔS° rxn =  S°(products) -  S°(reactants)

ΔS universe and Chemical Reactions ΔS universe = ΔS system + ΔS surroundings For a system of reactants and products, ΔS universe = ΔS rxn – ΔH rxn /T If ΔS universe > 0, the reaction is spontaneous If ΔS universe < 0, the reaction is not spontaneous –The reverse reaction is spontaneous If ΔS universe = 0, the reaction is at equilibrium –Neither the forward nor the reverse reaction is favored

C 6 H 12 O 6(s) + 6 O 2(g)  6 CO 2(g) + 6 H 2 O (g) Compound C 6 H 12 O 6(s) O 2(g) CO 2(g) H 2 O (g) ΔH° f (kJ/mol) S° (J/mol K) ΔS universe = ΔS rxn – ΔH rxn /T ΔS° rxn =  S°(products) -  S°(reactants) = [6 S°(CO 2(g) ) + 6 S°(H 2 O (g) )] – [S°(C 6 H 12 O 6(s) ) + 6 S°(O 2(g) )] = [6(214) + 6(189)] – [(212) + 6(205)] J/K ΔS° rxn = 976 J/K ΔH° rxn =  ΔH° f (products) -  ΔH° f (reactants) = [6 ΔH° f (CO 2(g) ) + 6 ΔH° f (H 2 O (g) )] – [ΔH° f (C 6 H 12 O 6(s) ) + 6 ΔH° f (O 2(g) )] = [6(-393.5) + 6(-242)] – [(-1275) + 6(0)] kJ ΔH° rxn = kJ

C 6 H 12 O 6(s) + 6 O 2(g)  6 CO 2(g) + 6 H 2 O (g) Compound C 6 H 12 O 6(s) O 2(g) CO 2(g) H 2 O (g) ΔH° f (kJ/mol) S° (J/mol K) ΔS universe = ΔS rxn – ΔH rxn /T ΔS° rxn = 976 J/K(per mole of glucose) ΔH° rxn = kJ(per mole of glucose) At 298 K, ΔS° universe = kJ/K – (-2538 kJ/298 K) ΔS° universe = 9.5 kJ/K

–ΔG means +ΔS univ A process (at constant T, P) is spontaneous if free energy decreases Gibbs Free Energy (G) Josiah Gibbs G = H – TS At constant temperature, ΔG = ΔH – TΔS (system’s point of view) ΔG = ΔH – TΔS Divide both sides by –T -ΔG/T = -ΔH/T + ΔS ΔS universe = ΔS – ΔH/T

ΔG and Chemical Reactions ΔG = ΔH – TΔS If ΔG < 0, the reaction is spontaneous If ΔG > 0, the reaction is not spontaneous –The reverse reaction is spontaneous If ΔG = 0, the reaction is at equilibrium –Neither the forward nor the reverse reaction is favored ΔG is an extensive state function

Ba(OH) 2(s) + 2NH 4 Cl (s)  BaCl 2(s) + 2NH 3(g) + 2 H 2 O (l) ΔH° rxn = 50.0 kJ (per mole Ba(OH) 2 ) ΔS° rxn = 328 J/K (per mole Ba(OH) 2 ) ΔG = ΔH - TΔS ΔG°= 50.0 kJ – 298 K(0.328 kJ/K) ΔG° = – 47.7 kJSpontaneous At what T does the reaction stop being spontaneous? The T where ΔG = 0. ΔG = 0 = 50.0 kJ – T(0.328 J/K) 50.0 kJ = T(0.328 J/K) T = 152 K  not spontaneous below 152 K

Effect of ΔH and ΔS on Spontaneity ΔH–+–+ΔH–+–+ ΔS++––ΔS++–– Spontaneous? Spontaneous at all temps Spontaneous at high temps Reverse reaction spontaneous at low temps Spontaneous at low temps Reverse reaction spontaneous at high temps Not spontaneous at any temp ΔG = ΔH – TΔS ΔG negative  spontaneous reaction

1. ΔG° =  ΔG° f (products) -  ΔG° f (reactants) ΔG° f = free energy change when forming 1 mole of compound from elements in their standard states 2. ΔG° = ΔH° - TΔS° 3. ΔG° can be calculated by combining ΔG° values for several reactions Just like with ΔH° and Hess’s Law Ways to Calculate ΔG° rxn

2H 2(g) + O 2(g)  2 H 2 O (g) 1. ΔG° =  ΔG° f (products) -  ΔG° f (reactants) ΔG° f (O 2(g) ) = 0 ΔG° f (H 2(g) ) = 0 ΔG° f (H 2 O (g) ) = -229 kJ/mol ΔG° = (2(-229 kJ) – 2(0) – 0) kJ = -458 kJ 2. ΔG° = ΔH° - TΔS° ΔH° = -484 kJ ΔS° = -89 J/K ΔG° = -484 kJ – 298 K( kJ/K) = -457 kJ

2H 2(g) + O 2(g)  2 H 2 O (g) 3. ΔG° = combination of ΔG° from other reactions (like Hess’s Law) 2H 2 O (l)  2H 2(g) + O 2(g) ΔG° 1 = 475 kJ H 2 O (l)  H 2 O (g) ΔG° 2 = 8 kJ ΔG° = - ΔG° 1 + 2(ΔG° 2 ) ΔG° = -475 kJ + 16 kJ = -459 kJ Method 1: -458 kJ Method 2: -457 kJ Method 3: -459 kJ

What is Free Energy, Really? NOT just “another form of energy” Free Energy is the energy available to do useful work If ΔG is negative, the system can do work (w max = ΔG) If ΔG is positive, then ΔG is the work required to make the process happen –Example: Photosynthesis –6 CO H 2 O  C 6 H 12 O O 2 –ΔG = 2870 kJ/mol of glucose at 25°C –2870 kJ of work is required to photosynthesize 1 mole of glucose