Chemistry - Thermodynamics

Introduction to Entropy : Introduction to Entropy by Mike Roller

Slide 2 : Entropy (S) = a measure of randomness or disorder

Slide 3 : Entropy: Time’s Arrow

Slide 4 :

Second Law of Thermodynamics : In any spontaneous process, the entropy of the universe increases. ?Suniverse > 0 Another version of the 2nd 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 : Entropy of the Universe ?Suniverse = ?Ssystem + ?Ssurroundings Positional disorder Energetic disorder ?Suniverse > 0 ? spontaneous process Both ?Ssys and ?Ssurr positive Both ?Ssys and ?Ssurr negative ?Ssys negative, ?Ssurr positive ?Ssys positive, ?Ssurr negative spontaneous process. nonspontaneous process. depends depends

Entropy of the Surroundings(Energetic Disorder) : 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) ?Hsys < 0 ?Hsys > 0 ?Ssurr > 0 ?Ssurr < 0

Positional Disorder and Probability : Positional Disorder and Probability Probability of 1 particle in left bulb = ½ " 2 particles both in left bulb = (½)(½) = ¼ " 3 particles all 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 = 1/1048576 " a mole of " all " = (½)6.02?1023 The arrangement with the greatest entropy is the one with the highest probability (most “spread out”).

Entropy of the System: Positional Disorder : Ssolid < Sliquid << Sgas Entropy of the System: Positional Disorder Ludwig Boltzmann Ordered states Disordered states Low probability (few ways) High probability (many ways) Low S High S Ssystem ? Positional disorder S increases with increasing # of possible positions Ludwig Boltzmann

The Third Law of Thermodynamics : 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 : Entropy Curve Solid Gas Liquid S (qrev/T) (J/K) Temperature (K) 0 0 ? fusion ? vaporization S° (absolute entropy) can be calculated for any substance

Entropy Increases with... : Entropy Increases with... Melting (fusion) Sliquid > Ssolid ?Hfusion/Tfusion = ?Sfusion Vaporization Sgas > Sliquid ?Hvaporization/Tvaporization = ?Svaporization Increasing ngas in a reaction Heating ST2 > ST1 if T2 > T1 Dissolving (usually) Ssolution > (Ssolvent + Ssolute) Molecular complexity more bonds, more entropy Atomic complexity more e-, protons, neutrons

Recap: Characteristics of Entropy : 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 ?S°rxn = ?nS°products - ?nS°reactants Raise T ? increase S Increase ngas ? increase S More complex systems ? larger S

Entropy and Gibbs Free Energy : Entropy and Gibbs Free Energy by Mike Roller

Entropy (S) Review : Entropy (S) Review ?Suniverse > 0 for spontaneous processes ?Suniverse = ?Ssystem + ?Ssurroundings ? 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 ?S°rxn: ?S°rxn = ?S°(products) - ?S°(reactants)

?Suniverse and Chemical Reactions : ?Suniverse and Chemical Reactions ?Suniverse = ?Ssystem + ?Ssurroundings For a system of reactants and products, ?Suniverse = ?Srxn – ?Hrxn/T If ?Suniverse > 0, the reaction is spontaneous If ?Suniverse < 0, the reaction is not spontaneous The reverse reaction is spontaneous If ?Suniverse = 0, the reaction is at equilibrium Neither the forward nor the reverse reaction is favored

C6H12O6(s) + 6 O2(g) ? 6 CO2(g) + 6 H2O(g) : C6H12O6(s) + 6 O2(g) ? 6 CO2(g) + 6 H2O(g) Compound C6H12O6(s) O2(g) CO2(g) H2O(g) ?H°f (kJ/mol) -1275 0 -393.5 -242 S° (J/mol K) 212 205 214 189 ?Suniverse = ?Srxn – ?Hrxn/T ?S°rxn = ?S°(products) - ?S°(reactants) = [6 S°(CO2(g)) + 6 S°(H2O(g))] – [S°(C6H12O6(s)) + 6 S°(O2(g))] = [6(214) + 6(189)] – [(212) + 6(205)] J/K ?S°rxn = 976 J/K ?H°rxn = ??H°f (products) - ??H°f(reactants) = [6 ?H°f(CO2(g)) + 6 ?H°f(H2O(g))] – [?H°f(C6H12O6(s)) + 6 ?H°f(O2(g))] = [6(-393.5) + 6(-242)] – [(-1275) + 6(0)] kJ ?H°rxn = -2538 kJ

C6H12O6(s) + 6 O2(g) ? 6 CO2(g) + 6 H2O(g) : C6H12O6(s) + 6 O2(g) ? 6 CO2(g) + 6 H2O(g) Compound C6H12O6(s) O2(g) CO2(g) H2O(g) ?H°f (kJ/mol) -1275 0 -393.5 -242 S° (J/mol K) 212 205 214 189 ?Suniverse = ?Srxn – ?Hrxn/T ?S°rxn = 976 J/K (per mole of glucose) ?H°rxn = -2538 kJ (per mole of glucose) At 298 K, ?S°universe = 0.976 kJ/K – (-2538 kJ/298 K) ?S°universe = 9.5 kJ/K

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

?G and Chemical Reactions : ?G and Chemical Reactions ?G = ?H – T?S If ?G < 0, the reaction is spontaneous If ?G > 0, the reaction is not spontaneous The reverse reaction is spontaneous If ?G = 0, the reaction is at equilibrium Neither the forward nor the reverse reaction is favored ?G is an extensive state function

Ba(OH)2(s) + 2NH4Cl(s) ? BaCl2(s) + 2NH3(g) + 2 H2O(l) : Ba(OH)2(s) + 2NH4Cl(s) ? BaCl2(s) + 2NH3(g) + 2 H2O(l) ?H°rxn = 50.0 kJ (per mole Ba(OH)2) ?S°rxn = 328 J/K (per mole Ba(OH)2) ?G = ?H - T?S ?G° = 50.0 kJ – 298 K(0.328 kJ/K) ?G° = – 47.7 kJ Spontaneous At what T does the reaction stop being spontaneous? The T where ?G = 0. ?G = 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 ?H and ?S on Spontaneity : Effect of ?H and ?S on Spontaneity ?H – + – + ?S + + – – 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 ?G = ?H – T?S ?G negative ? spontaneous reaction

Ways to Calculate ?G°rxn : 1. ?G° = ??G°f(products) - ??G°f(reactants) ?G°f = free energy change when forming 1 mole of compound from elements in their standard states 2. ?G° = ?H° - T?S° 3. ?G° can be calculated by combining ?G° values for several reactions Just like with ?H° and Hess’s Law Ways to Calculate ?G°rxn

2H2(g) + O2(g) ? 2 H2O(g) : 2H2(g) + O2(g) ? 2 H2O(g) 1. ?G° = ??G°f(products) - ??G°f(reactants) ?G°f(O2(g)) = 0 ?G°f(H2(g)) = 0 ?G°f(H2O(g)) = -229 kJ/mol ?G° = (2(-229 kJ) – 2(0) – 0) kJ = -458 kJ 2. ?G° = ?H° - T?S° ?H° = -484 kJ ?S° = -89 J/K ?G° = -484 kJ – 298 K(-0.089 kJ/K) = -457 kJ

2H2(g) + O2(g) ? 2 H2O(g) : 2H2(g) + O2(g) ? 2 H2O(g) 3. ?G° = combination of ?G° from other reactions (like Hess’s Law) 2H2O(l) ? 2H2(g) + O2(g) ?G°1 = 475 kJ H2O(l) ? H2O(g) ?G°2 = 8 kJ ?G° = - ?G°1 + 2(?G°2) ?G° = -475 kJ + 16 kJ = -459 kJ Method 1: -458 kJ Method 2: -457 kJ Method 3: -459 kJ

What is Free Energy, Really? : What is Free Energy, Really? NOT just “another form of energy” Free Energy is the energy available to do useful work If ?G is negative, the system can do work (wmax = ?G) If ?G is positive, then ?G is the work required to make the process happen Example: Photosynthesis 6 CO2 + 6 H2O ? C6H12O6 + 6 O2 ?G = 2870 kJ/mol of glucose at 25°C 2870 kJ of work is required to photosynthesize 1 mole of glucose