Chapter 19: Chemical Thermodynamics Spontaneous processes… …happen without outside help …are “product favored”

Chapter 19: Chemical Thermodynamics Spontaneous processes at 25 o C H 2 O (s) → H 2 O (l)

Chapter 19: Chemical Thermodynamics Spontaneous processes CO 2 (s) → CO 2 (g) at 25 o C ‘dry ice’

Chapter 19: Chemical Thermodynamics Spontaneous processes 4 Fe (s) + 3 O 2 (g) → 2 Fe 2 O 3 (s) at 25 o C rust

Chapter 19: Chemical Thermodynamics Spontaneous processes …occur in a definite direction: towards the formation of product H 2 O (s) → H 2 O (l) CO 2 (s) → CO 2 (g) 4 Fe (s) + 3 O 2 (g) → 2 Fe 2 O 3 (s) at 25 o C

Chapter 19: Chemical Thermodynamics The direction of a spontaneous processes may depend on temperature at -10 o C H 2 O (l) → H 2 O (s)

Chapter 19: Chemical Thermodynamics H 2 O (l) → H 2 O (s) CO 2 (s) → CO 2 (g) 2 Fe 2 O 3 (s) → 4 Fe (s) + 3 O 2 (g) at 25 o C At a given temperature and pressure, processes are spontaneous only in one direction If a processes is spontaneous in one direction it is non-spontaneous in the other direction

Chapter 19: Chemical Thermodynamics Which reactions are spontaneous? many spontaneous reactions are exothermic (  H < 0) … but not all! Some reactions are endothermic (  H > 0) and still spontaneous NH 4 NO 3 (s) → NH 4 + (aq) + NO 3 - (aq)  H > 0

Chapter 19: Chemical Thermodynamics Which reactions are spontaneous? Reactions proceed towards a more probable state In general, the more probable state is associated with more disorder

Chapter 19: Chemical Thermodynamics Entropy can be thought of as a measure of disorder Ludwig Boltzmann ( ) S = k log W k = 1.38 x J / K W = Wahrscheinlichkeit (probability)

Chapter 19: Chemical Thermodynamics The change in entropy for any process is:  S = S final - S initial What is the sign of  S for the following processes at 25 o C ? H 2 O (s) → H 2 O (l) CO 2 (g) → CO 2 (s)

Chapter 19: Chemical Thermodynamics Second Law of Thermodynamics For any spontaneous process, the entropy of the universe increases  S o universe =  S o system +  S o surroundings > 0

Chapter 19: Chemical Thermodynamics S solid < S liquid < S gas gas liquidsolid Of all phase states, gases have the highest entropy

Chapter 19: Chemical Thermodynamics Larger Molecules generally have a larger entropy S small < S medium < S large

Chapter 19: Chemical Thermodynamics Often, dissolving a solid or liquid will increase the entropy dissolves

Chapter 19: Chemical Thermodynamics Dissolving a gas in a liquid decreases the entropy dissolves

Chapter 19: Chemical Thermodynamics The entropy of a substance increases with temperature

Chapter 19: Chemical Thermodynamics S = k ln W W is a measure for how many different ways there are of arranging a molecule or an ensemble of molecules (the system) W reflects the number of microstates of a system The larger the possible number of microstates the higher the entropy

Chapter 19: Chemical Thermodynamics S solid < S gas gas solid Of all phase states, gases have the highest entropy

Chapter 19: Chemical Thermodynamics S solid < S gas gas solid Of all phase states, gases have the highest entropy

Chapter 19: Chemical Thermodynamics Larger Molecules generally have a larger entropy S small < S large

Chapter 19: Chemical Thermodynamics Often, dissolving a solid or liquid will increase the entropy

Chapter 19: Chemical Thermodynamics Often, dissolving a solid or liquid will increase the entropy

Chapter 19: Chemical Thermodynamics Dissolving a gas in a liquid decreases the entropy

Chapter 19: Chemical Thermodynamics } } } }

What is the sign of  S for the following reactions? FeCl 2 (s) + H 2 (g) → Fe (s) + 2 HCl (g) Ba(OH) 2 (s) → BaO (s) + H 2 O (g) 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g) Ag + (aq) + Cl - (aq) → AgCl (s)

Chapter 19: Chemical Thermodynamics For each of the following pairs, which substance has a higher molar entropy at 25 o C ? HCl (l) HCl (s) C 2 H 2 (g) C 2 H 6 (g) Li (s) Cs (s) Pb 2+ (aq) Pb (s) O 2 (g) O 2 (aq)HCl (l) HBr (l) CH 3 OH (l) CH 3 OH (aq)N 2 (l) N 2 (g)

Chapter 19: Chemical Thermodynamics  S o rxn = Σ n S o (products) – Σ m S o (reactants) If you know the standard molar entropies of reactants and products, you can calculate  S for a reaction:

Chapter 19: Chemical Thermodynamics substanceS o (J/K-mol) H C 2 H 4 (g)219.4 C 2 H 6 (g)229.5 What is  S o for the following reaction? C 2 H 4 (g) + H 2 (g) → C 2 H 6 (g)  S o rxn = Do you expect  S to be positive or negative? S o for elements are NOT zero

Chapter 19: Chemical Thermodynamics Which reactions are spontaneous? We need to know the magnitudes of both  S and  H ! GoGo = Gibbs free energy… … is a measure of the amount of “useful work” a system can perform  G o =  H o - T  S o

Chapter 19: Chemical Thermodynamics  G o =  H o - T  S o A reaction is spontaneous if  G o is negative J. Willard Gibbs (1839 – 1903)

Chapter 19: Chemical Thermodynamics 2 Na (s) + 2 H 2 O (l) → 2 NaOH (aq) + H 2 (g) The reaction of sodium metal with water: Is the reaction spontaneous? What is the sign of  G o ? What is the sign of  o ? What is the sign of  S o ?

Chapter 19: Chemical Thermodynamics  G o reaction is spontaneous (“product favored”)  G o > 0=> reaction is non-spontaneous  G o = 0=> reaction is at equilibrium “exergonic” “endergonic”

Chapter 19: Chemical Thermodynamics  G o =  H o - T  S o  H o  S o + - GoGo

Chapter 19: Chemical Thermodynamics  G o =  H o - T  S o H 2 O (s) → H 2 O (l)  G o < 0 at 25 o C (298K): spontaneous: HoHo > 0 SoSo  S o > 0 >  H o  =>  G o < 0 but: if T becomes very small: at 298K:  S o > 0 <  H o  =>  G o > 0

Chapter 19: Chemical Thermodynamics A diamond left behind in a burning house reacts according to 2 C (s) + O 2 (g) → 2 CO (g) What is the value of  G o for the reaction at 298K ? substance  H f o (kJ/mol)  G f o (kJ/mol) S o (J/K-mol) O C (diamond, s) C (graphite, s) CO 2 (g)

Chapter 19: Chemical Thermodynamics There are two possible ways to calculate  G o : I)  G o =  H o - T  S o II)  G o = Σ n  G f o (products) – Σ m  G f o (reactants) calculate  G o from  H o and S o :

Chapter 19: Chemical Thermodynamics I) calculate  G o from  H o and  S o :  G o =  H o - T  S o substance  H f o (kJ/mol)  G f o (kJ/mol) S o (J/K-mol) O C (diamond, s) C (graphite, s) CO (g) C (s) + O 2 (g) → 2 CO (g)  G o = kJ

Chapter 19: Chemical Thermodynamics substance  H f o (kJ/mol)  G f o (kJ/mol) S o (J/K-mol) O C (diamond, s) C (graphite, s) CO (g) C (s) + O 2 (g) → 2 CO (g) II)  G o = Σ n  G f o (products) – Σ m  G f o (reactants)  G o = kJ

Chapter 19: Chemical Thermodynamics Consider the following reaction: 2 H 2 (g) + O 2 (g) → 2 H 2 O (l)  S = J/K,  H = kJ, and  G = kJ at 25 o C. At what temperature does the reaction become spontaneous in the opposite direction?  G =  H - T  S The reaction “switches” direction if  H = T  S, i.e. if  G = 0  H < 0  S < 0  G < 0 at 298 K  G > 0 at high T

Chapter 19: Chemical Thermodynamics 0 =  H - T  S T  S =  H T = x 1 kJ / 1000J T = T = K kJ J/K At temperatures greater than K liquid water will spontaneously decompose into H 2 and O 2 !!

Chapter 19: Chemical Thermodynamics At the normal melting point, the Gibbs free energies of the solid and liquid phase of any substance are equal: H 2 O (s) → H 2 O (l) at 0 o C  G o = 0 H 2 O (l) → H 2 O (g) at 100 o C  G o = 0 At the normal boiling point, the Gibbs free energies of the liquid and gas phase of any substance are equal:

Chapter 19: Chemical Thermodynamics At a phase change:  G o = 0  G o =  H o - T  S o 0 =  H o - T  S o if we know  H o fus (or  H o vap ), we can calculate  S o at the melting (or boiling point):  H o = T  S o  S o =  H o T

Chapter 19: Chemical Thermodynamics  S o fus =  H o fus T The entropy of melting for H 2 O = 6.02 kJ/mol K = 22.0 J/mol-K  S o vap =  H o vap T The entropy of vaporization for H 2 O = 40.7 kJ/mol K = 109 J/mol-K units! melting temp boiling temp