Chapter 17 Free Energy and Thermodynamics

2 Thermodynamics vs. Kinetics

CHEMICAL THERMODYNAMICS The first law of thermodynamics: Energy and matter can be neither created nor destroyed; only transformed from one form to another. The energy and matter of the universe is constant. The second law of thermodynamics: In any spontaneous process there is always an increase in the entropy of the universe. The entropy is increasing. The third law of thermodynamics: The entropy of a perfect crystal at 0 K is zero. There is no molecular motion at absolute 0 K.

STATE FUNCTIONS A property of a system which depends only on its present state and not on its pathway.  H = Enthalpy = heat of reaction = q p A measure of heat (energy) flow of a system relative to its surroundings.  H° standard enthalpy  H f ° enthalpy of formation  H° =  n  H f ° (products) -  m  H f ° (reactants)  H =  U + P  V U represents the Internal energy of the particles, both the kinetic and potential energy.  U = q + w

HEAT VS WORK energy transfer as aenergy expanded to result of a temperaturemove an object against differencea force q p w = F x d endothermic (+q)work on a system (+w) exothermic (-q)work by the system (-w) q c = -q h w = -P  V

SPONTANEOUS PROCESSES A spontaneous process occurs without outside intervention. The rate may be fast or slow. Entropy A measure of randomness or disorder in a system. Entropy is a state function with units of J/K and it can be created during a spontaneous process.  S univ =  S sys +  S surr The relationship between  S sys and  S surr  S sys  S surr  S univ Process spontaneous? Yes No (Rx will occur in opposite direction) + - ? + - ?Yes, if  S sys >  S surr - + ? - + ?Yes, if  S surr >  S sys

7 Comparing Potential Energy The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.

8 Reversibility of Process any spontaneous process is irreversible –it will proceed in only one direction a reversible process will proceed back and forth between the two end conditions –equilibrium –results in no change in free energy if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction

9 Diamond → Graphite Graphite is more stable than diamond, so the conversion of diamond into graphite is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into pencil lead in your lifetime (or through many of your generations).

Predicting Relative S 0 Values of a System 1. Temperature changes 2. Physical states and phase changes 3. Dissolution of a solid or liquid 5. Atomic size or molecular complexity 4. Dissolution of a gas S 0 increases as the temperature rises. S 0 increases as a more ordered phase changes to a less ordered phase. S 0 of a dissolved solid or dissolved liquid is usually greater than the S 0 of the pure solute. However, the extent depends upon the nature of the solute and solvent.dissolved solid dissolved liquid A gas becomes more ordered when it dissolves in a liquid or solid. In similar substances, increases in mass relate directly to entropy. In allotropic substances, increases in complexity (e.g. bond flexibility) relate directly to entropy.

11 Factors Affecting Whether a Reaction Is Spontaneous The two factors that determine the thermodynamic favorability are the enthalpy and the entropy.enthalpyentropy The enthalpy is a comparison of the bond energy of the reactants to the products. –bond energy = amount needed to break a bond. – H– H The entropy factors relates to the randomness/orderliness of a system – S– S The enthalpy factor is generally more important than the entropy factor

12 Enthalpy related to the internal energy  H generally kJ/mol stronger bonds = more stable molecules if products more stable than reactants, energy released –exothermic –  H = negative if reactants more stable than products, energy absorbed –endothermic –  H = positive The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions. Hess’ Law  H° rxn =  (  H° prod ) -  (  H° react )

13 Entropy entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, S  S generally J/mol S = k ln W –k = Boltzmann Constant = 1.38 x J/K –W is the number of energetically equivalent ways, unitless Random systems require less energy than ordered systems

Entropy Changes in the System  S 0 rxn - the entropy change that occurs when all reactants and products are in their standard states.  S 0 rxn =  S 0 products -  S 0 reactants The change in entropy of the surroundings is directly related to an opposite change in the heat of the system and inversely related to the temperature at which the heat is transferred.  S surroundings = -  H system T

Entropy  S = S f - S i  S > q/T  S =  H/T For a reversible (at equilibrium) process  H - T  S < 0 For a spontaneous reaction at constant T & P  H - T  S If the value for  H - T  S is negative for a reaction then the reaction is spontaneous in the direction of the products. If the value for  H - T  S is positive for a reaction then the reaction is spontaneous in the direction of the reactants. (nonspontaneous for products)

A spontaneous endothermic chemical reaction. water Ba(OH) 2 8H 2 O( s ) + 2NH 4 NO 3 ( s ) Ba 2+ ( aq ) + 2NO 3 - ( aq ) + 2NH 3 ( aq ) + 10H 2 O( l ).  H 0 rxn = kJ

APPLICATION OF THE 3RD LAW OF THERMODYNAMICS S° = standard entropy = absolute entropy S is usually positive (+) for Substances, S can be negative (-) for Ions because H 3 O+ is used as zero Predicting the sign of  S° The sign is positive if: 1. Molecules are broken during the Rx 2. The number of moles of gas increases 3. solid  liquid liquid  gas solid  gas an increase in order occurs 1. Ba(OH) 2 8H 2 O + 2NH 4 NO 3(s)  2NH 3(g) + 10H 2 O (l) + Ba(NO 3 ) 2(aq) 2. 2SO (g) + O 2(g)  2SO 3(g) 3. HCl (g) + NH 3(g)  NH 4 Cl (s) 4. CaCO 3(s)  CaO (s) + CO 2(g)

Ex. 17.2a – The reaction C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g) has  H rxn = kJ at 25°C. Calculate the entropy change of the surroundings. combustion is largely exothermic, so the entropy of the surrounding should increase significantly  H system = kJ, T = 298 K  S surroundings, J/K Check: Solution: Concept Plan: Relationships: Given: Find: SST,  H

 S°=  n S°(product)-  m S°(reactant) 1. Acetone, CH 3 COCH 3, is a volitale liquid solvent. The standard enthalpy of formation of the liquid at 25°C is kJ/mol; the same quantity for the vapor is kJ/mol. What is  S when 1.00 mol liquid acetone vaproizes? 2.Calculate  S° at 25° for: a. 2 NaHCO 3 (s)  CO 2 (g) + Na 2 CO 3 (s) + H 2 O (g) b. 2 AgBr (s) + I 2(s)  Br 2(l) + 2 AgI (s) Table of S f

S°, S°, S°, Formula J/(molK) Formula J/(molK) Formula J/(molK) Nitrogen SulfurBromine N 2 (g) S 2 (g) Br - (aq) 80.7 NH 3 (g) 193 S(rhombic) 31.9 Br 2 (l) NO(g) S(monoclinic) 32.6Iodine NO 2 (g) SO 2 (g) I - (aq) HNO 3 (aq) 146 H 2 S(g) I 2 (s) Oxygen Fluorine Silver O 2 (g) F - (aq) -9.6 Ag + (aq) 73.9 O 3 (g) F 2 (g) Ag(s) 42.7 OH - (aq) HF(g) AgF(s) 84 H 2 O(g) Chlorine AgCl(s) 96.1 H 2 O(l) 69.9 Cl - (aq) 55.1 AgBr(s) Cl 2 (g) AgI(s) 114 HCl(g) Next page Ex 2

S°, S°, S°, Formula J/(molK) Formula J/(molK) Formula J/(molK) Hydrogen CarbonCarbon (continued) H + (aq)0 C(graphite) 5.7 HCN(l) H 2 (g) C(diamond) 2.4 CCl 4 (g) Sodium CO(g) CCl 4 (l) Na + (aq) 60.2 CO 2 (g) CH 3 CHO(g) 266 Na(s) 51.4 HCO 3 - (aq) 95.0 C 2 H 5 OH(l) 161 NaCl(s) 72.1 CH 4 (g) Silicon NaHCO 3 (s) 102 C 2 H 4 (g) Si(s) 18.0 Na 2 CO 3 (s) 139 C 2 H 6 (g) SiO 2 (s) 41.5 Calcium C 6 H 6 (l) SiF 4 (g) 285 Ca 2+ (aq) HCHO(g) 219 Lead Ca(s) 41.6 CH 3 OH(l) 127 Pb(s) 64.8 CaO(s) 38.2 CS 2 (g) PbO(s) 66.3 CaCO 3 (s) 92.9 CS 2 (l) PbS(s) 91.3 HCN(g) back

 G 0 system =  H 0 system - T  S 0 system  G 0 rxn =  m  G 0 products -  n  G 0 reactants Gibbs Free Energy (G)  G, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it.  G < 0 for a spontaneous process  G > 0 for a nonspontaneous process  G = 0 for a process at equilibrium

STANDARD FREE ENERGY OF FORMATION  G° f The free energy change that occurs when 1 mol of substance is formed from the elements in their standard state. Calculate  G° for: 2 CH 3 OH(l) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(g) Table of  G f

 G f °  G f °  G f ° Formula kJ/molFormula kJ/mol Formula kJ/mol NitrogenSulfur Bromine N 2(g) 0S 2(g) 80.1 Br - (aq) NH 3(g) -16 S (rhombic) 0 Br 2(l) 0 NO (g) S (monoclinic) 0.10 Iodine NO 2(g) 51 SO 2(g) I - (aq) HNO 3(aq) H 2 S (g) -33 I 2(s) 0 OxygenFluorine Silver O 2(g) 0F - (aq) Ag + (aq) 77.1 O 3(g) 163 F 2(g) 0 Ag (s) 0 OH - (aq) HF (g) -275 AgF (s) -185 H 2 O (g) Chlorine AgCl (s) H 2 O (l) Cl - (aq) AgBr (s) Cl 2(g) 0 AgI (s) HCl (g) Next PageEX 2EX 3

 G f °  G f °  G f ° Formula kJ/molFormula kJ/mol Formula kJ/mol Hydrogen CarbonCarbon (cont.) H + 0C (graphite) 0HCN (l) 121 H 2(g) 0C (diamond) 2.9CCl 4(g) SodiumCO (g) CCl 4(l) Na + (aq) CO 2(g) CH 3 CHO (g) Na (s) 0 HCO 3 - (aq) C 2 H 5 OH (l) NaCl (s) CH 4(g) Silicon NaHCO 3(s) C 2 H 4(g) 68.4Si (s) 0 Na 2 CO 3(s) C 2 H 6(g) -32.9SiO 2(s) CalciumC 6 H 6(l) 124.5SiF 4(g) Ca 2 + (aq) HCHO (g) -110Lead Ca (s) 0 CH 3 OH (l) Pb (s) 0 CaO (s) CS 2(g) 66.9PbO (s) -189 CaCO 3(s) CS 2(l) 63.6PbS (s) HCN (g) 125 BACK

INTERPRETING  G° FOR SPONTANEITY 1. When  G° is very small (less than -10 KJ) the reaction is spontaneous as written. Products dominate.  G°  G°(P) 2. When  G° is very large (greater than 10 KJ) the reaction is non spontaneous as written. Reactants dominate.  G° > 0  G°(R) <  G°(P) 3. When  G° is small (+ or -) at equilibrium then both reactants and products are present.  G° = 0 Ba(OH 2 )8 H 2 O (g) + 2 NH 4 NO 3(g)  2 NH 3(g) +10 H 2 O (l) + Ba(NO 3 ) 3(aq)

27 Free Energy Change and Spontaneity

Ex. 17.3a – The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has  H = kJ and  S = J/K at 25°C. Calculate  G and determine if it is spontaneous. Since  G is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.  H = kJ,  S = J/K, T = 298 K  G, kJ Answer: Solution: Concept Plan: Relationships: Given: Find: GGT,  H,  S

Ex. 17.3a – The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has  H = kJ and  S = J/K. Calculate the minimum temperature it will be spontaneous. The temperature must be higher than 673K for the reaction to be spontaneous  H = kJ,  S = J/K,  G < 0  Answer: Solution: Concept Plan: Relationships: Given: Find: T  G,  H,  S

GIBBS FREE ENERGY : G G = H - TS describes the temperature dependence of spontaneity Standard conditions (1 atm, if soln=1M & 25°):  G° =  H° - T  S° A process ( at constant P & T) is spontaneous in the direction in which the free energy decreases. 1. Calculate  H°,  S° &  G° for  S°  G° 2 SO 2(g) + O 2(g)  2 SO 3(g) at 25°C & 1 atm

31 Example -  G Calculate  G at 427°C for the reaction below if the P N2 = 33.0 atm, P H2 = 99.0 atm, and P NH3 = 2.0 atm N 2 (g) + 3 H 2 (g)  2 NH 3 (g)

 G AND EQUILIBRIUM The equilibrium point occurs at the lowest free energy available to the reaction system. When a substance undergoes a chemical reaction, the reaction proceeds to give the minimum free energy at equilibrium.  G =  G° + RT ln (Q) at equilibrium:  G = 0  G° = -RT ln (K)  G° = 0thenK = 1  G° 1  G° > 0thenK < 1 Q: Corrosion of iron by oxygen is 4 Fe (s) + 3 O 2(g)  2 Fe 2 O 3(s) calculate K for this Rx at 25°C.  G =  G° + RT ln (Q)

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Free Energy, Equilibrium and Reaction Direction If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (  G < 0) If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (  G > 0) If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (  G = 0)  G = RT ln Q/K = RT lnQ - RT lnK Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and ln Q = 0 so  G 0 = - RT lnK

FORWARD REACTION REVERSE REACTION Table 1 The Relationship Between  G 0 and K at 25 0 C  G 0 (kJ) KSignificance x x x x x x10 1 6x10 8 3x x10 35 Essentially no forward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to same extent Forward reaction goes to completion; essentially no reverse reaction

36 Example - K Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C N 2 (g) + 3 H 2 (g)  2 NH 3 (g)

Practice Problems on  G o & K 1.Calculate  Gº at 25ºC AgI (s)  Ag + (aq) + I - (aq) BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq) What is the value for K sp at 25ºC? 2.Calculate K at 25ºC for Zn (s) + 2H + (aq)  Zn 2+ (aq) + H 2 (g). Table

38 Temperature Dependence of K for an exothermic reaction, increasing the temperature decreases the value of the equilibrium constant for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant

 Gº & Spontaneity is dependent on Temperature  Hº  Sº  Gº Spontaneous at all T Non spontaneous at all T - -+/- At Low T= Spontaneous At High T= Nonspontaneous + ++/- At low T= Nonspontaneous At High T= Spontaneous Q. Predict the Spontaneity for H 2 O(s)  H 2 O(l) at (a) -10ºC, (b) 0ºC & (c) 10ºC.

Practice Problems on Temperature Relationships 1. At what temperature is the following process spontaneous at 1 atm? Br 2 (l)  Br 2 (g) What is the normal boiling point for Br 2 (l)? 2. Calculate  Gº & Kp at 35ºC N 2 O 4 (g)  2 NO 2 (g) 3. Calculate  Hº,  Sº &  Gº at 25ºC and 650ºC. CS 2 (g) + 4H 2 (g)  CH 4 (g) + 2H 2 S (g) Compare the two values and briefly discuss the spontaneity of the R x at both temperature.