18-1 CHEM 102, Fall LA TECH Instructor: Dr. Upali Siriwardane Office: CTH 311 Phone Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu,Th,F 8: :00 am. Test Dates : September 23, October 21, and November 16; Comprehensive Final Exam: November 18, 2010 Exam: 10:0-10:15 am, CTH 328. September 23, 2010 (Test 1): Chapter 13 October 21, 2010 (Test 2): Chapters 14 & 15 November 16, 2010 (Test 3): Chapters 16, 17 & 18 Comprehensive Final Exam: November 18, 2010 : Chapters 13, 14, 15, 16, 17 and 18 Chemistry 102(01) Fall 2010

18-2 CHEM 102, Fall LA TECH Review of Chapter 6. Energy and Chemical Reactions 6.1 The Nature of Energy 6.2 Conservation of Energy 6.3 Heat Capacity 6.4 Energy and Enthalpy 6.5 Thermochemical Equations 6.6 Enthalpy change for chemical Rections 6.7 Where does the Energy come from? 6.8 Measuring Enthalpy Changes: Calorimetry 6.9 Hess's Law 6.10 Standard Enthalpy of Formation 6.11 Chemical Fuels for Home and Industry 6.12 Food Fuels for Our Bodies

18-3 CHEM 102, Fall LA TECH Thermochemistry Heat changes during chemical reactions Thermochemical equation. eg. H 2 (g) + O 2 (g) ---> 2H 2 O(l)  H =- 256 kJ;  is called the enthalpy of reaction. if  H is + reaction is called endothermic if  H is - reaction is called exothermic

18-4 CHEM 102, Fall LA TECH Why is it necessary to divide Universe into System and Surrounding Universe = System + Surrounding system surroundings universe

18-5 CHEM 102, Fall LA TECH Types of Systems Isolated system no mass or energy exchange Closed system only energy exchange Open system both mass and energy exchange

18-6 CHEM 102, Fall LA TECH Universe = System + Surrounding Why is it necessary to divide Universe into System and Surrounding

18-7 CHEM 102, Fall LA TECH What is the internal energy change (  U) of a system?  U is associated with changes in atoms, molecules and subatomic particles  U is associated with changes in atoms, molecules and subatomic particles E total = E ke + E pe +  U  U = heat (q) + w (work)  U = q + w  U = q -P  V; w =- P  V

18-8 CHEM 102, Fall LA TECH What forms of energy are found in the Universe? mechanicalthermalelectricalnuclear mass: E = mc 2 others yet to discover

18-9 CHEM 102, Fall LA TECH What is 1 st Law of Thermodynamics Eenergy is conserved in the Universe All forms of energy are inter-convertible and conserved Energy is neither created nor destroyed.

18-10 CHEM 102, Fall LA TECH What exactly is  H? Heat measured at constant pressure q p Chemical reactions exposed to atmosphere and are held at a constant pressure. Volume of materials or gases produced can change. Volume expansion work = -P  V Volume expansion work = -P  V  U = q p + w;  U = q p -P  V  U = q p + w;  U = q p -P  V q p =  U + P  V;w = -P  V  H =  U + P  V;q p =  H(enthalpy )  H =  U + P  V;q p =  H(enthalpy )

18-11 CHEM 102, Fall LA TECH Heat measured at constant volume q v Chemical reactions take place inside a bomb. Volume of materials or gases produced can not change. ie: work = -P  V= 0  U = q v + w qv =  U + o;w = 0  U = q v =  U(internal energy ) How do you measure  U?

18-12 CHEM 102, Fall LA TECH What is Hess's Law of Summation of Heat? To heat of reaction for new reactions. Two methods? 1st method: new  H is calculated by adding  Hs of other reactions. 2nd method: Where  H f (  H of formation) of reactants and products are used to calculate  H of a reaction.

18-13 CHEM 102, Fall LA TECH Method 1: Calculate  H  H for the reaction: SO 2 (g) + 1/2 O 2 (g) + H 2 O(g) ----> H 2 SO 4 (l)  H = ? Other reactions: SO 2 (g) > S(s) + O 2 (g) ;  H = 297kJ H 2 SO 4 (l)------> H 2 (g) + S(s) + 2O 2 (g);  H = 814 kJ H 2 (g) +1/2O 2 (g) -----> H 2 O(g);  H = -242 kJ Other reactions: SO 2 (g) > S(s) + O 2 (g) ;  H = 297kJ H 2 SO 4 (l)------> H 2 (g) + S(s) + 2O 2 (g);  H = 814 kJ H 2 (g) +1/2O 2 (g) -----> H 2 O(g);  H = -242 kJ

18-14 CHEM 102, Fall LA TECH SO 2 (g) > S(s) + O 2 (g);  H 1 = 297 kJ - 1 H 2 (g) + S(s) + 2O 2 (g) > H 2 SO 4 (l);  H 2 = -814 kJ - 2 H 2 O(g) ----->H 2 (g) + 1/2 O 2 (g) ;  H 3 = +242 kJ - 3 ______________________________________ SO 2 (g) + 1/2 O 2 (g) + H 2 O(g) -----> H 2 SO 4 (l);  H = ? SO 2 (g) + 1/2 O 2 (g) + H 2 O(g) -----> H 2 SO 4 (l);  H = ?   H =  H 1 +  H 2 +  H 3   H =  H = -275 kJ Calculate  H for the reaction

18-15 CHEM 102, Fall LA TECH Calculate Heat (Enthalpy) of Combustion: 2nd method C 7 H 16 (l) + 11 O 2 (g) -----> 7 CO 2 (g) + 8 H 2 O(l) ;  H o = ? C 7 H 16 (l) + 11 O 2 (g) -----> 7 CO 2 (g) + 8 H 2 O(l) ;  H o = ?  H f (C 7 H 16 ) = kJ/mol  H f (CO 2 ) = kJ/mol  H f (H 2 O) = kJ/mol  H f O 2 (g) = 0 (zero) What method?  H o =  n   H f o products –  n   H f o reactants n = stoichiometric coefficients 2 nd method

18-16 CHEM 102, Fall LA TECH  H = [  n (  H o f ) Products] - [  n (  H o f ) reactants]  H = [ 7( ( )] - [ (0)] = [ ] - [-198.8] = [ ] - [-198.8] = = = kJ = kJ = kJ = kJ Calculate  H for the reaction

18-17 CHEM 102, Fall LA TECH Why is  H o f  H o f of elements is zero?  H o f, Heat formations are for compounds Note:  H o f of elements is zero

18-18 CHEM 102, Fall LA TECH Chapter 18. Thermodynamics: Directionality of Chemical Reactions 18.1Reactant-Favored and Product-Favored Processes 18.2Probability and Chemical Reactions 18.3Measuring Dispersal or Disorder: Entropy 18.4Calculating Entropy Changes 18.5Entropy and the Second Law of Thermodynamics 18.6Gibbs Free Energy 18.7Gibbs Free Energy Changes and Equilibrium Constants Constants 18.8Gibbs Free Energy, Maximum Work, and Energy Resources 18.9Gibbs Free Energy and Biological Systems 18.10Conservation of Gibbs Free Energy 18.11Thermodynamic and Kinetic Stability

18-19 CHEM 102, Fall LA TECH Chemical Thermodynamics spontaneous reaction – reaction which proceed without external assistance once started chemical thermodynamics helps predict which reactions are spontaneous

18-20 CHEM 102, Fall LA TECH ThermodynamicsThermodynamics Will the rearrangement of a system decrease its energy? If yes, system is favored to react — a product-favored system. Most product-favored reactions are exothermic. Often referred to as spontaneous reactions. “Spontaneous” does not imply anything about time for reaction to occur. Kinetic factors are more important for certain reactions.

18-21 CHEM 102, Fall LA TECH Thermodynamics and Kinetics Diamond  graphite Thermodynamically product favored Slow Kinetics Paper burns Thermodynamically product favored Fast Kinetics In this chapter we only look into thermodynamic factors

18-22 CHEM 102, Fall LA TECH Bases on Energy: Product- Favored Reactions In general, product-favored reactions are exothermic. (Negative  H) In general, reactant-favored reactions are endothermic. (Positive  H)

18-23 CHEM 102, Fall LA TECH Product-Favored Reactions But many spontaneous reactions or processes are endothermic or even have  H = 0. NH 4 NO 3 (s)  NH 4 NO 3 (aq);  H = +

18-24 CHEM 102, Fall LA TECH Direction of Chemical Reaction Product favored reactions are always a transformation of a reactants favored reaction. Product Favored Reaction 2Na (s) + 2Cl 2(g) => 2NaCl (s) Reactant Favored Reaction 2NaCl (s) => 2Na (s) + 2Cl 2(g) However, non spontaneous reactions could be made to take place by coupling with energy source: another reaction or electric current Electrolysis 2NaCl (l) => 2Na (s) + 2Cl 2(g)

18-25 CHEM 102, Fall LA TECH Expansion of a Gas The positional probability is higher when particles are dispersed over a larger volume Matter tends to expand unless it is restricted

18-26 CHEM 102, Fall LA TECH Gas Expansion and Probability

18-27 CHEM 102, Fall LA TECH Entropy, S The thermodynamic property related to randomness is ENTROPY, S. Product-favored processes: final state is more DISORDERED or RANDOM than the original. Spontaneity is related to an increase in randomness. Reaction of K with water

18-28 CHEM 102, Fall LA TECH S[H 2 O(l)] > S[H 2 O(s)] at 0  C.

18-29 CHEM 102, Fall LA TECH Entropies of Solid, Liquid and Gas Phases S (gases) > S (liquids) > S (solids) S (gases) > S (liquids) > S (solids)

18-30 CHEM 102, Fall LA TECH Entropy, S Entropies of ionic solids depend on coulombic attractions. S o (J/Kmol) MgO26.9 NaF51.5 S o (J/Kmol) MgO26.9 NaF51.5

18-31 CHEM 102, Fall LA TECH Entropy and Molecular Structure

18-32 CHEM 102, Fall LA TECH Entropy and Dissolving

18-33 CHEM 102, Fall LA TECH Qualitative Guidelines for Entropy Changes Entropies of gases higher than liquids higher than solids Entropies are higher for more complex structures than simpler structures Entropies of ionic solids are inversely related to the strength of ionic forces Entropy increases when making solutions of pure solids or pure liquids in a liquid solvent Entropy decrease when making solutions of gases in a liquid

18-34 CHEM 102, Fall LA TECH Entropy of a Solution of a Gas

18-35 CHEM 102, Fall LA TECH Phase Transitions H 2 O (s) => H 2 O (l)  H > 0;  S > 0 H 2 O (l) => H 2 O (g)  H > 0;  S > 0 spontaneous at high temperatures H 2 O (l) => H 2 O (s)  H H 2 O (s)  H < 0;  S < 0 H 2 O (g) => H 2 O (l)  H H 2 O (l)  H < 0;  S < 0 spontaneous at low temperatures

18-36 CHEM 102, Fall LA TECH Entropy Changes for Phase Changes For a phase change,  S SYS = q SYS /T (q = heat transferred) Boiling Water H 2 O (liq)  H 2 O(g)  H = q = +40,700 J/mol

18-37 CHEM 102, Fall LA TECH Phase Transitions Heat of Fusion energy associated with phase transition solid-to- liquid or liquid-to-solid  G fusion = 0 =  H fusion - T  S fusion 0 =  H fusion - T  S fusion  H fusion = T  S fusion Heat of Vaporization energy associated with phase transition gas-to- liquid or liquid-to-gas  H vaporization = T  S vaporization

18-38 CHEM 102, Fall LA TECH Qualitative prediction of  S of Chemical Reactions Look for (l) or (s) --> (g) Look for (l) or (s) --> (g) If all are gases: calculate  n If all are gases: calculate  n  n =  n (gaseous prod.) -  n(gaseous reac.) N 2 (g) + 3 H 2 (g) > 2 NH 3 (g)  n = = -2 If  n is -  S is negative (decrease in S) If  n is +  S is positive (increase in S)

18-39 CHEM 102, Fall LA TECH Entropy Change Entropy (  S) normally increase (+) for the following changes: i) Solid ---> liquid (melting) + ii) Liquid ---> gas + iii) Solid ----> gas most + iv) Increase in temperature + v) Increasing in pressure(constant volume, and temperature) + vi) Increase in volume +

18-40 CHEM 102, Fall LA TECH Predict  S! 2 C 2 H 6 (g) + 7 O 2 (g)--> 4 CO 2 (g) + 6H 2 O(g) 2 CO(g) + O 2 (g)-->2 CO 2 (g) 2 CO(g) + O 2 (g)-->2 CO 2 (g) HCl(g) + NH 3 (g)-->NH 4 Cl(s) HCl(g) + NH 3 (g)-->NH 4 Cl(s) H 2 (g) + Br 2 (l) --> 2 HBr(g) H 2 (g) + Br 2 (l) --> 2 HBr(g)

18-41 CHEM 102, Fall LA TECH 2 H 2 (g) + O 2 (g)  2 H 2 O(liq)  S o = 2 S o (H 2 O) - [2 S o (H 2 ) + S o (O 2 )]  S o = 2 mol (69.9 J/Kmol) – [2 mol (130.7 J/Kmol) + 1 mol (205.3 J/Kmol)]  S o = J/K There is a decrease in S because 3 mol of gas give 2 mol of liquid. Calculating  S for a Reaction Based on Hess’s Law second method: S o =  S o (products) -  S o (reactants)  S o =  S o (products) -  S o (reactants) Based on Hess’s Law second method: S o =  S o (products) -  S o (reactants)  S o =  S o (products) -  S o (reactants)

18-42 CHEM 102, Fall LA TECH Third Law of Thermodynamics Provides reference point for absolute entropy Entropy of a perfectly crystalline substance at absolute zero (T= 0 K) is zero. Unlike  H entropy values are positive above temperatures above absolute zero.

18-43 CHEM 102, Fall LA TECH Standard Molar Entropy Values

18-44 CHEM 102, Fall LA TECH Substance S o (J/K.mol) C (diamond) 2.37HBr (g) C (graphite) 5.69HCl (g) CaO (s) 39.75HF (g) CaCO3 (s ) 92.9HI (g) C2H2 (g) ` H2O (l) C2H4 (g) 219.4H2O (g) C2H6 (g) 229.5NaCl (s) CH3OH (l) 127O2 (g) CH3OH (g) 238SO2 (g) CO (g) SO3 (g) Standard Entropies at 25 o C

18-45 CHEM 102, Fall LA TECH Entropy & Spontaneity How can water boil and freeze spontaneously? Enthalpy change can not predict spontaneity! Some endothermic processes are spontaneous Need another thermodynamic property.

18-46 CHEM 102, Fall LA TECH Laws of Thermodynamics Zeroth: Thermal equilibrium and temperature First : The total energy of the universe is constant Second : The total entropy (S) of the universe is always increasing Third : The entropy(S) of a pure, perfectly formed crystalline substance at absolute zero is zero Zeroth: Thermal equilibrium and temperature First : The total energy of the universe is constant Second : The total entropy (S) of the universe is always increasing Third : The entropy(S) of a pure, perfectly formed crystalline substance at absolute zero is zero

18-47 CHEM 102, Fall LA TECH Dissolving NH 4 NO 3 in water—an entropy driven process. 2nd Law of Thermodynamics NH 4 NO 3 (s)  NH 4 NO 3 (aq);  H = +

18-48 CHEM 102, Fall LA TECH Second Law of Thermodynamics In the universe the ENTROPY cannot decrease for any spontaneous process The entropy of the universe strives for a maximum in any spontaneous process, the entropy of the universe increases for product-favored process  S universe = ( S sys + S surr ) > 0  S univ = entropy of the Universe  S sys = entropy of the System  S surr = entropy of the Surrounding  S univ =  S sys +  S surr

18-49 CHEM 102, Fall LA TECH Entropy of the Universe  S univ =  S sys +  S surr  s univ  S sys  S surr (  S sys >  S surr) (  S surr >  S sys)

18-50 CHEM 102, Fall LA TECH Can calc. that H o rxn = H o system = kJ Can calc. that  H o rxn =  H o system = kJ 2 H 2 (g) + O 2 (g)  2 H 2 O(liq)  S o sys = J/K Entropy Changes in the Surroundings = J/K 2nd Law of Thermodynamics

18-51 CHEM 102, Fall LA TECH 2 H 2 (g) + O 2 (g)  2 H 2 O(liq)  S o sys = J/K  S o surr = J/K  S o uni = J/K The entropy of the universe is increasing, so the reaction is product-favored. 2nd Law of Thermodynamics

18-52 CHEM 102, Fall LA TECH Gibbs Free Energy, G  S univ =  S surr +  S sys Multiply through by (-T) -T  S univ =  H sys - T  S sys -T  S univ =  G system Under standard conditions —  G o =  H o - T  S o  S univ =  H sys T +  S

18-53 CHEM 102, Fall LA TECH Gibbs Free Energy, G   G o =  H o - T  S o Gibbs free energy change = difference between the enthalpy of a system and the product of its absolute temperature and entropy predictor of spontaneity Total energy change for system - energy lost in disordering the system Total energy change for system - energy lost in disordering the system

18-54 CHEM 102, Fall LA TECH Predict the spontaneity of the following processes from  H and  S at various temperatures. a)  H = 30 kJ,  S = 6 kJ, T = 300 K b)  H = 15 kJ,  S = -45 kJ,T = 200 K

18-55 CHEM 102, Fall LA TECH a)  H = 30 kJ  S = 6 kJT = 300 K  G =  H sys -T  S sys or  G =  H - T  S.  H = 30 kJ  S= 6 kJ T = 300 K  G = 30 kJ - (300 x 6 kJ) = kJ  G = kJ b)  H = 15 kJ  S = -45 kJT = 200 K  G =  H sys -T  S sys or  G =  H - T  S.  H = 15 kJ  S = -45 kJ T = 200 K  G = 15 kJ -[200 (-45 kJ)] = 15 kJ -(-9000) kJ  G = kJ = 9015 kJ

18-56 CHEM 102, Fall LA TECH  G The sign of  G indicates whether a reaction will occur spontaneously. +Not spontaneous 0 At equilibrium -Spontaneous  S  G The fact that the effect of  S will vary as a function of temperature is important. This can result in changing the sign of  G. Free energy

18-57 CHEM 102, Fall LA TECH Predicting Whether a Reaction is Product Favored using  G Sign of  H system Sign of  S system Product-favored? Negative (exothermic)PositiveYes Negative (exothermic)NegativeYes at low T; no at high T high T Positive (endothermic)PositiveNo at low T; yes at high T yes at high T Positive (endothermic)Negative No

18-58 CHEM 102, Fall LA TECH Predict  G at different  H,  S, T  G =  H - T  S. - - all T all T – - /+ + high/low T + -/+ - low/high T -

18-59 CHEM 102, Fall LA TECH Gibbs Free Energy, G  G o =  H o - T  S o  G o =  H o - T  S o  H o  S o  G o Reaction exo(-)increase(+)-Prod-favored endo(+)decrease(-)+React-favored exo(-)decrease(-)?T dependent endo(+)increase(+)?T dependent

18-60 CHEM 102, Fall LA TECH  G o =  H o - T  S o

18-61 CHEM 102, Fall LA TECH  G f o Free energy change that results when one mole of a substance if formed from its elements will all substances in their standard states.  G values can then be calculated from:  G o =  np  G f o products –  nr  G f o reactants Standard free energy of formation,  G f o

18-62 CHEM 102, Fall LA TECH Substance  G f o Substance  G f o C (diamond) 2.832HBr (g) CaO (s) HF (g) CaCO3 (s) HI (g) 1.30 C2H2 (g) 209H2O (l) C2H4 (g) 86.12H2O (g) C2H6 (g) NaCl (s) CH3OH (l) O (g) CH3OH (g) SO2 (g) CO (g) SO3 (g) All have units of kJ/mol and are for 25 o C Standard free energy of formation

18-63 CHEM 102, Fall LA TECH How do you calculate  G There are two ways to calculate  G for chemical reactions. i)  G =  H - T  S. ii)  G o =   G o f (products) -   G o f (reactants)

18-64 CHEM 102, Fall LA TECH Gibbs Free Energy, G  G o =  H o - T  S o  G o =  H o - T  S o Two methods of calculating  G o (a) Determine  H o rxn and  S o rxn and use Gibbs equation. (b) Use tabulated values of free energies of formation,  G f o.  G o rxn =   G f o (products) -   G f o (reactants)

18-65 CHEM 102, Fall LA TECH Calculating  G o rxn Is the dissolution of ammonium nitrate product- favored? If so, is it enthalpy- or entropy-driven? NH 4 NO 3 (s) + heat  NH 4 NO 3 (aq)

18-66 CHEM 102, Fall LA TECH Calculating  G o rxn Method (a) : From tables of thermodynamic data we find  H o rxn = kJ  H o rxn = kJ  S o rxn = J/K or kJ/K  G o rxn = kJ - (298 K)( J/K) = -6.7 kJ = -6.7 kJ Reaction is product-favored in spite of negative  H o rxn. Reaction is “entropy driven” NH 4 NO 3 (s) + heat  NH 4 NO 3 (aq)

18-67 CHEM 102, Fall LA TECH Calculating  G o rxn Combustion of carbon C(graphite) + O 2 (g) --> CO 2 (g) Method (b) :  G o rxn =  G f o (CO 2 ) - [  G f o (graph) +  G f o (O 2 )]  G o rxn = kJ - [ 0 + 0] Note that free energy of formation of an element in its standard state is 0.  G o rxn = kJ Reaction is product-favored G o rxn =  G f o (products) -  G f o (reactants)  G o rxn =   G f o (products) -   G f o (reactants)

18-68 CHEM 102, Fall LA TECH We can calculate  G o values from  Ho and  S o values at a constant temperature and pressure.Example. Determine  G o for the following reaction at 25 o C EquationN 2 (g) + 3H 2 (g) 2NH 3 (g)  H f o, kJ/mol S o, J/K.mol Calculation of  G o

18-69 CHEM 102, Fall LA TECH Example. Calculate the  S o rxn at 25 o C for the following reaction. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) Substance S (J/K.mol) Substance S o (J/K.mol) CH 4 (g) O 2 (g) CO 2 (g) H 2 O (g) Calculation of standard entropy changes

18-70 CHEM 102, Fall LA TECH Calculate the  S for the following reactions using  S o =   S o (products) -   S o (reactants) a) 2SO 2 (g) + O 2 (g) > 2SO 3 (g)  S o [SO 2 (g)] = 248 J/K mole ;  S o [O 2 (g)] = 205 J/K mole;  S o [SO 3 (g)] = 257 J/K mole b) 2NH 3 (g) + 3N 2 O (g) > 4N 2 (g) + 3 H 2 O (l)  S o [ NH 3 (g)] = 193 J/K mole ;  S o [N 2 (g)] = 192 J/K mole;  S o [N 2 O(g)] = 220 J/K mole;  S[ H 2 O(l)] = 70 J/K mole

18-71 CHEM 102, Fall LA TECH a)2SO 2 (g) + O 2 (g------> 2SO 3 (g)  S o [SO 2 (g)] = 248 J/K mole ;  S o [O 2 (g)] = 205 J/K mole;  S o [SO 3 (g)] = 257 J/K mole  S o  S o =   S o (products) -   S o (reactants)  S o = [514] - [ ]  S o =  S o = -187 J/K mole

18-72 CHEM 102, Fall LA TECH Calculate the  G value for the following reactions using:  G o =   G o f (products) -   G o f (reactants) N 2 O 5 (g) + H 2 O(l) > 2 HNO 3 (l) ;  G o = ?  G f o [ N 2 O 5 (g) ] = 134 kJ/mole ;  G f o [H 2 O(g)] = -237 kJ/mole;  G f o [ HNO 3 (l) ] = -81 kJ/mole N 2 O 5 (g) + H 2 O(l) > 2 HNO 3 (l) ;  G o = ?  G f o 1 x x (-237) 2 (-81)  G o =  G o f (products) - 3  G o f (reactants)  G o = [-162] - [134 + (-237)]  G o =  G o = -59 kJ/mole The reaction have a negative  G and the reaction is spontaneous or will take place as written.

18-73 CHEM 102, Fall LA TECH Free Energy and Temperature 2 Fe 2 O 3 (s) + 3 C(s) ---> 4 Fe(s) + 3 CO 2 (g)  H o rxn = kJ  S o rxn = J/K  G o rxn = kJ Reaction is reactant-favored at 298 K At what T does  G o rxn change from (+) to (-)? Set  G o rxn = 0 =  H o rxn - T  S o rxn

18-74 CHEM 102, Fall LA TECH Effect of Temperature on Reaction Spontaneity

18-75 CHEM 102, Fall LA TECH How do you calculate  G at different T and P  G =  G o + RT ln Q Q = reaction quotient Q = reaction quotient at equilibrium  G =   =  G o + RT ln K  G o = - RT ln K If you know  G o you could calculate K

18-76 CHEM 102, Fall LA TECH Concentrations, Free Energy, and the Equilibrium Constant Equilibrium Constant and Free Energy  G =  G o + RT ln Q Q = reaction quotient 0 =  G o + RT ln K eq  G o = - RT ln K eq

18-77 CHEM 102, Fall LA TECH K eq is related to reaction favorability and so to G o rxn. The larger the (-) value of  G o rxn the larger the value of K.  G o rxn = - RT ln K  G o rxn = - RT ln K where R = 8.31 J/Kmol Thermodynamics and K eq

18-78 CHEM 102, Fall LA TECH For gases, the equilibrium constant for a reaction can be related to  G o by:  G o = -RT lnK For our earlier example, N 2 (g) + 3H 2 (g) 2NH 3 (g) At 25oC,  Go was kJ so K would be: ln K = = ln K = 13.27; K = 5.8 x 10 5  Go -RT kJ -( kJ.K-1mol-1)(298.2K) Free energy and equilibrium

18-79 CHEM 102, Fall LA TECH Calculate the  G for the following equilibrium reaction and predict the direction of the change using the equation:  G =  G o + RT ln Q ; [  G f o [ NH 3 (g) ] = -17 kJ/mole N 2 (g) + 3 H 2 (g) 2 NH 3 (g);  G = ? at 300 K, P N2 = 300, P NH3 = 75 and P H2 = 300 N 2 (g) + 3 H 2 (g) 2 NH 3 (g);  G = ?

18-80 CHEM 102, Fall LA TECH To calculate  G o Using  G o =   G o f (products) -   G o f (reactants)  G f o [ N 2 (g)] = 0 kJ/mole;  G f o [ H 2 (g)] = 0 kJ/mole;  G f o [ NH 3 (g)] = -17 kJ/mole Notice elements have  G f o = 0.00 similar to  H f o N 2 (g) + 3 H 2 (g) 2 NH 3 (g);  G = ?  G f o 002 x (-17)  G o =   G o f (products) -   G o f (reactants)  G o = [-34] - [0 +0]  G o = -34  G o = -34 kJ/mole

18-81 CHEM 102, Fall LA TECH To calculate Q Equilibrium expression for the reaction in terms of partial pressure: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) p 2 NH3 K=_________ p N2 p 3 H2 p 2 NH3 Q=_________ ; p N2 p 3 H2 Q is when initial concentration is substituted into the equilibrium expression 75 2 Q=_________ ; p 2 NH3 = 75 2 ; p N2 =300; p 3 H2 = x Q =6.94 x 10 -7

18-82 CHEM 102, Fall LA TECH To calculate  G o  G =  G o + RT ln Q  G o = -34 kJ/mole R = J/K mole or x kJ/Kmole T = 300 K Q= 6.94 x  G = (-34 kJ/mole) + ( x kJ/K mole) (300 K) ( ln 6.94 x )  G = ln 6.94 x  G = x (-14.18)  G =  G = kJ/mole

18-83 CHEM 102, Fall LA TECH Calculate K (from  G 0 ) N 2 O 4 --->2 NO 2  G o rxn = +4.8 kJ  G o rxn = J = - (8.31 J/K)(298 K) ln K G o rxn = - RT lnK  G o rxn = - RT lnK K = 0.14 G o rxn > 0 : K 0 : K < 1 G o rxn 1  G o rxn 1 K = 0.14 G o rxn > 0 : K 0 : K < 1 G o rxn 1  G o rxn 1 Thermodynamics and K eq

18-84 CHEM 102, Fall LA TECH Concentrations, Free Energy, and the Equilibrium Constant The Influence of Temperature on Vapor Pressure H 2 O (l) => H 2 O (g) K eq = p water vapor p water vapor = K eq = e - G'/RT

18-85 CHEM 102, Fall LA TECH  G as a Function of the Extent of the Reaction

18-86 CHEM 102, Fall LA TECH  G as a Function of the Extent of the Reaction when there is Mixing

18-87 CHEM 102, Fall LA TECH Maximum Work  G = w system = - w max (work done on the surroundings)

18-88 CHEM 102, Fall LA TECH Coupled Reactions How to do a reaction that is not thermodynamically favorable? Find a reaction that offset the (+)  G Thermite Reaction Fe 2 O 3(s) => 2Fe (s) + 3/2O 2(g) 2Al (s) + 3/2O 2(g)  Al 2 O 3(s)

18-89 CHEM 102, Fall LA TECH ADP and ATP

18-90 CHEM 102, Fall LA TECH Acetyl Coenzyme A

18-91 CHEM 102, Fall LA TECH Gibbs Free Energy and Nutrients

18-92 CHEM 102, Fall LA TECH Photosynthesis: Harnessing Light Energy

18-93 CHEM 102, Fall LA TECH Using Electricity for reactions with (+)  G: Electrolysis Non spontaneous reactions could be made to take place by coupling with energy source: another reaction or electric current Electrolysis 2NaCl (l) => 2Na (s) + 2Cl 2(g)