17-1 CHEM 102, Fall 15 LA TECH Instructor: Dr. Upali Siriwardane Office: CTH 311 Phone Office Hours: M.W &F, 8:00-9:00 & 11:00-12:00 and Tu,Th 8: :00 am. am. or by appointment Test Dates Chemistry 102 Fall 2015 Sept. 29, 2015 (Test 1): Chapter 13 Oct. 27, 2015 (Test 2): Chapter 14 &15 Nove. 17, 2015 (Test 3): Chapter 16 &17 November 19, 2015 (Make-up test) comprehensive: Chapters 13-17

17-2 CHEM 102, Fall 15 LA TECH Chapter 6. Thermochemistry 6.1 Chemical Hand Warmers The Nature of Energy: Key Definitions The First Law of Thermodynamics: There Is No Free Lunch Quantifying Heat and Work Measuring for Chemical Reactions: Constant-Volume Calorimetry Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Constant-Pressure Calorimetry: Measuring Relationships Involving Determining Enthalpies of Reaction from Standard Enthalpies of Formation Energy Use and the Environment 263

17-3 CHEM 102, Fall 15 LA TECH Chapter 17. Free Energy and Thermodynamics 17.1 Nature’s Heat Tax: You Can’t Win and You Can’t Break Even Spontaneous and Nonspontaneous Processes Entropy and the Second Law of Thermodynamics Heat Transfer and Changes in the Entropy of the Surroundings Gibbs Free Energy Entropy Changes in Chemical Reactions: Calculating Free Energy Changes in Chemical Reactions: Calculating Free Energy Changes for Nonstandard States: The Relationship between and Free Energy and Equilibrium: Relating to the Equilibrium Constant (K)

17-4 CHEM 102, Fall 15 LA TECH Chapter 6. Thermochemistry 6.1 Chemical Hand Warmers The Nature of Energy: Key Definitions The First Law of Thermodynamics: There Is No Free Lunch Quantifying Heat and Work Measuring for Chemical Reactions: Constant-Volume Calorimetry Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Constant-Pressure Calorimetry: Measuring Relationships Involving Determining Enthalpies of Reaction from Standard Enthalpies of Formation Energy Use and the Environment 263

17-5 CHEM 102, Fall 15 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

17-6 CHEM 102, Fall 15 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

17-7 CHEM 102, Fall 15 LA TECH 1) Calculate entropy change for the reaction: 2 C(s) + 1/2 O 2 (g) + 3 H 2 (g)--> C 2 H 6 O(l); ∆H = ? (ANS kJ/mol) Given the following thermochemical equations: C 2 H 6 O(l) + 3 O 2 (g) ---> 2 CO 2 (g) + 3 H 2 O(l); ∆H = kJ/mol 1/2 O 2 (g) + H2(g) ----> H 2 O(l);∆H = kJ/mol C(s) + O 2 (g) ----> CO 2 (g);∆H = kJ/mol

17-8 CHEM 102, Fall 15 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

17-9 CHEM 102, Fall 15 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

17-10 CHEM 102, Fall 15 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

17-11 CHEM 102, Fall 15 LA TECH 2) Calculate enthalpy change given the ∆Hf o [SO 2 (g)] = -297 kJ/mole and ∆Hf o [SO 3 (g)] = -396 kJ/mole 2SO 2 (g) + O 2 (g) -----> 2 SO 3 (g); ∆H= ? ANS -198 kJ/mole)

17-12 CHEM 102, Fall 15 LA TECH What is relation of  H of a reaction to covalent bond energy?  H =  bonds broken  -   bonds formed  How do you calculate bond energy from  H? How do you calculate  H from bond energy?

17-13 CHEM 102, Fall 15 LA TECH

17-14 CHEM 102, Fall 15 LA TECH 3) Use the table of bond energies to find the ∆H o for the reaction: H 2 (g) + Br 2 (g)  2 HBr(g); H-H = 436 kJ, Br-Br= 193 kJ, H-Br = 366 kJ

17-15 CHEM 102, Fall 15 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

17-16 CHEM 102, Fall 15 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

17-17 CHEM 102, Fall 15 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

17-18 CHEM 102, Fall 15 LA TECH 2 H 2 (g) + O 2 (g)  2 H 2 O(liq) 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)

17-19 CHEM 102, Fall 15 LA TECH 4) Calculate the ∆S for the following reaction using: a) 2SO 2 (g) + O2 (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

17-20 CHEM 102, Fall 15 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,  G

17-21 CHEM 102, Fall 15 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

17-22 CHEM 102, Fall 15 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

17-23 CHEM 102, Fall 15 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)

17-24 CHEM 102, Fall 15 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 positive  H o rxn. Reaction is “entropy driven” NH 4 NO 3 (s) + heat  NH 4 NO 3 (aq)

17-25 CHEM 102, Fall 15 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)

17-26 CHEM 102, Fall 15 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

17-27 CHEM 102, Fall 15 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

17-28 CHEM 102, Fall 15 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

17-29 CHEM 102, Fall 15 LA TECH 5) Predict the spontaneity of the following processes from ∆H and ∆S at various temperatures. a) ∆H = 30 kJ ∆S = 6 kJT = 300 K b) ∆H = 15 kJ ∆S = -45 kJ T = 200 K

17-30 CHEM 102, Fall 15 LA TECH 6) Calculate the ∆G o for the following chemical reactions using given ∆H o values, ∆S o calculated above and the equation ∆G = ∆H - T∆S. 2SO 2 (g) + O 2 (g) > 2 SO 3 (g) ; ∆G o = ∆ H o = -198 kJ/mole; ∆S o = -187 J/K mole; T = 298 K ∆ G o system ∆ H o system ∆ S o system T

17-31 CHEM 102, Fall 15 LA TECH 7) Which of the following condition applies to a particular chemical reaction, the value of ∆H° = 98.8 kJ and ∆S° = J/K. This reaction is ∆G system ∆H system ∆ S system T a)Negative always Negative (exothermic) PositiveYes a)Negative at low T Positive at high T Negative (exothermic) Negative ∆ G =-,at low T; ∆ G= +, at high T a)Positive at low T Negative at high T Positive (endothermic) Positive ∆ G = +,at low T; ∆ G= -, at high T a)Positive alwaysPositive (endothermic) Negative ∆ G= +, at any T

17-32 CHEM 102, Fall 15 LA TECH Effect of Temperature on Reaction Spontaneity

17-33 CHEM 102, Fall 15 LA TECH  G o =  H o - T  S o

17-34 CHEM 102, Fall 15 LA TECH 8) At what temperature a particular chemical reaction, with the value of ∆H° = 98.8 kJ and ∆S° = J/K becomes a) Equilibrium: b) Spontaneous:

17-35 CHEM 102, Fall 15 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

17-36 CHEM 102, Fall 15 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

17-37 CHEM 102, Fall 15 LA TECH 9) Calculate the non standard ∆G for the following equilibrium reaction and predict the direction of the change using the equation: ∆G= ∆G o + RT ln Q Given ∆G f o [NH 3 (g)] = -17 kJ/mole N 2 (g) + 3H 2 (g) → 2NH 3 (g); ∆G=? at 300K, P N2 = 300, P NH3 = 75 and P H2 = 300

17-38 CHEM 102, Fall 15 LA TECH 10) The K a expression for the dissociation of acetic acid in water is based on the following equilibrium at 25°C: HC 2 H 3 O 2 (l) + H 2 O ⇄ H + (aq) + C 2 H 3 O 2 - (aq) What is ∆G° if K a =1.8 x ?

17-39 CHEM 102, Fall 15 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.

17-40 CHEM 102, Fall 15 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

17-41 CHEM 102, Fall 15 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

17-42 CHEM 102, Fall 15 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

17-43 CHEM 102, Fall 15 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 = ?

17-44 CHEM 102, Fall 15 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

17-45 CHEM 102, Fall 15 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

17-46 CHEM 102, Fall 15 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

17-47 CHEM 102, Fall 15 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

17-48 CHEM 102, Fall 15 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

17-49 CHEM 102, Fall 15 LA TECH  G as a Function of the Extent of the Reaction

17-50 CHEM 102, Fall 15 LA TECH  G as a Function of the Extent of the Reaction when there is Mixing

17-51 CHEM 102, Fall 15 LA TECH Maximum Work  G = w system = - w max (work done on the surroundings)

17-52 CHEM 102, Fall 15 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)

17-53 CHEM 102, Fall 15 LA TECH ADP and ATP

17-54 CHEM 102, Fall 15 LA TECH Acetyl Coenzyme A

17-55 CHEM 102, Fall 15 LA TECH Gibbs Free Energy and Nutrients

17-56 CHEM 102, Fall 15 LA TECH Photosynthesis: Harnessing Light Energy

17-57 CHEM 102, Fall 15 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)