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Unit V - Chemical Thermodynamics Chapters 6 & 18 18.1. Determine H° for a chemical reaction from Hf° values of reactants and products. [Readings 6.8 Problems 76, 78, 80, 81, 82, & 102] Review Hºrxn = Hºf(prod) - Hºf(react) CO(g) + NO(g) --> CO 2(g) + 1/2 N 2(g) Hºf = -110.5 kJ +90 kJ mol mol x 1 mol x 1 mol -110.5 kJ +90 kJ |__________| -20.5 kJ -393.5 kJ 0 kJ mol mol x 1 mol x 1/2 mol -393.5 kJ 0 kJ |__________| -393.5 kJ Hºrxn = (-393.5 kJ) - (-20.5 kJ) = -373 kJ Since the natural tendency is for spontaneous processes to be exothermic -- then the reaction is spontaneous??? Page ‹#› 18.2. Recognize favorable conditions for spontaneous reactions. [Readings 18.4 Problems 15, 16, 17, 18, 56, & 60] Spontaneity in Chemical Reactions Chemical and physical processes that occur without continued external influence -Spontaneous Other processes occur only with external influence -- Nonspontaneous Spontaneous might be slow or fast. Nonspontaneous reactions do occur. The reverse of a spontaneous process is a Nonspontaneous process (v.v.). Examples of Spontaneous Processes Ice Melting: H2O(s) --> H 2O(l) @25ºC Iron Rusting: 4Fe(s) + 3O2(g) + 6H2O(l) --> 4Fe(OH) 3(s) Page ‹#› What drives a chemical reaction? Enthalpy, H Entropy, S Enthalpy, H (see chapter 8) Flow of energy (heat) CH4(g) + 2 O2(g) --> CO2(g) +H 2O (l) + heat CH4(g) + 2 O2(g) --> CO2(g) +H 2O (l)) H = - heat kJ Many exothermic reactions are spontaneous The natural tendency is to lower energy states Enthalpy, H Water freezing is an exothermic process. Is it a spontaneous process? At – 10°C? At + 10°C? Page ‹#› Enthalpy, H NH 4NO3 (s) ––> NH4+ (aq) + NO 3– (aq) H° = + 26.2 kJ/mol The reaction is endothermic Is it spontaneous? Entropy, S A measure of “randomness” or molecular disorder (degrees of freedom). A gas has more disorder than a liquid, thus more entropy. evaporation: H2O(l) --> H2O(g) at 25ºC Natural tendency is to increased disorder S = Sproducts - Sreactants Increased dissorder give positive S Factors Affecting Entropy Volume increasing volume increases entropy Temperature increases the kinetic energy (motion) Physical State compare the entropy of a gas to a solid. Molecular complexity Cl 2 (g) ––> 2 Cl (g) KHT (s) ––> K+ (aq) + HT– (aq) Page ‹#› Predicting whether a reaction is spontaneous Example: Ba(OH) 2•8 H 2O (s) + 2 NH 4Cl (s) ––> BaCl 2 (aq) + 2 NH 3 (aq) + 10 H 2O (l) H° = + 80.3 kJ unfavorable S° = + 428 J favorable Predicting whether a reaction is spontaneous Spontaneous process: Favored by decrease in H ( H< 0) Favored by an increase in S ( S > 0) Nonspontaneous process: Favored by increase in H ( H > 0) Favored by an decrease in S ( S< 0) 18.3. Determine S° for a chemical reaction from S° values of reactants and products. [Readings 18.5 Problems 62, 64, & 66] Page ‹#› Standard Molar Entropies S° • The entropy of 1 mol of a pure substance at: • 1 atm • 25°C • Units are J K-1 mol-1 • See values in table 17.1 and Appendix B Sºrxn = Sº(prod) - Sº(react) CO(g) + NO(g) --> CO 2(g) + 1/2 N 2(g) Sº = +198 J +211 J +214 J +192 J mol-K mol-K mol-K mol-K x 1 mol x 1 mol x 1 mol x 1/2 mol +198 J +211 J +214 J +96 J K K K K |__________| |__________| +409 J/K +310 J/K Sºrxn = (+310 J/K) - (+409 J/K) = -99 J/K Since the natural tendency is for spontaneous processes to increase entropy -- then is the reaction nonspontaneous??? CO(g) + NO(g) --> CO 2(g) + 1/2 N2(g) Hºrxn = -373 kJ Sºrxn = -99 J/K Driving Forces Oppose Page ‹#› 18.4. Determine G° for a chemical reaction from H° and S° values. Also calculate the temperature at which a process becomes spontaneous. [Readings 18.4 – 18.6 Problems 62, 64, 66, 68, 70, 72, & 74] Gibbs Free–energy • Gibbs free-energy predicts the spontaneity of a chemical or physical process • G= H–T S • G<0 The process is spontaneous • G=0 The process is at equilibrium • G>0 The process is nonspontaneous Example problem • Determine whether a process will be spontaneous at 500K if H = – 128 kJ and S = + 35 J/K • G = – 128 kJ – 500K * .035 kJ/K • G = – 144.5 kJ • Will the reaction ever be nonspontaneous? Page ‹#› •Example problem • Determine temperature at which a process will become spontaneous if H = – 50 kJ and S = – 35 J/K • At equilibrium, G = 0 • H=T S •Example problem • Determine temperature at which a process will become spontaneous if H = – 50 kJ and S = – 35 J/K • At equilibrium, G = 0 • H=T S T= ∆H −50kJ = −1 = 1430K ∆S −.035kJK Free Energy: energy available to do work Standard State Conditions 1 atm pressure 1 M concentration Temperature of 25°C G° = Hº - T Sº Page ‹#› CO(g) + NO(g) --> CO 2(g) + 1/2 N2(g) Hºrxn = -373 kJ Sºrxn = -99 J/K Gº = Hº - T Sº Gº = (-373 kJ) - {298 K(-99 J/K)} Gº = (-373 kJ) - (- 29,500 J) Gº = (-373 kJ) - (- 29.5 kJ) Gº = -343.5 kJ 18.5. Predict whether a chemical reaction, as written, is spontaneous, non-spontaneous, or at equilibrium. [Readings 18.4 Problems] Consider both driving forces Hº Sº – + + – + + – – Page ‹#› Consider both driving forces Hº Sº – ---> + ---> + <--- – <--- + <--- + ---> – ---> – <--- Consider both driving forces Hº Sº – ---> + ---> + <--- – <--- + <--- + ---> – ---> – Spontaneous <--- Consider both driving forces Hº Sº – ---> + ---> Spontaneous + <--- – <--- Nonspontaneous + <--- + ---> – ---> – <--- Page ‹#› Consider both driving forces Hº Sº – ---> + ---> Spontaneous + <--- – <--- Nonspontaneous + <--- + ---> Spontaneous @ high Temp – ---> – <--- Consider both driving forces Hº Sº – ---> + ---> Spontaneous + <--- – <--- Nonspontaneous + <--- + ---> – ---> – Spontaneous @ high Temp Spontaneous @ low Temp <--- 18.6. Determine G° for a chemical reaction from Gf° values of reactants and products. [Readings 18.69 Problems 70, 71, 72, 73, & 74] Page ‹#› Gºrxn = Gºf(prod) - Gºf(react) CO(g) + NO(g) --> CO 2(g) + 1/2 N 2(g) Gºf = -137 kJ +87 kJ mol mol x 1 mol x 1 mol -137 kJ +87 kJ |__________| -50 kJ -394 kJ 0 kJ mol mol x 1 mol x 1/2 mol -394 kJ 0 kJ |__________| -394 kJ Gºrxn = (-394 kJ) - (-50 kJ) = -344 kJ Three Possible Conditions Gº < 0 reaction is spontaneous R --> P Gº > 0 reaction is nonspontaneous Gº = 0 reaction is at equilibrium R <-- P R <==> P 18.7. Calculate the standard free energy for a chemical reaction from the equilibrium constant (vv). [Readings 18.9 Problems 86, 87, 88, 90, 92, & 94] Page ‹#› Nonstandard conditions of concentration G = Gº + RT ln Q Nonstandard conditions of concentration G = Gº + RT ln Q @ equilibrium, G = 0, Q = K G = Gº + RT ln Q 0 = Gº + RT ln K Gº = -RT ln K G° and K Consider the equation at standard conditions (unit concentration) A + B <==> C + D 1M 1M 1M 1M Kc = [C] [D] / [A] [B] if Gº < 0; reaction shifts to right; K c > 1 if Gº > 0; reaction shifts to left; if Gº = 0; reaction does not shift; Kc = 1 Kc < 1 The free energy drives a chemical reaction toward equilibrium Page ‹#› G° and K G° = -RT ln K (R = 8.31 J / mol K) CO(g) + NO(g) --> CO 2(g) + 1/2 N2(g) Gºrxn = -344 kJ / mol -344 kJ / mol = -(8.31 J / mol K) (298 K) ln K 344 kJ / mol / (2480 J / mol) (1kJ / 10 3 J)= ln K 139 = ln K K = 2 x 10 60 CO(g) + NO(g) --> CO 2(g) + 1/2 N2(g) Hºrxn = -373 kJ Sºrxn = -99 J/K Gºrxn = -343.5 kJ K = 2 x 10 60 18.8. Determine the equilibrium temperature, Te, for a chemical reaction from H° and S° (vv). [Problems web page] Page ‹#› Consider the hypothetical reaction: A <==> B Hºrxn = -150 kJ drives reaction R --> P Sºrxn = -600 J/K drives reaction R <-- P Therefore, the reaction is spontaneous at low temperatures. @ 25°C Gºrxn = 28.8 kJ @ 0°C Gºrxn = 13.8 kJ @ -25°C Gº rxn = -1.2 kJ At Equilibrium, G = 0 kJ and T = T e Gº = Hº - T Sº 0 = Hº - Te Sº Hº = Te Sº Hº / Sº = Te -150 kJ / -600 J/K = T e Te = -150 kJ / -0.600 kJ/K Te = 250 K or -23°C CO(g) + NO(g) --> CO 2(g) + 1/2 N2(g) Hºrxn = -373 kJ Sºrxn = -99 J/K Hº / Sº = T e -373 kJ / -99 J/K = T e -373 kJ / -0.099 kJ/K = T e Te = 3770 K or 3500°C Page ‹#› What is the melting point of NaCl(s)? NaCl (s) <==> NaCl (l) Hºrxn = 30.3 kJ Sºrxn = .0282 kJ/K Hº / Sº = T e Te = 1074 K or 801°C Page ‹#›