1 Second Law of Thermodynamics As the reaction goes to products our system becomes more disordered and the entropy of our system increases. One driving forces for a chemical reaction is one of the products is a gas. 2 19.3 GIBBS FREE ENERGY О”Suniverse= О”Ssurroundings+ О”Ssystem = - О”Hsystem / T + О”Ssystem -T О”Suniverse = О”Hsystem - T О”Ssystem = О”Gsystem О”G is the change in free energy for the system. О”Gosystem = О”Hosystem - T О”Sosystem (standard state) О”Gorxn from О”Gof and from О”Horxn and О”Sorxn О”Gorxn= S О”Gfo(products) - S О”Gfo(reactants О”Gorxn = О”Horxn - T О”Sorxn (Don't forget to change the entropy term from J to kJ) Gibbs Free Energy, G DSuniv = DSsurr + DSsys -D H sys D Suniv = + D Ssys T Multiply through by -T -T О”Suniv = О”Hsys - T О”Ssys -T О”Suniv = change in Gibbs free energy for the system = О”Gsystem Under standard conditions вЂ” О”Go = О”Ho - T О”So 3 Gibbs Free Energy, G 4 О”Go = О”Ho - T О”So Gibbs free energy change = total energy change for system - energy lost in disordering the system If reaction is exothermic (О”Ho negative) and entropy increases (О”So is +), then О”Go must be negative and reaction product-favored in the standard state. If reaction is endothermic (О”Ho is +), and entropy decreases (О”So is -), then О”Go must be + and reaction is reactant-favored in the standard state. 5 Product-Favored or Reactant-Favored? вЂў The reaction is product-favored if О”G is negative. вЂў We can see that this is always true if О”H is negative and О”S is positive. вЂў If О”H is positive and О”S is negative, О”G is always positive. 6 Product-Favored or Reactant-Favored? вЂў The other two cases are temperature dependent with a positive О”S favoring spontaneity at high temperature, thus overcoming the positive О”H, and a negative О”H favoring spontaneity at a low temperature when О”S is negative. вЂў Table next slide and Figure 20.8 7 Gibbs Free Energy, G О”Go = О”Ho - T О”So О”Ho О”So О”Go Reaction exo(-) increase(+) - Prod-favored endo(+) decrease(-) + React-favored exo(-) decrease(-) ? T dependent endo(+) increase(+) ? T dependent 8 Figure 20.8 9 Gibbs Free Energy, G О”Go = О”Ho - T О”So Two methods of calculating О”Go a) Determine О”Horxn and О”Sorxn and use Gibbs equation. b) Use tabulated values of free energies of formation, О”Gfo. DGorxn = S DGfo (products) - S DGfo (reactants) Calculating D o G rxn 10 Combustion of acetylene C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g) Use enthalpies of formation to calculate О”Horxn = -1238 kJ Use standard molar entropies to calculate О”Sorxn = -97.4 J/K or -0.0974 kJ/K О”Gorxn = -1238 kJ - (298 K)(-0.0974 kJ/K) = -1209 kJ Reaction is product-favored (in the standard state) in spite of negative О”Sorxn. Reaction is вЂњenthalpy drivenвЂќ. 11 Calculating D o G rxn Is the dissolution of ammonium nitrate productfavored? If so, is it enthalpy- or entropy-driven? NH4NO3(s) + heat NH4NO3(aq) Calculating D NH4NO3(s) + heat 12 o G rxn NH4NO3(aq) From tables of thermodynamic data: О”Horxn = +25.7 kJ О”Sorxn = +108.7 J/K or +0.1087 kJ/K О”Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K) = -6.7 kJ Reaction is product-favored (in the standard state) in spite of negative О”Horxn. Reaction is вЂњentropy drivenвЂќ. Calculating DGorxn DGorxn = S DGfo (products) - S DGfo (reactants) 13 Combustion of carbon C(graphite) + O2(g) --> CO2(g) О”Gorxn= О”Gfo(CO2) - [О”Gfo(graph) + О”Gfo(O2)] О”Gorxn = -394.4 kJ - [ 0 + 0 ] Note that free energy of formation of an element in its standard state is 0. О”Gorxn = -394.4 kJ Reaction is product-favored as expected. 14 Free energy and Temperature вЂў If we assume that О”Ho and О”So are relatively independent of temperature, we can calculate О”Go at any particular temperature we choose. О”GTorxn = О”Horxn - T О”Sorxn 15 Free Energy and Temperature 2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g) О”Horxn = +467.9 kJ О”Sorxn = +560.3 J/K О”Gorxn = +300.8 kJ Reaction is reactant-favored at 298 K At what T does О”Gorxn just change from being (+) to being (-)? When О”Gorxn = 0 = О”Horxn - T О”Sorxn. DHrxn 467.9 kJ T = = = 835.1 K DSrxn 0.5603 kJ/K 16 Free Energy and Temperature вЂў For the reaction below: Calculate О”Go at 298.15 K two ways, explain the sign of О”So, determine if the reaction in the standard state is product-favored at 298.15 K, determine which term or terms favor spontaneity, and calculate the temperature at which the reaction would first become reactant favored in the standard state. 3 H2(g) + CO(g) ====> CH4(g) + H2O(g) 20.4 THERMODYNAMICS AND THE EQUILIBRIUM CONSTANT 17 At equilibrium, О”GT = 0, and О”GTo = -RT lnKT вЂў Figure 20.11, shows the relationship between Q and K, which comes from the concentration dependence of Free Energy. вЂў One cannot calculate a new K by simply changing the T in the equation since О”GTo is a function of temperature. О”GTorxn = О”Horxn - TО”Sorxn вЂў A useful combined form of these equations is: О”GTo = О”Ho - TО”So = -RT ln KT. 18 Figure 20.11 Thermodynamics and Keq Keq is related to reaction favorability. When О”Gorxn < 0, reaction moves energetically вЂњdownhillвЂќ О”Gorxn is the change in free energy as reactants convert completely to products. But systems often reach a state of equilibrium in which reactants have not converted completely to products. In this case О”Grxn is < О”Gorxn , so state with both reactants and products present is more stable than complete conversion. 19 Thermodynamics and Keq 20 Product-favored reaction. 2 NO2 ---> N2O4 О”Gorxn = -4.8 kJ Here О”Grxn is less than О”Gorxn , so the state with both reactants and products present is more stable than complete conversion. 21 Thermodynamics and Keq Reactant-favored reaction. N2O4 --->2 NO2 О”Gorxn = +4.8 kJ Here О”Gorxn is greater than О”Grxn , so the state with both reactants and products present is more stable than complete conversion. 22 Thermodynamics and Keq Keq is related to reaction favorability and so to О”Gorxn. The larger the value of О”Gorxn the larger the value of K. o О”G rxn = - RT lnK where R = 8.31 J/KвЂўmol 23 Thermodynamics and Keq DGorxn = - RT lnK Calculate K for the reaction N2O4 --->2 NO2 О”Gorxn = +4.8 kJ О”Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K lnK = - 4800 J = - 1.94 (8.31 J/K)(298K) K = 0.14 When О”Gorxn > 0, then K < 1 24 THERMODYNAMICS AND THE EQUILIBRIUM CONSTANT вЂў Let us consider the derivation of these equations to further our understanding of them and their interrelationships. вЂў This derivation starts with the definition of G, G = H - TS вЂў At constant temperature, with H = E + PV, and wext = 0,and w = -PdV, we end up with, О”G = nRT О”P/P. 25 THERMODYNAMICS AND THE EQUILIBRIUM CONSTANT G = Go + nRT ln a, where a, is the activity, unitless concentration. вЂў For a reaction, we arrive at: О”GT = О”GTo + RT вЂў Where Q = K at equilibrium. ln QT 26 DERIVATIONS вЂў Derive the relationships between О”GTo and KT. вЂў Derive the relationships between О”Ho , KT , and T. 27 THE EQUILIBRIUM CONSTANT вЂў The temperature dependence of K can be calculate using the equation: О”GTo = О”Ho - T О”So = - RT ln KT. вЂў This equation can be written at two temperatures, T1 and T2, and combined to eliminate О”So. вЂў This produces the very useful equation: пѓ© KT2 пѓ№ ln пѓЄ K пѓє пѓЄпѓ« T1 пѓєпѓ» - D H пѓ©1 - 1 пѓ№ = R пѓЄT2 T1пѓє пѓ« пѓ» 28 THE EQUILIBRIUM CONSTANT A plot of ln K vs. 1/T yields a straight line with a slope of - О” Ho/R. 29 20.5 THERMODYNAMICS AND TIME вЂў The first and second Laws of Thermodynamics cannot be proven, they are laws of experience and tell us the direction of time in any given "picture". 30 Real World Examples of: Chapter 20: Second Law of Thermodynamics вЂњThe total entropy of the universe is always increasingвЂќ liq u id va p o riza tio n co n d en sa tio n gas -1 H 2 O (l) 40.7 kJm ol - 40.7 kJm ol -1 H 2O ( g ) 31 Faster molecules can break intermolecular forces like H-bonding to escape a solution 32 Kinetic Molecular Theory of Gases 33 Experiment: Two Florence flasks with H2O, one closed and the other open. This experiment demonstrates: 1. There is sufficient heat in the surroundings to allow liquid water to escape to the atmosphere; 2. Air currents and gas diffusion prevent the gaseous water from making contact with the water surface. 34 The Clausius-Clapeyron equation is a method for obtaining enthalpy of vaporization, at any temperature. пѓ¦ D H vap пѓ¶ ln Pvap = - пѓ§ пѓ·пЂ«C пѓЁ RT пѓё 35 In a closed systemвЂ¦Questions 1. Can we change the equilibrium in this system? 2. Is there any reason for wanting to change the equilibrium in this system? 36 Closed system equilibrium H 2 O (l) Kp = H 2O ( g ) PH 2O Pw a ter va p o r Q < K Reaction favors reactants to products Q > K Reaction favors products to reactants Q = K Reaction is at equilibrium 37 Frost-free freezers Kp = PH 2O Pw a ter va p o r Q < K favors reactants to products Air in the freezer is warmed then dried. The vapor pressure of ice is 4.579 torr. Warm, desiccated air can remove water vapor. 38 Second Law of Thermodynamics As the reaction goes to products our system becomes more disordered and the entropy of our system increases. One driving forces for a chemical reaction is one of the products is a gas. 39 Gibbs free energy H 2 O (l) п‚® H 2 O ( g ) D G = G rxn пЂ« R T ln Q п‚° D G rxn = - R T ln K p п‚° D G rxn = D H rxn - T D S rxn пѓћ 44.0 kJ - 35.5 kJ = 8.58 kJ п‚° K p = .0 3 1 3 п‚° п‚° 40 Gibbs free energy D G rxn = 8 .5 8 kJ п‚° K p = .0 3 1 3b ar DG > 0 reactant favored DG = 0 equilibrium DG < 0 product favored, reaction proceeds to products 41 If Q < K product favored What are the conditions necessary for the spontaneous formation of products? D G = G rxn пЂ« R T ln Q п‚° Q = PH 2O Pw a ter va p o r A Q of .0313 allows water vapor to evaporate. The atmosphere has a PH2O of .001%-4% water vapor (3%-100% humidity), in the winter Pvap can be as high as 20 torr. 42 Summary Water will Evaporate even though the process takes energy вЂўKinetic Molecular Theory Kp = PH 2O вЂўEvidence that P under normal conditions is easily manipulated w a ter va p o r вЂўHighly ordered H2O(l) has a large DS, this is the main driving force for producing a higher partial pressure of water. 43 As a rule of thumb the rate of an rxn doubles, or triples for every 10 degrees increase in temperature. What factors affect the rate of a reaction? пѓј The concentration of the reactants. The more concentrated the faster the rate; пѓ� Temperature. Usually reactions speed up with increasing temperature; пЃ± Physical state of reactants; пЃ± Catalyst (or inhibitor). A catalyst speeds up a reaction, an inhibitor slows it down. 44 ,H E ster + H 2 O п‚ѕ О”п‚ѕ п‚® carboxylic acid + alcohol + Exp Temp K k [L/molп‚ґs] DK K2/K1 1 288 .0521 2 298 .101 10 1.93 3 308 .184 10 1.82 4 318 .332 10 1.80 Concentrations of the Ester and H2O are held constant, only the temperature changes. 45 Arrhenius equation k = Ae - Ea RT Ea пѓ¦ 1 пѓ¶ ln k = ln A пѓ§ пѓ· R пѓЁT пѓё 46 Testing the вЂњRule of ThumbвЂќ пѓ¦ K 2 пѓ¶ пѓ¦ Ea пѓ¶ пѓ¦ 1 1 пѓ¶ ln пѓ§ пѓ·пѓ§ пѓ·=пѓ§ пѓ· пѓЁ K 1 пѓё пѓЁ R пѓё пѓЁ T1 T 2 пѓё Setting K2/K1 = 2, T1 = 273, T2 = 283 Solve for Ea, Ea = 44.5kJmol-1 47 Theoretical Results T1 T2 k2/k1 273 283 2.00 373 383 1.45 473 483 1.26 573 583 1.17 673 683 1.12 773 783 1.09 48 When can relying on the rule be dangerous H 2 ( g ) пЂ« C l2 ( g ) п‚® 2 H C l heat пЂ« C l 2 ( g ) п‚® 2 C l ( g ) Cl(g ) пЂ« H 2 (g ) п‚® HCl(g ) пЂ« H (g ) H ( g ) пЂ« C l2 ( g ) п‚® H C l ( g ) пЂ« C l ( g ) Once the rxn is initiated, no further heat is needed.