advertisement

Energy Changes in Chemical Reactions -- Chapter 17 E = internal energy = sum of all potential and kinetic energy of the system: DE = Efinal - Einitial 1. First Law of Thermodynamics (Conservation of energy) DE = q + w where, q = heat absorbed by system w = work done on system Surroundings Sign Conventions: Heat in Work out W<0 Q>0 Heat out Q<0 SYSTEM Work in W>0 Q(+) --> System absorbs heat W (+) --> work done on system E is a “state function” -- independent of pathway, but, q and w are not! (they do depend on pathway) Entropy and Spontaneity (usually small) Two factors affect any change or reaction (a) Enthalpy (H) (Recall that DH = DE + PDV) exothermic processes (negative DH) tend to be spontaneous, but not always (b) Entropy (S) – degree of disorder or randomness Change in entropy: DS = Sfinal - Sinitial For a reaction: DS = Sproducts - Sreactants Positive DS means: an increase in disorder as reaction proceeds products more disordered (random) than reactants For a reaction, positive DS favors spontaneity Entropy and Second Law Entropy (S) increases with the number of energetically equivalent ways to arrange the components of a system to achieve a particular state. Trends: • • • • In general: Sgas >> Sliquid > Ssolid larger mass => more entropy more complex structure => more entropy more atoms in molecule => more entropy Second Law of Thermodynamics “…for any spontaneous process, the overall entropy of the universe increases…” A spontaneous process can have a negative DS for the system only if the surroundings have a larger positive DS. DSuniv = DSsys + DSsurr where DSsurr = –DHsys T Third Law Third Law of Thermodynamics “…the entropy of a pure crystalline substance equals zero at absolute zero…” S = 0 at T = 0 K (a) Standard Entropy (at 25 ºC) = Sº [Table 17.2, p789] (note the units!) Entropy change for a reaction: (e.g. J/mole K) DSº = S Sº(products) - S Sº(reactants) Gibbs Free Energy (b) Gibbs Free Energy = G Defined as: G = H - TS – A combination of enthalpy and entropy effects – Related to maximum useful work that system can do For a process (e.g. a reaction), DG = DH - TDS For any spontaneous change, DG is negative!!! (i.e. the free energy must decrease) Standard Free Energy DGº (at 1 atm) generally used to decide if a reaction is spontaneous (Yes, if negative) Three ways to obtain DGº for a reaction (a) from DHº and DSº requires DHºf and Sº data for all reactants and products DGº = DHº - TDSº DHº = SDHºf(products) - SDHºf(reactants) DSº = SSº(products) - SSº(reactants) Determining DGº Three ways to obtain DGº for a reaction…. (b) from standard “Free Energies of Formation” DGºf DGº = SDGºf(products) - SDGºf(reactants) where DGºf is the free energy change for the formation of one mole of the compound from its elements, e.g. DGºf for Al2(SO3)3(s) equals DGº for the following reaction 2 Al(s) + 3 S(s) + 9/2 O2(g) --> Al2(SO3)3(s) Manipulating Chemical Equations to Get DGº If reaction is reversed, change sign of DG°. If reaction is multiplied or divided by a factor, apply same factor to DG°. DG° for overall reaction = sum of DG° values for individual reactions. Problem Given the following thermochemical reactions: (eq 1) C2H4(g) + 3 O2(g) --> 2 CO2(g) + 2 H2O(l) DG° = -1098.0 kJ/mol (eq 2) C2H5OH(l) + 3 O2(g)--> 2 CO2(g)+ 3 H2O(l) DG° = -975.2kJ/mol Calculate DG° for the following reaction: C2H4(g) + H2O(l) --> C2H5OH(l) Example Problem, cont. Reverse 2nd reaction to put C2H5OH on product side then rewrite 1st equation and add them together. (eq 2) 2 CO2(g) + 3 H2O(l) --> C2H5OH(l) + 3 O2(g) DH° = + 975.2 kJ/mol (note the sign change!!!) (eq 1) C2H4(g) + 3 O2(g) --> 2 CO2(g) + 2 H2O(l) DH° = -1098.0 kJ/mol Net: C2H4(g) + H2O(l) --> C2H5OH(l) {note: 3 O2, 2 CO2, and 2 H2O cancel out} DH° = DH°1 + DH°2 = 975.2 + (-1098.0) = -122.8 kJ/mol Effect of Temperature on DG (a) Effect of Temperature on DG DG depends on DH and DS: DG = DH – TDS but DH and DS are relatively independent of temperature, so DG at some temperature T can be estimated. DGºT ≈ DHº298 - TDSº298 Free Energy and Equilibrium (b) for a system at equilibrium: Gproducts = Greactants and DG = 0 since DG = DH - TDS = 0 DH = T DS or T = DH/DS Example Problem Given the following, determine the normal boiling point of mercury (Hg). DHvaporization of Hg = 60.7 kJ/mole entropies: liquid Hg: gaseous Hg: Equilibrium: Hgliq Sº = 76.1 J/mole K Sº = 175.0 J/mole K Hggas DG = 0 T = DH/DS so, DH - TDS = 0 or T = [60.7 x 103 J/mole]/[(175.0 - 76.1) J/mol K] = 614 K = 341 ºC Free Energy and Equilibrium Constant (c) Relationship between DGº and Equilibrium Constant (K) For any chemical system: DG = DGº + (RT) ln Q If DG is not zero, then the system is not at equilibrium. It will spontaneously shift toward the equilibrium state. At Equilibrium: ∴ DG = 0 and Q=K DGº = -RT ln K for gaseous reactions: K = Kp for solution reactions: K = Kc {units of DG must match those of R value} Free Energy and Equilibrium (cont.) DGº = – RT ln K K values can be determined from thermodynamic data ! when K > 1 DGº is negative ∴ spontaneous reactions have large K and negative DGº values Sample Problem 1. Consider the gas-phase reaction: N2O5(g) + H2O(g) --> 2 HNO3(g) and the following thermodynamic data. Compound DHºf (kJ/mole) Sº (J/mole K) N2O5(g) 11.0 356 H2O(g) - 242 189 HNO3(g) - 174 156 (a) Decide whether or not the above reaction is spontaneous at 25 ºC by calculating the value of the appropriate thermodynamic quantity. (b) Calculate the temperature (in ºC) at which the above reaction should have an equilibrium constant (Kp) equal to 1.00. Sample Problem 1. Consider the gas-phase reaction: N2O5(g) + H2O(g) --> 2 HNO3(g) and the following thermodynamic data. Compound DHºf (kJ/mole) Sº (J/mole K) N2O5(g) 11.0 356 H2O(g) - 242 189 HNO3(g) - 174 156 (a) Decide whether or not the above reaction is spontaneous at 25 ºC by calculating the value of the appropriate thermodynamic quantity. (b) Calculate the temperature (in ºC) at which the above reaction should have an equilibrium constant (Kp) equal to 1.00. (a) DG = – 48 kJ/mol, which is negative, so spontaneous. (b) 229 ºC Sample Problem For a certain reaction, DH° = -95.2 kJ and DS° = -157 J/K. Determine DSsurr and DSuniv (in J/K) for this reaction at 850 K. Based on these results, is the reaction spontaneous at this temperature? Sample Problem For a certain reaction, DH° = -95.2 kJ and DS° = -157 J/K. Determine DSsurr and DSuniv (in J/K) for this reaction at 850 K. Based on these results, is the reaction spontaneous at this temperature? • Answer: DSsurr = +112 J/K • DSuniv = -45 J/K • Not spontaneous, because DSuniv would decrease, which is against the 2nd Law of Thermodynamics.