Thermodynamics vs. Kinetics Equilibrium? On-line HW due Thursday Nov. 11 th at 11:59 PM

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1 Tro Chapter 17 - Free Energy and Thermodynamics Spontaneous and non-spontaneous processes Entropy and the second law of thermodynamics Gibbs free energy (ΔG ), entropy (ΔS ) and enthalpy (ΔH ) Gibbs free energy and K Thermodynamics vs. Kinetics Equilibrium? On-line HW due Thursday Nov. 11 th at 11:59 PM Perpetual motion, energy transactions and nature s heat tax Tro 17.2 Spontaneous vs. nonspontaneous change

2 Tro 17.2 Predicting spontaneous change CH O 2 CO 2 + 2H 2 O ΔH = -802 kj 2Fe (s) + 3/2O 2 Fe 2 O 3 ΔH = -826 kj H 2 O (l) H 2 O (s) ΔH = kj (spontaneous at T < 0 C) Tro 17.3 Entropy and the second law of thermodynamics S = K ln W S = entropy K =Boltzmann s constant (1.38e -23 J/K) W is the number of energetically equivalent ways to arrange the components of a system H 2 O (s) NaCl (s) H 2 O (l) ΔH = kj (spontaneous at T > 0 C) Na + (aq) + Cl - (aq) ΔH = +3.9 kj Entropy and Temperature/Phase Change

3 Tro 17.3 and 17.4 Entropy and the second law of thermodynamics The entropy of the universe is always increasing ΔS univ = ΔS sys + ΔS surr > 0 Entropy and the third law of thermodynamics A perfect crystal has zero entropy at absolute zero S sys = 0 at 0 K Predicting relative S 0 values of a system 1. Temperature changes and S for copper metal T (K) S (J/molK): Physical states and phase changes and entropy Na H 2 O C (graphite) S (s or l) 51.4 (s) 69.9 (l) 5.7 (s) S (g) Dissolution of a solid or liquid NaCl AlCl 3 CH 3 OH S (s or l) 72.1 (s) 167 (s) 127 (l) S (aq) Tro 17.4 Temperature and its effect on spontaneity e.g. consider the freezing of water - spontaneous at low temperatures, nonspontaneous at high temperatures Tro 17.4 Temperature and its effect on spontaneity cont d Product favored reactions for exothermic processes with net increase in entropy (+ ΔS) Reactant favored reactions for endothermic processes with a net decrease in entropy (- ΔS) But, what happens with endothermic processes with a net increase in entropy Or What happens with exothermic processes and a decrease in entropy

4 Tro 17.5 Entropy, free energy and work Gibb s free energy, G, combines a system s enthalpy and entropy G = H - TS ΔS univ = ΔS sys + ΔS surr ΔS surr = -ΔH sys /T ΔS univ = ΔS sys -ΔH sys /T -TΔS univ = -TΔS sys +ΔH sys ΔG sys = ΔH sys -TΔS sys = -TΔS univ Remembering the second law ΔS univ >0 spontaneous ΔS univ = equilibrium ΔS univ < 0 non-spontaneous Since absolute temperature is always positive, -TΔS univ < 0 for spontaneous processes and, since ΔG = -TΔS univ ΔG < 0 for a spontaneous process ΔG = 0 for a process at equilibrium ΔG > 0 for a non-spontaneous process Gibbs Free energy - A decrease in G corresponds to a spontaneous process

5 Tro 17.6 Calculating the change in entropy of a reaction (using table 17.2 or Appendix IIB) ΔS rxn = ΣmS products - ΣnS reactants N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔS rxn = [(2 mol NH 3 )(S of NH 3 )] - [(1 mol N 2 )(S of N 2 ) + (3 mol H 2 )(S of H 2 )] = [2(192.8)]-[1(191.6)+3(130.7)] = J/K Tro 17.7 Entropy, free energy, and work: Calculating Standard Free Energy changes (see Appendix II B) 4KClO 3 (s) Δ 3KClO 4 + KCl ΔH f = kj/mol ΔH f = kj/mol ΔH f = kj/mol S = J/molK S = J/molK S = 82.6 J/molK Calculate ΔG at 25 C 1) Calculate ΔH (must know ΔH f for each species) 2) Calculate ΔS (must know S for each species) 3) Calculate ΔG using ΔG = ΔH - TΔS Calculating ΔG rxn from ΔG f values 4KClO 3 (s) Δ 3KClO 4 + KCl ΔG f = kj/mol ΔG f = kj/mol ΔG f = kj/mol ΔG rxn = ΣmG products - ΣnG reactants ΔG rxn = [3(ΔG f KClO 4 ) + ΔG f KCl] - [4 ΔG f KClO 3 ] Calculating ΔG rxn at elevated temperatures 2SO 2 + O 2 2SO 3 At 298 K, ΔG = kj, ΔH = kj and ΔS = J/K Is the reaction spontaneous at 25 C? Is the reaction spontaneous at 900 C (assume ΔH and ΔS don t change much with temperature)? Tro 17.8 and 17.9 Free energy for Non-standard states: ΔG rxn and ΔG rxn Sign of ΔG allows us to predict spontaneity and thus reaction direction If Q<K, ln Q/K < 0; ΔG < 0 If Q>K, ln Q/K > 0; ΔG > 0 If Q=K, ln Q/K = 0; ΔG = 0 ΔG = RT ln Q/K = RT ln Q - RT ln K ΔG = -RT ln K (by choosing standard conditions for Q) ΔG = ΔG + RT ln Q

6 Tro 17.9 Free Energy and Equilibrium ΔG = -RT ln K Free Energy and Equilibrium Calculating ΔG rxn at non-standard conditions At 298 K, ΔG = kj 2SO 2 + O 2 2 SO 3 a) Calculate K at 298 K and at 973 K (ΔG 298 = kj/mol and ΔG 973 = kj/mol) b) Which direction would the above reaction proceed if we had atm of SO 2, atm of O 2 and atm of SO 3 at 25 C? At 700 C? Useless expressions: ΔG = -RTlnK ΔG = ΔG + RTlnQ ΔG = kj mol 298 K ΔG = kj mol 973 K

7 1) For each of the following pairs, choose (circle) the substance with the higher entropy per mole at a given temperature: a) Ar (l) or Ar (g) b) He (g) at 3 atm pressure or He (g) at 1.5 atm pressure S C 2 H 4 = Jmol -1 K -1 S N 2 O 4 = Jmol -1 K -1 S C 2 H 6 = Jmol -1 K -1 S NO 2 = Jmol -1 K -1 S H 2 = Jmol -1 K -1 2) Using S values provided above, calculate ΔS values for the following reactions. In each case, account for the sign of ΔS. a) C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) c) 1 mol of Ne (g) in 15.0 L or 1 mol of Ne (g) in 1.50 L d) CO 2 (g) or CO 2 (s) b) N 2 O 4 (g) 2NO 2 (g) 3). For a certain chemical reaction, ΔH = kj and ΔS = J/K. Is the reaction exothermic or endothermic Does the reaction lead to an increase or decrease in the disorder of the system? The value of K a for nitrous acid (HNO 2 ) at 25 C is 4.5x10-4 By using the value of K a, calculate ΔG for the dissociation of nitrous acid in aqueous solution. Calculate ΔG for the reaction at 298 K What is the value of ΔG at equilibrium Is the reaction spontaneous at 298 K What is the value of ΔG when [H + ] = 5.0 x 10-2 M, [NO 2- ] = 6.0 x 10-4 M, and [HNO 2 ] = 0.20 M?