Entropy and Free Energy

Transcription

1 Entropy and Free Energy How to predict if a reaction can occur, given enough time? THERMODYNAMICS 1 Thermodynamics If the state of a chemical system is such that a rearrangement of its atoms and molecules would decrease the energy of the system--- AND the K is greater than 1, 2 then this is a product-favored system. How to predict if a reaction can occur at a reasonable rate? KINETICS Most product-favored reactions are exothermic but this is not the only criterion Thermodynamics 3 Thermodynamics and Kinetics 4 Both product- and reactant-favored reactions can proceed to equilibrium in a spontaneous process. AgCl(s) Ag + (aq) + Cl (aq) K = 1.8 x Reaction is not product-favored, but it moves spontaneously toward equilibrium. Spontaneous does not imply anything about time for reaction to occur. Diamond is thermodynamically favored to convert to graphite, but not kinetically favored. Paper burns a product-favored reaction. Also kinetically favored once reaction is begun. 5 6 Spontaneous Reactions Spontaneous Reactions In general, spontaneous reactions are exothermic. But many spontaneous reactions or processes are endothermic or even have H = 0. Fe 2 O 3 (s) + 2 Al(s) ---> 2 Fe(s) + Al 2 O 3 (s) H = kj NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq) Page 1

2 7 8 One property common to spontaneous processes is that the final state is more DISORDERED or RANDOM than the original. Spontaneity is related to an increase in randomness. The thermodynamic property related to randomness is ENTROPY, S. Reaction of K with water The entropy of liquid water is greater than the entropy of solid water (ice) at 0 C. Directionality of Reactions How probable is it that reactant molecules will react? PROBABILITY suggests that a spontaneous reaction will result in the dispersal 9 Directionality of Reactions Probability suggests that a spontaneous reaction will result in the dispersal of energy or of matter or both. Matter Dispersal 10 *ofenergy * or of matter * or of energy & matter Directionality of Reactions Probability suggests that a spontaneous reaction will result in the dispersal of energy or of matter or both. Energy Dispersal Directionality of Reactions Energy Dispersal Exothermic reactions involve a release of stored chemical potential energy to the surroundings. The stored potential energy starts out in a few molecules but is finally dispersed over a great many molecules. The final state with energy dispersed is more probable and makes a reaction spontaneous. Page 2

3 Page 3 13 Entropy and States of Matter 14 S o (J/K mol) H 2 O(liq) H 2 O(gas) S (gases) > S (liquids) > S (solids) S (Br 2 liq) < S (Br 2 gas) S (H 2 O sol) < S (H 2 O liq) Entropy of a substance increases with temperature. Molecular motions of heptane, C 7 H 16 Molecular motions of heptane at different temps. Increase in molecular complexity generally leads to increase in S Entropies of ionic solids depend on coulombic attractions. Entropy usually increases when a pure liquid or solid dissolves in a solvent. S o (J/K mol) MgO 26.9 NaF 51.5 Mg 2+ & O 2- Na + & F -

4 Page 4 Standard Molar Entropies Entropy Changes for Phase Changes For a phase change, S = q/t where q = heat transferred in phase change For H 2 O (liq) ---> H 2 O(g) H = q = +40,700 J/mol S = q T 40, 700 J/mol = K = J/K mol Entropy and Temperature 21 Calculating S for a Reaction 22 S o = Σ S o (products) - Σ S o (reactants) Consider 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 increases slightly with T S increases a large amount with phase changes S o = 2 mol (69.9 J/K mol) - [2 mol (130.7 J/K mol) + 1 mol (205.3 J/K mol)] S o = J/K Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid. 2nd Law of Thermodynamics 23 2nd Law of Thermodynamics 24 A reaction is spontaneous if S for the universe is positive. S universe = S system + S surroundings S universe > 0 for spontaneous process Dissolving NH 4 NO 3 in water an entropy driven process. First calc. entropy created by matter dispersal ( S system ) Next, calc. entropy created by energy dispersal ( S surround ) S universe = S system + S surroundings

5 Page 5 2nd Law of Thermodynamics 25 2nd Law of Thermodynamics 26 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o system = J/K S o surroundings = q surroundings T = - H system T Can calc. that H o rxn = H o system = kj S o surroundings = S o surroundings - ( kj)(1000 J/kJ) K = J/K 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o system S o surroundings S o universe = J/K = J/K = J/K The entropy of the universe is increasing, so the reaction is product-favored. Spontaneous or Not? 27 Gibbs Free Energy, G 28 S univ = S surr + S sys S univ = H sys T + S sys Multiply through by -T -T S univ = H sys - T S sys -T S univ = change in Gibbs free energy for the system = G system Under standard conditions Remember that H sys is proportional to S surr An exothermic process has S surr > 0. G o sys = H o sys - T S o sys Gibbs free energy change = total energy change for system - energy lost in disordering the system If reaction is exothermic (negative H o ) (energy dispersed) and entropy increases (positive S o ) (matter dispersed) then G o must be NEGATIVE reaction is spontaneous (and productfavored). 29 Gibbs free energy change = total energy change for system - energy lost in disordering the system If reaction is endothermic (positive H o ) and entropy decreases (negative S o ) then G o must be POSITIVE reaction is not spontaneous (and is reactantfavored). 30

6 Page 6 Gibbs Free Energy, G 31 Gibbs Free Energy, G 32 H o S o G o Reaction Two methods of calculating G o exo( ) increase(+) Prod-favored endo(+) decrease(-) + React-favored exo( ) decrease(-)? T dependent a) Determine H o rxn and S o rxn and use GIbbs equation. b) Use tabulated values of free energies endo(+) increase(+)? T dependent of formation, G f o. G o rxn = Σ G f o (products) - Σ G fo (reactants) Free Energies of Formation Note that G f for an element = 0 33 Calculating G o rxn Combustion of acetylene C 2 H 2 (g) + 5/2 O 2 (g) --> 2 CO 2 (g) + H 2 O(g) Use enthalpies of formation to calculate H o rxn = kj Use standard molar entropies to calculate S o rxn = J/K or kj/k G o rxn = kj - (298 K)( J/K) = kj Reaction is product-favored in spite of negative S o rxn. Reaction is enthalpy driven 34 Calculating G o rxn Calculating G o rxn NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq) NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq) Is the dissolution of ammonium nitrate productfavored? If so, is it enthalpy- or entropy-driven? From tables of thermodynamic data we find H o rxn = kj S o rxn = J/K or kj/k G o rxn = kj - (298 K)( J/K) = -6.7 kj Reaction is product-favored in spite of negative H o rxn. Reaction is entropy driven

7 Page 7 Gibbs Free Energy, G Two methods of calculating G o a) Determine H o rxn and S o rxn and use GIbbs equation. b) Use tabulated values of free energies of formation, G o f. G o rxn = Σ G f o (products) - Σ G fo (reactants) 37 Calculating G o rxn G o rxn = Σ G f o (products) - Σ G fo (reactants) Combustion of carbon C(graphite) + O 2 (g) --> CO 2 (g) G o rxn = G fo (CO 2 ) - [ G fo (graph) + G fo (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 as expected. 38 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 39 More thermo? You betcha! 40 At what T does G o rxn just change from being (+) to being (-)? When G o rxn = 0 = H o rxn -T S o rxn T = H rxn S rxn = kj kj/k = K Thermodynamics and K eq FACT: G o rxn is the change in free energy when pure reactants convert COMPLETELY to pure products. FACT: Product-favored systems have K eq > 1. Therefore, both G rxn and K eq are related to reaction favorability. Thermodynamics and K eq K eq is related to reaction favorability and so to G o rxn. The larger the value of K the more negative the value of G o rxn G o rxn where R = 8.31 J/K mol = - RT lnk

8 Page 8 Thermodynamics and K eq G o rxn = - RT lnk Calculate K for the reaction 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 4800 J lnk = - = (8.31 J/K)(298K) K = 0.14 When G o rxn > 0, then K < 1 G is change in free energy at nonstandard conditions. G is related to G G = G + RT ln Q where Q = reaction quotient When Q < K or Q > K, reaction is spontaneous. When Q = K reaction is at equilibrium When G = 0 reaction is at equilibrium Therefore, G = - RT ln K Product favored reaction G o and K > 1 In this case G rxn is < G o rxn, so state with both reactants and products present is MORE STABLE than complete conversion. Figure Product-favored reaction. 2 NO 2 ---> N 2 O 4 G o rxn = 4.8 kj Here G rxn is less than G o rxn, so the state with both reactants and products present is more stable than complete conversion. Reactant-favored reaction. N 2 O 4 --->2 NO 2 G o rxn = +4.8 kj Here G o rxn is greater than G rxn, so the state with both reactants and products present is more stable than complete conversion.

9 Page 9 Thermodynamics and K eq K eq is related to reaction favorability. 49 When G o rxn < 0, reaction moves energetically downhill G o rxn is the change in free energy when reactants convert COMPLETELY to products.