Page 1 Entropy and Free Energy How to predict if a reaction can occur at a reasonable rate? KINEICS Chapter 17 How to predict if a reaction can occur, given enough time? HERMODYNAMICS 1 Objectives Spontaneity of Reactions Define Show how entropy affects molecules Show how entropy affects phase changes Second Law of hermodynamics Enthalpy, Entropy and Free Energy 2 Ch. 17.1 - Spontaneous Processes and Entropy If the state of a chemical system is such that a rearrangement of its atoms and molecules would decrease the energy of the system--- then this is a product-favored system. Most product-favored reactions are exothermic but this is not the only criterion 3 hermodynamics Both product- and reactant-favored reactions can proceed to equilibrium in a spontaneous process. AgCl(s) Ag + (aq) + Cl (aq) Reaction is not product-favored, but it moves spontaneously toward equilibrium. Spontaneous does not imply anything about time for reaction to occur. 4 hermodynamics and Kinetics 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 Spontaneous Reactions In general, spontaneous reactions are exothermic. Fe 2 O 3 (s) + 2 Al(s) 2 Fe(s) + Al 2 O 3 (s) r H = - 848 kj 6
Page 2 Spontaneous Reactions But many spontaneous reactions or processes are endothermic or even have H = 0. H = 0 NH 4 NO 3 (s) + heat NH 4 NO 3 (aq) 7 One property common to spontaneous processes is that the energy of the final state is more dispersed. In a spontaneous process energy goes from being more concentrated to being more dispersed. he thermodynamic property related to energy dispersal is ENROPY, S. 2nd Law of hermo a spontaneous process results in an increase in the entropy of the universe. Reaction of K with water 8 Directionality of Reactions Probability suggests that a spontaneous reaction will result in the dispersal of energy. Energy Dispersal 9 Directionality of Reactions Energy Dispersal Exothermic reactions involve a release of stored chemical potential energy to the surroundings. he stored potential energy starts out in a few molecules but is finally dispersed over a great many molecules. he final state with energy dispersed is more probable and makes a reaction spontaneous. 10 Each of the ways to distribute energy is called a microstate. hese microstates give a particular arrangement (State). Energy Dispersal 11 Directionality of Reactions Matter & energy dispersal As the size of the container increases, the number of microstates accessible to the system increases, and the density of states increases. Entropy increases. 12
Page 3 he entropy of liquid water is greater than the entropy of solid water (ice) at 0 C. Energy is more dispersed in liquid water than in solid water. 13 S o (J/K mol) H 2 O(liq) 69.95 H 2 O(gas) 188.8 14 S (solids) < S (liquids) < S (gases) Energy dispersal Entropy and States of Matter 15 Ch. 17.2-2nd Law of hermodynamics 16 A reaction is spontaneous if S for the universe is positive. S (Br 2 liq) < S (Br 2 gas) S (H 2 O sol) < S (H 2 O liq) S universe = S system + S surroundings S universe > 0 for spontaneous process Calculate the entropy created by energy dispersal in the system and surroundings. If S universe < 0, then the process is spontaneous in the opposite direction. Ch. 17.3 emperature and Spontaneity 17 Entropy and emperature 18 Discussion Questions! For the process A(l) A(s), which direction involves an increase in energy dispersal? Positional randomness? Explain your answer. As temperature increases/decreases (answer for both), which takes precedence? Why? At what temperature is there a balance between energy randomness and positional randomness? S increases slightly with S increases a large amount with phase changes Copyright 2009 Brooks/Cole Cengage Learning. - Cengage All rights reserved 17
Page 4 ΔS surr 19 ΔS surr 20 he sign of S surr depends on the direction of the heat flow. he magnitude of S surr depends on the temperature. = Ssurr H Heat flow (constant P) = change in enthalpy = ΔH 21 Entropy Changes for Phase Changes 22 For a phase change, S = q/ where q = heat transferred in phase change For H 2 O (liq) H 2 O(g) H = q = +40,700 J/mol q 40,700 J/mol S = = = +109 J/K mol 373.15 K Copyright 2009 Brooks/Cole Cengage Learning. - Cengage All rights reserved 21 Definition of Entropy Practice Determine whether entropy increases or decreases for the following reactions. a) HCl(g) + NH 3 (g) NH 4 Cl(s) b) 2H 2 O 2 (l) 2H 2 O(l) + O 2 (g) c) Cooling of nitrogen gas from -20⁰C to -50⁰C d) NaCl(s) Na + (aq) + Cl - (aq) e) CaCO 3 (s) CaO(s) + CO 2 (g) 23 2nd Law of hermodynamics S universe = S system + S surroundings Dissolving NH 4 NO 3 in water an entropy driven process. 24
Page 5 2nd Law of hermodynamics 25 2nd Law of hermodynamics 26 2 H 2 (g) + O 2 (g) 2 H 2 O(liq) S o system = -326.9 J/K S o surroundings = q surroundings = - H system Can calc. that H ro = H o system = -571.7 kj S o surroundings S o surroundings - (-571.7 kj)(1000 J/kJ) = 298.15 K = +1917 J/K 2 H 2 (g) + O 2 (g) 2 H 2 O(liq) S o system S o surroundings = -326.9 J/K = +1917 J/K S o universe = +1590. J/K he entropy of the universe is increasing, so the reaction is product-favored. Ch. 17.4 - Free Energy, G S univ = S surr S univ = H sys + S sys Multiply through by - - S univ = H sys - S sys - S univ + S sys J. Willard Gibbs 1839-1903 = change in Gibbs free energy 27 his means the a process is spontaneous in the direction in which the free energy decreases. Remember that S univ = - G/ 28 for the system = G system Under standard conditions G o sys = H o sys - S o sys Free Energy and Spontaneity Example H 2 O(s) H 2 O(l) H = 6.03x10 3 J/mol and S =22.1 J/K 29 Free Energy and Spontaneity Example H 2 O(s) H 2 O(l) H = 6.03x10 3 J/mol and S =22.1 J/K 30 At 10 C, the G = 6030 J 283K(22.1 J/K) = 6030 J 6254 J = -224 J So, ice melts spontaneously at 10 C, even though it s an endothermic reaction. At -10 C, the G = 6030 J 263K(22.1 J/K) = 6030 J 5800 J = 230 J So, ice does not melt spontaneously at -10 C. However, water would freeze spontaneously at - 10 C because opposite is more thermodynamically favorable.
Page 6 Free Energy and Spontaneity Example H 2 O(s) H 2 O(l) H = 6.03x10 3 J/mol and S =22.1 J/K At 0 C, the G = 6030 J 273K(22.1 J/K) = 6030 J 6030 J = 0 J So, neither process is favorable 31 Gibbs free energy change = total energy change for system - energy lost in energy dispersal If reaction is exothermic (negative H o ) and entropy increases (positive S o ) then G o must be NEGAIVE reaction is spontaneous (and product-favored). 32 Gibbs free energy change = total energy change for system - energy lost in energy dispersal If reaction is endothermic (positive H o ) and entropy decreases (negative S o ) then G o must be POSIIVE reaction is not spontaneous (and is reactantfavored). 33 Gibbs Free Energy, G H o S o G o Process Exo,<0 >0 <0 Spontaneous at all End,>0 <0 >0 Never Spontaneous Exo, <0 <0? dependent, spontaneous at lower s End, >0 >0-? dependent, spontaneous at higher s 34 Spontaneous or Not? Practice Use the table on the previous slide to classify the following reactions as one of the 4 types. 1. CH 4 (g) + 2O 2 (g) 2H 2 O(l) + CO 2 (g) H = -891 kj/mol S = -242 J/K 2. 2Fe 2 O 3 (s) + 3C(s, graphite) 4Fe(s) + 3CO 2 (g) H = +468 kj/mol S = +561 J/K 3. C(s, graphite) + O 2 (g) CO 2 (g) H = -394 kj/mol S = +3 J/K 35 Ch. 17.5 Entropy Changes in Reactions he 3 rd Law of hermodynamics: he entropy of a pure crystal at 0 K is 0 (S=0) Gives us a starting point. At all other temperatures, entropy must be > 0. (As you raise, entropy increases) 36
Page 7 Standard Molar Entropies 37 Entropy of a substance increases with temperature. Molecular motions of heptane, C 7 H 16 Molecular motions of heptane at different temps. 38 39 40 Increase in molecular complexity generally leads to increase in S. Entropies of ionic solids depend on coulombic attractions. S o (J/K mol) MgO 26.9 NaF 51.5 Mg 2+ & O 2- Na + & F - 41 Calculating ΔS for a Reaction 42 Liquids or solids dissolve in a solvent in a spontaneous process owing to the increase in entropy. Matter (and energy) are more dispersed. S o = S o (products) - 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 o = 2 mol (69.9 J/K mol) - [2 mol (130.7 J/K mol) + 1 mol (205.3 J/K mol)] S o = -326.9 J/K Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid.
Page 8 Entropy of Reaction Practice Calculate S for the following reaction: Al 2 O 3 (s) + 3H 2 (g) 2Al(s)+3H 2 O(g) S⁰ 51 131 28 189 S⁰ rxn = 2(28) + 3(189) 51 3(131) = 56 + 567 51 393 = 179 J/K 43 Ch. 17.6 Free Energy and Chemical Reactions Really want to determine the SANDARD FREE ENERGY CHANGE, G for a reaction his represents the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states. Note: G is not measured directly. Use it to compare relative tendency for reactions to occur. he more negative the G of reaction, the greater tendency of product to form 44 Gibbs Free Energy, G wo methods of calculating G o a) Determine r H o and r S o and use Gibbs equation. b) Use tabulated values of free energies of formation, G o f. 45 Free Energies of Formation 46 G ro = G fo (products) - G fo (reactants) Note that G f for an element = 0 Calculating ΔG rxn o 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 rxno = -1238 kj Use standard molar entropies to calculate S rxno = -97.4 J/K or -0.0974 kj/k G rxno = -1238 kj - (298 K)(-0.0974 kj/k) = -1209 kj Reaction is product-favored in spite of negative S rxno. Reaction is enthalpy driven 47 Calculating ΔG rxn o 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? 48
Page 9 Calculating Δ r G o 49 Gibbs Free Energy, G 50 NH 4 NO 3 (s) + heat NH 4 NO 3 (aq) From tables of thermodynamic data we find r H o = +26 kj r S o = +109 J/K or +0.109 kj/k r G o = +26 kj - (298 K)(+0.109 J/K) = -6 kj Reaction is product-favored in spite of positive H rxno. Reaction is entropy driven wo methods of calculating G o a) Determine r H o and r S o and use Gibbs equation. b) Use tabulated values of free energies of formation, f G o. r G o = f G o (products) - f G o (reactants) Calculating ΔG o rxn G ro = G fo (products) - G fo (reactants) Combustion of carbon C(graphite) + O 2 (g) CO 2 (g) G ro = G fo (CO 2 ) - [ G fo (graph) + G fo (O 2 )] r G o = -394.4 kj - [ 0 + 0] Note that free energy of formation element in its standard state is 0. of an G ro = -394.4 kj Reaction is product-favored as expected. 51 Free Energy and emperature 2 Fe 2 O 3 (s) + 3 C(s) 4 Fe(s) + 3 CO 2 (g) H ro = +468 kj S ro = +560 J/K G ro = +300.8 kj Reaction is reactant-favored at 298 K At what does G ro just change from being (+) to being (-)? When G ro = 0 = H ro - S r o = r H r S = 467.9 kj 0.5603 kj/k = 835.1 K 52 Practice - Calculate the G for this Reaction 2SO 2 (g) + O 2 (g) 2SO 3 (g) at 227 o C H o = -196.6 kj; S o = -189.6 J/K G o = -196.6 500 (-0.1896) kj G o = -101.8 kj 53 Free Energy, Enthalpy and Entropy Practice 2000 AP Exam #6a-c 2003 AP Exam #7b-d 2004 AP Exam #2d-e 2005 AP Exam #8a-c 2005B AP Exam #7c 2006 AP Exam #2a-d (Best one) 54