Entropy, Free Energy and the Direction of Chemical Reactions

Thermodynamics: Entropy, Free Energy and the Direction of Chemical Reactions Dr.ssa Rossana Galassi 320 4381420 rossana.galassi@unicam.it 20-1

Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions 1 The Second Law of Thermodynamics: Predicting Spontaneous Change 2 Calculating the Change in Entropy of a Reaction 3 Entropy, Free Energy, and Work 4 Free Energy, Equilibrium, and Reaction Direction 20-2

Limitations of the First Law of Thermodynamics DE = q + w E universe = E system + E surroundings DE system = -DE surroundings DE system + DE surroundings = 0 = DE universe The total energy-mass of the universe is constant. However, this does not tell us anything about the direction of change in the universe. 20-3

A spontaneous endothermic chemical reaction. water. Ba(OH) 2 8H 2 O(s) + 2NH 4 NO 3 (s) Ba 2+ (aq) + 2NO 3- (aq) + 2NH 3 (aq) + 10H 2 O(l) DH 0 rxn = +62.3 kj 20-4

The Concept of Entropy (S) Entropy refers to the state of order. A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process. more order less order solid liquid gas more order crystal + liquid more order crystal + crystal less order ions in solution less order gases + ions in solution 20-5

Spontaneous expansion of a gas stopcock closed 1 atm evacuated stopcock opened 0.5 atm 0.5 atm 20-6

Expansion of a gas and the increase in number of microstates. 20-7

1877 Ludwig Boltzman S = k ln W where S is entropy, W is the number of ways of arranging the components of a system, and k is a constant (the Boltzman constant), R/N A (R = universal gas constant, N A = Avogadro s number. A system with relatively few equivalent ways to arrange its components (smaller W) has relatively less disorder and low entropy. A system with many equivalent ways to arrange its components (larger W) has relatively more disorder and high entropy. DS universe = DS system + DS surroundings > 0 This is the second law of thermodynamics. 20-8

Random motion in a crystal The third law of thermodynamics. A perfect crystal has zero entropy at a temperature of absolute zero. S system = 0 at 0 K 20-9

Predicting Relative S 0 Values of a System 1. Temperature changes S 0 increases as the temperature rises. 2. Physical states and phase changes S 0 increases as a more ordered phase changes to a less ordered phase. 3. Dissolution of a solid or liquid S 0 of a dissolved solid or liquid is usually greater than the S 0 of the pure solute. However, the extent depends upon the nature of the solute and solvent. 4. Dissolution of a gas A gas becomes more ordered when it dissolves in a liquid or solid. 5. Atomic size or molecular complexity In similar substances, increases in mass relate directly to entropy. In allotropic substances, increases in complexity (e.g. bond flexibility) relate directly to entropy. 20-10

The increase in entropy from solid to liquid to gas. 20-11

The entropy change accompanying the dissolution of a salt. pure solid MIX pure liquid solution 20-12

The small increase in entropy when ethanol dissolves in water. Ethanol Water Solution of ethanol and water 20-13

The large decrease in entropy when a gas dissolves in a liquid. O 2 gas O 2 gas in H 2 O 20-14

Entropy and vibrational motion. NO NO 2 N 2 O 4 20-15

Sample Problem 20.1: Predicting Relative Entropy Values PROBLEM: Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]: PLAN: (a) 1mol of SO 2 (g) or 1mol of SO 3 (g) (b) 1mol of CO 2 (s) or 1mol of CO 2 (g) (c) 3mol of oxygen gas (O 2 ) or 2mol of ozone gas (O 3 ) (d) 1mol of KBr(s) or 1mol of KBr(aq) (e) Seawater in midwinter at 2 0 C or in midsummer at 23 0 C (f) 1mol of CF 4 (g) or 1mol of CCl 4 (g) In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature. SOLUTION: (a) 1mol of SO 3 (g) - more atoms (b) 1mol of CO 2 (g) - gas > solid (c) 3mol of O 2 (g) - larger #mols (d) 1mol of KBr(aq) - solution > solid (e) 23 0 C - higher temperature (f) CCl 4 - larger mass 20-16

Entropy Changes in the System DS 0 rxn - the entropy change that occurs when all reactants and products are in their standard states. DS 0 rxn = DS 0 products - DS 0 reactants The change in entropy of the surroundings is directly related to an opposite change in the heat of the system and inversely related to the temperature at which the heat is transferred. DS surroundings = - DH system T 20-17

Sample Problem 20.2: Calculating the Standard Entropy of Reaction, DS 0 rxn PROBLEM: Calculate DS 0 rxn for the combustion of 1mol of propane at 25 0 C. C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) PLAN: Use summation equations. It is obvious that entropy is being lost because the reaction goes from 6 mols of gas to 3 mols of gas. SOLUTION: Find standard entropy values in the Appendix or other table. DS = [(3 mol)(s 0 CO 2 ) + (4 mol)(s 0 H 2 O)] - [(1 mol)(s 0 C 3 H 8 ) + (5 mol)(s 0 O 2 )] DS = [(3 mol)(213.7j/mol*k) + (4 mol)(69.9j/mol*k)] - [(1 mol)(269.9j/mol*k) + (5 mol)(205.0j/mol*k)] DS = - 374 J/K 20-18

Sample Problem 20.3: Determining Reaction Spontaneity PROBLEM: At 298K, the formation of ammonia has a negative DS 0 sys; N 2 (g) + 3H 2 (g) 2NH 3 (g) DS 0 sys = -197 J/K Calculate DS 0 rxn, and state whether the reaction occurs spontaneously at this temperature. PLAN: DS 0 universe must be > 0 in order for this reaction to be spontaneous, so DS 0 surroundings must be > 197 J/K. To find DS 0 surr, first find DH sys ; DH sys = DH rxn which can be calculated using DH 0 f values from tables. DS 0 universe = DS 0 surr + DS 0 sys. SOLUTION: DH 0 rx = [(2 mol)(dh 0 fnh 3 )] - [(1 mol)(dh 0 fn 2 ) + (3 mol)(dh 0 fh 2 )] DH 0 rx = -91.8 kj DS 0 surr = -DH 0 sys/t = -(-91.8x10 3 J/298K) = 308 J/K DS 0 universe = DS 0 surr + DS 0 sys = 308 J/K + (-197 J/K) = 111 J/K DS 0 universe > 0 so the reaction is spontaneous. 20-19

Sample Problem 20.4: Calculating DG 0 from Enthalpy and Entropy Values PROBLEM: Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated. Note that the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation). +5 4KClO 3 (s) D +7-1 3KClO 4 (s) + KCl(s) SOLUTION: 20-20 Use DH 0 f and S 0 values to calculate DG 0 sys (DG 0 rxn) at 25 0 C for this reaction. PLAN: Use Appendix B values for thermodynamic entities; place them into the Gibbs Free Energy equation and solve. DH 0 rxn = mdh 0 products - ndh 0 reactants DH 0 rxn = (3 mol)(-432.8 kj/mol) + (1 mol)(-436.7 kj/mol) - (4 mol)(-397.7 kj/mol) DH 0 rxn = -144 kj

Sample Problem 20.4: Calculating DG 0 from Enthalpy and Entropy Values continued DS 0 rxn = mds 0 products - nds 0 reactants DS 0 rxn = (3 mol)(151 J/mol*K) + (1 mol)(82.6 J/mol*K) - (4 mol)(143.1 J/mol*K) DS 0 rxn = -36.8 J/K DG 0 rxn = DH 0 rxn - T DS 0 rxn DG 0 rxn = -144 kj - (298K)(-36.8 J/K)(kJ/10 3 J) DG 0 rxn = -133 kj 20-21

Lavoisier studying human respiration as a form of combustion. A whole-body calorimeter. 20-22

Components of DS 0 universe for spontaneous reactions exothermic endothermic system becomes more disordered exothermic system becomes more disordered system becomes more ordered 20-23

Gibbs Free Energy (G) DG, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it. DG 0 system = DH 0 system - TDS 0 system DG < 0 for a spontaneous process DG > 0 for a nonspontaneous process DG = 0 for a process at equilibrium DG 0 rxn = mdg 0 products - ndg 0 reactants 20-24

Sample Problem 20.6: Determining the Effect of Temperature on DG 0 PROBLEM: An important reaction in the production of sulfuric acid is the oxidation of SO 2 (g) to SO 3 (g): 2SO 2 (g) + O 2 (g) 2SO 3 (g) At 298K, DG 0 = -141.6kJ; DH 0 = -198.4kJ; and DS 0 = -187.9J/K (a) Use the data to decide if this reaction is spontaneous at 25 0 C, and predict how DG 0 will change with increasing T. (b) Assuming DH 0 and DS 0 are constant with increasing T, is the reaction spontaneous at 900. 0 C? PLAN: The sign of DG 0 tells us whether the reaction is spontaneous and the signs of DH 0 and DS 0 will be indicative of the T effect. Use the Gibbs free energy equation for part (b). SOLUTION: (a) The reaction is spontaneous at 25 0 C because DG 0 is (-). Since DH 0 is (-) but DS 0 is also (-), DG 0 will become less spontaneous as the temperature increases. 20-25

Sample Problem 20.6: Determining the Effect of Temperature on DG 0 continued (b) DG 0 rxn = DH 0 rxn - T DS 0 rxn DG 0 rxn = -198.4kJ - (1173K)(-187.9J/mol*K)(kJ/10 3 J) DG 0 rxn = 22.0 kj; the reaction will be nonspontaneous at 900. 0 C 20-26

DG and the Work a System Can Do For a spontaneous process, DG is the maximum work obtainable from the system as the process takes place: DG = work max For a nonspontaneous process, DG is the maximum work that must be done to the system as the process takes place: DG = work max An example 20-27

Reaction Spontaneity and the Signs of DH 0, DS 0, and DG 0 DH 0 DS 0 -TDS 0 DG 0 Description - + - - + - + + + + - + or - - - + + or - Spontaneous at all T Nonspontaneous at all T Spontaneous at higher T; nonspontaneous at lower T Spontaneous at lower T; nonspontaneous at higher T 20-28

Sample Problem 20.5: Calculating DG 0 rxn from DG 0 f Values PROBLEM: Use DG 0 f values to calculate DG rxn for the reaction in Sample Problem 20.4: D 4KClO 3 (s) 3KClO 4 (s) + KCl(s) PLAN: Use the DG summation equation. SOLUTION: DG 0 rxn = mdg 0 products - ndg 0 reactants DG 0 rxn = (3mol)(-303.2kJ/mol) + (1mol)(-409.2kJ/mol) - (4mol)(-296.3kJ/mol) DG 0 rxn = -134kJ 20-29

The effect of temperature on reaction spontaneity. 20-30

The coupling of a nonspontaneous reaction to the hydrolysis of ATP. 20-31

The cycling of metabolic free enery through ATP 20-32

Why is ATP a high-energy molecule? 20-33

Free Energy, Equilibrium and Reaction Direction If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (DG < 0) If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (DG > 0) If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (DG = 0) DG = RT ln Q/K = RT lnq - RT lnk Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and ln Q = 0 so DG 0 = - RT lnk 20-34

The Relationship Between DG 0 and K at 25 0 C DG 0 (kj) K Significance 200 100 9x10-36 3x10-18 Essentially no forward reaction; reverse reaction goes to completion 20-35 50 10 1 0-1 -10-50 -100-200 2x10-9 2x10-2 7x10-1 1 1.5 5x10 1 6x10 8 3x10 17 1x10 35 Forward and reverse reactions proceed to same extent FORWARD REACTION Forward reaction goes to completion; essentially no reverse reaction REVERSE REACTION

Sample Problem 20.7: Calculating DG at Nonstandard Conditions PROBLEM: The oxidation of SO 2, which we considered in Sample Problem 20.6 2SO 2 (g) + O 2 (g) 2SO 3 (g) is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature. (a) Calculate K at 298K and at 973K. (DG 0 298 = -141.6kJ/mol of reaction as written using DH 0 and DS 0 values at 973K. DG 0 973 = -12.12kJ/mol of reaction as written.) (b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500atm of SO 2, 0.0100atm of O 2, and 0.100atm of SO 3 and kept at 25 0 C and at 700. 0 C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature? (c) Calculate DG for the system in part (b) at each temperature. PLAN: Use the equations and conditions found on slide. 20-36

Sample Problem 20.7: Calculating DG at Nonstandard Conditions continued (2 of 3) SOLUTION: (a) Calculating K at the two temperatures: DG 0 = -RTlnK so K e (DG 0 / RT) (-141.6kJ/mol)(10 3 J/kJ) At 298, the exponent is -DG 0 /RT = - (8.314J/mol*K)(298K) = 57.2 K e (DG 0 / RT) = e 57.2 = 7x10 24 At 973, the exponent is -DG 0 /RT K e (DG 0 / RT) = e 1.50 = 4.5 (-12.12kJ/mol)(10 3 J/kJ) (8.314J/mol*K)(973K) = 1.50 20-37

Sample Problem 20.7: Calculating DG at Nonstandard Conditions continued (3 of 3) (b) The value of Q = pso 3 2 (pso 2 ) 2 (po 2 ) = (0.100) 2 (0.500) 2 (0.0100) = 4.00 Since Q is < K at both temperatures the reaction will shift right; for 298K there will be a dramatic shift while at 973K the shift will be slight. (c) The nonstandard DG is calculated using DG = DG 0 + RTlnQ DG 298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/10 3 J)(298K)(ln4.00) DG 298 = -138.2kJ/mol DG 973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/10 3 J)(973K)(ln4.00) DG 298 = -0.9kJ/mol 20-38

The relation between free energy and the extent of reaction. DG 0 < 0 K >1 DG 0 > 0 K <1 20-39