20-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 20 Thermodynamics: Entropy, Free Energy and the.

20-2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions 20.1 The Second Law of Thermodynamics: Predicting Spontaneous Change 20.2 Calculating the Change in Entropy of a Reaction 20.3 Entropy, Free Energy, and Work 20.4 Free Energy, Equilibrium, and Reaction Direction

20-3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. First law of thermodynamics  Law of conservation of energy: Energy can be neither created nor destroyed.  ∆E= q + w (q=heat, w=work, E=internal energy)  E univ= E sys + E surr  Heat gained by system is lost by surroundings and vice-versa.  Total energy of Universe is constant ∆E sys + ∆E surr=0=∆E univ

20-4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Limitations of the First Law of Thermodynamics  E = q + w E universe = E system + E surroundings  E system = -  E surroundings The total energy-mass of the universe is constant. However, this does not tell us anything about the direction of change in the universe.  E system +  E surroundings = 0 =  E universe

20-6 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 20.1 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 ).  H 0 rxn = +62.3 kJ

20-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. solid liquid gas more orderless order crystal + liquid ions in solution more orderless order more orderless order crystal + crystal gases + ions in solution

20-9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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.  S universe =  S system +  S surroundings > 0 This is the second law of thermodynamics.

20-10 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Entropy:  S, A measure of molecular randomness or disorder.  Thermodynamic function that describes number of arrangements that are available to a system existing in a given state.  Probability of occurrence of a particular arrangement(state) depends on the number of ways(microstates) in which it can be arranged.  S=k ln W, k= Boltzmann constant=R/Av number=1.38 x 10-23J/K.  Un its of S are J/K

20-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Entropy  More the microstates available, higher is the entropy. Ex: solid →liquid→gas  Entropy is a state function depends only on present state and not its path.  ∆ Ssys= Sfinal – Sinitial, ∆ Ssys > 0, as entropy increases,  ∆ Ssys < 0, as entropy decreases.

20-12 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Entropy  If there are n number of molecules, relative probability of finding all molecules in one place is: (1/2)n  ∆ Ssys= 5.76 J/K for 1 mol  ∆ Ssys=qrev/T, T=temp, q=heat absorbed, rev=reversible process

20-13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 20.2 1 atmevacuated Spontaneous expansion of a gas stopcock closed stopcock opened 0.5 atm Spontaneous expansion of gas

20-15 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Entropy and Second Law of Thermodynamics:  Second Law of Thermo: In any spontaneous process there is always an increase in the entropy of the universe. OR The entropy of the universe is increasing.  ∆Suniv= ∆Ssys + ∆Ssurr  If ∆Suniv= +ve, process is spontaneous in direction written.  ∆Suniv= -ve, process is spontaneous in the opposite direction.  ∆Suniv= 0, process has no tendency to occur, system is at equilibrium

20-17 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 20.4 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

20-18 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Predicting Relative S 0 Values of a System 1. Temperature changes 2. Physical states and phase changes 3. Dissolution of a solid or liquid 5. Atomic size or molecular complexity 4. Dissolution of a gas S 0 increases as the temperature rises. S 0 increases as a more ordered phase changes to a less ordered phase. 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. A gas becomes more ordered when it dissolves in a liquid or solid. In similar substances, increases in mass relate directly to entropy. In allotropic substances, increases in complexity (e.g. bond flexibility) relate directly to entropy.

20-20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 20.6 The entropy change accompanying the dissolution of a salt. pure solid pure liquid solution MIX

20-21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 20.7 Ethanol WaterSolution of ethanol and water The small increase in entropy when ethanol dissolves in water.

20-24 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Answer the following:  1. Will solid or gaseous CO2 have higher positional entropy?  2. Will N2 gas at 1 atm or 1.0 x 10-2 atm have higher positional entropy?  3. When solid sugar is dissolved in water to form a solution, what will be the sign of entropy change?  4. When iodine vapors are condensed to form crystals, what will be the sign of entropy change?

20-25 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.1: SOLUTION: 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)]: (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 ) PLAN:In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature. (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

20-26 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Entropy Changes in the System  S 0 rxn - the entropy change that occurs when all reactants and products are in their standard states.  S 0 rxn =   S 0 products -   S 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.  S surroundings = -  H system T

20-27 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.2: Calculating the Standard Entropy of Reaction,  S 0 rxn PROBLEM: Calculate  S 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.  S = [(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 )]  S = [(3 mol)(213.7J/mol*K) + (4 mol)(69.9J/mol*K)] - [(1 mol)(269.9J/mol*K) + (5 mol)(205.0J/mol*K)]  S = - 374 J/K

20-28 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Entropy changes in the surrounding:  Exothermic change: Heat is lost by the system and gained by the surroundings.  q sys 0, ∆ Ssurr > 0  Endothermic Change: Heat gained by the system is lost by the surroundings.  q sys> 0, q surr 0, q surr <0, ∆ Ssurr < 0  The change is entropy of the surroundings is α opposite charge in the heat of a system 1/α temperature at which heat is transferred.

20-29 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Entropy changes in the surrounding  ∆Ssurr = -q sys/T  ∆Ssurr = -∆H sys/T  Sign of ∆Ssurr depends on the direction of heat flow. When an exothermic process occurs ∆Ssurr is + ve. Opposite is true for endothermic process. Nature tends to seek the lowest possible energy.  Magnitude of ∆Ssurr depends on the temperature. Transfer of heat at low temperature produces a greater percent change in the randomness of the surroundings.

20-30 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.3: SOLUTION: Determining Reaction Spontaneity PROBLEM: At 298K, the formation of ammonia has a negative  S 0 sys ; Calculate  S 0 rxn, and state whether the reaction occurs spontaneously at this temperature. N 2 ( g ) + 3H 2 ( g ) 2NH 3 ( g )  S 0 sys = -197 J/K PLAN:  S 0 universe must be > 0 in order for this reaction to be spontaneous, so  S 0 surroundings must be > 197 J/K. To find  S 0 surr, first find  H sys ;  H sys =  H rxn which can be calculated using  H 0 f values from tables.  S 0 universe =  S 0 surr +  S 0 sys.  H 0 rx = [(2 mol)(  H 0 f NH 3 )] - [(1 mol)(  H 0 f N 2 ) + (3 mol)(  H 0 f H 2 )]  H 0 rx = -91.8 kJ  S 0 surr = -  H 0 sys /T = -(-91.8x10 3 J/298K)= 308 J/K  S 0 universe =  S 0 surr +  S 0 sys = 308 J/K + (-197 J/K) = 111 J/K  S 0 universe > 0 so the reaction is spontaneous.

20-31 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Problem  P.3. Sb2 S3(s) + 3Fe(s)______ 2Sb (s) + 3FeS (s) ∆H=-125kJ  Sb4 O6(s) + 6C(s)______ 4Sb (s) + 6 CO (s) ∆H= 778 kJ  Calculate ∆Ssurr for both of these reactions at 25 C and 1 atm

20-32 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B20.1 Lavoisier studying human respiration as a form of combustion. Figure B20.2 A whole-body calorimeter.

20-33 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 20.10 Components of  S 0 universe for spontaneous reactions exothermic system becomes more disordered exothermic system becomes more ordered endothermic system becomes more disordered ∆Ssurr must be larger ∆Ssurr must be smaller

20-34 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  G 0 system =  H 0 system - T  S 0 system  G 0 rxn =  m  G 0 products -  n  G 0 reactants Gibbs Free Energy (G)  G, 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.  G < 0 for a spontaneous process  G > 0 for a nonspontaneous process  G = 0 for a process at equilibrium

20-35 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Free Energy  G= H-TS, H= enthalpy, T=temp in K, S= entropy.  Also called as Gibbs Free Energy.  Free energy change (∆G) is a measure of the spontaneity of a process and of the useful energy available from it.  ∆ G= ∆H-T∆S(constant temp)  ∆Suniv = -∆G/T  Process is spontaneous in the direction in which the free energy decreases. (constant T and P)

20-36 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Free Energy  ∆S univ =0, at equilibrium  ∆S univ > 0, for a spontaneous process.  ∆S univ < 0, for a nonspontaneous process.  ∆G=0, at equilibrium  ∆G > 0, for a nonspontaneous process.  ∆G< 0, for a spontaneous process.

20-37 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Standard free energy of formation  ∆Gf0 is defined as a change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states.  Standard free energy of formation of an element in its standard state is zero.  An equation coefficient x ∆Gf0 by the number.  Reversing a reaction changes ∆Gf0

20-38 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Problem  P.4. For the equation 2CH3OH (g) + 3 O2(g) _____ 2 CO 2(g) + 4 H2O (g)  Calculate ∆G0 -163 0 -394 -229 ∆Gf0 kJ/mol

20-39 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.6: PROBLEM: PLAN: SOLUTION: Determining the Effect of Temperature on  G 0 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,  G 0 = -141.6kJ;  H 0 = -198.4kJ; and  S 0 = -187.9J/K (a) Use the data to decide if this reaction is spontaneous at 25 0 C, and predict how  G 0 will change with increasing T. (b) Assuming  H 0 and  S 0 are constant with increasing T, is the reaction spontaneous at 900. 0 C? The sign of  G 0 tells us whether the reaction is spontaneous and the signs of  H 0 and  S 0 will be indicative of the T effect. Use the Gibbs free energy equation for part (b). (a) The reaction is spontaneous at 25 0 C because  G 0 is (-). Since  H 0 is (-) but  S 0 is also (-),  G 0 will become less spontaneous as the temperature increases.

20-40 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.6: Determining the Effect of Temperature on  G 0 continued (b)  G 0 rxn =  H 0 rxn - T  S 0 rxn  G 0 rxn = -198.4kJ - (1173K)(-187.9J/mol*K)(kJ/10 3 J)  G 0 rxn = 22.0 kJ; the reaction will be nonspontaneous at 900. 0 C

20-41 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Problems P.6. Use ∆Gf0 values to calculate free energy change at 25 C for the following: 2NO (g) + O2(g) → 2NO2 (g) P.7. For the equation 2CH3OH (g) + 3 O2(g) _____ 2 CO 2(g) + 4 H2O (g) Calculate ∆G0 -163 0 - 394 -229 ∆Gf0 kJ/mol P.8. Is the reaction C2H4 (g) + H2O (l) _______ C2H5OH (l) spontaneous under standard conditions?

20-42 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  G and the Work a System Can Do For a spontaneous process,  G is the maximum work obtainable from the system as the process takes place:  G = - work max For a nonspontaneous process,  G is the maximum work that must be done to the system as the process takes place:  G = work max An example

20-43 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ∆G and work a system can do:  ∆G < 0, for a spontaneous process., ∆G=-wmax, maximum useful work obtainable from the system.  ∆G > 0, for a nonspontaneous process., minimum work that must be done to the system to make the process take place.  In any real process work is performed irreversibly and some of the free energy is lost to the surroundings as heat.  You can never obtain maximum possible work.  You need more than minimum energy to start the process for a nonspontaneous type.  At equilibrium a reaction can no longer do any work.

20-44 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Effect of Temperature on Reaction spontaneity 1.Reaction is spontaneous at all temperatures: ∆H 0, -T ∆S= -ve, ∆G =-ve 2. Reaction is nonspontaneous at all temperatures: 2. Reaction is nonspontaneous at all temperatures: ∆H > 0, ∆S 0, ∆S <0, -T ∆S= +ve,∆G =+ve 3. Reaction is spontaneous at higher temperatures: 3. Reaction is spontaneous at higher temperatures: ∆H > 0, ∆S >0, -T ∆S favors spontaneity at high temp, ∆G =-ve and reaction is spontaneous. 4. Reaction is spontaneous at lower temperatures: 4. Reaction is spontaneous at lower temperatures: ∆H < 0, ∆S <0, ∆H favors spontaneity at low temp, Reaction occurs if -T ∆S< ∆H

20-45 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 20.1 Reaction Spontaneity and the Signs of  H 0,  S 0, and  G 0 H0H0 S0S0 -T  S 0 G0G0 Description -+-- +-++ ++-+ 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

20-46 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 20.11 The effect of temperature on reaction spontaneity. Reaction between Cu2O and C. Relatively const ∆H and steadily increasing T∆S

20-47 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Free Energy, Equilibrium and Reaction Direction If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (  G < 0) If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (  G > 0) If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (  G = 0)  G = RT ln Q/K = RT lnQ - RT lnK Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and ln Q = 0 so  G 0 = - RT lnK

20-48 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. FORWARD REACTION REVERSE REACTION Table 20.2 The Relationship Between  G 0 and K at 25 0 C  G 0 (kJ) KSignificance 200 100 50 10 1 0 -10 -50 -100 -200 9x10 -36 3x10 -18 2x10 -9 2x10 -2 7x10 -1 1 1.5 5x10 1 6x10 8 3x10 17 1x10 35 Essentially no forward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to same extent Forward reaction goes to completion; essentially no reverse reaction

20-49 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Free Energy and Equilibrium:   G =  G  + RT ln(Q)  Q = reaction quotient from the law of mass action.  Equilibrium point occurs at the lowest value of free energy available to the reaction system.   G  =  RT ln(K)

20-50 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.7: PROBLEM: Calculating  G at Nonstandard Conditions 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. (  G 0 298 = -141.6kJ/mol of reaction as written using  H 0 and  S 0 values at 973K.  G 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  G for the system in part (b) at each temperature. PLAN:Use the equations and conditions found on slide.

20-51 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SOLUTION: Sample Problem 20.7: Calculating  G at Nonstandard Conditions continued (2 of 3) (a) Calculating K at the two temperatures:  G 0 = -RTlnK so At 298, the exponent is -  G 0 /RT = - (-141.6kJ/mol)(10 3 J/kJ) (8.314J/mol*K)(298K) = 57.2 = e 57.2 = 7x10 24 At 973, the exponent is -  G 0 /RT (-12.12kJ/mol)(10 3 J/kJ) (8.314J/mol*K)(973K) = 1.50 = e 1.50 = 4.5

20-52 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 20.7: Calculating  G 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  G is calculated using  G =  G 0 + RTlnQ  G 298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/10 3 J)(298K)(ln4.00)  G 298 = -138.2kJ/mol  G 973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/10 3 J)(973K)(ln4.00)  G 298 = -0.9kJ/mol

20-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 20 Thermodynamics: Entropy, Free Energy and the.

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20-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 20 Thermodynamics: Entropy, Free Energy and the.

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