Chapter

Chapter

Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Chapter 17 Free Energy and Thermodynamics Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall Reversible Tro, Chemistry: A Molecular Approach 7 Comparing Potential Energy The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.

Tro, Chemistry: A Molecular Approach 8 Thermodynamics vs. Kinetics Tro, Chemistry: A Molecular Approach 10 Diamond Graphite Tro, Chemistry: A Molecular Approach 11 Substance H kJ/mol Substance

H kJ/mol Al(s) Br2(l) C(diamond) CO(g) Ca(s) Cu(s) Fe(s) H2(g) H2O(g) HF(g) HBr(g) I2(s) N2(g) NO(g) Na(s) S(s) 0 0 +1.88

-110.5 0 0 0 0 -241.82 -268.61 -36.23 0 0 +90.37 0 0 Al2O3 Br2(g) C(graphite) CO2(g) CaO(s) CuO(s) Fe2O3(s) H2O2(l) H2O(l)

HCl(g) HI(g) I2(g) NH3(g) NO2(g) O2(g) SO2(g) -1669.8 +30.71 0 -393.5 -635.5 -156.1 -822.16 -187.8 -285.83 -92.30 +25.94 +62.25 -46.19 +33.84 0

-296.9 4 particles partition in 16 microstates 1 mole of particles partition in 2 N (N~1023) 1/16 4/16 6/16 4/16 1/16 There is a tide in the affairs of men, Which taken at the flood, leads on to fortune. Omitted, all the voyage of their life is bound in shallows and in miseries. On such a full sea are we now afloat. And we must take the current when it serves, or lose our ventures. William Shakspeare Letters put together form words, in their specific organization, convey timeless meaning! Tro, Chemistry: A Molecular Approach 17

Increases in Entropy Tro, Chemistry: A Molecular Approach 19 The 2 Law of Thermodynamics nd Suniverse = Ssystem + Ssurroundings Tro, Chemistry: A Molecular Approach 20 Entropy Change in State Change when materials change state, the number of macrostates it can have changes as well for entropy: solid < liquid < gas because the degrees of freedom of motion increases solid liquid gas

Tro, Chemistry: A Molecular Approach 21 Entropy Change and State Change Tro, Chemistry: A Molecular Approach 22 Ex. 17.2a The reaction C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) has Hrxn = -2044 kJ at 25C. Calculate the entropy change of the surroundings. Given: Find: Concept Plan: Hsystem = -2044 kJ, T = 298 K Ssurroundings, J/K T, H

Ssurr Relationships: Solution: S Ssurr Ssurr H sys T H sys T 6.86 kJ K 2044 kJ 298 K 6.86 103 KJ

Check: combustion is largely exothermic, so the entropy of the surrounding should increase significantly G, H, and S Tro, Chemistry: A Molecular Approach 29 Ex. 17.3a The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has H = +95.7 kJ and S = +142.2 J/K at 25C. Calculate G and determine if it is spontaneous. Given: Find: Concept Plan: H = +95.7 kJ, S = 142.2 J/K, T = 298 K G, kJ T, H, S

G G H TS Relationships: Solution: G H TS 95.7 103 J 298 K 142.2 KJ 5.33 10 4 J Answer: Since G is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature. Ex. 17.3a The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has H = +95.7 kJ and S = +142.2 J/K. Calculate the minimum temperature it will be spontaneous.

Given: Find: Concept Plan: H = +95.7 kJ, S = 142.2 J/K, G < 0 G H TS Relationships: Solution: T G, H, S 3 J K

95.7 10 J T 142.2 3 J K 673 K T G H TS 0 3 95.7 10 J T 142.2 0 95.7 10 J T 142.2 J K Answer: The temperature must be higher than 673K for the reaction to be spontaneous

Substance S J/mol-K Substance S J/mol-K Al(s) Br2(l) C(diamond) CO(g) Ca(s) Cu(s) Fe(s) H2(g) H2O(g) HF(g) HBr(g) I2(s) N2(g)

NO(g) Na(s) S(s) 28.3 152.3 2.43 197.9 41.4 33.30 27.15 130.58 188.83 173.51 198.49 116.73 191.50 210.62 51.45 31.88 Al2O3(s) Br2(g)

C(graphite) CO2(g) CaO(s) CuO(s) Fe2O3(s) H2O2(l) H2O(l) HCl(g) HI(g) I2(g) NH3(g) NO2(g) O2(g) SO2(g) 51.00 245.3 5.69 213.6 39.75 42.59 89.96 109.6

69.91 186.69 206.3 260.57 192.5 240.45 205.0 248.5 Relative Standard Entropies States the gas state has a larger entropy than the liquid state at a particular temperature the liquid state has a larger entropy than the solid state at a particular temperature Substance S, (J/molK) H2O (g) 70.0

H2O (l) 188.8 Tro, Chemistry: A Molecular Approach 35 Relative Standard Entropies Molar Mass Tro, Chemistry: A Molecular Approach 36 Relative Standard Entropies Allotropes Tro, Chemistry: A Molecular Approach 37

Relative Standard Entropies Molecular Complexity Molar S, Substance Mass (J/molK) Ar (g) 39.948 154.8 NO (g) 30.006 Tro, Chemistry: A Molecular Approach 210.8 38 Relative Standard Entropies Dissolution Substance

S, (J/molK) KClO3(s) 143.1 KClO3(aq) 265.7 Tro, Chemistry: A Molecular Approach 39 Ex. 17.4 Calculate S for the reaction 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(l) Given: standard entropies from Appendix IIB Find: S, J/K Concept Plan: S, J/molK

NH3(g) 192.8 O2(g) 205.2 NO(g) 210.8 H2O(g) 188.8 SNH3, SO2, SNO, SH2O, S

S n pSproducts nr Sreactants Relationships: Solution: Substance S n pSproducts nrSreactants [4(S NO( g ) ) 6(SH 2 O( g ) )] [4(S NH 3 ( g ) ) 5(SO 2 ( g ) )] J J

J J [4(210.8 K ) 6(188.8 K )] [4(192.8 K ) 5(205.2 K )] J 178.8 K Check: S is +, as you would expect for a reaction with more gas product molecules than reactant molecules Calculating G at 25C: Goreaction = nGof(products) - nGof(reactants) at temperatures other than 25C: assuming the change in Horeaction and Soreaction is negligible Greaction = Hreaction TSreaction Tro, Chemistry: A Molecular Approach

41 Substance Gf kJ/mol Substance Gf kJ/mol Al(s) Br2(l) C(diamond) CO(g) Ca(s) Cu(s) Fe(s) H2(g) H2O(g) HF(g)

HBr(g) I2(s) N2(g) NO(g) Na(s) S(s) 0 0 +2.84 -137.2 0 0 0 0 -228.57 -270.70 -53.22 0 0 +86.71 0 0

Al2O3 Br2(g) C(graphite) CO2(g) CaO(s) CuO(s) Fe2O3(s) H2O2(l) H2O(l) HCl(g) HI(g) I2(g) NH3(g) NO2(g) O2(g) SO2(g) -1576.5 +3.14 0 -394.4 -604.17

-128.3 -740.98 -120.4 -237.13 -95.27 +1.30 +19.37 -16.66 +51.84 0 -300.4 42 Substance CH4(g) O2(g) CO2(g) H2O(g) O3(g) Ex. 17.7 Calculate G at 25C for the reaction

CH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4 O3(g) Gf, kJ/mol -50.5 0.0 -394.4 -228.6 163.2 Given: standard free energies of formation from Appendix IIB Find: G, kJ Concept Plan: Relationships: Gf of prod & react G

G n p G f products nr G f reactants Solution: G n p G f products nr G f reactants [(G f CO 2 ) 2(G f H 2 O) (G f O3 )] [(G f CH 4 ) 8(G f O 2 )] [( 394.4 kJ) 2( 228.6 kJ) (163.2 kJ)] [( 50.5 kJ) 8(0.0 kJ)] 148.3 kJ Ex. 17.6 The reaction SO2(g) + O2(g) SO3(g) has H = -98.9 kJ and S = -94.0 J/K at 25C. Calculate G at 125C and determine if it is spontaneous. Given: Find:

H = -98.9 kJ, S = -94.0 J/K, T = 398 K G, kJ Concept Plan: T, H, S G H TS Relationships: Solution: G G H TS J

98.9 103 J 398 K 94.0 K 61.5 103 J 61.5 kJ Answer: Since G is -, the reaction is spontaneous at this temperature, though less so than at 25C G Relationships if a reaction can be expressed as a series of reactions, the sum of the G values of the individual reaction is the G of the total reaction G is a state function if a reaction is reversed, the sign of its G value reverses if the amounts of materials is multiplied by a factor, the value of the G is multiplied by the same factor the value of G of a reaction is extensive

Tro, Chemistry: A Molecular Approach 45 Free Energy and Reversible Reactions the change in free energy is a theoretical limit as to the amount of work that can be done if the reaction achieves its theoretical limit, it is a reversible reaction Tro, Chemistry: A Molecular Approach 46 Real Reactions in a real reaction, some of the free energy is lost as heat if not most therefore, real reactions are irreversible Tro, Chemistry: A Molecular Approach

47 G under Nonstandard Conditions G = G only when the reactants and products are in their standard states there normal state at that temperature partial pressure of gas = 1 atm concentration = 1 M under nonstandard conditions, G = G + RTlnQ Q is the reaction quotient at equilibrium G = 0 G = RTlnK Tro, Chemistry: A Molecular Approach 48 Tro, Chemistry: A Molecular Approach

49 Tro, Chemistry: A Molecular Approach 50 Example - G Calculate G at 427C for the reaction below if the PN2 = 33.0 atm, PH2= 99.0 atm, and PNH3= 2.0 atm N2(g) + 3 H2(g) 2 NH3(g) Q= PNH32 = (2.0 atm)2 = 1.2 x 10-7 (33.0 atm)1 (99.0)3 PN21 x PH23 H = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J

S = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K G = -92380 J - (700 K)(-198.2 J/K) G = +46400 J G = G + RTlnQ G = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7) G = -46300 J = -46 kJ 51 G and K Because Grxn = 0 at equilibrium, then G = RTln(K). When K < 1, G is positive and the reaction is spontaneous in the reverse direction under standard conditions. Nothing will happen if there are no products yet! When K > 1, G is negative and the reaction is

spontaneous in the forward direction under standard conditions. When K = 1, G is 0 and the reaction is at equilibrium under standard conditions. Tro, Chemistry: A Molecular Approach 53 Example - K Estimate the equilibrium constant and position of equilibrium for the following reaction at 427C N2(g) + 3 H2(g) 2 NH3(g) H = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J S = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K G = -92380 J - (700 K)(-198.2 J/K) G = +46400 J G = -RT lnK +46400 J = -(8.314 J/K)(700 K) lnK lnK = -7.97 K = e-7.97 = 3.45 x 10-4 since K is << 1, the position of equilibrium favors reactants

54 Why Is the Equilibrium Constant Temperature Dependent? Combining these two equations G = H TS G = RTln(K) It can be shown that This equation is in the form y = mx + b. The graph of ln(K) versus inverse T is a straight line with slope and y-intercept .

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