Lecture 4. The Second Law of Thermodynamics

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1 Lecture 4. The Second Law of Thermodynamics

2 LIMITATION OF THE FIRST LAW: -Does not address whether a particular process is spontaneous or not. -Deals only with changes in energy. Consider this examples: -Drop a rock from waist-high height, rock will fall spontaneously -Plunger of a spray is presses, gas comes out spontaneously -Metallic sodium is placed in a jar with chlorine gas, reaction occurs

3 Spontaneous Processes: Why does the color spread when placing a drop of dye in a glass of clean water?

4 SPONTANEOUS PROCESSES: Spontaneous processes are those that can proceed without any outside intervention. The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously go back to B.

5 SPONTANEOUS PROCESSES: Processes that are spontaneous in one direction are not spontaneous in the reverse direction.

6 SPONTANEOUS PROCESSES Processes that are spontaneous at one temperature may be not spontaneous at other temperatures. Above 0C it is spontaneous for ice to melt. Below 0C the reverse process is spontaneous

7 SPONTANEOUS PROCESSES: Changes in the extent of disorder. When we spill a bowl of sugar, why do the grains go everywhere and cause such a mess? natures s way to seek disorder. It is easy to create disorder; difficult to create order.

8 ENTROPY: Entropy can be thought of as a measure of the randomness or disorder of a system. It is related to the various modes of motion in molecules.

9 Spontaneity and the Sign of S Why does a room with a fragrance bottle at the other end of the room is suddenly filled with the aroma? Limonene (l) Limonene (g) S(process) = S (final state) - S(initial state) A spontaneous process is accompanied by S of positive sign.

10 Reaction Spontaneity by Inspection Why do damp clothes become dry when hung outside? S(g) >> S(l) > S(s) By inspection alone, decide whether the sublimation of solid carbon dioxide is spontaneous or not. How about the condensation of water?

11 Spontaneity by inspection. By inspection alone, decide whether the following reaction is spontaneous or not. SOCl (l) + H O(g) HCl(g) + SO (g) N (g) + 3H (g) NH 3 (g)

12 ENTROPY: Statistical Definition when an ideal gas expands isothermally from V 1 to V S V nr ln V 1 where n is the number of moles of the gas present

13 Sample Problem: Calculate the entropy change when.0 moles of an ideal gas are allowed to expand isothermally from an initial volume of 1.5 L to.4 L.

14 THERMODYNAMIC DEFINITION OF ENTROPY: In isothermal process, q -w for a reversible process q rev q T rev V nrtln V q T V nr ln V rev S 1 1 Although for the expansion of gases, they are applicable to all types of processes at constant temperature

15 THE CARNOT CYCLE: 1. Reversible Isothermal Expansion. work U done w heat absorbed q 0 -RT V ln V 1 V RT ln V 1

16 THE CARNOT CYCLE:. Reversible Adiabatic Expansion. q 0 work done U C v (T 1 -T ) 3. Reversible isothermal Compression U 0 work done w heat released 1 q 1 -RT RT V ln V V ln V 4 3

17 CARNOT CYCLE: 4. REVERSIBLE ADIABATIC COMPRESSION q 0 work done U C v (T -T 1 )

18 CARNOT CYCLE SUMMARY: U(cycle) 0 work done q(cycle) w(cycle) q -RT q 1 V ln V 1 V RT ln V 4 3

19 THERMODYNAMIC EFFICIENCY: Efficiency net work done ny heat engine heat absorbed by the engine w q T T T 1 1 T T 1

COMPRESSOR (gas compressed, heated) RED gas at high pressure and temperature PINK gas at high

20 REFRIGERATOR: 1. CONDENSER (hot side heat exchange). EXPANSION VALVE (gas expands, cools and liquifies) 3. EVAPORATOR (cold side heat exchanger) 4. COMPRESSOR (gas compressed, heated) RED gas at high pressure and temperature PINK gas at high pressure reduced temperature BLUE liquid at low pressure and greatly reduced temperature LIGHT BLUE gas at low pressure and warmer temperature

21 REFRIGERATORS Heat Source T - q q 1 w q HEAT ENGINE work w Coefficient of Performance,COP q1 COP w q 1 T1 T -T 1 Cold Reservoir T 1

22 AIR CONDITIONERS; HEAT PUMPS AIR CONDITIONERS: OPERATION IS SIMILAR TO REFRIGERATORS: - q q 1 w HEAT PUMPS: USE TO HEAT A ROOM RATHER THAN LOWER THE TEMPERATURE: Coefficient of Performance,COP COP q w T T -T 1

23 SAMPLE PROBLEM: Calculate the coefficient of performance of a reversible refrigerator operating between and interior temperature of 4 o C and an exterior temperature of o C.

24 Change in entropy of the universe : for reversible processes; qsys qsurr T T nrt ln(v / V ) nrt ln(v / V ) 0 T T for irreversible processes S S S S S nr ln(v / V1) 0 Thus, univ univ univ S S S sys sys sys S surr surr surr 0 The Second Law of Thermodynamics: The entropy of an isolated system increases in an irreversible processes and remain unchanged in a reversible process.

25 ENTROPY CHANGES DUE TO MIXING: For For Gas A; Gas B; S S n n A B V R ln V R ln A A V V A V V B B B mix S where R(n x A and A ln x x B A n B ln x are mole fractions of B ) gas A and B

26 ENTROPY CHANGES PHASE CHANGE: Consider : constant H O ( s ) pressure 0 C,1atm H fus vap S S o process so, q fush T H vap T O ( l ) rev fus H

27 SAMPLE PROBLEM The enthalpy of vaporization of methanol is 35.7 kj/mol at its normal boiling point of 64.1 o C. Calculate (a) the entropy of vaporization of methanol at this temperature and (b) the entropy change of the surrounding

28 ENTROPY and changes in Temperature: ds dq T rev dq dh nc p dt (closed system, constant pressure ) S T T 1 C T p dt nc p T ln T 1

29 Sample Problem: Calculate S and S surr for reversibly heating.0 moles of liquid water from 0.00 o C to 100 o C at constant pressure of 1.0 atm. Specific heat capacity of liquid water is J/K-g.

30 SAMPLE PROBLEM: Calculate the change in entropy when 1.0 mole of ice at -10 o C is heated until it is a superheated steam at 10 o C. C C C fus vap p p p S.0 J/K mol S J/K - mol H H H O(s) 37.7 J/K - mol O(l) 75.3 J/K - mol O(g) 33.6 J/K - mol

31 THIRD LAW OF THERMODYNAMICS: Suppose we lower the temperature to zero; S S S k k b b T Cp dt 0 T lnw at 0 K, W 1 ln1 0 Every substance has a finite positive entropy, but at absolute zero the entropy maybe come zero, and it does in case of pure perfect crystalline substance.

32 THIRD LAW OF THERMODYNAMICS: The entropy of a perfect crystal at 0 K is zero. It is impossible to reach a temperature of absolute zero It is impossible to have a (Carnot) efficiency equal to 100% (this would imply Tc = 0).

33 THIRD LAW OF THERMODYNAMICS: T = 0, S = 0 T > 0, S > 0

34 ENTROPY OF CHEMICAL REACTIONS: For a hypothetical reaction, aa bb cc dd the entropy of reaction is given by : r S o cs o (C) ds o (D) - as o (A) - bs o (B) r S o vs o ( products ) vs o (reac tan ts )

35 SAMPLE PROBLEM: Calculate the value of the standard molar entropy changes for the following reactions at 98 K. a) b) c) S H N o CaCO : : CaCO H N values (98 K, 3 : (g) 3 (g) 130.6, 191.5, (s) 9.9, O O O O 1bar) : (J/K mol) : : (g) (g) CaO : 05, 05, CaO(s) 39.8, H NO : H NO(g) O : CO O(l) CO (g) : 13.6

36 The GIBBS Energy: H Consider the reaction, (g) O (g) O(g l) Is spontaneous or not? dsuniv dssys dssurr 0 dqsurr dssys 0 T multiply by T dhsys -TdSsys 0 we now define a function : G H -TS Gibbs Energy : G H -TS H

37 EQUILIBRIUM and SPONTANEITY CRITERIA: dg sys 0 G G G1 0; system is at equilibrium G G G1 0; spontaneous process from 1 to (at constant temperature and pressure) Significance of GIBBS energy : - incorporates both enthalpy and entropy. - entropy and enthalpy contributions can reinforce each other.

38 THE GIBBS ENERGY: Factors affecting H S G + + Positive at low Temp; negative at high Temp; Reaction is spontaneous at forward at high T and spontaneous in reverse direction at low temperature. + - Positive at all temperatures, Reaction is spontaneous in the reverse reaction at ll temperatures - + Negative at all temperatures. Reaction is spontaneous in the forward direction at all T. - - Negative at low temperatures; positive at high temperatures. Reaction is spontaneous at Low temperatures. Tends to reverse at high temperatures.

39 The HELMHOLTZ Energy: for A U -TS Helmholtz Energy constant temperature and volume processes. criteria for spontaneity and equilibrium : A 0 sys

40 SAMPLE PROBLEM: Calculate the value of G for the melting of ice at a) 0 o C b) 10 o C c) -10 o C. The molar enthalpy and entropy of fusion of water are 6.01 and.0 J/K-mol.

41 STANDARD MOLAR GIBBS ENERGY OF FORMATION the standard For a hypothetical reaction, aa bb cc dd molar Gibbs Energy of reaction is given by : r G o cg o (C) dg o (D) - ag o (A) - bg o (B) r G o o vg ( products ) vg o (reac tan ts )

42 GIBBS FREE ENERGY, G G o = H o - T S o 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 rxn = G o f (products) - G o f (reactants) G o rxn = G f o (products) - G fo (reactants)

43 FREE ENERGIES OF FORMATION Note that G f for an element = 0

44 SAMPLE CALCULATION, G o rxn For the combustion of acetylene C H (g) + 5/ O (g) --> > CO (g) + H O(g) a) by inspection is the reaction spontaneous or not? b) Calculate the G o rxn using standard molar enthalpies and entropies. c) Is the reaction spontaneous or not? Is it entropy or enthalpy driven?

45 CALCULATING G o rxn NH 4 NO 3 (s) + heat ---> > NH 4 NO 3 (aq) Is the dissolution of ammonium nitrate product-favored? If so, is it enthalpy- or entropy-driven?

46 CALCULATING G o rxn From tables of thermodynamic data we find: = -5.7 kj H o rxn S o rxn G o rxn NH 4 NO 3 (s) + heat ---> > NH 4 NO 3 (aq) = J/K or kj/k 5.7 kj - (98 K)( J/K) rxn = -5.7 kj = -6.7 kj Reaction is spontaneous in spite of negative H o rxn Reaction is entropy driven rxn.

47 Gibbs Free Energy, G G o = H o - T S o Two methods of calculating G o a) Determine H o rxn equation. and S o rxn and use Gibbs b) Use tabulated values of free energies of G o rxn rxn = G o f (products) - G o f (reactants) Gformation, o G o rxn = f o (products) f. - G fo (reactants)

48 G o rxn Calculating G o rxn rxn = G o f (products) - G o f (reactants) G o rxn = G f o (products) - G fo (reactants) Combustion of carbon C(graphite) ) + O (g) --> > CO (g) G o rxn = G o f (CO ) - [ G o f (graph) ) + G o f (O )] 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 spontaneous.

49 FREE ENERGY AND TEMPERATURE Iron metal can be produced by reducing its ore (Iron(III( Iron(III) oxide with graphite: Fe O 3 (s) + 3 C(s) ---> > 4 Fe(s) ) + 3 CO (g) H o rxn = kj S o rxn = J/K G o rxn = kj A) Is the reaction spontaneous or not? B) At what temperature will the reaction become spontaneous? 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

50 Thermodynamics and Keq 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.

51 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 = - RT lnk rxn where R = 8.31 J/K mol

52 THERMODYNAMICS and K eq G o rxn = - RT lnk Calculate K for the reaction N O 4 ---> NO G o rxn = +4.8 kj G o rxn = J = - (8.31 J/K)(98 K) ln K ln K = J (8.31 J/K)(98 K) K = 0.14 When G o rxn > 0, then K < 1 =

53 Free Energy and Chemical Equilibrium The sign of G tells the direction of spontaneous reaction when both reactants and products are present at standard state conditions. Under nonstandard conditions, G becomes G. G = G + RT lnq The reaction quotient is obtained in the same way as an equilibrium expression

54 Figure G, G,, and Keq

55 Product favored reaction (spontaneous) G o and K > 1 G, G,, and K eq In this case G rxn is < G o rxn, so state with both reactants and products present is MORE STABLE than complete conversion.

56 G, G,, and K eq Spontaneous reaction. NO ---> > N 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.

57 G, G,, and Keq Non spontaneous reaction. N O 4 ---> NO 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.

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

59 END OF CHAPTER

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