# Thermodynamics and Equilibrium

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1 Thermdynamics and Equilibrium

2 Thermdynamics Thermdynamics is the study f the relatinship between heat and ther frms f energy in a chemical r physical prcess. We intrduced the thermdynamic prperty f enthalpy, H, in Chapter 6. We nted that the change in enthalpy equals the heat f reactin at cnstant pressure. In this chapter we will define enthalpy mre precisely, in terms f the energy f the system. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 2

3 First Law f Thermdynamics T state the laws f thermdynamics, we must first understand the internal energy f a system and hw yu can change it. Heat is energy that mves int r ut f a system because f a temperature difference between system and surrundings. Wrk, n the ther hand, is the energy exchange that results when a frce F mves an bject thrugh a distance d; wrk (w) = Fd Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 3

4 First Law f Thermdynamics T state the laws f thermdynamics, we must first understand the internal energy f a system and hw yu can change it. Remembering ur sign cnventin. Wrk dne by the system is negative. Wrk dne n the system is psitive. Heat evlved by the system is negative. Heat absrbed by the system is psitive. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 4

5 First Law f Thermdynamics T state the laws f thermdynamics, we must first understand the internal energy f a system and hw yu can change it. In general, the first law f thermdynamics states that the change in internal energy, U, equals heat plus wrk. U q w Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 5

6 Heat f Reactin and Internal Energy When a reactin is run in an pen vessel (at cnstant P), any gases prduced represent a ptential surce f expansin wrk. It fllws therefre, that w PV Yu can calculate the wrk dne by a chemical reactin simply by multiplying the atmspheric pressure by the change in vlume, V. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 6

7 Heat f Reactin and Internal Energy When a reactin is run in an pen vessel (at cnstant P), any gases prduced represent a ptential surce f expansin wrk. Relating the change in internal energy t the heat f reactin, yu have U U q p w ( kj) U kj ( 2.47 kj) Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 7

8 Enthalpy and Enthalpy Change In Chapter 6, we tentatively defined enthalpy in terms f the relatinship f H t the heat at cnstant pressure. We nw define enthalpy, H, precisely as the quantity U + PV. Because U, P, and V are state functins, H is als a state functin. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 8

9 Spntaneus Prcesses and Entrpy A spntaneus prcess is a physical r chemical change that ccurs by itself. Examples include: A rck at the tp f a hill rlls dwn. Heat flws frm a ht bject t a cld ne. An irn bject rusts in mist air. These prcesses ccur withut requiring an utside frce and cntinue until equilibrium is reached. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 9

10 Entrpy and the Secnd Law f Thermdynamics The secnd law f thermdynamics addresses questins abut spntaneity in terms f a quantity called entrpy. Entrpy, S, is a thermdynamic quantity that is a measure f the randmness r disrder f a system. The SI unit f entrpy is jules per Kelvin (J/K) and, like enthalpy, is a state functin. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 10

11 Entrpy, Enthalpy, and Spntaneity Nw yu can see hw thermdynamics is applied t the questin f reactin spntaneity. Recall that the heat at cnstant pressure, q p, equals the enthalpy change, H. The secnd law fr a spntaneus reactin at cnstant temperature and pressure becmes qp H S (Spntaneus reactin, cnstant T and P) T T Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 11

12 Entrpy, Enthalpy, and Spntaneity Nw yu can see hw thermdynamics is applied t the questin f reactin spntaneity. Rearranging this equatin, we find H TS 0 (Spntaneus reactin, cnstant T and P) This inequality implies that fr a reactin t be spntaneus, H-TS must be negative. If H-TS is psitive, the reverse reactin is spntaneus. If H-TS=0, the reactin is at equilibrium Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 12

13 Standard Entrpies and the Third Law f Thermdynamics The third law f thermdynamics states that a substance that is perfectly crystalline at 0 K has an entrpy f zer. When temperature is raised, hwever, the substance becmes mre disrdered as it absrbs heat. The entrpy f a substance is determined by measuring hw much heat is required t change its temperature per Kelvin degree. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 13

14 Standard Entrpies and the Third Law f Thermdynamics The standard entrpy f a substance r in (Table 19.1), als called its abslute entrpy, S, is the entrpy value fr the standard state f the species. Standard state implies 25 C, 1 atm pressure, and 1 M fr disslved substances. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 14

15 Standard Entrpies and the Third Law f Thermdynamics The standard entrpy f a substance r in (Table 19.1), als called its abslute entrpy, S, is the entrpy value fr the standard state f the species. Nte that the elements have nnzer values, unlike standard enthalpies f frmatin, H f, which by cnventin, are zer. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 15

16 Standard Entrpies and the Third Law f Thermdynamics The standard entrpy f a substance r in (Table 19.1), als called its abslute entrpy, S, is the entrpy value fr the standard state f the species. The symbl S, rather than S, is used fr standard entrpies t emphasize that they riginate frm the third law. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 16

17 Entrpy Change fr a Reactin Yu can calculate the entrpy change fr a reactin using a summatin law, similar t the way yu btained S. S ns (prducts) ms (reactants) Even withut knwing the values fr the entrpies f substances, yu can smetimes predict the sign f S fr a reactin. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 17

18 Entrpy Change fr a Reactin Yu can calculate the entrpy change fr a reactin using a summatin law, similar t the way yu btained S. The entrpy usually increases in the fllwing situatins: 1. A reactin in which a mlecule is brken int tw r mre smaller mlecules. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 18

19 Entrpy Change fr a Reactin Yu can calculate the entrpy change fr a reactin using a summatin law, similar t the way yu btained S. The entrpy usually increases in the fllwing situatins: 2. A reactin in which there is an increase in the mles f gases. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 19

20 Entrpy Change fr a Reactin Yu can calculate the entrpy change fr a reactin using a summatin law, similar t the way yu btained S. The entrpy usually increases in the fllwing situatins: 3. A prcess in which a slid changes t liquid r gas r a liquid changes t gas. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 20

21 A Prblem T Cnsider Calculate the change in entrpy, S, at 25 C fr the reactin in which urea is frmed frm NH 3 and CO 2. The standard entrpy f NH 2 CONH 2 is 174 J/(ml.K). See Table 19.1 fr ther values. 2NH3(g) CO2(g) NH2CONH2(aq) H2O(l) The calculatin is similar t that used t btain H frm standard enthalpies f frmatin. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 21

22 A Prblem T Cnsider Calculate the change in entrpy, S, at 25 C fr the reactin in which urea is frmed frm NH 3 and CO 2. The standard entrpy f NH 2 CONH 2 is 174 J/(ml.K). See Table 19.1 fr ther values. 2NH3(g) CO2(g) NH2CONH2(aq) H2O(l) S : 2 x It is cnvenient t put the standard entrpies (multiplied by their stichimetric cefficients) belw the frmulas. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 22

23 A Prblem T Cnsider Calculate the change in entrpy, S, at 25 C fr the reactin in which urea is frmed frm NH 3 and CO 2. The standard entrpy f NH 2 CONH 2 is 174 J/(ml.K). See Table 19.1 fr ther values. 2NH3(g) CO2(g) NH2CONH2(aq) H2O(l) S We can nw use the summatin law t calculate the entrpy change. ns (prducts) ms (reactants) Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 23

24 A Prblem T Cnsider Calculate the change in entrpy, S, at 25 C fr the reactin in which urea is frmed frm NH 3 and CO 2. The standard entrpy f NH 2 CONH 2 is 174 J/(ml.K). See Table 19.1 fr ther values. 2NH3(g) CO2(g) NH2CONH2(aq) H2O(l) We can nw use the summatin law t calculate the entrpy change. S [(174 70) ( )]J / K 356 J/K Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 24

25 Free Energy Cncept The American physicist J. Willard Gibbs intrduced the cncept f free energy (smetimes called the Gibbs free energy), G, which is a thermdynamic quantity defined by the equatin G=H-TS. This quantity gives a direct criterin fr spntaneity f reactin. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 25

26 Free Energy and Spntaneity Changes in H an S during a reactin result in a change in free energy, G, given by the equatin G H TS Thus, if yu can shw that G is negative at a given temperature and pressure, yu can predict that the reactin will be spntaneus. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 26

27 Standard Free-Energy Change The standard free energy change, G, is the free energy change that ccurs when reactants and prducts are in their standard states. The next example illustrates the calculatin f the standard free energy change, G, frm H and S. G H TS Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 27

28 A Prblem T Cnsider What is the standard free energy change, G, fr the fllwing reactin at 25 C? Use values f H f and S, frm Tables 6.2 and N2(g) 3H2(g) 2NH3(g) H f : x (-45.9) kj S : x x 193 J/K Place belw each frmula the values f H f and S multiplied by stichimetric cefficients. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 28

29 A Prblem T Cnsider What is the standard free energy change, G, fr the fllwing reactin at 25 C? Use values f H f and S, frm Tables 6.2 and N2(g) 3H2(g) 2NH3(g) Yu can calculate H and S using their respective summatin laws. H f nh (prducts) mh [ 2( 45.9) 0]kJ 91.8 kj f (reactants) Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 29

30 A Prblem T Cnsider What is the standard free energy change, G, fr the fllwing reactin at 25 C? Use values f H f and S, frm Tables 6.2 and S N2(g) 3H2(g) 2NH3(g) Yu can calculate H and S using their respective summatin laws. ns (prducts) ms [ 2193 ( )]J/K (reactants) -197 J/K Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 30

31 A Prblem T Cnsider What is the standard free energy change, G, fr the fllwing reactin at 25 C? Use values f H f and S, frm Tables 6.2 and N2(g) 3H2(g) 2NH3(g) Nw substitute int ur equatin fr G. Nte that S is cnverted t kj/k. G H TS 91.8kJ (298 K)( kj/k) 33.1kJ Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 31

32 Standard Free Energies f Frmatin The standard free energy f frmatin, G f, f a substance is the free energy change that ccurs when 1 ml f a substance is frmed frm its elements in their stablest states at 1 atm pressure and 25 C. By tabulating G f fr substances, as in Table 19.2, yu can calculate the G fr a reactin by using a summatin law. G ng f (prducts) mg f (reactants) Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 32

33 A Prblem T Cnsider Calculate G fr the cmbustin f 1 ml f ethanl, C 2 H 5 OH, at 25 C. Use the standard free energies f frmatin given in Table C2H5OH(l) 3O2(g) 2CO2(g) 3H2O(g) G f : (-394.4) 3(-228.6)kJ Place belw each frmula the values f G f multiplied by stichimetric cefficients. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 33

34 A Prblem T Cnsider Calculate G fr the cmbustin f 1 ml f ethanl, C 2 H 5 OH, at 25 C. Use the standard free energies f frmatin given in Table C2H5OH(l) 3O2(g) 2CO2(g) 3H2O(g) G f : (-394.4) 3(-228.6)kJ Yu can calculate G using the summatin law. G G ng f (prducts) mg f (reactants) [2( 394.4) 3( 228.6) ( 174.8)]kJ Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 34

35 A Prblem T Cnsider Calculate G fr the cmbustin f 1 ml f ethanl, C 2 H 5 OH, at 25 C. Use the standard free energies f frmatin given in Table C 2 H 5 OH(l) 3O 2 (g) 2CO 2 (g) 3H 2 O(g) G f : (-394.4) 3(-228.6)kJ Yu can calculate G using the summatin law. G ng f G kj (prducts) mg f (reactants) Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 35

36 G as a Criteria fr Spntaneity The fllwing rules are useful in judging the spntaneity f a reactin. 1. When G is a large negative number (mre negative than abut 10 kj), the reactin is spntaneus as written, and the reactants transfrm almst entirely t prducts when equilibrium is reached. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 36

37 G as a Criteria fr Spntaneity The fllwing rules are useful in judging the spntaneity f a reactin. 2. When G is a large psitive number (mre psitive than abut +10 kj), the reactin is nnspntaneus as written, and reactants d nt give significant amunts f prduct at equilibrium. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 37

38 G as a Criteria fr Spntaneity The fllwing rules are useful in judging the spntaneity f a reactin. 3. When G is a small negative r psitive value (less than abut 10 kj), the reactin gives an equilibrium mixture with significant amunts f bth reactants and prducts. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 38

39 Free Energy Change During Reactin As a system appraches equilibrium, the instantaneus change in free energy appraches zer. Figure 19.9 illustrates the change in free energy during a spntaneus reactin. As the reactin prceeds, the free energy eventually reaches its minimum value. At that pint, G = 0, and the net reactin stps; it cmes t equilibrium. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 39

40 Figure 19.9 Free-energy change during a spntaneus reactin.

41 Relating G t the Equilibrium Cnstant The free energy change when reactants are in nn-standard states (ther than 1 atm pressure r 1 M) is related t the standard free energy change, G, by the fllwing equatin. G G RTlnQ Here Q is the thermdynamic frm f the reactin qutient. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 41

42 Relating G t the Equilibrium Cnstant The free energy change when reactants are in nn-standard states (ther than 1 atm pressure r 1 M) is related t the standard free energy change, G, by the fllwing equatin. G G RTlnQ G represents an instantaneus change in free energy at sme pint in the reactin appraching equilibrium. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 42

43 Relating G t the Equilibrium Cnstant The free energy change when reactants are in nn-standard states (ther than 1 atm pressure r 1 M) is related t the standard free energy change, G, by the fllwing equatin. G G RTlnQ At equilibrium, G=0 and the reactin qutient Q becmes the equilibrium cnstant K. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 43

44 Relating G t the Equilibrium Cnstant The free energy change when reactants are in nn-standard states (ther than 1 atm pressure r 1 M) is related t the standard free energy change, G, by the fllwing equatin. 0 G RTlnK At equilibrium, G=0 and the reactin qutient Q becmes the equilibrium cnstant K. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 44

45 Relating G t the Equilibrium Cnstant This result easily rearranges t give the basic equatin relating the standard free-energy change t the equilibrium cnstant. G RTlnK When K > 1, the ln K is psitive and G is negative. When K < 1, the ln K is negative and G is psitive. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 45

46 A Prblem T Cnsider Find the value fr the equilibrium cnstant, K, at 25 C (298 K) fr the fllwing reactin. The standard free-energy change, G, at 25 C equals 13.6 kj. 2NH3(g) CO2(g) NH2CONH2(aq) H2O(l) Rearrange the equatin G =-RTlnK t give lnk G RT Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 46

47 A Prblem T Cnsider Find the value fr the equilibrium cnstant, K, at 25 C (298 K) fr the fllwing reactin. The standard free-energy change, G, at 25 C equals 13.6 kj. 2NH3(g) CO2(g) NH2CONH2(aq) H2O(l) Substituting numerical values int the equatin, 3 lnk J/(ml K) 298 K J 5.49 Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 47

48 A Prblem T Cnsider Find the value fr the equilibrium cnstant, K, at 25 C (298 K) fr the fllwing reactin. The standard free-energy change, G, at 25 C equals 13.6 kj. 2NH3(g) CO2(g) NH2CONH2(aq) H2O(l) Hence, K e Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 48

49 Spntaneity and Temperature Change All f the fur pssible chices f signs fr H and S give different temperature behavirs fr G. H S G Descriptin + Spntaneus at all T + + Nnspntaneus at all T + r Spntaneus at lw T; Nnspntaneus at high T r Nnspntaneus at lw T; Spntaneus at high T Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 49

50 Calculatin f G at Varius Temperatures In this methd yu assume that H and S are essentially cnstant with respect t temperature. Yu get the value f G T at any temperature T by substituting values f H and S at 25 C int the fllwing equatin. G T H TS Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 50

51 A Prblem T Cnsider Find the G fr the fllwing reactin at 25 C and 1000 C. Relate this t reactin spntaneity. CaCO3(s) CaO(s) CO2(g) H f : kj S : J/K Place belw each frmula the values f H f and S multiplied by stichimetric cefficients. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 51

52 A Prblem T Cnsider Find the G fr the fllwing reactin at 25 C and 1000 C. Relate this t reactin spntaneity. CaCO3(s) CaO(s) CO2(g) H f : kj S : J/K Yu can calculate H and S using their respective summatin laws. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 52

53 A Prblem T Cnsider Find the G fr the fllwing reactin at 25 C and 1000 C. Relate this t reactin spntaneity. H CaCO3(s) CaO(s) CO2(g) H f : kj S : J/K nh f (prducts) mh [( ) ( )]kJ f (reactants) kj Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 53

54 A Prblem T Cnsider Find the G fr the fllwing reactin at 25 C and 1000 C. Relate this t reactin spntaneity. CaCO3(s) CaO(s) CO2(g) H f : kj S : J/K S ns (prducts) [( ) (92.9)] ms (reactants) J / K Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 54

55 A Prblem T Cnsider Find the G fr the fllwing reactin at 25 C and 1000 C. Relate this t reactin spntaneity. CaCO3(s) CaO(s) CO2(g) H f : kj S : J/K Nw yu substitute H, S (= kj/k), and T (=298K) int the equatin fr G f. G H TS T Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 55

56 A Prblem T Cnsider Find the G fr the fllwing reactin at 25 C and 1000 C. Relate this t reactin spntaneity. CaCO3(s) CaO(s) CO2(g) H f : kj S : J/K Nw yu substitute H, S (= kj/k), and T (=298K) int the equatin fr G f. G T 178.3kJ (298 K)(0.1590kJ / K) Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 56

57 A Prblem T Cnsider Find the G fr the fllwing reactin at 25 C and 1000 C. Relate this t reactin spntaneity. CaCO3(s) CaO(s) CO2(g) H f : kj S : J/K Nw yu substitute H, S (= kj/k), and T (=298K) int the equatin fr G f. G T kj S the reactin is nnspntaneus at 25 C. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 57

58 A Prblem T Cnsider Find the G fr the fllwing reactin at 25 C and 1000 C. Relate this t reactin spntaneity. CaCO3(s) CaO(s) CO2(g) H f : kj S : J/K Nw we ll use 1000 C (1273 K) alng with ur previus values fr H and S. G T 178.3kJ (1273 K)(0.1590kJ / K) Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 58

59 A Prblem T Cnsider Find the G fr the fllwing reactin at 25 C and 1000 C. Relate this t reactin spntaneity. CaCO3(s) CaO(s) CO2(g) H f : kj S : J/K Nw we ll use 1000 C (1273 K) alng with ur previus values fr H and S. G 24.1kJ T S the reactin is spntaneus at 1000 C. Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 59

60 Operatinal Skills Calculating the entrpy change fr a phase transitin Predicting the sign f the entrpy change f a reactin Calculating S fr a reactin Calculating G frm H and S Calculating G frm standard free energies f frmatin Interpreting the sign f G Writing the expressin fr a thermdynamic equilibrium cnstant Calculating K frm the standard free energy change Calculating G and K at varius temperatures Cpyright Hughtn Mifflin Cmpany.All rights reserved. Presentatin f Lecture Outlines, 19 60

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