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, HTS must be negative. If HTS is psitive, the reverse reactin is spntaneus. If HTS=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=HTS. 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 FreeEnergy 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 Freeenergy change during a spntaneus reactin.
41 Relating G t the Equilibrium Cnstant The free energy change when reactants are in nnstandard 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 nnstandard 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 nnstandard 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 nnstandard 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 freeenergy 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 freeenergy 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 freeenergy 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 freeenergy 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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