Chapter Eighteen. Thermodynamics

Transcription

1 Chapter Eighteen Thermodynamics 1

2 Thermodynamics Study of energy changes during observed processes Purpose: To predict spontaneity of a process Spontaneity: Will process go without assistance? Depends on entropy & enthalpy Measurement of Free energy, G Entropy (S): Measurement of disorder How is energy spread out in system? Free Energy ( G = H -T S) Relates enthalpy with entropy Predict if process will occur 2

3 Ch.6 Enthalpies of Formation, H f Need defined conditions to compare reaction energies Substance formed from elements in natural phase 25 C, 1.0 atm, 1 mole of product, 1.0 M solution General reaction: H o = H o (prod) - H o (react) Get numbers from table Remember to multiply by moles of each species The enthalpy defined as Enthalpy of Formation, H f o H fo tabulated (see back of book Appendix C) 3

4 Table of Enthalpies of Formation at 25 C All H f are for standard state 25 C, 1atm, 1M aqueous solution, 1 mole product 4

5 Three Laws of Thermodynamics 1st Law of Thermodynamics (Ch. 6) Energy cannot be created or destroyed Introduces enthalpy, H 2nd Law of Thermodynamics Reactions favor 1 direction Introduces entropy (S) 3rd Law of Thermodynamics Provides starting point for determining S 5

6 Spontaneous Change Spontaneous process Occurs by itself No outside action needed Water flows downhill naturally Non-spontaneous process Cannot take place by itself. Water does not naturally flow uphill But it can be pumped uphill If a process is spontaneous, the reverse process is non-spontaneous and vice versa. The spontaneous signifies nothing about how fast a process occurs. (Reaction Rate) 6

7 Spontaneity and Relation to Enthalpy Enthalpy of a Fuel Cell Reaction is exothermic ΔH = -286kJ/mol Favors H 2 O production Reality: H 2 and O 2 remain separate No water formed E act = +112kJ/mol Nonspontaneous reaction 2H 2 (g) + O 2 (g) 2H 2 O(l) Enthalpy determines equilibrium state. Amounts of heat generated or required for a reaction Does not determine spontaneity! 7

8 Phase Changes and Entropy S = S final -S initial for a spontaneous reaction This only describes disorder, not practical spontaneity Overall reaction is often temperature dependent Melting: S >0: Spontaneous Liquid more disordered Water> 0 C Vaporization: S >0: Spontaneous Gas more disordered Water> 100 C 8

9 Second Law of Thermodynamics 9

10 Second Law of Thermodynamics ΔS o univ = ΔS sys + ΔS surr >0 for a spontaneous rxn ΔS o univ >0 if number of microstates increase Moles of products > moles of reactants More particles after reaction than before Especially gases! More complex molecules > simple molecules More atoms to vibrate in molecule Larger atoms: more electron movement in orbitals A change to a more disordered phase from s to l, l to g, or s to g At equlibrium: ΔS o univ = ΔS sys + ΔS surr =0 No further changes seen 10

11 Standard Molar Entropies, S Entropy of 1 mole of a substance in its standard state. Method identical to that of finding H 1. Add up S, of all products 2. Subtract S, of all reactants 3. ΔS rxn = S o (products) -S o (reactants) For phase change: H 2 O(l) H 2 O(g) 1. Products: H 2 O(g) 188.7J/molK Reactants: H 2 O(l) 69.9J/molK 1. ΔS rxn = J/molK J/molK ΔS rxn =118.8J/molK (spontaneous) 11

12 Third Law of Thermodynamics The Third Law defines entropy in terms of statistics. At T=0K; a substance will be a perfectly ordered crystal Only 1 way to arrange atoms in a perfect substance No molecular movement, no energy causing disorder So, at T = 0K, S = 0 12

13 Third Law of Thermodynamics and S Used to find S for any substance Cool the substance to as close to 0K as possible Absolutely no molecular movement S initial = 0 Warm up to a specific temperature and calculate H S is proportional to - H S is temperature dependent S= H/T All energy due to entropy change S substance = S final S initial = S final 0= S final 13

14 Gibb s Free Energy 14

15 Determination of Spontaneity For chemical reactions at const. P: S univ = S surr + S sys and S surr = H /T S univ = H /T +S sys Multiply by T: T S univ = H -T S sys Define T S univ as Gibb s Free Energy, G G = H -T S H -T S balances disorder, temperature and enthalpy H -T S must be negative for a spontaneous reaction 15

16 Gibb s Free Energy, G G sys = H -T S sys If G < 0 (negative), a process is spontaneous. If G > 0 (positive), a process is non-spontaneous. If G = 0, no net change, and the process is at equilibrium. Thermodynamic function that relates enthalpy and entropy to spontaneity.. 16

17 Standard Free Energy, G o Free energy change when reactants and products are in their standard states. Standard free energy of formation, G o f Free energy change that occurs in the formation of 1 mol of a substance in its standard state from its elements in their standard states G fo = G o (moles all products) - G o (moles all reactants) At equilibrium, G = 0 so H = T S, S = H / T & T= H / S Usually used to find phase change temperatures 17

18 Find H o, S o, and G o for the solubilization of calcium phosphate at 25 C. Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3 (aq) Look up table information: H fo (Ca 3 (PO 4 ) 2 (s)) H fo (Ca 2+ (aq)) H fo (PO 3 4 (aq)) S o (Ca 3 (PO 4 ) 2 (s)) S o (Ca 2+ (aq)) S o (PO 3 4 (aq)) = 4121 kj/mol = kj/mol = 1277 kj/mol = J/molK = 55.2 J/molK = 222 J/molK 18

19 Find H o, S o, and G o for the solubilization of calcium phosphate at 25 C. Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3 (aq) Find the for the H o for the reaction: H o = [3 Ca PO 4 3 ] [Ca 3 (PO 4 ) 2 ] H o = [3( ) + 2( 1277)] [ 4121] H o = 62 kj/mol Find S o changes for the reaction: S o = [3( 55.2) + 2( 222)] [236] S o = 846 J/mol K = kj/mol K 19

20 Find H o, S o, and G o for the solubilization of calcium phosphate at 25 C. Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3 (aq) Find G o for the reaction: G o = H o T S o H o = 62 kj / mol T = = 298 K S o = kj/mol K T S o = 298 K x ( 0.846kJ/mol K ) = kj/mol G o = 62 kj/mol (-252kJ/mol ) = +190 kj/mol Reaction is non-spontaneous at room temperature. 20

21 What is G for this reaction at 75 C? Do H and S change with temperature? H and S are relatively constant with changing temp. To find G o at other T: Assume H o & S o constant With this assumption: G o = H o T S o H o = 62 kj / mol T = = 348 K S o = kj/mol K G o = 62 kj/mol (348K) x ( kj /mol K) = +232 kj/mol Reaction less spontaneous at higher temperature. The salt becomes less soluble 21

22 G and Temperature Dependence G sys = H sys T S sys 22

23 Gibb s Free Energy and Equilibrium 23

24 Free Energy and Phase Changes Phase changes are reactions at equilibrium: G o = 0 H o =T S o Can solve for H o knowing T and S o, but usually you can look up H o values so unnecessary. S o = H o /T (H 2 O (l) H 2 O (s) ) H o = 6.01 kj / mol (exothermic) T = 273K (room temperature) S o = 6010J/mol/273K= -22.0J/molK (nonspontaneous) T= H o / S o (H 2 O (l) H 2 O (g) ) H o = kJ/mol-(-285.8kJ/mol)= 44kJ / mol (endothermic) S o = 188.7J/molK-69.9J/molK= 118.8j/molK (spontaneous) T = 44000J/mol/ 118.8j/molK = 370K 24

25 Free Energy and Phase Changes Calculate G o for evaporation of water: H 2 O(l) H 2 O(g) Calculate H o S o and G o from table data Assume no temp dependence for H o S o G o at room temperature: 298K H o = [ kj/mol] S o = [+188.7J/K] [+69.91J/K] G o = 44.0 kj - (298K)(0.1188KJ/molK) G o at the boiling point of water: 373K G o = kj - (373K)( kj/k) = -0.3kJ H o = kj/mol S o = J/molK G o = kj/mol Experimentally, G o = 0 for a phase transition Error due to temperature dependence is 0.3 kj 25

26 Gibb's Energy and Equilibrium G determines spontaneity: G = G o + RT ln Q R = = J/mol K T = temperature in K Q = reaction quotient G o is found from tabulated data Determines G at any composition or temperature At equilibrium, G = 0 and Q = K eq, so G o = RT ln K eq 26

27 Compare the molar solubility of lead chloride at room temperature (25 C). PbCl 2 (s) Pb 2+ (aq) + 2Cl (aq) PbCl 2 (s) Pb 2+ (aq) + 2Cl (aq) Initial 0 0 Change + x + 2x Equilibrium x 2x K sp = [Pb 2+ ][Cl ] 2 = = [x][2x] 2 = 4x 3 x = The molar solubility of lead(ii) chloride is M at 25 C 27

28 Find the Molar Solubility at 90 C Look up in table H o f (Pb 2+ aq) = 1.7 kj/mol H o f (Cl - aq) = kj/mol H o f (PbCl 2s ) = 359 kj/mol PbCl 2 (s) Pb 2+ (aq) + 2Cl (aq) S o (Pb 2+ aq) = J/mol K S o (Cl - aq) = J/mol K S o (PbCl 2s ) = J/mol K Calculate G o H o = [ 1.7 kj/mol + 2( kj/mol)] [ 359 kj/mol] =+ 23 kj/mol T = = 363K S o = [ (56.5)] [136] = 13 J/molK G o = H o T S o = j/mol 363K( 13J/mol K)=28000J/mol Calculate K sp G o = R T ln K eq = R T ln K sp K sp =e - G /RT = e /(8.314)(363) = 9.3 x

29 Find the Molar Solubility at 90 C PbCl 2 (s) Pb 2+ (aq) + 2Cl (aq) Make a table: [Pb 2+ ] [Cl ] Initial 0 0 Change + x +2x Equilibrium x 2x Calculate molar solubility K sp = [Pb 2+ ] e [Cl ] e2 = = [x][2x] 2 = 4x 3 x = Molar solubility of PbCl 2 at 90 C = > Solubility about double at higher T Most salts dissolve more at higher temperatures 29