Chemistry 163B. Thermochemistry. Chapter 4 Engel & Reid

Transcription

1 Chemistry 163B Thermochemistry Chapter 4 Engel & Reid 3

2 heats of reactions (constant volume; bomb calorimeter) ΔU V = q V ΔU= U products -U reactants 4

3 heats of reactions (constant volume; fig 4.3 E&R) ΔU V = q V ΔU= U products -U reactants 5

4 heats of reactions (constant pressure; coffee cup calorimeter ) ΔH P = q P ΔH= H products -H reactants 6

5 heats of reactions (constant pressure; fig 4.4 E&R) ΔH P = q P ΔH= H products -H reactants 7

6 most reactions at P constant, ΔH P =q P C(gr) + O 2 (g) CO 2 (g) ΔH 298 =-393 kj ATP + H 2 O ADP +H 3 PO 4 SiO 2 (α quartz) SiO 2 (β quartz) ΔH kj ΔH 848 =-907 kj NaCl(s) + H 2 O ( ) Na + (aq) + Cl (aq) ΔH 298 =+3.9 kj 8

7 topics for thermochemistry, parts of Ch. 4 Engel & Reid HW#3 16,17,18 HW#3 16 Calculate H reaction Hess s Law, standard heats of formation H reaction vs U reaction HW#3 18, *19 Temperature (and later pressure) dependence of H reaction HW#3 20, 21 Calorimetry Heats of solution H reaction from bond enthalpies MIDTERM 1 9

8 Hess s Law ΔH reaction = H products -H reactants 10

9 ΔH State Function Hess s Law O 2 (g) + C (gr) I ΔH I =? C(dia) + O 2 (g) ΔH II =ΔH combustion (gr) II III ΔH III = ΔH combustion (dia) CO 2 (g) ΔH I =ΔH II +ΔH III ΔH I ΔH II + ΔH III ΔH kj C(gr) C (dia)? C(gr) + O 2 (g) CO 2 (g) CO 2 (g) C (dia) + O 2 (g) ΔH I = ΔH II +ΔH III =1.90 kj 11

10 factors affecting H reaction stoichiometry H is extensive ; H is intensive physical state phase or crystal form of reactants and products temperature and pressure does NOT depend on path 12

11 notation: H (etc) in terms of molar enthalpies and stoichiometric coefficients n A A +n B B n C C + n D D Hreaction nc HC ndh D nah A nb H B total enthalpy of reactants H n H H n H i prods i i reacts i i i prods i reacts H n H n H H reaction i i i i i prods i reacts reaction i i i i i H is stoichiometric coefficient of i n if i is product species i n if i is reactant species i th molar enthalpy of reactant i number of moles of i in stoichiometry reactant / product 13

12 H reaction vs U reaction at fixed temperature T H U PV H H H prods reacts H U ( PV ) U ( PV ) ( PV ) prods assume: i) PV and PV is small for solids and liquids ii) Gasses follow ideal gas law PV ( PV ) ( PV ) prods reacts PV ( n R) ( n R) PV n RT gas prod gas react gas reacts Hreaction Ureaction ngasrt E&R eqn

13 standard states and enthalpies of formation (done on board) standard state H ( T, P 1atm [ or 1bar]) often T is assumed 298K for standard state H 298 define standard states molar enthalpy ( H298 ) formation H f of formation define enthalpy of formation 15

14 ΔH rxn from H f reactants products (-ΔH f ) reactants (ΔH f ) products elements in most stable state 0 ( H ) rxn i ( H ) f i i do HW#3 problem 16b (E&R 4.20b) 16

15 topics for thermochemistry, parts of Ch. 4 E&R Calculate H reaction Hess s Law, standard heats of formation H reaction vs U reaction Temperature (and pressure) dependence of H reaction Calorimetry Heats of solution H reaction from bond enthalpies MIDTERM 1 17

16 temperature dependence of H of substance at constant P change in enthalpy of substance T 1 T 2 H dh ncpdt dp P T T P P P P T 2 2 T 1 1 P H( T ) H( T ) C dt 2 1 H( T ) H( T ) C dt 2 1 T T T T P P T for a given reactant or product at constant P dh nc dt dh C dt dh C dt 18

17 temperature dependence of H reaction (reaction carried out at constant P,T) H H reaction i i i H ( T ) H ( T ) ( C ) dt i 2 i 1 P i T T 2 1 H H rxn rxn T, P 1 T, P 2 H ( T ) H ( T ) reaction 2 i i 2 j H ( T ) H ( T ) ( C ) dt i i 2 i i 1 i P i j j i T H ( T ) H ( T ) ( C ) dt reaction 2 reaction 1 i P i T i H ( T ) H ( T ) C dt reaction 2 reaction 1 P T where C ( C ) "( C ) ( C ) " T T P i P i P products P reactants i T 2 1 change of H rxn T 1 T 2 19

18 example problems : H from H f and H T=298K 398K H f H C 2 H 2 (g) C 6 H 6 ( ) calc Hº kj mol 3( ) 1( 49. 1) kj [ per mol C H ( )] 6 6 (mol) (kj mol -1 ) + (mol)(kj mol -1 ) = kj C P CP C 2 H 2 (g) C 6 H 6 ( ) calc Hº J K mol 3 ( 44. 0) 1( ) ( H ) ( H ) CPdT 298 J 1 1 ( H ) C 298 P T C P independent of T -3-1 ( H ) kj + ( kj K ) ( 100K ) = kj 398 K 1 20

19 topics for thermochemistry, parts of Ch. 4 E&R Calculate H reaction Hess s Law, standard heats of formation H reaction vs U reaction Temperature (and pressure) dependence of H reaction Calorimetry Heats of solution H reaction from bond enthalpies MIDTERM 1 21

20 heats of reactions (constant volume) ΔU V = q V ΔU= U products -U reactants 22

21 heats of reactions (constant pressure) ΔH P = q P ΔH= H products -H reactants 23

22 DSC- differential scanning calorimetry (enrichment, don t FRET) useful for small samples (often biological) 24

23 DTA differential thermal analysis DSC differential scanning calorimetry T of reference rises linearly with time heat input 25

24 DTA differential thermal analysis DSC differential scanning calorimetry T rise due to q from calorimeter into sample PLUS ΔH from process (e.g. chemical reaction, phase change or protein denaturation) ΔT T rise due to q from calorimeter into reference 26

25 DTA differential thermal analysis DSC differential scanning calorimetry denaturation of protein Intermediates during denaturation T sample lower than T reference since some of q in goes to denaturation t 27

26 how H and ΔH change with pressure (don t FRET now) will prove later: H P 1 V H T reaction P V V T V TV T V T P the coefficient of thermal expandion P T i ( V TV ) i i i i Will show later: dependence of H reaction on pressure is usually weak 28

27 heats of formation of ions (heat of solutions), Example 4.4 p 73 0 H f importance: how to assign for an individual ion in solution since ions come in pairs 0 resolution: assign (H + (aq, 1M) ideal )=0 H f 0 and measure for other ions relative to H + H f 29

28 heats of formation of ions (heat of solutions) HCl( g, ) H measured 0 H Cl aq f 0 - HOW TO GET H f for Cl (aq) HO bar H (aq, 1M) +Cl (aq, 1M) ( ( )) = kj kj = - ideal H HCl g H H aq H Cl aq ( ( )) ( ( )) ( ( )) kj = -(-92.3kJ) H (Cl (aq)) = f f f kj mol 1 f ideal GOT 0 H f for Cl NOW ( aq) 30

29 heats of formation of ions (heat of solutions) GET 0 H f for Na ( aq) NaCl() s H HO 2 measured Na (aq, 1M) +Cl (aq, 1M) = kj + - ideal f f f kJ = -(-411.2kJ) + H ( Na ( aq)) + ( ideal kJ = -H (NaCl(s)) + H (Na (aq)) + H (Cl (aq)) H ( Na ( aq)) 0 1 f = kj mol f kj) Then yada yada yada etc. 31

30 ΔH reaction from bond enthalpies (p. 69) reactants ΔH reaction products ΔH vap ΔH sublim -ΔH vap -ΔH sublim reactants (gas phase) products (gas phase) ~ΔH bond enthalpies 32

31 ΔH reaction from bond enthalpies Similar bonds (C-H, C-C, C=C, C=O, etc) in similar molecule have similar enthalpies (energies) Use bond enthalpies (averaged over experimental data from several molecules) to approximate the enthalpies of the bonds broken in reactants and bonds formed in products to approximate gas phase ΔH reaction 33

32 ΔH reaction from bond enthalpies reactants (gas phase) products (gas phase) H enthalpy bonds broken H enthalpy bonds formed H 0 endothermic H 0 exothermic atoms in gas phase 34

33 ΔH reaction from bond enthalpies 35

34 example ΔH f CH 3 CH=CH 2 (g) ΔH f : 3C(gr) + 3H 2 (g) CH 3 CH=CH 2 (g) 3ΔH sub (C)=3(717) kj ΔH gas phase : 3C(g) + 3H 2 (g) CH 3 CHCH 2 (g) ΔH=3BE(H 2 ) = 3 (436kJ) ΔH= 6BE(C-H) BE(C-C) BE(C=C) = 6 (413kJ) (348kJ) (614kJ) 3C(g) + 6H (g) ΔH f = ( )kj= 19 kj 36

35 example ΔH f CH 3 CH=CH 2 (g) ΔH f : 3C(gr) + 3H 2 (g) CH 3 CH=CH 2 (g) 3ΔH sub (C)=3(717) kj ΔH gas phase : 3C(g) + 3H 2 (g) CH 3 CHCH 2 (g) ΔH=3BE(H 2 ) = 3 (436kJ) ΔH= 6BE(C-H) BE(C-C) BE(C=C) = 6 (413kJ) (348kJ) (614kJ) 3C(g) + 6H (g) ΔH f = ( )kj= 19 kj 37

36 example ΔH f CH 3 OCH 3 (g) ΔH f : 2C(gr) + 3H 2 (g) + ½O 2 (g) CH 3 OCH 3 (g) 2ΔH sub (C)=2(717) kj ΔH gas phase : 2C(g) + 3H 2 (g) + ½O 2 (g) CH 3 OCH 3 (g) ΔH=3BE(H 2 ) +½BE(O 2 ) = 3 (436kJ)+ ½ (495kJ) ΔH= 6BE(C-H) 2BE(C-O) = 6 (413kJ) 2 (358kJ) 2C(g) + 6H (g) + O (g) ΔH f = ( ½ )kj= 204 kj 38

37 bond enthalpy vs bond energy often [mis]used interchangeably Usually both meant to mean bond enthalpy bond enthalpy: thermodynamic heat measured at const P bond energy: the bond strength from quantum mechanical calculation can be interconverted by the ΔH= ΔU+ Δn gas RT relation (p. 68; example problem 4.1 for O-H bond bond energy= 461 kj mol -1 vs bond enthalpy=463.5 kj mol -1 ) Table 4.3 E&R is weird (hard to read) 39

38 table 4.3 E&R 40

39 topics for thermochemistry, parts of Ch. 4 E&R Calculate H reaction Hess s Law, standard heats of formation H reaction vs U reaction Temperature (and pressure) dependence of H reaction Calorimetry Heats of solution H reaction from bond enthalpies MIDTERM 1 41

40 end of thermochemistry section!!! on to the 2 nd Law 42

41 E&R prob 4.20b HW3 #16 [for part b. use Appendix A (4.1) only; no peeking at A(4.2)!!] combustion: C 2 H 5 OH( ) + 3O 2 (g) 2CO 2 (g) + 3H 2 O( ) a. q U H?? (assume pure liquids, gases 1 bar partial pressure) 0 combustion 43

42 E&R prob 4.20b HW3 #16 (cont) C 2 H 5 OH( ) + 3O 2 (g) 2CO 2 (g) + 3H 2 O( ) 0 : H f 0 H comb = H f EtOH 0 ( ) kj mol kj kj [(-1 mol)h f ( EtOH) + (-3 mol) (0) + (2 mol) ( ) + (3 mol)( )] mol mol SOLVE FOR H f (EtOH) H ( ) = [ + (2) (-393.5) + 3 (-285.8) - H ] kj mol f EtOH combustion H ( ) = kj mol 0 1 f EtOH 44

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44 E&R prob 4.20b HW3 #16 46

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47 standard states and standard heats of formation standard state (º): gas partial pressure 1 bar liquid or solid pure substance at 1 bar solute in soln 1 M standard molar heat of formation (H f º): (ΔH T º) reaction where 1 mole of substance is produce from elements in their most stable form at given temperature 0 H f at 298K in kj/mol C(gr)=0 O (g)=0 C(dia)=1.89 H O(g)= H O( )= F (g)=0 Cl (g)=0 I (g)=62.4 I (s)=