Difference of 2 kj per mole of propane! E = kj

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1 Ethaly, H Fr rcesses measured uder cstat ressure cditi, the heat the reacti is q. E = q + w = q P ext V he subscrit remids is that the heat measured is uder cstat ressure cditi. hermdyamics Slve r q q = E + P ext V Chage i Ethaly, H he heat measured uder cstat ressure cditi is the sum the chage i iteral eergy lus PV wrk. 2 Ethaly, H At this it, we deie Ethaly, H. H = E + PV H = E + (PV) = E + PV + VP 0 Fr a cstat ressure rcess, VP =0 H = E + PV H = q Nw we ca relate H t E. he dierece is PV wrk! ecall this quatity is q rm the revius slide. Ethaly, H Hw big is the dierece betwee E ad H? Let s cmare the cmbusti rae gas, C 3 H 8. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) Dierece 2 kj er mle rae! 6 mles gas 7 mles gas Cstat Vlume N PV wrk Eergy released is 2045 kj er mle rae E = kj Cstat Pressure here is PV wrk Eergy released is 2043 kj er mle rae H = kj H = E + PV = PV PV = 2 kj er mle rae H = E + PV he system des 2 kj/mle wrk i the cstat ressure exasi rm 6 mles gas t 7 mles gas. 3 4 Ethaly, H H = E + PV Ethaly, H H = E + PV H is a state ucti. his meas H is ath ideedet. H reresets the amut heat released/absrbed at cstat P r a reacti as writte by the balaced chemical equati. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) I the reacti is carried ut with 0.5 mle rae, the (0.5)(-2043 kj) = kj r 1022 kj eergy is realeased. Physical states ad rducts must be seciied as slid (s), liquid (l), gas (g), aqueus (aq) whe ethaly chages are rerted. H = kj Ethaly chage r a reacti is equal i magitude but site i sig t the H r the reverse reacti. Ethaly is a extesive rerty. 3 CO 2 (g) + 4 H 2 O (g) C 3 H 8 (g) + 5 O 2 (g) H = kj 5 Mre eergy H r the cmbusti rae is released r reacti (ii) H = kj whe H 2 O is rduced as a gas H = kj whe H 2 O is rduced ad cdesed t a liquid (i) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) (ii) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) H = kj H = kj he dierece betwee the tw H r the abve reactis is 176 kj eergy. It requires 176 kj eergy t cvert 4 mles water rm a liquid t a gas. Less eergy H 2 O (l) H 2 O (g) H = 44.0 kj 4 H 2 O (l) 4 H 2 O (g) H = 4 (44.0) = 176 kj released r reacti (i) because 176 kj is absrbed t cvert water t steam. 6 1

2 Ethaly, H H = E + PV Ethaly, H H = E + PV Ethaly eacti: H = H ial H iitial = H rducts H Fr may reactis, the H the reacti cat be measured directly. Hess Law We ca g hme, r we must thik usig alterative ath t attai the H the reacti that is iterest. Whe a reacti is carried ut i a series stes, H r the reacti will be equal t the sum the ethaly chages r each ste. Path 1 is the direct ath. Path 2 is the idirect ath. H > 0 whe H rducts > H Edthermic H < 0 whe H rducts < H Exthermic Get used t seeig reacti t writte rm let t right! A C Ha C F Hb H = Ha + Hb 7 8 Ethaly, H Ethaly, H Hess Law i acti r Exerimet 7 Examle: H hydrati Na 2 CO 3 (s) + 10 H 2 O (l) Na 2 CO 3 10 H 2 O (s) Path 1 he cmlete cmbusti ctae i a autmbile egie yields carb dixide ad water. Direct ath cat be de! H a Path 1 is the direct ath. Path 2 is the idirect ath. Path 2 Na 2 CO 3 (aq) + 10 H 2 O (l) H b Na 2 CO 3 (s) + 10 H 2 O (l) Na 2 CO 3 (aq) + 10 H 2 O (l) Na 2 CO 3 (aq) + 10 H 2 O (l) Na 2 CO 3 10 H 2 O (s) H hydrati = H a + (-H b ) Fid a alterative ath! H a - H b (i) C 8 H 18 (l) + 25 / 2 O 2 (g) 8 CO 2 (g) + 9 H 2 O (l) H 1 = kcal Health icials are ccered abut a side-reacti which uts a dagerus llutat, carb mxide, it the autmbile exhaust. (ii) C 8 H 18 (l) + 17 / 2 O 2 (g) 8 CO (g) + 9 H 2 O (l) H 2 =? H 2 cat be measured directly sice reacti (i) will always ccur simultaeusly t sme extet. Hw ca H 2 be determied? What might be als iterestig is hw much eergy yu are lsig i the uel i yur car is t udergig cmlete cmbusti Ethaly, H Examle: What is the heat reacti r (ii) C 8 H 18 (l) + 17 / 2 O 2 (g) 8 CO (g) + 9 H 2 O (l) H 2 =? Direct ath cat be de! H 1 = kcal C 8 H 18 (l) + 25 / 2 O 2 (g) 8 CO 2 (g) + 9 H 2 O (l) H 2 =? H 2 = H 1 -(H 3 ) = H 1 -(8 H 4 ) = (-67.6) = kcal/mle H 3 8 CO (g) + 9 H 2 O (l) + 4 O 2 (g) H 3 is the cmbusti carb mxide Dable! CO (g) + ½ O 2 (g) CO 2 (g) H 4 = kcal Fid a alterative ath! Eergetically, r every mle ctae that uderges icmlete cmbusti, ( / )100 = 58.5% yu ly get 58.5% the eergy. 11 Examle: Calculate the heat cmbusti carb t rm carb mxide, CO (i) C (s) + 1 / 2 O 2 (g) CO (g) H =? Give that (ii) C (s) + O 2 (g) CO 2 (g) H = kj (iii) CO (g) + 1 / 2 O 2 (g) CO 2 (g) H = kj Aswer: eacti (i) is t dable, but the alterate aths are reactis (ii) ad (iii). earrage reactis (ii) ad (iii) t give reacti (i). Add reacti (ii) t the reverse (iii) (ii) C (s) + O 2 (g) CO 2 (g) H = kj + (iii) CO 2 (g) CO (g) + 1 / 2 O 2 (g) H = kj C(s) + 1 / 2 O 2 (g) CO (g) H = kj 2

3 H ad Heat Caacity H ad Heat Caacity q = C ecall rm the discussi heat caacity, i a islated system ctaiig Substace A q A where A is the umber mles substace A = 2 1 ( 2 = ial temerature, 1 = iitial temerature) Itrduce a rrtiality cstat, C A, such that q = A C A C A is the mlar heat caacity; it is characteristic the substace. Mlar heat is the quatity heat ecessary t raise the temerature e mle the material by e degree Celsius. 13 Fr a cstat ressure rcess, Sice H = q, q = C C H Fr iiitesimal it i chages, dh C d dh C d Itegrate, ad assume that C is cstat ver 1 t 2. 2H dh Cd 1H 1 H H 2 H 1 C( 2 1) H C ( 2 1) 2 he subscrit remids is that: 1. heat measured is uder cstat ressure cditi 2. heat caacity used i the equati is the heat caacity determied uder cstat ressure cditi. Whe C varies with temerature, it is exressed as a series exasi. C = a + b + c 2 + where a, b, ad c are cstats determied emirically. 14 H ad Heat Caacity q = C Whe C is t cstat ver 1 t 2, C cat be take ut the Itegral sig. Whe C varies with temerature, it is exressed as a series exasi. C = a + b + c 2 + where a, b, ad c are cstats determied emirically. 22H dh Cd 11H 22H 2dH (a b c )d 11H H a d b d c d b c H a( 2 1) 2( 2 1) 3( 2 1) Fr examle, C water vaur varies with temerature, C (H 2 O, g) = x x emerature is i Kelvi! 15 H ad Heat Caacity: elate C t C v ecall rm the deiiti ethaly, H, Fr iiitesimal chages, dh = de + d(pv) where Substitute dh, de, ad d(pv), we have C d = C v d + d H = E + PV dh=c d de=c v d d(pv)=d C d = C v d + d (C -C v )d = d 22 (C C ) d d v 1 1 (C C v )= C -C v = Itegrate, t ad assumig that t C ad C v are ideedet temerature ver 1 ad 2. C is greater tha C v by the amut, the gas cstat. his is the PV wrk assciated with maitaiig the system at cstat ressure. Fr matmic ideal gases, the mlar heat caacity C = Cv + = 3 / 2 + = 5 / 2 = J deg -1 mle Stadard Ethaly Frmati, H he ethaly chage whe e mle the cmud is rmed rm the elemets i their stable rms at 25 C ad 1 atm. ables H are available r varius cmuds. Examle: he stadard ethaly rmati r CO 2 is kcal/mle. What that meas is, C (s, grahite) + O 2 (g) CO 2 (g) H (CO 2, g) = kcal/mle Stadard Ethaly Frmati, H Ethaly rmati a substace ca be calculated rm kwig the ethaly ther reactis. Examle: Calculate the ethaly rmati ethal. Give (i) he cmlete cmbusti ethal is H = kj/mle (ii) he decmsiti water is H = kj/mle (iii) he heat rmati CO 2 (g) is H = kj/mle he reactis crresdig t the abve heats reacti are: I geeral, H meas the chage i ethaly uder stadard cditis 25 C ad 1 atm. By deiiti, the ethaly rmati ay elemet i its mst thermdyamically stable rm is zer. H (O 2, g) = 0 H (Na, s) = 0 H (Br 2, l) = 0 etc. (i) C 2 H 5 OH (l) + 3 O 2 (g) 2 CO 2 (g) + 3 H 2 O (l) H = kj/mle (ii) H 2 O (l) H 2 (g) + 1 / 2 O 2 (g) H = kj/mle (iii) C (grahite) + O 2 (g ) CO 2 (g) H = kj/mle he reacti that we are lkig r is 2 C (grahite) + ½ O 2 (g) + 3 H 2 (g) C 2 H 5 OH (l) H =? everse (i) + 3(everse (ii)) + 2(iii) 17 Check that H = kj/mle 18 3

4 Use Hess Law t calculate H Use Hess Law t calculate H Hess' Law states that H r a reacti ca be ud idirectly by summig H values r ay set reactis which sum t the desired reacti. H reacti rducts rducts H ( rducts) H ( ) As a result, we ca use the tabulated Heats Frmati t calculate H. H reacti rducts rducts H ( rducts) H ( ) he ethaly a reacti is the sum the ethaly rmati each rduct multilied by its stichimetric ceiciet i the balaced equati subtracted rm the ethaly rmati each reactat multilied by its stichimetric ceiciet i the balaced equati. Examle Determie the heat reacti r ZO (s) + 2 HCl (g) ZCl 2 (s) + H 2 O (l) H =? Give, H (ZO) =-83.2 kcal/mle H (HCl) = kcal/mle H (ZCl 2 ) = kcal/mle H (H 2 O) = kcal/mle H = H (ZCl 2 ) + H (H 2 O) (H (ZO) + 2 (H (HCl)) = ( (-22.1)) = kcal/mle Usig Bd Ethaly t determie H Usig Bd Ethaly t determie H he bd ethaly (r bd eergy) is the eergy required t break a chemical bd. It is usually exressed i uits kcal/mle r kj/mle, measured at 298 K. he ethaly dissciati the O-H bd i a water mlecule i the gas hase is H 2 O (g) H (g) + OH (g) H = 120. kcal he exact bd ethaly a articular chemical bd deeds u the mlecular evirmet i which the bd exists. Bd ethaly values give i chemical data bks are averaged values. he ethaly dissciati the O-H bd i the hydrxyl radical i the gas hase is OH (g) O (g) + H (g) H = 101 kcal he bd eergy is deied as the average the tw values. Bd eergy (O-H) = ½ ( ) = 111 kcal/mle Usig Bd Ethaly t determie H Fr the diatmic elemet, H 2, the bd eergy is H 2 (g) H (g) + H (g) H = 104 kcal everse the rcess, we get the Heat Bd Frmati H (g) + H (g) H 2 (g) H = -104 kcal H>0 r bd dissciati H<0 r bd rmati he Heat Bd Frmati is the ethaly chage whe e mle a articular tye bd is rmed rm gaseus atms. Cl (g) + Cl (g) Cl 2 (g) H = -58 kcal H (g) + Cl (g) HCl (g) H = -103 kcal Usig Bd Ethaly t determie H he reacti betwee hydrge gas, H 2 (g), ad chlrie gas, Cl 2 (g), rms HCl (g). he heat reacti H 2 (g) + Cl 2 (g) 2 HCl (g) H = -44 kcal Csider the abve reacti rm bd breakig ad bd rmati. ΔH = ΔH (bds breakig) - ΔH (bds rmig) Bd breakig: H 2 (g) H (g) + H (g) H = 104 kcal Cl 2 (g) Cl (g) + Cl (g) H = 58 kcal Bd rmig: 2 H (g) + 2 Cl (g) 2 HCl (g) H = 2 (-103) = -206 kcal ΔH = ΔH (bds breakig) - ΔH (bds rmig) ΔH = ΔH = -44 kcal

5 Usig Bd Ethaly t determie H Usig Bd Ethaly t determie H Mea (sigle) Bd Eergies at 298K (kcal er mle bds) H C N O F S Cl Br I I Br Cl S F O N C H 104 Mea (multile) Bd Eergies at 298K (kcal er mle bds) C=C; 147 C=N; 147 C=O; 192 N=N; 100 C C; 200 C N; 210 C O; 256 N N; 226 O=O; 119 S=S; 100 S=O; 125 Bd ethaly values: 1. Sigle Bds: geerally, H < 100 kcal/mle eg sigle carb-carb bd H = 83 kcal/mle 2. Duble Bds: geerally, H < 200 kcal/mle eg duble carb-carb bd H = 147 kcal/mle A duble bd is usually t as strg as 2 sigle bds. 3. rile Bds: geerally, H < 300 kcal/mle eg trile carb-carb bd H = 200 kcal/mle A trile bd is usually t as strg as 3 sigle bds. 4. Usually sigle bds with dieret atms are strger tha thse betwee like atms. eg X-Y bd is strger tha X-X ad Y-Y bds Usig Bd Ethaly t determie H Usig Bd Ethaly t determie H Fr a reacti that is exthermic, H < 0 ΔH = ΔH (bds breakig) - ΔH (bds rmig) weak bds strg bds Fr a reacti that is edthermic, H > 0 Examle Fid H r the llwig reacti rm bd eergies 2 H 2 (g) + O 2 (g) 2 H 2 O (g) H =? Fr the llwig reacti, strg bds weak bds H 2 (g) + Cl 2 (g) 2 HCl (g) H = -44 kcal he reacti ges rm with weaker bds t the rduct that is rmed with a strger bd. 27 Bd breakig: 2 H-H H = 2 (104) = 208 kcal 1 O=O H = 119 kcal Bd rmig: 4 O-H H = 4 (-111) = -444 kcal ΔH = ΔH (bds breakig) - ΔH (bds rmig) ΔH = ΔH = -117 kcal Or, Kee the sigs assciated with bd breakig (+) ad bd rmig (-) ad simly add them u t get ΔH. Eg: ΔH= ( )+(-444)=-117 kcal I yu use this rmula, remve the sigs assciated with bd breakig ad bd rmig. 28 Usig Bd Ethaly t determie H ΔH = ΔH (bds breakig) - ΔH (bds rmig) Usig Bd Ethaly t determie H ΔH = ΔH (bds breakig) - ΔH (bds rmig) Examle Fid H r the cmbusti rae rm bd eergies Examle Shw, by bd eergies, that CH 3 CH 2 CH 3 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) H =? C (grahite) + ½ O 2 (g) CO (g) H = kcal Bd breakig: 2 C-C H = 283 = 166 kcal 8 C-H H = 899 = 792 kcal 5O=O O H = 5119= 595 kcal Bd rmig: 8 O-H H = 8(-111) = -888 kcal 6 C=O H = 6(-192) = kcal ΔH = ΔH (bds breakig) - ΔH (bds rmig) ΔH = ΔH = -487 kcal (-2038 kj) Cmared t the measured heat cmbusti rae H = kj 29 Heat varizati carb is kcal/mle Bd rmig: 1 C=O H = -256 = -256 kcal Bd breakig: C (grahite) C (g) H = kcal ½ O=O H = ½ 119= 59.5 kcal ΔH = ΔH (bds breakig) - ΔH (bds rmig) ΔH = ΔH = -25 kcal Bd ethaly calculati the reacti is clse t the rerted H value. 30 5

6 Variati H with emerature Suse we wat t determie the ethaly chage at -stadard emerature,. ( is the stadard temerature) H eactats at Prducts at H C d H 1 H H 2 C d 1eactats at Prducts at H I >, the H 1 is the heat ecessary t warm the rm t at cstat ressure. H 2 is the rducts udergig the reverse temerature chage. H = H 1 + H + H 2 H = C + H d + C rducts d rducts231 Variati H with emerature Examle (a) Calculate the ethaly chage i the rmati 1 mle ethal at 25 C by the reacti shw. hermdyamic data are give belw r each substace. H C 2 H 4 (g) + H 2 O (l) C 2 H 5 OH (l) Aswer: (a) H (kcal/mle) C (cal/mle l deg) H = H (ethal) - ( H (ethee) + H (water)) = ( ( )) = kcal/mle H = H + C C d rducts 32 Variati H with emerature H (kcal/mle) C (cal/mle deg) C 2 H 4 (g) + H 2 O (l) C 2 H 5 OH (l) (b) What is the H whe the are iitially at 15.0 C ad the rduct, at the ed the reacti, is at 100. C? H eactats at 15.0 C Prducts at 100 C H 28 C d H 298C 110H 1 ( 1H )(. cal 0. ) H 1 H 2 eactats at 25.0 C Prducts at 25.0 C H (rm (a)) = kcal/mle 226H 2 ( H 2. )( rducts18cal d 0. ) he Clausius-Claeyr Equati he varizati curves mst liquids have similar shae. he var ressure icrease as the temerature icreases. = b e a Exerimets shwed that the ressure P, ethaly varizati, H varizati, ad temerature are related. his is the Clausius-Claeyr equati. P A e H varizati where is the gas cstat ad A is a cstat r the liquid-gas system. H = H 1 + H + H 2 H = H = kcal he Clausius-Claeyr Equati = b e a Cmare k = b e a he Clausius-Claeyr Equati = b e a he bilig its a ure cmud was measured at tw dieret ressures. Estimate the heat varizati the cmud. (K) P (kpa) P A e H varizati k Ae Ea 211 H 1 va 1k 2 E 1 a k1 P A e H varizati 1 H 2 va Check yur aswer: H va = 74.4 kj/mle Clausius-Claeyr Equati Arrheius Equati he Clausius-Claeyr equati allws us t estimate the var ressure (P 2 ) at ather temerature ( 2 ), i the var ressure (P 1 ) is kw at sme temerature ( 1 ), ad i the ethaly varizati, H va, is kw. I the hase trasiti is rm slid t gas, the ethaly sublimati. (H sub ) is the chage i ethaly whe e mle slid varizes t rm e mle gas. 1 2 H sub

7 he Clausius-Claeyr Equati Cmare he Clausius-Claeyr Equati he var ressures ice at tw temeratures are summarized belw. Estimate the heat sublimati ice. Sle = - H va / Arrheius Plt Sle = -E a / (K) P (trr) 760 trr = 1 atm k trr = atm trr = atm /, K 211 H 1 va 1k 2 E 1 a Clausius-Claeyr Equati Arrheius Equati k1 2 H 1H sub Check yur aswer: H sub = J ml -1 he Clausius-Claeyr equati allws us t estimate the var ressure (P 2 ) at ather temerature ( 2 ), i the var ressure (P 1 ) is kw at sme temerature ( 1 ), ad i the ethaly varizati, H va, is kw Summary: Chage i ethaly a system, H, ca be determied by: he subscrit remids us H = q that the heat measured is elatig H t E, uder cstat ressure cditi. H = E + PV I terms heat caacity, assumig that C is cstat 1 ad 2. H = C ( 2-1 ) Fr a matmic gas, C v = 3 / 2 Hess Law H reacti rducts rducts C = + C v H ( rducts) H ( ) Clausius-Claeyr equati 1 H 2 va