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Z* H  0޽h ? ̙3380___PPT10.T[04 0 E=0(    0p,$D  0 ( Chapter 2 Chemical Thermodynamics >) 2$&$0  0u ,$D 0 W#Chemical reactions: (1) Whether a reaction will occur or not (2) How far it will get, the equilibrium Thermodynamics (3)Heat of Reaction -exothermic /endothermic (4) Its rate ---- Kinetics (5) The relationship between substance compositions, structure and its properties (2U 2$$$$$ $ ;$$$$$! $ $$$($( ,$,S0$044  0`,$D 0H  0޽h ? ̙33I A ___PPT10! +UD'  = @B DP' = @BA?%,( < +O%,( < +D' =%(D' =%(DR' =A@BBB%B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B%barn(outVertical)*<3<*DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D4' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(+p+0+0 ++0+0 +  0 @$M(  $ $ 0&@@ 3Two issues: Heat of reaction Direction of reaction f 2' 2 2 $$  H $ 0޽h ? ̙33` 0 P(  h  0P0  1 Thermochemistry -N_S^p ㉳Q`(ur`Qpev9eSϑ%U0%Hegaϑ/{S^p `W,gi_8^(u/g  pRf[,{N[_ S^pv[IN Vy{ell 2{(2& _`ZH  0޽h ? ̙33I 0 `((  (Q ( 0X`"  _1 Thermochemistry 1-1 Concepts 1-2 The first law of thermodynamics 1-3 Heat of reaction L_(2 2& L&"NH ( 0޽h ? ̙33  0 '0(  0r 0 S PPp    k p  '0 #"" 0 0 ?pz  K  @` 0 0 ?z p Z & "  @`  0 0 ?z  pisolated system .  @`  0 0 ?p z  L"  @`  0 0* ? pz  Z & "  @`  0 0# ? z  m close system .   @`  0 0X" ?ph  L"  @` 0 0F ?hp  \" & "  @` 0 0\R ?h  l open system .   @` 0 0[ ?ph ME"  @` 0 0`d ?ph Nm "  @` 0 0pm ?h R  @``B 0 0o ?ZB 0 s *1 ?hhZB 0 s *1 ?  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(2 2&(H < 0޽h ? ̙332 0 @r(  @: @ 0H P- "(4) Process and path Process means how a system changes from one state to another. There are different kinds of process, such as these below: Isothermal process Isobaric process Isovolume process Adiabatic process Cyclic process Path refers to the different way to complete a process. z(29 2&&&3 cH @ 0޽h ? ̙33C 0 ]UD(  D' D 0  h e(5) Internal energy The sum of the energy in a system is called internal energy, its symbol is U. F(2O 2&M D 0 { ,$D 0 VWhat is the relationship among %U and Q0W? +(2 2&&  &H D 0޽h ? ̙33___PPT10f+ x D' E = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*D%(D' =-6B'blinds(horizontal)*<3<*D+8+0+D0 +J  0 d\H(  H H 0  `,$D 0 d$1-2 The First Law of Thermodynamics $% 2# N H 0 P W ,$D 0 initial state ! final state U1 U2 % U = U2 -U1 = Q  W :(2 2&  & . & . &              f H 0  p,$D 0 The sign rule: When the system is endothermic, Q is positive +, When the system do work on the surroundings, W is positive +. 6(2B$=H H 0޽h ? ̙33___PPT10f+_D' E = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*H%(D' =-6B'blinds(horizontal)*<3<*H+8+0+H0 +X 0 L(  L` L 00    hThe tragic life of the First Law s discoverer.----After studying the work habits, food intake and color of the blood of sailors in the tropics compared with those in northern Europe, the young German doctor I. R. Von Mayer concluded that amount of energy in food is used both to heat the body and to do work, and thus heat and work represent energy in different forms. When he published these ideas, they were ridiculed, and Von Mayer became despondent. Soon thereafter, James Joule , an English brewer and amateur scientist, demonstrated the equivalence of heat and work experimentally but gave Von Mayer no credit for the idea& & Finally Von Mayer received recognition for his great insight. | 2- H L 0޽h ? ̙33L  0   \ (  \L  ! \#   BB \ 3 jJ  !BB \ 3 jJ !1!BB \ 3 jJ11!BB \ 3 jJ 1f > b  \S @ABC S  \ @AB-C2DE4FjJ -^ J!l2-Y@ 3"`> k A   \ @AB:C DE4FjJ Au;":::D; BZ.@ 3"`HXb   \ p@ABCEFjJ@3"`\ H  \ p@ABCEFjJ@"`\ WBB  \ 3 jJ= X= BB  \ 3 jJ= XuXBB \B 3 jJ 8 p BB \ 3 jJ 8 p BB \ 3 jJp BB \ 3 jJ)BB \ 3 jJ}}BB \ 3 jJ} \ Z  ??   _ vaccum  \ T  ??@` d&q3[_[ňn 2 d \ Th  ?? p  A water bath 2 H \ 0޽h ? ̙33" 0 `b(  `* ` 0X P@  <Sb_e 󁾏s^a0T4lneSS sS Q = 0 sSlSOMRT)n^*gS S 5"W = 0 4" %U = Q W = 0 sSlSO-NQ NS sSU = f (T) C4l_'Y 4"_\v>epNKmϑ FO[flSOSegvP\ N_cknx0S_P ! 0e N_[hQcknx sS:t`lSOvQUN:N)n^TvQpe0 (2 2 $,H$$,2$$f$f      H ` 0޽h ? ̙33 0 PD(  P  P 0.  1-3 Heat of reaction 1-3-1 Definition Considering a closed system which do volume work only, when the Treactants = Tproducts the absorbing heat / giving off heat is called heat of reaction. )(2 2# & & &&A D>h  CH P 0޽h ? ̙33< 0 T|(  TD T 0A    > 1 Quantity of heat at constant volume, Qv %V = 0 W = P%V = 0 %U=Qv Its physical meaning is that in a constant volume reaction, the heat that the system absorbed is totally used to increase its internal energy. (2 2&.&  && & $$(&(,,0&00&044h%H T 0޽h ? ̙33  0 N F X(  X X 0X  F82 Quantity of heat at constant pressure, Qp Psurroundings = P1 = P2 W = Psurroundings%V %U = Qp P%V Qp =%U + P%V = U2 U1 + P (V2 V1) = U2 + PV2 (U1 + PV1) Supposing H a" U + PV which is called enthalpy(q). There are some explanations on enthalpy. `$ It has no clear physical meaning, just for convenience a$ Its absolute value can not be obtained. b$ For ideal gas,its enthalpy obey H = f (T). o(2. 2$0$  $ ,$,$,$ , $ $$($((,(,,0$0448$8<<$$,  $,$,$,$ $  ,  $  , $$($((,(($((,( ($(($(,,0$0448$8 8$8<<$,$8$  )$,'    H X 0޽h ? ̙33 0 ZR d(  d d 0 `0 RH = U +PV TN[ %U = 0 PV = constant %(PV) = 0 4" %H = %U + %(PV) = 0 H = f (T) Qp =H2 -H1=%H It means that at constant pressure, the heat that the system absorbed is totally used to increase its enthalpy. (2 2 &&   && & $$(&(,,0&00&00.00&00.00&0448&88.88&8<<&&D +H d 0޽h ? ̙33 0 0hT(  h h 0D ` 3 Relation between Qp and Qv: `$constant pressure %H1 %U1 ! products`! n20P10V20T reactants n10P10V10T b$ ! %H 3 %U3 a$constant volume %H2 %U2 ! productsa! n20P20V10T @(2 2$ $  ,  $  ,  $ &&.  $&$$.$((,&,,&,004&44.488<&<<.<&. &   &  . &.&.  $&$$&$((((,&,004&44.488<&<<.<<&<& &  . &..&&  $&$$.$((,&,,.,004&44.488<&<0' h 0 ,$D 0 5"%H1 = %H2 + %H3 = %U2 + %(PV)2 + %H3 = %U2 + %(PV)2 = %U2 + P%V + %PV 4" %H1 = %U2 + %PV %H1 = %U2 + %n(products reactants)RT 4" Qp = Qv + %n(products reactants)RT (2( 2$,$ $  ,  $ $,$$,$  $$$$,$$$$((,$,,,,,$,004$44,44$488<$<<,<<$<$,$ $ $$  $$$((,$,,,,,$,004$44,44$488<$<$ $  ,  $ $,$$ , $,$((,$,000 0000 0000 000VH h 0޽h ? ̙33___PPT10f+ x D' >= @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*h%(D' =-6B'blinds(horizontal)*<3<*h+8+0+h0 +$ 0 @ld(  l\ l 0 T e 1-3-2 Reaction rate of progress and the molar enthalpy change of reaction 1 Reaction rate of progress, A A + B B ! G G + F F t = 0 nA nB nG nF t nA nB nG nF Define unitmol (2 2J&$  $$ ,  $ $$(,(($(,,0$0448,88$8<<,$,$  ,$,$ ,  $  ,  $  ,  $  ,  $  ,  $ $$($(,,0$044J$ R l s *8n l 2A ?? P  H l 0޽h ? ̙33 0 "  Pp (  pT p c $x ` ,$ 0 Example Put 10mol N2 (g) and 20mol H2 (g) together when 2mol NH3(g) were produced, calculate the reaction rate of progress according to the following two equations. (a) N2(g) + 3H2(g) ! 2NH3 (g) (b) 1/2 N2(g) + 3/2 H2(g) ! NH3(g)(2$ $,$, $  $,n$,$,$$,$, $,$ $  ,  $ $$ p c $@ @,$D 0 Solution (a) N2(g) + 3H2(g) ! 2NH3 (g) (b) 1/2 N2(g) + 3/2 H2(g) ! NH3(g) t = 0 10 20 0 10 20 0 t 9 17 2 9 17 2 a = (2 0) / 2 = 1(mol) b = (2 0) / 1 = 2 (mol) }(2 2  "*"*"  "*"* "*""*"# " $$("(,,0"044H p 0޽h ? ̙33___PPT10b+l D'  = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DP' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*p%(D' =-6B#blinds(vertical)*<3<*p+8+0+p0 + 0 `t((  t t 0 P  2 The molar enthalpy change of reaction When= 1mol its unit is KJ/mol. %rHm = %rH / It is obviously that %rHm is corresponding to reaction equation. The %rHm will mean nothing if there is no corresponding reaction equation. (2 2&* & &&  $.$$&$$.$$&$((,.,,&,004&488<.<<&<<.<+<&<.&.C&1-CH t 0޽h ? ̙33 0 ZRpx(  x x 0\  R(3) Thermodynamic standard statestandard state of substance U0H0G0S/fr`Qpe TN|~v N Tr` ^wQ g N Tvpe

+B#style.visibility<*|%(D' =-6B'blinds(horizontal)*<3<*|+8+0+|0 +p  0   (  x  0M  `  `$ The heat of reactions has connections with the physical state of substances. Hence, we must specify that state in writing a thermochemical equation. For a pure substance, the symbols(s), (l), or (g) are written after the formula. For a species in aqueous solution, the symbols (aq) is used. a$ The value of %H depends, at least slight, upon the temperature at which a reaction is carried out. Hence, that temperature should be specified. Unless indicated otherwise, the temperature is 25!. b$ The magnitude of %rHmis directly proportional to the amount of reactant or product. %H for a reaction is equal in magnitude but opposite in sign to %H for the reverse reaction. That is, heat of the reverse reaction = -heat of the reaction.(2$$$$$$Q$$ $ $ $$  $$$((,,,,$,,,,00?4$488?<$<E$$H|H  0޽h ? ̙33 0 8(    0L{ @`C S^pQp (Wpe

+B#style.visibility<*%(D' =-6B#blinds(vertical)*<3<*+8+0+0 + 0 (    c $d] ,$D 0 xExample  (1) CH3COOH(l) + 2O2(g) = 2CO2(g.) + 2H2O(l) %rHm (1) = 874.5 KJ /mol (2) C(graphite) + O2g = CO2(g) %rHm (2) = 393.5 KJ /mol (3) 2H2(g) + O2(g) = 2H2O(l) %rHm (3) = 571.6 KJ /mol Calculate%rHm (4) of the following reaction: (4) 2C(graphite + 2H2(g) + O2(g) = CH3COOH(l).  (2/ 20(2$$&. &. &. &.&  .&.&.&! &  . $$(&(,,0&00.00&0448.88&88.8<<&.&&.&.&.&  .&.&.& &  $$$$$$$$$$$$$$(&(,,,,,,,,,,,, ,,/"D Rl  c $P$,$D 0 "Solution 5"(4) =(2)2 + (3) (1) %rHm4 =2%rHm2 +%rHm3 %rHm1 = 393.5 2 571.6 ( 874.5) = 484.1 (KJ /mol)x(2    ( ( ( $$( (,,0(00 00(044448(8<< ( (   (( ( ( $$(((( (,, 0 0448 8<<    & H  0޽h ? ̙33___PPT10f+ x D' &= @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*+8+0+0 + 0 H(    0P@ ,2 Heats of formation " 2$  00,$D 0 pnIt is prescribed that at standard state1atm and 298K , the enthalpy of the most stable form of element equals to zero. Eg. O2(g) ,HO2 = 0; Br2(l) , HBr2 = 0; Cgraphite , HC = 0|y(2?(2!'$ $  G$,$, $,$,$$ $  ,  $ $$b' B*   0 | ,$D  0 rjDefinition: Standard molar heat of formation of a compound is equal to the enthalpy change when one mole of compound is formed from the stable elements at 298K and 1atm. %fHm (2! $$ $$  $"H  0޽h ? ̙33___PPT10+#D' = @B DD' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*+8+0+0 + 0 (    c $P    %rHm reactants ! products !"i%fHmreactants !"(i%fHmproducts) ! elements involved in the reaction! %rHm + "i%fHmreactants = "i%fHmproducts %rHm= "i%fHmproducts "i%fHmreactants (2( (    $($$ $$($(( ,(, , ,004 488< <( ( (   !$*"*  $"$((,",004*44"44*488 <*<<"<" *  "  * *"       (*H  0޽h ? ̙33  0 tl (    c $(DP{ 8<uEg. Cgraphite + O2(g) = CO2(g) %rHm = 393.5 KJ /mol %rHm = %fHm(CO2) = "H products "H reactants = HCO2 (g) 0 = 393.5 KJ /mol 2(2 ""  "*"*"*"* " $$("(,,0(00 00(0448 8<<( (( (  ( (   $$*8  c $m ,$D 0 Eg. Cgraphite = Cdiamond %rHm = 1.88 KJ /mol %fHmdiamond = 1.88 KJ /mol JV(2 2 (   ( $$(((( ((((,,0 0448(88 8<<<<( ! H  0޽h ? ̙33___PPT10+'c:DO' &= @B D ' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(+8+0+0 +1  0 0(  x  c $P $Example : H2(g) + I2(g) = 2HI(g) %fHm (KJ /mol) 0 62.438 26.5 Calculate%rHml(2$ $ (  ( ( ( +  ( (,.,   c $p ,$D 0 ^fSolution: %rHm= 26.52 62.438 = 9.438 ( KJ /mol) 4(2 ( (     :   H  0޽h ? ̙33___PPT10+'c:DO' "= @B D ' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(+8+0+0 + 0 @0(    0L %fHm/fN~Nh` N3z[US(vq:N0vTyir(vv[q

  0PP 3 Heats of combustion %cHm = "i %cHmreactants "i %cHm products I 2             j  f  0"PP` F4 estimated by bond energy Bond energy refers to the energy needs to destroy 1 mole gas molecule chemical bond at 298K and 1atm. %rHm = "Ereactants "Eproducts B 2(2 2g& .  &  . && .  $&$$.$$&$((Z  H  0޽h ? ̙33 0 kc`(    0|"P c1-3-5 Measurement of heat flow ( 2H  0޽h ? ̙33Q 0 p@(    T" ??P0P___PPT100(___PPT9h` tt S_  reversible path t`-Nve_ OYePgP 244-246 , t`lSOI{)n :(2 2&3&3&3 & 3&3&3$F    Z!" ??@p j, Y` n=0.646mol T = 298K ~` P1= 16105Pa V1 = 1.010 3m3 ! P2 = 1.0105Pa V2 = 1610 3m3 JM K&3.3&3.3&3.3 &3.3&3.3&3.3 &3.3&3.3&3.3&3.3&3t)   #   T " ?? Y ,$D 0 _1Nek PY = 1.0105Pa W = PY%V = 1.01051510 3 = 1.5103 (J)R(2 $, $,$,$,$, $,$h #>  T8S" ??\ P,$D  0 _2Nek PY = 2.0105Pa PY = 1.0105Pa W = PY%V + PY %V = 2.0105710 3 + 1.0105810 3 = 2.2103 (J) (2 2 $, $,$, $, $,$,$,$, $,$,$,$ $!H  0޽h ? 3fȺdfI A ___PPT10! +G7-D' p"= @B DP' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*+p+0+0 ++0+0 + 0 rj(  d  Txv" ??  &ePYek PY = PQ P` 2 $,$,$6^  T܂" ?? | [INSO|ePY!k0Rs^a` sSkek;RePcяs^a` ُ/f@b g_-NgyrkvNy y:NS_ /ft`v0[ES0RvgP0 D(2D$CH  0޽h ? 3fȺdf  0 R(    T" ??p ~%U = Q W t`lSO I{)n N %U = 0 R Q = W [Oir(vvS H2O(l.) ! H2O(g.) (%rHm (373K) = Qp = Qr = 44.0KJ/mol)g(2#$ $3,3 $3,3 $3,3$3,3 $3,3$3,3$3     T" ?? p ,$D 0 XZI{)nǏ z %rSm = Qr / T irINR`)neSO|vqSI{NS_vp)nq0 (2 2$,$,$,$ @H  0޽h ? 3fȺdf___PPT10f+ x D'  = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*+8+0+0 + 0   f (  x  H"?? p ekpe R103 J S)R 103 J N 1.5 24 N 2.2 12.8 N 2.8 8.0 ePYek 4.4 4.4N(2  ( (  B  ZD??pB  ZD??p B  ZD??  T" ??X ,$D 0 J`$ gPYek | R | < | S)R | sSSO|TsXZPvR < sXTSO|ZPvR sXYu Nu  NS_ 8A(2 2$A$l   T$" ??( ,$D 0 Na$ePYek | R | = | S)R | sS~ǏN!kePbavTS)_sKNT SO|TsXb` YS` l gYu NNUOq_TSS   S_ 8L(2 2L$$Z*H  0޽h ? 3fȺdf___PPT10+qaD'  = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*D' =%(D' =%(DP' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B#blinds(vertical)*<3<*+p+0+0 ++0+0 + 0 &(    T" ??p  jS_vyrp `$ S_/f(WePcяNs^avr` N[b a$ (WS^Ǐ z-N _Segv_ SO|TsXGWV0RSegvr`S'`  b$ R`)ne NUOS_-N SO|[sXZPg'YR sX[SO|ZPg\RN[(u‰p w ُyǏ zg~Nm0Hesgؚ <C(2: 2& 3u$|H  0޽h ? 3fȺdfJ 0 (  R  T" ??P   N0Sf[S^eT -N_Sf[S^SۏLeTv$Rnc ㉳Qel(ur`Qpe9eSϑ%S0%Geg㉳Q `Sf[S^SۏLvq_TV } ; q$Rncd[zSO| ; 1u$RncI{)nI{S N ; :(2 2 $3`*  H  0޽h ? 3fȺdf  0 `(    Tl" ??P@x8___PPT10N___PPT90( "N Q[S^eTvV } ϑSNOD(2 &$3&$3&3Z  T# ?? ,$D 0 xx-NTH+ + OH = H2O qpS^C + O2 = CO2 $NS^%rHm < 0 FO%rHm N\O:N$Rnc NN8TpS^_NSۏL \&(2) 2$,$,$, $,$,$,$, $,$,$ H  0޽h ? 3fȺdf___PPT10f+KD'  = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*+8+0+0 +X 0 r j  (  L  T(# ?? ,$D 0 O: Ba(OH)28H2O(s.) + NH4SCN(s.)!Ba(SCN)2(s.) + 2NH3(g.) + 10H2O(l.) %rHm > 0 CuSO45H2O(s.) ! CuSO4(s.) + 5H2O(l.) (510K) %rHm (298K) = 78.96 KJ/mol NH4HCO3(s.) ! NH3(g.) + H2O(g.) + CO2(g.) (389K) %rHm (298K) = 185.57KJ/mol J(2 2 &.&. &.&. &. &.&.&. &.&. &. &.&.&.&.&. &.&. &.&.&.&PF7Lv  T J# ??P`,$D 0 Rg Nb N*NS^ qQ TpS ! l.0g. sS S^ir-NlSOir(vϑ < ubir-Nir(vϑ0 NS݋ S^SO|-NRP[v;mRVX'YN b;mRV'YvRP[XYN0b_a0W sSSO|vmqN z^'YN0S SO|mqN^S'Y/fS^ۏLvSNR0 V{(2 2j$$ $ N QH  0޽h ? 3fȺdf___PPT10f+ x D'  = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*+8+0+0 + 0 0((    T\# ??@ V)RNSf[S^ۏLv$N*NV } ϑSNO%H < 0 mqN^X'Y%S > 0 ^+(2 2 3,H  0޽h ? 3fȺdf  0 (    Td# ??L xRN q(entrophy) 10q [INcSO|mqN^vr`Qpe (uSh:y0 (2 2$3&$3&33 & 3Pb  Tw# ??_,$D 0 O TI{)nǏ z Y ! ~ S1 S2 %S = S2 S1 e" Q/T 5"Qre"Q 4"%S = Qr/T e" Q/T USMOJ/K &o(2 21   r #  H  0޽h ? 3fȺdf___PPT10f+KD'  = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*+8+0+0 + 0 nf(    TЎ# ??p 3 P___PPT100(___PPT9h` L :NN[ϑir(vqS RUSMO:NJ/mol %SUSMO:NJ/mol.K :N NI{)nǏ z %Sv{lQ_(WirS-N QN_ gsQ reversible path-NvQrSck0S [‰sa_NSO|v_‰r`pe gsQ| }(2 2$$#$ $ $$3$  H  0޽h ? 3fȺdf  0  B(  N  T# ??Pp-   O N*N{USSO| `$ N*N|P[A0B0C S(W N*NMOnKNNQs Svr` P33 = 321 = 6 a$ N*N|P[A0B0C S(WV*NMOnKNNQs Svr` P43 = 432 = 24 b$$N*N|P[A0B S(WV*NMOnKNNQs Svr` P42 = 43 = 12(2($,,*$,,)$,, $  0`# ,$D 0 k`$0a$ |P[;mRV'Y SO|v_‰r`peY ka$0b$ |P[Y SO|v_‰r`peY ~_‰r`peY mqN^'Y sSS'Y0 D2(2 22$$3 $DH  0޽h ? 3fȺdf___PPT10+'c:DO'  = @B D ' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(+8+0+0 + 0 0>(    T# ??  1 S = k3 l3yQf8^pe k = 1.3810 23 J/K _‰r`pe    eϑ~ S   J/K wQRT'` c TNir( S(g) > S(l) > S(s) T{|ir( RP[ϑ'Y/~g YBg S'Y SI2 (g) > SBr2 (g) > SCl2 (g) > SF2 (g) SC4H10 > SC3H8 > SC2H6 > SCH4 SNO2(g) > SNO(g) > SN2(g) (2 2 $3$, $$3$,$,$,$,$,$,$,$,$,$,$,$  fH  0޽h ? 3fȺdf 0 )4Hr(  H  S% H# #" ,$D 0NB H S D< < H  ~BCDEF@<S%NB H S DD% D%B2 H 3 X !aB2 H 3 !a(#B2  H 3 !)(#B2  H 3 aX B2  H 3 #aD%NB  H S D<D%NB  H S D<D%NB H S D! !~T  t" H#  t"NB H S D NB H S DNB H S D B2 H 3 )$ B2 H 3 pa B2 H 3  @ aB2 H 3  @ )B2 H 3  m NB H S DNB H S DNB H S D NB H S D NB H S Dt" t":   'H# #"  @`#,$D 0B2 (H 3  p LN   )H   *H  ~BCeDEFe@#"   NB +H S D  NB ,H S D B2 -H 3 p B2 .H 3   B2 /H 3 @ B2 0H 3  @ NB 1H S D  NB 2H S D  NB 3H S D   NB 4H S D   H H 0޽h ? ̙33~___PPT10^+aDB'  = @B D' = @BA?%,( < +O%,( < +D4' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*'H%(+ 0 @8(    T$ ?? Pz2 pRf[,{ N[_ 0 Ke NUO[tevfSOvSP[bRP[S gNycRb__ sSS gNyX[(Wr` vQqS{Qir((WP = 1.013105PaT)n^Tevq 2 6H  0޽h ? 3fȺdf ___PPT10+OD' $= @B DC' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*Dn' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*+8+0+0 +! 0 / ' P (    c $!$P` 00 <$D 0  $ G a   N ??[,$D 0f  T$$ ??G @"$ 0T___PPT104,___PPT9ld 30ir(vhQq SS^vhQid\qS Standard entrophy of substances Standard molar entrophy change of reactionsa(2&3&3&3  & 3  $ 3$3$3$3>   01$pT,$D 0 h1 hQqir((Wh`P=1.013105 1mol evq

+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*+8+0+0 +  0 `F(    T\$ ??  N2 S^vhQid\qS %rSm = "Sm N "Sm S '(2 2 3 t  Tt$ ??``  2 aA + bB ! dD + eE Sm(J/moL " K) Sm(A) Sm(B) Sm(D) Sm(E) %rSm = dSm(D) + eSm(E) aSm(A) bSm(B) |m(2- 2<  H  0޽h ? 3fȺdfl  0 ~p(    T$ ??,$ 0 4~OBlS^ 2HCl (g) = H2(g) + Cl2(g)vhQid\qS00 n@(23333 3$   T<$ ??PS ,$D 0 2 2HCl (g) = H2(g) + Cl2(g) Sm(J/mol" K) 187 130 223 %rSm = 130 + 223 2187 = 21(J/mol" K)[(2) 2 # V) H  0޽h ? 3fȺdf___PPT10f+ x D'  = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*+8+0+0 +( 0 h(  0  T$ ??p ~S^vqS%rSmNT gsQ0 5"kNir(vSm(T)T! ! S^ir0ubir! 4"vQ]

0 SO|SSSۏLN NSe_ %S = 0 SO|s^aNSe_ pRf[,{N[_(Wd[zSO|vNUOSǏ z-N SO|vq T2 4"%Sd[zSO| > 0 sSpϑ;`/f1uؚ)npnS0W ONO)npn0 d(2 2$$, $,$,$, $,$,$,$,$,$,$,$,$,$,$,$,$ -H  0޽h ? 3fȺdf___PPT10f+ x D' ,= @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*+8+0+0 +!  0 #$w(  x  c $($&PP` `  &  P $ #"&  ,$D 0  NH0&?? P w? < 2cchh  N:&??   i+0 2cchh  NDC&??0  ] +  2chh  NL&?? 0 U4chh  NU&??" P  W?chh   N`&?? "   z >(2cchh   Nj&??0"   ^ "(2chh   Nt&??" 0  U3chh   Nd&??C P"  Wchh   N4&?? C "  z >(2cchh  N&??0C "  \ +"(2chh  N&??C 0"  U2chh  N&??dPC  X"chh  N&?? dC  x+ >(2cchh  NH&??0d C  \ "(2chh  NP&??d0C  U1chh  N`&??Pd | S^SۏLchh  N&?? d |%S >(2cchh  N&??0 d ^%H"(2chh  NH&??0d ^c hhB  Zo ??PB  T1 ??dPdB  T1 ??C PC B  T1 ??" P" B  T1 ?? P B  Zo ??PB  Zo ??B  T1 ??00B   T1 ??  B ! T1 ??B " Zo ??PP # TX& ??P0,$D 0 N<4"N*Nb%rHmN%rSmޏ(WNwvevr`Qpe. 2$,$,$,$,$J H  0޽h ? 3fȺdf___PPT10.+ gDO' ,= @B D ' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*#%(+8+0+#0 += 0   , (    T@) ??  @m$D 0P___PPT100(___PPT9h`  N T^e1u 10Gibbs1uN1u$Rnc c[I{)nI{S N pRf[,{N[_ %S = Qr / T %U = Q WSO W^ F(2 &$3&$3&$3 & $3 &$3&$3%$,$,$|    TC& ??p ,$D 0 HQ = %U + P%V + W^ = %H + W^ b%(2$,$,$$ (  TU) ??@ P ,$D 0 pHQr e" Q = %H + W^ T%S e" %H + W^ (2 2$3,3$3 $,$$ $,$$6l  Tb) ??P ,$D  0 $ (%H T%S) e" W^ 8 2$, 4 H  0޽h ? 3fȺdfI A ___PPT10! +G7-D' p)= @B DP' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*+p+0+0 ++0+0 + 0 | t  (  "  TPv) ?? " (%H T%S) e" W^. 2$,* h  TT) ??0y,$D 0 j [(H2 H1) T(S2 S1)] e"W^ [H2 TS2 (H1 TS1)] e" W^ z(2 2$,$,$,$,$,$,$,$,$,$,  :  T) ??0`f,$D 0 >N G a" H TS [G2 G1] e" W^ %G e" W^ 1 %G > W^ SۏL spontaneous processes %G = W^ YNs^ar` be in equilibrium %G < W^ S^ NۏL nonspontaneous processes @(2 2 $3 $,$,$, $,$,%$,"$,"$     H  0޽h ? 3fȺdf___PPT10+LLD'  = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*D' =%(D' =%(DP' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B#blinds(vertical)*<3<*+p+0+0 ++0+0 +k 0 H@(    T@) ??q tS_S^NSe_ۏLes^ae SO|ZPg'YvW^ ; GvQ\ϑ %G/fSO|ZP^SOyRvg'YP^ ُ*Ng'YP^(WS_-N_0R[s0 ~F(2H   T`) ??d ,$D 0 FGibbs1uvirINI{)nI{S N /fSO|@bwQ gvZP^SOyRvR 0d$ 2&33H  0޽h ? 3fȺdf___PPT10+'c:DO' )= @B D ' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(+8+0+0 + 0 nf (      T ) ??P I{)nI{S N S^vW^ = 0 R %G e" 0 %G d" 0 %G < 0 SۏL spontaneous processes %G = 0 YNs^ar` be in equilibrium %G > 0 NۏL nonspontaneous processesF(2 &.&   X   Tp) ?? P`D ,$D 0 RirINI{)nI{S N SO| T^e1uQ\veTsS:N NZP^SOyRvSf[S^ۏLveT0 *(2*& *   TL) ??3 0,$D  0 bpRf[,{N[_vSyhI{)nI{S W^ = 0agN N NUOSǏ zvQ T^e1u;`/fQ\v0 b2 2&3.3&3$3$36H   0޽h ? 3fȺdfI A ___PPT10! +aD' )= @B DP' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-6B'blinds(horizontal)*<3<* D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<* D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<* +p+0+ 0 ++0+ 0 + 0 tl$(  $ $ T, ??@` HRBl%rGm 1uG a" H TS I{)nI{S ! %G = %H T%St*(2$^ H $ 0޽h ? 3fȺdf 0 4 , ( (  (6 ( TH, ??PoL___PPT10,$z___PPT9\T h20S^v%rGm (1)ĉ[h` N1.013105 c[T 3z[US(v T^e1u:N0  (2* 2& 3& 3. 3& 3. 3 &  3& 3$,$$b V ( T#, ??P ,$D 0 2 ir(vhQid\ ub T^e1ug)n^ NYNh`vTyCQ }vg3z[vUS(ub1molg~ir(v T^e1uv9eSϑ (u%fGmKJ/mol h:y0 nQ(2 2A$,$,$j) ( T;, ?? P@ $D  08___PPT10N___PPT90( z3 S^v%rGmhQid\1uS  %rGm = ("i %fGm)N "i %fGm S Z(2% 2$$$,$,$,$, $,$, $,$,$  H ( 0޽h ? 3fȺdfI A ___PPT10! +AR^cD' R,= @B DP' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*(%(D' =-6B'blinds(horizontal)*<3<*(D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*(%(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<*(D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*(+p+0+(0 ++0+(0 + 0 rj,(  , , T|], ??P|  FP4 The Gibbs-Helmholtz Equation 1u G a" H TS I{)nI{S ! %G = %H T%S h` ! %rGm = %rHm T%rSm %rHm0%rSmTSS N'Y %rGmSS'Y T = 298 K %rGm (298K) Na)n^T , %rGm (T) `(2 2! 333333333333333333 r     H , 0޽h ? 3fȺdf 0 LD 0(  0 0 N,??@`  R`S N )n^[S^S'`vq_T %H %S %G = %H T%S O - + - NUOT " H2O2(l) ! H2O(l) + 1/2O2(g) + - + NUOT CO(g) ! C(s) + 1/2O2(g) + +  ؚ)n- " CaCO3 (s) ! CaO (s) + CO2 (g) -  NO)n- " NH3(g) + HCl (g) ! NH4Cl(s)  I -3"    < 3   3  K,B 0 ZD??  B 0 ZD??ppB 0 ZD?? ` H 0 0޽h ? 3fȺdf  0  4(  4p 4 H<,?? O(u$Nyel{H2O2RS^v%rGm298K v^fS^&TSۏLH2O2(l) ! H2O(l) + 1/2O2(g)J(2   X    4 H,?? 6,$D 0 0v H2O2(l) ! H2O(l) + 1/2O2(g) %fGm(KJ/mol) 120.42 237.18 0 %fHm(KJ/mol) 187.78 285.83 0 Sm(J/mol K) 109.6 69.91 205.03(2   ''. "  5* 4 H,??   ,$D 0 ~%rGm = ("i %fGm)N ("i%fGm)S = 237.18 ( 120.42)= 116.76(KJ/mol) O(2      4 H@-?? ,$D  0  %rGm = %rHm T%rSm = ( 285.83 + 187.78) 298( 69.91 + 205.03/2 109.6) 10 3 = 98.05 29862.8210 3 = 116.77 (KJ/mol) (2;      F 4 T0'- ??`P,$D  0 .NUO)n^ NS^ckTۏL S^ NSۏL. * 233H 4 0޽h ? 3fȺdfH@___PPT10 +QF]D$' 4-= @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*4%(D' =-6B'blinds(horizontal)*<3<*4D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*4%(D' =-6B'blinds(horizontal)*<3<*4D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*4%(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<*4D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*4D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*4%(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<*4D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*4++0+40 ++0+40 ++0+40 ++0+40 + 0   08 (  8 8 H4;-??  ,$D 0 RSolution: 2CH4 (g) ! CH3CH3 (g) + H2 (g) %fGm( KJ/mol ) -50.75 -32.89 0 %fHm( KJ/mol ) -74.81 -84.68 0 Sm( J/molK) 186.15 229.49 130.57 8(2A 2       0  8 .>7077f 8 H|R-??^ Example For the reaction 2CH4 (g) ! CH3CH3 (g) + H2 (g) (a) What is the value of %rGm? Is this reaction spontaneous under standard conditions? (b)Is it spontaneous under any conditions? Would it be worth looking for a catalyst? (2V 2  *N 8 H|g-?? ,$D  0 ((a) %rGm (298k) = -32.89 -2 (-50.75) = 68.69 ( KJ/mol ) nonspontaneous (b) %rHm> 0, %rSm= < 0 It is nonspontaneous under any conditions. So it would not be worth looking for a catalyst.(2  2"    dl3 HH 8 0޽h ? 3fȺdfI A ___PPT10! +*tD' -= @B DP' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*8%(D' =-6B'blinds(horizontal)*<3<*8D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*8%(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<*8D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*8+p+0+80 ++0+80 + 0   @<j (  <Z < H܅-??  Eample Estimate the normal boiling point of methyl alcohol ,CH3OH. With the known values of the standard enthalpy of formation and standard entropy of methyl alcohol liquid and gas: CH3OH (l) ! CH3OH (g) %fHm KJ / mol 238.64 201.2 Sm J / (molK) 127 238(2 2?       , (^ ) < Hț-??p,$D 0 $Solution: CH3OH (l) = CH3OH (g) %rHm= 201.2 ( 238.64) = 37.4 (KJ / mol) %rSm= 238 127 = 111 (J /mol / K) %rGm (T) = %rHm T%rSm d" 0 T%rSme" %rHm T e" (37.4103 / 111) = 337 (K) = 64! sS bp. = 64! The actual boiling point of CH3OH is 65!.$(2  & '  $ '6H < 0޽h ? 3fȺdf___PPT10b+l D' -= @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DP' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*<%(D' =-6B#blinds(vertical)*<3<*<+8+0+<0 +  0 P@8(  @ @ H-??   lExample The normal boiling point of benzene, C6H6, is 80!(353.2 K) and %rHm is 30KJ / mol . Calculate %rSmfor benzene. | 2/       3<@  @ H.?? o ,$D 0 hSolution  C6H6 (l) = C6H6 (g) %rGm (353.2K) = %rHm 353.2 %rSm = 0 %rSm= (30 103) / 353.2 = 87.2 (J / mol /K) I(2. 2              x *H @ 0޽h ? 3fȺdf___PPT10f+ x D' -= @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*@%(D' =-6B'blinds(horizontal)*<3<*@+8+0+@0 + 0 JB`D(  D D TtR ??P 08  Calculation of %G for nonstandard conditions The free-energy charges, %G, accompanying changes carried out under nonstandard condition can be calculated from equation (1) %rGm = %rGm+ RT lnQ (1) Qreaction quotient S^q Equation (1) is derived in most physical chemistry and thermodynamics texts, which is beyond the scope of this text. 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