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C2H5OH(g.) (298K) %rGm = 6.2 (KJ/mol) S_P1 = 1/2 P P2 = 1/10 P P3 = 1/20 Pe S^&TSۏL Pl(2$3,3$3,3 $3,3$3,3$3,3$3,3$3,3 $3,3 $3,3$3     T ??PPp,$D 0 B   %rGm (1)= %rGm + RT3() = 6.2 + 8.3142983(1/2) = 6.2 1.7 = 4.5 (KJ/mol) > 0 %rGm(2) = 6.2 + 8.31429810 33(1/10) = 6.2 5.7 = 0.495 (KJ/mol) > 0 %rGm(3) = 6.2 + 8.31429810 33(1/20) = 6.2 7.4 = 1.2 (KJ/mol) < 0 SSSۏL ^(2 2 "*"*"*"*b"*"*"*."*"*"*3";A9H  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 ^ (   H  T ??P` 2O2Bl298Ke FeuvS^SuvagN0 & 2$3 H   T" ??,$D 0 N 2Fe (s.) + 3/2 O2 (g.) ! Fe2O3(s.) %fGm (KJ/mol) 0 0 741.0 4" %rGm = 741.0 KJ/mol {(2 2!  *^'     T1 ??  ,$D  0 傁TۏL %rGm < 0 sS %rGm = %rGm + RT3Q = -741.0 + 8.31429810 33(PO2/P)  3/2> O = -741.0  3.723(PO2 / P) > 0 3(PO2 / P) < 199.4 PO2 / P< 2.5210 87 PO2 < 2.5510 82 Pa )  *J*  TU ??pp@,$D 0 r\ NSSu . 2aH  0޽h ? 3fȺdf ___PPT10 +D ' b= @B D ' = @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<* D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<* D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* ++0+ 0 ++0+ 0 ++0+ 0 +b  0   $ (  $ $ T\j ??` DO3Bl NS^S^SuvagN0 Zn(s) + Cu2+(aq) = Zn2+(aq) + Cu(s) x(20 2&$3,3 $3,3 $3 $' $ T؊ ??0Pb R Zn(s) + Cu2+(aq) = Zn2+(aq) + Cu(s) t=0 1mol/dm3 0 t x 1-x %fGm (KJ/mol) 0 64.77 153.89 0 %rGm = %fGm (Zn2+) %fGm (Cu2+)= 153.89 64.77 = 218.66 (KJ/mol) %rGm = %rGm + RT3 = 218.66 + 8.31429810 33([Zn2+]/[Cu2+]) > 0 3([Zn2+]/[Cu2+])>(218.66/2.48) > 0 ([Zn2+]/[Cu2+]) > 2.13 1038 sS`OS^TۏL [Zn2+] > 2.131038[Cu2+] (2 2"  (  (   "*@"*"*."*"*"*"*"*"*"*"*$"*"*"*"*!"*"*"* "*"*"*"* "*"* "*"*"   X v $ N ??q -!q $ 8A 9??"` X  9H $ 0޽h ? 3fȺdf 0 (R(  ( ( T ?? `@  Npf `$ S_%rGmpe 41.84 KJ/mol a$ S_%rGmpef6q T7h{|WvS^ v T)n^e K'Y hfS^ckTۏLv z^'Y0 (2 2$3 ( (  ( ( ( ( ( ( ( (  ( ( ( ( ( ( ( (   $3,3$3,3$3 ( (E      H ( 0޽h ? 3fȺdf 0  8(     0< `. 2.Chemical equlibrium 2.1 Characteristics of an equilibrium 2.2 Equlibrium constants and equlibrium constant expressionsK, Kc, Kp 2.3 Transfer rates^alSs = (CR Cs^) / CR = %C / CR 3. Shifts in chemical equilibium 3.1 Effect of changes in concentrations 3.2 Effect of changes in pressure 3.3 Effect of changes in temperature   3.4 Effect of catalystg(2 l !  0   H  0޽h ? ̙33*  0 **@I, *(  ,x , c $l' `}   h)~  I,# #"* ` -, <t?P  P  @` ,, <t?YP  P  @` +, < &? P Y P  @` *, <'? P  P  @` ), <(?'P  P  @` (, <v?kP ' P  @` ', <PF?P k P  @` &, <N? P  P  @` %, <@V?Y P  P  @` $, <w? YP  P  @` #, <f? P  P  @` ", <o?' P  P  @` !, <x?k 'P  P  @`  , <y? kP  P  @` , <Ȉ?   P  @` , <<?Y   P  @` , <Ԓ? Y  P  @` , <?  P  @` , <?'  P  @` , <IJ?k '  P  @` , <\? k  P  @` , <t?@  P  @` , <?Y@  P  @` , <? @Y  P  @` , <? @  P  @` , <X?'@  P  @` , <p?k@'  P  @` , <?@k  P  @` , < ?@ P  @` , < ?Y@ P  @` , <? Y@ P  @` , < ?  @ P  @`  , <?' @ P  @`  , <(?k'@ P  @`  , <01?k@ P  @`  , <2? P  @`  , <A?Y P  @` , <C? Y P  @` , <R?   P  @` , <PT?'  P  @` , <hc?k' P  @` , <k?k P  @`ZB /, s *1 ?ZB 0, s *1 ?@@ZB 1, s *1 ?  ZB 2, s *1 ?  ZB 3, s *1 ?P P `B 4, 0o ?`B 5, 0o ?ZB 6, s *1 ?kkZB 7, s *1 ?''ZB 8, s *1 ?  ZB 9, s *1 ?  ZB :, s *1 ?YYZB ;, s *1 ?`B <, 0o ?ZB C, s *1 ?' `B ., 0o ?'ZB F, s *1 ? Y`B D, 0o ?  `B G, 0o ?YH , 0޽h ? ̙33h  0  (    0* `* ,Lu Each component concentration will not change with time. Although different initial concentrations give different equilibium concentrations, the quantity, [HI]2eq/ [H2]eq[I2] eq, is constant . Kc= [HI]2eq/ [H2]eq[I2] eq The constant is called equilibrium constant for the equilibrium and is reprensented by the symbol Kc. @M(2 h|y 3H   0J P,$D  0 ; Dynamic equilibrium. The forward reaction rate equals to the reverse reaction rate. It is a conditional equilibrium. When the concentration, pressure and temperature change, the equilibrium position will change. D(2 2YH  0޽h ? ̙33___PPT10+ADD' R= @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 +l 0 0#(  0T 0 0Z`i 2.2 Equlibrium constants and Equlibrium constant expressionsK Kc Kp aA + bB = gG + hH Z`(2H$3$3   0 0gp ,$D 0 d 2.2.1 Experimental equilibrium constants (1) Concentration equilibrium constants Kc P9(2, 2   M$Zv 0 s ,A =??"`P  = 0 0r `@  nKc= 4$ $$  0 0x`   C C: mol/L  2H 0 0޽h ? ̙33y___PPT10Y+D='  = @B +v 0 4-(  4 4 0 @ .(2) Partial pressure equilibrium constants Kp */ 2,+ 4 0$@p eKp = *   v 4 s ,A ;??"`p ; 4 0 p  L PPa & 2H 4 0޽h ? ̙33y___PPT10Y+D='  = @B +  0 8(  86 8 0 P' :(3) Heterogeneous equilibium CaCO3(s) = CaO(s) + CO2(g) L; 2#     ,  T 8 0ȣ  Kc = [CO2] Kp = P CO2 (2 2    ,   8 0 W ,$D 0 $6CaCO3(s) + H2O (l) + CO2(g) = Ca2+ ( aq)+ 2HCO3-(aq) 7 2         &H 8 0޽h ? ̙33___PPT10f+_D'  = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*8%(D' =-6B'blinds(horizontal)*<3<*8+8+0+80 +~ 0 .&<(  < < 0<@ :2.2.2 Standard equlibrium constants (1) In solution, K= 0:(2 2% " !R < s *Dv < s ,A ???"` d= ? < 0    ^(2) In gas phase, K= , 2R < s *}' }v < s ,A A??"`= #   AH < 0޽h ? ̙33 0 |t@ (  @ @ 0p@ N(3) In heterogeneous phases  2F @ 0@@p  0CO2(g) + H2O +Ca2+ = CaCO3 (s) + 2H+(aq) K= 0(2 2  R @ s *  v @ s ,A  E??"`G  EH @ 0޽h ? ̙33_ 0 D(  D D s *` ,tExample 1: At the temperature of 1000K, put 1.00mol SO2 and 1.00mol O2 in a closed container of 5.00dm3 when equilibium is reached , 0.85mol SO3 is formed. 1 Calculate K,Kp and Kc of the reaction 2SO2 + O2 = 2SO3 at the temperature of 1000K; 2 Use thermodynamic data to calculate Kat the temperature of 298K.;(26 (! Bg % @$ D s * ,$D  0 >`Solution(1) 2SO2 (g) + O2 (g) = 2SO3 (g) t=0 (mol) 1.00 1.00 0 teq 1.00 0.85 1.00 0.85/2 0.85 = 0.15 = 0.575p(2  1SJR 7H D 0޽h ? ̙33___PPT10+b=D' = @B DD' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*D%(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<*DD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*D+8+0+D0 +;  0  H{(  H6 H s *# PSO2 = n RT /V = 0.158.3141000 /(510 3)=2.49105 (Pa) PO2=0.5758.3141000 /(510 3)=9.56105 (Pa) PSO3=0.858.3141000 /(510 3)=1.41106(Pa) 4"Kp = PSO32 / (PSO22PO2)=(1.41106)/[(2.49105) 9.56105] = 3.410 5 (Pa)(2"       Z',,@( H s *F  xK= = 3.410 51.013105 = 3.4 :=(2(  R H s *q ( qv H s ,A G??"`f  G H s *R  cKc=[SO3]2/([SO2]2[O2]) = (0.85/5)2 / (0.15/5)2(0.575/5) = 0.8525 /(0.1520.575) = 279.2 (dm3/mol) d(2   bH H 0޽h ? ̙33d 0  0L(  Ll L s *kp1 `(2) 2SO2 (g) + O2 (g) = 2SO3 (g) %fGm(KJ/mol) 300.7 0 370.37 4"%rGm = 2%fGmSO3(g) 2%fGmSO2(g) = 370.372 + 300.372 = 140 (KJ/mol) 5"%rGm = RT3K 4"lgK= %rGm/2.303RT = 14010 3 / (2.3038.314298) = 240 K(298K) = 3.441024 0(2 2!  0@-@   2H L 0޽h ? ̙33  0 @Pf(  P P 0(@  Comments: `$ Equilibrium constants depend on the temperature, and have nothing to do with concentrations or pressures. a$ Equilibrium constants express how far away the equilibium proceeds. The bigger the value of K, the more complete the reaction proceeds. b$ For the same reaction written in different ways, the corresponding equilibrium constants are different. Zx(2  k  i" 8 P 0ȫ ,$D 0 H2 + I2 = 2HI K1 1/2 H2 + 1/2 I2 = HI K2 2HI = H2 +I2 K3 K1 = K22 = (K3) 1 H(2 2  33 333 3 333 33 3 3 3H P 0޽h ? ̙33___PPT10f+ x D' 7= @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*P%(D' =-6B'blinds(horizontal)*<3<*P+8+0+P0 + 0 PT7(  T T 0,  ` mw If Equation(3) = Equation(1) + Equation(2), K3 = K1 K2 Equation(3) = Equation(1) - Equation(2), K3 = K1 /K2 ?(29 2 .+z T s *TP  H2S = H+ + HS K1 = HS = H+ + S2 K2 = H2S = 2H+ + S2 K = 3(2 2 Z   R T s * : v T s ,A I??"`  IR T s *\ v T s ,A K??"`` "  M  KH T 0޽h ? ̙33 0 IA`X(  X X 0DP`. AIf there are two or more than two equilibria existing in a reaction system, it is called multiple equilibria. A component has only one concentration when it appears in different equations. 4 2Y3R," 6 RH X 0޽h ? ̙33  0 >6p\(  \ \ s * ` Example 2: (1)NO (g) + 1/2Br2(l) = NOBr(g) (25!) nSN]xp K1 = 3.61015 vapor pressure of Br2(l) is28.4KPa at the temperature of 25!2 Br2(l.) = Br2(g.) K2= P / P = 28.4 / 101.325 = 0.28 Please calculate the Kof the following reaction: (3)NO(g) + 1/2Br2(g) = NOBr(g)(2 - d#<K1 \ s * &  ,$D  0 Solution(2 ) 1/2 Br2(l) = 1/2 Br2(g) K2 = K21/2 = 0.529 (1) (2 ) = (3) K3 =K1 / K2 = 3.610 15/0.529 = 6.810 15 }(2 2  H \ 0޽h ? ̙33___PPT10+wD' != @B DD' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*\%(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<*\D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*\+8+0+\0 +$ 0 `d(  `, ` 0@C  0  According to the definition, lgK = %rGm/2.303RT According the rule of multiple equilibria, from the known K t(2 2""      H  <1 H ` 0޽h ? ̙33 0 d6(  d d 0V s   2.3 Equilibirium transform rates^alSs Def: = (nini neq)/nini = %n /nini 100% when volume is constant = Cini Ceq /Cini 100% Compared with equilibrium constant, changes with the initial and equilibrium concentrations, while equilibrium constant only changes with the temperature and has nothing to do with concentrations. K(2 2 & "   $   H d 0޽h ? ̙33 0 {h(  h h 0wP [!3. Shifts of chemical equilibrium"(2"  h 0 ~ PV  %rGm = %rGm + RT3Q = RT3K + RT3Q = RT3(Q/K) %rGm = RT3(Q/K) %rGm < 0 Q < K shift to the right, forward reaction %rGm = 0 Q = K at equilibrium %rGm > 0 Q > K shift to the left, reverse reaction (2&20 &$H h 0޽h ? ̙33. 0 ln(  l6 l 0  0 v3.1 Effect of Changes in Concentration Changes in concentrations do not change the value of the equilibrium constant. Dw(2"MH l 0޽h ? ̙33 0 LDp(  p p s * JExample 3 Reaction CO (g) + H2O (g) = H2 (g) + CO2 (g) Kc = 9 t = 0(mol/L) 0.020 0.020 0 0 t = 0(mol/L) 0.020 1.00 0 0 teqmol/L 0.020 x 0.020 x x x teqmol/L 0.020 y 1.00 y y y Kc = [H2][CO2] / ([CO][H2O] = x / (0.02 x)2 = 9 x = 0.015 (mol/L) COvlSs = 0.015/0.02100% = 75% In the same way Kc = y2 /[(0.02 y)(1.00 y)] = 9 y = 0.01995 COvlSs = 0.01995/0.020 100% = 99.8% (2 !   z1. IK> Ay %) H p 0޽h ? ̙33 0 tB(  t  t s *T`  6Example 4 At temperature of 298K mix equal volume of 0.100mol/L AgNO3 0.100 mol/L Fe(NO3)2 and 0.0100 mol/L Fe(NO3)3 together. Fe2+ + Ag+ = Fe3+ + Ag (s) K= 2.98 Calculate to solve 1 which direction will the reaction shift? 2 when equilibrium is reached, concentrations of each component (3) Transform efficiency of Ag+ (4) [Ag+]0[Fe3+]remain the same [Fe2+] = 0.300 mol/dm 3 Transform efficiency of Ag+^(2 :H t 0޽h ? ̙33 0 phx(  x x s *aP `XP___PPT92*  .Solution: = 0.0100/(0.1000.100) = 1 < K The equilibrium will shift to the right. Forward reaction will take place. d (2@(2(2L(2 2 AM @`R x s *8 v x s ,A M??"`" MH x 0޽h ? ̙33~ 0 .&|(  | | s *@  ,(2) Fe2+ + Ag+ = Fe3+ + Ag t = 0(mol/L) 0.100 0.100 0.100 t eq 0.100 x 0.100 x 0.100+x K = (0.0100+x)/(0.100 x) = 2.98 x = 0.0127 (mol/L) 4" [Fe2+] = [Ag+] = 0.0837 mol/L [Fe3+] = 0.0227 mol/L (2 2 6aRh !SH | 0޽h ? ̙33  0 L(    s *( 0a  (3) Ag+ vlSs  = 0.0127 / 0.100 = 12.7% (4)dkeAg+vlSs:N R Fe2+ + Ag+ = Fe3+ + Ag t = 0 0.300 0.100 0.0100 teq 0.300 0.100 0.100(1 ) 0.0100+0.0100 K = (0.0100+0.100)/{(0.300 0.100)[0.100(1 )]} = 2.98  = 0.381 =38.1% (2 2 () ( (  (  (1 (w M/H  0޽h ? ̙33) 0 i(  1  03P o3.2 Effect of changes in pressure Changing the pressure does not change the value of the equilibrium constant. F"(2N 2!KH  0޽h ? ̙33 0 NF (  v  s ,A  O??"`  Ov  s ,A  Q??"` s Q   s *`A 0@ LExample 5N2 (g) + 3H2 (g) = 2NH3 (g) ^' 2     s *J@`  p>When the total pressure is two times of the original pressure,?(2?  s * O@  V$Equilibrium will shift to the right.%(2%H  0޽h ? ̙33 0 nf0(  v  s ,A  S??"`R  Sv  s ,A  U??"` h "  U  s * YPp .Example 6:CO (g) + H2O (g) = CO2 (g) + H2 (g) T/ 2     s *4`0  p>When the total pressure is two times of the original pressure,?(2?H  0޽h ? ̙33J  0 d\@(    0jh rO1 Tb(l N2 (g) + 3H2 (g) = 2NH3 (g) SelAir _sO(u S_`lTy/}ǏYe q_T(lvub dke >ezz eEQelhP(2      0H#  0w  ,$D 0 NO2YNpˆ6RYNp C2H6 (g) = C2H4 (g) + H2 (g) R`T0Pe Ǒ(uReQH2O (g.)velcؚYNpvNs0 0*(2 2       h  H  0޽h ? ̙33___PPT10f+ x D' 7= @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 P(  v  s ,A W??"` :7  Wv  s ,A Y??"` vJ 0 Yl  0<&  <3.3 Effect of changes in temperature 5" %rGmT =%rHm T %rSm %rGmT = RT 3K 4" RT 3K =%rHm T %rSm 3K = [ %rHm298K / RT ] + [%rSm298K / R ] D(2 2*3 33 33 33 33 33 33 33 33 33 3 3   h+    H  0޽h ? ̙33 0 `D(     0hP {  %rHm < 0 T2 > T1 K2 < K1 %rHm < 0 T2 > T1 K2 < K1 To increase the temperature, reaction will shift to the endothermic reaction.  2                                    N $t [H  0޽h ? ̙33( 0 h(  0  s *5  Example 7:Calculate: 1 %rGm at 427!700K for the reaction N2g + 3H2(g) ! 2NH3(g) If PN2= 33.0atm PH2 = 99.0atm and PNH3 = 2.0atm, (%rHm = 92.22 KJ/mol %rSm = 198.53 J/molK %rGm= 32.96KJ/mol) predict the direction of spontaneous change. 2 K298K , K700K , K773K 3 Discuss the suitable reaction conditions. (2. 2    ^+"  ;+H  0޽h ? ̙335  0 u(  R  s *}   s *+P  8%rGm = %rGm + RT3Q = 46.75 + 8.31470010 33 = 46.75 + 8.31470010 33 = 46.75 + 5.82( 15.896) = 46.75 92.51 = 45.76 (KJ/mol) < 0 8Z2 C|   s *;  4Solution (1) %rGm(700K) = %rHm T%rSm = 92.22 700( 198.5310 3) = 46.7I5 (KJ / mol) > 0 The forward reaction is nonspontaneous under standard state.(2 W 3`  s *"`k v  s ,A c??"`6 @ ? I  c   `@Y~ 1? [%The forward reaction is spontaneous. &0&  8A a??"`@   aH  0޽h ? ̙33H 0 (    s *4Sp 0  &lgK773 "lgK298 = N(2  s *`Z (2) 5"%rGm = RT3K 4"3K298K = 13.30 K(298K) = 5.99105 3K(700K)= 8.30 4"K(700K) = 3.0510 4^`(2.$d&R  s *3 v  s ,A e??"`w  eR  s *bv  s ,A g??"` Z  g  s *h ` NjK773K = 7.010 5 %rHm < 0 increaseT! K decreases.L6(2t H  0޽h ? ̙33 0 rj(    s *vPP` p^(3) Low temperature and high pressure will be helpful for the forward reaction. However, when the temperature is below 673K the reaction speed is very slow while high temperature will cause dissociation of NH3. Thus the practical reaction condition is 300105Pa -700105Pa 773K with a catalyst of iron.0(2!3 1   !b{;8H  0޽h ? ̙33 0 @8(    0hP 8vKey Words and Concepts Reversible reaction forward reaction reverse reaction equlibrium constants equlibrium constant expressions Shifts in chemical equilibium Le Chtelier s Principle >(23`N  )   H  0޽h ? ̙33@ 0 (  H  0 ^  R Summary \~KvBll u 1uKv[IN T~Rs^aSm^ u Y͑s^aSt(1) + (2) = (3) K3 = K1K2 K3 = K1 / K2 u %rGm = RT 3K u _ +YvI{Se z_ u _ ['lS؏SNz] (2 2$3 3            $ $ ((((, , 0000 : H  0޽h ? ̙33 0 A(     04  ]3.4 Effect of catalyst Catalysts change the reaction rates, and not change the equilibrium. 0(2F 2GH  0޽h ? ̙33@ 0 (  H  0PV  `If a change is made in a system at equilibrium, the equilibrium will shift in such a way as to reduce the effect of the change. This is Le Chtelier s Principle found in 1884. 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