ࡱ> nT _ewPNG  IHDR;~n^sBIT&C`PLTEUf""U"f"DDff"ww3fffDDUUww""33U"fDDUUffwẅ̈ݙ*F tRNS\\bKGDHIDATxv0U/`P1wKKN8 BcE3R ;sOp7O.XA9?e8c[ mUB'j`"K,5 |-U3ByG={81YYWbZEhCu,gg(Ќ#cus$ c _5hi @ qc<o荏l.,(>])o*R РKh+UWV luNˑrpP bYBägz7zǀy!1o0w6̓%96SE]cC~iy:LY~H*cauK2LӒ̷X'7W؟˴LZ7N>Zs>9U{UKue'zHWUe^W\Pfil8.v8քjL,ӡd;WIJzN:kиJ_]+,,mPb`Yw*YR>tnZ~Q@m6pO*X+vDtgIu8o %;t/ukʑֽoGXuDƿ #RE]30Eg) [86}Y̝q*GM68}se!D:1Xׅ#3*GLMe-R`3wݭWœO]+CunǞ횕tvgu~Ou' >m4"JFIFKKMSO Palette C   ")$+*($''-2@7-0=0''8L9=CEHIH+6OUNFT@GHEC !!E.'.EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE"!!1AQqa!1 ?F3]'IƓe5,J$3?O 5LXyYL4tO%fxM\DXL`: a2HS=rtd+f\UtV5m7L?cf7 `Zx5\kX(   l   & - 4 ; T X ^ bD/ 0|DTimes New Romantt. 0DArial Blackmantt. 0" DTahomalackmantt. 0"0DAriallackmantt. 0@DCourier Newmantt. 01@ .  @n?" dd@  @@``    zw*   "    J@ii^h^QSQGRGSFp $    (        : ` +           " !  ) . #  ) &&%$$$ DK NS (BC\EFGHIJ QPORUTVWX!$$@`$$$$$$b$bpe# zt OR$FE-Qڛb>" 0AA ff̙f3f@d g4KdKd|; 0php8 <4BdBd@w 0t.g4:d:dm 0p@ pp<4!d!d@w 0t.ʚ;ʚ;<4dddd@ x 0@r0___PPT10 pp2___PPT9/ 0? %O ==<hij]nopqrs^_u`vwxyaz{|}~mP4   0` ̙33` ` ff3333f` 333MMM` f` f` 3>?" dd@,|?" dd@   " @ ` n?" dd@   @@``PR    @ ` ` p>>L0 VN (    6T P  T Click to edit Master title style! !  0   RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  0詍 ``  X*  0 `   Z*  0< `   Z*z  bA޽h @ ?Parchment ̙33 Default Design 0 zr@ ( )   0w P   w P*    0w    w R*  d  c $ ?  w  0Pw  @ w RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  6fw `P  w P*    6]: `  w R*  H  0޽h ? ̙3380___PPT10. 0X(   $+ X X 0Dw P   w X*  X 0w    w Z*  X 6w `P  w X*  X 6w `  w Z* H X 0޽h ? ̙3380___PPT10.1 0L0 (     0 &FURTHER TOPICS ON CHEMICAL EQUILIBRIUMT(2 82 f0f<f   0 OKNOCKHARDY PUBLISHING(2fJ   C "A KTRE: .  <̙? x @b 0 0  <̙? x @^ 0 0  0"`>B 2008 SPECIFICATIONS>G$G   H  0޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   ` (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0W   0a INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... www.knockhardy.org.uk/sci.htm Navigation is achieved by... either clicking on the grey arrows at the foot of each page or using the left and right arrow keys on the keyboard (2 2(2 2  75, dB   <D?( (    0D{0P( OKNOCKHARDY PUBLISHING(2    08X fCHEMICAL EQUILIBRIUM (2)2(2$f<fB  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   p  (  v  0mu   CONTENTS The Equilibrium Law The equilibrium constant Kc Calculations involving Kc Calculations involving gases Mole fraction and partial pressure calculations Calculations involving Kp 202 2f  l >OkdB  <D)P0Q  0Pp@ 0 0dB  @ <D)PXP   0X0@ 0 0  0{ @ 0 0  6~h !@ 0 0  6p @& 0 0  6Kh @- 0 0  6>  @4 0 0  6!  @; 0 0  0DX fCHEMICAL EQUILIBRIUM (2)2(2$f<fB  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   La ( w L8 L 0|0  Simply states  If the concentrations of all the substances present at equilibrium are raised to the power of the number of moles they appear in the equation, the product of the concentrations of the products divided by the product of the concentrations of the reactants is a constant, provided the temperature remains constant There are several forms of the constant; all vary with temperature. Kc the equilibrium values are expressed as concentrations in mol dm-3 Kp the equilibrium values are expressed as partial pressures The partial pressure expression can be used for reactions involving gases@d@9ffF  (  (Rf@  J   dB L <D)P0Q L 0Pp@ 0 0dB L@ <D)PXP L 0X0@ 0 0 L 0Tq~ aTHE EQUILIBRIUM LAW2(2ffB L s *޽h ? ̙33y___PPT10Y+D=' = @B +z  0L0   P (  Pv P 0쳉0  .for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C]c . [D]d = a constant, (Kc) [A]a . [B]b where [ ] denotes the equilibrium concentration in mol dm-3 Kc is known as the Equilibrium Constant@PPPv@, f 3f3f 3f3         0    $ /          D  \  + dB P <D)P0Q P 0Pp@ 0 0dB P@ <D)PXP P 0X0@ 0 0 P 0߉q~ THE EQUILIBRIUM CONSTANT KcF(2fffF ? P  N 8N   P fB  P 6D3olB  P <BD3o{>T   P# L)?fB  P 6D3olB P <BD3o{^B P 6Do a B P s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 0Ti(  TB T 0D8' 0 VALUE OF Kc AFFECTED by a change of temperature NOT AFFECTED by a change in concentration of reactants or products a change of pressure adding a catalyst HX(@ ffffb &    v T 0t+80  .for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C]c . [D]d = a constant, (Kc) [A]a . [B]b where [ ] denotes the equilibrium concentration in mol dm-3 Kc is known as the Equilibrium Constant@PPPv@, f 3f3f 3f3         0    $ /          D  \  + dB T <D)P0Q T 0Pp@ 0 0dB T@ <D)PXP T 0X0@ 0 0 T 0S8q~ THE EQUILIBRIUM CONSTANT KcF(2fffF ? T  N 8N   T fB  T 6D3olB  T <BD3o{>T   T# L)?fB  T 6D3olB T <BD3o{^B T 6Do a B T s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 :2@(  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0k  0``8  construct the balanced equation, including state symbols (aq), (g) etc. determine the number of moles of each species at equilibrium divide moles by volume (in dm3) to get the equilibrium concentrations in mol dm-3 (If no volume is quoted, use a V; it will probably cancel out) from the equation constructed in the first step, write out an expression for Kc. substitute values from third step and calculate the value of Kc with any units @D@@@ 1  B  Z<  2  A     0x8s CALCULATIONS INVOLVING KcF(2fffB  s *޽h ? ̙33 0L0 ZRPx( w xdB x <D)@0A x 0Pp@ 0 0dB x@ <D)@X@ x 0X0@ 0 0 x 0<8)  + construct the balanced equation, including state symbols (aq), (g) etc. determine the number of moles of each species at equilibrium divide moles by volume (in dm3) to get the equilibrium concentrations in mol dm-3 (If no volume is quoted, use a V; it will probably cancel out) from the equation constructed in the first step, write out an expression for Kc. substitute values from third step and calculate the value of Kc with any units Example 1 One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc. @D@@@ 1  B  j3!<  2  A  )      x 08s CALCULATIONS INVOLVING KcF(2fffB x s *޽h ? ̙33  0L0   `* (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  08   construct the balanced equation, including state symbols (aq), (g) etc. determine the number of moles of each species at equilibrium divide moles by volume (in dm3) to get the equilibrium concentrations in mol dm-3 (If no volume is quoted, use a V; it will probably cancel out) from the equation constructed in the first step, write out an expression for Kc. substitute values from third step and calculate the value of Kc with any units Example 1 One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc. CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)  @D@@@J^1B j3&        <  2  A  )    M F ?   H 7 8N    fB   6DolB   <BDo{>T   # L)?fB   6DolB  <BDo{  08s CALCULATIONS INVOLVING KcF(2fffB  s *޽h ? ̙33  0L0 pN( w dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0i  0H8   construct the balanced equation, including state symbols (aq), (g) etc. determine the number of moles of each species at equilibrium divide moles by volume (in dm3) to get the equilibrium concentrations in mol dm-3 (If no volume is quoted, use a V; it will probably cancel out) from the equation constructed in the first step, write out an expression for Kc. substitute values from third step and calculate the value of Kc with any units Example 1 One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc. CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) moles (initially) 1 1 0 0 moles (at equilibrium) 1 - 2/3 1 - 2/3 2/3 2/3 @D@@@J?1B j3&        <  2  A  )    ; F ?   H 7 8N    fB   6DolB   <BDo{>T   # L)?fB   6DolB  <BDo{  0:  <6Initial moles of CH3COOH = 1 moles reacted = 2/3 equilibrium moles of CH3COOH = 1/3 For every CH3COOH that reacts; a similar number of C2H5OH s react (equil moles = 1 - 2/3) a similar number of CH3COOC2H5 s are produced a similar number of H2O s are produced @ff fffff*ffff fffffff+fff&  t   0:s CALCULATIONS INVOLVING KcF(2fffB  s *޽h ? ̙33 0L0 og( w dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  0':) d^ construct the balanced equation, including state symbols (aq), (g) etc. determine the number of moles of each species at equilibrium divide moles by volume (in dm3) to get the equilibrium concentrations in mol dm-3 (If no volume is quoted, use a V; it will probably cancel out) from the equation constructed in the first step, write out an expression for Kc. substitute values from third step and calculate the value of Kc with any units Example 1 One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc. CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) moles (initially) 1 1 0 0 moles (at equilibrium) 1 - 2/3 1 - 2/3 2/3 2/3 equilibrium concs. 1/3 / V 1/3 / V 2/3 / V 2/3 / V V = volume (dm3) of the equilibrium mixture :@D@@@1EPB j3&        v  <  2  A  )    G   F ?   H 7 8N    fB   6DolB   <BDo{>T   # L)?fB   6DolB  <BDo{  0S:s CALCULATIONS INVOLVING KcF(2fffB  s *޽h ? ̙33 0L0 2*( w dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0!  0o:  construct the balanced equation, including state symbols (aq), (g) etc. determine the number of moles of each species at equilibrium divide moles by volume (in dm3) to get the equilibrium concentrations in mol dm-3 (If no volume is quoted, use a V; it will probably cancel out) from the equation constructed in the first step, write out an expression for Kc. substitute values from third step and calculate the value of Kc with any units Example 1 One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc. CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) moles (initially) 1 1 0 0 moles (at equilibrium) 1 - 2/3 1 - 2/3 2/3 2/3 equilibrium concs. 1/3 / V 1/3 / V 2/3 / V 2/3 / V V = volume (dm3) of the equilibrium mixture Kc = [CH3COOC2H5] [H2O] [CH3COOH] [C2H5OH] B@D@@@&@@1EP@ j3&        v           <  2  A  )    G    : F ?   H 7 8N    fB   6DolB   <BDo{>T   # L)?fB   6DolB  <BDo{^B  6Do    0:s CALCULATIONS INVOLVING KcF(2fffB  s *޽h ? ̙33 0L0 K( w dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0>  0:  construct the balanced equation, including state symbols (aq), (g) etc. determine the number of moles of each species at equilibrium divide moles by volume (in dm3) to get the equilibrium concentrations in mol dm-3 (If no volume is quoted, use a V; it will probably cancel out) from the equation constructed in the first step, write out an expression for Kc. substitute values from third step and calculate the value of Kc with any units Example 1 One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc. CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) moles (initially) 1 1 0 0 moles (at equilibrium) 1 - 2/3 1 - 2/3 2/3 2/3 equilibrium concs. 1/3 / V 1/3 / V 2/3 / V 2/3 / V V = volume (dm3) of the equilibrium mixture Kc = [CH3COOC2H5] [H2O] = 2/3 / V . 2/3 / V = 4 [CH3COOH] [C2H5OH] 1/3 / V . 1/3 / V"@D@@@N@,@1@ j3&        v       3   <  2  A  )    G    w F ?   H 7 8N    fB   6DolB   <BDo{>T   # L)?fB   6DolB  <BDo{^B  6Do  ^B  6DoW   0 :s CALCULATIONS INVOLVING KcF(2fffB  s *޽h ? ̙33i 0L0 (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0z  0 r0  Example 2 Consider the equilibrium P + 2Q R + S (all species are aqueous) One mole of P and one mole of Q are mixed. Once equilibrium has been achieved 0.6 moles of P are present. How many moles of Q, R and S are present at equilibrium ? P + 2Q R + S Initial moles 1 1 0 0 At equilibrium 06 02 04 04 (04 reacted) (2 x 04 reacted) (get 1 R and 1 S for every P that reacts) 1- 06 remain 1- 08 remain Explanation " if 0.6 mol of P remain of the original 1 mol, 0.4 mol have reacted " the equation states that 2 moles of Q react with every 1 mol of P " this means that 0.8 (2 x 0.4) mol of Q have reacted, leaving 0.2 mol " one mol of R and S are produced from every mol of P that reacts " this means 0.4 mol of R and 0.4 mol of S are present at equilibriumy@r@  #fMf f*f6 g  F ?   8N    fB   6DfolB   <BDfo{>T   # L)?fB   6DfolB  <BDfo{F ?   x ]8N   fB  6DfolB  <BDfo{>T  # L)?fB  6DfolB  <BDfo{  0rs CALCULATIONS INVOLVING KcF(2fffB  s *޽h ? ̙33 0L0 yq  (      0 rs XCALCULATIONS INVOLVING GASES (2fdB   <D)@0A   0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0   00r Method carried out in a similar way to those involving concentrations one has the choice of using Kc or Kp for the equilibrium constant when using Kp only take into account gaseous species for the expression use the value of the partial pressure of any gas in the equilibrium mixture pressure is usually quoted in Nm-2 or Pa - atmospheres are sometimes used as with Kc, the units of the constant Kp depend on the stoichiometry of the reaction @@@a  ,  4  /f    +          B   s *޽h ? ̙33  0L0 + # 0(    0;rs XCALCULATIONS INVOLVING GASES (2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0;  0xCr Method carried out in a similar way to those involving concentrations one has the choice of using Kc or Kp for the equilibrium constant when using Kp only take into account gaseous species for the expression use the value of the partial pressure of any gas in the equilibrium mixture pressure is usually quoted in Nm-2 or Pa - atmospheres are sometimes used as with Kc, the units of the constant Kp depend on the stoichiometry of the reaction USEFUL RELATIONSHIPS total pressure = sum of the partial pressures partial pressure = total pressure x mole fraction mole fraction = number of moles of a substance number of moles of all substances presentL@@@Hq@5@=@a  ,  4  1f    +           ^B  6Do  B  s *޽h ? ̙33 0L0 C;(    0 ers ^"MOLE FRACTION AND PARTIAL PRESSURE #(2#fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  0T   # L)?fB   6DfolB  <BDfo{B  s *޽h ? ̙33  0L0 b Z P ( w   0Hus CALCULATIONS INVOLVING KpF(2fffdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  0Ruk  WExample When nitrogen (1 mole) and hydrogen (3 moles) react at constant temperature at a pressure of 8 x 106 Pa, the equilibrium mixture was found to contain 0.7 moles of ammonia. Calculate Kp . N2(g) + 3H2(g) 2NH3(g) moles (initially) 1 3 0 moles (equilibrium) 1 - x 3 - 3x 2x (x moles of N2 reacted) mole fractions (1-x)/(4-2x) (3-3x)/(4-2x) 2x/(4-2x) (total moles = 4 - 2x) partial pressures P.(1-x)/(4-2x) P.(3-3x)/(4-2x) P.2x/(4-2x) (total pressure = P ) x@nVfffffff   e^&   F ?   8N    fB   6DfolB   <BDfo{>T   # L)?fB   6DfolB  <BDfo{B  s *޽h ? ̙33  0L0 $  ` ( w   0vus CALCULATIONS INVOLVING KpF(2fffdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0y  0ui   Example When nitrogen (1 mole) and hydrogen (3 moles) react at constant temperature at a pressure of 8 x 106 Pa, the equilibrium mixture was found to contain 0.7 moles of ammonia. Calculate Kp . N2(g) + 3H2(g) 2NH3(g) moles (initially) 1 3 0 moles (equilibrium) 1 - x 3 - 3x 2x (x moles of N2 reacted) mole fractions (1-x)/(4-2x) (3-3x)/(4-2x) 2x/(4-2x) (total moles = 4 - 2x) partial pressures P.(1-x)/(4-2x) P.(3-3x)/(4-2x) P.2x/(4-2x) (total pressure = P ) at equilibrium there are 0.7 moles of ammonia, so 2x = 0.7 and x = 0.35 the total pressure (P) = 8 x 106 Pa. C@nVfffffff   e^G  &  F F ?   8N    fB   6DfolB   <BDfo{>T   # L)?fB   6DfolB  <BDfo{B  s *޽h ? ̙33{ 0L0  p (    0us CALCULATIONS INVOLVING KpF(2fffdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  0 u Example When nitrogen (1 mole) and hydrogen (3 moles) react at constant temperature at a pressure of 8 x 106 Pa, the equilibrium mixture was found to contain 0.7 moles of ammonia. Calculate Kp . N2(g) + 3H2(g) 2NH3(g) moles (initially) 1 3 0 moles (equilibrium) 1 - x 3 - 3x 2x (x moles of N2 reacted) mole fractions (1-x)/(4-2x) (3-3x)/(4-2x) 2x/(4-2x) (total moles = 4 - 2x) partial pressures P.(1-x)/(4-2x) P.(3-3x)/(4-2x) P.2x/(4-2x) (total pressure = P ) at equilibrium there are 0.7 moles of ammonia, so 2x = 0.7 and x = 0.35 the total pressure (P) = 8 x 106 Pa. applying the equilibrium law Kp = (PNH3)2 with units of Pa-2 (PN2) . (PH2)3 C@i@@nVfffffff   e^G  7        @  w  R F ?   8N    fB   6DfolB   <BDfo{>T   # L)?fB   6DfolB  <BDfo{^B  6Do  H B  s *޽h ? ̙33p 0L0 ( w   0`us CALCULATIONS INVOLVING KpF(2fffdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  0 w< Example When nitrogen (1 mole) and hydrogen (3 moles) react at constant temperature at a pressure of 8 x 106 Pa, the equilibrium mixture was found to contain 0.7 moles of ammonia. Calculate Kp . N2(g) + 3H2(g) 2NH3(g) moles (initially) 1 3 0 moles (equilibrium) 1 - x 3 - 3x 2x (x moles of N2 reacted) mole fractions (1-x)/(4-2x) (3-3x)/(4-2x) 2x/(4-2x) (total moles = 4 - 2x) partial pressures P.(1-x)/(4-2x) P.(3-3x)/(4-2x) P.2x/(4-2x) (total pressure = P ) at equilibrium there are 0.7 moles of ammonia, so 2x = 0.7 and x = 0.35 the total pressure (P) = 8 x 106 Pa. applying the equilibrium law Kp = (PNH3)2 with units of Pa-2 (PN2) . (PH2)3 Substituting in the expression gives Kp = 1.73 x 10-14 Pa-2 @C@i@e@nVfffffff   e^G  7        .Z  w  |   F ?   8N    fB   6DfolB   <BDfo{>T   # L)?fB   6DfolB  <BDfo{^B  6Do  H B  s *޽h ? ̙33 0L0 X(  X  X 00wfM  STHE END0(2f<f  X 0 4w &FURTHER TOPICS ON CHEMICAL EQUILIBRIUMT(2 82 f0f<f  X 0;wp u) JONATHAN HOPTON & KNOCKHARDY PUBLISHING0*(2&f(fJ  X C "A KTRE: .  X <̙? x @X 0 0 X <̙? x @T 0 0H X 0޽h ? ̙33y___PPT10Y+D=' = @B +r ` & 0;%1]PWoYh0EF R_mpcwό֮xuE 3"v-H9E_T&+iOh+'0T hp   No Slide TitleHOPTONJONATHAN HOPTON903Microsoft PowerPoint@MMK&@@hNx@ GSg  )'    """)))UUUMMMBBB999|PP3f333f3333f3ffffff3f̙3ff333f333333333f33333333f33f3ff3f3f3f3333f33̙33333f333333f3333f3ffffff3f33ff3f3f3f3fff3ffffffffff3ffff̙fff3fffff3fff333f3f3ff3ff33f̙̙3̙ff̙̙̙3f̙3f333f3333f3ffffff3f̙3f3f3f333f3333f3ffffff3f̙3f3ffffffffff!___www4'A x(xKʦ """)))UUUMMMBBB999|PP3f3333f333ff3fffff3f3f̙f3333f3333333333f3333333f3f33ff3f3f3f3333f3333333f3̙33333f333ff3ffffff3f33f3ff3f3f3ffff3fffffffff3fffffff3f̙ffff3ff333f3ff33fff33f3ff̙3f3f3333f333ff3fffff̙̙3̙f̙̙̙3f̙3f3f3333f333ff3fffff3f3f̙3ffffffffff!___wwwýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýüýýýýýýýýýýýýüýýýüýýý____________________________________ýýüýýýýýýýüýüýüýýýýýý__________________________ý_______ýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýüýýýýýýýýýýýrsrýýýýýýýýýýüýýýýýýýýýýýýýýýKnýýýýýýýýýýýýýüýýýýýtnýýýýýýýüýýýýýýýýýýýnüýýüýüýýüýüýýýýüýüýýýüýýýüýýüýüýýrxKýýýýüýüýýýüýýýüýýüýüýýüýüýýýýüýüýýýýýýýýx-Onnýýýýýýýýýýýýýýýýýýýýýýýýýýýý-wnD-ýýýýýýýýýýýýýýýýýýýýýýýIsKEPOýýýýüýüýýýýýýýýýýüýýýýýýýýýýüýüÓtJVOýýýýýýýüýýýýýýýýýýüýüýýýýýýýýýýüýýxJ('Orýýýýýýýýýýýýýýýüýýýýýýýýý-xP'-OýýýýüýýýýýýýýýýýýýýýýýüxOO(OOüýýýýýýýýýýýüýüýüýýýýýýýüýýýýýýýý-OOP-ýüýüýýýýýýýüýýýýýýýýýýýüýüýüýýýýýýýrxOO'O'O'ýýýýýýýýýüýýýýýýýýýýüýýýýýýýýýüýx-ONOxrýýýýýüýýýýýýýýýüýýýýýýýýýýüýýO-O'-'O-ýýýýüýýýüýüýüýýýüýýýýýýýüýüýüýýýüýýýüýüýr--O'---OýýýýýýýüýüýüýýýüýýýüýüýüýýýüýýýýýýýüýýýýýýýO'-'-'-OýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýP---'--xýýýýýýýýýýýýýýýýýýýýýýýýýýýO-'-'-OýýýýýýýýüýýýüýýýýýýýüýüýýýüýýýýýýýýüýýsU-UOýýýýýýüýüýýýüýýýýýýýýüýýýüýýýýýýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýüýýýýýýýýýüýüýýüýýýüýýüýüýýüýüýýýýüýüýýüýýýüýýüýüýýüýüýýýýüýüýýüýýýüýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýüýýýýýýýüýýýýüýüýýýýýüýýýýüýýýýýýýüýýýýüýüýýýýýüýýýýüýýýýýýýüýýýýüýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýüýüýýýýýýýýýüýýýýýýýýýýüýüýýýýýýýýýüýýýýýýýýýýüýüýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýüýýýýýýýýüýýýýýýýýýüýýýýýýýýüýýýýýýýýýüýýýýýýýýüýýýýüýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýüýýýýýýýýýýýýüýýýüýýýýýýýüýüýýýýýýýýýýýýüýýýüýýýýýýýüýüýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýüýýýýýýýýýýýüýýüýüýýüýüýýýýüýüýýýüýýýüýýüýüýýüýüýýýýüýüýýýüýýýüýýüýüýýüýüýýýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýý_ýýýýýýýýýýýýýýý________________________________________________ýýýýüýüýýýýýýýý______ý___________________ü____________________________ýýüýüýýýýýýýýýýüýý______________________________________ýýýýýýýýýýýýý________ý__________ýý___________________ý__________ýýýýýýýýýýýü_____________________________________________üýýýýýýýýýýýüýüýüý______ü________ý______________ý________ýýýýýýýýüýüýüýýýýýý_______________________________________________ýýýýýýýüýýýýýýýýý______ý___ýý________ý__ý____________ý______ý__ý__ýýýüýýýýýýýýýýüýýýýýýüýýýüýüýüýýýüýýýýýýýüýüýüýýýüýýýüýüýüýýýüýýýýýýýüýüýüýýýüýýýüýüýüýýýüýýýýýýýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýý___________________________________ýýýýýýýýüýýýüýýýýýýýüýüýý__________________________________________ýýýýýýýüýýýüýýýýýýýüýüýýýýý_______________________________ýýýýýýýýýýýýýýýýýýýýýý__ý_________________________ýý____ý__ýýýýýýýýýýýýýýýýýý_______________________________ýýüýýýýýüýýýüýýýýý__________________________ý__ýýýüýýýýýüýýýüýýýýýý___________________________________ýýýýýýýýýýýýýýýýýýý______ý__________ý____ý_______ýý__ýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýüýýýýýýýýýüýüýýüýýýüýýüýüýýüýüýýýýüýüýýüýýýüýýüýüýýüýüýýýýüýüýýüýýýüýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýüýýýýýýýüýýýýüýýüýýýýýýýýüýýýüýýýýýýýüýýýýüýýý__________________ý_____________________ýýýýýýýýüýýýýýýý_____________ý____________________ýýýüýýýýýýýýýý_____________________________________________üýüýýýýýýýýýüýýýý_________________________________ýýýýýýýüýýýýýýýýýý___________________________________________________ýýýýýýýüýýýýýýýýüýýýýýýýýýüýýýýýýýýüýýýýýýýýýüýýýýýýýýüýýýýüýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýüýýýýýýýýýýýýüýýýüýýýýýýýüýüýýýýýýýýýýýýüýýýüýýýýýýýüýüýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýý՜.+,D՜.+,x4    On-screen Show z| "Times New Roman Arial BlackTahomaArial Courier NewDefault DesignSlide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9 Slide 10 Slide 11 Slide 12 Slide 13 Slide 14 Slide 15 Slide 16 Slide 17 Slide 18 Slide 19 Slide 20 Slide 21 Slide 22 Slide 23 Slide 24 Slide 25 Slide 26 Slide 27 Slide 28  Fonts UsedDesign Template Slide Titlesx 8@ _PID_HLINKSA0<489,4,Slide 4490,5,Slide 5478,7,Slide 7480,15,Slide 15481,17,Slide 17482,22,Slide 22316,3,Slide 3316,3,Slide 3316,3,Slide 3316,3,Slide 3'_ljJONATHAN HOPTONJONATHAN HOPTON  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root EntrydO)PicturesCurrent UserSummaryInformation(TPowerPoint Document(jDocumentSummaryInformation8