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RClick to edit Master text styles Second level Third level Fourth level Fifth level!      s */  ``  >*  s * F  `   @*  s *[  `   @*H  0޽h ? 3380___PPT10.Ws $Blank Presentation  (     s *  P    >*   s *p      @* d  c $ ?    s *   @  RClick to edit Master text styles Second level Third level Fourth level Fifth level!      0p#  `P   >*   0@9  `   @* H  0޽h ? 3380___PPT10.Ys]tp 3+ ( Pvdiroo   6=@@` %Chemical KineticsH  0޽h ? 3380___PPT10.Wsp @(  R  s *A?P1f 6  6p@ +` 6Rate of reaction = DX / Dt JAA A A,  6 8[ i%Phenolphthalein reacting with OH- ion(& H  0޽h ? 3380___PPT10.Xs~Dp `D( P|9 {  s *O  ! X = Concentration of Phenolphthalein (M) Time (s) 0.0050 0.0 0.0045 10.5 0.0040 22.3 0.0035 35.7 0.0030 51.1 0.0025 69.3 0.0020 91.6 0.0015 120.4 0.0010 160.9 0.00050 230.3 0.00025 299.6 0.00015 350.7 0.00010 391.2   6l 9 #M = moles/literH  0޽h ? 3380___PPT10.Xsp g_p(    s *6o  Use the data in the above table to calculate the rate at which phenolphthalein reacts with the OH- ion during each of the following periods: (a) During the first time interval, when the phenolphthalein concentration falls from 0.0050 M to 0.0045 M. (b) During the second interval, when the concentration falls from 0.0045 M to 0.0040 M. (c) During the third interval, when the concentration falls from 0.0040 M to 0.0035 M.  6в=    9 Problem 1 H  0޽h ? 3380___PPT10.Ys@O^ p SKP(  R  s *A?= H  6@ pk,$H___PPT10( f___PPT9H@^___PPTMac1180   hnamd` Arial&Monotype Typography    hnamd` Arial&Monotype Typography  bReaction rate varies with time. Instantaneous rate of reaction = dX/dt = the rate at each instant0c!'   6`  7@,$, g#Rate = k Xphenolphthalein .($  H  0޽h ? 33___PPT10.Xsh4+.ADO' D= @B D ' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*$%(+8+0+ + p ME((  (w ( s * P Calculate the rate constant for the reaction between phenolphthalein and the OH- ion if the instantaneous rate of reaction is 2.5 x 10-5 mole per liter per second when the concentration of phenolphthalein is 0.0025 M.(R ( 6BV    9 Problem 2  ( s *A?] ,$D, ( 6Z P' ,$, Wk = 0.010 s-1 .( H ( 0޽h ? 33#___PPT10.Zs{{5+| ZD' D= @B DF' = @BA?%,( < +O%,( < +D4' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+#.<*(%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*(%(+8+0+( + p   , ( 0X  , , 6Y  E@ OExpressing the rate of reaction  , 60~  2 HI = H2(g) + I2(g) RAIA I A , s *n ` Px___PPT10XP ___PPT9 ___PPTMac11x ,namd$ Times Roman  ,namd$ Times Roman  ,namd$ Times Roman  ,namd$ Times Roman  zProblem 4. Calculate the rate at which HI disappears at the moment when I2 is being formed at a rate of 1.8 x 10-6 moles per liter per second. Hint: HI disappears twice as fast as I2 is formed! 2IAI&A I AA&AI A $A$ , 6` @p` &k = rate constant for forming H2 or I2R'AIA I A , 60  zHk = rate constant for destroying HI%%AH , 0޽h ? 3380___PPT10.[s p 0g( {5 0 0 s *  & yGCan you figure out the rate of a reaction from its stoichiometry alone?HH` 0 6 G,$, CH3Br + OH- <=> CH3OH + Br -AIA I AIAI A  0 6  ; ,$, Rate = k (CH3Br) (OH-) R   0 6ɽ p ,$, +Experiments showed thatH 0 0޽h ? 33___PPT10.^sN**+D' D= @B D' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*0%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*0%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*0%(++0+0 ++0+0 ++0+0 +_ p 4/(  4 4 6` z GA very simlar reaction:= 4 6g P %(CH3)3CBr + OH- <=> (CH3)3COH + Br - &   $$ ((,,00  4 6@Q  h ,$, oRate = k [(CH3)3CBr] >    4 s *U p Jv ,$, R>The rate law for a reaction must be determined experimentally! 4 6߱ `  .One might expect a rate = k [(CH3)3CBr] [OH-] T/   H 4 0޽h ? 33h`___PPT10@.^s%g+gyY^D' D= @B DS' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*4%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*4?%(+p+0+4 ++0+4 + p 8A(  8 8 6  PWhy????? Why can t chemistry be easy????))F 8 6U PV,$,@___PPT10 V___PPT980J___PPTMac11$   hnamd` Arial&Monotype Typography  jSome reactions occur in a single step. Some don t!! &6' 8 s *s ,$, ]IThe rate of reaction will be determined by the slowest intermediate step. 8 s *r 0 ,$, ^. Rate-determining step 2 8 s *: r ,$, X( Rate-limiting step 2H 8 0޽h ? 33  ___PPT10 ._s{+Q'npDs ' D= @B D. ' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*8'%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*8'6%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*8J%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*8%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*8%(++0+8 ++0+8 ++0+8 ++0+8 +v p  <B(  <" < 6 @ 4Example: Decomposition of N2O5 into NO2 and O2 .l5    ; < 6 Pp,$, Step 1: N2O5 <=> NO2 + NO3 j    D < 6`OW,$, $Step 2: NO2 + NO3 <=> NO2 + NO + O2 j%     % < 6X; ,$, Step 3: NO + NO3 <=> 2 NO2T     < 6Pp,$, > unimolecular ,  < 6,$, = bimolecular ,  < 6   ,$, = bimolecular ,  < 6y f   molecularityH < 0޽h ? 33ld___PPT10D._st+v<D' D= @B Dw' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*<%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*<%%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<*<%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<* < %(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<* < %(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<* < %(+P+0+< ++0+< ++0+< ++0+ < ++0+ < ++0+ < + p .& @( ppt hDH @ @ 6{@@` N2O5 <=> NO2 + NO3 j    @ 6 P p = First order ,   @ 6P .p,$, pRate depends on (N2O5)>    @ 69_0 ZP ,$, (Decomposition of HI   @ 6R U 0 ,$, > second order ,  @ 6s  ,$, ~Rate depends on (HI)2 .HAI A H @ 0޽h ? 33j b ___PPT10B .as+7D* ' D= @B D ' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<* @%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<* @%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<* @ %(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+#.<* @%(++0+ @ ++0+ @ ++0+ @ ++0+ @ +" p D"(  D D 6} *What is the order of 2 NO + O2 <=> 2 NO2 ?>+   H D 0޽h ? 3380___PPT10.asA,p E=x(  xR x s *A?X x 6ͼ 6 \,Collision theory model of chemical reactions--y x s *k;PT @___PPT10 V___PPT980J___PPTMac11$   hnamd` Arial&Monotype Typography  The rate of any step in a reaction is proportional to the concentrations of the reagents consumed in that step. The rate law for a one-step reaction should therefore agree with the stoichiometry of the reaction.&qdH x 0޽h ? 3380___PPT10.asp 5-|( z | | s *@0PH@___PPT9" 6The rate laws for chemical reactions can be explained by the following general rules. The rate of any step in a reaction is directly proportional to the concentrations of the reagents consumed in that step. The overall rate law for a reaction is determined by the sequence of steps, or the mechanism, by which the reactants are converted into the products of the reaction. The overall rate law for a reaction is dominated by the rate law for the slowest step in the reaction.(WW @` | 6'7  7SummaryH | 0޽h ? 3380___PPT10.bsc "$ x$( PI  $R $ 3     ~ $ C Z   @    Question 1H $ 0޽h ? 3380___PPT10.Ys]t"$ bZ H(  HR H 3     ` H # ^  @   H H 0޽h ? 3380___PPT10.asfq"$ bZ0L( pl^p^ LR L 3     ` L # ^  @   H L 0޽h ? 3380___PPT10.asfq"$ bZ@P( pl^^ PR P 3     ` P # ^  @   H P 0޽h ? 3380___PPT10.asfq"$ bZPT( {5v TR T 3     ` T # P^  @   H T 0޽h ? 3380___PPT10.asfq"$ bZ`X( ^^ XR X 3     ` X # ^  @   H X 0޽h ? 3380___PPT10.asfq"$ bZp\( ^^v \R \ 3     ` \ # ^  @   H \ 0޽h ? 3380___PPT10.asfq"$ bZ`( {^^ `R ` 3     ` ` # ^  @   H ` 0޽h ? 3380___PPT10.asfq"$ bZd(  @^P^ dR d 3     ` d # p_  @   H d 0޽h ? 3380___PPT10.asfq "$ bZh( _P_  hR h 3     ` h # _  @   H h 0޽h ? 3380___PPT10.asfq "$ bZl( ^ ^P_ lR l 3     ` l # "_  @   H l 0޽h ? 3380___PPT10.asfq "$ bZp( {5v pR p 3     ` p # 00_  @   H p 0޽h ? 3380___PPT10.asfq "$ bZt(  _` _ tR t 3     ` t # >_  @   H t 0޽h ? 3380___PPT10.asfqdxp^RЀ3ÿ lHbP  @AL G@;b `B&VȀزB$DrO0hz7B$ YC,F{@#B$q ?ӤӉ`*'S ,@ʚ;2Nʚ;g4ddddn` ppp@ <4!d!d @gʚ;<4dddd @gʚ; h___PPT2001D<4X`___PPTMac11:@N  (namd Symbol"    (namd Symbol"    (namd Symbol"    (namd Symbol"    (namd Symbol"    (namd Symbol"  ? %          /     ! ": p E=x(  xR x s *A?X x 6ͼ 6 \,Collision theory model of chemical reactions--y x s *k;PT @___PPT10 V___PPT980J___PPTMac11$   hnamd` Arial&Monotype Typography  The rate of any step in a reaction is proportional to the concentrations of the reagents consumed in that step. The rate law for a one-step reaction should therefore agree with the stoichiometry of the reaction.&qdH x 0޽h ? 33___PPT10u.as+D=' n= @B +*p 5-|( z | | s *@0PH@___PPT9" 6The rate laws for chemical reactions can be explained by the following general rules. The rate of any step in a reaction is directly proportional to the concentrations of the reagents consumed in that step. The overall rate law for a reaction is determined by the sequence of steps, or the mechanism, by which the reactants are converted into the products of the reaction. The overall rate law for a reaction is dominated by the rate law for the slowest step in the reaction.(WW @` | 6'7  7SummaryH | 0޽h ? 33___PPT10u.bsc +D=' 4= @B + "$ bZ(  R  3     `  #