PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16In the Binary Collision Model we made a good case for theIn the Binary Collision Model we made a good case for therate expression:rate expression:R={R={((ABAB))2 2 <u<urelrel> e > e -E-EAA/RT/RT} (N} (NAA/V)(N/V)(NBB/V)/V)Often, NOften, NAA/V and N/V and NBB/V are concentrations in units of molecules /V are concentrations in units of molecules per ml. To get these in moles per liter, just multiply by 1000/Nper ml. To get these in moles per liter, just multiply by 1000/N00!!So the Binary Collision Model predicts R=kSo the Binary Collision Model predicts R=kRRCCAACCBBBasically, the Binary Collision Model predicts a reaction Basically, the Binary Collision Model predicts a reaction rate that is first order in A, first order in B and second rate that is first order in A, first order in B and second order overall.order overall.B) First Order Reactions B) First Order Reactions [will need a model later!][will need a model later!]Assume this is first order to get Assume this is first order to get −dCAdt= kC= kCAA if reaction is 1 if reaction is 1stst order orderlnClnCAAº = - k (0) + gº = - k (0) + gIntegrate both sides:Integrate both sides:++ [dC[dCAA/C/CAA] = - ] = -  kdtkdtNeed to find g from initial conditions.Need to find g from initial conditions.CCA A = C= CAA0 0 ee-kt-ktlnClnCAAoolnClnCAAt Slope = -kSlope = -kCCAACCAAoottSlope = kSlope = kttln(Cln(CAAo o / C/ CAA))Equations for first order reactions areEquations for first order reactions arevery important. In the laboratory almostvery important. In the laboratory almostALL reactions can be made to APPEAR ALL reactions can be made to APPEAR First Order.First Order.Bonus * Bonus * Bonus * Bonus * Bonus * BonusBonus * Bonus * Bonus * Bonus * Bonus * BonusHalf Life or Half TimeHalf Life or Half Time lnCACAo⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =−ktRate (mol L-1 s-1 )001.01.02.02.01 x 101 x 10-5-5[N[N22OO55] (mol L] (mol L-1-1))2 x 102 x 10-5-53 x 103 x 10-5-5NN22OO55  2NO 2NO22 + 1/2 O + 1/2 O22 Example of a first order reaction.Example of a first order reaction.Rate =Rate =-d[N-d[N22OO55]/dt =]/dt =k[Nk[N22OO55]]t = tt = t00 , C , CAA = C = CAA00Initial conditionsInitial conditions1CA=1CA0+k(t−t0),Take tTake t00 = 0 = 0 C) 2C) 2ndnd Order Kinetics: A Order Kinetics: A  Products Products- -  dC dCAA/[C/[CAA]]2 2 = k = k  dt dt 1/[C1/[CAA]] = kt + g= kt + gAt tAt t1/2 1/2 ,,CCAA = ( = (CCAA0 0 / 2/ 2) and) and1/[C1/[CAA]] = 1/[C= 1/[CAA00] + kt] + kttt1/21/2 = 1 / kC = 1 / kCAA002) Initial Rates Method2) Initial Rates MethodIf x<<cIf x<<c00 dx/dt = const = kcdx/dt = const = kc00 (sure to be true if t is (sure to be true if t is small enough!)small enough!)c = cc = c00 - x - xWhere cWhere c00 is the initial concentration and x is a is the initial concentration and x is a function of time, x = x(t). x is simply the amount reacted.function of time, x = x(t). x is simply the amount reacted.1st order reactiondCdCAA/dt = -kC/dt = -kCAACCAAooCCAAt Initial Initial Slope = Slope = -k-k CCAAooCCAACCAAoottSlope NOT ConstantSlope NOT ConstantBlow up first one percentBlow up first one percentof Cof CAA vs t curve vs t curveInitial part of curve will lookInitial part of curve will looklike a straight line if t is small!like a straight line if t is small!tcc/c/ t = -k t = -k CCAAooMeasure Measure c vs. c vs.  t for first 1% of reaction. Here, c t for first 1% of reaction. Here, c00>>x, >>x,  c c  dx and dx and  t t  dt dt Know cKnow c00 , measure c and t, ∆ ∆ , measure c and t, ∆ ∆ obtain k obtain knnthth Order Reactions Order ReactionsA + B + C A + B + C  products products a= initial conc of Ab= initial conc of Bc= initial conc of CBonus * Bonus * Bonus * Bonus * Bonus * BonusBonus * Bonus * Bonus * Bonus * Bonus * Bonusdxdt⎛ ⎝ ⎞ ⎠ 2=ka2n1b1n2c1n3Can do a similar trick for nCan do a similar trick for n22 , n , n33(Note, have kept b, c constant!)(Note, have kept b, c constant!)(dx/dt)(dx/dt)11 and (dx/dt) and (dx/dt)22 are measured in the laboratory, are measured in the laboratory, while awhile a11 and a and a22 are known quantities. are known quantities.1) Exponents in rate law do not depend on stoichiometric1) Exponents in rate law do not depend on stoichiometric coefficients in chemical reactions.coefficients in chemical reactions.2)What is the detailed way in which the reactants are converted2)What is the detailed way in which the reactants are convertedinto products? This is not described by the chemical into products? This is not described by the chemical equation, which just accounts for mass balance.equation, which just accounts for mass balance.3) Rate at which reaction proceeds and equilibrium is3) Rate at which reaction proceeds and equilibrium isachieved, depends on the achieved, depends on the MechanismMechanism by which by whichreactants form products.reactants form products.MechanismMechanism ConceptConceptElementary Reactions:Elementary Reactions: these are hypothetical these are hypothetical constructs, or our guess about how reactants are converted constructs, or our guess about how reactants are converted to products.to products.The The Mechanism Mechanism is a is a setset of of Elementary ReactionsElementary Reactions!!HH22ClCl22H HClClHClHCl+suppose reaction actually takes place during a collision of suppose reaction actually takes place during a collision of HH22 with Cl with Cl22 (this is the (this is the binary collision picturebinary collision