Empirical Chemical Kinetics time dependence of reactant and product concentrations e.g. for A + 2B → 3C + D d[A] 1 d[B] 1 d[C] d[D] rate =− =− = = d2d3ddtttt =ν In general, a chemical equation is 0∑ ii R i nt()− n (0 ) The extent of a reaction (the advancement) = ii ν i For an infinitesmal advancement dξ each reactant/product =ν ξ concentration changes by d[Ri ]i d . d1d[R]ξ By definition, the rate == i ν ddtti Reaction rates usually depend on reactant concentrations, e.g., rate= k [A]xy [B] order in B total order = x+y rate constant In elementary reaction steps the orders are always integral, but they may not be so in multi-step reactions. The molecularity is the number of molecules in a reaction step. Rate Laws d[A] Zero order: −=k dt 0 [A] = [A]00 -[A]t kt [A]0 t1 = 2 2k 0 0 t −=d[A] First order: k1[A] [A] ln[A] dt − = kt1 [A]t [A]0 e ln 2 0 t 0 t t1 = 2 k1 d[A] 2 Second order: −=2[A]k [A] dt 2 [A] [A] = 0 t + 12[A]kt20 0 t -1=+ -1 [A]t [A]02 2kt 1 t1 = 2 2[A]k20 [A]-1 0 t Second-order Kinetics: Two Reactants A + B k → products da rate=− =kab a = [A], b =[B] dt =− − ka()()00 x b x ddxa Since aa=− x, =− 0 ddtt dx So =−ka()() x b − x dt 00 x dx kt = ∫ ()()−− 0 axbx00 −111x =−dx ()−−−∫ ()() ab000 ax 0 bx 0 −1 ab = ln 0 ()− ab00 ab 0 or more usefully, aa=+−0 ln ln ()abkt00 bb0 Data Analysis “Classical” methods of data analysis are often useful to explore the order of reactions, or to display the results (e.g. A semi-log plot to demonstrate exponential decay). However, these methods should be avoided for quantitative data analysis, since errors (and thus weighting) can be distorted. e.g. [A] [A]-1 0 t 0 t Modern data analysis uses computer for direct curve fitting, e.g. by chi-square minimization. 2 χ2 ʈexpt-theory =  Á˜ data points ˯exptl. error A good fit has χ2 / degs. freedom= 1 The errors in model parameters are found by taking each in turn χ2 χ2 and finding the variation necessary to change min to min +1. To avoid correlation the fit must be re-optimized with all the other adjustable parameters variable. Temperature Dependence of Complex Reactions kk Assume some complex reaction A + B 12→→ C + D for which the overal reaction rate constant can be expressed in terms of the elementary steps: kknn12… k = 12 overall n3 k3 … -Ei / RT If each rate constant obeys the Arrhenius expression, kAii= e nn12 ʈAA12… knEnEnERToverall=+--exp - 1 1 2 2 3 3 … / Á˜n3 {}() ˯A3 … i.e. The Arrhenius parameters are kA= ni overall ’i i EnE= Âi ii The overall “activation energy” may be negative, if ni is negative and the corrresponding Ei is large enough. Temperature Dependence of Rate Constants − The Arrhenius expression,kA= e,Ea / RT often expressed as E lnkA=- lna , is empirical. RT Curvature in the Arrhenius plot is sometimes attributed to tunnelling, but other reasons include complex reactions, non-equilibrium kinetics, and temperature dependence of the pre-exponential factor, which arises naturally in most theories. dlnk The activation energy is defined by ERT= 2 a dT Various theories predict an expression of the form − = mERTth / kATth e E whence lnkAmT=+ ln ln −th th RT The empirical parameters are then related to theory by: =+ EEmRTath =−m A ATthexp{}() E a E th / RT Theoretical T Dependence of Rate Constants Simple Collision Theory A ∝ T1/2 −−8RT kpZT==πσ()eeEmin/2RT p() E min / RT πµ Transition State Theory A ∝ T ==RT‡// RT 1 ∆−∆SR‡‡ HRT kKc ee Lh Lh c0 This gives =∆‡‡ + =∆ −∆ν + EURTHa RTRTfor gases =∆‡ + Unimolecular gas reaction EHRTa e RT ∆ ‡ Ae= SR/ hL =∆‡ + Bimolecular gas reaction EHRTa 2 2 e RT ∆ ‡ Ae= SR/ hLc0 =∆‡ + Liquid phase reaction EHRTa e RT ∆ ‡ Ae= SR/ hLc0 T Dependence of State to State Reactants Rate constants have been determined for the reaction of OH with different vibrational states of H2. → OH + H2 (v) H2O + H =×−12 − 3 -1 -1 kRTv=0 9.3 10 exp( 18000 / ) cm molecule s =×−11 − 3 -1 -1 kRTv=1 6.0 10 exp( 11000 / ) cm molecule s The Boltzmann populations for these states of H2 are ==−∆ fQvv==01/ vib and f 1exp( ERTQ / ) / vib =− −∆−1 ∆= -1 where QERTEvib []1 exp( / ) and 49.6 kJ mol and the thermal rate constant can be expressed as =+ kT() fvv==00k f vv == 11k ln k(T) 1000/T.