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ChE 452 Lecture 09 Mechanisms & Rate Equations 1 Objective Describe relationship between rate equation and mechanism (Quasi steady state approximation) Why does it arise When does it work When does it fail 2 Previous Lectures: Measurement Of Rate Laws For Reactions Rate laws are hard to determine experimentally. • • • Accurate rate laws only from direct rate measurements Even then, more than one rate law will fit a given set of data. (see example 3.C) The optimal parameters and the quality of the fit depend on the way the data is fit as well as the rate law. • An accurate rate law fit by a poor procedure can look worse than the wrong rate law fit by a better procedure. 3 Today: How Else Can We Get A Rate Law? • • • Start with mechanisms Derive fundamental equations Fit rate data 4 Method To Find A Rate Law Find the mechanism of the reaction Computationally Experimentally Use the quasi-steady state approximation to derive a rate equation Generally more accurate-but we need a mechanism 5 Idea That Reactions Follow Mechanisms First Proposed By Van’t Hoff Experimentally no correlation between reaction and rate law • Thermo, MD suggested simple relationship between rate law and mechanism • Van’t Hoff suggested there must be extra reactions • 6 Examples Considered By Van’t Hoff Reaction 4PH 3 P4 6H 2 2AsH 3 As 2 3H 2 2PH 3 4O 2 P2 O 5 + 3H 2 O Rate equation (2.T.1) r AsH 3 = - k 4 (2.T.5) r PH 3 = - k 5 PH 3 C12 H 22 O 11 H 2 O C 6 H 12 O 6 C 5 H 9 O 5 CH 2 O H (2.T.7) H+ (2.T.9) H+ CH 3 COOH ROH CH 3 COOR H 2 O AsH 3 (2.T.3) H+ CH 3 COOR H 2 O CH 3 COOH ROH rPH3 k 3 PH 3 (2 .T.11) ClCH 2 COOH H 2 O HOCH 2 COOH HCl O 21/ 2 rS - k 6 [sucrose] [ H + ] rAc k 7 CH 3COOR H (2.T.2) (2.T.4) (2.T.6) (2.T.8) (2.T.10) rAc k 8 CH 3 COOH ROH H rC 2 H 3ClO 2 k 9 C 2 H 3 ClO 2 (2.T.12) (2.T.14) (2.T.13) 7 All Reactions Actually Occur By A Series of Chemical Transformations Called Elementary Reactions Elementary reaction – a reaction which goes from reactants to products without going through any stable intermediates Mechanism – the sequence of elementary steps which occur when the reactants come together to form products. H CH 3CH 2 HC = CH 2 H CH 3CH 2 HC `CH 2 H + CH 3CH 2 HC `CH 2 CH 3 HC = CHCH 3 H H + CH 3CH 2 HC `CH 2 CH 3CH 2 HC = CH 2 H H + CH 3CH 2 HC `CH 2 H 2 C = CHCH 2 CH 3 H + (a) (b) (c) (d) (4.13) 8 Other Key Definitions Continued • • • • • • • Reactive intermediate Molecularity Unimolecular Bimolecular Termolecular Overall reaction Stoichiometric reaction 9 All Elementary Reactions Have At Least Two Reactants And Two Products H + HBr H2 + Br (4.11) X + H2 2H + X (4.12) 10 Kinetics Of Elementary Reactions 2 A+BP+Q 4 2A P Q (4.19) r2 = k2[A][B] r4 k 4[A] rA = -k2[A][B] rA 2k4 [A]2 (4.20) (4.21) 2 (4.23) (4.24) 11 Multiple Reactions 5 rA ri A ,i i =1 (4.25) Memorize this equation 12 Next: Example Illustrating Rates Of Overall Reactions In terms Of Elementary Rates; H CH3CH 2 HC = CH 2 H + CH3CH 2 HC `CH 2 H + CH CH H C C H CH HC = CHCH H ` 3 2 2 3 3 H + CH3CH 2 HC `CH 2 CH3CH 2 HC = CH 2 H H + CH CH H C C H H C = CHCH CH H ` 3 2 2 2 2 3 (a) (b) (c) (d) (4.13) 1 + A + H+ I P + H 2 (4.44) 13 Differential Equations For Each Specie 5 rA riA,i d[ I ] r1 r2 r3 dt (4.45) i 1 (4.25) d[A] r2 - r1 dt (4.46) 14 Next Substitute Rate Laws A 1 3 + + + H I P + H 2 r2 k 2 [I] (4.48) (4.44) r3 k 3[I] (4.49) 15 Resultant Differential Equations d[A] + k 2[ I ] - k1[A][H ] dt (4.50) d[ I ] + k1[A][H ] - (k 2 + k3 )[ I ] dt (4.51) 16 Integration Of The Rate Equation Three approaches: • Analytical integration of the differential equations • Numerical integration of the differential equations • Approximate integration of the rate equation 17 Analytical Solution: d[A] + k 2[ I ] - k1[A][H ] dt (4.50) d[ I ] + k1[A][H ] - (k 2 + k3 )[ I ] dt (4.51) 2 equations and 2 unknowns 18 Analytical Solution 0 (k 4 k 6 )exp(-k 5t) - (k 5 - k 6 )exp(-k 4 t) (k 4 k 5 ) (4.53) k 6 )(k 6 - k 5 ) exp(-k 5t) - exp(-k 4 t) k 2 (k 4 k 5 ) (4.54) [A] = [A] 0 (k 4 [I] = [A] 1 k 4 k 6 k 2 k3 2 k6 k 2 k3 2 4k6k3 (4.55) 1 k5 k 6 k 2 k3 2 k6 k 2 k3 2 4k6k3 (4.56) k 6 k1 H (4.57) Derivation 21 Plot Of Result 1 Concentration, Mol/Lit [A] 0.8 [I] S 0.6 0.4 [P] 0.2 [I] 0 0 1 2 3 4 Time, Min 22 Next: The Pseudo-Steady State Approximation According to the pseudo-steady-state approximation, one can compute accurate values of the concentrations of all of the intermediates in a reaction by assuming that the net rate of formation of the intermediates is negligible (i.e., the derivatives with respect to time of the concentrations of all intermediates are negligible compared to other terms in the equation.) 23 For Our Example d[I] 0 k1[A][H ] (k 2 k3 )[I] dt (4.61) Or k [ H ] X 1 [I] [A ] k k 3 2 (4.60) 24 7E-09 7E-09 6E-09 6E-09 Concentration, Mol/Lit Concentration, Mol/Lit Plot Of [I]X Calculated By Steady State To Exact 5E-09 4E-09 Key 3E-09 Exact 2E-09 Steady State Approx 1E-09 0 5E-09 4E-09 Key 3E-09 Exact 2E-09 Steady State Approx 1E-09 0 5E-08 1E-07 1.5E-07 Time, Mins 2E-07 0 0 1 2 3 4 Time, Mins 25 (4.61) 0.25 0.25 0.2 0.2 Rate, Mole/Lit-Min d[ I ] k1[A][H + ] - (k 2 + k3 )[ I ] dt Rate, Mole/Lit-Min Why Does The Steady State Approximation Work? 0.15 Key 0.1 k1 [A][H +] 0.15 Key 0.1 k1 [A][H +] (k 2+ k 3)[I] (k 2+ k 3)[I] 0.05 0.05 Deriv 0 0 5E-08 1E-07 1.5E-07 Time, Mins Deriv 2E-07 0 0 1 2 3 4 Time, Mins Derivative much smaller than other terms in the equation. 26 Steady-State Approximation Does Not Always Work • Assumes derivatives small compared to other terms in equation • Generally holds only when the concentration of intermediate small compared to reactants and products • Need rate constants for destruction of intermediates to be more than a factor of 100 larger than the rate constants for formation of intermediates • Common examples of failures include: • Cyclic reactions, biological reactions • Most reactions in plasma reactors • Explosions, combustion 27 Next: Example H 2 Br2 2HBr (4.67) 1 X + Br2 2Br + X 2 Br + H 2 HBr + H 3 H + Br2 HBr + Br 4 X + 2Br Br2 + X 5 H + HBr H 2 + Br Calculate as rate equation. 28 Solution rBr 2 r1 r3 r4 (4.69) r1 k1[X][Br2 ] r3 k3[H][Br2 ] r4 k1[X][Br]2 1 X + Br2 2Br + X 2 Br + H 2 HBr + H 3 H + Br2 HBr + Br 4 X + 2Br Br2 + X 5 H + HBr H 2 + Br rBr2 k1[X][Br2 ] k 3[H][Br2 ] k 4[X][Br] 2 29 Solution Continued d[Br] 2 r1 r2 r3 2 r4 r5 dt (4.74) d[Br] rBr 2 k1[X][Br2 ] - k 2 [H 2 ][Br] + k 3[H][Br2 ] dt 2 k 4 [ Br]2 [X] + k 5[H][HBr] (4.75) d[H] rH = k 2 [H 2 ][Br] - k 3[H][HBr2 ] - k5[H][HBr] dt (4.76) 30 Next Apply S.S. Approximation 0 2k1[X][Br 2 ] - k 2 [H 2 ][Br] + k 3[H][Br 2 ] 2k 4 [Br] [X] + k 5 [H][HBr] 2 (4.77) 0 k 2[H 2 ][Br] - k 3[H][Br2 ] - k 5[H][HBr] (4.78) Adding the two equations together 0 2k1[X][Br2 ] 2k 4[Br] [X] 2 (4.79) 31 Solution Continued Repeating the equation 0 2k1[X][Br2 ] 2k 4[Br]2[X] (4.79) Solving for [Br] 1/ 2 k [Br] = 1 k4 [Br2 ]1/ 2 (4.80) 32 Solving for [H] 0=k 2[H2 ][Br]-k3[H][HBr2 ]-k5[H][HBr] (4.78) Moving the k3 and k4 terms to the other side 1/ 2 k1 k 3[H][Br2 ] k 5 [H][HBr] k 2 [H 2 ][Br]=k 2 [H 2 ] [Br2 ]1/2 k4 Substituting or [Br] 1/ 2 k1 k 3[H][Br2 ] k 5 [H][HBr] k 2 [H 2 ][Br]=k 2 [H 2 ] [Br2 ]1/2 k4 1/ 2 k1 Answer: k 2 [H 2 ][Br2 ]1/2 k4 [H] = k 3 [Br2 ] + k 5 [HBr] (4.82) 33 We Can Extend This Procedure To Any Reaction. 1) Set up the differential equation for the species of interest in terms of rate of all of the elementary reactions using equation (4.25) to keep track of the coefficients. 2) Substitute the expression for the rate of each of the elementary reactions using equations from section 4.3. 3) Set the derivatives of the intermediate concentrations to zero. 34 General Steps Continued 4) Eliminate terms in the expression in (1) which contain the concentrations of unstable intermediates other than the species of interest. (Usually done by adding equations together) 5) Solve the reactant expression for the concentration of the species of interest. 35 Discussion Problem Derive an equation for [H] 1 2OH H 2 + O2 2 H 2O + H OH H 2 3 OH +O : H +O 2 4 OH +H O : +H 2 5 wall H 6 HO 2 +X H O2 + X 7 wall HO 2 +X 8 H 2O 2 + X 2HO + X (4.85) 36 After Considerable Algebra 2k1[H2 ][O2 ] [H]= k 5 +(k 6 [X]-2k 3 ) O2 (4.98) Derivation Notice that denominator goes to zero at a critical value of O2 [H] grows to ! (Not physical) 37 The Key Thing To Remember When Simplifying Rate Expressions • • • • Write down the differential equations in terms of the rates and then substitute in the rate equations. Keep track of what terms you want to eliminate and eliminate them. Adding equations together helps. Steady-state approximation sometimes fails. 40 Class Question What did you learn new today? 41