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Reaction Thermodynamics Review

Reaction Thermodynamics Review. D G rx – indicates direction of reaction given the current distribution of reactants and products. D G ˚ rx – indicates the free energy difference between reactants and products at their standard state concentrations).

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Reaction Thermodynamics Review

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  1. Reaction Thermodynamics Review DGrx – indicates direction of reaction given the current distribution of reactants and products. DG˚rx – indicates the free energy difference between reactants and products at their standard state concentrations). The intrinsic‘favorability’ of a reaction. Q – indicates the current distribution of reactants and products. DG = DG˚ + RT ln Q K – indicates the equilibrium distribution of reactants and products. DG˚ = -RT ln K

  2. Reaction Kinetics r – The rate of the reaction – M s-1 k – The rate constant for the reaction – Indicates the ‘intrinsic’ speed of a reaction. Ea –The activation energy for a reaction – Indicates the free energy difference between the reactants and the transition state. Rate law – Indicates the dependence of r on k and the concentrations of reactants and any other reagentthat influences the rate of a reaction.

  3. Reaction Kinetics aA + bBcC + dD Rate (r) = 1/nidci/dt = -(1/a) (d[A]/dt) or ..... + (1/c) (d[C]/dt) etc. rate = k [A]a [B]b [L]l k = rate constant (if you determine k using the change in a reagent for which the stoichiometric coefficient ≠ 1 you must also adjust for this. A and B are reactants. L = catalyst, intermediate a/b/l = reaction orders with respect to A, B, L, respectively. overall order, n = a + b + l.

  4. Elementary Reactions A + B C + D and r = k [A] [B] Partial orders of reactants = stoichiometric coefficient i.e. a = a and b = b. 2. no catalysts or intermediates in rate law. 3. reverse reaction is also elementary 4. Represents the actual ‘collision’ that takes place resulting in the change in molecular arrangement. 5. Typically n will be 2 or less for an elementary reaction (and its reverse reaction).

  5. e.g.... A + C  I + D I + B  F + C Mechanism – A reaction is represented as a series of elementary steps that add up to overall stoichiometry and represent the actual collision order in the reaction. slow fast stoichiometry: A + B  D + F I is intermediate and C is catalyst r = k[A][C] Experimental Goals Determine rate law – Find n and k e.g. r = k [A] [C], n = 2, and k = r/([A][C]). • Determine mechanism(s) consistent with rate expression 3. Determine Activation energy by T dependence of k.

  6. Most Common Reaction Orders 1st order: rate = k [A] rate = k [A]2 or.... 2nd order: rate = k [A][B] rate expression can include ...... orders > 2, half-integral orders, inverse dependency – [X] in denominator

  7. 1st Order Reactions half-life ln([A]/[Ao]) = -kt r = -d[A]/dt = k[A] if [A] = [Ao]/2 then t = t1/2 & ..... -d[A]/[A] = k dt [Ao][A]d[A]/[A] = -k 0tdt ln 0.5 = -kt½ t½= ln 2/k ln [A] - ln[Ao] = ln([A]/[Ao]) = -kt Linear: ln [A] = -kt + ln [Ao] t½= 0.693/k [A] = [Ao] e(-kt) t½is independent of [Ao]

  8. 2nd Order Reactions half-life r = - d[A]/dt = k[A]2 1/[A] – 1/[Ao] = kt -d[A]/[A]2= -k dt 2/[Ao] - 1/[Ao] = kt½ 1/[Ao] = kt½ [Ao][A] [A]-2 d[A] = -k 0tdt 1/[A] – 1/[A0] = kt linear 1/[A] = kt + 1/[A0] t½= 1/(k[Ao]) [A] = [Ao]/(1 + kt[Ao]) as [Ao]  t½

  9. 0 Order Reactions (rare – some free radical reactions) r = -d[A]/dt = k ∫AoAd[A] = k ∫0tdt [A] = kt + [Ao] [Ao] Plot [A] vs. t slope = k Yint = [Ao] [A] t

  10. Determining Reaction Order, n Half-life Method 1st order – t½ is constant regardless of [A]0. 2nd order – t½ doubles as [A]0↓ by ½. plot lnt½vs. ln [Ao]  slope = 1-n 115 (115) 230 (115) 345 167(333)500 Advantage: Single experiment Disadvantage: Requires reaction integrity over multiple half-lives unless ‘fraction’ < ½ used.

  11. Determining Partial Order (a) Initial rate method ― r = k [A]a [B]b 1. vary [Ao] while holding [Bo] etc. cst. 2. find initial rate r from plot of [A] vs. t r2/r1 = ([A0,2]/[A0,1])a log (r2/r1) = a log ([A0,2]/[A0,1]) two data points: a = log(r2/r1)/log([A0,2]/[A0,1]) Multiple data points: plot log (ro) vs. log [Ao]: slope = a 3. repeat for other reagents in rate expression 20.4 - p726

  12. AB* A + B G C + D Transition State theory (collision theory) A + B → C + D Boltzmann Distribution N2/N1 =exp(-Ea/RT) Ea where N2 = # collisions leading to reaction & N1= total # collisions DG Eadetermines rate - DG° determines Equil.

  13. F2(g) + 2ClO2(g) 2FClO2(g) Determining the Rate Law X = 1 rate = k [F2]x[ClO2]y r2 = k•(2[F2])x•[ClO2]y = k•2x•[F2]x•[ClO2]y = 2x = 2 r1 k•[F2]x•[ClO2]y k• [F2]x•[ClO2]y r2 = k•[F2]x•(4•[ClO2])y = k•[F2]x•4y•[ClO2]y = 4y = 4 r1 k•[F2]x•[ClO2]y k•[F2]x• [ClO2]y y = 1 rate = k [F2][ClO2] ― Note that partial orders ≠ reaction coefficients. Problem 13.70

  14. Arrhenius Activation Energy The T dependence of reaction rates is due to the dependence of k on T. This in turn is due to the dependence of Ea on T. Through empirical observation Arrhenius determined that …. k = A exp(-Ea/RT) ln k = (-Ea/R)(1/T) + ln A plot ln k vs. 1/T slope = -Ea/RYint = ln A or .. ln (k2/k1)/(1/T2 – 1/T1) = -Ea/R

  15. H2 + I2 2HI ln k = (-Ea/R)(1/T) + ln A Ea = 160 kJ mol-1

  16. REACTION MECHANISM 1. List of Elementary Steps 2. Must add to overall stoichiometry 3. Must be “consistent” with Rate Law 1. Rate Determining Step Method (RDS) 2. Steady State Method (SS)

  17. A +BC + D C+ EB+ F A + ED + F C is an intermediate Product in an early step, reactant in a later step. Doesn’t appear in stoichiometry. May appear in rate law B is a catalyst Reactant in an early step, product in a later step. Doesn’t appear in stoichiometry. Must appear in the rate law.

  18. H+ + HNO2 + FNH2 FN2+ + 2H2O Rate Law: r = k [H+] [HNO2] [Br-] Br- must be catalyst or intermediate and must show up in mechanism. FNH2 is reactant that is not in the rate law. It must show up in the mechanism in a later step. If an RDS mechanism is sufficient to explain rate law then it must be a reactant in a step after the rate-determining step.

  19. k1 H+ + HNO2 H2NO2+ fast k-1 k2 H2NO2+ + Br-  ONBr + H2O slow k3 ONBr + FNH2 FN2+ + H2O + Br- fast H+ + HNO2 + FNH2 FN2+ + 2H2O Rate Law: r = k [H+] [HNO2] [Br-] r = k2 [H2NO2+] [Br-] [H2NO2+] = k1/k-1 [H+][HNO2] r = k [H+][HNO2][Br-] k = (k2k1/k-1)

  20. k1 H+ + HNO2 H2NO2+ k-1 k2 H2NO2+ + Br-  ONBr + H2O k3 ONBr + FNH2 FN2+ + H2O + Br- [ONBr] = k2 [H2NO2+][Br-]/(k3[NH2]) r =k3k2 [H2NO2+][Br-] [FNH2]/(k3[NH2]) apply ss assumption to H2NO2+ H+ + HNO2 + FNH2 FN2+ + 2H2O Rate Law: r = k [H+] [HNO2] [Br-] r = k3 [ONBr][FNH2] d[ONBr]/dt = 0 k2 [H2NO2+][Br-] = k3 [ONBr][NH2]

  21. k1 H+ + HNO2 H2NO2+ k-1 k2 H2NO2+ + Br-  ONBr + H2O r = k2k1 [Br-][H+][HNO2] k-1 + k2[Br-] k1[H+][HNO2] k-1 + k2[Br-] [H2NO2+] = r =k2[H2NO2+][Br-]/[NH2] apply ss assumption to H2NO2+ H+ + HNO2 + FNH2 FN2+ + 2H2O Rate Law: r = k [H+] [HNO2] [Br-] r = k1[H+][HNO2] = k-1[H2NO2+] + k2 [H2NO2+][Br-]

  22. r = k2k1 [Br-][H+][HNO2] k-1 + k2[Br-] H+ + HNO2 + FNH2 FN2+ + 2H2O Rate Law: r = k [H+] [HNO2] [Br-] SS RDS r = k [H+][HNO2][Br-] k = (k2k1/k-1) same as RDS mech. when ... k2[Br-] << k-1

  23. D  2M 2 (M + B  P) RDS Half Orders in Rate Law ..... D + 2B  2P Reactant split in first step - prior to RDS r = k2[M][B] & [M] = (k1/k-1 [D])1/2 r = k[B][D]1/2

  24. k1 Hg22+ Hg2+ + Hg fast k-1 k2 Hg + Tl3+ Hg2+ + Tl+ slow Term in denominator of rate law Hg22+ + Tl3+ 2Hg2+ + Tl+ Look for P that is co-product prior to RDS … or I that is P in first step & R after RDS r = k2 [Hg][Tl3+] [Hg] = k1[Hg22+]/(k-1[Hg2+]) r = k [Hg22+][Tl3+]/[Hg2+]

  25. Unimolecular Reactions Still involve some type of collision. A  B (+ C) A + M  A* + M A*  B (+ C) r = k[A]

  26. Enzyme Kinetics Mechanism: k1 1. E + S  ES k2 k3 2. ES  E + P r = k3 [ES] k1[E][S] = (k2 + k3) [ES] [ES] = k1/(k2 + k3) [E][S] [ES] = [E][S]/KM(KM = (k2+k3)/k1 r = k3/KM [E][S] If [S] >> KM then [ES] = [E]tot and ….. r = k3 [E]tot

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