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Reaction kinetics: 1 st order reactions [A] t Decay reactions, like radio-activity; S N 1 reactions Rate: - Rewriting: - Integration gives: So: ln[A] t – ln[A] 0 = -ktor: -kt Since [A] 0 is constant we can also write: ln[A] t = -k 1 t + C

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Plotting ln[A] t against t gives a straight line with slope -k: The halflife, t 1/2, is defined as the time that is needed to reduce the concentration of the reactant to 50% of its original value. In formula: =-kt 1/2 so -kt

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Reaction kinetics: 2 nd order reactions So: -= - When [A] [B], this equation is mathematically rather complicated. A simplification reads as follows: take [P] = x, then [A] = [A] 0 – x and [B] = [B] 0 – x The rate then becomes: = k 2 ([A] 0 -x)([B] 0 -x) so Integration gives: Plottingagainst t gives a straight line with slope k 2 ([B] 0 -[A] 0 )

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Special cases: [A] 0 >>[B] 0 (pseudo-first order kinetics) Example: -in which k'=k 2 [H 2 O] This is a pseudo-first order reaction, since [H 2 O] is constant. The second-order rate constant k 2 can be calculated from k' and [H 2 O]. In a dilute aqueous solution, [H 2 O]=55 M.

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Special cases: [A] 0 = [B] 0 Integration gives: Plotting ofagainst t gives k 2 as the slope.

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Reversible reactions Take the simplest possibility: On t = 0: [A] = [A] 0 [B] = 0 t = t: [A] = [A] 0 -x [B] = x = k 1 [A] – k -1 [B] = k 1 ([A] 0 – x) – k -1 x = k 1 [A] 0 – (k 1 + k -1 )x Integration gives:(1) At equilibrium, the net reaction rate = 0, so [B] t is constant (=[B] e = x e ), so: k 1 [A] e = k -1 [B] e = k -1 x e [A] 0 xexe t [A] t [B] t

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There is an equilibrium constant: so:(2) Combining eq (1) with (2) gives: This is the rate equation for a first order process! Determination of (k 1 + k -1 ) by plottingagainst t Eq (2) gives2 equations, 2 unknowns Individual values of k 1 and k -1 can be determined

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Preequilibria Very complicated kinetics, unless you assume that [A·B] is constant during a large part of the reaction (steady state approach) k 1 [A][B] = k -1 [A·B] + k 2 [A·B] = (k -1 + k 2 )[A·B] So the rate equation now becomes: = k 2 [A·B] =

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Two possibilities: - rapid breakdown of A·B, k 2 >>k -1, so = k 1 [A][B] - slow breakdown of the complex: k 2 <

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Interpretation of rate constants: the Arrhenius equation E act E S P Reaction coordinate X (TS) Every reaction has to overcome an energy barrier: the transition state (TS, X ). At higher temperature, more particles are able to overcome the energy barrier. Arrhenius equation: E a can be determined by measuring k obs at two different temperatures:

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Idem, from statistical mechanics (collision theory) P = probability factor (not every collision is effective) Z = collision number (number of collisions per second) Arrhenius:

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Idem, from transition state theory: or[X ] = K [A][B] = k [X ] = k K [A][B] = k[A][B], so k = k K Statistical mechanics gives us the following relation: so k B = Boltzmanns constant; h = Plancks constant E act E S P Reaction coordinate X (TS)

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For all equilibria we can write: G 0 = - RT ln K, so for our case we get: G = - RT ln K Expressing K in terms of G and RT gives the following equation for k: Since G = H - T S, we can also write: (1) (2) Eq (1) and (2) are called the Eyring equations

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The Eyring and Arrhenius equations resemble each other: Arrhenius: so: Eyring: so E a = H + RT In order to determine H and S it is easier to differentiate ln (k/T) to 1/T: ln k 1/T ln A slope = 1/T slope =

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So, the procedure to determine activation parameters is: - determine k at different temperatures - plotting ln(k/T) against 1/T gives H - then gives S and when you have H and S, you also have G since G = H-T S

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Interpretation of activation parameters G, the Gibbs free energy of activation, determines at which rate a certain reaction will run at a given temperature H is a measure for the amount of binding energy that is lost in the transition state relative to the ground state (including solvent effects) S is a measure for the difference in (dis)order between the transition state and the ground state –for monomolecular reactions: S 0 J/mol.K –for a bimolecular reaction: S << 0 J/mol.K (two particles have to come together in the transition state to form one particle, demanding a much greater order)

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Example: G = 62.8 kJ/mol(very fast rx) H = 33.0 kJ/mol(rel. low, compensation of C-H bond cleavage by hydration TS) S = -100 J/mol.K(bimolecular rx)

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Another example: H = 85 kJ/mol (relatively high: no new bonds to be formed, no compensation for the partial cleavage of the C-C bond in the transition state; acetonitrile is aprotic, compensation of H by solvation will be less than in water S = 0 J/mol.K (monomolecular reaction)

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Application of activation parameters for the elucidation of reaction mechanisms: A S of +12 J/mol.K was found monomolecular process A S of -117 J/mol.K was found bimolecular process; rate determining step in this case is the attack of water on the carbonyl group. Look in your course book for the exact reaction mechanisms!

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Solvation (solvent effects) Influence of solvation on the reaction rate: k(H 2 O) = 10 -7 l.mol -1.s -1 ; k(DMF) = 10 -1 l.mol -1.s -1 so G ~ 30 kJ/mol H2OH2O DMF G H2O G DMF E reaction progress G DMF < G H2O

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What is the background of this strong solvent effect? In H 2 O there is more solvation than in DMF, due to hydrogen bonds. Note the changes in entropy: loss of S because of orientation of the substrates, gain of S because of the liberation of water (less solvated transition state). The balance is not easy to predict! In general, in case of ions, the ground state is more solvated than the transition state: TS ( ) is hardly solvated due to the spreading of charge. Again a strong solvent effect here: k(H 2 O)= 7.4x10 -6 s -1 ; k(DMF) = 37 s -1

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Solvation effects in (bio)polymers Polymers or enzymes may have apolar pockets, which leads to: - less solvation and therefore higher reaction rates; - changes in pK a s of acidic/basic groups: R = CH 3 :pK a = 9.7 R = polymer:pK a = 7.7 K a = [PyN][H + ] [PyNH + ] E.g. lysine, R-NH 2 + H + R-NH 3 + pK a (H 2 O) = 10.4, in some enzymes pK a = 7 !

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The energy diagram Consider the gas phase chlorination of methane: Reaction course is via the route of lowest energy (mountain pass) CH 4 + Cl H 3 C···H···Cl CH 3 + HCl r H-Cl r C-H

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A cross-section of this mountain landscape gives the well- known energy diagram: H 3 C···H···Cl

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What does the transition state look like? Hammond postulate: The TS closely resembles the species with the highest energy content Exothermic reaction ( a ) has a low E a, TS resembles ground state; Endothermic reaction ( b ) has a high E a, TS resembles the product

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Kinetic isotope effects Difference in effectivity of C-H and C-D bond cleavage: primary isotope effect The background is the difference in bond strength, caused by the difference in mass between H and D: E 0 of covalent bond is given by: 1/2h = 1/2hc(1/ ); 1/ (C-H) 3000 cm -1, E 0 (C-H) 18 kJ/mol; 1/ (C-D) 2200 cm -1, E 0 (C-D) 13 kJ/mol

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k obs = 7.5 This is the maximum primary kinetic isotope effect at ~25ºC. The isotope effect tells us something about the transition state. For this, we have to look at the stretching vibrations of the C-H(D) bond: Antisymmetrical stretching vibration: leads to reaction Symmetrical stretching vibration: involvement of H(D) depends on the structure of the transition state In this case: E a = D x

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Symmetrical stretching vibration: involvement of H(D) depends on the structure of the transition state: H exactly in the middle between C and Cl: symmetrical transition state, kinetic isotope effect is maximum (~7.5). H not in the middle: isotope effect < 7.5 E reactantsproducts E 0 EaDEaD EaHEaH E 0 C-D E 0 C-H H 3 C · · · · ·H · · · · ·Cl a H 3 C · · H · · · · · · · ·Cl E 0 C-D E 0 C-H E 0 EaHEaH E 0 H-Cl E 0 D-Cl EaDEaD b E 0 E 0 C-D E 0 C-H EaDEaD EaHEaH c H 3 C · · · · · · · · H · ·Cl symmetrical TSearly TS late TS

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Rule of thumb: k H /k D 4-7: bond cleavage, symmetrical transition state k H /k D 1-4: bond cleavage, asymmetrical transition state; or no bond cleavage (secondary isotope effect) Example: Maximum isotope effect at symmetrical TS, so when pK a (acid) = pK a (HB)

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Some more examples:

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