Reaction Kinetics: Understanding Order, Molecularity, and Steady State - Prof. Stefan Fran, Study notes of Physical Chemistry

Download Reaction Kinetics: Understanding Order, Molecularity, and Steady State - Prof. Stefan Fran and more Study notes Physical Chemistry in PDF only on Docsity! Lecture 26 Kinetics Elementary Reactions Reaction Order Multi-step processes Intermediates , Steady State Approximation Reaction Order • The power to which a the concentration of a species is raised in the rate law is the reaction d h ll d h f ll f hor er. T e overa or er is t e sum o a o t e powers of all reactants. Examples: 1. v = k[NO]2[O2] First order in O2, Second order in NO, Third order overall. 2. v = k[A]1/2[B]2 Half order in A, Second order in B, 2 1/2 order overall. • The order does not need to be an integer. Q ti R ti O dues on: eac on r er What is the order of the reaction: 2 H2 + O2 2 H2O A. First order B. Second order C. Third order D. Unknown Q ti R ti O dues on: eac on r er What is the order of the reaction: 2 H2 + O2 2 H2O A. First order B. Second order C. Third order D. Unknown Q ti 2 R ti O dues on : eac on r er The detailed mechanism of many enzymes has been shown to include a rate determining step: E + S ES What is the order of the overall reaction? A. First order B. Second order C. Third order kD. Un nown P ll l Fi O d R iara e rst r er eact ons h l• T e so utions are: k A [B] = k 1[A]0 1 exp{ (k + k )t }1 B k 1 + k 2 – – 1 2 [C] = k 2[A]0 k + k 1 – exp{ – (k 1 + k 2)t } • The production of B and C k2 1 2 occurs with a constant proportion: C [B] = k 1[C] k 2 Sequential first-order reactions C i l ionsecut ve e ementary react ons • A → B → C rate equations are:k1 A • d[A]/dt = -k1[A] • d[B]/dt = k1[A] - k2[B] B • d[C]/dt = k2[B] • Either k1 or k2 can be the rate k2 limiting step.C Sequential first-order reactions C i l ionsecut ve e ementary react ons • First solve eqn. for A k1 A [A] = [A]oe– k 1t • Substitute into eqn. for BB d[B] = k [A] k [B] = k [A] e– k 1t k [B] k2 dt 1 – 2 1 o – 2 [B] = k 1[A]o k 2 – k 1 e– k 1t – e– k 2t • Similarly for CC [C] = [A]o 1 – 1k 2 – k 1 (k 2e– k 1t – k 1e– k 2t) St d t t i tiea y-s a e approx ma on • Equations representing kinetic networks of more than three states are not soluble analytically. • One means of pushing the techniques as far as possible using analytical solutions is to set the derivatives of intemediates equal to zero: d[Intermediate]/dt = 0 • The build up of intermediate B shown in the figure - (two slides back) implies that the steady state approximation does not work for the system shown there. Application of the t d t t i tis ea y-s a e approx ma on • The steady state approximation can be applied to the consecutive reaction scheme k1 k2 A → B → C. if th t ti f B i f i l t t e concen ra on o s a r y cons an . • The result of setting d[B]/dt = 0 is: k1[A] - k2[B] = 0 and d[C]/dt = k1[A]. Since d[A]/dt = - k1[A] we see that d[C]/dt d[A]/dt d [C] (1 { k t}) = - an = - exp - 1 hQuestion: Rate sc emes Which rate equation describes the rate of disappearance of A? k2k1 B A C D k3 k-1 A. d[A]/dt = -(k1+ k2 + k3)[A] + k-1[B] B d[A]/dt = (k1+ k2 + k3)[A] - k 1[B]. - C. d[A]/dt = - k1[B] - k2[C] - k3[D] + k-1[A] D. d[A]/dt = k1[B] + k2[C] + k3[D] - k-1[A] Example using the steady state i tiapprox ma on Assuming that you have information that allows you to use the steady state approximation determine the form of d[F]/dt in terms of [U]. k1 k2 U ↔ I → F k-1 A d[F]/dt [U] k k /(k + k ). = 1 2 2 -1 B. d[F]/dt = [U] k1 k2 C d[F]/dt = [U] (k + k )/k k. 2 -1 1 2 D. d[F]/dt = [U] (k2 + k-1) Example using the steady state i tiapprox ma on Assuming that you have information that allows you to use the steady state approximation determine the form of d[F]/dt in terms of [U]. k1 k2 U I F ↔ → k-1 A d[F]/dt = [U] k k /(k + k ). 1 2 2 -1 B. d[F]/dt = [U] k1 k2 C d[F]/dt = [U] (k2 + k )/k k2. -1 1 D. d[F]/dt = [U] (k2 + k-1) Example using the steady state approximation The equations below show the reasoning behind the solution: k1 k2 U ↔ I → F k-1 d[I]/dt = k1[U] - (k2 + k-1)[I] = 0 k [U] (k k )[I]1 = 2 + -1 [I] = [U] k1/(k2 + k-1) d[F]/dt k [I] [U] k k /(k + k ) = 2 = 1 2 2 -1 Photochemistry • Beer-Lambert law is: I = I 10– A A = εlc c = [E] • The concentration is c or [E] • The intensity of absorbed light is I – I o , , o . • The extinction coefficient ε has units of M-1cm-1 The exctinction coefficient. depends on the frequency ε(ω). Often ε is tabulated at the peak of absorption spectrum. Photochemistry • Therefore, the rate of excitation of the excited state of molecule E is: – d[E]dt = φ(Io – I) 1000 l = φIo(1 – 10 – A)1000l • The quantum yield is φ. • 1000 converts from cm3 to L Photochemistry • We use the approximation: 10– A ≈ 1 – 2.303A • Thus, d[E] 2303φI ε[E] • Photoexcitation is a pseudo-first order – dt = o NA process where the excitation rate constant is: 2303φI εk = oNA