Elementary reactions 1/21 stoichiometry = mechanism (Cl. + H 2 HCl + H. ) monomolecular reactions (decay: N 2 O 4 some isomerisations) 2 NO 2 ; radioactive decay; bimolecular reactions (collision; most common) ClCH 3 + CN Cl + CH 3 CN trimolecular reactions O + O 2 + N 2 O 3 + N 2 (N 2 carries out the surplus energy)
Reaction mechanisms 2/21 A (general) reaction is a sequence of elementary reactions = reaction mechanism. Example: 2 H 2 + O 2 2 H 2 O H 2 + O 2 HO 2. + H. HO 2. + H2 H 2 O + OH. H. + O 2 O. + OH. H 2 + OH. H 2 O + H. H 2 + O. H. + OH. radical A. activated molecule A (energy-rich, local energy minimum) activated complex (transition state) AB, AB # (saddle point)
Reaction mechanisms 3/21 We need to get rid of unstable (unknown) intermediates. rate-determining step fastest (parallel reactions) slowest (consecutive reactions) Bodenstein principle of (quasi)stationary state intermediates fast reach (almost) constant concentrations e.g.: A A B dc A dτ 0 pre-equilibrium reversible reaction part of chain e.g.:... slow A + B fast C + D slow... fast c C c D c A c B K can be derived from the above principle (for γ = 1)
Lindemann(-Hinshelwood) mechanism 4/21 A(g) B(g) Inellastic collisions in the gas phase activate molecules: A + A k 1 k 1 A k 2 A + A B c A c A stationary state dc A dτ = 0 A dτ = B dτ = k k 2 A 2 k 2 + k 1 c A dc 1 c dc A /dτ = dc B /dτ k 1 c A k 2 (ambient pressures): dc B dτ = k 2k 1 k 1 c A 1st order k 1 c A k 2 (low pressures): dc B dτ = k 1c 2 A 2nd order E.g.: cyclopropane propene, N 2 O 5 NO 2 + NO 3., dimethyldiazene (azomethane) CH 3 -N=N-CH 3 C 2 H 6 + N 2
Chain reactions 5/21 initiation (typically free radicals are produced) heat chemical (peroxides) light (UV) propagation (cyclic reaction with radical recovery) chain transfer (no branching) chain branching termination recombination (of radicals) reaction (low-reactive radical inhibition) deactivation at walls
Example (not in detail) 6/21 H 2 + Cl 2 2 HCl initiation: Cl 2 k 1 2 Cl. propagation: Cl. + H 2 k 2 HCl + H. H. + Cl 2 k 3 HCl + Cl. cycle (up to 10 6 ) termination: 2Cl. k 4 Cl 2 dc HCl dτ steady state k = k 2 c Cl.c H2 + k 3 c H.c Cl2 = 2k 1 2 c 1/2 k 4 Cl 2 c H2
[xoctave; xcat../octave/michaelismentenova.m Michaelis Menten] 7/21 Enzyme catalysis: Michaelis Menten kinetics Mechanism of Michaelis and Menten (Enzyme, Substrate, Product): E + S k 1 ES k 2 k 1 stationary state (because c E, c ES c S ): E + P balance: c E + c ES = c E0 Eliminating c E ( c ES ) from dc P dτ : dc ES dτ = k 1c E c S (k 1 + k 2 )c ES = 0 also from: dc P dτ = dc S dτ = k 1c E c S k 1 c ES dc P dτ c S c = k 2 c ES = k E0 2 K M /c S + 1 = v max K M + c S where K M = k 2 + k 1 k 1 = Michaelis constant a v max = k 2 c E0 c S K M dc S dτ = v max (zeroth order, most of E is saturated, ES) c S K M dc S dτ = v max K M c S (first order, most of E is free, E)
Michaelis Menten kinetics II Experimentally available: K M and v max = k 2 c E0 (often not both c E0 and k 2 simultaneously) Integrated form dc S dτ = v 1 max K M /c S + 1 K M ln c S0 c S + c S0 c S = v max τ 8/21 credit: wikipedia cannot solve for c S (τ) (using elem. functions) numerical solution credits: pitt.edu, Wikipedie
Metabolism of alcohols Alcohol dehydrogenase, various types In liver and the lining of the stomach Further oxidation to acids and H 2 O + CO 2 CH 3 CH 2 OH CH 3 CHO hangover human: CH 2 OH CH 2 OH... (COOH) 2 (kidney stones) CH 3 OH HCHO ( ) HCOOH ( ) 9/21 credit: wikipedia Example. Calculate the time needed to metabolize c S0 = 1 wt. o / oo of ethanol to c S = 0.1 o / oo Data: v max = 0.12 g L 1 hod 1, K M = 0.06 g L 1. ρ blood = 1.06 g cm 3 1 wt. o / oo = 1.06 g L 1 0th order: τ = c S0 c S v max = More accurate: τ = K M ln c S0 c S (1 0.1) 1.06 0.12 + c S0 c S v max h = 7.7 h = 9.2 h c w /(g dm -3 ) 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 12 τ/h Other data: K M = 0.02 0.05 g L 1, include absorption of ethanol in the body,...
Michaelis Menten kinetics III 10/21 Rate: v = dc S dτ = v 1 max K M /c S + 1 Linear in (1/c S, 1/v) (Lineweaver & Burk): 1 v = K M v max 1 c S + 1 v max 0.15 v max v/promile.h -1 0.1 0.05 1/v K M /v max 0 0 0.25 0.5 0.75 1 c s /promile 0 1 K M 0 1 v max 1/c S
Example carbonic anhydrase [plot/anhydrase.sh] 11/21 CO 2 + H 2 O HCO 3 + H + [CO 2 ]/mmol dm 3 v/mol dm 3 s 1 1.25 2.78 10 5 2.5 5.00 10 5 5.0 8.33 10 5 20.0 16.7 10 5 credit: chemistry.umeche.maine.edu KM = 0.01mol dm 3, vmax = 25 10 5 mol dm 3 s 1 [according to DeVoe, Kistiakowski, JACS 83, 274 (1961)]
Babel of units 12/21 Enzyme activity unit (amount of substance / time) Deprecated units: SI: mol s 1 (katal) 1 M = 1 mol dm 3 1 m = 1 mol kg 3 more common: µmol/min ( enzyme unit, U) (1 mol dm 3 ) Specific activity (per kg of enzyme) 1 Da = 1 g mol 1 SI: mol s 1 kg 1 1 bar = 10 5 Pa 1 Å = 10 10 m µmol min 1 mg 1 1 cal th = 4.184 J Turnover number (per mole), (thermochemical) SI: mol s 1 mol 1 = s 1 1 cal it = 4.1868 J (intl./iapws) often min 1 etc. 1 cal IUNS = 4.182 J Molar mass: g mol 1 = Da (dalton) (food) or 1 g mol 1 /N A = 12 1 m(12 C) = 1 u = 1.660539 10 27 kg = 1 Da Example: 1 µg of enzyme (M = 40 kda) in the excess of substrate provides the reaction rate of 6 µmol of substrate/min. What is the turnover number (in s 1 )? 4000 s 1
Inhibition 13/21 reversible irreversible reversible inhibition: the inhibitor is bound non-covalently (H-bonds, etc.), decreases the turnover E + S k 1 k 1 ES + + I I k 2 E + P k i k i 1 k i k i 1 EI + S k i 1 k i 1 ESI k i 2 E + P irreversible inhibition: catalyst poisoning, usually covalently bound inactive complex EI
Competitive reversible inhibition 14/21 E + S k 1 k 1 ES k 2 E + P + I k i k i 1 EI The inhibitor binds to the same site as the substrate ( competes with the substrate)
Uncompetitive reversible inhibition 15/21 E + S k 1 k 1 ES k 2 E + P + I k i k i 1 ESI k i 2 E + P The inhibitor binds to the enzyme-substrate complex Also anti-competitive often partial (slows down the reaction)
Mixed (non-competitive) reversible inhibition 16/21 E + S k 1 k 1 ES + + I I k 2 E + P k i k i 1 k i k i 1 EI + S k i 1 k i 1 ESI k i 2 E + P Mixed inhibition: the inhibitor bound both to E and ES (Pure) non-competitive inhibition: inhibitor affects a different part of the enzyme, k i = k i, k i 1 = k i 1
Reversible inhibition: some math 17/21 E + S k 1 k 1 ES + + I I k 2 E + P k i k i 1 EI k i k i 1 ESI stationary state: dc ES dτ = k 1c E c S (k 1 + k 2 )c ES k i c Ic ES + k i 1 c ESI = 0 pre-equilibrium: c EI = k i c k E c I, c ESI = k i i 1 k i 1 c ES c I balance: c E + c ES + c EI + c ESI = c E0 we assume c I c E, c I c I0 is known (no balance of I needed)
Reversible inhibition: Lineweaver Burk 18/21 1 v = αk M 1 + α, α = 1 + k i c v max c S v max k I, α = 1 + k i i 1 k i 1 c I 1/v αk M /v max 0 α αk M 0 α v max 1/c S
Reversible inhibition 19/21 time vs. substrate conc. Lineweaver Burk 1 0.9 0.8 0.7 α=1 α =1 (no inhibition) α=2 α =1 (competitive) α=1 α =2 (uncompetitive) α=2 α =2 (non-competitive) 8 7 6 α=1 α =1 (no inhibition) α=2 α =1 (competitive) α=1 α =2 (uncompetitive) α=2 α =2 (non-competitive) 0.6 5 c S 0.5 1/v 4 0.4 0.3 0.2 3 2 0.1 1 0 0 1 2 3 4 5 6 7 8 9 10 τ 0-1 -0.5 0 0.5 1 1.5 2 2.5 3 1/c S
Reversible inhibition summary 20/21 1 v = αk M 1 + α, α = 1 + k i c v max c S v max k I, α = 1 + k i i 1 k i 1 c I No inhibition: α = α = 1 Competitive: α > 1 inhibitor binds to the free enzyme in the L-B diagram: greater K M the same v max Uncompetitive: α > 1 inhibitor binds to the enzymesubstrate complex in the L-B diagram: smaller K M smaller v max Mixed (non-competitive): α, α > 1 inhibitor binds to both the free enzyme and enzyme-substrate complex in the L-B diagram: the same K M smaller v max
Photochemistry instant 21/21 Photon energy = hν = energy source for the reaction Planck constant: h = 6.62607 10 34 J s Frequency ν, wave number ν = 1/λ, wave length λ. It holds: c = λν. Quantum yield Φ = # of molecules transformed/decomposed/... # of photons absorbed Chain reactions: Φ > 1. Example: 2 HI H 2 + I 2 HI + hν H. + I. H. + HI H 2 + I. 2 I. I 2 Φ = 2 Example: How much HI decomposes by absorbing energy of 100 J in the form of light of wave length 254 nm? 0.42 mmol