CHEM 3340 Physical Chemistry II Reaction kinetics PDF

CHEM 3340 Physical Chemistry II Reaction kinetics Chemical Kinetics Chemical reactions have directions, and some of them are fast (e.g., explosion) and some of them are slow (2H2+O2 = 2H2O) Chemical Thermodynamics: describes state of chemical reactions t→∞, when they reach equilibrium Chemical kinet...

CHEM 3340 Physical Chemistry II Reaction kinetics Chemical Kinetics Chemical reactions have directions, and some of them are fast (e.g., explosion) and some of them are slow (2H2+O2 = 2H2O) Chemical Thermodynamics: describes state of chemical reactions t→∞, when they reach equilibrium Chemical kinetics: study of rates of chemical reactions How does the amount (concentration) of substances vary with time? Important step: use of analytical chemistry (e.g., molecular spectroscopy) to determine concentrations vs. time Rates of Reactions Take reaction: R →P ◊ Rate of formation of a product: lim Δt→ 0 Δ[P]/Δt=d[P]/dt ◊ Rate of consumption of a reactant: lim Δt→ 0 -Δ[R]/Δt=-d[R]/dt For more complex reaction A+2B →3C+D The -d[A]/dt will be different from -d[B]/ dt, however -d[A]/dt = 1/2 -d[B]/dt = 1/3 d[C]/dt =d[D] /dt = v v: rate of a chemical reaction [mol/ (dm3s)] When equilibrium is reached: d[A]/dt = 0, thus v = 0 Rate can be better expressed with the term: extent of a reaction Extent of Reaction Initial state Final state A B nA,0 nB,0 nA,0 + dnA nB,0 + dnB The changes are related -dnA=dnB Change of extent of reaction: dξ=-dnA=dnB Extent of reaction: ξ [mol] For finite changes: ∆ξ=-∆nA=∆nB Final state nA,0 - ∆ξ nA,0 + ∆nA nB,0 + ∆ξ nB,0 + ∆nB Extent of reaction 2A + B -> 3C + D The numbers: stoichiometric coefficients, >0 Alternative way of writing a reaction 0 = -2A - B + 3C + D The numbers: stoichiometric numbers positive: products negative: reactants 0= JJ In general J Change of extent of reaction d⇥ = dnJ / J dnA/(-2) = -dnB = dnC/3 = dnD= dξ Rate expressed with extent of reaction 1 d[J] 1 v= = J dt J Rate: 1 dnJ V dt V: volume of the reactor (constant) First derivative of extent of reaction: d⇥ = dnJ / J d⇥ 1 = dt J Rate with extent of reaction: dnJ dt 1d v= V dt For surface reactions we used surface concentration, σJ = nJ/A 1 d⇥J v= J dt Rate law The rate of chemical reaction can be determined experimentally. It depends on ◊ concentrations of chemical substances, [J] ◊ physical parameters: temperature (T), pressure (p), light intensity (I) Rate law: a functional form of the dependence v = f([A], [B],...., T, p, I) Once we know the rate law, we can predict the composition of the mixture from initial concentrations at any time: ultimate goal of a chemist Dependence of rate law on concentrations A + B →P v = k [A]a [B]b k: rate constant. Always depends on temperature a: order of rate law with respect to A b: order of rate law with respect to B In general, a and b are NOT the stoichiometric numbers Order of the reaction with respect to a species: power to which the concentration is raised (can be negative, can be not integer(e.g., 1/2, 0.123232), can be zero). If the concentration of a species occurs more than once in the equation, the order is indefinite. Overall order: sum of the individual orders: in this case, a+b Example: 4 NO2(g) + O2(g) → 2N2O5(g) v = k [NO2]2 [O2] Third order rate law with second-order in NO2, first order in O2 Elementary Reactions Elementary reactions occur in single step: single event collision of reactant molecules that produces products Molecularity: number of molecules coming together to react in an elementary reaction Unimolecular reaction: molecularity is one, a single molecule falls apart or rearranges the atoms Bimolecular reaction: a pair of molecules collide In ELEMENTARY reactions the rate can be written with order as stoichiometric coefficient: Unimolecular: A →P v = k [A] Bimolecular: A + B→P v = k [A] [B] However, if the order is the same as the stoichiometric coefficient, that DOES NOT indicate elementary reaction Reaction mechanism: sum of elementary reactions that explains the rate law First order reactions A →P v = k [A] unit of k: 1/s Assume at t=0, [A] = [A]0 What is the concentration of A as a function of time? Step 1. Obtain d[A]/dt from rate law(s) v = - d[A]/dt = k[A] Step 2. Integrate equation (separation of variables) 1 1 d[A] = d[A] = kdt [A] [A] Definite integral: starting point: t =0, [A]=[A]0 end point: t=t, [A]=[A] [A] [A]0 1 d[A] = [A] [A] [ln[A]][A]0 ln[A] kdt t kdt 0 = [ kt]t0 ln[A]0 = kt First order reactions A →P ln[A] ln[A]0 = ln([A]/[A]0 ) = kt [A] = [A]0 e kt kt Product concentration t=0, [P] =[P]0 [A]0 - [A] = [P] -[P]0 consumed A produced P [P] = [P]0 + [A]0 - [A] = [P]0 + [A]0 - [A]0 e-kt [P] = [P]0 + [A]0 (1- e-kt) First order reactions t1/2, half life: time taken for the concentration of the reactant to decrease to half of its original value ln ([A]/ [A]0 )= -kt ln ([A]0/2 / ln [A]0 )= -kt1/2 -ln 2 = -kt1/2 t1/2=(ln2)/k Independent of concentration! τ, time constant: time taken for the concentration of the reactant to decrease to 1/e of its original value ln ([A]0/e / ln [A]0 )= -kτ -ln e = -kτ τ = 1/k Physical meaning of rate constant: 1/k defines a time constant First order reaction Example CH3N2CH3(g) → C2H6 (g) + N2 (g) azomethane For gas phase reactions, very often pressures or partial pressures are given pV = nRT p= RT n/V= RT c Partial pressures can be used instead of concentrations! Rate laws can also be written with partial pressures for gas phase reactions! t (s) 0 1000 2000 3000 4000 p (Pa) 10.9 7.63 5.32 3.71 2.59 Is the reaction first order? If yes, ln p/p0 = -kt , i.e., -(ln p/p0 )/t= k constant First order reaction Example CH3N2CH3(g) → C2H6 (g) + N2 (g) azomethane t (s) 0 1000 2000 3000 4000 p (Pa) 10.9 7.63 5.32 3.71 2.59 k=-ln(p/p0)/t (1/s) - 3.6x10-4 3.6x10-4 3.6x10-4 3.6x10-4 Yes, the reaction is first order, k = 3.6x10-4 1/s Graphical method ln p/p0 = -kt the plot of ln p/p0 vs. t shall be a straight line with a slope of k p0=10.9 Pa Radioactive decay: half life Radioactivity is a process in which unstable atomic nuclei fall apart following first order kinetics by emitting high-energy electromagnetic waves (γ photon), or particles (electron and antineutrino - β decay, He nuclei-α decay) Typically it takes place through first order reaction: they use halflifes (t1/2 = ln 2 / k) instead of rate constants 238U→234Th, α decay with t1/2 = 4.5 billion years Not very dangerous 210Po→206Pb, α decay with t1/2 = 139 days Very dangerous, but can be blocked by a sheet of paper 250000 times more toxic than HCN Second-order reactions A →P v = k [A]2 unit of k: mol-1 dm3 s-1 Assume at t=0, [A] = [A]0 What is the concentration of A as a function of time? v = - d[A]/dt = k[A]2 1 d[A] = 2 [A] 1 d[A] = [A]2 kdt Definite integral: starting point: t =0, [A]=[A]0 end point: t=t, [A]=[A] [A] [A]0 [ [A]0 [A] = 1 + kt[A]0 kdt 1 d[A] = 2 [A] [A] 1/[A]][A]0 1 [A] t kdt 0 = [ kt]t0 1 = kt [A]0 Second order reaction 1 [A] 1 = kt [A]0 Plot of 1/[A] vs. t: straight line Decay faster than first order reaction Half life: 1 [A]0 /2 1 = kt1/2 [A]0 t1/2 = 1/(k[A]0) The half-life depends on initial concentration Second order reaction A+B →P v = k[A] [B] unit of k: mol-1 dm3 s-1 Assume at t=0, [A] = [A]0, [B]=[B]0 What is the concentration of A as a function of time? [B] = [B]0 - ([A]0 - [A]) v = - d[A]/dt = k[A][B] Integration result: ln [B]/[B]0 [A]/[A]0 ⇥ = ([B]0 [A]0 )kt plot of ln{([B]/[B]0 )/ ([A]/[A]0)} vs. t should give a straight line First order reaction close to equilibrium So far we neglected the reaction in the opposite direction: kinetically irreversible reactions (the opposing reaction is very slow, or the reaction is far from equilibrium) Reversible reaction A B A →B v = k[A] v = - d[A]/dt = k[A] d[A]/dt = -k[A] Overall rate of change of [A] when equilibrium is reached in dilute solutions: Keq = [B]/[A] B →A v = k’[B] v = d[A]/dt = k’[B] d[A]/dt = k’[B] - k[A] Definition of equilibrium: no more change, d[A]/dt = 0 d[A]/dt = 0 = k’[B]eq - k[A]eq [B]eq /[A]eq = k/k’ = K k’[B]eq = k[A]eq In a reversible reaction the ratio of the forward and reverse rate constants give the equilibrium constant First order reaction close to equilibrium A B Assumption: t=0, [A] = [A]0, [B]=0 [B]=[A]0 - [A] d[A]/dt = k’([A]0-[A]) - k[A] Integration: k ⇥ + ke (k+k )t [A] = [A]0 ⇥ k+k Important part: e-(k+k’)t The exponential decay is with a time constant: 1/(k+k’) Relaxation method A B We can determine the rate constants k and k’ from: ◊ Equilibrium constant: K = [B]eq/[A]eq = k/k’ ◊ Close to equilibrium the reactant will decay with a time constant τ=1/(k+k’) Two equations, two unknown-> k and k’ can be expressed For more complicated mechanisms, the time constant needs to be determined. Implementation: ◊ Wait until equilibrium is reached ◊ Apply a small change in parameters: temperature jump, pressure jump ◊ Measure relaxation to the new equilibrium state Accounting for the rate laws Usually reactions take place through a series of elementary reactions Reaction mechanism: set of elementary reactions and rate constants We would expect extremely complicated rates for complicated mechanism experimentally measured rate laws are often simple There are simplifying concepts: quasi-steady state approximation rate determining step pre-equilibrium Consecutive Reactions ka kb A →I →P va =-d[A]/dt = ka [A] vb =d[P]/dt = kb [I] d[I]/dt = ka [A] - kb [I] t= 0, [A] = [A]0, [I]=0, [P]=0 Quasi-steady-state approximation (QSSA): I. there is an induction period during which the concentration of the intermediate rises from zero II. After the induction period, the lifetime if intermediate species will be small: after they form from A, they react quickly to give products → their concentration changes very slowly d[I]/dt ≈0 Rates of consecutive reactions with QSSA ka kb A →I →P d[I]/dt = ka [A] - kb [I] ≈0 [I]≈ ka [A]/kb [A] simply decays with first order kinetics [A] = [A]0 e-ka t Intermediate Product [P ] 0 [I]≈ ka [A]/kb= ka /kb x [A]0 e-ka t d[P]/dt = kb [I] = kb ka [A]/kb= ka A d[P]/dt = ka [A]0 e-ka t t d[P ] = [P ] = ka [A]0 e ka t dt = ka [A]0 [ e ka t 0 [P] = [A]0 (1-e-ka t) /ka ]t0 = [A]0 (1 e ka t ) Validity of QSSA ka kb A →I →P ◊In this case: good approximation when ka > kb) A+B ka k’a I Keq = ka/k’a= [I]/([A][B]) kb I →P v=d[P]/dt = kb [I] [I] = Keq[A][B] d[P]/dt = kb Keq[A][B] k[A][B] Second order rate law with composite rate constant: k = kb Keq = kb ka/k’a Temperature dependence of reaction rates Rate law: a functional form of the dependence v = f([A], [B],...., T, p, I) Temperature dependence occurs in the rate Swante Arrhenius constant v = k [A]a[B]b General empirical observation: increase of temperature of 10 K causes in increase of rate constant by 2-3 times Arrhenius: founder of physical chemistry Hydrolysis of sucrose: ln k = ln A - Ea/(RT) A: pre-exponential factor Ea: activation energy R: Gas constant Ea, A: Arrhenius parameters Nobel Prize: 1903 Arrhenius Equation ln k = ln A - Ea/(RT) k = A exp(-Ea/(RT)) k increases with temperature for every elementary reaction Larger the activation energy → larger increase If Ea is small: little dependence (Ea=0, k is independent of T ) For reaction with complicated mechanism: k might decrease, but very rare Example The rate of reaction at T = 700 K is 0.011 1/s, at T = 730K is 0.035 1/s. What is the activation energy and the pre-exponential factor? T1 = 700 K, k1 = 0.011 1/s ln k1 = ln A - Ea/(RT1) T2 = 730 K, k2 = 0.035 1/s ln k2 = ln A - Ea/(RT2) ln k2 - ln k1 = ln 0.035 - ln 0.011 =1.1575= -Ea/R (1/T2 - 1/T1) = -Ea/R (1/730 - 1/700) = -Ea/R x -5.871x10-5 1.1575= -Ea/R x -5.871x10-5 = -Ea/8.314 J/molK x -5.871x10-5 1/K= 7.0614 x 10-6Ea J/mol Ea = 1.1575/7.0614 x 10-6 J/mol = 164 kJ/mol ln k1 = ln A - Ea/(RT1) ln A = ln k1 + Ea/(RT1) = ln (0.011) + 164x103 J/mol /(8.314 J/molK) / 700 K = 23.67 A = exp(23.67) =1.9x1010 1/s A has the same unit as k. A is a large number! Activation energy and the pre-exponential factor More data: fit line to ln k vs. 1/T Interpretation of parameters During a chemical reaction the potential energy of a molecule changes A+B -> P Reaction coordinate: ‘microscopic’, collection of motions (changes in interatomic distances and bond angles) that are directly involved in the formation of products) There is usually a maximum: activated complex: transition state of the reaction Peak of energy: Activation energy is the minimum kinetic energy that reactants must have in order to form products k = A exp(-Ea/(RT)) Potential energy diagram for consecutive reactions RDS: rate determining step Arrhenius Interpretation A number of collisions between A and B exp(-Ea/(RT)) fraction of collisions with energy excess of Ea k = A exp(-Ea/(RT)) Number of successful collision! Rate of reaction Rate Theories Goal 1: interpret rate equations v = k [A]a[B]b Goal 2: interpret temperature dependence of rate constant k = A exp(-Ea/(RT)) Collision theory: ◊ builds on Arrhenius interpretation: reaction takes place through collision of molecules with excess kinetic energy Ea ◊ Uses kinetic theory of gases to calculate: number of collisions with a kinetic energy of at least Ea Collision theory A+B → P rate = number of overall collisions in unit volume x fraction of effective collisions between molecule A and B Z: Number of collisions / volume: collision density Kinetic theory of gases: when we have pure molecule A, the number of collisions / molecule c̄rel z =NA [A] c̄rel z=N ZAA = 1/2 NA [A] z = 1/2 NA2[A]2 c̄rel overall number of collisions N =number of molecules/volume N =NA[A] NA: Avogadro number σ = d2 π, collision cross section d: diameter of molecule Collision Density When we have only molecules A: ZAA = 1/2 NA2[A]2 σ = d2 π c̄rel = 8RT ⇥µ μ = MA/2 c̄rel When we have mixture of A and B: ZAB = NA2[A][B] σ = d2 π d = 1/2 (dA + dB) 8RT c̄rel = ⇥µ MA MB µ= MA + M B c̄rel Fraction of ‘effective’ collisions Boltzmann distribution for number of molecules having energies between ε and ε+dε N( , + d ) 1 = exp( ) N kT kT dε Fraction of molecules between energies εa and ∞ N( a ,⇥) N = 1 kT ⇥ a exp( kT )d = exp( a kT ) exp[ εa / (kT)] = exp[ Ea / (RT)] Ea: energy limit above which reaction takes place (independent of temperature) Collision theory A+B → P rate = number of overall collisions in unit volume x fraction of effective collisions this would give number of product molecules/time to get moles of product molecules /time: divide by NA rate= ZAB x exp[-Ea/(RT)] / NA ZAB = NA2[A][B] c̄rel v = NA c̄rel exp[ Ea /(RT )] ⇥ [A][B] k v = k x [A][B] k = NA ⇤c̄rel exp[ Ea /(RT )] = NA ⇤ 8kT exp[ Ea /(RT )] ⇥µ Collision Theory v = k x [A][B] k = NA ⇤ 8kT exp[ Ea /(RT )] ⇥µ Collision theory predicts: Bimolecular reactions, the rate will proportional: [A][B] Temperature dependence of rate constant: k∝ T1/2 exp(-Ea/RT) (This function is dominated by the exponential part for large value of Ea) Collision theory: comparison with experiments Problems with collision theory: ◊ cannot interpret the value of Ea ◊ often overestimates, sometimes underestimates the preexponential factor Steric factor: P Not all collisions with energy larger than Ea are effective There is a geometric requirement Only a fraction of collisions are effective: P=Aexp/Acoll k = P NA ⇤ 8kT exp[ Ea /(RT )] ⇥µ Based on this argument P≤1 Example Let’s take reaction 6 dm3mol-1 s-1 A = 1.24x10 exp H2 (g) + C2H4 (g)→ C2H6 (g) T=628 K What is the steric factor? Acoll = NA ⇤ P = Aexp/Acoll 8kT ⇥µ μ = M(H2) x M(C2H4)/[M(H2)+M(C2H4)] = 2x10-3 kg/NA x 28 x10-3 kg/NA / [2x10-3 kg/NA + 28 x10-3 kg/NA] = 3.12 x10-27 kg From Table 1B.2: σ(H2) = 0.27 nm2 = d2 π -> d(H2) = 0.293 nm σ(C2H4) = 0.64 nm2 = d2 π -> d(C2H4) = 0.451 nm σ = [1/2 (d(H2) + d(C2H4))]2 π = [0.5x(0.293nm+0.451 nm0 ]2 π =0.435 nm2 Acoll = 6x1023 x 0.435x10-18 m2 x [8x1.38x10-23x628K/3.1415/3.12x20-27 kg]0.5 = 6.94 x108 m3 mol-1s-1= 6.94 x1011 dm3 mol-1s-1 P = Aexp/Acoll = 1.24x106 dm3mol-1 s-1 / (6.94 x1011 dm3 mol-1s-1 ) = 1.79 x10-6 , very small number! Steric Factor Steric factor > 1? Ionic product: during reaction, K and Br2 can react from much larger distance than 1/2(d(K) + d(Br2)) because the Coulombic interactions that exist during formation of KBr can bring together K and Br atoms: → larger σ → large k How to calculate Ea? Transition State Theory Activated Complex Theory Tries to solve deficiencies of collision theory: ◊ Predict Ea ◊ Better description of steric factor During the course of the reaction, there is special state: transition state, with maximal potential energy, C‡ Not a real molecule (real molecules exist in minima): ◊ If kinetic energy is slightly smaller than Ea, then molecule goes back to reactants ◊ If kinetic energy is slightly larger than Ea, products are formed Transition state theory: simplified mechanism A+B C‡ C‡ → P ‡ [C ] ‡ Kc = [A][B] [C‡]= K‡ [A][B] v =k‡ [C‡] =k‡ K‡ [A][B] = k2 [A][B] k2 = k‡ x K‡ Next step is estimation of k‡ and K‡ These steps are not really reactions ◊ Second step: slow vibration along a bond that is being formed with frequency ν ◊ Rate constant: number of P / time, will be proportional to the number of vibration/time along the reaction coordinate, k‡∝ ν ◊ Not every vibration goes to products, k‡ = κν κ: transmission coefficient Estimation of A+B ‡ K ‡ [C ] ‡ Kc = [A][B] C‡ Not real equilibrium ◊ Real equilibrium is established throughout collisions and energy exchanges among all modes of motion (rotation, translation, vibration) ◊ In this equilibrium, vibration along the bond is formed does not give back A and B → the equilibrium constant shall be modified ◊ A new quasi-equilibrium constant is introduced that contains all motion but the one corresponding to the formation of product K̄ ‡ c ‡ ◊ ‘Kind of’ equilibrium constant, but one vibration mode of C is discarded ◊ Statistical mechanics calculations show that Kc‡ kT ‡ = K̄c h Eyring-Polanyi equation v = k‡ K‡ [A][B] k2 = k‡ K‡ Kc‡ k‡ = κν H. Eyring k2 = kT h kT ‡ = K̄c h ‡ K̄C M. Polanyi κ is often taken as 1 Transition state theory: calculating the equilibrium constant One can approximate the structure of the transition state, and approximate theoretically (e.g. Schrödinger equation) or spectroscopically the equilibrium constant Thermodynamic aspects k2 = kT h ‡ K̄C Equilibrium constant can be expressed with Gibbs energy of activation: ‡G ‡ ‡ ‡ RT G= K =e RT ln K kT k2 = e h Thermodynamics ‡ G= ‡ H T ∆‡H: enthalpy of activation kT k2 = e h ‡G RT ‡ S ∆‡S: entropy of activation ‡S R e ‡H RT Activation Energy kT k2 = e h ‡S R e ‡H RT Activation energy (Arrhenius) comes from the slope (m=Ea/R→ Ea = -m R ) of ln k2 vs. 1/T Ea = dlnk2 dlnk2 dlnT d 2 2 R⇥ = RT ⇥ = RT ⇥ [ + d(1/T ) dT dT 1/T This is for liquid phase reactions For gas phase: because of the expansion work Ea =∆‡H + RT Ea =∆‡H + 2RT H/(RT ) ] dT ‡ ∆‡H/(RT2) Pre-exponential Factor kT k2 = e h ‡S R e ‡H RT Ea =∆‡H + RT k2 = A exp[-Ea/(RT)] = A exp(-RT/RT) exp(-∆‡H/(RT) ) k2 = A exp(-1)exp(-∆‡H/(RT) ) kT A/e = e h kT A=e e h A=e 2 ‡S R kT RT e o h p ‡S R For liquid ‡S R For gas Pre-exponential factor kT A=e e h ‡S R Important term: Activation entropy, ∆‡S Typically negative: two molecules come together to give C‡ More complex structure of C → larger decrease in entropy, smaller value of A and k2 Unimolecular Reactions "You can hear the sound of two hands when they clap together," said Mokurai. "Now show me the sound of one hand." Zen How unimolecular reactions are really possible? Reaction is collision of two molecules → bimolecular reaction A →P It should really be A + A→P v = k [A][A] Experimental observation: ◊ at low [A]0 we do observe second order kinetics at ◊ at large [A]0 the rate law is often of first order v = k [A] Lindemann-Hinshelwood mechanism Many gas phase isomerization reactions cyclo-C3H6 → CH3CH=CH2 A + A → A* + A v = k [cyclo-C3H6] d[A*]/dt =ka [A]2 A* + A → A + A d[A*]/dt = -ka’ [A][A*] d[A*]/dt = -kb [A*] A* → P A*: energized molecule short lifetime →QSSA approximation d[A*]/dt =ka [A]2 - ka’ [A][A*] -kb [A*] ≈0 ka [A]2 [A ] = kb + ka⇥ [A] Rate of reaction v = d [P] /dt = kb[A*] kb ka [A]2 v = kb [A ] = kb + ka⇥ [A] If kb > ka’ [A] v = ka [A] 2 Second order reaction k = ka Low [A] Activation energy of composite reaction kb ka [A] = k[A] v= ka k = kbka/ka’ ka A* A+A→ ka’ A* + A → A + A kb A* → P Aa exp( Ea (a)/RT ) ⇥ Ab exp( Ea (b)/RT ) ka kb = k= ka Aa exp( Ea (a )/RT ) Aa Ab k= exp([ Ea (a) Ea (b) + Ea (a )]/RT ) Aa Overall activation energy: Ea = Ea(a) + Ea(b) - Ea(a’) If Ea(a’) > Ea(a) + Ea(b), the overall activation energy will be negative: the reaction will slow down with increasing temperature! Temperature compensation in biological systems Rate of photosynthesis vs. temperature Overall activation energy: Ea = Ea(a) + Ea(b) - Ea(a’) It is possible to balance these to Ea ≈ 0 Many biological organisms do not have body temperature control, therefore, it is beneficial to have reactions with little temperature dependence Ruoff et al., FEBS, 2007 Explains temperature adaptation of biological systems Catalysis Catalyst: substance that accelerates a reaction but undergoes no net chemical change Formal kinetics: A → B, v = k[A][C] Interpretation: A+C →B+C Homogeneous catalyst: catalyst in the same phase as the reaction mixture Heterogeneous catalyst: catalyst is in different phase than the reaction mixture (Pt or Ni for hydrogenation of double bonds in alkenes. Liquid vegetable oil → solid margarine ) J. J. Berzelius Also coined the word ‘protein’ Potential Energy Diagram Mechanism: Catalyst decreases the activation energy by opening a new reaction route H2O2 (l) → H2O(l) + 1/2 O2 (g) Ea= 76 kJ/mol with I- as catalyst: Ea = 57 kJ/mol → 2000x increase in rate with an enzyme (catalase): 8kJ/mol → 1015 increase Acids and bases are often catalyst: only the protonated (or deprotonated) form of the molecule reacts Catalyst does not change the equilibrium constant (K= k+/k-), therefore, it accelerates both the forward and the backward step Enzymes Enzymes are ‘homogeneous’ biological catalysts Usually consist of proteins and nucleic acid Reactant(s): substrate They contain an active site (‘catalyst’) and a special 3D structure that will make the enzyme extremely specific for certain target substrates Two mechanisms: ◊ Lock and key ◊ Induced fit Kinetics of Enzyme Catalysis For long time it was a mystery They measure the initial rate (d[P]/dt) S→P Characteristics: ◊ At fixed [S]0, rate linearly increases with enzyme concentration, [E]0 ◊ For low values of [S]0 the rate linearly increases with [S]0 ◊ For large values of [S]0 maximum velocity Michaelis-Menten mechanism Maud Menten E+S ka’ ka ES ES → P kb ES: enzyme-substrate complex Rate of overall reaction: v = kb [ES] Let’s use quasi-steady-state approximation for [ES] d [ES]/dt = ka[E][S] - ka’ [ES] - kb[ES] = 0 ka [S] ≈ [S]0 [ES] = 0 [E][S] = [E][S]/Km 1/KM = ka/(ka’ + kb) ka + kb [E] + [ES] = [E]0 [ES] = ([E]0 [ES])[S0 ]/Km [E]0 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 [ES] = 1 + KM /[S]0 Michaelis Menten mechanism E+S ka’ ka ES ES → P KM =(ka’ + kb)/ka kb Michaelis constant [E]0 [ES] = 1 + KM /[S]0 kb [E]0 v = d[P ]/dt = kb [ES] = 1 + KM /[S]0 Michaelis-Menten equation At low [S]0, 1> KM/[S]0: v = vmax = kb [E]0 Lineweaver-Burk plot kb [E]0 vmax v= = 1 + KM /[S]0 1 + KM /[S]0 1 1 KM 1 = + v vmax vmax [S]0 For characterization of an enzyme: determine KM, and vmax (kb) Further experiments are required to calculate ka and ka’ by following [S] or [P] as the reaction proceeds vmax = kb [E]0 Catalytic efficiency A+C→P Often in catalysis v = k[C] [A] = kcat [C] kcat = k[A] The efficiency of the catalyst: v/[C] =kcat kcat: [1/s], catalytic constant, turnover frequency the number of times a catalyst enters reaction to make products / 1s Catalytic constant of enzymes: vmax = kb [E]0 → kcat = kb This is not very good for enzymes, [S]0 is usually small kb v= [E]0 [S]0 = kef f [S]0 KM Catalytic efficiency, ε ε = keff/[E]0 = kb/KM The enzyme catalase: ε = 4x108 dm3 mol-1s-1 (practically decomposes every H2O2 it can catch) Adsorption Surface of a solid substance has special properties, different from the bulk solution Particles (molecules) can attach to the surface: adsorption The substance that adsorbs: adsorbate Solid substance: adsorbent or substrate Reverse of adsorption: desorption Physiosorption Physiosorption (physical adsorption): there is van der Waals interaction between the adsorbate and the substrate Weak interaction, and the released energy when the particle is physiosorbed is of the same order of magnitude as the enthalpy of condensation: -20 kJ/mol − -100 kJ/mol The molecule retains its identity on the surface Chemisorption Chemisorption (’chemical adsorption’): molecules stick to the surface by forming a covalent bond Enthalpy of chemisorption: ◊ larger (in absolute value) than that of physiosorption ◊ on the order of 100 kJ/mol or larger The molecule can even dissociate: dissociative adsorption O2 (g)→ 2 O (ad) Physiosorption vs. Chemisorption Extent of Adsorption Picture of solid surface: there is a certain number of available adsorption sites Fractional coverage, θ: number of adsorption sites occupied/number of adsorption sites available Alternative view: imagine a complete coverage of a surface with adsorbates. Volume of monolayer coverage: V∞ θ = V / V∞ , where V is the volume of adsorbate Rate of adsorption: r = d θ / dt Adsorption Isotherms The fractional coverage depends: ◊pressure of gas ◊ temperature The variation of θ with pressure p at a chosen temperature: adsorption isotherm Langmuir isotherm Assumptions 1. Adsorption cannot proceed beyond monolayer 2. All sites are equivalent and the surface is uniform 3. The ability of a molecule to adsorb at a given site is independent of the occupation of neighboring sites Langmuirk Isotherm A(g) + M(surface) d ka ka: adsorption rate constant kd: adsorption rate constant AM(surface) Adsorption Rate of adsorption: v = dθ/ dt v proportional to ◊ (partial) pressure of A, p ◊ Number of vacant sites: N(1-θ), N: number of sites dθ/ dt = ka p N (1-θ) Langmuir isotherm A(g) + M(surface) kd ka AM(surface) Desorption Rate proportional to the number of adsorbed species: Nθ dθ/ dt = -kd N θ dθ/ dt = ka p N (1-θ) -kd N θ Isotherm: equilibrium, no net change dθ/ dt = ka p N (1-θ) -kd N θ = 0 Kp = 1 + Kp K = ka/kd Langmuir Isotherm A(g) + M(surface) kd ka Kp = 1 + Kp Irving Langmuir Temperature dependence of the isotherm: through K K depends on T the same way as equilibrium constant does d ln K / dT = ∆adHo/ (RT2) AM(surface) K = ka/kd Heterogeneous Catalysis Surface acts as a catalyst Decomposition of PH3: at high pressures: zeroth order at low pressures: first order PH3(g) + M(surface) PH3(surface) PH3(surface)→ P +3/2 H2 k rate of reaction: proportional to the surface coverage of PH3 kKp v=k = 1 + Kp Low pressures: v =kKp High pressures: v = k K Langmuir-Hinshelwood mechanism A+B →P A K A pA = 1 + K A pA + K B pB B v = kθAθB K B pB = 1 + K A pA + K B pB kKA KB pA pB v= (1 + KA pA + KB pB )2 Catalytic Converter Example: CO + O2 → CO2 kKA KB pA pB v= (1 + KA pA + KB pB )2 Catalyst: Pt CO oxidation Example: CO + O2 → CO2 Catalyst: Pt Nobel prize 2007: Gerhard Ertl