1 Ejemplos que involucran de transferencia de masa Flujo de oxígeno en los capilares con reacción en los tejidos Flujo de oxígeno en los capilares con reacción en los tejidos X

2 Capilares  Un sistema de flujo X x1x1 x2x2 En un sistema de flujo, transporte por convección >> Transporte por difusión

3 Ejemplos que involucran la transferencia de masa Transferencia de masa en gotas con reacción en la interfase CC C int R gota RR

4 Ejemplos que involucran la transferencia de masa Difusión de i a una superficie donde ocurre la reacción A+3B  2C Difusión de i a una superficie pasivada donde ocurre la reacción A+3B  2C Difusión de i a una superficie pasivada donde ocurre la reacción A+3B  2C UNREACTED SOLID LIQUID FILM BULK AQUEOUS SOLUTION CACA CBCB CCCC At surface, A+3B  2C

5 Lecture 17 * THE CONCEPT OF REACTION RATE *all slides in white were elaborated by K. Osseo-Asare

6 CONCEPT OF REACTION RATE Origins of Slow Rates Origins of Slow Rates Homogeneous and Heterogeneous Systems Homogeneous and Heterogeneous Systems Definitions of Reaction Rate Definitions of Reaction Rate Empirical Rate Laws Empirical Rate Laws Rate Constant and Reaction Order Rate Constant and Reaction Order Activation Energy Activation Energy

7 RATE PROCESSES IN AQUEOUS SYSTEMS Recall: Thermodynamic feasibility of reactions:  G r < 0 But note: Favorable  G r does not necessarily lead to favorable reaction rate. e.g., H 2 (g) + O 2 (g) = H 2 O (l)  G r  = -237.1 kJ mol -1 Reaction does not proceed without external assistance, e.g., ignition 8 Fe 2+ + ClO 4 - + 8H + = 8Fe 3+ + Cl - + 4H 2 O  G r  = -477.6 kJ mol -1 Percholorate ion acts as oxidant only towards a very small group of metal ions, e.g., V(II), V(III), Ti(III)

8 ORIGINS OF SLOW REACTION RATES Chemical kinetics: Rates of (a)making and breaking of chemical bonds (b) electron transfer leading to changes in oxidation states Transport Processes: Rates at which reactants move to reaction sites and products leave reaction sites. Very important for heterogeneous systems.

9 WHY STUDY KINETICS? We need answers to the following questions: (i)Fast or slow reactions? (ii)What are the key variables that influence reaction rates (iii)How to express rates mathematically? (iv)How to predict reaction rates? (v)How to rationalize (explain) observed reaction rates Science: Rate Mechanisms Science: Rate Mechanisms Engineering: Empirical Rate Equations Engineering: Empirical Rate Equations

10 CLASSIFICATION OF RATE PROCESSES (a) According to number of phases: homogeneous/heterogeneous (b) According to nature of rate-determining step: chemical kinetics/transport Classification of Chemical Kinetics (a) Atom-transfer Fe(H 2 O) 6 3+ + Cl - = Fe(H 2 O) 5 Cl 2+ + H 2 O (b) Group-transfer R 3 Si - OR + H 2 O = R 3 Si – OH + ROH (c) Electron-transfer Cu 2+ + H 2 = Cu(s) + 2H + Note: Cu 2+ + 2e - = Cu(s); H 2 = 2H + + 2e -

11 DEFINITIONS OF REACTION RATE Consider a species A, n A moles at time t, fluid volume = V, reactor volume = V r 1. Homogeneous System (a)Basis: unit volume of fluid r A = (1/V)dn A /dt   i (b)Basis: unit volume of reactor r A = (1/V r )dn A /dt 2. Heterogeneous System (a)Basis: unit interfacial (S/L, L/L) area r A  = (1/S)dn A /dt   i ’ (b)Basis: unit mass of solid in solid-liquid system r A  = (1/W)dn A /dt

12 Interconversion of rates: Recall:r A = (1/V)dn A /dt (volume basis) r A  = (1/S)dn A /dt (area basis) r A = (S/V)r A    i = (S/V)  I ’ r A = (S/V)r A    i = (S/V)  I ’ Factors influencing chemical kinetics: For reaction aA + bB  pP r A = f (System variables) r A = f (System variables) = f (T, C A, C B,...) Note: For (homogeneous) reaction where V (fluid volume) does not change with time, r A = (1/V)dn A /dt r A = (1/V)dn A /dt = (1/V)d(C A V)dt = (1/V)d(C A V)dt = dC A /dt DEFINITIONS OF REACTION RATE

13 RATE LAWS AND REACTION ORDERS: Consider overall reaction: 4Fe 2+ + O 2 + 4H + = 4Fe 3+ + 2H 2 O (1) Empirical rate obtained: d[Fe 3+ ]/dt = k[Fe 2+ ] 2 P [H + ] -1/4 (2) What is reaction order with respect to?:What is reaction order with respect to?: (i) Fe 3+, (ii) Fe 2+, (iii) O 2, (iv) H +, (v) overall Expressions for:Expressions for: (i) d[Fe 2+ ]/dt, (ii) dP /dt, (iii) d[H + ]/dt O2O2 O2O2

14 RATE LAWS AND REACTION ORDERS: Recall: aA + bB  pP -(1/a)dC A /dt = -(1/b)dC B /dt = (1/p)dC P /dt = kC A  C B  Recall: 4Fe 2+ + O 2 + 4H + = 4Fe 3+ + 2H 2 O d[Fe 3+ ]/dt = k[Fe 2+ ] 2 P /[H + ] 1/4 Reaction order: [Fe 3+ ] , [Fe 2+ ] 2, P 1, [H + ] -1/4 Overall: 0 + 2 + 1 – 1/4 = 11/4 Rate Laws: -1/4 d[Fe 2+ ]/dt = -d P /dt = -1/4 d[H + ]/dt = -1/4 d[H + ]/dt = 1/4 d[Fe 3+ ]/dt = 1/4 d[Fe 3+ ]/dt O2O2 O2O2 O2O2

15 Table 16.4 Rate laws for redox reactions between metal complexes ReactionRate LawRate constants Cr(II) + Co(III) = Cr(III) + Co(II)-d[Cr(II)]/dt = k 1 [Cr(II)][Co(III)] Cr 2+ /ct(NH 3 ) 5 X 2+, k 1 (M -1 s -1 ) = 0.5 (X = H 2 O), = 1.5 x 10 6 (X = OH-) = 10 3 (X = Cl-) Cr(II) + Fe(III) = Cr(III) + Fe(II)k 1 [Cr 2+ ] + k 2 [Cr 2+ ][FeOH 2+ ] +k 3 [Cr 2+ ][FeCl 2+ ] + k 4 [Cr 2+ ][Fe 3+ ][Cl - ] k 1 = 2.3 x 10 3 M -1 s -1, k 2 = 3.3 x 10 6 M -1 s -1, k 3 = 2.0 x 10 7 M -1 s -1, k 4 = 2.0 x 10 4 M -2 s -2 Fe(II) + Co(III) = Fe(III) + Co(II)k 1 [Fe 2+ ][Co 3+ ] + k 2 [Fe 2+ ][CoOH 2+ ] k 1 = 10 M -1 s -1, k 2 = 6500 M -1 s -1 at 0  C

16 APPARENT ACTIVATION ENERGY aA + bB  pP -(1/a)dC A /dt = -(1/b)dC B /dt = (1/p)dC P /dt = kC A  C B  Overall reaction order: n =  Temperature dependence of rate constant: k = k o exp (-E a /RT) Reaction orderRate constant activation energy ln k Slope = -E a /R 1/T

17 El Orígen de la Energía de Activación La mayoría de las reacciones incluye el rompimiento de enlaces y/o son reacciones muy complicadas O 2 + 4H + + 4e - = 2H 2 OEh˚ = 1.23V La reducción de oxígeno molecular involucra la transferencia de 4 electrones, El proceso de la reducción de oxígeno en términos de una serie de transferencias consecutivas de un electrón con los intermediarios de reacción: dos radicales libres (HO 2, OH) y peróxido de hidrógeno (H 2 O 2 ): O 2 + H + + e - = HO 2 Eh˚ = -0.32V HO 2 + H + + e - = H 2 O 2 Eh˚ = 1.68V HO 2 + H + + e - = H 2 O 2 Eh˚ = 1.68V H 2 O 2 + H + + e - = OH + H 2 O Eh˚ = 0.80V H 2 O 2 + H + + e - = OH + H 2 O Eh˚ = 0.80V OH + H + + e - = H 2 OEh˚ = 2.74V OH + H + + e - = H 2 OEh˚ = 2.74V Total =  /4 = 1.23V