Transcript Slides
ME 475/675 Introduction to Combustion Lecture 17 Relationship between kinetics and equilibrium, Reaction mechanisms, Steady state approximation Announcements β’ HW 6 Due Monday, 10/5/15, X1, Ch 4 (16A, include the forward and reverse of both reactions, π1π , π1π , π2π , π2π ) β’ Return midterm Monday? β’ College of Engineering's Distinguished Lecture Series β’ Tesla Co-founder JB Straubel will speak about β’ Nevadaβs clean energy future β’ Oct. 11, 2015 at 5 p.m., 4th Floor of the Joe Crowley Student Union β’ free and open to the public; reservations are required (sold out?). β’ www.unr.edu/tesla-lecture Chapter 4 Chemical Kinetics (from last lecture) β’ Global (apparent) Reaction (this is what is observed) β’ πΉ + πππ₯ β πππ β’ Made up of many intermediate steps that are not seen β’ ππΉ ππ‘ = βππΊ π πΉ π ππ₯ π, ππΊ π global rate constant (black box) β’ Elementary bimolecular reaction (break one and form one bond): β’ π΄+π΅ βπΆ+π· ππ΄ β’ = βππππππππ π΄ 1 π΅ 1 ππ‘ β’ Using collision theory we predicted β’ π π = 1 2 2 8πππ΅ π πππ΄π ππ΄π΅ ππ₯π π β’ but π and πΈπ΄ = ? β πΈπ΄ π π’ π (function of temperature alone, not pressure) β’ βSemi-empiricalβ three parameter form: β’ π π = π΄π π ππ₯π β πΈπ΄ π π’ π , π΄, π and πΈπ΄ values are tabulated pp. 112 Multistep Mechanisms Reaction Rates β’ A sequence of intermediate (elementary) reactions leading from overall-reactants to overall-products β’ L steps, i = 1, 2,β¦ L β’ N species, j = 1, 2,β¦ N; some are intermediate (not in overall products for reactants) β’ Example (forward and reverse reactions) β’ R1: π»2 + π2 β’ R2: π» + π2 ππΉ1 ,ππ 1 ππΉ2 ,ππ 2 β’ R3: ππ» + π»2 ππΉ3 ,ππ 3 β’ R4: H + π2 + π π»π2 + π» π=1 ππ» + π π=2 π»2 π + π» π=3 ππΉ4 ,ππ 4 π»π2 + π π = 4 β’ Number of steps: L = 4 β’ Number of Species (j = π»2 , π2 , π»π2 , π», ππ», π, π»2 π, π): N = 8 β’ 8 time-dependent unknowns molar concentration: π π‘ β’ Need 8 equations (constraints) Species net reaction rates β’j=1 β’ π π2 ππ‘ = ππ 1 π»π2 π» + ππ 2 ππ» π + ππΉ3 ππ» π»2 βππΉ1 π»2 π2 β ππΉ2 π» π2 β ππΉ4 π» π2 π β’j=2 β’ ππ» ππ‘ = ππΉ1 π»2 π2 + ππ 2 ππ» π + ππΉ3 ππ» π»2 + ππ 4 π»π2 π βππ 1 π»π2 π» β ππΉ2 π» π2 β ππ 3 π»2 π π» β ππΉ4 π» π2 π β’ i = 3, 4, β¦8 β’β¦ β’ Book describes compact notation What happens at equilibrium? β’ A general bimolecular reaction (not necessarily elementary) β’ ππ΄ + ππ΅ ππ ππ ππΆ + ππ· β’ Consumption β Generation of A β’ ππ΄ ππ‘ = π βππ π΄ β’ At equilibrium β’ ππ π΄ β’ ππ π ππ π π π΅ π΅ ππ΄ ππ‘ β’ πΎπΆ π = ππ π π΅ π + ππ πΆ π π· π = 0, so = ππ πΆ = πΎπΆ π = ππ π π π π· π πΆ ππ·π π΄ππ΅π = Rate Coefficient based on molar concentrations β’ Looks like the Equilibrium Constant based on partial pressures πΎπ π from Chapter 2 Relationship between Rate Coefficients and Equilibrium Constant (Chapter 2) β’ ππ΄ + ππ΅ ππ ππ ππΆ + ππ· ππΆ π ππ· π ππ ππ ππ΄ π ππ΅ π π π ππΆ ππ ππ· ππ ππ ππ ππ ππ΄ π ππ΅ π π π’ π ππ ππ β’ Equilibrium Constant (based on partial pressures) : πΎπ π = β’ Rate constant: πΎπΆ π = β’ Using π = ππ π β’ πΎπΆ π = πΎπ π = ππΉ π ππ π = ππ π π’ π ππ π+πβ(π+π) π π’ π π π’ π π+πβ(π+π) ππ πΆ ππ·π π΄ππ΅π = ππΆ π ππ· π π π’ π π π’ π ππ΄ π ππ΅ π π π’ π π π’ π = , or β’ πΎπ π = πΎπΆ π β’ Note: If ππ = π + π = π + π = ππ , then πΎπ π = πΎπΆ π β’ So if we know ππ then we can find ππ ! π+πβ(π+π) Example 4.2 (page 120, turn in next time) β’ In their survey of experimental determinations of rate coefficients for the N-H-O system, Nanson and Salimian [reference 10 in book] recommend the following rate coefficient for the reaction β’ ππ + π β π + π2 . β’ ππ = 3.80 β β20,820 109 π 1.0 ππ₯π π = ππ3 ππππβπ β’ These coefficients generally use [cgs] (centimeter, gram, seconds)] units, not [mks] β’ Determine the rate coefficient at 2300 K for the reverse reaction, i.e. β’ π + π2 β ππ + π Reaction Mechanisms (and approximations) β’ Describes concepts used to find global (apparent) rate β’ Steady State Approximation β’ Slow creation and rapid consumption of radicals cause them to reach steady-state quickly β’ Example: Zelovich two-step system π1 β’ π + π2 β ππ + π β’ This is a relatively βslowβ reaction, but N is highly reactive and is consumed βas soon asβ it is created π2 β’ π + π2 β ππ + π (this is the reverse of reaction in last example) β’ Very fast consumption of N β’ Production β Consumption of N β’ ππ ππ‘ π β’ Since π consumption is very fast, π reaches steady-state almost right away β’ π π‘ = π1 π π2 β π2 π π2 ππ = π1 π π2 π2 π2 ππ ππ‘ β0 [algebraic equation (instantaneous), not differential (dynamic)] β’ This molar concentration is small and changes almost immediately when the other concentrations change β’ π π ππ ππ‘ π π1 π π2 π2 π2 = ππ‘ Uni-molecular Reaction β’ Apparent (global) β’ π΄ β πππππ’ππ‘; β’ Find π πππππ’ππ‘π ππ‘ ππππ = ππππ π΄ β’ Measurements show that, for uni-molecular reactions β’ At high pressures reaction rate is not a function of pressure β’ At low pressure, reaction rate increases with pressure β’ This can be explained using a βthree-stepβ mechanism ππ β’ π΄ + π β π΄β + π (π΄β is an energized state of π΄ and highly reactive) β’ π΄β +π π β π’ππ β’ π΄ πππ π΄ + π (De-energization of A) πππππ’ππ‘π β’ π΄β is assumed to be very reactive and reach steady state conditions quickly π Product production rate π πππππ’ππ‘π ππ‘ π΄β β’ = ππ’ππ β’ Molecular Balance for π΄β β’ π π΄β ππ‘ β β’ π΄ β’ β’ ππ β ππ π΄ π πππ π + ππ ππ’ππ ππ π β’ Recall π = ππ π = ππ π π β’ ππππ = π΄ = ππππ π΄ = ππ π β’ For π need π΄β ) = ππ π΄ π β πππ π΄β π β ππ’ππ π΄β β 0 (since π΄β is consumed as fast as it is produced) π πππππ’ππ‘π = ππ‘ πππ π +ππ’ππ π π π ππππ = π’ππ π πππ π +ππ’ππ β’ For π ππ’ππ β (π΄ πππππ’ππ‘π , π π π’ π ππππ π ππ’ππ ππ ππ π π π’ π πππ ππ π π+ππ’ππ π’ ππ’ππ π π’ π β« , ππππ πππ ππ π π π βͺ π’ππ π’ , ππππ πππ ππ β β ππ’ππ ππ πππ ππ ππ π π π’ π β ππ π (consistent with observations) = ππ π ~π1 (also consistent with observations) End 2015 Hypothetical Chain reactions (example) β’ Globally: π΄2 + π΅2 β 2π΄π΅ (π΄ and π΅ are general atoms) β’ π π΄π΅ ππ‘ = 2ππππ π΄2 π΅2 β’ Proposed intermediate steps β’ Slow creation of free radicals π΄ (and π΅) π1 β’ π΄2 + π β π΄ + π΄ + π β’ Fast consumption of π΄ and π΅ (neglect reverse because π΄ and π΅ are small) π2 β’ π΄ + π΅2 β π΄π΅ + π΅ π3 β’ π΅ + π΄2 β π΄π΅ + π΄ β’ De-energization ter-molecular reaction is slow π4 β’ π΄ + π΅ + π β π΄π΅ + π β’ Assume π2 and π3 are much greater than π1 and π4 Production - Consumption equations β’ Reactants β’ β’ π π΄2 ππ‘ π π΅2 ππ‘ = 0 β π1 π΄2 π β π3 π΅ π΄2 = 0 β π1 π΄ π΅2 β’ Products β’ π π΄π΅ ππ‘ = π2 π΄ π΅2 + π3 π΅ π΄2 + π4 π΄ π΅ π β’ Intermediates (fast) β’ β’ ππ΄ ππ‘ ππ΅ ππ‘ = 2π1 π΄2 π β π2 π΄ π΅2 + π3 π΅ π΄2 β π4 π΄ π΅ π β 0 = π2 π΄ π΅2 β π3 π΅ π΄2 β π4 π΄ π΅ π β 0 β’ π΅ = π2 π΄ π΅2 π3 π΄2 βπ4 π΄ π β’ 2π1 π΄2 π β π2 π΄ π΅2 + β’ 2π1 π΄2 π β π2 π΄ π΅2 π3 π΄2 βπ4 π΄ π π2 π΄ π΅2 π3 π΄2 βπ4 π΄ π π3 π΄2 β π4 π΄ π =0 + π3 π΄2 β π4 π΄ π π2 π΄ π΅2 = 0