Transcript Slides
ME 475/675 Introduction to Combustion Lecture 17 Relationship between kinetics and equilibrium, Reaction mechanisms, Steady state approximation Announcements • HW 6 Due Monday, 10/5/15, X1, Ch 4 (16A, include the forward and reverse of both reactions, 𝑘1𝑓 , 𝑘1𝑟 , 𝑘2𝑓 , 𝑘2𝑟 ) • Return midterm Monday? • College of Engineering's Distinguished Lecture Series • Tesla Co-founder JB Straubel will speak about • Nevada’s clean energy future • Oct. 11, 2015 at 5 p.m., 4th Floor of the Joe Crowley Student Union • free and open to the public; reservations are required (sold out?). • www.unr.edu/tesla-lecture Chapter 4 Chemical Kinetics (from last lecture) • Global (apparent) Reaction (this is what is observed) • 𝐹 + 𝑎𝑂𝑥 → 𝑏𝑃𝑟 • Made up of many intermediate steps that are not seen • 𝑑𝐹 𝑑𝑡 = −𝑘𝐺 𝑇 𝐹 𝑛 𝑂𝑥 𝑚, 𝑘𝐺 𝑇 global rate constant (black box) • Elementary bimolecular reaction (break one and form one bond): • 𝐴+𝐵 →𝐶+𝐷 𝑑𝐴 • = −𝑘𝑏𝑖𝑚𝑜𝑙𝑒𝑐 𝐴 1 𝐵 1 𝑑𝑡 • Using collision theory we predicted • 𝑘 𝑇 = 1 2 2 8𝜋𝑘𝐵 𝑇 𝑝𝑁𝐴𝑉 𝜎𝐴𝐵 𝑒𝑥𝑝 𝜇 • but 𝑝 and 𝐸𝐴 = ? − 𝐸𝐴 𝑅𝑢 𝑇 (function of temperature alone, not pressure) • “Semi-empirical” three parameter form: • 𝑘 𝑇 = 𝐴𝑇 𝑏 𝑒𝑥𝑝 − 𝐸𝐴 𝑅𝑢 𝑇 , 𝐴, 𝑏 and 𝐸𝐴 values are tabulated pp. 112 Multistep Mechanisms Reaction Rates • A sequence of intermediate (elementary) reactions leading from overall-reactants to overall-products • L steps, i = 1, 2,… L • N species, j = 1, 2,… N; some are intermediate (not in overall products for reactants) • Example (forward and reverse reactions) • R1: 𝐻2 + 𝑂2 • R2: 𝐻 + 𝑂2 𝑘𝐹1 ,𝑘𝑅1 𝑘𝐹2 ,𝑘𝑅2 • R3: 𝑂𝐻 + 𝐻2 𝑘𝐹3 ,𝑘𝑅3 • R4: H + 𝑂2 + 𝑀 𝐻𝑂2 + 𝐻 𝑖=1 𝑂𝐻 + 𝑂 𝑖=2 𝐻2 𝑂 + 𝐻 𝑖=3 𝑘𝐹4 ,𝑘𝑅4 𝐻𝑂2 + 𝑀 𝑖 = 4 • Number of steps: L = 4 • Number of Species (j = 𝐻2 , 𝑂2 , 𝐻𝑂2 , 𝐻, 𝑂𝐻, 𝑂, 𝐻2 𝑂, 𝑀): N = 8 • 8 time-dependent unknowns molar concentration: 𝑗 𝑡 • Need 8 equations (constraints) Species net reaction rates •j=1 • 𝑑 𝑂2 𝑑𝑡 = 𝑘𝑅1 𝐻𝑂2 𝐻 + 𝑘𝑅2 𝑂𝐻 𝑂 + 𝑘𝐹3 𝑂𝐻 𝐻2 −𝑘𝐹1 𝐻2 𝑂2 − 𝑘𝐹2 𝐻 𝑂2 − 𝑘𝐹4 𝐻 𝑂2 𝑀 •j=2 • 𝑑𝐻 𝑑𝑡 = 𝑘𝐹1 𝐻2 𝑂2 + 𝑘𝑅2 𝑂𝐻 𝑂 + 𝑘𝐹3 𝑂𝐻 𝐻2 + 𝑘𝑅4 𝐻𝑂2 𝑀 −𝑘𝑅1 𝐻𝑂2 𝐻 − 𝑘𝐹2 𝐻 𝑂2 − 𝑘𝑅3 𝐻2 𝑂 𝐻 − 𝑘𝐹4 𝐻 𝑂2 𝑀 • i = 3, 4, …8 •… • Book describes compact notation What happens at equilibrium? • A general bimolecular reaction (not necessarily elementary) • 𝑎𝐴 + 𝑏𝐵 𝑘𝑓 𝑘𝑟 𝑐𝐶 + 𝑑𝐷 • Consumption – Generation of A • 𝑑𝐴 𝑑𝑡 = 𝑎 −𝑘𝑓 𝐴 • At equilibrium • 𝑘𝑓 𝐴 • 𝑘𝑓 𝑇 𝑘𝑟 𝑇 𝑎 𝐵 𝐵 𝑑𝐴 𝑑𝑡 • 𝐾𝐶 𝑇 = 𝑘𝑟 𝑇 𝐵 𝑏 + 𝑘𝑟 𝐶 𝑐 𝐷 𝑑 = 0, so = 𝑘𝑟 𝐶 = 𝐾𝐶 𝑇 = 𝑘𝑓 𝑇 𝑎 𝑐 𝐷 𝑑 𝐶 𝑐𝐷𝑑 𝐴𝑎𝐵𝑏 = Rate Coefficient based on molar concentrations • Looks like the Equilibrium Constant based on partial pressures 𝐾𝑃 𝑇 from Chapter 2 Relationship between Rate Coefficients and Equilibrium Constant (Chapter 2) • 𝑎𝐴 + 𝑏𝐵 𝑘𝑓 𝑘𝑟 𝑐𝐶 + 𝑑𝐷 𝑃𝐶 𝑐 𝑃𝐷 𝑑 𝑃𝑜 𝑃𝑜 𝑃𝐴 𝑎 𝑃𝐵 𝑏 𝑜 𝑜 𝑃𝐶 𝑃𝑐 𝑃𝐷 𝑃𝑑 𝑃𝑜 𝑃𝑜 𝑃𝑜 𝑃𝐴 𝑎 𝑃𝐵 𝑏 𝑅𝑢 𝑇 𝑃𝑜 𝑃𝑜 • Equilibrium Constant (based on partial pressures) : 𝐾𝑃 𝑇 = • Rate constant: 𝐾𝐶 𝑇 = • Using 𝑖 = 𝑁𝑖 𝑉 • 𝐾𝐶 𝑇 = 𝐾𝑃 𝑇 = 𝑘𝐹 𝑇 𝑘𝑅 𝑇 = 𝑃𝑖 𝑅𝑢 𝑇 𝑃𝑜 𝑐+𝑑−(𝑎+𝑏) 𝑅𝑢 𝑇 𝑅𝑢 𝑇 𝑐+𝑑−(𝑎+𝑏) 𝑃𝑜 𝐶 𝑐𝐷𝑑 𝐴𝑎𝐵𝑏 = 𝑃𝐶 𝑐 𝑃𝐷 𝑑 𝑅𝑢 𝑇 𝑅𝑢 𝑇 𝑃𝐴 𝑎 𝑃𝐵 𝑏 𝑅𝑢 𝑇 𝑅𝑢 𝑇 = , or • 𝐾𝑃 𝑇 = 𝐾𝐶 𝑇 • Note: If 𝑁𝑅 = 𝑎 + 𝑏 = 𝑐 + 𝑑 = 𝑁𝑃 , then 𝐾𝑃 𝑇 = 𝐾𝐶 𝑇 • So if we know 𝑘𝑓 then we can find 𝑘𝑟 ! 𝑐+𝑑−(𝑎+𝑏) Example 4.2 (page 120, turn in next time) • 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 • 𝑁𝑂 + 𝑂 → 𝑁 + 𝑂2 . • 𝑘𝑓 = 3.80 ∗ −20,820 109 𝑇 1.0 𝑒𝑥𝑝 𝑇 = 𝑐𝑚3 𝑔𝑚𝑜𝑙∗𝑠 • These coefficients generally use [cgs] (centimeter, gram, seconds)] units, not [mks] • Determine the rate coefficient at 2300 K for the reverse reaction, i.e. • 𝑁 + 𝑂2 → 𝑁𝑂 + 𝑂 Reaction Mechanisms (and approximations) • Describes concepts used to find global (apparent) rate • Steady State Approximation • Slow creation and rapid consumption of radicals cause them to reach steady-state quickly • Example: Zelovich two-step system 𝑘1 • 𝑂 + 𝑁2 → 𝑁𝑂 + 𝑁 • This is a relatively “slow” reaction, but N is highly reactive and is consumed “as soon as” it is created 𝑘2 • 𝑁 + 𝑂2 → 𝑁𝑂 + 𝑂 (this is the reverse of reaction in last example) • Very fast consumption of N • Production – Consumption of N • 𝑑𝑁 𝑑𝑡 𝑁 • Since 𝑁 consumption is very fast, 𝑁 reaches steady-state almost right away • 𝑁 𝑡 = 𝑘1 𝑂 𝑁2 − 𝑘2 𝑁 𝑂2 𝑆𝑆 = 𝑘1 𝑂 𝑁2 𝑘2 𝑂2 𝑑𝑁 𝑑𝑡 ≈0 [algebraic equation (instantaneous), not differential (dynamic)] • This molar concentration is small and changes almost immediately when the other concentrations change • 𝑑 𝑁 𝑆𝑆 𝑑𝑡 𝑑 𝑘1 𝑂 𝑁2 𝑘2 𝑂2 = 𝑑𝑡 Uni-molecular Reaction • Apparent (global) • 𝐴 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡; • Find 𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑑𝑡 𝑘𝑎𝑝𝑝 = 𝑘𝑎𝑝𝑝 𝐴 • Measurements show that, for uni-molecular reactions • At high pressures reaction rate is not a function of pressure • At low pressure, reaction rate increases with pressure • This can be explained using a “three-step” mechanism 𝑘𝑒 • 𝐴 + 𝑀 → 𝐴∗ + 𝑀 (𝐴∗ is an energized state of 𝐴 and highly reactive) • 𝐴∗ +𝑀 𝑘 ∗ 𝑢𝑛𝑖 • 𝐴 𝑘𝑑𝑒 𝐴 + 𝑀 (De-energization of A) 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 • 𝐴∗ is assumed to be very reactive and reach steady state conditions quickly 𝑃 Product production rate 𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑑𝑡 𝐴∗ • = 𝑘𝑢𝑛𝑖 • Molecular Balance for 𝐴∗ • 𝑑 𝐴∗ 𝑑𝑡 ∗ • 𝐴 • • 𝑆𝑆 ≈ 𝑘𝑒 𝐴 𝑀 𝑘𝑑𝑒 𝑀 + 𝑘𝑒 𝑘𝑢𝑛𝑖 𝑘𝑒 𝑀 • Recall 𝑀 = 𝑁𝑀 𝑉 = 𝜒𝑀 𝑁 𝑉 • 𝑘𝑎𝑝𝑝 = 𝐴 = 𝑘𝑎𝑝𝑝 𝐴 = 𝜒𝑀 𝑃 • For 𝑃 need 𝐴∗ ) = 𝑘𝑒 𝐴 𝑀 − 𝑘𝑑𝑒 𝐴∗ 𝑀 − 𝑘𝑢𝑛𝑖 𝐴∗ ≈ 0 (since 𝐴∗ is consumed as fast as it is produced) 𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 = 𝑑𝑡 𝑘𝑑𝑒 𝑀 +𝑘𝑢𝑛𝑖 𝑘 𝑘 𝑀 𝑘𝑎𝑝𝑝 = 𝑢𝑛𝑖 𝑒 𝑘𝑑𝑒 𝑀 +𝑘𝑢𝑛𝑖 • For 𝑃 𝑘𝑢𝑛𝑖 ∗ (𝐴 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠, 𝑃 𝑅𝑢 𝑇 𝑘𝑎𝑝𝑝 𝑃 𝑘𝑢𝑛𝑖 𝑘𝑒 𝜒𝑀 𝑅 𝑇 𝑢 𝑃 𝑘𝑑𝑒 𝜒𝑀 𝑅 𝑇+𝑘𝑢𝑛𝑖 𝑢 𝑘𝑢𝑛𝑖 𝑅𝑢 𝑇 ≫ , 𝑘𝑎𝑝𝑝 𝑘𝑑𝑒 𝜒𝑀 𝑘 𝑅 𝑇 ≪ 𝑢𝑛𝑖 𝑢 , 𝑘𝑎𝑝𝑝 𝑘𝑑𝑒 𝜒𝑀 → → 𝑘𝑢𝑛𝑖 𝑘𝑒 𝑘𝑑𝑒 𝑘𝑒 𝜒𝑀 𝑃 𝑅𝑢 𝑇 ≠ 𝑓𝑛 𝑃 (consistent with observations) = 𝑘𝑒 𝑀 ~𝑃1 (also consistent with observations) End 2015 Hypothetical Chain reactions (example) • Globally: 𝐴2 + 𝐵2 → 2𝐴𝐵 (𝐴 and 𝐵 are general atoms) • 𝑑 𝐴𝐵 𝑑𝑡 = 2𝑘𝑎𝑝𝑝 𝐴2 𝐵2 • Proposed intermediate steps • Slow creation of free radicals 𝐴 (and 𝐵) 𝑘1 • 𝐴2 + 𝑀 → 𝐴 + 𝐴 + 𝑀 • Fast consumption of 𝐴 and 𝐵 (neglect reverse because 𝐴 and 𝐵 are small) 𝑘2 • 𝐴 + 𝐵2 → 𝐴𝐵 + 𝐵 𝑘3 • 𝐵 + 𝐴2 → 𝐴𝐵 + 𝐴 • De-energization ter-molecular reaction is slow 𝑘4 • 𝐴 + 𝐵 + 𝑀 → 𝐴𝐵 + 𝑀 • Assume 𝑘2 and 𝑘3 are much greater than 𝑘1 and 𝑘4 Production - Consumption equations • Reactants • • 𝑑 𝐴2 𝑑𝑡 𝑑 𝐵2 𝑑𝑡 = 0 − 𝑘1 𝐴2 𝑀 − 𝑘3 𝐵 𝐴2 = 0 − 𝑘1 𝐴 𝐵2 • Products • 𝑑 𝐴𝐵 𝑑𝑡 = 𝑘2 𝐴 𝐵2 + 𝑘3 𝐵 𝐴2 + 𝑘4 𝐴 𝐵 𝑀 • Intermediates (fast) • • 𝑑𝐴 𝑑𝑡 𝑑𝐵 𝑑𝑡 = 2𝑘1 𝐴2 𝑀 − 𝑘2 𝐴 𝐵2 + 𝑘3 𝐵 𝐴2 − 𝑘4 𝐴 𝐵 𝑀 ≈ 0 = 𝑘2 𝐴 𝐵2 − 𝑘3 𝐵 𝐴2 − 𝑘4 𝐴 𝐵 𝑀 ≈ 0 • 𝐵 = 𝑘2 𝐴 𝐵2 𝑘3 𝐴2 −𝑘4 𝐴 𝑀 • 2𝑘1 𝐴2 𝑀 − 𝑘2 𝐴 𝐵2 + • 2𝑘1 𝐴2 𝑀 − 𝑘2 𝐴 𝐵2 𝑘3 𝐴2 −𝑘4 𝐴 𝑀 𝑘2 𝐴 𝐵2 𝑘3 𝐴2 −𝑘4 𝐴 𝑀 𝑘3 𝐴2 − 𝑘4 𝐴 𝑀 =0 + 𝑘3 𝐴2 − 𝑘4 𝐴 𝑀 𝑘2 𝐴 𝐵2 = 0