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