1. ME 475/675 Introduction to Combustion
Lecture 16
Chemical kinetics

2. Announcements
HW 6 Monday,
Thank you for coming early to the midterm.

3. Chapter 4 Chemical Kinetics
Describes rates at which:
Reactants are consumed and products are produced,
Thermal energy is produced (exothermal) or consumed (endothermal)
“Global” Combustion Reaction (molar based)
1 𝐹 𝑓𝑢𝑒𝑙 +𝑎𝑂𝑥 𝑜𝑥𝑖𝑑𝑖𝑧𝑒𝑟 →𝑏𝑃𝑟(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)
On what does reaction rate depend? In general it:
Increases with reactant molar concentrations and temperature
Decreases with product molar concentrations
Define Molar Concentration of species 𝑖, 𝑖 (number density)
𝑖 = 𝑁 𝑖 𝑉 = 𝜒 𝑖 𝑁 𝑉 = 𝜒 𝑖 𝑃 𝑅 𝑢 𝑇 = 𝑃 𝑖 𝑅 𝑢 𝑇 𝑁 𝑉 = 𝑃 𝑅 𝑢 𝑇
= 𝜒 𝑖 𝑚 𝑀𝑊 𝑀𝑖𝑥 𝑉 = 𝜒 𝑖 ρ 𝑀𝑊 𝑀𝑖𝑥
Units: 1 𝑘𝑚𝑜𝑙𝑒 𝑖 𝑚 𝑚 100𝑐𝑚 𝑚𝑜𝑙𝑒 𝑖 1 𝑘𝑚𝑜𝑙𝑒 𝑖 = 𝑚𝑜𝑙𝑒 𝑖 𝑐𝑐
Number of molecules
n = N*NAV
Avogadro's Number
𝑁 𝐴𝑉 =6.022∗ 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑘𝑚𝑜𝑙𝑒
Number of molecules in 12 kg of C12
𝑁 𝐴 =6.022∗ 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑚𝑜𝑙𝑒
Number of molecules in 12 g of C12

4. Global Rate Equations 𝑑 𝐹 𝑑𝑡 =− 𝑘 𝐺 𝑇 𝐹 𝑛 𝑂𝑥 𝑚 “Black Box” Approach
𝐹+𝑎𝑂𝑥→𝑏𝑃𝑟
𝑑 𝐹 𝑑𝑡 =− 𝑘 𝐺 𝑇 𝐹 𝑛 𝑂𝑥 𝑚
𝑘 𝐺 𝑇 = Global rate coefficient, strongly dependent on temperature T
𝑚,𝑛= Reaction order of Oxidizer and Fuel
𝑚+𝑛= Overall reaction order
Different 𝑘 𝐺 𝑇 and 𝑚,𝑛 for different temperature and pressure ranges
pp (Chapter 5)
“Black Box” Approach
Based on measurements and correlations (empirical)
Not based on
“causality”
understanding individual reaction steps, which may involve intermediate species
Overall global reaction may involve many intermediate reaction steps

5. Multi-step Reaction
Example Overall Reaction: 2 𝐻 2 + 𝑂 2 →2 𝐻 2 𝑂 (hydrogen combustion)
H-H O-O H-O-H
Many possible intermediate, elementary reactions steps
Each step breaks one bond, and forms one bond
𝐻 2 + 𝑂 2 →𝐻 𝑂 2 +𝐻 (𝐻 𝑂 2 = hydroperoxy radical)
H-H O-O H-O-O H
𝐻+ 𝑂 2 →𝑂𝐻+𝑂
H O-O O-H O
𝑂𝐻+ 𝐻 2 → 𝐻 2 𝑂+𝐻
O-H H-H H-O-H H
𝐻+ 𝑂 2 +𝑀→𝐻 𝑂 2 +𝑀 (𝑀= passive species, process termination step)
H O-O H-O-O
Free Radicals = reactive, short-lived molecules with unpaired electron
Unpaired with proton, Charged, in this example: 𝐻 𝑂 2 , 𝐻, 𝑂𝐻, 𝑂

6. Bimolecular Reactions
Two molecules react, form two new molecules
𝐴+𝐵→𝐶+𝐷
For example: one step of multi-step reaction: 𝐻 2 + 𝑂 2 →𝐻 𝑂 2 +𝐻
Rate for this elementary reaction
𝑑 𝐴 𝑑𝑡 =− 𝑘 𝑏𝑖𝑚𝑜𝑙𝑒𝑐 𝐴 1 𝐵 1 = 𝑑 𝐵 𝑑𝑡 =− 𝑑 𝐶 𝑑𝑡 =− 𝑑 𝐷 𝑑𝑡
𝑘 𝐵𝑖𝑚𝑜𝑙𝑒 =𝑓𝑛 𝑇 , but is theoretically-based
Units 𝑘 𝑏𝑖𝑚𝑜𝑙𝑒𝑐 = 𝑚 3 𝑘𝑚𝑜𝑙∗𝑠
Use collision theory to find 𝑘 𝐵𝑖𝑚𝑜𝑙𝑒
During time 𝑡, an A particle, of diameter 𝜎, moving at speed 𝑣, “sweeps” volume
𝜋 𝜎 𝑣𝑡
If it is in a region of stationary B particles, also of diameter 𝜎, and number density 𝑛 𝐵 𝑉 , then the number of collisions per time is: 𝑛 𝐵 𝑉 𝜋 𝜎 2 𝑣 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛𝑠 𝑡𝑖𝑚𝑒

7. Collision Rate between all A and B particles
If all A and B particles are actually moving randomly (with a Maxwellian velocity distribution) with average speed 𝑣 (which depends on temperature)
Can be shown (CBS) that the rate at which a single moving particle A (diameter 𝜎 𝐴 ) collides with a field of randomly moving B ( 𝜎 𝐵 ) particles is 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛𝑠 𝑠𝑒𝑐
𝑍 𝑐 = 2 𝑛 𝐵 𝑉 𝜋 𝜎 𝐴𝐵 2 𝑣 𝐴
2 bigger than might be expected from last expression, due to motion
Where 𝜎 𝐴𝐵 = 𝜎 𝐴 + 𝜎 𝐵 2 and CBS 𝑣 𝐴 = 8 𝑘 𝐵 𝑇 𝜋 𝑚 𝐴
Now consider many A particles, with number density 𝑛 𝐴 𝑉
The number of collisions between all A’s and B’s per volume and time is
𝑍 𝐴𝐵 𝑉 = 𝑛 𝐴 𝑉 𝑛 𝐵 𝑉 𝜋 𝜎 𝐴𝐵 𝑘 𝐵 𝑇 𝜋𝜇
𝑛 𝐴 = 𝑁 𝐴 𝑁 𝐴𝑉 , 𝑛 𝐵 = 𝑁 𝐵 𝑁 𝐴𝑉 ,and CBS 𝜇= 𝑚 𝐴 𝑚 𝐵 𝑚 𝐴 + 𝑚 𝐵

8. Relation to reaction rate
− 𝑑 𝐴 𝑑𝑡 = 𝑍 𝐴𝐵 𝑉 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 𝑙𝑒𝑎𝑑𝑠 𝑡𝑜 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑘𝑚𝑜𝑙𝑒𝑠 𝐴 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐴 = 𝑍 𝐴𝐵 𝑉 𝒫 1 𝑁 𝐴𝑉
𝒫=𝑝∗𝑒𝑥𝑝 − 𝐸 𝐴 𝑅 𝑢 𝑇
𝑝= Steric factor (collision of geometry) < 1
𝑒𝑥𝑝 − 𝐸 𝐴 𝑅 𝑢 𝑇 = Fraction of all collisions with energy greater than activation energy 𝐸 𝐴 CBS
− 𝑑 𝐴 𝑑𝑡 = 𝑁 𝐴 𝑁 𝐴𝑉 𝑉 𝑁 𝐵 𝑁 𝐴𝑉 𝑉 𝜋 𝜎 𝐴𝐵 𝑘 𝐵 𝑇 𝜋𝜇 𝑝∗𝑒𝑥𝑝 − 𝐸 𝐴 𝑅 𝑢 𝑇 𝑁 𝐴𝑉
− 𝑑 𝐴 𝑑𝑡 =𝑝 𝑁 𝐴𝑉 𝜎 𝐴𝐵 2 𝐴 𝐵 8𝜋 𝑘 𝐵 𝑇 𝜇 𝑒𝑥𝑝 − 𝐸 𝐴 𝑅 𝑢 𝑇 =𝑘 𝑇 𝐴 𝐵
𝑘 𝑇 =𝑝 𝑁 𝐴𝑉 𝜎 𝐴𝐵 𝜋 𝑘 𝐵 𝑇 𝜇 𝑒𝑥𝑝 − 𝐸 𝐴 𝑅 𝑢 𝑇
but 𝑝 and 𝐸 𝐴 = ?

9. Arrhenius Form For a limited temperature range Three parameter form:
𝑘 𝑇 =𝐴𝑒𝑥𝑝 − 𝐸 𝐴 𝑅 𝑢 𝑇
𝐴= pre-exponential factor
Three parameter form:
𝑘 𝑇 =𝐴 𝑇 𝑏 𝑒𝑥𝑝 − 𝐸 𝐴 𝑅 𝑢 𝑇
𝐴, 𝑏 and 𝐸 𝐴 values are tabulated pp 112 (Chapter 4)

10. Other Elementary Reactions
Uni-molecular (isomerization or decomposition)
𝐴→𝐵 (change in structure)
𝐴→𝐵+𝐶 (examples 𝑂 2 →2𝑂; 𝐻 2 →2𝐻)
Measurements show that
At high pressures − 𝑑 𝐴 𝑑𝑡 = 𝑘 𝑢𝑛𝑖 𝐴 (first order in pressure)
At low pressures − 𝑑 𝐴 𝑑𝑡 = 𝑘 𝑢𝑛𝑖 𝐴 𝑀 (Explain this later):
𝑀= other molecules with which A may collide
Ter-molecular
𝐴+𝐵+𝑀→𝐶+𝑀
Recombination (𝐻+𝐻+𝑀→ 𝐻 2 +𝑀; 𝐻+𝑂𝐻+𝑀→ 𝐻 2 𝑂+𝑀)
− 𝑑 𝐴 𝑑𝑡 = 𝑘 𝑡𝑒𝑟 𝐴 𝐵 𝑀
Third order
𝑀= third body, caries energy away
If A = B (i.e. 𝐴+𝐴+𝑀→𝐶+𝑀), then − 𝑑 𝐴 𝑑𝑡 = 2𝑘 𝑡𝑒𝑟 𝐴 𝐴 𝑀

11. Multi-step Mechanism Reaction Rates
A sequence of reactions leading from Reactants to Products
Example: hydrogen combustion (forward and reverse reactions)
L steps, i = 1, 2,… L
N species, j = 1, 2,… N
R1: 𝐻 2 + 𝑂 2 𝑘 𝐹1 , 𝑘 𝑅1 𝐻 𝑂 2 +𝐻 𝑖=1
R2: 𝐻+ 𝑂 2 𝑘 𝐹2 , 𝑘 𝑅2 𝑂𝐻 +𝑂 𝑖=2
R3: 𝑂𝐻+ 𝐻 2 𝑘 𝐹3 , 𝑘 𝑅3 𝐻 2 𝑂+𝐻 𝑖=3
R4: H+ 𝑂 2 +𝑀 𝑘 𝐹4 , 𝑘 𝑅4 𝐻 𝑂 2 +𝑀 𝑖=4
Number of steps: L = 4
Number of Species (𝑗= 𝐻 2 , 𝑂 2 , 𝐻 𝑂 2 ,𝐻, 𝑂𝐻, 𝑂, 𝐻 2 𝑂,𝑀): N = 8
8 time-dependent unknown molar concentrations: 𝑗 𝑡
Need 8 equations (constraints)

12. Species net reaction rates
j = 1
𝑑 𝑂 2 𝑑𝑡 = 𝑘 𝑅1 𝐻 𝑂 2 𝐻 + 𝑘 𝑅2 𝑂𝐻 𝑂 + 𝑘 𝐹3 𝑂𝐻 𝐻 2
− 𝑘 𝐹1 𝐻 2 𝑂 2 − 𝑘 𝐹2 𝐻 𝑂 2 − 𝑘 𝐹4 𝐻 𝑂 2 𝑀
j = 2
𝑑 𝐻 𝑑𝑡 = 𝑘 𝐹1 𝐻 2 𝑂 2 + 𝑘 𝑅2 𝑂𝐻 𝑂 + 𝑘 𝐹3 𝑂𝐻 𝐻 2 + 𝑘 𝑅4 𝐻 𝑂 2 𝑀
− 𝑘 𝑅1 𝐻𝑂 2 𝐻 − 𝑘 𝐹2 𝐻 𝑂 2 − 𝑘 𝑅3 𝐻 2 𝑂 𝐻 − 𝑘 𝐹4 𝐻 𝑂 2 𝑀
j = 3, 4, …8
Book describes compact notation

13. What happens at equilibrium?
A general reaction
𝑎𝐴+𝑏𝐵 𝑘 𝑓 𝑘 𝑟 𝑐𝐶+𝑑𝐷
𝑑 𝐴 𝑑𝑡 =𝑎 − 𝑘 𝑓 𝐴 𝑎 𝐵 𝑏 + 𝑘 𝑟 𝐶 𝑐 𝐷 𝑑
At equilibrium 𝑑 𝐴 𝑑𝑡 =0, so
𝑘 𝑓 𝐴 𝑎 𝐵 𝐵 = 𝑘 𝑟 𝐶 𝑐 𝐷 𝑑
𝑘 𝑓 𝑇 𝑘 𝑟 𝑇 = 𝐾 𝐶 𝑇 = 𝐶 𝑐 𝐷 𝑑 𝐴 𝑎 𝐵 𝑏
𝐾 𝐶 𝑇 = Equilibrium Constant based on molar concentration

14. End 2017

15. Relationship between Rate Coefficients and Equilibrium Constant (Chapter 2)
𝑎𝐴+𝑏𝐵 𝑘 𝑓 𝑘 𝑟 𝑐𝐶+𝑑𝐷
𝐾 𝑃 𝑇 = 𝑃 𝐶 𝑃 𝑜 𝑐 𝑃 𝐷 𝑃 𝑜 𝑑 𝑃 𝐴 𝑃 𝑜 𝑎 𝑃 𝐵 𝑃 𝑜 𝑏 𝑖 = 𝑃 𝑖 𝑅 𝑢 𝑇
𝐾 𝐶 𝑇 = 𝑘 𝐹 𝑇 𝑘 𝑅 𝑇 = 𝐶 𝑐 𝐷 𝑑 𝐴 𝑎 𝐵 𝑏 = 𝑃 𝐶 𝑅 𝑢 𝑇 𝑐 𝑃 𝐷 𝑅 𝑢 𝑇 𝑑 𝑃 𝐴 𝑅 𝑢 𝑇 𝑎 𝑃 𝐵 𝑅 𝑢 𝑇 𝑏 = 𝑃 𝐶 𝑃 𝑜 𝑐 𝑃 𝐷 𝑃 𝑜 𝑑 𝑃 𝐴 𝑃 𝑜 𝑎 𝑃 𝐵 𝑃 𝑜 𝑏 𝑃 𝑜 𝑅 𝑢 𝑇 𝑐+𝑑−(𝑎+𝑏)
𝑖 = 𝑃 𝑖 𝑅 𝑢 𝑇
𝐾 𝐶 𝑇 = 𝐾 𝑃 𝑇 𝑃 𝑜 𝑅 𝑢 𝑇 𝑐+𝑑−(𝑎+𝑏)
Or 𝐾 𝑃 𝑇 = 𝐾 𝐶 𝑇 𝑅 𝑢 𝑇 𝑃 𝑜 𝑐+𝑑−(𝑎+𝑏)
Note: If 𝑎+𝑏=𝑐+𝑑, then 𝐾 𝑃 𝑇 = 𝐾 𝐶 𝑇

16. Example 4.2 (page 120, turn in next time)
In their survey of experimental determinations of rate constants for the H-H-O system, Nanson and Salimian [reference 10 in book] recommend the following rate coefficient for the reaction
𝑁𝑂+𝑂→𝑁+ 𝑂 2 .
𝑘 𝑓 =2.80∗ 𝑇 1.0 𝑒𝑥𝑝 −20,820 𝑇 = 𝑐𝑚 3 𝑔𝑚𝑜𝑙∗𝑠
Determine the rate coefficient at 2300 K for the reverse reaction, i.e.
𝑁+ 𝑂 2 →𝑁𝑂+𝑂