handout21

# handout21 - EXAMPLE Now lets look at an unstable chain...

This preview shows pages 1–4. Sign up to view the full content.

EXAMPLE: Now lets look at an unstable chain reaction, for example the well-known explosive reaction: 2 H 2 + O 2 2 H 2 O . (5.207) The following mechanism is proposed: Initiation H 2 R I 2 H (5.208) The reaction is initiated e.g. by a spark. We consider this step to be independent of [ H 2 ] . Propagation H + O 2 k 1 OH + O (5.209) O + H 2 k 2 OH + H (5.210) OH + H 2 k 3 H + H 2 O (5.211) NOTE: Here, 2 OH molecules are being produced and 1 H 3 H . fi Termination H k 4 T H . . . wall (5.212) 5–65

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
H + O 2 + M k 5 T HO 2 + M (5.213) Eq. 5.212: at low pressure, H hits a wall of the container and is removed from the propagation loop. Eq. 5.213: at high pressure, a three-body interaction terminates the chain. NOTE: Our strategy for analysing this process is to again assume the SSA for the intermediates; however under certain conditions, this assumption will not be correct. Here we have three intermediates (three equations with three unknowns) Then we have d [ O ] d t = k 1 [ H ] [ O 2 ] - k 2 [ O ] [ H 2 ] (5.214) d [ OH ] d t = k 1 [ H ] [ O 2 ] + k 2 [ O ] [ H 2 ] - k 3 [ OH ] [ H 2 ] (5.215) d [ H ] d t = - k 1 [ H ] [ O 2 ] + k 2 [ O ] [ H 2 ] + k 3 [ OH ] [ H 2 ] + R I - k 4 T [ H ] - k 5 T [ H ] [ O 2 ][ M ] (5.216) Now applying the SSA and solving for [ H ] , we get - k 3 [ OH ] SS [ H 2 ] SS = - k 1 [ H ] SS [ O 2 ] SS + k 2 [ O ] SS [ H 2 ] SS + R I - k 4 T [ H ] SS - k 5 T [ H ] SS [ O 2 ] SS [ M ] SS (5.217) 5–66
0 = 2 k 2 [ O ] SS [ H 2 ] SS + R I - k 4 T [ H ] SS - k 5 T [ H ] SS [ O 2 ] SS [ M ] SS (5.218) From Eq. 5.214 we know that [ O ] SS = k 1 [ H ] SS [ O 2 ] SS k 2 [ H 2 ] SS (5.219) 0 = 2 k 1 [ H ] SS [ O 2 ] SS + R I - k 4 T [ H ] SS - k 5 T [ H ] SS [ O 2 ] SS [ M ] SS (5.220) and finally [ H ] SS = R I k 4 T + k 5 T [ O 2 ] SS [ M ] SS - 2 k 1 [ O 2 ] SS . (5.221) Lets see if we can understand this result. In essence we have three cases depending on the relative sizes of the three terms in the denominator: at low pressure (not many interparticle collisions, the H hits a wall) k 4 T >> k 5 T [ O 2 ][ M ] , k 1 [ O 2 ] [ H ] will stay small (the SSA is valid) This will result in a stable chain that terminates without an explosion.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern