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

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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
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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
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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.
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