Page 192 - Introduction to chemical reaction engineering and kinetics

174 Chapter 7: Homogeneous Reaction Mechanisms and Rate Laws

(c) At temperatures near 700°C and pressures near 1 bar, the overall reaction rate is observed
to be first-order in ethane pressure with a rate constant k = 1.1 x 1015 exp(-306000/RT).
How well does this model reproduce these results?
(d) Now improve the model and test the importance of other reactions by including them in
the computer model and examining the results. Use the following cases.
(dl) Reversible reaction steps.
(i) Include the reverse of step (3) in the mechanism and rerun the simulation-does
it affect the calculated rates?

(6) C21& + H’ + C2H; ; ks = 1013

(ii) How else might one estimate the significance of this reaction without running
the simulation again?
(d2) Steps involving energy transfer.
How many of the reactions in this mechanism might be influenced by the rate of
energy transfer? One of them is the termination step, which can be thought of as a
three-step process (reactions (7) to (9) below). As described in Section 6.4.3, there
are possible further complications, since two other product channels are possible
(reactions (10) and (11)).

(7) CzHs + H* + C2H; ; k7 = 6 x 1013
(8) CzH; + M + CzHs + M ; kg = 3 x 1013
(9) C2H; --f C2H; + H* ; k9 = 2 x 1013
(10) &Hi + 2CH; ; klo = 3 x 10 12
(11) C2H; --z C2H4 + H2 ; kll = 3 x 10 12

Include these reactions in the original model in place of the original reaction (5).
(You can assume that M is an extra species at the initial ethane concentration for
this simulation.) Use the values of the rate constants indicated, and run the model
simulation. What influence does this chemistry have on the conversion and selec-
tivity? How would you estimate the rate constants for these reactions?
(d3) The initiation step.
The initiation step also requires energy input.

(12) C& + M + C2H;i + M ; kn = 2 X 1013 exp(-340,00O/RT)

The other reactions, (8) and (lo), have already been included.
At 1 bar and 700°C is this reaction limited by energy transfer (12) or by decompo-
sition (lo)?
(d4) Termination steps.
Termination steps involving two ethyl radicals are also ignored in the original mech-
anism. Include the following reaction:

(13) 2 C2H; + C2H.4 + C2H6 ; k13 = 6 x 10”

Does this make a significant difference? Could you have predicted this result from
the initial model calculation?
(d5) Higher molecular-weight products.
Higher molecular-weight products also are made. While this is a complex pro-
cess, estimate the importance of the following reaction to the formation of higher
hydrocarbons by including it in the model and calculating the C4Hg product