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Sec. 7.2   Searching for a Mechanism                           347

                                  For  constant concentrations of  CS2 and the third body, M, we take a ratio of  Equa-
                                  tion (E7-1.10) to (E7-1.9):
                                                                                           (E7-1.11)


                                   which is of  the same form as that suggested by  Figure E7-1.1. Equation (E7-1.11)
                                   and similar equations involving scavengers are called Stem-Volmer  equations.

                                     Now, let us proceed to some slightly more complex examples involving
                                chain reactions. A chain reaction consists of the following sequence:
                                     1.  Initiation: formation of  an active intermediate.
                   Steps in a chain   :2.  Propagation  or chain transfer:  interaction  of  an  active intermediate
                         reaction       with the reactant or product to produce another active intermediate.
                                     3.  Termination: deactivation of  the active intermediate.

                                   Example 7-2  PSSH Applied to Thermal Cracking of Ethane

                                   The thermal decomposition of  ethane to ethylene, methane, butane, and hydrogen is
                                   believed to proceed in the following sequence:
                                   Init(iation:

                                                      >
                                   (1)      C2H6  bH6 2CH3*               - r1C2H6  = klC2H6  [C2H61
                                                                            Let kl  = kIC2H6
                                   Propagation:
                                                    k2
                                   (2) CH,.  +CZH6  * CH,+C,H,*           -r2~~~~ I[c2H[61
                                                                                     [CH3*
                                                                                   k2
                                                                                 =
                                   (3)    C2H5*    k3   )  C,H4+H*          ‘3C2H,  = k3  [C2H50  1
                                                    k4
                                   (4)   He fClH6 * C2H5.  +H2            -r4C2H6  = k4  cHo  I[c2H6jl
                                   Temnination:

                                                     kS
                                   (5)     2C2 H,       ’ C4H10          -r5C2Hj.  = kSCZHj[C1H50  I-’
                                                                            Let k5   k5C2Hj.
                                     (a)  Use the, PSSH to derive a rate law for the rate of  formation of  ethylene.
                                      (Ib)  Compare the PSSH solution in Part (a) to that obtained by  solving the com-
                                         plete set of ODE mole balances.

                                   Solution
                                   Part (a)  Developing the Rate Law
                                   The rate of formation of ethylene is

                                                                                             (E7-2.1)
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