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Physical Chemistry 3: — Chemical Kinetics — - Christian-Albrechts ...

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Appendix C 268<br />

C.2 Application to two consecutive first-order reactions<br />

Consider the reaction system<br />

A 1<br />

−→ B<br />

B 2<br />

−→ C<br />

(C.11)<br />

(C.12)<br />

which is described by the following rate equations<br />

[A]<br />

<br />

[B]<br />

<br />

[C]<br />

<br />

= − 1 [A] (C.13)<br />

=+ 1 [A] − 2 [B]<br />

(C.14)<br />

=+ 2 [B]<br />

(C.15)<br />

The solution for Eq. C.13 is<br />

[A] = [A] 0<br />

− 1 <br />

(C.16)<br />

Thus we have to find the solution of the inhomogeneous DE for [B] (Eq. C.14)<br />

[B]<br />

<br />

+ 2 [B] = 1 [A] 0<br />

− 1 <br />

(C.17)<br />

• Solution of the homogeneous DE by separation of variables:<br />

y<br />

[B]<br />

<br />

= − 2 [B] (C.18)<br />

[B] <br />

= × − 2 <br />

(C.19)<br />

• Determination of a particular solution by variation of constant:<br />

y<br />

[B] <br />

<br />

= ()<br />

[B] <br />

= () × − 2 <br />

= ()<br />

<br />

× − 2 − () × 2 − 2 <br />

(C.20)<br />

(C.21)<br />

(C.22)<br />

Insertion of [B] <br />

and [B] <br />

<br />

into Eq. C.17 gives<br />

()<br />

<br />

which simplifies to<br />

× − 2 − () × 2 − 2 + () × 2 − 2 = 1 [A] 0<br />

− 1 <br />

()<br />

<br />

= 1 [A] 0<br />

( 2− 1 )<br />

This DE is easily integrated to obtain ():<br />

(C.23)<br />

(C.24)

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