CHAP. 9: CHEMICAL KINETICS [CONTENTS] 276
Example
The dimerization
2 A→A 2 (9.35)
is a second-order reaction whose rate constant at a certain temperature equals k = 5 ×
10 −^5 dm−^3 mol−^1 s−^1. The initial monomer concentration iscA0= 0.7 mol dm−^3 and the initial
dimer concentration is zero. Calculate the time in which the dimer concentration will decrease
to 0.1 mol dm−^3 , and the time in which all monomer will be consumed.
Solution
We first carry out material balance. From equations (9.10) we obtain
cA=cA0− 2 x , cA 2 =x =⇒ cA=cA0− 2 cA 2 = 0. 7 − 2 · 0 .1 = 0. 5 mol dm−^3.
We calculate the time needed for the formation of 0. 1 mol dm−^3 of dimer, i.e. the time at which
the monomer concentration equals 0. 5 mol dm−^3 , from equation (9.32)
τ=
1
5 · 10 −^5
( 1
0. 5
−
1
0. 7
)
= 11 429s= 3. 175 hours.
The time needed for all of the monomer to be consumed is calculated as a limit of relation (9.32),
withcAapproaching zero from the right (from positive values)
lim
cA→ 0 +
τ=
1
k
lim
cA→ 0 +
( 1
cA
−
1
cA0
)
= +∞.
All of the monomer will be consumed after an infinite time. This result applies in general for all
one-way reactions regardless of their reaction order. The only exception is the zero-order reaction
where the reactant is consumed after a finite time given by equation (9.21).
9.2.3.5 Type.
A + B→products. (9.36)
