Bihar Board 12th Chemistry Answer Key 7.2.2025


Kinetics of Parallel Reactions

Kinetics of Parallel Reactions

Kinetics of Parallel Reactions

Parallel Reactions

Reactions in which initial species react to give multiple products simultaneously.
Let's consider a reactant 'A' reacts to form product 'B' and 'C' simultaneously is given below-
Consecutive Reactions
Where k1 and k2 are the rate contents.

If the initial concentration of 'A' is'[A]o' and that of 'B' and 'C' be zero. After time 't', the concentration of A, B and C be [A], [B,] and [C] respectively.
If we assume that both the reactions are first order, then both the rates R1 and R2 can be given as-
Rate R1 = −d[A]/dt = d[B]/dt = k1[A]1     ---Eq.-1
Rate R1 = −d[A]/dt = d[C]/dt = k2[A]1     ---Eq.-2
Now total rate of reaction-
−d[A]/dt = R1 + R2     ---Eq.-3
or, −d[A]/dt = k1[A] + k2[A]     ---Eq.-4
or, −d[A]/dt− = (k1 + k2)[A]     ---Eq.-5
or, d[A]/dt = −(k1 + k2)[A]     ---Eq.-6
on integrating the above equation with respect to 't', we get the following equation-

[A] = [A]o e−(k1 + k2)t      ---Eq.-6


Similarly,
d[B]/dt = k1[A] = k1[A]oe−(k1 + k2)t      ---Eq.-7
[B] = − k1[A]o/(k1 + k2)(e−(k1 + k2)t) + C      ---Eq.-8
When 't' = 0, [B] = 0
Putting the value of 't' and [B] in Eq.-8 we get-
C = k1[A]o/(k1 + k2)      ---Eq.-9
Now putting the value of 'C' in Eq.-8 we get-

[B] = k1[A]o/(k1 + k2)(1 − e−(k1 + k2)t)      ---Eq.-10

Likewise [B]-

[C] = k2[A]o/(k1 + k2)(1 − e−(k1 + k2)t)      ---Eq.-11

Now the ratio of [B] to [C]-
[B]/[C] = k1/k2      ---Eq.-12
Percentage Yield of [B] = k1/(k1 + k2) x 100      ---Eq.-13
Percentage Yield of [C] = k2/(k1 + k2) x 100      ---Eq.-14
Consecutive Reactions

If we assume that one is 1st order and other is 2nd order reactions, then-
d[B] = k1[A]1
d[C] = k2[A]2
So, − d[A]/dt = k1[A] + k2[A]2
or, [A] = k1/ek1t(k1 + k2[A]o) − k2[A]o
Case-1
When k2[A]o <<< k1
then, [A] = [A]oe − k1t
Case-2
When k2[A]o >>> k1
then, 1/[A] = 1/[A]oe + k2t

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B.Sc. 2nd Semester Chemistry

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