V. Irreversible Mixed 2nd-Order Reaction
A. Given:
!A + B
k
P
B. Rate Equation:
1. If [A0] = [B0]:
Since...
-d[A]/dt = -d[B]/dt = d[P]/dt = k[A][B]
The rate expression reduces to...
-d[A]/dt = k[A]2
• Described in Section III.
(1)
V. Irreversible Mixed 2nd-Order Reaction
2. If [A0] ≠ [B0]:
[A0] - [A] = [B0] - [B]
Consequently...
[B] = [A] - [A0] + [B0]
(2)
Substituting [B] from eq (2) into eq (1)...
-d[A]/dt = k[A]{[A] - [A0] + [B0]}
(3)
Integrate... (CRC #37)
# [Ao ]( [Bo ] " [Ao ] + [A]) &
1
ln$
' = kt
B
"
A
A
B
[ o] [ o] %
[ ][ o ]
(
• This expression is undefined at [A0] = [B0].
!
(4)
V. Irreversible Mixed 2nd-Order Reaction
C. Graphics:
f(x)
(see below)
Let...
linear,
slope = k
t (sec)
f(x) = ax
such that...
# [Ao ]( [Bo ] " [Ao ] + [A]) &
1
f(x) =
ln$
'
[Bo ] " [Ao ] %
[A][Bo ]
(
!
x=t
a=k
• [A0] and [B0] are known, and we measure [A] vs. t. Problems arise
(particularly at high conversions) from a hypersensitivity of the rates to the
accuracy in measuring [A0] and [B0].
V. Irreversible Mixed 2nd-Order Reaction
D. Flooding Techniques:
1. Let [B0] >> [A0]
Then [B] ≈ [B0] and eq (4) reduces to...
(1/[B0])ln{[A0]/[A]} = kt
Rearranging...
ln{[A]/[A0]} = -k[B0]t = -kobsdt
such that...
kobsd = k[B0] = constant
• "pseudo-first-order" conditions.
2. Note: Ideally...
[B0] > 20[A0])
If the reaction is high order in [B], even a 10% consumption of B can
distort the data.