Transcript Document

Reactions
In previous lectures materials flows were analyzed as steady-state
processes. Time was not a variable. In many processes time
variability is important.
Reaction Rates
A mathematical expression describing the rate at which the mass
or volume of some material A is changing with time T is:
dA/dt = r
r = reaction rate
Zero Order Reaction
Reaction rate, r, is a constant:
r=k
So: dA/dt = k
First Order Reactions
The change of the component is proportional to the amount of
the component present, or:
r = kA
k is called the reaction rate constant
the units on k are time-1 in the case of a first order reaction
Second Order Reaction
The change of the component is proportional to the square of the
amount of the component present
r = k A2
So: dA/dt = k A2
In terms of concentration:
Zero Order
Units on k
First Order
Units on k
Second Order
Units on k
dC/dt = k
(mass/length3)/time or
concentration/time
dC/dt = kC
(dC/dt)/C
time-1
dC/dt = k C2
(dC/dt)/C2
or
concentration-1 time-1
length3/(mass.time)
Zero Order Reactions
If A = A0 when t = 0 then the reaction rate equation
can be solved by integration:
dC/dt = k
A
t
dt
dC = k
A0
0
If the concentration is increasing k is positive
If the concentration is decreasing k is negative
Increasing concentration:
A
dC = k
A0
t
dt
0
C = C0 + kt
Decreasing concentration:
A
t
dC = - k
A0
dt
0
C = C0 – k t
Example
Nitrogen gas is used to strip oxygen from water before an aeration
test is performed. The following data were obtained during a similar
test. How long will it take before the oxygen concentration is less than
0.5 mg/L?
Time, s
0
15
30
45
60
90
Concentration, mg/L
9.5
8.375
7.25
6.125
5
2.75
From the graph k = 0.075 mg/(L.s)
C = C0 – k
0.5 = 9.5 – (0.075)*t
t = 120 s
First Order Reactions
dC/dt = r = k C
If A = A0 when t = o:
A
t
dC/C = k
A0
dt
0
ln(A/A0) = k t
or A/A0 = ekt
or lnA – lnA0 = kt
Again k is positive when the concentration increases
k is negative when the concentration decreases
ln A – ln A0 = k t
ln A = k t + ln A0
y
= mx+ b
So, if we plot (ln A) on the y-axis and (t) on the x-axis
The slope is the reaction rate constant, k, and the
intercept is ln A0
Example
When the BOD of a wastewater is caused by particulate material
its removal may often be described as first order. Consider the
following data which were obtained from a batch test of a food
processing waste water:
time, hr concentration, mg/L
0
220
12
190
24
170
48
130
96
75
144
45
How long will it take to reduce the BOD to 30 mg/L?
What will the BOD be after 5 days?
So, k = 0.011 hr-1
Now:
ln(A/A0) = -kt
t = [ln(A/A0)]/-k
t = [ln(30/220)]/(-0.011)
= 181 hr = 7.55 days
Also:
[BOD] = [BOD0] e-kt
= (220) e-(0.011)(5x24)
= 58.8 mg/L