1. Reactions
In previous lectures materials flows were analyzed as steady-state
processes. Time was not a variable. In many processes time
variability is important.

2. 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

3. First Order Reactions Second Order Reaction
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

4. In terms of concentration:
Zero Order dC/dt = k
Units on k (mass/length3)/time or
concentration/time
First Order dC/dt = kC
Units on k (dC/dt)/C time-1
Second Order dC/dt = k C2
Units on k (dC/dt)/C2 concentration-1 time-1
or length3/(mass.time)

5. Zero Order Reactions
If A = A0 when t = 0 then the reaction rate equation
can be solved by integration:
dC/dt = k A0 A
dC = k t dt
If the concentration is increasing k is positive
If the concentration is decreasing k is negative

6. Increasing concentration:
dC = k t dt
C = C0 + kt
Decreasing concentration: A0 A
dC = - k t dt
C = C0 – k t

7. 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?
From the graph k = mg/(L.s)
C = C0 – k
0.5 = 9.5 – (0.075)*t
t = 120 s

8. First Order Reactions dC/dt = r = k C If A = A0 when t = o: dC/C = kt dt
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

9. ln A – ln A0 = k t ln A = k t + ln A0 y = m x + 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

10. 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:
How long will it take to reduce the BOD to 30 mg/L?
What will the BOD be after 5 days?

11. So, k = 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