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 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

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)

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

Increasing concentration: dC = k t dt C = C0 + kt Decreasing concentration: A0 A dC = - k t dt 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? 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: dC/C = k t 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

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

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?

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