VOLUME 2 Post-Treatment, Reuse, and Disposal
2.5 Reaction Order Data Analysis and Design
For design of water and wastewater treatment facilities, the physical, chemical and biological processes are used. Most commonly, the laboratory experiments are conducted to model the performance of reactors.
Valuable information such as order of reaction, reaction rate constants, and range and limits of these reactions are developed through analysis of experimental data. Numerous examples on zero-,first-, and second-order reactions are provided below to describe the procedures used in these analyses.
2.5.1 Zero-Order Reaction
Ifn=0, the generic Equation 2.5 is reduced to Equation 2.11 with respect to the concentration of reactant.
−dC
dt =k (2.11)
where
C=concentration remaining of reactant after reaction timet, mg/L t =reaction time, variable time unit (t)
k=zero-order reaction rate constant, mg/L·t
This is the rate law that describes the zero-order reaction. In the zero-order system, the reaction pro- ceeds at a rate independent of the concentration of the reactants. This is often the case when the reactant is present in high concentration. Rearranging Equation 2.11 and integrating it, the concentration ofCcan be expressed by Equation 2.12.
dC= −kdt C
C0
dC= −k t
0
dt
C=C0−kt (2.12)
whereC0 =initial concentration or reactant att=0, mg/L
Equation 2.12 is a linear relationship. If concentration data are plotted with respect to time, the result is a straight line. The slope of the line is the zero-order rate constantk, which has unit of concentration/time (e.g., mole/L·d). It is often useful to express a reaction in terms ofhalf-life. The half-life,t1/2, is defined as the time required for the concentration to decrease to one-half orC=(1/2)C0. Substituting this rela- tionship in Equation 2.12, it is reduced to Equation 2.13.
1
2C0O=C0−kt1/2 t1/2=1
2 C0
k (2.13)
EXAMPLE 2.15: REACTION ORDER, AND RATE CONSTANT
In a batch reactor, an industrial waste is exposed to UV radiation in the presence of a catalyst. The initial concentration of an organic dye in the industrial waste was 3000 mg/L. Samples were withdrawn from the reactor at different time intervals and the concentrations of organic dye remaining were measured. Fol- lowing results were obtained. Determine (a) the order of the reaction, (b) calculate the reaction rate cons- tant, (c) half-life, and (d) dye concentration remaining after 24 h of UV exposure.
Time, h Dye Concentration
Remaining, mg=L
0 3000
4 2800
8 2590
12 2380
16 2150
20 1990
Solution
1. Plot the concentration of dye remaining with time; draw the best-fit line and indicate the reaction order. The graphical relationship is shown inFigure 2.2.
Since the dye concentration remaining represents a linear relationship, the reaction is zero order.
2. Determine the slope andk.
The slope of the line is the reaction rate constant.
Slope=k=–51.5 mg/L·h 3. Determine the half-life.
C0=3000 mg/L
Use Equation 2.13 to calculate the half-life.
t1/2=1
2× 3000 mg/L
51.5 mg/L·h=29.1 h
C = −51.5 t + 3000
0 1000 2000 3000 4000
0 5 10 15 20
C, mg/L
t, h FIGURE 2.2 Zero-order destruction of organic dye (Example 2.15).
EXAMPLE 2.16: CONCENTRATION REMAINING
In the fermentation of sugar, an enzymatic reaction is utilized. Initial sugar solution was 0.15 mole/L.
After 4 hours of reaction, 10% sugar was converted into alcohol. Assuming no side reactions and zero-order reaction rate, calculate (a) reaction rate constant, (b) concentration remaining after 12 h, and (c) half-life.
Solution
1. Calculate the concentration of sugar remaining after 4 h.
C=(1−0.1)×0.15 mole/L=0.135 mole/L 2. Calculate the reaction rate constant.
Use Equation 2.12 to calculate the reaction rate constant.
C=C0−kt
0.135 mole/L=0.15 mole/L−k×4 h k= 1
4 h×(0.15 mole/L−0.135 mole/L)=3.75×10−3mole/L·h 3. Calculate the concentration of sugar remaining after 24 h.
Use Equation 2.12 to calculate the concentration of sugar remaining.
C=C0−kt=0.15 mole/L−3.75×10−3mole/L·h×24 h
=0.15 mole/L−0.09 mole/L=0.06 mole/L 4. Calculate the half-life.
Use Equation 2.13 to calculate the half-life.
t1/2=1
2× 0.15 mole/L
3.75×10−3mole/L·h=20 h
EXAMPLE 2.17: REACTION RATES OF REACTANTS
Gaseous ammonia is produced in a catalytic experiment utilizing the Haber process: N2+3 H2→ 2 NH3. The measured generation rate of ammonia (dCNH3/dt) was 2.0×10−4mole/L·s. If there were no side reactions, calculate the rate of reaction of (a) N2and (b) H2expressed in terms of NH3. Solution
1. Calculate the reaction rate of N2.
The balanced equation N2+3 H2→2 NH3indicates that ammonia is produced at twice the utili- zation rate of N2. This means that the reaction rate of N2is half that of NH3.
−dCN2
dt O=1 2
dCNH3
dt =1
2×2.0×10−4mole/L·s=1.0×10−4mole/L·s dCN2
dt = −1.0×10−4mole/L·s
4. Determine the concentration of dye remaining after 24 h of UV exposure from Equation 2.12.
Use Equation 2.12 to calculate the concentration of dye remaining.
C=C0−kt=3000 mg/L−51.5 mg/L·h×24 h=1764 mg/L
2. Calculate the reaction rate of H2.
−dCH2
dt =3 2
dCNH3
dt =3
2×2.0×10−4mole/L·s=3.0×10−4mole/L·s dCH2
dt = −3.0×10−4mole/L·s
2.5.2 First-Order Reaction
Ifn=1, the generic Equation 2.5 becomes Equation 2.14.
−dC
dt =kC (2.14)
This is the rate law for thefirst-order reaction. This suggests that the reaction rate is dependent upon the concentration of the chemical remaining. As reaction continues, the concentrationCremaining decreases with time. Therefore, thefirst-order reaction rate slows down over time. Equation 2.14 is rearranged and integrated with respect to time to give Equations 2.15a and 2.15b.
1
CdC= −kdt C
C0
1
CdC= −k t
0
dt
ln C
C0 = −kt
C=C0e−kt (2.15a)
ln( ) =C ln( ) −C0 kt or C=C010−kt (2.15b) where
k=thefirst-order reaction rate constant to the basee, t−1 K=thefirst-order reaction rate constant to the base 10, t−1
Equations 2.15a and 2.15b express an exponential decay. The semi-log plot of this equation will give a linear relationship, and slope of the line is the reaction rate constant. The half-life for thefirst-order relationship is expressed by Equations 2.16a and 2.16b.
t1/2=0.693
k (2.16a)
t1/2=0.301
K (2.16b)
EXAMPLE 2.18: DETERMINATION OF FIRST-ORDER REACTION RATE CONSTANT FROM GRAPHICS
The following COD data have been collected from a batch biological reactor study with an acclimated culture of microorganisms. Prepare arithmetic and semi-log plot of the data. Determine the reaction rate constants to the base 10, and to the basee.
EXAMPLE 2.19: DETERMINATION OF REACTION RATE CONSTANT BY CALCULATION
Determine the reaction rate constant from the data in Example 2.18.
Solution
1. Select the reaction rate equation.
Assumefirst-order reaction rate. Determinek(basee) from Equation 2.15a.
C=C0e−kt
t, h 0 3 5 7 10 12 15
C, mg=L COD 350 178 110 75 35 20 13
Solution
1. Prepare the arithmetic and semi-log plot of COD data.
The arithmetic and semi-log plots of COD data are illustrated inFigure 2.3a and b.
2. Determine the reaction order.
The arithmetic plot is nonlinear while the semi-log plot shows a linear relationship. Therefore, the data represents thefirst-order reaction rate.
3. Determine the slope from the semi-log plot.
Slope= −0.098 h−1
4. Determine the reaction rate constants to the base 10 and to the basee.
The slope of the line is the reaction rate constantK(base 10)=0.098 h−1.
Based on the relationship given in Equations 2.15a and 2.15b,k(basee) is calculated.
10−Kt=e−kt
−Ktlog(10)= −ktlog(e)
−Kt×1= −kt×0.434 or K=0.434k or k=2.03K k= K
0.434=0.098
0.434=0.23 h−1 0
100 200 300
(a)400 (b)
C, mg/L COD Log(C) = –0.098 t + 2.5
0 1 2 3
Log(C)
0 5 10 15
t, h 0 5 10 15
t, h
FIGURE 2.3 Graphical presentation of COD data: (a) arithmetic plot and (b) semi-log plot (base 10) (Exam- ple 2.18).
ln C
C0 = −kt k= −1
tln C C0
2. Apply the data of Example 2.18.
t, h C, mg=L C
C0 ln C
C0 1
tln C C0 , h1
0 350 1.000 0
3 178 0.509 0.676 0.23
5 110 0.314 1.157 0.23
7 75 0.214 1.540 0.22
10 35 0.100 2.303 0.23
12 20 0.057 2.862 0.24
15 13 0.037 3.293 0.22
Average¼0.23 3. The reaction rate constants for every data point are approximately the same. Therefore, the plot is
linear, and the assumption for thefirst-order reaction rate is correct.
4. The average value ofk(basee)=0.23 h−1.
EXAMPLE 2.20: CALCULATIONS FOR HALF-LIFE
A storage basin contains toluene and dieldrin. Thefirst-order removal constantskfor toluene and dieldrin are 0.067 h−1and 2.67×10−5h−1, respectively. Calculate the storage time for both chemicals to reduce their concentrations to one-half of their initial values.
Solution
1. Determine the half-life of toluene from Equation 2.16a.
ktoulene=0.067 h−1 ttoulene,1/2= 0.693
ktoulene= 0.693
0.067 h−1=10.3 h
2. Determine the half-life of dieldrin from Equation 2.16a.
kdieldrin=2.67×10−5h−1 tdieldrin,1/2= 0.693
kdieldrin= 0.693
2.67×10−5h−1=26,000 h or 26,000 h× 1
24 h/d×365 d/yr=3 yr
EXAMPLE 2.21: DETERMINATION OF INITIAL CONCENTRATION
An industrial sludge contains benzene. The allowable benzene limit for disposal of sludge in a landfill is 15 ppb. A 30-day-old sludge had benzene concentration of 1.1 ppb. If thefirst-order removal rate cons- tantkfor benzene is 0.00345 h−1, calculate if fresh sludge would have met the allowable concentration for disposal in the landfill.
Solution
1. Calculate the concentration of benzene in the fresh sludge from Equation 2.15a.
C=C0e−kt ln C
C0 = −0.00345 h−1×(30 d×24 h/d)= −2.48 C
C0=0.0837 C0= C
0.0837=1.1 ppb
0.0837=13.1 ppb 2. Compare the limit.
The initial concentration of benzene in fresh sludge is 13.1 ppb, which is lower than 15 ppb. There- fore, the fresh sludge would have met the allowable limit for disposal in the landfill.
EXAMPLE 2.22: CONCENTRATION RATIO REMAINING, AND CONCENTRATION RATIO REMOVED
A hazardous waste has half-life (t1/2) of 12 h. Calculate the time to reach 40% and 80% concentrations remaining, and the time to achieve 40% and 80% decay of the original hazardous waste. Plot the percent decay and percent concentration remaining versus time.
Solution
1. Calculate the reaction rate constantkfrom Equation 2.16a.
t1/2=0.693 k k=0.693
t1/2 =0.693
12 h =0.058 h−1
2. Calculate the time to achieve 40% and 80% concentration remaining from Equation 2.15a.
C=C0e−kt or C C0=e−kt 0.4=e−0.058 h−1×t40%removal
ln(0.4)= −0.058 h−1×t40%removal t40%removal= ln(0.4)
−0.058 h−1= −0.916
−0.058 h−1=15.8 h The time for 40% concentration remaining is 15.8 h.
0.8=e−0.058 h−1×t80% removal
ln(0.8)= −0.058 h−1×t80% removal
t80% removal= ln(0.8)
−0.058 h−1= −0.223
−0.058 h−1=3.8 h The time for 80% concentration remaining is 3.8 h.
3. Calculate the time to achieve 40% and 80% decay.
The decay or destruction=(1−C/C0) To achieve 40% decay, 0.4=(1−C/C0)
EXAMPLE 2.23: REACTION RATE CONSTANT OBTAINED FROM HALF-LIFE Chemical contamination of groundwater aquifer has been reported. The contaminants are benzene, tri- chloroethane (TEC), and toluene. The half-lives of these chemicals are 69, 231, and 12 days, respectively.
Calculate thefirst-order reaction rate constants of all these chemicals.
C/C0=0.6 (60% concentration remaining) t40%decay=t60%removal= ln(0.6)
−0.058 h−1= −0.511
−0.058 h−1=8.8 h The time for 40% concentration decay is 8.8 h.
To achieve 80% decay, 0.8=(1−C/C0) C/C0=0.2 (20% concentration remaining)
t80%decay=t20%removal= ln(0.2)
−0.058 h−1= −1.61
−0.058 h−1=27.8 h The time for 80% concentration decay is 27.8 h.
4. Tabulate the concentration ratio of decay (1−C/C0), and concentration ratio of remaining (C/C0) with time.
The concentration ratios of decay and remaining are derived from each other. These ratios are:
(1C=C0) (decay) 0.2 0.4 0.6 0.8
C=C0(remaining) 0.8 0.6 0.4 0.2
Time, h 3.8 8.8 15.8 27.8
5. Plot the percent decay and percent remaining concentration.
The plot of percent concentration destroyed and percent concentration remaining are shown in Figure 2.4.
0.0
0 5 10 15 20 25 30
0.2 0.4 0.6 0.8 1.0
% Remaining or decay, %
Time, h
% Remaining % Decay
CC0= 1 – e–kt
= e–kt CC0 Decay 1 –
Remaining
FIGURE 2.4 The graphical representation of percent destruction and percent remaining concentrations of a hazardous waste (Example 2.22).
Solution
The reaction rate constants are calculated from Equation 2.16a.
t1/2=0.693 k kbenzene=0.693
69 d =0.01 d−1 kTCE=0.693
231 d=0.003 d−1 ktoluene=0.693
12 d =0.058 d−1
EXAMPLE 2.24: RADIOACTIVE DECAY
Strontium-90 (90Sr) is a radioactive nuclide that has a half-life of 28 years. Like most radioactive nuclides, it exhibitsfirst-order decay. If a groundwater source is contaminated by90Sr, how long will it take to reduce the radioactivity by 99%.
Solution
1. Calculate thefirst-order reaction rate from Equation 2.16a.
k=0.693
28 yr =0.0248 yr−1
2. Calculate the reaction period from Equation 2.15a.
To reduce the radioactivity by 99%, 0.99=(1−C/C0) C/C0=1−0.99=0.01 (radioactivity remaining) C=C0e−kt or C
C0=e−kt 0.01=e−0.0248 yr−1×t ln(0.01)= −0.0248 yr−1×t t= ln(0.01)
−0.0248 yr−1= −4.61
−0.0248 yr−1=186 yr
The time to reduce the radioactivity by 99% is 186 years.
EXAMPLE 2.25: NITRIFICATION REACTION
The simplified consecutive reaction that describes the biological nitrification is given by equation:
NH3-N−O2
k1
NO2-N−O2
k2
NO3-N.
Write the decomposition equations that describe the rates of decomposition and formation of the reactants and products. Also, give the solution expressing the concentrations of each constituent with respect to time.
Solution
1. Develop the differential equations.
Assume that the rate of consecutive reactions arefirst order. The differential equations expressing the concentrations (as NH4-N) of NH3-N, NO2-N, and NO3-N are given by Equations 2.17 through 2.19.
dCNH3-N
dt = −k1CNH3-N (2.17)
dCNO2-N
dt =k1CNH3-N−k2CNO2-N (2.18)
dCNO3-N
dt =k2CNO2-N (2.19)
2. Express the solutions of the differential equations.
It is assumed that att=0, the initial concentration of NH3-N, NO2-N, and NO3-N are, respectively, C0NH3-N,C0NO2-N, andCNO0 3-N.
The solutions of the differential equations are given by Equations 2.20 through 2.22.
CNH3-N=C0NH3-N·e−k1t (2.20)
CNO2-N=k1C0NH3-N k2−k1
e−k1t−e−k2t
+C0NO2-N·e−k2t (2.21)
CNO3-N=CNH0 3-N 1−k2·e−k1t−k1·e−k2t k2−k1
+C0NO2-N 1−e−k2t
+CNO0 3-N (2.22)
EXAMPLE 2.26: GRAPHICAL REPRESENTATION OF NITRIFICATION REACTIONS
In a wastewater treatment plant, nitrification is carried out by oxidation of NH3. The initial concentration of ammonia nitrogen (NH3-N) is 25 mg/L, and the initial concentration of nitrite nitrogen (NO2-N) and nitrate nitrogen (NO3-N) are zero. Plot the concentration profiles of NH3-N, NO2-N, and NO3-N with respect to time in days. The reaction rate constantsk1andk2at 20◦C are 0.6 d−1and 0.3 d−1, respectively.
The operating temperature in the nitrification facility is 15◦C. Assumeθ=1.047.
Solution
1. Determine the reaction rate constants at 15◦C from Equation 2.10.
(k1)15=0.6 d−1×(1.047)(15−20)=0.48 d−1 (k2)15=0.3 d−1×(1.047)(15−20)=0.24 d−1
2. Calculate the concentrations ofCNH3-N,CNO2-N, andCNO3-Nat different time intervalstin days.
Calculate these concentrations from Equations 2.20 through 2.22.
Att=0, NH3-N=25 mg/L, and NO2-N and NO3-N=0.
The calculation steps att=1 day are given below.
CNH3-N=C0NH3-N·e−k1t=25 mg/L×e−(0.48 d−1)×1 d=15.5 mg/L CN02-N=k1C0NH3-N
k2−k1
e−k1t−e−k2t
+CNO0 2-N·e−k2t
=0.48 d−1×25 mg/L 0.24−0.48
( )d−1 e−(0.48 d)×1 d−e−(0.24 d)×1 d +0
= −50 mg/L×(0.619−0.787)
=8.4 mg/L
CNO3-N=C0NH3-N 1−k2·e−k1t−k1·e−k2t k2−k1
+CNO0
2-N 1−e−k2t +C0NO
3-N
=25 mg/L× 1−0.24 d−1×e−(0.48 d)×1 d−0.48 d−1×e−(0.24 d)×1 d
0.24−0.48 ( )d−1
+0+0
= −25 mg/L× 1−0.15−0.38 0.24−0.28
=1.1 mg/L
It may be noted that the sum of the concentrations of NH3-N, NO2-N, and NO3-N at all the points is 25 mg/L. This is because all concentrations are expressed in mg/L as NH4-N.
3. Assume different values oftand calculate the concentrations of NH3-N, NO2-N, and NO3-N at differ- ent time intervals. The calculated values are summarized below.
t, d NH3-N, mg=L NO2-N, mg=L NO3-N, mg=L
0 25.0 0 0
1 15.5 8.4 1.1
2 9.6 11.8 3.6
3 5.9 12.5 6.6
4 3.7 11.8 9.5
5 2.3 10.5 12.2
6 1.4 9.0 14.6
7 0.9 7.6 16.5
8 0.5 6.3 18.2
9 0.3 5.1 19.6
10 0.2 4.1 20.7
11 0.1 3.3 21.6
12 0.1 2.6 22.3
13 0.0 2.1 22.8
14 0.0 1.7 23.3
4. Plot the concentration profiles.
The concentration profiles of NH3-N, NO2-N, and NO3-N with respect to time at 20◦C are shown inFigure 2.5.
0 5 10 15 20 25 30
0 2 4 6 8 10 12 14
C, mg/L
t, d
NO3-N
NO2-N NH3-N
NO3-N NO2-N
NH3-N
FIGURE 2.5 Concentration profiles of NH3-N, NO2-N, and NO3-N with time (Example 2.26).
EXAMPLE 2.27: PARALLEL IRREVERSIBLE FIRST-ORDER REACTIONS
In a batch reactor, a reactant A produces two products B and C in a competitive reaction. Both reactions followfirst-order kinetics with respect to A. Also, the concentrations of B and C at any time are expressed as A. Develop the rate equations and solve the differential equations. Plot the concentration profiles. The initial concentration of A is 45 mg/L and that of B and C are zero. Assume that the reaction coefficientsk1
andk2at 20◦C are 0.35 and 0.21 per day when all concentrations of reactant and products are expressed in mg/L as reactant A. The reaction vessel temperature is 15◦C andθ=1.135. The parallel reaction is as follows.
A C k1 B
k2
Solution
1. Develop the rate equations.
−dCA
dt =k1CA+k2CA=CA(k1+k2) dCB
dt =k1CA
dCC
dt =k2CA
2. Integrate the rate equation and develop concentration equations Equations 2.23 through 2.25.
CA=CA0e−(k1+k2)t (2.23)
CB= k1
k1+k2
CA0 1−e−(k1+k2)t
(2.24) CC= k2
k1+k2
CA0 1−e−(k1+k2)t
(2.25)
3. Determine the reaction rate constantsk1andk2at 15◦C from Equation 2.10.
(k1)15=0.35 d−1×(1.135)(15−20)=0.19 d−1 (k2)15=0.21d−1×(1.135)(15−20)=0.11 d−1 4. Set up the initial conditions.
Att=0, CA0 =45 mg/L, CB0=0,andCC0 =0.
5. Calculate the concentrations ofCA,CB, andCCat 0.5 d from Equations 2.23 through 2.25.
CA=45 mg/L×e−(0.19+0.11)d−1×0.5d=45 mg/L×0.86=38.7 mg/L CB= 0.19 d−1
(0.19+0.11)d−145 mg/L×1−e−(0.19+0.11)d−1×0.5d
=0.63×45 mg/L×(1−0.86)=4.0 mg/L
CC= 0.11 d−1
(0.19+0.11) d−145 mg/L×1−e−(0.19+0.11)d−1×0.5 d
=0.37×45 mg/L×(1−0.86)=2.3 mg/L
6. Assume different values oftand tabulate the concentrations ofCA,CB, andCCwith respect to time. The concentrations ofCBandCCare expressed asCA.
t, d CA, mg=L CB, mg=L CC, mg=L
0 45 0.0 0
0.5 38.7 4.0 2.3
1 33.3 7.4 4.3
2 24.7 12.9 7.4
3 18.3 16.9 9.8
4 13.6 19.9 11.5
5 10.0 22.1 12.8
6 7.4 23.8 13.8
7 5.5 25.0 14.5
8 4.1 25.9 15.0
9 3.0 26.6 15.4
10 2.2 27.1 15.7
11 1.7 27.4 15.9
12 1.2 27.7 16
13 0.9 27.9 16.2
14 0.7 28.1 16.3
Note:The sum of concentrations ofCA,CB, andCCat any timetis 45 mg=L since all concentrations are expressed in mg=L as reactant A.
EXAMPLE 2.28: PSEUDO FIRST-ORDER OR AUTO-CATALYTIC REACTION A waste stabilization pond has hydrogen sulfide (H2S) concentration of 17 mg/L. It was found experi- mentally that the hydrogen sulfide odors are completely eliminated at a residual concentration of 0.034 mg/L. A decision therefore was made to aerate the lake so that H2S is oxidized to nonodorous sul- fate ion in accordance with the following oxidation reaction: H2S+2 O2SO2−4 +2H+. Experimen- tally it was determined that the overall reaction follows second-order kinetics with respect to both oxygen and hydrogen sulfide concentrations,d[H2S]/dt= −k[O2][H2S]. It is expected that 2 mg/L oxygen con- centration will be maintained in the lake. Therefore, the reaction can be modified to pseudofirst order.
The rate constantkfor the reaction was determined experimentally and was 1000 L/mole·d. If the aera- tion completely inhibits the anaerobic respiration, how long would it take for H2S concentration to reach the desired residual concentration for odor control.
Solution
1. Determine the reaction rate equation.
Since the concentration of oxygen is maintained constant, the reaction can be modified to pseudo first order. The modified equation is expressed by Equation 2.14 and the H2S concentration is obtained from Equation 2.15a.
d[H2S]
dt = −k′[H2S] [H2S] = [H2S]0e−k′t where
k′ =modified reaction rate constant to expressk′=k[O2], L/mole·d [H2S] =equilibrium concentration of H2S, mole/L
[H2S]0=initial concentration of H2S, mole/L
2. Calculate the value of the modified reaction rate constantk′. k′=k[O2] =1000 L
mole·d×2 mg/L O2× 1 g
1000 mg× 1
32 g/mole O2=0.0625 d−1 7. Plot the concentration profiles ofCA,CB, andCC(Figure 2.6).
10 20 30 40 50
00 2 4 6 8 10 12 14
C, mg/L
CA t, dCB CC
CA
CB
CC
FIGURE 2.6 Time-dependent concentrations of reactant and products in parallel irreversible reaction (Example 2.27).
3. Calculate the molar concentration of initial and equilibrium H2S concentrations.
Initial concentration,[H2S]0 =17 mg/L H2S× 1 g
1000 mg× 1
34 g/mole H2S
=5×10−4mole/L H2S Equilibrium concentration,[H2S] =0.034 mg/L H2S× 1 g
1000 mg× 1
34 g/mole H2S
=10−6mole/L H2S 4. Calculate the time to reach the equilibrium concentration of H2S.
10−6mole/L=5×10−4mole/L×e−0.0625 d−1×t e−0.0625 d−1×t=0.002
t= − 1
0.0625 d−1ln(0.002)= − −6.21
0.0625 d−1=99.4 d≈100 d
2.5.3 Second-Order Reaction The second-order reactions are of two types:
a. Type I reactions (2 A→P) are expressed by the following equation:
−d[A]
dt =k[A]2
b. Type II reactions (A+B→P) are expressed by the following equation:
−d[A]
dt =k[A][B] or −d[B]
dt =k[A][B], wherek =reaction rate constant, L/mole·t
Several examples on both types of second-order reactions are given below.
EXAMPLE 2.29: SECOND-ORDER TYPE I EQUATION DERIVATION AND REACTION RATE CONSTANT
Consider the following reaction carried out in a batch reactor: A+A→P.
Develop the following: (a) equation expressing the concentration with respect to time, (b) equation expressing half-life, and (c) determination of kinetic coefficientskand the initial concentrationCA.
t, h 0 1 2 3 4 5 6
CA, mg=L 100 80 67 57 50 44 40
1=CA, L=mg 0.01 0.0125 0.0149 0.0175 0.02 0.0227 0.025
Solution
1. Develop the concentration equation.
r= −dCA
dt =kC2A
whereCA=concentration of A at timet, mg/L dCA
C2A = −k dt
Integrate and solve forCA. This is represented by Equation 2.26.
CA
CA0
dCA
C2A =t
0
−k dt
1 CA− 1
CA0
=kt (2.26)
2. Develop the half-life expression.
The half-life expression (Equation 2.27) is obtained by substitutingCA=(1/2)CA0. t1/2= 1
kCA0
(2.27) 3. DeterminekandCA0.
A plot of 1/CAversustgives a linear relationship as shown inFigure 2.7.
The slope of the line iskand the intercept is 1/CA0. CA0= 1
0.01 L/mg=100 mg/L k=0.0025 L/mg·h
4. Determine the half-life from Equation 2.27.
t1/2= 1
0.0025 L/mg·h×100 mg/L=4h 1/CA, L/mg
t, h 1
k 1/CA = 0.0025 t + 0.01
1/CA0 0.000 0.005 0.010 0.015 0.020 0.025 0.030
0 1 2 3 4 5 6
FIGURE 2.7 Determination of reaction rate constant of second-order Type I reaction (Example 2.29).
EXAMPLE 2.30: SECOND-ORDER DECAY
Benzene concentration in an industrial sludge is 10 ppb. What will be the concentration remaining after 90 days storage? Assume second-order decay, and reaction rate constant is 0.00085 L/µg·d.
Solution
Determine the concentration remaining from Equation 2.26.
1 CA= − 1
CA0
+kt 1
CA= 1
10μg/L+0.00085 L
μg·d×90 d=(0.1+0.0765)L
μg=0.1765L μg CA=5.7μg/L(ppb)
EXAMPLE 2.31: REACTION ORDER AND REACTION RATE CONSTANT
A chemical reaction was carried out in a batch reactor. The experimental data are given below. Determine the order of reaction and the reaction rate constant.
Time, min 0 1 2 3 4 5 6
CA, mg=L 100 80 67 57 50 44 40
Solution
1. Use Equation 2.5 for two data points.
−dCA1
dt =kCnA1
−dCA2
dt =kCnA2 2. Take log on both sides.
log −dCA1
dt
=log(k)+nlog(CA1)
log −dCA2
dt
=log(k)+nlog(CA2)
3. Equate logkin both equations to solve forn.
n=
log −dCA1 dt
−log −dCA2 dt log(CA1)−log(CA2)
4. Select two data pointst=2 andt=4 to determine the differentials.
−dCt=2
dt = −Ct=2+1−Ct=2−1
t2+1−t2−1 = −57−80 3−1 =11.5
−dCt=4
dt = −Ct=4+1−Ct=4−1
t4+1−t4−1 = −44−57 5−3 =6.5 5. Calculatenand reaction order.
n=
log −dCt=2
dt
−log −dCt=4
dt
log(Ct=2)−log(Ct=4) =log(11.5)−log(6.5)
log(67)−log(50) =1.061−0.813 1.826−1.699=0.248
0.127=1.95≈2 The reaction is Type I second-order and Equation 2.5 withn=2 is valid.
EXAMPLE 2.32: EFFECT OF REACTION ON SPECIES CONCENTRATION
Many photochemical reactions occur in the atmosphere that are related with smog formation. Following are three example reactions:
NO2+OH−−−−k1 HNO3 (Reaction 1) 2 HO2−−−−k2 H2O2+O2 (Reaction 2) NO2+O3−−−−k3 NO−3 +O2 (Reaction 3)
Write differential equations to describe the net effect of these three reactions on the mole fractions of nitrogen dioxide (NO2) and hydroperoxy radical (HO†2). Assume that reactions occur under well-mixed condition and are second order.
Solution
1. Write the rate law expressions.
r1=k1CNO2COH†
r2=k2CHO†2
2
r3=k3CNO2CO3
2. Write the differential equations of NO2in Reaction 1.
NO2and OH†are consumed, and nitric acid (HNO3) is produced.
dCNO2
dt = −r1= −k1CNO2×COH†
3. Write the differential equation of HO†2in Reaction 2.
HO†2is consumed and converted to hydrogen peroxide (H2O2) and oxygen (O2).
dCHO†2
dt = −2r2= −2k2 CHO†2
2
In this reaction, two moles of HO†2are consumed. The exponent 2 expresses second-order reaction of HO†2.
4. Write the differential equation of NO2in Reaction 3.
NO2reacts with ozone (O3) to produce nitrate (NO−3) and oxygen (O2).
dCNO2
dt = −r3= −k3CNO−3 ×CO3
5. Write the differential equation of NO2in Reactions 1 and 3.
NO2is consumed in both Reactions 1 and 3. Therefore, the effects of each reaction are added.
dCNO2
dt = −r1−r3= −k1CNO2COH†−k3CNO2CO3
EXAMPLE 2.33: TYPE II SECOND-ORDER REACTION, DERIVATION OF EQUATION
Consider the following reaction is carried out in a batch reactor. A+B−−−−k P. Develop the equation expressing the concentration of reactants A and B with respect to time.1
Solution
1. Write the Type II second-order reaction.
−dCA
dt = −dCB
dt =kCACB
2. Use the stoichiometric relationships.
dCA=dCB
CA
CA0
dCA=CB
CB0
dCB
CA0−CA=CB0−CB
whereCA0andCB0 are initial concentrations.
CB=CB0−CA0+CA
−dCA
dt =kCA(CB0−CA0+CA)
− dCA
CA(CB0−CA0+CA)=kdt
3. Using the method of partial fraction, integrate the equation.
1
CA(CB0−CA0+CA); p
CA+ q
CB0−CA0+CA
, wherepandqare constants, dimensionless
Note:The identity symbol (≡) is used to indicate an equality between the two sides of the equation for all values ofCA.
CA(CB0−CA0+CA) CA(CB0−CA0+CA); p
CA
CA(CB0−CA0+CA) 1
+q (CB0−CA0+CA)CA
(CB0−CA0+CA)
1;p(CB0−CA0−CA)+qCA