Presentation on theme: "FE Review for Environmental Engineering"— Presentation transcript:
1FE Review for Environmental Engineering Problems, problems, problemsPresented by L.R. Chevalier, Ph.D., P.E.Department of Civil and Environmental EngineeringSouthern Illinois University Carbondale
2FE Review for Environmental Engineering Chemical Foundations
3ProblemStrategySolutionCalculate the molecular weight, equivalent weight, molarity and normality of the following:a. 200 mg/L HClb. 150 mg/L H2SO4c. 100 mg/L Ca(HCO3)2
4Use periodic table to get molecular weight Convert mg/L to mol/L ProblemStrategySolutionUse periodic table to get molecular weightConvert mg/L to mol/LDetermine n for each compoundApply equationsEW = MW/nN = Mn
5Problem Strategy Solution “in an Acid/Base reaction, n is the # of hydrogen ions that a molecule transfers”
16What is the pOH if [OH-] = 10-8? What is the pH if [OH-] = 10-8? ExampleSolutionWhat is the pH if [H+] = 10-3?pH = 3What is the pOH if [OH-] = 10-8?pOH = 8What is the pH if [OH-] = 10-8?pH = = 6What is the [H+] if [OH-] = 10-5?[H+]= = 10-9 mol/L
17ProblemStrategySolutionDerive a proof that in a neutral solution, the pH and the pOH are both equal to 7.
18ProblemStrategySolutionEvaluate the governing equation
29Consider the problem of removing nitrogen from municipal wastewater StrategySolutionConsider the problem of removing nitrogen from municipal wastewaterRemove nitrogen to prevent the stimulation of algae growthPrevent excessive nitrate [NO3-] level in drinking water from causing a potentially lethal condition in babies known as methemoglobinemia
30One way to remove is a process known as ammonia stripping ProblemStrategySolutionOne way to remove is a process known as ammonia strippingWhen organic matter decomposes, nitrogen is first released in the form of ammoniaNH3 - low solubility in water (ammonia)NH4+ - highly soluble in water (ammonium ion)
31ProblemStrategySolutionBy driving the equilibrium toward the right, less soluble gas is formed and encouraged to leave the solution and enter air stream in a gas stripping tower.This technique has been adapted for use in removing VOC’s (volatile organic chemicals) from groundwater.How can the reaction be driven to the formation of ammonia (NH3)?Need to decrease [H+] or increase the pH.
32Highly Low Soluble Solubility Want to consider [NH3]/[NH4+] ProblemStrategySolutionHighlySolubleLowSolubilityWant to consider[NH3]/[NH4+]Should we decrease this or increase this?
33Highly Low Soluble Solubility Increase it. How can we do this? Problem StrategySolutionHighlySolubleLowSolubilityIncrease it.How can we do this?
40Summary Of Example Problem Nitrogen, in the form of ammonia (NH3) is removed chemically from the water by raising the pHThis converts ammonium ion (NH4+) into ammoniaNH3 is then stripped from the water by passing large quantities of air through the water
41ProblemStrategySolutionA sample of water at pH 10 has 32.0 mg/L of carbonate and 56.0 mg/L of bicarbonate ion. Find the alkalinity as CaCO3.
42Problem Strategy Solution Determine the MW of HCO3- and CO3-2 Determine the EW of HCO3- and CO3-2Convert the concentrations of HCO3- , CO3-2, H+ and OH- to mg/L as CaCO3Add the concentrations in mg/L as CaCO3 of HCO3- , CO3-2, and OH-, and subtract H+
43Now we need to convert to mg/L CaCO3 ProblemStrategySolutionNow we need to convert to mg/L CaCO3
44Problem Strategy Solution I will leave it up to you to check calculations for H+ and OH-end of problem
45The solubility product for the dissociation of ProblemStrategySolutionThe solubility product for the dissociation ofMg(OH)2 is 9 x Determine the concentration of Mg2+ and OH- at equilibrium.
46Problem Strategy Solution Write the equation for the reaction Write the solubility product equationRecognize from Eqn. 1 the relationship between the number of moles of Mg2+ and the number of moles of OH- resulting from the dissociation of Mg(OH)2, and how this relates to Eqn 2
47Problem Strategy Solution 1. Write the equation for the reaction. 2. The solubility product equation is:
48Problem Strategy Solution 3. If x is the amount of Mg2+ resulting from the dissociation is given as x, then the amount of OH- is equal to 2x......end of example
49ProblemStrategySolutionMagnesium is removed from an industrial waste stream by hydroxide precipitation at a pH = 10. Determine the solubility of Mg2+ in pure water at 25° C and pKsp of
50Problem Strategy Solution 1. Identify the two governing equations (Ksp and Kw)2. Recognize that [OH-] = pH3. Substitute to derive an equation [Mg2+] = f(pH)
511. What are your two governing equations? ProblemStrategySolution1. What are your two governing equations?2. Two unknowns, and two equations.
523. Given the pH, we know [H+]. ProblemStrategySolution3. Given the pH, we know [H+].4. Solve for [OH-]2
535. Substitute into 1st governing equation, and solve for [Mg2+]. ProblemStrategySolution5. Substitute into 1st governing equation, and solve for [Mg2+].
54ProblemStrategySolution6. Substitute value of pH given in the problem statement, then convert to mg/L. NOTE: units in [ ] are moles per liter!
55ProblemStrategySolution7. For a pH of 11, the solubility is mg/L. For a pH of 12 the solubility is mg/L. Work these solutions on your own...... end of example.
56ProblemStrategySolutionThe chemical 1,4-dichlorobenzene (1,4-DCB) is used in an enclosed area. At 20C (68F) the saturated vapor pressure of 1,4-DCB is 5.3 x 10-4 atm. What would be the concentration in the air of the enclosed area (units of g/m3) at 20C ? The molecular weight of 1,4-DCB is 147 g/mol.
57Problem Strategy Solution Rearrange the ideal gas law to solve for n/V [mol/L] and apply the appropriate conversions.
58ProblemStrategySolutionRearrange the Ideal Gas Law to solve for the concentration of 1,4-DCB in the air
59ProblemStrategySolutionAnaerobic microorganisms metabolize organic matter to carbon dioxide and methane gases. Estimate the volume of gas produced (at atmospheric pressure and 25° C) from the anaerobic decomposition of 2 moles of glucose. The reaction is:
60Problem Strategy Solution Recognize that each mole of glucose produces 3 moles of methane and 3 moles of carbon dioxide gases, for a total of 6 moles. Therefore, 2 moles of glucose produces a total of 12 moles.Use the ideal gas law to solve for V given n=12 moles
61Problem Strategy Solution Each mole of glucose produces 3 moles of methane and 3 moles of carbon dioxide gases, for a total of 6 moles. Therefore, 2 moles of glucose produces a total of 12 moles. The entire volume is thenNote: The volume of 1 mole of any gas is the same. Thus, 1 mole of carbon dioxide gas is the same volume of 1 mole of methane gas.
62Note: STP is 273.13°K and 101.325 kPa (0°C and 1 atm). ExampleSolutionShow that one mole of any ideal gas will occupy L at standard temperature and pressure (STP).Note: STP is °K and kPa (0°C and 1 atm).
63Use the ideal gas law to solve for volume. Note: J = N. m Pa = N/m2 ExampleSolutionUse the ideal gas law to solve for volume.Note: J = N. m Pa = N/m2......end of example
64Similarly, if we consider the volume at 25°C ExampleSolutionSimilarly, if we consider the volume at 25°C......end of example
65Convert 80 mg/m3 of SO2 in 1 m3 of air, 25° C, 103.193 kPa to ppm ExampleSolutionConvert 80 mg/m3 of SO2 in 1 m3 of air, 25° C, kPa to ppm
67ProblemStrategySolutionA 1 m3 volume tank contains a gas mixture of moles of oxygen, moles of nitrogen and 6.15 moles of carbon dioxide. What is the partial pressure of each component in the gas mixture at 25°C and kPa?
68Problem Strategy Solution Convert temperature Use the ideal gas law to determine the pressure of each gasApply Dalton’s Law
70The Henry’s law constant for oxygen at 25° C is 1.29 x 10-3 mol/L-atm. ExampleSolutionCalculate the concentration of dissolved oxygen (units of mol/L and mg/L) in a water equilibrated with the atmosphere at 25° C.The Henry’s law constant for oxygen at 25° C is 1.29 x 10-3 mol/L-atm.Note: The partial pressure of oxygen in the atmosphere is 0.21 atm.
71ExampleSolutionwhich you can convert to 8.7 mg/L
72Problem Strategy Solution A constant volume, batch chemical reactor achieves a reduction of compound A from 120 mg/L to 50 mg/L in 4 hours. Determine the reaction rate for both zero- and first-order kinetics. Clearly indicate the units of k.
73Problem Strategy Solution Using the two boundary conditions A = 120 mg/L at t=0A = 50 mg/L at t=4 hrsDetermine k using:C = Co - kt (zero order)C = Coe-kt (first order)
75First-Order Problem Strategy Solution Note the difference in units. The units associated with rate constants are specific for the reaction order.
76Problem Strategy Solution Consider how the choice of a rate constant effects the design of a treatment facility.For Q = 0.5m3/s and an initial concentration of 150 mg/L, what size of reactor is required to achieve 95% conversion assumingZero-order reactionFirst order reactionUse the values of k from the previous problemZero – order k = 17.5 mg/L·hrFirst order k = hr-1
77Problem Strategy Solution Note that 95% conversion means C = 0.05Co Solve for t in each caseRecognize that Q = [L3/T] = Volume/timeSolve for volume V= Qt for each case
80ProblemStrategySolutionHow long will it take the carbon monoxide (CO) concentration in a room to decrease by 99% after the source of carbon monoxide is removed, and the windows opened? Assume the first-order rate constant for CO removal (due to dilution by the in coming clean air) is 1.2 h-1.
811. First order reaction is C=Coe-kt 2. If 99% is removed, C=0.01Co ProblemStrategySolution1. First order reaction is C=Coe-kt2. If 99% is removed, C=0.01Co
82Problem Strategy Solution This is a first-order reaction, so use [CO]=[COo]e-ktWhen 99% of the CO is removed, [CO] = 0.01[COo]0.01[COo] = [COo]e-ktwhere k = 1.2 h-1Solve for t = 3.8 h
83ProblemStrategySolutionAn engineer is modeling the transport of a chemical contaminant in groundwater. The individual has a mathematical model that only accepts first-order degradation rate constants and a handbook of with a table for “subsurface chemical transformation half-lifes”. Subsurface half-lives for benzene, TCE and toluene are listed as 69, 231, and 12 days respectively. What are the first-order rate constants for all three chemicals?
84Problem Strategy Solution Apply this equation to each individual compound
86Problem Strategy Solution After the Chernobyl nuclear accident, the concentration of 137Cs in milk was proportional to the concentration of 137Cs in the grass that cows consumed. The concentration in the grass was, in turn, proportional to the concentration in the soil. Assume that the only reaction by which 137Cs was lost from the soil was through radioactive decay, and the half-life for this isotope is 30 years. Calculate the concentration of 137Cs in cow’s milk after 5 years if the concentration in milk shortly after the accident was 12,000 Bq/L. (Note: A Bequerel is a measure of radioactivity. One Bequerel equals one radioactive disintegration per second).
87Problem Strategy Solution 1. Determine k 2. Apply the equation C=Coe-kt
89Problem Strategy Solution A biological wastewater treatment process is known to exhibit first-order kinetics with a temperature correction factor equal to For 20ºC, k=6.0 day-1.Determine the required reaction time required to meet 75% conversion in the summer and winter. Assume an average summer and winter temperature of 30ºC and 0ºC respectively.
90Solve for t given C=Coe-kt For 75% conversion, C=0.25Co ProblemStrategySolutionCorrect kSolve for t given C=Coe-ktFor 75% conversion, C=0.25Co