Presentation on theme: "FE Review for Environmental Engineering Problems, problems, problems Presented by L.R. Chevalier, Ph.D., P.E. Department of Civil and Environmental Engineering."— Presentation transcript:
FE Review for Environmental Engineering Problems, problems, problems Presented by L.R. Chevalier, Ph.D., P.E. Department of Civil and Environmental Engineering Southern Illinois University Carbondale
CHEMICAL FOUNDATIONS FE Review for Environmental Engineering
Calculate the molecular weight, equivalent weight, molarity and normality of the following: a. 200 mg/L HCl b. 150 mg/L H 2 SO 4 c. 100 mg/L Ca(HCO 3 ) 2
Use periodic table to get molecular weight Convert mg/L to mol/L Determine n for each compound Apply equations EW = MW/n N = Mn
“in an Acid/Base reaction, n is the # of hydrogen ions that a molecule transfers”
a)Convert 300 ppm Mg 2+ to mg/L as CaCO 3 b)Convert 30 mg/L Mg 2+ as CaCO 3 to mg/L Note: MW Mg 2+ is 24.31 g/mol
Determine the molecular weight of the species Determine n Equate EW=MW/n Apply equation
a)Convert 300 ppm Mg 2+ to mg/L as CaCO 3 300 ppm = 300 mg/L EW Mg 2+ = 24.31/2 = 12.16 g/eq (300)(50/12.16) = 1233.55 mg/L as CaCO 3
b) Convert 30 mg/L Mg 2+ as CaCO 3 to mg/L (30 mg/L as CaCO 3 )(12.16/50) = 7.3 mg/L
Balance the following chemical equations: CaCl 2 + Na 2 CO 3 CaCO 3 + NaCl C 6 H 12 O 6 + O 2 CO 2 + H 2 O NO 2 +H 2 O HNO 3 + NO
CaCl 2 + Na 2 CO 3 CaCO 3 + 2NaCl C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O 3NO 2 +H 2 O 2HNO 3 + NO
What is the pH if [H+] = 10 -3 ? pH = 3 What is the pOH if [OH - ] = 10 -8 ? pOH = 8 What is the pH if [OH - ] = 10 -8 ? pH = 14 - 8 = 6 What is the [H + ] if [OH - ] = 10 -5 ? [H+]=10 5-14 = 10 -9 mol/L
Derive a proof that in a neutral solution, the pH and the pOH are both equal to 7.
Remove nitrogen to prevent the stimulation of algae growth Prevent excessive nitrate [NO 3 - ] level in drinking water from causing a potentially lethal condition in babies known as methemoglobinemia Consider the problem of removing nitrogen from municipal wastewater
When organic matter decomposes, nitrogen is first released in the form of ammonia NH 3 - low solubility in water (ammonia) NH 4 + - highly soluble in water (ammonium ion) One way to remove is a process known as ammonia stripping
By 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 (NH 3 )? Need to decrease [H + ] or increase the pH.
Highly Soluble Low Solubility Want to consider [NH 3 ]/[NH 4 + ] Should we decrease this or increase this?
Highly Soluble Low Solubility Increase it. How can we do this?
Summary Of Example Problem Nitrogen, in the form of ammonia (NH 3 ) is removed chemically from the water by raising the pH This converts ammonium ion (NH 4 + ) into ammonia NH 3 is then stripped from the water by passing large quantities of air through the water
A 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 CaCO 3.
1.Determine the MW of HCO 3 - and CO 3 -2 2.Determine the EW of HCO 3 - and CO 3 -2 3.Convert the concentrations of HCO 3 -, CO 3 -2, H + and OH - to mg/L as CaCO 3 4.Add the concentrations in mg/L as CaCO 3 of HCO 3 -, CO 3 -2, and OH -, and subtract H +
...... end of problem I will leave it up to you to check calculations for H + and OH -
The solubility product for the dissociation of Mg(OH) 2 is 9 x 10 -12. Determine the concentration of Mg 2+ and OH - at equilibrium.
1.Write the equation for the reaction 2.Write the solubility product equation 3.Recognize from Eqn. 1 the relationship between the number of moles of Mg 2+ and the number of moles of OH - resulting from the dissociation of Mg(OH) 2, and how this relates to Eqn 2
1. Write the equation for the reaction. 2. The solubility product equation is:
3. If x is the amount of Mg 2+ resulting from the dissociation is given as x, then the amount of OH - is equal to 2x......end of example
Magnesium is removed from an industrial waste stream by hydroxide precipitation at a pH = 10. Determine the solubility of Mg 2+ in pure water at 25° C and pK sp of 10.74.
1. Identify the two governing equations (K sp and K w ) 2. Recognize that [OH - ] = 10 -14+pH 3. Substitute to derive an equation [Mg 2+ ] = f(pH)
1. What are your two governing equations? 2. Two unknowns, and two equations.
4. Solve for [OH - ] 2 3. Given the pH, we know [H + ].
5. Substitute into 1st governing equation, and solve for [Mg 2+ ].
6. Substitute value of pH given in the problem statement, then convert to mg/L. NOTE: units in [ ] are moles per liter!
7. For a pH of 11, the solubility is 0.442 mg/L. For a pH of 12 the solubility is 0.004 mg/L. Work these solutions on your own...... end of example.
The chemical 1,4-dichlorobenzene (1,4-DCB) is used in an enclosed area. At 20 C (68 F) 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/m 3 ) at 20 C ? The molecular weight of 1,4-DCB is 147 g/mol.
Rearrange the ideal gas law to solve for n/V [mol/L] and apply the appropriate conversions.
Rearrange the Ideal Gas Law to solve for the concentration of 1,4-DCB in the air
Anaerobic 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:
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
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 then Note: 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.
Show that one mole of any ideal gas will occupy 22.414 L at standard temperature and pressure (STP). Note: STP is 273.13°K and 101.325 kPa (0°C and 1 atm).
......end of example Use the ideal gas law to solve for volume. Note: J = N. m Pa = N/m 2
......end of example Similarly, if we consider the volume at 25°C
Convert 80 mg/m 3 of SO 2 in 1 m 3 of air, 25° C, 103.193 kPa to ppm
A 1 m 3 volume tank contains a gas mixture of 18.32 moles of oxygen, 16.40 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 101.3 kPa?
Convert temperature Use the ideal gas law to determine the pressure of each gas Apply Dalton’s Law
Calculate 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.
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.
Using the two boundary conditions A = 120 mg/L at t=0 A = 50 mg/L at t=4 hrs Determine k using: C = C o - kt (zero order) C = C o e -kt (first order)
First-Order Note the difference in units. The units associated with rate constants are specific for the reaction order.
Consider how the choice of a rate constant effects the design of a treatment facility. For Q = 0.5m 3 /s and an initial concentration of 150 mg/L, what size of reactor is required to achieve 95% conversion assuming a)Zero-order reaction b)First order reaction Use the values of k from the previous problem Zero – order k = 17.5 mg/L·hr First order k = 0.219 hr -1
1.Note that 95% conversion means C = 0.05C o 2.Solve for t in each case 3.Recognize that Q = [L 3 /T] = Volume/time 4.Solve for volume V= Qt for each case
1. 95% conversion means C= 0.05C o 2. Zero-order
How 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.
1. First order reaction is C=C o e -kt 2. If 99% is removed, C=0.01C o
This is a first-order reaction, so use [CO]=[CO o ]e -kt When 99% of the CO is removed, [CO] = 0.01[CO o ] 0.01[CO o ] = [CO o ]e -kt where k = 1.2 h -1 Solve for t = 3.8 h
An 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?
Apply this equation to each individual compound
After the Chernobyl nuclear accident, the concentration of 137 Cs in milk was proportional to the concentration of 137 Cs 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 137 Cs was lost from the soil was through radioactive decay, and the half-life for this isotope is 30 years. Calculate the concentration of 137 Cs 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).
1. Determine k 2. Apply the equation C=C o e -kt
A biological wastewater treatment process is known to exhibit first-order kinetics with a temperature correction factor equal to 1.023. 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.
Correct k Solve for t given C=C o e -kt For 75% conversion, C=0.25C o