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Environmental Geochemistry Grand Prismatic Spring, Yellowstone

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1 Environmental Geochemistry Grand Prismatic Spring, Yellowstone
89.315  Grand Prismatic Spring, Yellowstone

2 Atomic Structure Review:
What is the atomic weight of this atom? What is the atomic number of this atom? What is the ionic charge of this atom? If we remove an electron from this atom, what then would be it’s charge? What is the name of this atom?

3 Potential Energy = Kinetic Energy or
Centrifugal force keeps the negatively-charged electrons from spiraling into the positively charged nucleus by electrostatic forces? With a stable atom… Potential Energy = Kinetic Energy or Electrostatic Force = Centrifugal Force Electrons exist in stable orbits at certain discrete distances from the nucleus. The allowable distances were determined by restricting the angular momentum of the electrons to multiples of: h/2π h= x10-34Js (Plank’s Constant)

4 E = total energy of an atom (PE + KE) k = proportionality constant
The energy (E) of an atom is the sum of the kinetic and potential energies. The kinetic component = the revolution of the electrons around the nucleus. The potential component = the electrostatic attraction between positively charged protons and negatively charged electrons. - 2π2mk2e4 E = n2h2 E = total energy of an atom (PE + KE) k = proportionality constant e = charge of the electron (*If you want to follow the mathematical path to this point, see page 2 of the text book.)

5 What happens if an electron moves from one orbit to another
What happens if an electron moves from one orbit to another? How does this happen? If an electron moves from a lower orbit to a higher orbit, it must gain energy to overcome the electrostatic forces. If an electron moves from a higher orbit to a lower orbit, it releases energy in the form of a photon. (Electromagnetic radiation that behaves as a particle.) E = hc/λ Where c = speed of EM radiation; λ = wavelength of photon

6 m = mass of particle; v = velocity of particle
deBroglie discovered that particles can have wave-like properties. λ = h/mv m = mass of particle; v = velocity of particle

7 Spectra and Elemental Analysis
For neutral atoms, transitions between orbitals release different amounts of energy, and the resulting emission spectra are different and characteristic for each element. If the atoms have been ionized, the sequence of emission lines will be slightly shifted because the electrostatic attraction between the nucleus and electrons will have changed.

8 Emission spectrum: the spectrum produced when electrons move from higher orbitals to lower orbitals. This gives rise to light-lines of specific wavelength appearing, and the other wavelengths not characteristic of the specific atoms, remaining dark. Absorption spectrum: the spectrum produced when white light travels through a cold, dilute gas, and atoms in the gas absorb at characteristic frequencies. This gives rise to dark lines (absence of light) in the otherwise continuous spectrum.

9 Review of Emission and Absorption Spectra



12 Going back to the electron…
Let’s consider electrons as apartment dwellers. Electrons prefer the smallest apartment, closest to the ground floor. Electrons are antisocial (likes repel), and prefer to live one to a room until each room in an apartment has one occupant. Each room in an apartment can hold no more than 2 occupants. The apartment building has only 7 habitable floors. The floors of the apartment are called shells. Each shell (or floor) has one or more apartments called subshells.

13 s → single room; maximum occupancy = 2
p → three rooms; maximum occupancy = 6 d → five rooms; maximum occupancy = 10 f → seven rooms; maximum occupancy = 14 Each “room” in a subshell is called an orbital. Remember each orbital can hold no more than 2 electrons.

14 To summarize in more scientific terms:
An electron will enter the available orbital with the lowest energy. The overall energy of the atom is minimized. For each set of orbitals (s, p, d, f) the electrons will first be added singly to each available orbital. After all the orbitals in a set have a single electron, subsequent electrons can enter these orbitals if they have the opposite spin. Atoms attain their maximum stability when the available orbitals are either completely filled, half-filled or empty.

15 *Remember: stable atoms may have full, half full or empty orbitals
Let’s do a couple of examples… A neutral Na atom has 11 electrons. What is its configuration? 1s2 2s2 2p6 3s1 A neutral Cl atom has 17 electrons. What is its configuration? 1s2 2s2 2p6 3s2 3p5 *Remember: stable atoms may have full, half full or empty orbitals

16 Ionization and Valences
Valence: the combining capacity of atoms. Valence electrons: those electrons occupying the outer-most shell. Valence charge: the net charge of an atom or oxidation number. Ions: atoms that have gained or lost electrons. Anions: an atom that has gained an electron(s) giving the atom a net negative charge. Cations: an atom that has lost an electron(s) giving the atom a net positive charge.

17 Ionization potential: the energy required to remove an electron from an atom and place it at an infinite distance. In any given row of the periodic table, as we move from left to right, the ionization potential tends to increase… It becomes more difficult to remove electrons from the atom. Hence, elements on the left-hand side of the periodic table tend to form cations and those on the right-hand side (excluding the noble gases for the moment) tend to form anions. Now, lets look closely at the first 3 rows of the periodic table


19 Chemical Bonding Two or more atoms may combine to form a compound. The compound is held together by chemical bonds. There are four basic types of chemical bonding: Ionic Bonding, Covalent Bonding, Metallic Bonding Hydrogen Bonding

20 Ionic Bonding: occurs when cations and anions combine by electrostatic attraction.

21 Covalent Bonding: occurs when two or more atoms combine by sharing valence electrons.

22 Metallic Bonding: occurs in the case of pure metals in which electrons are freely shared among all of the atoms. These compounds are good conductors of electricity. Metals generally form cations (ions with a positive charge) and are capable of forming more than one oxidation number. For example: Iron (Fe) may have a 2+ charge (ferrous) or a 3+ charge (ferric).

23 Water is a covalently-bonded molecule
Water is a covalently-bonded molecule. Although the electrons are shared, they are not shared evenly. They tend to spend more time around the oxygen molecule. This results in a polarized molecule that is slightly positive on the H side and slightly negative on the oxygen side.

24 Hydrogen Bonding: a specific type of intermolecular bonding where at least one of the atoms is H, and the other atom(s) is something other than H. The hydrogen side of the molecule invariably has a slight positive charge bias and the other atom(s) side has a slight negative charge bias.

25 How does one determine the atomic weight of an element
How does one determine the atomic weight of an element? See the following example. Carbon as two stable isotopes and may have either 6 or 7 neutrons. To calculate the atomic weight of carbon you need to know the mass and the occurrence of the isotopes. Element Isotope Mass of Isotope Proportion in element Mass X Proportion Sum 12C + 13C Carbon 12C 12.000amu 0.989 11.868 12.011amu 13C 13.003amu 0.011 0.143

26 Mole: the number of carbon atoms in exactly 12 grams of pure 12C.
Avogadro’s number: the number of atoms in a mole (6.022X1023.) Gram-atomic weight: the atomic weight of a mole of an element in grams. Gram-molecular weight: the weight of a mole of a compound in grams. Gram-equivalent weight of an ion: the molecular or atomic weight divided by the valence. In the case of an acid or base, it is the number of H+ or OH- ions that can be produced when the acid or base is dissolved in water. (We will cover this more closely in Chapter 3.)

27 Measurements of Concentration
Absolute Mass or weight per weight would be either in SI units or units such as ppt respectively. Concentrations of Solutions Solute: the substance that is being dissolved. Solvent: the material in which the solute is dissolved. Molarity (M): the number of moles of solute per volume of solution in liters. Molality (m): the number of moles of solute per kilogram of solvent. Normality( N): the number of equivalents (often gram-equivalents) per liter of solution. Mole fraction: the ratio of the number of moles of a given component to the total number of moles of solution.

28 Example of weight percent calculation, Page 11.
Plagioclase feldspars form a solid solution series with end members albite (NaAlSi3O8) and anorthite (CaAl2Si2O8).

29 Example 1-5 (page 11) A particular plagioclase contains 5wt% Ca. Calculate the mole fraction of anorthite in the plagioclase. The first thing we must do is weight % of anorthite. Wt% CaAl2Si2O8 = [(5)(40 + (2)(27) + (2)(28) + (8)(16))]/40 = 34.8 We then subtract the weight percent of anorthite from 100% to give the weight percent of albite. Wt% NaAlSi3O8 = 100 – 34.8 = 65.2

30 Next, determine the relative number of moles of anorthite by diving the weight percentage of anorthite by the molecular weight of the anorthite molecule. Relative number of moles (An) = 34.8 / 278 = 0.13 Relative number of moles (Ab) = 65.2 / 262 = 0.25 Moles (An) Mole fraction (An) = Moles (An) + Moles (Ab) = 0.13 / ( ) = 0.34

31 A particular plagioclase contains 5wt% Ca
A particular plagioclase contains 5wt% Ca. Calculate the weight percent of Anorthite and Albite, and name the plagioclase. Next, calculate the mole fraction of anorthite in the plagioclase. The molecular weights of Ca, Al, Si and O are, 40, and 16, respectively. 5((40+(2 X 27)+(2 X 28)+(8 X 16)) wt% CaAl2Si2O8 = 40 = 34.8 wt% Anorthite wt% NaAlSi3O8 = 100 – 34.8 = 65.2 wt% Albite Name that mineral. Andesine

32 Chemical Reactions

33 Species: substances in a chemical reaction that can be either, ions, molecules, solids, liquids, gases, etc. Reactant: the substances that are the starting materials in a chemical reaction and are found on the left side of the chemical equation. Product: the substances produced as a result of a chemical reaction and are found on the right side of the chemical equation. CH4 + 2O2 → CO2 + 2H2O Reactants Products

34 Ag+ + NO3- + Na+ + Cl- → AgCl(s) + Na+ + NO3-
There are 3 main types of chemical reactions Precipitation Acid-Base Oxidation-Reduction Precipitation Reaction: occurs when 2 solutions are mixed and a solid, called a precipitate, forms. Ag+ + NO3- + Na+ + Cl- → AgCl(s) + Na+ + NO3- The above is a complete ionic equation. It includes the spectator ions of NO3-, and Na+. The net ionic equation is as follows. Ag+ + Cl- → AgCl(s)

35 H+ + Cl- + K+ + OH- → H2O(aq) + K+ + Cl-
Acid-Base Reactions: involve the transfer of protons. Acids are proton donors. Bases are proton acceptors. An example of a complete ionic equation of an acid-base reaction is as follows. H+ + Cl- + K+ + OH- → H2O(aq) + K+ + Cl- The net ionic equation is written: H+ + OH- → H2O(aq)

36 Oxidation-Reduction Reactions: occur when there is a transfer of electrons.
For example: CH4 + 2O2 → CO2 + 2H2O In this example, the oxidation state of the carbon changes from –4 to +4; 8 electrons are transferred. The oxygen reactant was neutral, but the oxygen in the products carries a –2 charge. There are 4 oxygen molecules. (4 X –2 = –8) The electrons from the carbon atom were transferred to the four oxygen atoms. (Note: the oxidation state of the H atoms remain unchanged.)

37 Balancing a Chemical Equation
H2 + O2 → H2O 2 2

38 When given a chemical equation, you must be certain that the number of atoms of each element on the product side of the equation is equal to that of the reactant side of the equation. 2 Al O2 → A2O3 3/2 Multiply both sides by 2 4 Al + O2 → A2O3 3 2

39 Balancing equations - a summary
You may only put numbers in front of molecules, never altering the formula itself. (i.e.H2O is not the same asH2O2.) Don't worry if the numbers turn out to be fractions - you can always double or triple all the numbers at a later stage. Balance complicated molecules with lots of different atoms first. Putting numbers in front of these may mess up other molecules, so use the simpler molecules to adjust these major changes. If you recognize the atoms making up a standard group such as sulphate, nitrate, phosphate, ammonium etc. that survive unscathed throughout the chemical reaction, treat them as an indivisible item to be balanced as a whole. This makes life easier and helps understanding of the chemistry. Leave molecules representing elements until last. This means that any numbers you put in front of those molecules won't unbalance any other molecule.

40 Determining a chemical equation from examination of precipitate
A 100 g precipitate forms from a reaction between AgNO3 and NaCl. The precipitate contains 75.3% silver and 24.7% chloride, by weight. First determine the number of moles of each element in the product, using the following formula. Moles = 100g X %composition/molecular weight of element Moles(Ag) = 100 g X 0.753/107.9 g•mol-1 = mol Moles(Cl) = 100 g X 0.247/35.45 g•mol-1 = mol From this information we can determine that the equation reads: Ag+ + Cl- → AgCl(s) Why?

41 Ca3(PO4)2 + 2H+ + SO2-4 → CaSO4 + H3PO4
How can you determine balanced chemical equations for reactions? Example 1-8 In the production of phosphate fertilizer, calcium phosphate rock is reacted with sulfuric acid to produce calcium sulfate and phosphoric acid. Ca3(PO4)2 + H2SO4 → CaSO4 + H3PO4 Because Sulfuric acid is a strong acid and phosphoric acid is a weak acid the equation is more accurate written thusly: Ca3(PO4)2 + 2H+ + SO2-4 → CaSO4 + H3PO4 More on this subject will be discussed in Chapter 3

42 Ca3(PO4)2 + 2H+ + SO2-4 → CaSO4 + H3PO4

43 Photosynthesis equation
A moment for a deep thought……. Photosynthesis equation This implies CO2 fertilization of plants when exposed to elevated atmospheric CO2. What other nutrients does a plant require and from where would these nutrients be found? Let’s look at nitrogen and phosphorus…



46 Ok…. Let’s get back to chapter 1

47 Empirical Formula: the formula that represents the simplest whole-number ratio of the atoms that make up a compound. Molecular formula: the actual number of each kind of atom in the compound. See page 14, example1-7. The empirical formula of a compound is CH3 and its molecular weight is 30g. Calculate the molecular formula. The formula weight is 15g. (1C X 12g mol-1 + 3H X 1g mol-1). The molecular weight is twice that of the formula weight. Thus, the molecular formula of the compound is C2H6 (Ethane—more about organic molecules will be seen in Chapter 5).

48 Gases Ideal Gases consist of molecules that move completely independently of one another and occupy a volume much greater than the volume of the molecule. The Ideal Gas Law combines Boyle’s (P1V1 = P2V2), Charles’s (V = nT), and Avogadro’s (V = an) laws. PV = nRT Where P = pressure, V = volume, n = number of moles, R = universal (or ideal) gas constant, and T = temperature (K). R combines the standard temperature (273 K), pressure (1 atm) and the fact that 1 mol of gas occupies a volume of 22.4 liters at STP

49 Ideal gas laws are handy and seem to reflect common sense, however, most gases (especially at high pressure and/or low temperature) deviate from ideal behavior. This occurs for 2 related reasons. Gas molecules do have a finite volume. When P is increased, the number of molecules/volume of gas will increase. Therefore, gas molecules comprise a significant portion of the volume. Gas molecules do interact with each other, and as the number of molecules/VGas increases, the number of interactions increases.

50 Johannes van der Waals introduced two numbers to correct for non-ideal gases.
V-nb Where n is the number of moles of gas and b is an empirical constant that depends on the gas. V, of course, is the volume of gas. This number corrects for the volume of the gas molecules. a(n/V)2 Where a is an empirical constant, n is the number of moles and V is the volume of gas. This number corrects for the interactions between the gas molecules.

51 The ideal gas equation with the corrected numbers :
Pobs = a(n/V)2 Table 1-6 (pg.18) lists van der Waals constants for some common gases. Example 1-11 A cylinder of compressed nitrogen has a volume of 100L and contains 500 mol of N2. At a temperature of 25°C, calculate the pressure exerted by the gas on the cylinder. nRT V -nb (500mol)( L·atm·K-1·mol-1)(298.15K) _ Pobs = (100L) – (500mol)(0.0387L·mol-1) (1.37atm·L2·mol-2)(500mol/100L)2 = 117.4atm = psi

52 Water Water is arguably the most important substance on Earth. Without water, life, as we know it, would not exist. Water is the only substance that occurs naturally, in all three states (solid, liquid and gas.)

53 The shape of the water molecule by VSEPR theory…
Oxygen contains 8 electrons (1s2, 2s2, 2p4) Hydrogen contains 1 electron (1s1) Lone electron pairs Shared electron pairs

54 Water has relatively high transparency for visible light

55 Water Strider (Gerris remigis)
Water has the highest surface tension of all common liquids Water Strider (Gerris remigis)


57 Heat Capacity (specific heat): the amount of energy required to raise the temperature of 1 g of a substance by 1°C. For water, the heat capacity is 1 cal·g-1·°C-1. Latent heat of fusion: The heat absorbed as a substance changes phase from solid to liquid. For water, the latent heat of fusion is 80cal·g-1. Latent heat of vaporization: The heat absorbed when a substance changes phase from liquid to gas. For water, the latent heat of vaporization is 540cal·g-1.

58 Why does ice float? Shouldn’t a solid sink?
Liquid water Water vapor Water is most dense at 4°C. Looking at the above models, can you guess why?

59 Box Models and Geochemical Cycles
A box model consists of several boxes showing the reservoirs for a particular substance and the rate at which material is transferred between the reservoirs. In a steady-state system: the total amount of a substance in each reservoir remains constant; thus the rate of addition of a material into and removal of a material from a reservoir must be the same. Residence time: is the average length of time a particular substance will reside in a reservoir. Residence time = Amount of material in reservoir Rate of addition (or removal)

60 This is a simplified box model of the hydrologic cycle
This is a simplified box model of the hydrologic cycle. Using the data from this figure, you can calculate the residence time for water in the various reservoirs.

61 A box model may not be pretty, but it is much more useful than this colorful diagram.

62 Example 1-12 Calculate the residence time of water in the atmospheric reservoir.
Remember: Residence time = Amount of material in reservoir 0.13 X 1017kg / 4.49 X 1017kg·yr-1 = year or 10.6 days Rate of addition (or removal)

63 dAi/dt = Finput – Foutput = Finput – kAi
What happens when the steady state is perturbed? First-order kinetics model: dAi/dt = Finput – Foutput = Finput – kAi Where dAi/dt is the rate of change of the amount of substance A in reservoir i. k is a rate constant and can be solved for by k = Finput / Ai ( where Ai is the amount of substance A in reservoir I at time zero.) When the system is at a steady state, Finput = Foutput These formulas become important when dealing with negative and positive feedback mechanisms.

64 Positive feedback is a feedback system in which the system responds to the perturbation in the same direction as the perturbation (It is sometimes referred to as cumulative causation). Negative feedback is a feedback system in which the system responds to the perturbation in the opposite direction as the perturbation. Lets consider the global warming and how it is influenced by positive and negative feedback.


66 As listed on the syllabus, the following problems from Chapter 1 are due this Monday September 17th. There are no late assignments.  From Chapter 1: problems 2, 4, 6, 10, 11, 12, 20, 29, 32, 34, 36, 37, 44, 45, 46, 49 If there are multiple sections to a problem, you are expected to complete all of the sections to the problem. Please let me know if you need help sooner rather than after the problems are due.

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