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OME General Chemistry Lecture 6: Chemical Reactions Dr. Hartwig Pohl Office: Beyer-Bau 122e Email: [email protected] Phone: +49 351 463 42576 1 Empirical & Molecular Formulas For any new chemical, chemists determine its elemental composition, the empirical formula, and its structure to understand its properties Empirical formula, the simplest positive integer ratio of atoms present in a compound, can be determined from the mass percentage Molecular formula, gives the actual number of atoms of each kind present per molecule Law of definite proportions, a chemical compound always contains the same proportion of elements by mass Mass percentage, divide mass of each atom in the molecule by the total molecular mass Combustion analysis, common way to determine the elemental composition of an unknown hydrocarbon 2 The Mole & Molar Masses Molecular Mass (amu), sum of the average masses of the atoms in one molecule of the substance, calculated by summing the atomic masses of the elements in the substance Formula Mass (amu), used for ionic compounds because they do not have an identifiable molecular unit, sum of the atomic masses of all the elements in the empirical formula The Mole (mol), Avogadro’s number, the number of constituent particles, atoms or molecules, of a given substance per mol (6.022e23 mol–1) Molar Mass (g mol–1), the mass of 6.022e23 atoms, molecules, or formula units of that substance 3 Chemical Equations A chemical equation is an expression that gives the identities and quantities of the substances in a chemical reaction Chemical formulas and other symbols are used to indicate the starting material(s) or reactant(s), which are written on the left side of the equation, and the final compound(s) or product(s), which are written on the right side. An arrow, read as yields or reacts to form, points from the reactants to the products. A balanced chemical equation is when both the numbers of each type of atom and the total charge are the same on both sides. A chemical reaction represents a change in the distribution of atoms but not in the number of atoms Stoichiometry is a collective term for the quantitative relationships between the masses, numbers of moles, and numbers of particles (atoms, molecules, and ions) of the reactants and products in a balanced reaction. 4 Mass Relationships in Chemical Eqns A balanced chemical equation gives the identity of the reactants and products and the accurate number of molecules or moles of each that are consumed or produced. A stoichiometric quantity is the amount of product or reactant specified by the coefficients in a balanced chemical equation. Converting amounts of substances to moles, and vice versa, is the key to all stoichiometry problems. 5 Limiting Reactants & Percent Yield A limiting reactant is the reactant that restricts the amount of product obtained. The reactant that remains after a reaction has gone to completion is present in excess • Determines the maximum amount of product that can be formed from the reactants Theoretical yield, the amount you would obtain if the reaction occurred perfectly and your method of purifying the product was 100% efficient Actual yield, the measured mass of products obtained from a reaction, is less than the theoretical yield Percent yield, the ratio of the actual yield to the theoretical yield 6 General Reaction Types Synthesis (Combination), two or more simple substances combine to form a more complex substance. A + B → AB where A and B are elements or compounds, and AB is a compound consisting of A and B. Examples: 2Na + Cl2 → 2NaCl formation of table salt S + O2 → SO2 formation of sulfur dioxide 4Fe + 3O2 → 2Fe2O3 iron rusting CO2 + H2O → H2CO3 carbon dioxide reacting with water to form carbonic acid 7 General Reaction Types Decomposition, opposite of a synthesis reaction, the separation of a chemical compound into elements or simpler compounds. AB → A + B where A and B are elements or compounds, and AB is a compound consisting of A and B. Examples: 2H2O → 2H2 + O2 electrolysis of water H2CO3 → H2O + CO2 carbonic acid forming CO2 in sodas MCO3 → MO + CO2 metal carbonates + heat gives metal oxides and CO2 2MClO3 → 2MCl + 3O2 metal chlorates + heat gives metal chlorides and oxygen 8 General Reaction Types Single Displacement, one element is replaced by another in a compound. A + BC → AC + B where A, B, and C are elements or compounds; A and B must be different metals, hydrogen, or halogens; C is either an anion or cation and is called a spectator ion. Examples: Mg + 2HCl → MgCl2 + H2 Cl2 + 2NaBr → NaCl + Br2 Reaction proceeds if A is more reactive than B, if A is more likely to donate its electrons than B, see reactivity series chart. For common metals: slide 11. For halogens: F > Cl > Br > I. Zn + 2AgNO3 → Zn(NO3)2 + 2Ag Ag + Cu(NO3)2 → no reaction 9 General Reaction Types Double Displacement, a chemical process involving the exchange of bonds between two reacting chemical species, which results in the creation of products with similar or identical bonding affiliations. AB + CD → AD + CB Precipitation, usually when AB and CD are aq. ionic compounds (or acids), the dissolved cations and anions switch partners resulting in the formation of two new ionic compounds AD and CB, one of which is in the solid state Pb(NO3)2 (aq) + 2KCl (aq) → 2KNO3 (aq) + PbCl2 (s) Neutralization, when AB is an acid and BC is a base, cations and anions switch partners, resulting in the formation of water and a new ionic compound (salt) H2SO4 (aq) + 2LiOH (aq) → Li2SO4 (aq) + 2H2O 10 Solubility Rules & Reactivity Series Solubility Rules Reactivity Series 1. Alkali metal compounds, acetates, nitrates, and ammonium compounds are all soluble. K Na Ca Mg Al C Zn Fe Sn Pb H Cu Ag Au Pt 2. Hydroxides of alkali metals and NH4+1, Ca+2, Sr+2, and Ba+2 are soluble. 3. All halides are soluble except for those containing Ag+1, Pb+2, and Hg2+2. 4. Most sulfates are soluble, except for BaSO4, SrSO4, Ag2SO4, PbSO4, and CaSO4. 5. Most phosphates, carbonates, chromates and sulfides are insoluble. 6. All acids are soluble! Potassium Sodium Calcium Magnesium Aluminum Carbon Zinc Iron Tin Lead Hydrogen Copper Silver Gold Platinum 11 Special Cases Acid-Base, can have different definitions depending on the concept employed, generally involves a transfer of protons (H+) from one species (the acid) to another (the base). Oxidation-Reduction, reactions in which a transfer of electrons occurs from one species to another resulting in a change in oxidation state. Substitution Reaction, a single replacement reaction in which one functional group in an organic compound is replaced by another (will be addressed later!). Complexation, a reaction in which several ligands react with a metal atom to form a coordination complex. Catalysis, a reaction that does not proceed directly but through a third substance that is not consumed during the reaction. 12 Definition of Acids & Bases Arrhenius definition: An acid is a substance that dissociates in water to produce H+ ions (protons), and a base is a substance that dissociates in water to produce OH– ions (hydroxide); an acid-base reaction involves the reaction of a proton with the hydroxide ion to form water Brønsted–Lowry definition: An acid is any substance that can donate a proton, and a base is any substance that can accept a proton; acid-base reactions involve two conjugate acid-base pairs and the transfer of a proton from one substance (the acid) to another (the base) Lewis definition: A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor 13 Water: Acid & Base Water is amphiprotic: it can act as an acid by donating a proton to a base to form the hydroxide ion (OH–), or as a base by accepting a proton from an acid to form the hydronium ion (H3O+), one water molecule can react with another: 2H2O(l) ⇋ H3O+ (aq) + OH– (aq) equilibrium constant K = [H3O+] [OH–] [H2O]2 By treating [H2O] as a constant, a new equilibrium constant, the ion-product constant of liquid water (Kw), can be defined: Kw = K[H2O]2 = [H3O+] [OH–] = 1.00e–14 (at 25 ºC) The acidic, basic, or neutral condition of an aqueous solution depends on the relative concentrations of H3O+ and OH– ions. Aqueous solutions are neutral when [H+] = [OH–] , acidic when [H+] > [OH–] , and basic when [H+] < [OH–]. 14 The pH Scale The pH scale provides a convenient way of expressing the hydrogen ion (H+) concentration of a solution and describes acidity or basicity quantitatively. pH = – log [H+] (pOH = – log [OH–], pH = 14 – pOH) Pure water is neutral at pH 7 on a scale from 0 to 14. pH decreases with increasing [H+] adding an acid to pure water increases the [H+] and decreases the [OH–], pH < 7 = acidic pH increases with decreasing [H+] adding a base to pure water increases the [OH–] and decreases the [H+], pH > 7 = basic Can be measured with pH paper (paper dyed with colored molecules that change w/certain pH) or, more accurately, using a pH meter (glass electrode whose voltage depends on the H+ conc.) 15 Acids & Bases in Aqueous Solutions Strong acids/bases react essentially completely with water to give H+ and the corresponding anion (acids) or OH– and the corresponding cation (bases) acids donate a proton to a water molecule that acts as a proton acceptor or base. bases accept a proton from water, so small amounts of OH– are produced (water acts like an acid) Neutralization Reaction, a reaction in which a strong acid and a strong base react in stoichiometric amounts to produce water and a salt Weak acids/bases are only slightly dissociated in water to give ions that are at equilibrium with their molecular form, the dominant species in the case of weak acids/bases is the molecular form 16 Strength of Acids & Bases Strong acids undergo complete dissociation Weak acids undergo partial dissociation • Depends on strength of acid/base • Described by equilibrium constant 17 Acid-Base Equilibrium Constants The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases Ionization of a weak acid in water: HA(aq) + H2O(l) ⇋ H3O+(aq) + A–(aq) Equilibrium Constant: K = [H3O+] [A–] [H2O] [HA] The concentration of water is constant for all reactions in aqueous solution, so [H2O] can be incorporated into a new quantity, the acid ionization constant (Ka): Ka = K[H2O] = [H3O+] [A–] [HA] larger Ka = stronger acid and higher [H+] at equilibrium 18 Acid-Base Equilibrium Constants A very similar situation can be considered for bases… Ionization of a weak base in water: B(aq) + H2O(l) ⇋ BH+(aq) + OH–(aq) Equilibrium Constant: K = [BH+] [OH–] [H2O] [B] Base Ionization Constant: Kb = K[H2O] = [BH+] [OH–] [B] larger Kb = stronger base and higher [OH–] at equilibrium 19 Acid-Base Equilibrium Constants Equilibrium constants often expressed as… pKa = – log Ka and pKb = – log Kb Important Trends: 1. Smaller pKa larger acid ionization constants stronger acids • moderate/weak acids pKa ≈ 0 to 14, strong acids pKa < 0 2. Smaller pKb larger base ionization constants stronger bases • moderate/weak bases pKb ≈ 0 to 14, strong bases pKb < 0 3. Related by: pKa + pKb = 14 20 Conjugate Acid-Base Pairs Two species that differ by only a proton constitute a conjugate acid-base pair Example: In the reaction of HCl with water, HCl, the parent acid, donates a proton to a water molecule, the parent base, forming Cl–; HCl and Cl– constitute a conjugate acid-base pair In the reverse reaction, the Cl– ion in solution acts as a base to accept a proton from H3O+, forming H2O and HCl; H3O+ and H2O constitute a second conjugate acid-base pair HCl parent acid CH3COOH parent acid + + H2O ⇋ H3O+ + Cl– parent base conjugate acid conjugate base H2O ⇋ H3O+ + parent base conjugate acid CH3COO– conjugate base Any acid-base reaction must contain two conjugate acid-base pairs! 21 Strength of Conjugate Pairs There is an inverse relationship between the strength of the parent acid and the strength of the conjugate base the conjugate base of a strong acid is a weak base, and the conjugate base of a weak acid is a strong base • However, a weak acid/base will not necessarily have a strong conjugate base/acid One can use the relative strengths of acids and bases to predict the direction of an acid-base reaction by following a simple rule: An acid-base equilibrium always favors the side with the weaker acid/base Important relationship or conjugate pairs: Kw = KaKb 22 Structure & Acid/Base Strength The acid-base strength of a molecule depends strongly on its structure! The stronger the A–H or B–H+ bond, the less likely the bond is to break to form H+ ions, and thus the less acidic the substance. • The larger the atom to which H is bonded, the weaker the bond acid strengths of binary hydrides increase as we go down a column of the periodic table The conjugate base (A– or B) contains one more lone pair of electrons than the parent acid (AH or BH+)… • Any factor that stabilizes the lone pair on the conjugate base favors dissociation of H+ and makes the parent acid a stronger acid acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table 23 Polyprotic Acids & Bases Polyprotic acids contain more than one ionizable proton, and the protons are lost in a stepwise manner. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion; the fully deprotonated species is the strongest base The strengths of the conjugate acids and bases are related by pKa + pKb = pKw, and equilibrium favors formation of the weaker acid-base pair. Example: Sulfuric Acid, H2SO4 H2SO4(aq) + H2O(l) ⇋ H3O+(aq) + HSO4–(aq) HSO4–(aq) + H2O(l) ⇋ H3O+(aq) + SO42–(aq) pKa(H2SO4) < pKa(HSO4–) pKb(SO42–) < pKb(HSO4–) 24 Example: Acid/Base Strength What is the pH of the resulting solution if you add one drop (0.05 mL) of 2 M HCl (pKa = –7) to 100 mL of water? 1. Strong or weak acid? pKa = –7 strong acid! • Strong acid all HCl is dissociated into H3O+ 2. Determine amount amount of HCl added = [H3O+] 3. Calculate pH. 0.05 mL * 1 L * 2 mol * 1 * 1000 mL = 10–3 M = [H3O+] 1000 mL L 100 mL 1L pH = –log[H3O+] = 3 25 Example: Acid/Base Strength What is the pH of a 0.125 M HClO solution? Ka = 3.5e–8 Strong or weak acid? pKa = 7.46 weak acid! 1) Write the balanced equation and the expression for Ka 2) Solve for the hydronium ion concentration HClO(aq) + H2O(aq) ⇋ ClO–(aq) + H3O+(aq) [HClO] [ClO–] [H3O+ ] i 0.125 0 0 Δ –x +x +x Ka = [ClO–][H3O+] [HClO] Ka = x2 _ 0.125 f 0.125–x +x +x = Ka* 0.125 M = 0.438e–8 M weak acid x = [H3O+] = 6.6e–5 M x2 pH = –log[H3O+] = 4.18 26 Example: Conjugate Acid-Base Pairs The Ka of formic acid (HCOOH) is 1.8e–4. What is the pH of 0.35 M solution of sodium formate (NaHCOO)? 1. Is NaHCOO a strong or weak base? NaHCOO = conjugate base of HCOOH, Kb(NaHCOO) = Kw/Ka(HCOOH) = 5.6e–11 pKb (NaHCOO) = 10.3 weak base 2. Write the balanced equation and the expression for Kb NaHCOO(aq) + H2O(aq) ⇋ HCOOH(aq) + NaOH(aq) [NaHCOO] [HCOOH] [NaOH] Kb = [HCOOH][NaOH] = Kw = 1.0e–14 i 0.35 0 0 [NaHCOO] Ka 1.8e–4 Δ –x +x +x x2 / 0.35 = 5.6e–11 f 0.35–x +x +x – x = 4.4e–6 M = [OH ] weak base pOH = –log[OH–] = 5.36 pH = 14 – pOH = 8.64 27 Acid-Base Titration An acid–base titration is the determination of the concentration of an acid or base by neutralizing the acid or base with an acid or base of known concentration. This allows for quantitative analysis of the concentration of an unknown acid or base solution. It makes use of the neutralization reaction that occurs between acids and bases. Problems are a mix of stoichiometry and equilibrium… Stoichiometry: Add a strong base (acid) to a weak acid (base), and determine the extent of neutralization using stoichiometric concepts. Equilibrium: The concentration of weak base/acid conjugate species is determined, and the equilibrium expression used to determine [H3O+], or pH. 28 Titrations of Acids & Bases In acid-base titrations, a buret is used to deliver measured volumes of an acid or base solution of known concentration (the titrant) to a flask that contains a solution of a base or an acid of unknown concentration. Plotting the pH changes that occur during an acid-base titration against the amount of acid or base added produces a titration curve; the shape of the curve provides important information about what is occurring in solution during the titration. 29 Titrations: Strong Acid w/ Strong Base Before addition of any strong acid, the initial [H3O+] equals the concentration of the strong acid; addition of strong base decreases the [H3O+] because added base neutralizes some of the H3O+ present. Equivalence point, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present Addition of a strong base after the equivalence point causes an excess of OH– and produces a rapid increase in pH. For the titration of a monoprotic strong acid with a monobasic strong base, the volume of base needed to reach the equivalence point is: moles of base = moles of acid (volume)b(molarity)b = (volume)a(molarity)a VbMb = VaMa 30 Titrations: Weak Acids w/ Strong Bases The shape of the titration curve for a weak acid depends dramatically on the identity of the acid or base and the corresponding value of Ka or Kb. Generally: Before base is added, the pH of the weak acid is higher than the pH of the strong acid, pH changes more rapidly during the first part of the titration, due to the higher starting pH The pH changes much more gradually around the equivalence point, the pH of the weak acid at the equivalence point is greater than 7.00 Above the equivalence point, the two curves are identical; once acid has been neutralized, the pH of the solution is controlled only by the amount of excess of OH– present, regardless of whether the acid is weak or strong 31 Example: Titration Calculate the pH during the titration of 20 mL of 0.25 M nitrous acid (HNO2; Ka = 4.5e–4) after adding the following volumes of 0.25 M NaOH: (a) 0 mL, (b) 10 mL, (c) 20 mL. (a) Initial pH of the solution before addition of base. HNO2 + H2O ⇋ H3O+ + NO2– Ka = [NO2–][H3O+] [HNO2] Ka = x2 _ 0.25 [HNO2] [NO2–] [H3O+ ] i 0.25 0 0 Δ –x +x +x f 0.25–x +x +x x2 = Ka* 0.25 M = 1.13e–4 M x = [H3O+] = 0.01 M pH = –log[H3O+] = 1.97 32 Example: Titration (b) Stoichiometry of neutralization reaction… OH– from NaOH consumes initially added HNO2 converting it to H2O and NO2–. HNO2 + OH– → H2O +NO2– 10 mL NaOH * 0.25 mol NaOH = 0.0025 mol NaOH 103 mL NaOH 20 mL HNO2 * 0.25 mol HNO2 = 0.005 mol HNO2 103 mL HNO2 0.005 mol HNO2 – 0.0025 mol NaOH = 0.0025 mol HNO2 left unreacted 0.0025 mol NO2– produced New concentrations = 0.0025 mol / 30 mL = 0.083 M = [HNO2] = [NO2–] 33 Example: Titration (b) Determine new pH. HNO2 + H2O ⇋ H3O+ + NO2– Ka = [NO2–][H3O+] [HNO2] Ka = (0.083)x 0.083 [HNO2] [NO2–] [H3O+] i 0.083 0.083 0 Δ –x +x +x f 0.083–x 0.083+x +x x = [H3O+] = Ka = 4.5e–4 M pH = –log[H3O+] = 3.35 34 Example: Titration (c) Stoichiometry of neutralization reaction… OH– from NaOH consumes initially added HNO2 converting it to H2O and NO2–. HNO2 + OH– → H2O +NO2– 20 mL NaOH * 0.25 mol NaOH = 0.005 mol NaOH 103 mL NaOH 20 mL HNO2 * 0.25 mol HNO2 = 0.005 mol HNO2 103 mL HNO2 0.005 mol HNO2 – 0.005 mol NaOH = 0 mol HNO2 left unreacted 0.005 mol NO2– produced New concentration = 0.005 mol / 0.04 L = 0.123 M = [NO2–] pH is determined by the NO2– equilibrium now! 35 Example: Titration (b) Determine new pH. NO2– + H2O ⇋ OH– + HNO2 Kb = [NO2–][H3O+] [HNO2] Kb = x2 0.125 [NO2–] [HNO2] [OH–] i 0.125 0 0 Δ –x +x +x +x +x f 0.125–x Kb = Kw/Ka = 1.0e–14/4.5e–4 = 2.2e–11 x = [OH–] = 1.65e–6 M pOH = –log[OH–] = 5.78 pH = 14 – pOH = 8.22 36 Oxidation-Reduction Reactions The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides: • Metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule oxidation is the loss of electrons, oxygen atoms acquire a negative charge and form oxide ions (O2–) reduction is the gain of electrons Oxidation states of each atom in a compound is the charge that atom would have if all of its bonding electrons were transferred to the atom with the greater attraction for electrons (larger EN). Atoms in their elemental form are assigned an oxidation state of zero. An atom is oxidized when its oxidation number increases and an atom is reduced when its oxidation number decreases Oxidation-reduction reactions are called redox reactions, in which there is a net transfer of electrons from one reactant to another. The total number of electrons lost must equal the total number of electrons gained. 37 Example of a Redox Reaction The reaction between hydrogen and fluorine is an example of an oxidation-reduction reaction: H2 + F2 → 2HF The overall reaction may be written as two half-reactions: H2 → 2H+ + 2e− (the oxidation reaction) F2 + 2e− → 2F− (the reduction reaction) There is no net change in charge in a redox reaction so the excess electrons in the oxidation reaction must equal the number of electrons consumed by the reduction reaction. The ions combine to form hydrogen fluoride: H2 + F2 → 2H+ + 2F− → 2HF 38 Rules for Assigning Oxidation States 1. The OS of an individual atom is 0 2. The total OS of all atoms in: a neutral species is 0 and in an ion is equal to the ion charge 3. Group 1 metals have an OS of +1 and group 2 an OS of +2 4. The OS of fluorine is –1, when in compounds 5. Hydrogen generally has an OS of +1 in compounds 6. Oxygen generally has an OS of –2 in compounds 7. In binary metal compounds, group 17 elements (halogens) have an OS of –1, group 16 of –2, and group 15 of –3. 39 Examples: Oxidation States Determine the oxidation states of the elements in the following reactions: a) Fe(s) + O2(g) → Fe2O3(g) b) Fe2+ c) Ag(s) + H2S → Ag2S(s) + H2(g) ANSWERS: a) Fe and O2 are neutral elements, therefore they have an OS of "0" according to Rule #1. The product has a total OS equal to "0" and following Rule #6, O has an OS of –2, which means Fe has an OS of +3. b) The OS of Fe corresponds to its charge, therefore the OS is +2. a) Ag has an OS of 0, H has an OS of +1 according to Rule #5, S has an OS of –2 according to Rule #7 and hence Ag in Ag2S has an OS of +1. 40 Examples: Oxidation States Determine the OS of the bold element in each of the following: a) Na3PO3 b) H2PO4– ANSWERS: a) The oxidation numbers of Na and O are +1 and –2. Since sodium phosphite is neutral, the sum of the oxidation numbers must be zero. Letting x be the oxidation number of phosphorus then, 0= 3(+1) + x + 3(–2), x = oxidation number of P = +3 b) Hydrogen and oxygen have oxidation numbers of +1 and –2. The ion has a charge of –1, so the sum of the oxidation numbers must be –1. Letting y be the oxidation number of phosphorus, –1= y + 2(+1) +4(–2), y = oxidation number of P = +5 41 Oxidants & Reductants Oxidants (oxidizing reagents), compounds that are capable of accepting electrons, they can oxidize other compounds • An oxidant is reduced in the process of accepting electrons Reductants (reducing agents), compounds that are capable of donating electrons, they can cause the reduction of another compound • A reductant is oxidized in the process of donating electrons oxidant + reductant oxidation–reduction gains e– loses e– redox reaction is reduced is oxidized Remember: An atom is oxidized when its oxidation number increases and an atom is reduced when its oxidation number decreases. 42 Examples: Oxidation / Reduction Determine which element is oxidized and which element is reduced in the following reactions (be sure to include the OS of each): a) Zn + 2H+ → Zn2+ + H2 b) 2Al + 3Cu2+→2Al3+ +3Cu c) CO32– + 2H+→ CO2 + H2O ANSWERS: a) Zn is oxidized (oxidation number: 0 → +2); H+ is reduced (+1 → 0) b) Al is oxidized (oxidation number: 0 → +3); Cu2+ is reduced (+2 → 0) a) This is not a redox type because each element has the same oxidation number in both reactants and products: O = –2, H = +1, C = +4 43 Types of Redox Reactions Combination reactions are some of the simplest redox reactions and as the name suggests involves the "combining" of elements to form a chemical compound. As usual, oxidation and reduction occur together. General Equation: Example: Oxidation States: A + B → AB H2 + O2 → H2O 0 + 0 → (2)(+1)+(–2) = 0 Decomposition reactions are the reverse of combination reactions, meaning they are the breakdown of a chemical compound into the individual elements. General Equation: Example: Oxidation States: AB → A + B H2O → H2 + O2 (2)(+1)+(–2) → 0 + 0 44 Types of Redox Reactions (cont.) Combustion reactions always involve oxygen, in the form of O2 and are almost always exothermic, meaning they produce heat. General Equation: Example: Oxidation States: CxHy + O2 → CO2 + H2O CH4 + 2O2 → CO2 + 2H2O ((–4)+(4)(+1)) + 0 → ((+4)+(2)(–2)) + ((2)(+1)+(–2)) Disproportionation Reactions: In some redox reactions substances can be both oxidized and reduced. General Equation: Example: Oxidation States: 2A → A' + A'' 2H2O2(aq) → 2H2O(l) + O2(g) ((2)(+1)+(2)(–1)) → ((2)(+1)+(–2)) + 0 45 Types of Redox Reactions (cont.) A single replacement reaction involves the "replacing" of an element in the reactants with another element in the products. General Equation: Example: Oxidation States: A + BC → AB + C Cl2 + 2NaBr → 2NaCl + Br2 0 + ((+1)+(–1)) → ((+1)+(–1)) + 0 A double replacement reaction is similar to a single replacement reaction, but involves "replacing" two elements in the reactants, with two in the products. General Equation: Example: Oxidation States: AB + CD → AD + CB Fe2O3 + 6HCl → 2FeCl3 + 3H2O ((2)(+3)+(3)(–2)) + ((+1)+(–1)) → ((+3)+(3)(–1)) + ((2)(+1)+(–2)) Not actually a redox reaction... No change in oxidation state! 46 Complexation A coordination complex or metal complex consists of a central atom or ion, which is usually metallic and is called the coordination center, and a surrounding array of bound molecules or ions, that are in turn known as ligands or complexing agents. Many metal-containing compounds, especially those of transition metals, are coordination complexes. Coordination refers to the "coordinate covalent bonds" (dipolar bonds, the ligands are donating electrons from a lone electron pair into an empty metal orbital on the coordination center) between the ligands and the central atom. The final charge of the complex is the sum of all the charges of the individual components. Thus a complex in solution can have a positive, negative or zero charge. 47 Ligand Complexation Example A ligand that is commonly used in chemical analyses is ethylenediamine tetra acetic acid (EDTA). It possesses lone pairs on 4 oxygen atoms and 2 nitrogen atoms that it can share with metal atoms to for octahedral complexes. Ligands can have several different descriptors, however the commonality for all complexing agents is that they can contribute an electron pair to a vacant orbital of a central atom. 48 Properties of Ligands 1. They are Lewis bases (donate an electron pair). 2. They contain at least one atom that is not a hydrogen or carbon. 3. They can have multiple atoms and charge centers like EDTA or they can be single ions like chloride. 4. They can be negatively charged or neutral; for example, water is a ligand. 5. The charge density of the resulting complex will be small due to the increase of volume of the complex over the individual ions. 6. The concentration of the complex can be calculated using a thermodynamic constant called the formation constant. 49 Catalysts A catalysts is a substance that participates in a reaction and causes it to occur more rapidly but that can be recovered unchanged at the end of the reaction and reused Catalysts are not involved in the stoichiometry of the reaction and are usually shown above the arrow in a net chemical equation A biological catalyst is called an enzyme Note: chemical processes in industry rely heavily on catalysts, which are added to a reaction mixture in trace amounts Can be classified as either homogeneous (uniformly dispersed throughout the reaction mixture) or heterogeneous (in a different physical state from the reactants) 50