Electrochemistry: Oxidation numbers. EIT Review F2006 Dr. J.A. Mack. Electrochemistry: Oxidation numbers

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1 EIT Review F2006 Dr. J.A. Mack Electrochemistry: Oxidation numbers In the compound potassium bromate (KBrO 3 ), the oxidation number of bromine (Br) is? Part Oxidation Numbers Electrochemistry: Oxidation numbers Variable In the compound potassium bromate (KBrO 3 ), the oxidation number of bromine (Br) is? Metals take on their formal charge: Non-metals do the same The compound is neutral so the sum of the oxidation numbers should be zero. +1?? 5 3 (-2) = -6 KBrO 3 1+?? + (-6) = 0?? =

2 The Behavior of Solutes: Strong Electrolytes: complete dissociation into ions 1 M Na 3 PO 4 (aq) 3M Na + (a) + 1M PO 3 4 4M overall in ions Non Electrolytes: no dissociation into ions 1M CH 3 OH (aq) 1M overall in methanol molecules Weak Electrolytes: partial dissociation into ions 1M HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 (aq) Example: A solution of sodium phosphate is added to a solution of aqueous barium nitrate. A white ppt is observed. Unbalanced Equation: Na 3 PO 4 (aq) + Ba(NO 3 (aq) Ba 3 (PO 4 (s) + NaNO 3 (aq) Molecular: 2Na 3 PO 4 (aq) + 3Ba(NO 3 (aq) Ba 3 (PO 4 (s) + 6NaNO 3 (aq) Ionic: 6Na + (aq) + 2PO 3-4 (aq) + 3Ba 2+ (aq) + 6NO 3- (aq) / Net Ionic: / Ba 3 (PO 4 (s) + 6Na + (aq) + 6NO 3- (aq) / / between 1 & 2 M overall 42 2PO 4 3- (aq) + 3Ba 2+ (aq) Ba 3 (PO 4 (s) 43 Solutions and Concentration How many grams of sodium phosphate are in 35.0 ml of a 1.51 M Na 3 PO 4 solution? Molarity: Moles of solute per liter of solution. Molarity (M) = moles of solute L of Solution {units: mol/l} molarity is a conversion factor that transforms units of volume to mole and vise versa ml solution L mols Na 3 PO 4 g Na 3 PO 4 use M as a conversion factor L 1.51 mol Na PO 35.0mL 3 10 ml L g 1 mol Na PO 3 4 = 8.66g Na3PO

3 Dilutions: New Molarity = Rearranging: old Molarity old Volume new Volume M 1 V 1 M 2 = V 2 M 1 V 1 = M 2 V 2 The ph Scale: Since 1 ph = log H + 1 log = log(x) x [H + ] = Molarity of H + (or anything for that matter ) ph = -log[h + ] In a neutral solution, [H + ] = [OH - ] = 1.00 x 10-7 M at 25 oc ph = - log [H + ] = ] = -log (1.00 x 10-7 ) = - [0 + (-7)]( = What is the ph of the M HCl solution that Jane made? HCl (aq) Strong electrolyte! ph = -log[h + ] H + (aq) + Cl (aq) Therefore [H + ] = [HCl] + ph = log[h ] = log(0.0515) = 1.29 Reaction Quotient & Equilibrium Constant Equilibrium Established H 2 + I 2 2 HI H2 + I2 2HI 48 49

4 THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B c C + d D the following is a CONSTANT (at a given T) The Reaction Quotient, Q In general, ALL reacting chemical systems are characterized by their REACTION QUOTIENT, Q. Q a A + b B c C + d D K = [C] c [D] d [A] a [B] b equilibrium constant conc. of products conc. of reactants If K is known, then we can predict concentrations of products or reactants. If Q < K, then system will shift to the right, convert reactants to products. If Q > K, then system will shift to the left, products convert to reactants. If Q = K, then system is at equilibrium Forms of Chemical Bonds There are 2 extreme forms of connecting or bonding atoms: Ionic Ionic complete complete transfer of 1 or more electrons from one atom to another Covalent Covalent some valence electrons shared between atoms Most bonds are somewhere in between. When there exists a difference in the Electronegativity (EN) between the two bonding atoms, ( EN) the bond is said to be polar. Example: EN Bond Br 2 0 non-polar HCl 0.9 polar-covalent NaF 3.1 ionic 52 53

5 Electronegativity: The measure of the tendency for a given atom to polarize the electrons in a covalent bond is called the Electronegativity (EN) of an atom. Electronegativity is related to the ionization energy and the electron affinity of an atom. EN decreases EN increases Periodic Table Rank the following bonds by increasing strength: H F, H Cl and H I Each bond has hydrogen in common so we have a basis for comparison. From the periodic table, in terms of electronegativity: F > Cl > I One concludes that the H-F bond is the most polar, followed by H-Cl then H-I. Bond Strength H I < H Cl < H F inc. EN Multiple bonds in covalent molecules: The oxygen and nitrogen that makes up the bulk of the atmosphere also exhibits covalent bonding in forming diatomic molecules. O: + O: O = O double bond N. + N :N N: triple bond Polyatomic Molecules (More than two atoms) Carbon dioxide: CO 2 O=C=O Bond Strength and Bond Properties: Covalent bond strength increases with increasing EN example: HCl bond is stronger than the HBr bond Covalent bond strength increases with increasing bond order example: O=O bond in stronger than O O bond triple > double > single Bond Order: Bond length decreases with increasing bond order (Strength) example: O=O bond is shorter than O O bond 56 57

6 Heat Transfer: Surroundings H bonding in water brings about a network of interactions which explain phenomena such as: heat in q > 0 (+) E > 0 System heat out q < 0 ( ) E < 0 capillary action surface tension why ice floats Thermochemistry: Energy in E f E Final work in > E i Energy out E f < E i work out E initial Heat and the Specific Heat Capacity: When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant. The amount of heat (q) transfer is related to the mass and temperature by: q = m C T energy q = heat lost or gained (J) m = mass of substance (g) q in q out C = the Specific Heat Capacity of a compound J o g C E initial E system > 0 (+) E system < 0 ( ) E Final 60 T is the temperature change in degrees Celsius or Kelvin s 61

7 Enthalpy: H How many kj of energy are released when g of methane (CH 4 (g)) is combusted? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H = -802 kj Ideal Gas Law P V = constant for a set number of gas moles n T 128.5g CH 4 mol CH4 802 kj = g 1mol CH kj How many hours would this power a 100 W light bulb if one could use all of this energy? 100 W = 100 Js kj 10 3 J kj s 100J min hr 60s 60min = 17.8 hrs 62 P V = n R T Latm R = "gas constant" = mol K PV = nrt 63 Problems: How many grams of krypton (Kr) will it take to exert a pressure of 11.2 atm in an 18.5 L at 28.2 o C. PV = nrt L atm R = mol K What wavelengths correspond to FM radio (93.5 MHz) signals? λ υ= c c λ = υ PV n = = RT 11.2 atm 18.5 L L atm mol K ( ) K = mols Kr λ = m s Hz 1s 93.5 MHz 1 MHz Hz g mols Kr = 702 g Kr (3sf) mol = 3.21 m 64 65

8 What wavelengths correspond to FM radio (93.5 MHz) signals? λ υ= c c λ = υ λ = m s Hz 1s 93.5 MHz 1 MHz Hz = 3.21 m Each element s outermost electrons (valence) are related to the elements position on the periodic table. We can use the periodic table to determine the electron configuration by counting: N: 7 electrons : 1s 1 2: 1s 2 3: 2s 1 4: 2s 2 5: 2p 1 6: 2p 2 7: 2p 3 All of the subshells below the valence are full. 1s 2 2s 2 2p

9 Electron Configurations cont Orbital box notation: Periodic Trends 1s 2s 2p 3s 3p Aluminum: Al (13 electrons) 3s 3p Electron Configuration 1s 2 2s 2 2p 6 3s 2 3p 1 2s 2p 1s Kinetics: Rate Law & Reaction Order The reaction rate law expression relates the rate of a reaction to the concentrations of the reactants. Each concentration is expressed with an order (exponent). The rate constant converts the concentration expression into the correct units of rate (Ms 1 ). (It also has deeper significance, which will be discussed later) For the general reaction: aa+ bb cc+ dd x y Rate = k [A] [B] x and y are the reactant orders determined from experiment. x and y are NOT the stoichiometric coefficients. Reaction Orders: A reaction order can be zero, or positive integer and fractional number. Order Name 0 zeroth 1 first 2 second 0.5 one-half 1.5 three-half two-thirds 72 73

10 Recognizing a first order process: Whenever the conc. of a reactant falls off exponentially, the kinetics follow first order. A products Determining the Rate constant for a first order process Taking the log of the integrated rate law for a first order process we find: kt ( = [A] oe ) ln [A] ln[a] = ln[a] o k t [A] = [A] e o kt A plot of ln[a] versus time (t) is a straight line with slope -k and intercept ln[a] o A certain reaction proceeds through t first order kinetics. The half-life of the reaction is 180 s. What percent of the initial concentration remains after 900s? Step 1: Determine the magnitude of the rate constant, k. t 1 = ln 2 2 k ln 2 k = = ln 2 t 180s = k = s 1 A certain reaction proceeds through t first order kinetics. The half-life of the reaction is 180 s. What percent of the initial concentration remains after 900s? Using the integrated rate law, substituting in the value of k and 900s we find: [A] = e [A] o kt k = s -1 [A] [A] e = s 900 s = o Since the ratio of [A] to [A] 0 represents the fraction of [A] that remains, the % is given by: = 3.12% 77