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Monday, January 8, 2018 6.1 Enery and its Units Types of Energy Energy is the potential of capacity to move matter Kinetic energy is associated with matter that is in motion [ K = ½mv2 ] Potential energy is associated with matter that is held in place by a force [ P = mgh ] Internal energy is the sum of the kinetic and potentials of all particles in a substance Units of Energy Joule (kg x m2 / s2) is the SI unit for energy Calorie is the amount of energy required to raise 1g of water by 1oC 1 cal = 4.184 J Example 6.1 A pitcher can throw a baseball so that it travels between 60-90 miles per hour. A baseball weighing 0.143 kg travels 33.5 m/s. What is the kinetic energy of this baseball in joules and calories? Problem Strategy Kinetic energy is defined as K = ½mv2, substitute the given values into this formula Solution K = ½ (0.143 kg) (33.5 m/s)2 = 80.2 J, in cal, K = 80.2 J / 4.184 J = 19.2 cal Law of Conservation of Energy Energy can be converted from one form to another, but the total amount of energy is never-changing Wednesday, January 10, 2018 6.2 First Law of Thermodynamics; Work & heat Thermodynamic System vs. Surroundings The system is the substance in which the change occurs The surroundings are the environment in which the system is in Work and Heat Work (w) is the transfer of energy into or out of a system that causes the movement of matter When work is done on a system it is positive, when a system does work on its surroundings it is negative Heat (q) is the transfer of energy that causes a temperature difference between the system and its surroundings When heat is added to the system it is positive, when heat is given to the surroundings it is negative Change in Internal Energy How the energy of a system changes when stressors are applied to it, in reference to the initial state of the system A state function is a property of a system that depends on its initial state First Law of Thermodynamics The change in internal energy of a system is equal to its heat plus work [ U = q + w ] Friday, January 12, 2018 6.3 Heat of Reaction; Enthalpy of Reaction Heat of Reaction The heat (q) of the reaction is the heat absorbed or evolved to achieve a constant temperature of the system Exothermic processes evolve heat (-q) to the surroundings Endothermic processes absorb heat (+q) from the surroundings Pressure-Volume Work Pressure-volume work is done when a system changes volume or pressure [ w = -PV ] Enthalpy and Enthalpy of Reaction The heat of the reaction is equal to enthalpy when a constant pressure is applied to the system [ H = U + PV ] The change in enthalpy at a constant pressure and a given temperature is the enthalpy of the given reaction 6.4 Thermochemical Equations A thermochemical equation gives the molar ratio and the enthalpy of the reaction Example 6.2 Aqueous sodium hydrogen carbonate solution (baking soda solution) reacts with hydrochloric acid to produce aqueous sodium chloride, water and carbon dioxide gas. The reaction absorbs 12.7 kJ of heat at constant pressure for each mole of sodium hydrogen carbonate. Write the thermochemical equation for the reaction. Problem Strategy Translate the names of the substances to their formulas and balance the equation. The problem tells you that heat is absorbed so the enthalpy change is positive. Solution NaHCO3 (aq) + HCl (aq) NaCl (aq) + H2O (l) + CO2 (g) The equation is for 1 mol NaHCO3 with the absorption of 12.7 kJ of heat, so the thermochemical equation is: NaHCO3 (aq) + HCl (aq) NaCl (aq) + H2O (l) + CO2 (g); H = + 12.7 kJ The rules for manipulating thermochemical equations are: When a thermochemical equation is multiplied by any factor, the enthalpy is multiplied by the same factor When the chemical equation is reversed, the sign of the enthalpy is also reversed Example 6.3 When 2 mol H2 (g) and 1 mol O2 (g) react to give liquid water, 572 kJ of heat evolves. 2H2 (g) + O2 (g) 2H2O (l); H =-572 kJ. Write this equation for 1 mole of liquid water. Give the reverse equation, in which 1 mol of liquid water dissociates into hydrogen and oxygen. Problem Strategy Use the rules for manipulating thermochemical equations Solution Multiply the coefficients and enthalpy by ½ : H2 (g) + ½O2 (g) H2O (l); H = -286 kJ Reverse the equation and the sign of enthalpy : H2O (l) H2 (g) + ½O2 (g) ; H = +286 kJ 6.5 Applying Stoichiometry to heats of Reaction Example 6.4 How much heat is evolved when 9.07 x 105 g of ammonia is produced according to the following equation? (Assume that the reaction occurs at constant pressure.) N2 (g) + 3H2 (g) 2NH3 (g); H = -91.8 kJ Problem Strategy Convert grams of NH3 to moles of NH3 and then to kJ of heat Solution (9.07 x 105 g NH3) x (1 mol NH3 / 17.0 g NH3) x (-91.8 kJ / 2 mol NH3) = -2.45 x 106 kJ Thus 2.45 kJ of heat evolves 6.6 Measuring Heats of Reaction Heat Capacity and Specific Heat The heat capacity of a substance is the quantity of heat needed to raise the temperature of the substance one degree Celsius [ q = CT ] The specific heat capacity is the quantity of heat needed to raise 1g of the substance by one degree Celsius [ q = mCT ] Measurement of Heat of Reaction A calorimeter measures heat absorbed or evolved during a physical or chemical change It consists of an insulated container and a thermometer Bomb calorimeters involve gases and are closed systems Example 6.6 Suppose that 0.562g of graphite is placed in a bomb calorimeter with an excess of oxygen at 25.00oC and 1 atm pressure. Excess O2 ensures that all carbon burns to form CO2. The Graphite ignited, and it burns according to the equation C(graphite) + O2(g) CO2(g). On reaction, the calorimeter temperature rises to 25.89oC. The heat capacity of the calorimeter and its contents was determined to be 20.7 kJ/oC. What is the heat of reaction at 25.00oC and 1atm pressure? Express the answer as a thermochemical equation. Problem Strategy Determine the amount of heat that is absorbed by the calorimeter. Convert this to the heat released per mole of reaction. Solution qrxn = -CT = -20.7kJ/oC (25.89oC – 25.00oC) = -18.4 kJ 1molC x (12gC / 1molC) x (-18.4kJ / 0.562gC) = -3.9x102 kJ/mol C(graphite) + O2(g) CO2(g); H = -3.9x102 kJ/mol 6.7 Hess’s Law Hess’s law of heat summation states that for a chemical equation that be written in multiple steps, the enthalpy change for the overall reaction can be determined by the summation of the enthalpies of each individual step Example 6.7 What is the enthalpy of reaction for the formation of tungsten carbide, WC, from the elements? W(s) + C(graphite) WC(s) 2W(s) + 3O2(g) 2WO3(s); -1685.8 kJ C(graphite) + O2(g) CO2(g); -393.6 kJ 2WC(s) + 5O2(g) 2WO3(s) + 2CO2(g); -2391.9 kJ Problem Strategy Determine if any equations need to be reversed or multiplied. Apply the same manipulations to the enthalpies of each part of the reaction Solution Equation 1 must be multiplied by ½, equation 2 can be left as it is and equation three must be multiplied by –½. H = ½(-1685.8kJ) + -393.5kJ + -½(-2391kJ) = -40.5kJ 6.8 Standard Enthalpies of Formation Standard state is the thermodynamic conditions used to compare thermodynamic data (usually 1 atm and 25oC) An allotrope is multiple unique forms of an element in the same state The reference form of an element is its most stable form under thermodynamic conditions The standard enthalpy of formation is the change in enthalpy for the formation of one mol of the substance in its reference form from its elements in their reference form Example 6.8 Use values of Hof to calculate the heat of vaporization, of Hovap, of carbon disulfide at 25oC. The vaporization process is CS2(l) Cs2(g) Problem Strategy Subtract the enthalpy of the reactants from the products Solution 89.7kJ 116.9kJ Hovap = nHof (products) - nHof (reactants) = 27.2 kJ Example 6.9 Large quantities of ammonia are used to prepare nitric acid. The first step consists of the catalytic oxidation of ammonia to nitric acid, NO. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g). What is the standard enthalpy change for this reaction? Problem Strategy 4(-45.9kJ) + 5(0kJ) 4(90.3kJ) + 6(-241.8kJ) Hovap = nHof (products) - nHof (reactants) = -906 kJ 9.1 Describing ionic bonds An ionic bond is formed by the electrostatic attraction between opposing ions when one or more electrons are transferred between ions to fil their valence shells Lewis-Electron-Dot Symbols A diagram of valence electrons in which they are depicted by dots and the nucleus is represented by the elements symbol Example 9.1 Use Lewis electron-dot symbols to represent the transfer of electrons from magnesium to fluorine atoms to form noble-gas configurations 62230042735500Problem Strategy Write the symbols for the elements and their original valence electron configurations. Determine which element will donate electrons, and how many. Solution Energy Involved in Ionic Bonding When atoms bond to each other there is first the exchange of electrons in which the removal of an electron uses more energy than is given off when the other atom accepts it Once the ions have been formed their opposite charges allow for a bond to be formed Coulomb’s Law is that the potential energy produced by distancing oppositely charged ions is directly proportional to the product of the charges and inversely proportional to the distance between the ions Lattice energy is the change in energy when an ionic solid is separated into its ions in the gaseous state Lattice Energies from the Born-Harber Cycle The Born-Harber cycle is the process by which the lattice energy of an ionic solid can be indirectly determined through a thermochemical cycle Properties of Ionic Substances Ionic substances typically have high melting points because there are strong interactions between ions require more energy to be broken The liquid form of ionic substances is a good electric conductor because among molecules it contains many ions which can carry an electrical current 9.11 Bond Enthalpy Bond dissociation energy is the energy required to break a bond Bond enthalpy (energy) is the average strength of a bond between two specific elements Example 9.13 Polyethylene is formed by linking many ethylene molecules in long chains. Estimate the enthalpy change per mole of ethylene for this reaction using bond enthalpies Problem Strategy Break bonds in the reactants to form new bonds that will give the products. The bond enthalpy change is the sum of the bond enthalpies for the bonds broken minus the sum of the bond enthalpies for the bonds formed. Solution For every double bond broken between C, two single bonds are formed between C. H = 614kJ – (2 x 348kJ) = -82k 18.1 First Law of Thermodynamics: A Review The first law of thermodynamics is essentially conservation of energy applied to thermodynamic systems Mathematical Statement of the First Law The internal energy (U) of a thermodynamic system is the sum of its kinetic energy and its potential energy [ U = EK + EU ] It can only change with the transfer of energy into or out of the system There are two types of energy transfers between the system and the surroundings, heat (q) and work (w) [ U = q + w ] [ Pressure-Volume work = -PV ] Enthalpy of Reaction Enthalpy (H) is the systems internal energy plus pressure times volume [ H = U + PV ] [ q = H ] 18.2 Entropy and the Second Law of Thermodynamics Entropy Entropy (S) is a measure how dispersed energy is in a system When energy is concentrated in a few energy states, entropy is low When energy id dispersed over many energy states, entropy is high In a spontaneous process energy is dispersed, so the entropy is increasing The entropy of a gas is greater than that of a liquid, which is greater than that of a solid This is because when particles in these substances have more energy to move around, more energy means greater entropy Second Law of Thermodynamics The second law of thermodynamics states that the entropy for a system increases for a spontaneous process Unlike energy, entropy can be created or destroyed during a reaction Entropy and Molecular Disorder High entropy systems can also be described as disordered because the energy is dispersed so widely and across so many states Entropy Change for a Phase Transition When systems are at equilibrium entropy is virtually zero unless heat is absorbed into the system [ S = q / T ] Criterion for a Spontaneous Reaction For a reaction to be spontaneous, S > q / T 18.3 Standard Entropies and the Third Law of Thermodynamics Third Law of Thermodynamics The third law of thermodynamics states that substances of perfect crystalline structure at 0 K have an entropy of zero When the temperature of these substances is increased or decreased energy will be absorbed or evolved, changing the entropy The standard entropy (S0) is the substances entropy in its standard state Entropy Change for a Reaction Entropy usually increases when: A molecule is broken into more molecules A reaction where there is an increase in moles of gas A solid changes into a liquid or a liquid changes into a gas [ S0 = nS0 (products) - nS0 (reactants) ] 18.4 Free Energy and Spontaneity Gibbs free energy (G) is a thermodynamic quantity that relates to the spontaneity of a reaction The changes in enthalpy and entropy while a reaction occurs, impact the free energy [ G = H - TS ] Standard Free-Energy Change Standard state is at 1 atm for solutions of 1 M concentration and a specific temperature The standard free-energy change (G0) occurs when reactants in their standard state are converted to products in their standard state [ G0 = H0 - TS0 ] Standard Free Energies of Formation The standard free energy of formation (G0f) is the change in free energy when 1 mole is formed from elements in their reference form [ G0f = nG0f (products) - nG0f (reactants) ] Go as a Criterion for Spontaneity The spontaneity of a reaction can be judged on the following rules When G0 is a large negative number the reaction occurs spontaneously as written and continues to completion When G0 is a large positive number the reaction does not occur spontaneously as written and little products are produced When G0 is a small number (-10kJ to +10kJ) the reaction is at equilibrium and products and reactants are formed 18.5 Interpretation of Free Energy Maximum Work In theory, spontaneous reactions generate useful work, but this does not occur in reality For entropy to be produced the maximum useful work cannot be The free energy change is the maximum amount of free energy available to do work Free Energy Change During Reaction Not all free energy is changed into work, the leftovers are changed into entropy 18.6 Relating Go to the Equilibrium ConstANT The thermodynamic equilibrium constant (K) expresses the concentrations of gases as pressures and the concentration of solutions as molarities Kp is used for reactions only involving gases Kc is used for reactions only involving solutions To determine the free energy change under nonstandard conditions G is determined To determine the standard free energy change Go is determined [ G = Go + RT ln(Q) ] 18.7 Change of Free Energy with Temperature This method is used under the assumption that Ho and So are constant despite changes in temperature to determine an approximate answer [ GoT = Ho -TSo ] Spontaneity and Temperature Change Ho So Go Spontaneity - + - Always Spontaneous + - + Never Spontaneous - - +/- Spontaneous at low T + + +/- Spontaneous at high T Calculation of Go at Various Temperatures When Go = 0, [ T = Ho / So ] 19.1 Balancing Oxidation-Reduction Reactions in Acidic and Basic Solutions Skelaton Oxidation-Reduction Equations The skeleton equation has information regarding the reducing agent, oxidizing agent, the products and the solution the reaction occurs in Half-Reaction Method in Acidic and Basic Solutions The skeleton equation can be completed by applying the half-reaction method Steps in Balancing Ovidation-Reduction Equations in Acidic Solutions Assign oxidation numbers to each atom to identify which are oxidized and reduced Split the equation into two; one should contain species that are oxidized and the other should contain species that are reduced Balance each half-reaction Balance everything except O and H Balance O by adding H2O to either side Balance H by adding H+ to either side Balance the charges by adding electrons to the more positive side Combine the two reactions to give the final, balanced reaction Multiply the half reactions by a number that allows the electrons to cancel when the reactions are recombined Simplify the equation by cancelling and reducing coefficients Additional Steps for Balancing Oxidation-Reduction Equations in Bassic Solutions For every H+ ion OH- ions should be added to both sides of the equations Simplify by reducing H+ and OH- ions to H2O 19.2 Construction of Voltaic Cells Voltaic cells are made up of two half cells in which a half-reaction takes place A half cell can be made from a metal strip in a solution of its metal ion In voltaic cells electrons flow from one electrode to the other if there is an external circuit and ions flow from one solution to the other if there is an internal circuit A salt bridge is a tube of electrolyte gel that connects the two half cells and allows the flow of ions without the solutions mixing The oxidation reaction occurs at the anode (-) and reduction occurs at the cathode (+) Electrons are released from the anode and travel to the cathode The cell reaction is the sum of the reactions for each half cell 19.3 Notation for Voltaic Cells The oxidation reaction which occurs at the anode is written on the left The reduction reaction which occurs at the cathode is written on the right The salt bridge is denoted by two bars (||) The electrode terminal is separated by the electrode electrolyte by a single bar (|) An inert electrode is used when a reaction involves a gas (Pt, etc.) Example: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) anode terminal | anode electrolyte || cathode electrolyte | cathode terminal 19.4 Cell Potential Work is needed to move electrons from the anode to the cathode The amount of work needed to move a charge depends on the total charge moved and the potential difference Potential difference is the difference in electric potential (units: Volts) Electrical work (V) = charge (C) x potential difference (V) The Faraday constant (F) is the charge on one mole of electrons, 96,485 C/mole- In a voltaic cell the potential difference across the electrodes is less than the max voltage The cell potential is the maximum potential difference across the electrodes The anode should have a negative cell potential and the cathode a positive The maximum work from a voltaic cell can be calculated by Wmax = -nFEcell 19.5 Standard Cell Potentials and Standard Electrode Potentials A cell potential is the driving source of a cells reaction Half of the reaction occurs at the anode (oxidation) and the other half occurs at the cathode (reduction) The cell potential is a combination of the oxidation potential (tendency of a reduced species to donate electrons) and reduction potentials (tendency of an oxidized species to gain electrons), Ecell = oxidation potential + reduction potential Tabulating Standard Electrode Potentials Standard cell potential is measured at 1M, 1 atm and at a temp of 25C There is a long table of electrode potentials that can be used to calculate cell potentials Strengths of Oxidizing and Reducing Agents Standard electrode potentials are used to determine the strength of oxidizing or reducing agent, they are set up like: oxidized species + ne reduced species or the opposite The SRP of the strongest reducing agents are more negative while the strongest oxidizing agents are the most positive Calculating Cell Potentials It can be done by addition of half-reactions and standard potentials or using standard reduction potentials 19.6 Equilibrium Constants from Cell Potentials Free energy (G) is equal to the useful work (w) For a voltaic cell this work is equal to -nFEocell Combining the two free energy equations gives nFEocell = RT lnK This equation can be rearranged to give Eocell = 0.0592 log(K) / n 19.7 Dependance of Cell Potential on Concentration Cell potential depends on concentration and pressure Nernst Equation An equation that connects the cell potential and reaction quotient Ecell = Eocell – (RT/nF) lnQ OR Ecell = Eocell – (2.303RT/nF) logQ Determination of pH The Nernst equation can be used to relate cell potential and pH One of the electrons must be hydrogen submerged in bubbled in hydrogen By solving for Q the concentration of H protons can be determined The hydrogen electrode is often replaced with a glass electrode which is an ion-selective electrode that can be used to monitor certain ions 19.8 Some Commercial Voltaic Cells Lelanché (zinc-carbon) dry cells are composed of a zinc anode, a carbon cathode, a graphite rod in the middle surrounded by magnesium dioxide and ammonium and zinc chlorides Alkaline dry cells are the same as Lelanché cells, but they have potassium hydroxide instead of ammonium chloride so that they perform in cold weather Lithium-iodine batteries are composed of a lithium anode and an iodine cathode A reliable battery used in pacemakers Dry cells usually cannot be recharged because the reaction is difficult to reverse Lead storage cells are rechargeable. The anode is made of spongy lead and the cathode is composed of lead (IV) oxide It is recharged by an external electric current which reverse the half reactions Nickle-cadmium cells is a storage battery that can also be recharged and are used in calculators, portable power tools, shavers and toothbrushes A fuel cell is like a battery but it has a continuous supply of reactants A PEM (proton-exchange membrane) uses hydrogen and oxygen to catalyze the reaction 19.9 electrolysis of molten salts A Downs cell is a commercial electrochemical cell used to obtain sodium metal by the electrolysis of molten sodium chloride The products of the reaction are kept separate to keep them from reacting 19.10 Aqueous Electrolysis In the electrolysis of molten salt the half reactions involve ions from the salt Electrolysis in aqueous solutions can involve water in the half reactions The reduction half reaction of water is 2H2O(l) + 2e- H2(g) + 2OH-(aq) The oxidation half reaction of water is 2H2O(l) O2(g) + 4H+(aq) + 4e- Electrolysis of Sulphuric Acid Solutions For these reactions the half reactions of H2SO4 and water must be considered The possible reduction half-reactions at the cathode include: 2H+(aq) + 2e- 2H2(g) ; 2H2O(l) + 2e- H2(g) + 2OH-(aq) The possible oxidation half-reactions at the anode include: 2SO4-(aq) S2O82-(aq) + 2e- ; Eo = 2.01 V 2H2O(l) O2(g) + 4H+(aq) + 4e- ; Eo = 1.23 V Because water has a smaller reduction potential it will be oxidized more easily Standard reduction and oxidation potentials are measured at equilibrium, so overvoltage may be required for electrolysis conditions since it is far from equilibrium Overvoltage is more dramatic when gas is a product Electrolysis of Sodium Chloride Solutions For these reactions the half reactions of NaCl and water must be considered The possible reduction half-reactions at the cathode include: Na+(aq) + e- Na(s) ; Eo = -2.71 V 2H2O(l) + 2e- H2(g) + 2OH-(aq) ; Eo = -0.83 V The possible oxidation half-reactions at the anode include: 2Cl-(aq) Cl2(g) + 2e- ; Eo = 1.63 V 2H2O(l) O2(g) + 4H+(aq) + 4e- ; Eo = 1.23 V Given waters smaller reduction potential it would be expected to be oxidized more easily than Cl-, but the close potentials could alter this conclusion By applying the Nernst equation to the reaction of chorine ions it becomes apparent that when NaCl is present in large concentrations it becomes more likely to be oxidized Electroplating of Metals Electroplating protects metals from corrosion Zinc is used to electroplate steel because scratches don’t impact its effectiveness The steel is placed in a bath of zinc salts; Zn2+(aq) + 2e- Zn(s) Electrolysis can be used to purify metals 19.11 Stoichiometry of Electrolysis The quantity of charge passed through the circuit is measured instead of the weight of the substance One Faraday (96485 C) is the charge on one mole of electrons Electric charge (C) = current (A) x time (s) 13.1 Definition of Reaction Rate The rate of reaction is how much product, unit volume, is formed per unit time The reaction rate is the molar increase of products over unit time (mol/Ls or M/s) Example: 2N2O5(g) 4NO2(g) + O2(g) Rate of formation of O2(g) = [O2] / t This equation finds the instantaneous or average rate of formation depending on the length of the time being analyzed As the concentration of the products increases the reactants decreases Because there are twice as many moles of N2O5 as there is O2 the rate of formation of O2 = -½ the rate of decomposition of N2O5; [O2] / t = -½ [N2O5] / t