Chapter 11 Solutions PDF

Chapter 11 CHEM 0120  Chapter 11: Solutions Solution: a homogeneous mixture of two or more substances What are the components of a solution? Solute Solvent Solutions can exist as any of the 3 states of matter: Gases Liquids Solids...

Chapter 11 CHEM 0120  Chapter 11: Solutions Solution: a homogeneous mixture of two or more substances What are the components of a solution? Solute Solvent Solutions can exist as any of the 3 states of matter: Gases Liquids Solids 2  Components of a Solution Solute When the solution is a gas or a solid dissolved in a liquid, the solute is the gas or solid In remaining cases, the solute is the component in the smaller amount Solvent When the solution is a gas or solid dissolved in a liquid, the solvent is the liquid In the remaining cases, the solvent is the component in the greater amount 3  Dissolving a Solute in a Solution Let’s start with a solid in water. Go from (s) (aq) Is there a difference between dissolving a molecular and an ionic compound? 4  Dissolving a Solute in a Solution Let’s start with a solid in water. Go from (s) (aq) Is there a difference between dissolving a molecular and an ionic compound? C12H22O11(s) K2Cr2O7(s) 5 Common Types of Solutions 6  What favors the spontaneous formation of a solution? Spontaneous process: a process that occurs under specified conditions without the requirement of energy from some external source Two criteria that favor the spontaneous formation: Decrease in the internal energy of the system Increased dispersal of matter in the system 7  Solubility Solubility: the amount that dissolves in a given quantity of solvent at a given temperature to give a saturated solution The maximum amount of solute that can be dissolved in a given amount of solvent. When one substance does not dissolve in another, it is said to be insoluble. Oil is insoluble in water. The solubility of one substance in another depends on the following: Nature’s tendency toward mixing The types of intermolecular attractive forces 8  Solubility A solution that has the solute and solvent in dynamic equilibrium is said to be saturated. If you add more solute, it will not dissolve. The saturation concentration depends on the temperature and pressure of gases. A solution that has less solute than saturation is said to be unsaturated. More solute will dissolve at this temperature. A solution that has more solute than saturation is said to be supersaturated. 10 Solubility 11 Solubility: Supersaturated 12  Solubility Limit There is typically a limit to the solubility. Gases are always soluble in each other. Two liquids that are mutually soluble are miscible. Ethanol and water are miscible. Oil and gasoline are miscible. Oil and water are immiscible. Miscible Fluids: fluids that mix with or dissolve in each other in all proportions 13  Solubility: Will it Dissolve? (Already answered.) Why or why not? Ethanol and water are miscible. (CH3CH2OH & H2O) Both are polar substances. Both have –OH groups which have strong hydrogen bonding attractions. Oil and gasoline are miscible. Both are mixtures of hydrocarbons and are nonpolar. Oil and water are immiscible. Water is polar and oil is nonpolar. Mixing will not take place because the hydrogen bonding between the water molecules will not break and be replaced with London forces. “Like dissolves Like”. 14  “Like dissolves Like” Summary A chemical will dissolve in a solvent if it has a structure similar to that of the solvent. Compatible intermolecular forces Polar molecules and ionic compounds will be more soluble in polar solvents. Nonpolar molecules will be more soluble in nonpolar solvents. 15  Solubilities of Alcohols in Water: Why the Change? Although water and alcohols both have –OH groups, as the hydrocarbon end (–R) increases in size, the alcohol becomes less like water and the solubilities decrease. 16  Solubility of Ionic Substances in Water Ionic substances can have largely different solubilities in water. NaCl: 36 g / 100 mL (at RT) Ca3(PO4)2: 0.002 g / 100 mL Differences in solubility due to: Hydration energy Lattice energy 17  Ion-dipole interactions Ion-dipole force is responsible for the energy of attraction between an ion and a water molecule. When ions dissolve in water, they become hydrated. Hydration: the attraction of ions for water molecules. Each ion is surrounded by water molecules. 18 Hydration Example of polar water molecules orienting with respect to nearby ions. The O atom in water orients toward the cation (Li+). Each H atom in water orients toward the anion (F-). 19  Dissolving of LiF in H2O Ions on the surface initially hydrate. The ions at the corners are easier to remove due to fewer lattice forces. Upon hydration, ions are removed away from the crystal and move into the body of the liquid. 20  Lattice Energy Lattice Energy: the energy holding ions together in the crystal lattice Works against the solution process. Dependent on the charges of the ions. Alkali metal ions (Na+ and K+) are generally soluble, but phosphate ions (PO43-) are generally insoluble. It is inversely proportional to the distance between neighboring ions (the distance depends on the sum of the radii of the ions). Example: Alkaline earth hydroxides. The solubility ranking is: Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2 21  Solution Process of Ionic Solids Hydration energy pulls ions apart. Lattice energy keeps the ions Hydration energy together. If a solid has a large lattice + – energy, it is typically insoluble. Lattice energy 22  Effects of Temperature on Solubility The solubility of one substance in another varies with temperature and pressure. Solubility is generally given in grams of solute that will dissolve in 100 g of water. For most solids, the solubility of the solid increases as the temperature increases. Solubility curves can be used to predict whether a solution with a particular amount of solute dissolved in water is saturated (on the line), unsaturated (below the line), or supersaturated (above the line). 23  Effects of Pressure Cylinder containing carbon Change on Solubility dioxide gas and water saturated with carbon dioxide. Pressure change: Little to no effect on the solubility of liquids or solids in water. Has a large Piston effect on the solubility of a gas. Gaseous CO2 Aqueous solution CO2 is more soluble at higher of CO2 pressures since the partial pressure is increased. When the piston is pushed down, increasing the partial pressure of carbon dioxide, Once the partial pressure is more gas dissolves (which tends to reduce thecarbon reduced, the solubility decreases. dioxide partial pressure). 24  Henry’s Law It is possible to predict the effect of pressure on the solubility of a gas in a liquid using Henry’s Law. The solubility of a gas (Cgas) is directly proportional to its partial pressure, (Pgas). Cgas = kHPgas kH is called the Henry’s law constant. 25  Colligative Properties Colligative properties are properties whose value depends only on the number of solute particles and not on what they are. Value of the property depends on the concentration of the solution. Ways to express concentrations: Molarity Mass Percentage of Solute Molality Mole Fraction 26  Solutions and Concentrations Molarity is used to quantitatively discuss the concentration. It is the ratio of the number of moles of solute to the number of liters of solution moles of solute Molarity M = liters of solution Molarity becomes a way of expressing concentration as well as a conversion factor between volume and moles Can then further convert between moles and grams 27  Example 0.42 g NaNO3 is dissolved to make 50.0 mL of solution. What is the solution’s concentration (in M)? 28  Example How many mL of 1.75 M NaCl are needed to get 100.0 g of NaCl? 29  Molality Molality of a solution is the moles of solute per kilogram of solvent moles of solute molality = kilograms of solvent What are the differences between molarity and molality? 30  Example An aqueous solution is 0.273 m KCl. What is the molar concentration of KCl? The density of the solution is 1.011 x 103 g/L. (Molar mass of KCl = 74.6 g/mol) 31  Mass Percentage of Solute Mass percentage: the percentage by mass of solute contained in a solution. mass of solute Mass percentage of solute = × 100% mass of solution 32  Mole Fraction Mole fraction: the fraction of the moles of component substance in the total moles of solution moles of substance A = total moles of solution The total of all the mole fractions in a solution is 1. The mole fraction has no units. 33  Chapter 11 Part 2... Vapor Pressure of Solutions Raoult’s Law Freezing Point Depression and Boiling Point Elevation Review of Phase Diagrams 34 Chapter 11: Part 2 CHEM 0120  Vapor Pressure of Solutions The vapor pressure of a solvent above a solution is lower than the vapor pressure of the pure solvent. The solute particles replace some of the solvent molecules at the surface. 2  Vapor-pressure lowering Vapor-pressure lowering (of a solvent): a colligative property equal to the vapor pressure of the pure solvent minus the vapor pressure of the solution ° ∆𝑷 = 𝑷𝐬𝐨𝐥𝐯𝐞𝐧𝐭 − 𝑷𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 3  Raoult’s Law and Vapor Pressure Raoult’s Law: the partial pressure of solvent, PA, over a solution equals the vapor pressure of the pure solvent, P°A, times the mole fraction of solvent, XA, in the solution. 𝑃 = 𝑃° 𝑋 Because the mole fraction is always less than 1, the vapor pressure of the solvent in solution will always be less than the vapor pressure of the pure solvent. ∆𝑃 = 𝑃° 𝑋 4  Ideal vs Nonideal solution In ideal solutions, the made solute–solvent interactions are equal to the sum of the broken solute–solute and solvent– solvent interactions. Ideal solutions follow Raoult’s law. Effectively, the solute is diluting the solvent. If the solute–solvent interactions are stronger or weaker than the broken interactions, the solution is nonideal. 5  Phase Diagrams A graphical way to summarize the conditions under which the different states of a substance are stable  Colligative Properties Related to Vapor Pressure Lowering Vapor Pressure lowering occurs at all temperatures. The result is the temperature required to boil the solution is higher than the boiling point of the pure solvent. Another result is the temperature required to freeze the solution is lower than the freezing point of the pure solvent. 7  Boiling Point Elevation Boiling-point elevation (ΔTb): a colligative property of a solution equal to the boiling point of the solution minus the boiling point of the pure solvent. It is directly proportional to the molal concentration of the solution. ∆𝑇 = 𝐾 𝑐 The proportionality constant, Kb, is called the boiling- point-elevation constant It depends only on the solvent. 8  Freezing-Point Depression Freezing-Point Depression (ΔTf): a colligative property of a solution equal to the freezing point of the pure solvent minus the freezing point of the solution. It is directly proportional to the molal concentration of the solution. ∆𝑇 = 𝐾 𝑐 The proportionality constant, Kf, is called the freezing-point- depression constant It depends only on the solvent. 9 Table of Constants 10  Osmosis Osmosis: the phenomenon of solvent flow through a semipermeable membrane to equalize the solute concentrations on both sides of the membrane The flow of solvent from a solution of low concentration into a solution of high concentration. The semipermeable membrane allows solvent, but not solute, to flow through it. 11  Osmotic Pressure Osmotic pressure: a colligative property of a solution equal to the pressure that, when applied to the solution, stops osmosis. Osmotic pressure (π) is directly proportional to the molarity (M) of the solute particles. 𝜋 = 𝑀𝑅𝑇 R: gas constant T: temperature 12  Electrolyte Solutions Electrolytes undergo dissociation when dissolved in water. The van’t Hoff factor (i) accounts for this effect. actual number of particles in solution after dissociation i number of formulas units initially dissolved in solution ∆Tf = iKfm ∆Tb = iKbm π = iMRT 13  Electrolyte Solutions The van’t Hoff factor (i) is 1 for all nonelectrolytes: H2O C12H22O11(s) C12H22O11(aq) For strong electrolytes i should be equal to the number of ions: NaCl(s) H2O Na+(aq) + Cl–(aq) Na2SO4(s) H2O 2Na+(aq) + SO42–(aq) 14  Electrolyte Solutions The van’t Hoff factor (i) is usually smaller than predicted due to the formation of ion pairs. An ion pair is made up of one or more cations and one or more anions held together by electrostatic forces. 15  Colloids Colloid: it is a dispersion of particles of one substance (the dispersed phase) throughout another substance or solution (the continuous phase) Will appear homogeneous, but they are larger particles of one substance dispersed throughout another substance A colloid is not the same as a true solution. A colloid has the ability to scatter light at a detectable level. All gases and liquids scatter light, however the scattering from a true solution or pure substance is quite small and typically not detectable. Dispersed particles are larger than normal molecules (~1×103 pm - 2×105 pm) 16  Tyndall Effect Tyndall effect: the scattering of light by colloidal-size particles Which vessel contains the true solution and which contains the colloid? 17  Characterization of Colloids Characterized according to the state of the dispersed phase and of the continuous phase Some examples of types of colloids: Aerosol: Liquid droplets or solid particles dispersed throughout a gas Foam: A gas dispersed in a liquid. 18  Water as the Continuous Phase Colloids with water as the continuous phase are either: Hydrophilic colloids Hydrophobic colloids Hydrophilic colloid: a colloid in which there is a strong attraction between the dispersed phase and the continuous phase Hydrophobic colloid: a colloid in which there is a lack of attraction between the dispersed phase and the continuous phase 19  Association colloids Micelle: a colloidal-sized particle formed in water by the association of molecules or ions that each have a hydrophobic end and a hydrophilic end Association colloid: a colloid in which the dispersed phase consists of micelles Example: Soap Hydrophilic ends are on the outside of the micelle facing the water Hydrophobic ends point inward toward one another 20 Chapter 12 CHEM 0120  Reaction Rates The reaction rate is the speed of a chemical reaction. It is the increase in molar concentration of product of a reaction per unit time (or the decrease in molar concentration of reactant per unit time). The rate of a reaction is a measure of how fast the reaction makes products or uses reactants. The ability to control the speed of a chemical reaction is important. 2  Chemical Kinetics The study of reaction rates, how reaction rates change under varying conditions, and what molecular events occur during the overall reaction. The rate of any given reaction can be affected by numerous variables: Concentrations of Reactants or Catalyst Temperature Nature of Reactants 3  Reactant Concentration Affect on Reaction Rate Generally, the greater the concentration of reactant molecules, the faster the reaction. This increases the frequency of reactant molecule contact. Concentration of gases depends on the partial pressure of the gas. Higher pressure = higher concentration Concentrations of solutions depend on the solute-to-solution ratio (molarity). 4  Temperature Affect on Reaction Rate Increasing temperature increases the reaction rate for most reactions. The Arrhenius equation is a mathematical relationship between the absolute temperature and the speed of a reaction. This relationship will be examined in the next class. 5  Nature of the Reactants Affect on Reaction rate Nature of the reactants refers to what kind of reactant molecules there are and what physical condition they are in. Small molecules tend to react faster than large molecules. Gases tend to react faster than liquids, which react faster than solids. Powdered solids are more reactive than “blocks.” More surface area for contact with other reactants Certain types of chemicals are more reactive than others. For example, potassium metal is more reactive than sodium. Ions react faster than molecules. No bonds need to be broken. 6  Defining Rate Rate is how much a quantity changes in a given period of time. The speed at which you walked to chemistry lecture is a rate: The distance you walked (feet or miles) in a given period of time (1 minute or 1 hour) So, the rate at which you walked has units of ft/min or mi/hr. We can further break it down into different periods of time for average and instantaneous rates. 7  Defining Reaction Rate The rate of a chemical reaction is measured in terms of how much the concentration of a reactant decreases (or the product concentration increases) in a given period of time. Example: H2(g) + I2(g) 2 HI(g) For reactants, a negative sign is placed in front of the definition. The rate of chemical reaction over two time periods can be written as follows: 8  Concentrations as a Function of Time As time goes on, the rate of a reaction generally slows down because the concentration of the reactants decreases. At some time the reaction stops, either because the reactants run out or because the system has reached equilibrium. 9  Average Rate The average rate is the change in measured concentrations in any particular time period. Linear approximation of a curve The larger the time interval, the more the average rate deviates from the instantaneous rate. 10  Instantaneous Rate The instantaneous rate is the change in concentration at any one particular time. Slope at one point of a curve The instantaneous rate is determined by taking the slope of a line tangent to the curve at that particular point. In calculus: first derivative of the function 11  Reaction Rate and Stoichiometry In most reactions, the coefficients of the balanced equation are not all the same. H2(g) + I2(g) 2 HI(g) For these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another. For the above reaction, for every 1 mol of H2 used, 1 mol of I2 will also be used and 2 mol of HI made. Therefore, the rate of change will be different. To be consistent, the change in the concentration of each substance is multiplied by 1/coefficient. 12  Measuring Reaction Rates To measure the reaction rate you need to be able to measure the concentration of at least one component in the mixture at many points in time. There are two ways of approaching this problem: – For reactions that are complete in less than 1 hour, it is best to use continuous monitoring of the concentration. – For reactions that happen over a very long time, sampling of the mixture at various times can be used. 13  Finding the Rate Law The rate law must be determined experimentally. The rate law shows how the rate of a reaction depends on the concentration of the reactants. Changing the initial concentration of a reactant will therefore affect the initial rate of the reaction. 14  The Effect of Concentration on Reaction Rate The rate law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants and homogeneous catalysts as well. Rate = k [A]n The rate of a reaction is directly proportional to the concentration of each reactant raised to a power. For the reaction aA + bB , the rate law would have the form given below. n and m are called the for each reactant. k is called the. 15  Dependence of Rate on Concentration Rate Law: an equation that relates the rate of a reaction to the concentration of reactants (and catalysts) raised to various powers. C aA + bB dD + eE Rate = k[A]m[B]n[C]p m, n and p are called the for each reactant. k is called the – a proportionality constant in the relationship between rate and concentrations. 16  Reaction Order The exponent on each reactant in the rate law is called the with respect to that reactant, as experimentally determined. The sum of the exponents on the reactants is called the. The rate law for the reaction 2 NO(g) + O2(g) 2 NO2(g) is as follows: Rate = k[NO]2[O2] The reaction is second order with respect to [NO]; first order with respect to [O2]. The reaction is third order overall. 17  Reaction Orders If a reaction is zero order, the rate of the reaction does not change. Doubling [A] will have no effect on the reaction rate. Rate law: Rate = k[A]0 = k If a reaction is first order, the rate is directly proportional to the reactant concentration. Doubling [A] will double the rate of the reaction. Rate law: Rate = k[A]1 = k[A] If a reaction is second order, the rate is directly proportional to the square of the reactant concentration. Doubling [A] will quadruple the rate of the reaction. Rate law: Rate = k[A]2 18  Determining the Reaction Order The reaction order must be determined from the experimental data. 19  Determining the Reaction Order The reaction order must be determined from the experimental data. Example: 2N2O5 2(g) + O2(g) 2O5] 2O5 Experiment 1 1.0 x 10-2 mol/L -6 mol Experiment 2 2.0 x 10-2 mol/L 9.6 x 10-6 mol 20 Determining the Rate Law with Multiple Reactants Changing each reactant will affect the overall rate of the reaction. By changing the initial concentration of one reactant at a time, the effect of each reactant’s concentration on the rate can be determined. Example: 21  Change of Concentration with Time products, the rate law depends on the concentrations of A and B. An integrated rate law uses calculus to transform a rate law into a mathematical relationship between concentration and time. 22  Integrated Rate Law: First-Order Rate law: Rate = k[A]1 = k[A] Integrated rate law: t kt 0 [ ] or = [ ] A graph of first order: ln[A] versus time results in a straight line k y-intercept = ln[A]initial 23  Integrated Rate Law: Second-Order Rate law: Rate = k[A]2 Integrated rate law: = + [ ] [ ] A graph of second order: 1/[A] versus time results in a straight line where slope = k y-intercept = 1/[A]initial 24  Integrated Rate Law: Zero-Order Rate law: Rate = k[A]0 = k Integrated rate law: [A]t kt + [A]0 A graph of zero order: [A] versus time results in a straight line k y-intercept = [A]initial 25 Rate Law Relationships 26  Half-Life of a Reaction - (t1/2) of a reaction: the time it takes for the reactant concentration to decrease to one-half of its initial value. Reactions:. First-order: / = Second-order: / = [ ] [ ] Zero-order: / = 27  Relationship between Order and Half-Life For a reaction, the lower the initial concentration of the reactants, the shorter the half-life. t1/2 = [A]0/2k For a reaction, the half-life is independent of the concentration. t1/2 = 0.693/k For a , the half-life is inversely proportional to the initial concentration. Increasing the initial concentration shortens the half-life. t1/2 = 1/(k[A]0) 28  Effect of Temperature on Rate Changing the temperature changes the rate constant of the rate law This relationship is given by the Arrhenius equation: / = where T is temperature in Kelvin R mol) Ea is the , the extra energy needed to start the molecules reacting A is a constant referred to as the frequency factor The rate the reactant energy approaches the activation energy 29  Potential-Energy Curve Exothermic reaction. Heat energy is released when the reaction proceeds in the forward direction. 30  Activation Energy There is an energy barrier to almost all reactions. The is the amount of energy needed to convert reactants into the activated complex. The activated complex is also know as a transition state. The activated complex is a chemical species with partially broken and partially formed bonds. Always very high in energy because of its partial bonds 31  Energy Profile for the Isomerization of Methyl Isonitrile The is the difference in energy between the reactants and the activated complex. The is the number of molecules that begin to form the activated complex in a given period of time. As the reaction begins, the C—N bond weakens enough for the CN triple bonded group to start to rotate. The is a chemical species with partial bonds. 32  The Exponential Factor in the Arrhenius equation The exponential factor in the Arrhenius equation is a number between 0 and 1 It represents the fraction of reactant molecules with sufficient energy so they can make it over the energy barrier. The higher the energy barrier (larger activation energy), the fewer molecules that have sufficient energy to overcome it. That extra energy comes from converting the kinetic energy of motion to potential energy in the molecule when the molecules collide. Increasing the temperature increases the average kinetic energy of the molecules. Therefore, increasing the temperature will increase the number of molecules with sufficient energy to overcome the energy barrier; reaction rate. 33  A Summary of Temperature and Reaction Rate The frequency factor is the number of times that the reactants approach the activation barrier per unit time. The exponential factor is the fraction of the approaches that are successful in surmounting the activation barrier and forming products. The exponential factor increases with increasing temperature but decreases with increasing activation energy. 34  Arrhenius Plots: Experimental Measurements The Arrhenius equation can be algebraically solved to give the following form: ln = ln This equation is in the form y = mx + b, – where y = ln(k) and x = (1/T). A graph of ln(k) versus (1/T) is a straight line 35  Arrhenius Equation: Two-Point Form If you have only two (T,k) data points, the following forms of the Arrhenius equation can be used: 1 1 ln = 36  Example The rate constant for the formation of hydrogen iodide from the elements is 2.7 x 10- L/(mol s) at - L/(mol s) at 650 K. A) Find the activation energy. B) Calculate the rate constant at 700 K. 37  Collision Theory Collision Theory of Reaction Rates: a theory that assumes that, for a reaction to occur, reactant molecules must collide with an energy greater than some minimum value and with the proper orientation. The minimum energy of collision required for two molecules to react is the activation energy (Ea). The rate constant (k) for a reaction is dependent upon: 1. The collision frequency 2. The fraction of collisions having a higher energy than Ea The fraction of collision that occur with the reactant molecules properly oriented. 38  Effective Collisions: Kinetic Energy Factor For a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide the activated complex can form. 39  Effective Collisions Collision frequency: the number of collisions that happen per second The more collisions there are per second, the more collisions can be effective and lead to product formation. The higher the frequency of effective collisions, the faster the reaction rate. Collisions in which these two conditions are met (and therefore lead to reaction) are called. When two molecules have an effective collision, a temporary, high- energy (unstable) chemical species is formed: the activated complex. 40  Orientation Factor The proper orientation results when the atoms are aligned in such a way that the old bonds can break and the new bonds can form. The more complex the reactant molecules, the less frequently they will collide with the proper orientation. 41  Reaction Mechanisms Chemical reactions are generally described with a net chemical equation listing all the reactant molecules and product molecules. The probability of more than three molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible. Most reactions occur in a series of small reactions involving one, two, or (at most) three molecules. Reaction Mechanism: the set of elementary reactions whose overall effect is given by the net chemical equation. Knowing the rate law of the reaction helps us understand the sequence of reactions in the mechanism. 42  Reaction Mechanism Example Overall reaction: H2(g) + 2 ICl(g 2 HCl(g) + I2(g) (net chemical equation) Mechanism: H2(g) + ICl(g) HCl(g) + HI(g) (elementary reaction) HI(g) + ICl(g) HCl(g) + I2(g) (elementary reaction) The reactions in this mechanism are elementary reactions, meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps. 43  Reaction Intermediates H2(g) + 2 ICl(g 2 HCl(g) + I2(g) H2(g) + ICl(g) HCl(g) + HI(g) HI(g) + ICl(g) HCl(g) + I2(g) Notice that the HI is a product in step 1, but it is a reactant in step 2. Because HI is made but then consumed, HI does not show up in the overall reaction. Materials that are products in an early mechanism step but reactants in a later step are called intermediates. Reaction intermediates: a species produced during a reaction that does not appear in the net equation because it reacts in a subsequent step in the mechanism. 44  Determining the Rate Equation for an Elementary Reaction The number of reactant particles in an elementary step is called its molecularity. A unimolecular step involves one particle. A bimolecular step involves two particles. These two particles may be the same kind of particle. A termolecular step involves three particles. These are exceedingly rare in elementary steps. 45  Rate Equation for an Elementary Reaction H2(g) + 2 ICl(g) (g) + I2(g) 1) H2(g) + (g) (g) + HI(g) Rate = 2) HI(g) + (g) (g) + I2(g) Rate = Treat each step in the mechanism like its own reaction with its own Ea and rate law. The rate law for an overall reaction must be determined experimentally. But the rate law of an elementary step can be deduced from the equation of the step. 46 Rate Laws for Elementary Steps (from Pearson Education, INC.) 47  Rate-Determining Step Rate-determining step: the slowest step in the reaction mechanism When on step is slower than the other steps, the result is that product production cannot occur any faster than the slowest step; the step determines the rate of the overall reaction. The slowest step has the largest activation energy. The rate law of the rate-determining step determines the rate law of the overall reaction. 48  Reaction Mechanism Example NO2(g) + CO(g) (g) + CO2(g) Rate = k[NO2]2 Proposed Mechanism 1. NO2(g) + NO2(g) (g) + NO(g) Rate = k1[NO2]2 Slow 2. NO (g) + CO(g) 2(g) + CO2(g) Rate = k2[NO ][CO] Fast The first step is slower than the second step because its activation energy is larger. The first step in this mechanism is the rate-determining step: It is the slowest step. The rate law of the first step is the same as the rate law of the overall reaction. 49  Criteria for Validating a Mechanism To validate (not prove) a mechanism, two conditions must be met: 1. The elementary steps must sum to the overall balanced chemical reaction. 2. The rate law predicted by the rate-determining step of the mechanism must be consistent with the experimentally observed rate law. 50  Mechanisms with an Initial Fast Step When a mechanism contains a fast initial step, the rate- limiting step may contain intermediates. When a previous step is rapid and reaches equilibrium, the forward and reverse reaction rates are equal, so the concentrations of reactants and products of the step are related and the product is an intermediate. Substituting into the rate law of the rate-determining step will produce a rate law in terms of just reactants. 51  Example of a Mechanism with an Initial Fast Step 1) 2 NO(g) N2O2(g) Fast 2) H2(g) + N2O2(g) 2O(g) + N2O(g) Slow Rate = k2[H2][N2O2] H2(g) + N2O(g) 2O(g) + N2(g) Fast 52  Catalysts Catalysts are substances that affect the rate of a reaction Mechanism without catalyst: without being consumed. O3(g) + O(g) 2(g) Catalysts work by providing an alternative mechanism for Mechanism with catalyst: the reaction with a lower activation energy. (g) + O3(g) O2(g) + (g (g) + O(g) 2(g g Catalysts are consumed in an early mechanism step and then made in a later step. 53  Factors Affecting Reaction Rate Catalysts are substances that affect the speed of a reaction without being consumed. Most catalysts are used to speed up a reaction; these are called positive catalysts. – Catalysts used to slow a reaction are called negative catalysts. Homogeneous catalysts: all species present in same phase Heterogeneous catalysts: species present in different phase 54  Catalyst Types are in the same phase as the reactant particles. Cl(g) in the destruction of O (g) are in a different phase than the reactant particles. Solid catalytic converter in a car’s exhaust system 55 Catalyst Types 56  Catalytic Hydrogenation Reaction 1. Adsorption: The reactants are adsorbed onto metal surfaces. 2. Diffusion: The reactants diffuse on the surface until they approach each other. Reaction: The reactants react to form the products. Desorption: The products desorb from the surface into the gas phase. 57 Chapter 13: Part 1 CHEM 0120 2  Arrow Conventions Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions. A single arrow indicates that all the reactant molecules are converted to product molecules at the end. A double arrow indicates that the reaction reaches equilibrium with reactant and products present. 3  Reaction Dynamics When a reaction starts, the reactants are consumed and products are made. The reactant concentrations decrease and the product concentrations increase. Eventually, the products can react to re-form some of the reactants, assuming that the products are not allowed to escape. Processes that proceed in both the forward and reverse directions are said to be reversible. Reactants ⇌ Products 4  Example Catalytic methanation: CO(g) + 3H2(g) → CH4(g) + H2O(g) Steam reforming: CH4(g) + H2O(g) → CO(g) + 3H2(g) Consists of a forward and reverse reaction; therefore we can write it as: CO g + 3H g ⇌ CH g + H O(g) The favorability of products or reactants depends upon the conditions of the reaction. 5  Chemical Equilibrium Chemical equilibrium: the state reached by a reaction mixture when the rates of forward and reverse reactions have become equal 6  Equilibrium notes At equilibrium, the rates of the forward and reverse reactions are equal. This does not mean the concentrations of reactants and products are equal. Some reactions reach equilibrium after almost all of the reactant molecules are consumed The position of equilibrium favors the products. Other reactions reach equilibrium when only a small amount of the reactant molecules are consumed The position of equilibrium favors the reactants. 7  Equilibrium Constant, Kc 𝑎A + 𝑏B ⇌ 𝑐C + 𝑑D [C] [D] 𝐾 = [A] [B] Kc is the value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted Law of mass action: a relation that states that the values of the equilibrium-constant expression Kc are constant for a particular reaction at a given temperature, whatever equilibrium concentrations are substituted 8  Writing Equilibrium Constant Expressions So, for the reaction 2 N2O5(g) ⇌ 4 NO2(g) + O2(g) Write the expression for the equilibrium constant, Kc 9  Kinetics of Equilibrium Dynamic equilibrium: consists of a forward reaction and a reverse reaction occurring at the same speed. Once at equilibrium, the forward and reverse reactions continue. 𝑎A ⇌ 𝑏B At equilibrium, kf[A]a = kr[B]b (rate of forward reaction = rate of reverse reaction) Kc can be identified as the ratio of rate constants for the forward and reverse reactions 𝑘 𝐾 = 𝑘 This ratio is just a comparison of the rate constants. The rate is dependent on the concentrations, coefficients and rate constants. 10  K for Reactions involving Gases The concentration of a gas in a mixture is proportional to its partial pressure. Kp: an equilibrium constant for a gaseous reaction in terms of partial pressures For 𝑎A(𝑔) + 𝑏B(𝑔) ⇌ 𝑐C(𝑔) + 𝑑D(𝑔), 𝑃 𝑃 𝐾 = 𝑃 𝑃 11  Relationship between Kc and Kp The relationship between Kc and Kp is: Kp = Kc(RT)Δn Δn is the sum of the coefficients of gaseous products minus the sum of the coefficients of gaseous reactants in the chemical equation 12  Example Nitrogen monoxide, a pollutant in automobile exhaust, is oxidized to nitrogen dioxide in the atmosphere. 2NO g + O g ⇌ 2NO (g) Kp = 2.2 x 1012 at 298 K Solve Kc for this reaction. 13  Equilibrium Constant for the Sum of Reactions Useful rule: If a given chemical equation can be obtained by taking the sum of other equations, the equilibrium constant for the given equation equals the product of the equilibrium constants of the added equations. For the reactions (1) 𝒂𝐀 ⇌ 𝒃𝐁 and (2) 𝒃𝐁 ⇌ 𝒄𝐂, the equilibrium constant expressions are as follows: [ ] [ ] 𝐾 = 𝐾 = [ ] [ ] For the reaction 𝒂𝐀 ⇌ 𝒄𝐂, the equilibrium constant expression is as follows: [𝐶] 𝐾 𝐾 =𝐾 = [𝐴] 14  Heterogeneous Equilibria Heterogeneous equilibrium: an equilibrium involving reactants and products in more than one phase. The concentrations of pure solids and pure liquids do not change during the course of a reaction. Because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression. 15  Example for Heterogeneous Equilibria Write the expression for Kc for the following reactions: CaCO3(s) ⇌ CaO(s) + CO2(g) H2O(l) ⇌ H2O(g) 16  Using the Equilibrium Constant Ways to use the equilibrium constant: Qualitatively interpret the equilibrium constant. Does the equilibrium favor reactants or products? To predict the direction of the reaction. To calculate the equilibrium concentrations. 17  Qualitatively Interpreting Kc 𝑎A + 𝑏B ⇌ 𝑐C + 𝑑D [C] [D] 𝐾 = [A] [B] When Kc >> 1, the equilibrium mixture is mostly products. When Kc Kc, the reaction will go to the left. Indicates that the reaction mixture has more products. If Qc < Kc, the reaction will go to the right. Indicates that the reaction mixture has more reactants. If Qc = Kc, the reaction mixture is at equilibrium. 20  Reaction Quotient, Qc If Qc > Kc, the reaction will go to the left. If Qc < Kc, the reaction will go to the right. If Qc = Kc, the reaction mixture is at equilibrium. 21 Chapter 13 Part 2 CHEM 0120  Calculating Equilibrium Concentrations Steps to follow when solving for equilibrium concentrations: 1) Set up an ICE table with expressions in x Initial concentration or Starting concentration Change in concentration to reach equilibrium Equilibrium concentration 2) Substitute the expressions in x for equilibrium concentrations into the equilibrium-constant equation. 3) Solve the equilibrium-constant equation for the values of the equilibrium concentrations. 2  Change in Reaction Conditions Ways to alter the equilibrium composition of a gaseous reaction mixture: 1) Change the concentrations by adding reactants to the reaction vessel or by removing products from the reaction vessel. 2) Change the partial pressure of gaseous substances by changing the volume. 3) Change the temperature. 3 Removing Products or Adding Reactants Apply Le Chatelier’s principle When a system in chemical equilibrium is disturbed by a change of temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change of variable. Example: CO g + 3H g CH g + H O(g) At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same. 4  Effect of Concentration Changes on Equilibria The net reaction occurs left to right (the reaction occurs in the forward direction) to give a new equilibrium when reactant is added or product is removed from an equilibrium mixture. More products are produced. The net reaction occurs right to left (reverse direction) to give a new equilibrium when more product is added or reactant is removed from an equilibrium mixture. More reactants are produced. 5  Effect of Pressure Change If the volume of the container is decreased, the concentration of the gases increases. Partial pressure of the gases increases, which causes an increase of the total pressure in the container. According to Le Chatelier’s principle, the equilibrium will shift to remove that pressure. If the pressure is increased by decreasing the volume of a reaction mixture, the equilibrium shifts to the side with fewer gas molecules. 6 Chemical equilibrium change with pressure change CO g + 3H g CH g + H O(g) Comparison of gases: 1) At equilibrium (consider the amounts in the earlier example) 2) After compression 3) Compressed and at equilibrium (shifts equilibrium to the right) 7  Effect of Temperature: Endothermic Endothermic reactions absorb energy. Heat is “behaving” as a reactant. In an endothermic chemical reaction, heat is a reactant. Increasing the temperature causes an endothermic reaction to shift right (in the direction of the products); the equilibrium constant increases. Decreasing the temperature causes an endothermic reaction to shift left (in the direction of the reactants); the equilibrium constant decreases. 8  Effect of Temperature: Exothermic Exothermic reactions release energy. Heat is “acting” as a product. In an exothermic chemical reaction, heat is a product. Increasing the temperature causes an exothermic reaction to shift left (in the direction of the reactants); the value of the equilibrium constant decreases. Decreasing the temperature causes an exothermic reaction to shift right (in the direction of the products); the value of the equilibrium constant increases. 9 Chapter 14: Part 1 Acids and Bases Table of Common Acids 2 Table of Common Bases 3  Common Strong Acids and Bases There are three different concepts of acids and bases: Arrhenius Bronsted-Lowry Lewis 4  Definitions of Acids and Bases Arrhenius definition (simplest and most restrictive) Acids are substances that when dissolved in water produce a hydronium, H3O+ (hydrogen ion H+). Bases are substances that when dissolved in water produce a hydroxide ion, OH–. Brønsted–Lowry definition (based on reactions in water) Acids are substances that when dissolved in water donate hydronium, H3O+ (hydrogen ion H+). Bases are substances that accept a hydronium, H3O+ (hydrogen ion H+). Lewis definition (most expansive and based on electron donors and acceptors) Acids are substances that accept or need an electron pair. Bases are substances that donate an electron pair to another substance. 5  Arrhenius Concept Acids produce H+ ions in aqueous solution. HCl(aq) → H+(aq) + Cl−(aq) 6  Arrhenius Concept Bases produce OH− ions in aqueous solution. NaOH(aq) → Na+(aq) + OH−(aq) 7  Arrhenius Acid-Base Reactions The H+ from the acid combines with the OH− from the base to make a molecule of H2O. Think of H2O as H—OH. The cation from the base combines with the anion from the acid to make a salt. acid + base → salt + water Example: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) (acid) (base) (salt) (water) 8  Brønsted–Lowry Concept of Acids and Bases Viewed as proton (H+)-transfer reactions Acid: the species donating a proton in a proton-transfer reaction Base: the species accepting the proton in a proton-transfer reaction Any reaction involving an H+ (proton) that transfers from one molecule to another is an acid–base reaction, regardless of whether it occurs in aqueous solution or if there is OH− present. All reactions that fit the Arrhenius definition also fit the Brønsted–Lowry definition. 9  Brønsted–Lowry Theory The acid is an H+ donor. The base is an H+ acceptor. Base structure must contain an atom with an unshared pair of electrons. In a Brønsted–Lowry acid–base reaction, the acid molecule donates an H+ to the base molecule. H—A + :B → :A– + H—B+ (acid) (base) (conjugate) (conjugate) base acid 10  Conjugate acid-base pairs Consists of two species in an acid-base reaction, one acid and one base, that differ by the loss or gain of a proton. Each reactant and the product it becomes is called a conjugate acid-base pair. 11  Brønsted–Lowry Acids Brønsted–Lowry acids are H+ donors. Any material that has H can potentially be a Brønsted–Lowry acid. Because of the molecular structure, often one H in the molecule is easier to transfer than others. When HCl dissolves in water, the HCl is the acid because HCl transfers an H+ to H2O, forming H3O+ ions. Water acts as base, accepting H+. HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) (acid) (base) (conjugate) (conjugate) base acid 12  Brønsted–Lowry Bases Brønsted–Lowry bases are H+ acceptors. Any material that has atoms with lone pairs can potentially be a Brønsted–Lowry base. Because of the molecular structure, often one atom in the molecule is more willing to accept H+ transfer than others. When NH3 dissolves in water, the NH3(aq) is the base because NH3 accepts an H+ from H2O, forming OH–(aq). Water acts as acid, donating H+. NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq) (base) (acid) (conjugate) (conjugate) acid base 13  Brønsted–Lowry Concept Summary A base is a species that accepts protons. OH- is only one example of a base. Acids and bases can be molecular substances or ions. Acid-base reactions are not restricted to aqueous solutions. Some species are amphiprotic, the species can act as either an acid or a base, depending on what the other reactant is. 14  Lewis Concept of Acids and Bases Lewis Acid: a species that can form a covalent bond by accepting an electron pair from another species They are electron deficient Lewis Base: a species that can form a covalent bond by donating an electron pair to another species Must have a lone pair of electrons on it that it can donate 15  Strong Acid/Base vs Weak Acid/Base A strong acid is a strong electrolyte. Complete or near complete ionization of acid molecule in water A weak acid is a weak electrolyte. Only a small percentage or partial ionization of the acid molecule in water A strong base is a strong electrolyte. Complete or near complete ionization of base molecule in water Produces OH– ions, either through dissociation or reaction with water A weak base is a weak electrolyte. Only a small percentage or partial ionization of the base molecule in water Produces OH– ions, either through dissociation or reaction with water 16 Relative Strengths of Acids and Bases The strongest acids have the weakest conjugate bases. The strongest bases have the weakest conjugate acids. Reactions proceed in the direction of the weaker acid or weaker base. 17  Molecular Structure and Acid Strength Binary acids (H—Y) have acidic hydrogens attached to a nonmetal atom. – Example: HCl and HF Going across a row of elements, the electronegativity increases, the more polarized the bond (δ+H—Xδ−), the more acidic the bond. Going down a column of elements, the size of atom X increases, the H—X bond strength decreases, the acid strength increases. Binary acid strength increases to the right across a period. Acidity: H—C < H—N < H—O < H—F Binary acid strength increases down the column. Acidity: H—F < H—Cl < H—Br < H—I 18  Strengths of Oxoacids, H-O-Y- For a series of oxoacids of the same structure, differing only in atom Y, the acid strength increases with the electronegativity of Y HIO < HBrO < HClO For a series of oxoacids, (HO)mYOn, the acid strength increases with n, the number of O atoms bonded to Y HClO < HClO2 < HClO3 < HClO4 19  Autoionization of Water Pure water is considered a nonelectrolyte, a nonconductor of electricity Measurements show a very small conduction Results from autoionization A reaction in which two like molecules react to give ions H O 𝑙 + H O 𝑙 ⇌ H O 𝑎𝑞 + OH (𝑎𝑞) All aqueous solutions contain both H3O+ and OH–. The concentrations of H3O+ and OH– are equal in water. [H3O+] = [OH–] = 10−7 M at 25 °C 20  Equilibrium Constant Expressions Write an equilibrium constant expression for: H O 𝑙 + H O 𝑙 ⇌ H O 𝑎𝑞 + OH 𝑎𝑞 21  Ion-Product constant for Water, Kw Kw: the equilibrium value of the ion product [H3O+][OH –] H O 𝑙 + H O 𝑙 ⇌ H O 𝑎𝑞 + OH (𝑎𝑞) Kw = [H3O+][OH –] = 1.0 ×10-14 at 25 °C The product of the H3O+ and OH– concentrations is always the same number at room temperature, 1 × 10–14. If acid or base is added to water, [H3O+] and [OH –] will no longer be equal, but we can still use Kw = [H3O+][OH –] For example, if [H3O+] increases, the [OH–] must decrease so that the product stays constant. Inversely proportional 22  Solutions of a Strong Acid Strong acids donate practically all their hydrogen atoms. – Near 100% ionization of acid molecule occurs in water – Strong electrolyte Example: Calculate the hydroxide-ion concentration in 0.10 M HCl. (Hint: The autoionization equilibrium still exists.) 23  Example 15.4 Calculate the concentrations of hydronium ion and hydroxide ion at 25 °C in: a) 0.15 M HNO3 b) 0.010 M Ca(OH)2 24  Solutions of a Strong Acid or Base All aqueous solutions contain both H3O+ and OH– ions. Neutral solutions have equal [H3O+] and [OH–]. [H3O+] = [OH–] = 1.00 × 10−7 Acidic solutions have a larger [H3O+] than [OH–]. [H3O+] > 1.00 × 10−7; [OH–] < 1.00 × 10−7 Basic solutions have a larger [OH–] than [H3O+]. [H3O+] < 1.00 × 10−7; [OH–] > 1.00 × 10−7 25  The pH of a Solution The acidity of basicity of a solution depends on the hydronium-ion concentration and is expressed as pH. pH: the negative of the base 10 logarithm of the molar hydronium-ion concentration pH = – log[H3O+] Example: A solution has a hydronium-ion concentration of 1.0 x 10-3 M. What is the pH? 26  The pH scale pH < 7 is acidic, pH > 7 is basic; pH = 7 is neutral Normal range of pH is 0 to 14 27  What about the hydroxide-ion? Can calculate the pOH (a measure of hydroxide-ion concentration) similar to pH. pOH = Since Kw = [H3O+][OH –] = 1.0 ×10-14 at 25 °C, we can determine that pH + pOH = 28  Strong Acid and Strong Base Solutions There are two sources of H3O+ in an aqueous solution of a strong acid—the acid and the water. There are two sources of OH− in an aqueous solution of a strong acid—the base and the water. For a strong acid or base, the contribution of the water to the total [H3O+] or [OH−] is negligible. The [H3O+]acid is so large that [H3O+]water is too small to be significant. Except in very dilute solutions, generally < 1 × 10−4 M 29  Finding the pH of a Strong Acid There are six strong acids: Five are monoprotic and one is diprotic. –Monoprotic: HCl, HBr, HI, HClO4, and HNO3 –Diprotic: H2SO4 For a monoprotic strong acid, the acid concentration equals the hydronium concentration. –[H3O+] = [HAcid] –Example: 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00 30  Chapter 14 Part 2: Acid-Base Equilibria  Solutions of a Weak Acid Weak acids donate a small fraction (partial ionization) of their hydrogen atoms. Weak acid molecules do not donate much of their hydrogens to water. Much less than 1% ionized in water [H3O+] [acid]. Buffers that need to work mainly with added base generally have [acid] > [base]. 64  Titration In an acid–base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete. When the reaction is complete, we have reached the endpoint of the titration. An indicator may be added to determine the endpoint. An indicator is a chemical that changes color when the pH changes. When the moles of H3O+ = moles of OH−, the titration has reached its equivalence point. 65 Titration set-up for an acid-base reaction 66  Titration Curve It is a plot of pH versus the amount of added titrant. The inflection point of the curve is the equivalence point of the titration. Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH. The pH of the equivalence point depends on the pH of the salt solution. Equivalence point of neutral salt: pH = 7 Equivalence point of acidic salt: pH < 7 Equivalence point of basic salt: pH > 7 Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH. 67 Titration Curve Comparison Strong Acid versus Strong Base Weak Base versus Strong Acid Weak Acid versus 68 Strong Base Chapter 15: Equilibria  Solubility Equilibria All ionic compounds dissolve in water to some degree. However, many compounds have such low solubility in water that they are classified as insoluble. The concepts of equilibrium can be applied to salts dissolving, and using the equilibrium constant Ksp, their relative solubilities in water can be determined. 2  Solubility Product Constant (Ksp) The equilibrium constant for the solubility equilibrium of a slightly soluble (or nearly insoluble) ionic compound. The equilibrium constant for the dissociation of a solid salt into its aqueous ions. A dissociation reaction is written for the slightly soluble ionic compound: M X 𝑠 ⇌ 𝑝M 𝑎𝑞 + 𝑞X 𝑎𝑞 Solubility product: 𝐾 = [𝑀 ] [𝑋 ] 3 Table of Solubility Products: Ksp 4  Example Write the solubility product expressions for the following salts: AgCl Pb3(AsO4)2 5  Molar Solubility (s) Molar solubility: The moles of solute that will dissolve in a liter of solution. The molarity of the dissolved solute in a saturated solution. Molar solubility is related to Ksp. Ksp values of compounds cannot always be compared. For Ksp values to be compared, the compounds must have the same dissociation stoichiometry. 6  Example A liter of a solution saturated at 298 K with calcium oxalate, CaC2O4, is evaporated to dryness, giving a 0.0061-g residue of CaC2O4. Calculate the molar solubility of CaC2O4. 7  Example By experiment, it is found that 1.2 x 10-3 mol of lead(II) iodide, PbI2, dissolves in 1 L of aqueous solution at 298 K. What is the solubility product constant at this temperature? 8  Solubility and the Common-Ion Effect Addition of a soluble salt that contains one of the ions of the slightly soluble salt decreases the solubility of the slightly soluble salt. For example: The addition of Pb(NO3)2 to the solubility equilibrium of solid PbCrO4 decreases the solubility of PbCrO4. PbCrO4(s) ⇌ Pb2+(aq) + CrO42−(aq) Addition of Pb2+ shifts the equilibrium to the left. 9  Example What is the molar solubility of calcium oxalate in 0.15 M calcium chloride? The solubility product for calcium oxalate is 2.3 x 10-9. 10  Precipitation Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound. By comparing the ion product, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur. If Q = Ksp, the solution is saturated and no precipitation will occur. If Q < Ksp, the solution is unsaturated and no precipitation will occur. If Q > Ksp, the solution would be above saturation. The salt above saturation will precipitate. Note: Q for a solubility reaction is called the ion product instead of reaction quotient since it is the product of ion concentrations in a solution, each concentration raised to a power equal to the number of ions in the formula of the ionic compound. 11  Fractional Precipitation The technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth. A successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different. A solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions but not the others. (Selective precipitation) 12  Effect of pH on Solubility For insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide. And the lower the pH, the higher the solubility. Higher pH = increased [OH−] M(OH)n(s) ⇌ Mn+(aq) + nOH−(aq) For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility. M2(CO3)n(s) ⇌ 2 Mn+(aq) + nCO32−(aq) H3O+(aq) + CO32−(aq) ⇌ HCO3−(aq) + H2O(l) 13 Chapter 16 CHEM 0120  Thermodynamics Thermodynamics: the study of the relationship between heat and other forms of energy involved in a chemical or physical process Used to predict whether a process will occur under the specified conditions. Processes that will occur are spontaneous. Nonspontaneous processes require the application of an external force. 2  First Law of Thermodynamics The change in internal energy of a system, U, equals heat plus work (q + w). Total energy must be conserved. Energy is neither created nor destroyed, only transferred and transformed. = + Internal energy: Sum of potential and kinetic energy of all particles that compose the system. It is a state function (it is not affected by the path). 3  Pressure-Volume Work Pressure is a force that acts against the sides of a container. If a part of that container can move, then work can be done w = -P V Pressure-volume work – volume change against an external pressure  Enthalpy Enthalpy – the heat exchanged in a reaction under constant pressure Constant pressure generally means an open system It is a state function Enthalpy can also be defined as the internal energy plus the product of pressure and volume H = U + PV If we want to look at the change in enthalpy: H= U+P V  Enthalpy The value of H is the amount of heat absorbed or released by a reaction under constant pressure If the value of H is positive, then the system has absorbed energy from the surroundings, and the reaction is endothermic If the value of H is negative, then the system has given off energy to the surroundings (heat has evolved), and the reaction is exothermic  Spontaneity Determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction. If the system after the reaction has less potential energy than the system before the reaction, the reaction is thermodynamically favorable. 7  Spontaneous vs Reversible Processes A spontaneous process is irreversible because there is a net release of energy when it proceeds in that direction. Proceeds in one direction. If one direction is spontaneous, the opposite direction must be nonspontaneous. A reversible process will proceed back and forth between the two end conditions. The result is no change in free energy. 8  Spontaneous Processes Spontaneous processes occur because they release energy from the system. Most spontaneous processes proceed from a system of higher potential energy to a system at lower potential energy. Exothermic But there are some spontaneous processes that proceed from a system of lower potential energy to a system at higher potential energy. Endothermic 9  Determining Whether a Reaction is Spontaneous. Two factors determine whether a reaction is spontaneous: Enthalpy change and Entropy change of the system. The H, is the difference between the sum of the internal energy and PV work energy of the reactants and that of the products. The entropy change, S, is the difference between the randomness of the reactants and that of the products. 10  Entropy Entropy (S): a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that a system can contain energy It is a state function. In spontaneous processes, the entropy of the system plus its surroundings increases. *Entropy change entirely in the system for the flasks. 11  Entropy A state function that is a measure of the matter and/or energy dispersal within a system, determined by the number of system microstates; often described as a measure of the disorder of the system 12  Changes in Entropy ( S) S = Sfinal Sinitial Entropy change is favorable when the result is a more random system. S is positive. Changes that increase the entropy are as follows: Reactions whose products are in a more random state Solid more ordered than liquid; liquid more ordered than gas Reactions that have larger numbers of product molecules than reactant molecules Increase in temperature Solids dissociating into ions upon dissolving 13  Example of the Change in Entropy Ssolid < Sliquid < Sgas Calculate S for the melting of ice. Entropy of 1 mol of ice = 41 J/K Entropy of 1 mol of liquid water = 63 J/K H2O(s 2O(l) 14  Second Law of Thermodynamics The total entropy of a system and its surroundings always increases for a spontaneous process. Energy can be neither created nor destroyed during a spontaneous process, but the energy is dispersed, so entropy is produced. = entropy created + Second Law restated: For a spontaneous process at a given temperature (T), the change in entropy of the system is greater than the heat divided by the absolute temperature, q/T > 15  Entropy Change for a Phase Transition At equilibrium, entropy change results from the absorption of heat. = Repeat the calculation for change in entropy for the melting of ice. fus = 6.0 kJ / 1 mol of ice 16  Example #1 The enthalpy change when liquid methanol, CH3OH, vaporizes at 25 is 38.0 kJ/mol. What is the entropy change when 1.00 mol of vapor in equilibrium with liquid condenses to liquid at 25 ? 17  Standard States The state of a material at a defined set of conditions. Represented by a superscript degree sign on the symbol of the quantity For gases: pure gas at exactly 1 atm pressure For pure liquids and solids: 1 atm pressure Temperature is usually 298 K For solutions: 1 M concentration 18  Third Law of Thermodynamics A substance that is perfectly crystalline at 0 K has an entropy of zero. Therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy. Therefore, the absolute entropy of substances is always positive. 19  Standard Absolute Entropies, S° S° designates standard state conditions. Entropy is an extensive physical property of matter. Entropies are for 1 mol of a substance at 298 K for a particular state, a particular allotrope, a particular molecular complexity, a particular molar mass, and a particular degree of dissolution. 20  Relative Standard Entropies: States The gas state has a larger entropy than the liquid state at a particular temperature. The liquid state has a larger entropy than the solid state at a particular temperature. 21 Standard Molar Entropy Values 22  Relative Standard Entropies: Molar Mass The larger the molar mass, the larger the entropy. Available energy states are more closely spaced, allowing more dispersal of energy through the states. 23  Relative Standard Entropies: Allotropes The less constrained the structure of an allotrope is, the larger its entropy. The fact that the layers in graphite are not bonded together makes graphite less constrained. 24  Relative Standard Entropies: Dissolution Dissolved solids generally have larger entropy, distributing particles throughout the mixture. 25  Relative Standard Entropies: Molecular Complexity Larger, more complex molecules generally have larger entropy. More energy states are available, allowing more dispersal of energy through the states. 26  Free Energy (G) Free energy: a thermodynamic quantity defined by the equation G = H – TS Gives a direct relationship to the spontaneity of a reaction. = S is positive when spontaneous, so G must be negative. Standard free-energy change = 27  Standard Free Energy of Formation, Gf° Gf°: the free-energy change that occurs when 1 mole of substance is formed from its elements in their reference forms at 1 atmosphere and at a specified temperature For elements in their most stable states, the value is zero. = 28  Spontaneity of a Reaction using G° Useful rules in judging the spontaneity of a reaction: If G°is a large negative number, the reaction is spontaneous Reactants transform almost entirely to products at equilibrium. If G°is a large positive number, the reaction is nonspontaneous Reactants do not give significant amounts of products at equilibrium. When G°has a small negative or positive value, the reaction gives an equilibrium mixture with significant amounts of both reactants and products. 29  Why is it “free” energy? The free energy is the maximum amount of energy released from a system that is available to do work on the surroundings. For many exothermic reactions, some of the heat released as a result of the enthalpy change goes into increasing the entropy of the surroundings, so it is not available to do work. And even some of this free energy is generally lost to heating up the surroundings. 30  Free Energy Change During a Reaction A decrease in free energy can show up as work done (to give maximum work), but also appears as an increase in entropy. 31  Gibbs Free Energy, G A process will be spontaneous when G is negative. G will be negative under the following conditions: H is negative and S is positive. Exothermic and more random H is negative and large and S is negative but small. H is positive but small and S is positive and large. Or high temperature G will be positive under the following conditions: H is positive and S is negative. Never spontaneous at any temperature When G = 0, the reaction is at equilibrium. 32  Relating G° to the equilibrium constant, K G = G° only when the reactants and products are in their standard states. Their normal state at that temperature Partial pressure of gas = 1 atm Concentration = 1 M Under nonstandard conditions, G = G° + RT ln Q. Q is the reaction quotient. At equilibrium, G = 0. G° RT ln K 33  Thermodynamic Equilibrium Constant, K K is the equilibrium constant in which: The concentrations of gases are expressed in partial pressures in atmospheres; The concentrations of solutes in liquid solutions are expressed in molarities. 34  Spontaneity using G° and K Since G = 0 at equilibrium, = ln When K > 1, G° is negative and the reaction is spontaneous in the forward direction under standard conditions. When K < 1, G° is positive and the reaction is nonspontaneous as written. 35  Example #2 What is the standard free-energy change G°at 25 for the following reaction? (What information is required?) H2(g) + Br2(l HBr(g) What is the value of the thermodynamic equilibrium constant K? 36  Why is K temperature dependent? Combining these two equations, G° = H° – T S° and G° = – RT ln(K) it can be shown that – H°rxn 1 S°rxn ln(K) = + R T R This equation is in the form y = mx + b. The graph of ln(K) versus inverse T is a straight line. 37  Changes of Free Energy with Temperature Assume H°and S°are constant with respect to temperature; = 38  Example #3 Sodium carbonate can be prepared by heating sodium hydrogen carbonate. 2 NaHCO3(s 2CO3(s) + H2O(g) + CO2(g) Estimate the temperature at which NaHCO3 decomposes to products at 1 atm. (What info is required?) 39  Chapter 17: Electrochemistry  Electrochemistry Electrochemistry is the study of redox reactions that produce or require an electric current. The conversion between chemical energy and electrical energy is carried out in an electrochemical cell. Spontaneous redox reactions take place in a galvanic/voltaic cell. Nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy. 2  Oxidation-Reduction Reactions in which electrons are transferred from one atom to another are called oxidation–reduction reactions. Atoms that lose electrons are being oxidized; atoms that gain electrons are being reduced. 3  Oxidation-Reduction Oxidation is the process that occurs when the oxidation number of an element increases; an element loses electrons; a compound adds oxygen; a compound loses hydrogen; or a half-reaction has electrons as products. Reduction is the process that occurs when the oxidation number of an element decreases; an element gains electrons; a compound loses oxygen; a compound gains hydrogen; or a half-reaction has electrons as reactants. 4 A Spontaneous Redox Reaction: Zn(s) + Cu2+(aq) Zn2+ + Cu(s)  Voltaic (Galvanic) Cells: Spontaneous Redox Reactions Electrical current: The amount of electric charge that passes a point in a given period of time Whether as electrons flowing through a wire or as ions flowing through a solution Redox reactions involve the movement of electrons from one substance to another. Therefore, redox reactions have the potential to generate an electric current. A spontaneous redox reaction does not require external energy to proceed. The reaction’s G is negative. Voltaic (galvanic) cells produce an electrical current from spontaneous redox reactions. To use that current, we need to separate the place where oxidation is occurring from the place where reduction is occurring.  Electrochemical Cells Oxidation and reduction half-reactions are kept as separate in half-cells in an electrochemical cell. To constitute an electrical circuit: Electron flow through a wire along with Ions (electrolyte) flowing through a solution via the salt bridge. The flow of electrons require a conductive solid electrode to allow the transfer of electrons either through: An external circuit or Metal or graphite electrode An electrochemical cell requires the exchange of ions between the two half-cells of the system via a salt bridge.  Voltaic Cell The salt bridge is required to complete the circuit and maintain charge balance.  Voltage and Current Voltage is the difference in potential Current is the number of electrons that energy between the reactants and flow through the system per second. products. It is also called the potential Unit = ampere difference. Unit = volt 1 A of current = 1 coulomb of charge flowing each second 1 V = 1 J of energy per coulomb of 1 A = 6.242 × 1018 electrons per second charge The voltage needed to drive Electrode surface area dictates the electrons through the external number of electrons that can flow. circuit Larger batteries produce larger currents. The amount of force pushing the electrons through the wire is called the electromotive force, emf.  Cell Potential The difference in potential energy between the anode and the cathode in a voltaic cell is called the cell potential. The cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode. The cell potential under standard conditions is called the standard emf, E°cell. 25 °C, 1 atm for gases, 1 M concentration of solution Sum of the cell potentials for the half-reactions  Cell Notation Shorthand description of a voltaic cell is written as follows: electrode | electrolyte || electrolyte | electrode Oxidation half-cell on the left; reduction half-cell on the right Single line | = phase barrier If multiple electrolytes in same phase, a comma is used rather than | Often use an inert electrode Double line || = salt bridge Example:  Voltaic Cell Anode = Zn(s) Cathode = Cu(s) The anode is oxidized to Cu2+ ions are reduced at Zn2+ ions. the cathode.  Electrodes When the half-reaction involves a gas, an inert electrode can be used. An inert electrode, such as platinum, is one that does not participate in the reaction but provides a surface for the transfer of electrons to take place on. Hydrogen electrode written as a cathode: H+(aq) | H2(g) | Pt Hydrogen electrode written as an anode: Pt | H2(g) | H+(aq) 13  Electrochemical Cell Notation with an Inert Electrode The half-reaction involves reducing Mn oxidation state from +7 to +2. An inert electrode (Pt) will provide a surface for the electron transfer without reacting with the MnO4. Fe(s) | Fe2+(aq) || MnO4– (aq), Mn2+(aq), H+(aq) | Pt(s) 14  Cell potential A measure of the driving force of the cell reaction. Work is needed to move electrons in a wire or to move ions through a solution to an electrode. With electricity, an electric charge moves from a point at high electric potential to a point at low electric potential. Electrical work expended in moving a charge through a conductor: Electrical work = charge × potential difference Faraday constant (F): the magnitude of charge on one mole of electrons; 9.6485 ×104 C/mol e- 15  Cell potential Cell potential (Ecell): The maximum potential difference between the electrodes of a voltaic cell. Measured by an electronic digital voltmeter. Maximum work obtainable from a voltaic cell: = n: number of moles of electrons transferred in the overall cell equation F: Faraday constant 16  Standard Reduction Potential The absolute tendency of a half-reaction standard reduction potential cannot be measured. Only the potentials relative to another half-reaction can be measured. To overcome this limitation, a standard half-reaction for the reduction of H+ to H2 is selected and assigned a potential difference of 0 V. Standard hydrogen electrode, SHE 17 18  Half-cell Potentials Standard reduction potentials compare the tendency for a particular reduction half-reaction to occur relative to the reduction of H+ to H2. Half-reactions with a stronger tendency toward reduction than the SHE have a positive value for E°red. Half-reactions with a stronger tendency toward oxidation than the SHE have a negative value for E°red. For an oxidation half-reaction, E°oxidation E°reduction. 19 Table of Standard Reduction Potentials (from Pearson Education, Inc.) A redox reaction will be spontaneous when there is a strong tendency for the oxidizing agent to be reduced and the reducing agent to be oxidized. Higher on the table of standard reduction potentials = stronger tendency for the reactant to be reduced Lower on the table of standard reduction potentials = stronger tendency for the product to be oxidized  Calculating Cell Potentials under Standard Conditions Cell potentials are intensive properties of matter. Because cell potentials are intensive physical properties, when determining the cell potential, do not multiply the half-cell E° values, even if you need to multiply the half-reactions to balance the redox equation. The cell potential of an electrochemical cell (Eocell) is the difference between the electrode pote