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Chemistry for Biology Chapter 3 Chemical Equilibrium Btec , Biology dept, Guangdong institute of Education 2009, Bio-department Thermodynamics: Directionality of Chemical Reactions Josian W. Gibbs 1839-1903. Pioneered concepts of chemical thermodynamics and free energy. Ludwig Boltzmann 1844-1906. Famous for his equation statistically defining entropy. So far, we have tried to answer the following questions: (1) What are the energetics (heat) of a reaction? Is it exothermic (H= -) or endothermic (H= +)? (2) How fast (kinetics) and how (mechanism) does the reaction go? (3) To what extent does it go? (equilibrium) And finally now …….. (4) Does it go, i.e., is it spontaneous? This is the subject of this chapter. Spontaneous processes: defined descriptively as a process that occurs by itself (and the reverse does not occur by itself) hot heat cold The opposite: is spontaneous, hot cold heat is not spontaneous, but, it is possible (how does a refrigerator work?). Other spontaneous processes (ask yourself: does reverse ever occur by itself?) nail rusting eggs breaking (Humpty Dumpty) paper burning water freezing at -10oC ice melting at +10oC gases mix All of these spontaneous processes are also described as: irreversible Irreversibility ═ Spontaneity There are reversible processes, but the systems must be at equilibrium. heat + ice → water at +10oC Spontaneous, irreversible heat + ice ← water at -10oC Spontaneous, irreversible heat + ice water at 0oC Reversible; equilibrium Both ice and water coexist at 0oC Either process, → or ← can occur at equilibrium What makes a process spontaneous (irreversible)? G = H - TS Free energy Enthalpy Entropy A G<0 B Exothermic reactions tend to be spontaneous (exception, dissolving ammonium salts), and increasing entropy (randomness) tends to cause processes to be spontaneous; but overall Gibbs Free Energy must decrease in order for a process to be spontaneous. examples: Ice melting – similar to gas expansion – more randomness and disorder, even though process is endothermic Ink drops in water – ink becomes evenly distributed in water; increase in randomness and disorder. Decay of biological organisms – increase in randomness and disorder. Dissolving of salts in water – increase in randomness and disorder To Summarize: what contributes to spontaneity? 1. Exothermic processes (heat is evolved). 2. Any process which increases randomness and disorder. The thermodynamic quantity which describes randomness and disorder is called ENTROPY and denoted as S The SECOND LAW OF THERMODYNAMICS postulates the existence of entropy; it also states that the entropy of the universe is constantly increasing. It is not a conserved quantity. 1. Gases have more entropy than liquids, which have more entropy than solids. 2. Corollary: Melting, or vaporization, increases entropy. 3. Corollary: In a chemical rx., increasing the number of moles of a gas, increases the entropy (e.g., H2O(g) H2(g) + ½O2(g)). 4. Dissolving or mixing increases entropy. 5. Corollary: precipitation decreases entropy. 6. Increasing the temperature increases entropy. Thermodynamic Calculations What is ΔG for the oxidation of SO2 to SO3 at 25°C? Is the reaction spontaneous? exothermic? SO2(g) + ½O2(g) SO3(g) ΔHf° -296.8 0 -395.2 ΔS° 0.2485 ½(0.205) 0.2562 ΔH = -98.4 kJ ΔS = -0.0948 kJ ΔG = ΔH – TΔS = -98.4 – (298)(-0.0948) = -98.4 + 28.2 = -70.15 kJ The reaction is spontaneous. The reaction is exothermic. Note: Be sure you convert ΔS values from J to kJ What is ΔG for the decarboxylation of limestone at 25°C? Is the reaction spontaneous? exothermic? CaCO3(s) CaO(s) + CO2(g) ΔHf° -1207.1 -635.5 -393.5 ΔH = +178.1 kJ ΔS° 0.0929 0.0398 0.2136 ΔS = +0.1605 kJ ΔG = ΔH – TΔS = +178.1 – (298)(+0.1605) = + 178.1 - 47.83 = +130.27 The reaction is not spontaneous. The reaction is endothermic. Note: Be sure you convert ΔS values from J to kJ How do we make the decarboxylation of limestone spontaneous? Set ΔG = 0, the the crossing over point where the reaction converts from nonspontaneous to spontaneous. ΔG = 0 = ΔH – TΔS 0 = +178.1 –T(+0.1605) T = 1109K = 837℃ When the temperature falls below 837°C, CO2 begins spontaneously to react with CaO to form CaCO3: CaO(s) + CO2(g) CaCO3(s) At room temperature, ΔG = + 130.27 kJ Calculate the boiling point of methanol. CH3OH(l) CH3OH(g) ΔHf° -238.7 -200.7 ΔS +0.1268 +0.2398 ΔH = +38.0 kJ ΔS = +0.113 kJ At equilibrium, ΔG is always 0. ΔG = 0 = ΔH – TΔS =+38.0 –T(+0.113) Tb = 336K = 63.3℃ Note: Be sure you convert ΔS values from J to kJ Additional aspects of Free Energy • Even though a reaction has a negative G it may occur too slowly to be observed (i.e. combustion). • Thermodynamics gives us the direction of a spontaneous process, it does not give us the rate of the process. • A nonspontaneous process can be driven if coupled with a spontaneous process – this is very important in life processes (i.e., respiration to form ATP), and can be used in industrial processes, such as smelting. • To calculate K values, use ΔG° = -RT ln Keq. This refers to the ΔG difference of the standard states of compounds, before equilibrium is attained. Laws of Thermodynamics 1st Law. Energy is neither created nor destroyed. In chemistry, chemical energy can be converted into heat and vice versa. 2nd Law. Entropy increases spontaneous; i.e., the natural tendency is for randomization. 3rd Law. The entropy of a perfect crystal at 0K is zero (it is impossible to attain 0K). Reaction Rates • Some chemical reactions proceed rapidly. Like the precipitation reactions where the products form practically the instant the two solutions are mixed. • Other reactions proceed slowly. Like the decomposition of dye molecules of a sofa placed in front of a window. • The rate of a reaction is measured in the amount of reactant that changes into product in a given period of time. Generally moles of reactant used per second. Like miles per hour. • Chemists study ways of controlling reaction rates. 15 Reaction Rates, Continued Initially, only reactants are present After 15 seconds, the left reaction is 60% complete, but the right reaction is only 20% complete After 30 seconds, the left reaction is complete, whereas the right reaction is only 40% done. After 45 seconds, the right reaction is still not complete 16 Collision Theory • • In order for a reaction to take place, the reacting molecules must collide with each other. Once molecules collide they may react together or they may not, depending on two factors: 1. Whether the collision has enough energy to “start to break the bonds holding reactant molecules together." 2. Whether the reacting molecules collide in the proper orientation for new bonds to form. 19 Effective Collisions • Collisions in which these two conditions are met (and therefore the reaction occurs) are called effective collisions. • The higher the frequency of effective collisions, the faster the reaction rate. • There is a minimum energy needed for a collision to be effective. We call this the activation energy(活化能). The lower the activation energy, the faster the reaction will be. 20 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, it can form the activated complex(活化复合体). 21 Effective Collisions: Orientation Effect(定向效应) 22 Reaction Energy Diagram 23 Factors Effecting Reaction Rate: Reactant Concentration • The higher the concentration of reactant molecules, this increases the frequency of reactant molecule collisions the faster the reaction will generally go. • Since reactants are consumed as the reaction proceeds, the speed of a reaction generally slows over time. 24 Effect of Concentration on Rate Low concentrations of reactant molecules lead to fewer effective collisions, therefore a slower reaction rate. High concentrations of reactant molecules lead to more effective collisions, therefore a faster reaction rate. 25 Factors Effecting Reaction Rate: Temperature • Increasing the temperature increases the energy so that their collisions can overcome the activation energy. • And, increasing the temperature also increases the frequency of collisions. • Both these mean that increasing temperature increases the reaction rate. 26 Effect of Temperature on Rate Low temperatures lead to fewer molecules with enough energy to overcome the activation energy, and less frequent reactant collisions, therefore a slower reaction rate High temperatures lead to more molecules with enough energy to overcome the activation energy, and more frequent reactant collisions, therefore, a faster reaction rate. Activation Energy • The energy barrier that prevents any collision between molecules from being an effective collision is called the activation energy. • The larger the activation energy of a reaction, the slower it will be. (屏障) At a given temperature. 28 Relative potential energy Exothermic Reaction Activation energy, large Activation energy, small Reactants Hreaction Products Progress of reaction 29 Relative potential energy Endothermic Reaction Activation energy Products Hreaction Reactants Progress of reaction 30 Effect of Catalysts on Rate • A catalyst( ) is a substance that increases the rate of a reaction, but is not consumed in the reaction. • Catalysts lower the activation energy of a reaction. • Catalysts work by providing an easier pathway for the reaction. 催化剂 31 Catalyst Effect on Activation Energy 32 Catalyst Effect on Activation Energy 33 Enzymes • Enzymes( ) are protein molecules produced by living organisms that catalyze chemical reactions. • The enzyme molecules have an active site to which organic molecules bind. • When the organic molecule is bound to the active site, certain bonds are weakened • This allows a particular chemical change to occur with greater ease and speed. 酶 (催化) (活化位点) (削弱). i.e., the activation energy is lowered. 34 Chemical Equilibrium • When a reaction reaches equilibrium, the amounts of reactants and products in the system stay constant. • The forward and reverse reactions still continue. • Because they go at the same rate, the amounts of materials do not change. 37 Equilibrium Initially, we only have reactant molecules in the mixture. The reaction can only proceed in the forward direction, making products. Eventually, the forward and reverse rates are equal. At this time equilibrium is established. As the reaction proceeds, the forward reaction slows down as the reactants get used up. At the same time, the reverse reaction speeds up as product concentration increases. Once equilibrium is established, the concentrations of the reactants and products in the final mixture do not change, (unless conditions are changed). Equilibrium Equal • The rates of the forward and reverse reactions are equal at equilibrium. • But that does not mean the concentrations of reactants and products are equal. • Some reactions reach equilibrium only after almost all the reactant molecules are consumed— we say the position of equilibrium favors the products. • Other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed—we say the position of equilibrium favors the reactants. 42 An Analogy: Population Changes When Narnians feel overcrowded, some will emigrate to Middle Earth. However, as time passes, emigration will occur in both directions at the same rate, leading to populations in Narnia and Middle Earth that are constant, though not necessarily equal. 43 Equilibrium Constant • Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them. • For the reaction H2(g) + I2(g) 2HI(g) at equilibrium, the ratio of the concentrations raised to the power of their coefficients is constant. 2 K eq HI H 2 I 2 44 Equilibrium Constant • For the general equation aA + bB cC + dD, the relationship is given below: The lowercase letters represent the coefficients of the balanced chemical equation. Always products over reactants. • The constant is called the equilibrium constant, Keq. c d C D K eq a b A B 45 Writing Equilibrium Constant Expressions • For aA + bB cC + dD, the equilibrium constant expression is: C D A a Bb c K eq d • So for the reaction 4 2 N2O5 4 NO2 + O2, the K NO2 O2 eq N 2O5 2 equilibrium constant expression is: 46 Equilibrium Constants for Heterogeneous Equilibria (异质的) • Pure substances in the solid and liquid state have constant concentrations. Adding or removing some does not change the concentration because they do not expand to fill the container or spread throughout a solution. • Therefore, these substances are not included in the equilibrium constant expression. For the reaction CaCO3(s) + 2 HCl(aq) CaCl2(aq) + CO2(g) + H2O(l): K eq CaCl 2 CO 2 HCl2 47 Write the Equilibrium Constant Expressions, Keq, for Each of the Following: • 2 CO2(g) 2 CO(g) + O2(g) • BaSO4(s) Ba+2(aq) + SO4-2(aq) • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) 48 Write the Equilibrium Constant Expressions, Keq, for Each of the Following, Continued: • 2 CO2(g) 2 CO(g) + O2(g) [CO]2•[O2] Keq = [CO2]2 • BaSO4(s) Ba+2(aq) + SO4-2(aq) Keq =[Ba+2]•[SO4-2] • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) [CO2] •[H2O]2 Keq = [CH ]•[O ]2 4 2 49 What Does the Value of Keq Imply? • When the value of Keq > > 1, we know that when the reaction reaches equilibrium, there will be many more product molecules present than reactant molecules. The position of equilibrium favors products. • When the value of Keq < < 1, we know that when the reaction reaches equilibrium, there will be many more reactant molecules present than product molecules. The position of equilibrium favors reactants. 50 A Large Equilibrium Constant 51 A Small Equilibrium Constant 52 Write the Equilibrium Constant Expressions, Keq, and Predict the Position of Equilibrium for Each of the Following: • 2 HF(g) H2(g) + F2(g) Keq = 1 × 10-95 • 2 SO2(g) + O2(g) 2 SO3(g) Keq = 8 × 1025 • N2(g) + 2 O2(g) 2 NO2(g) Keq = 3 × 10-17 53 Write the Equilibrium Constant Expressions, Keq, and Predict the Position of Equilibrium for Each of the Following: • 2 HF(g) H2(g) + F2(g) H 2 F2 K eq 2 HF • 2 SO2(g) + O2(g) 2 SO3(g) SO3 2 Keq SO2 2 O2 • N2(g) + 2 O2(g) 2 NO2(g) NO 2 K eq 2 N 2 O2 2 Keq = 1 × 10-95 Favors reactants. Keq = 8 × 1025 Favors products. Keq = 3 × 10-17 Favors reactants. Calculating Keq • The value of the equilibrium constant may be determined by measuring the concentrations of all the reactants and products in the mixture after the reaction reaches equilibrium, then substituting in the expression for Keq. • Although you may have different amounts of reactants and products in the equilibrium mixture, the value of Keq will always be the same. (取代) The value of Keq depends only on the temperature. The value of Keq does not depend on the amounts of reactants or products with which you start. 55 Initial and Equilibrium Concentrations for H2(g) + I2(g) 2HI(g) Equilibrium Constant Equili brium Initial [HI] K [H ][I ] 2 [H2] [I2] [HI] [H2] [I2] [HI] eq 2 0.50 0.0 0.50 1.0 0.50 0.0 0.11 0.11 0.78 2 [0.78]2 50 [0.11][0.11] 0.50 0.055 0.055 0.39 [0.39]2 50 [0.055][0.055] 0.50 0.50 0.165 0.165 1.17 [1.17]2 50 [0.165][0.165] 0.0 0.5 0.0 0.53 0.033 0.934 [0.934]2 50 [0.53][0.033] 56 Example 3.3—Find the Value of Keq for the Reaction from the Given Concentrations: 2 CH4(g) C2H2(g) + 3 H2(g). Given: [CH4] = 0.0203 M, [C2H2] = 0.0451 M, [H2] = 0.112 M Find: K eq Solution Map: [CH ], [C H ], [H ] 4 2 2 2 K eq Relationships: Solve: Keq Check: Keq [C2H 2 ] [H 2 ]3 [CH4 ]2 [C2H 2 ] [H 2 ]3 0.04510.112 3 0.154 2 [CH4 ]2 0.0203 Keq is unitless. 57 Practice—Calculate Keq for the Reaction 2 NO2(g) N2O4(g) at 100 C if the Equilibrium Concentrations Are [NO2] = 0.0172 M and [N2O4] = 0.0014 M. 58 Practice—Find the Value of Keq for the Reaction from the Given Concentrations: 2 NO2(g) N2O4(g). Given: [NO2] = 0.0172 M, [N2O4] = 0.0014 M Find: K eq Solution Map: [NO2], [N2O4] K eq Relationships: Solve: K eq Check: [N 2O4 ] [NO 2 ]2 Keq [N 2O 4 ] 0.0014 4.7 2 2 [NO 2 ] 0.0172 Keq is unitless. Example 3.4—Find the Value of [HI] for the Reaction at Equilibrium from the Given Concentrations and Keq: H2(g) + I2(g) 2HI(g) Given: [I2] = 0.020 M, [H2] = 0.020 M, Keq = 69 Find: Solution Map: Relationships: Solve: [HI] [I2], [H2], Keq [HI] K eq [HI]2 [H 2 ] [I 2 ] [HI] [HI]22 2[H Keq [H [H [I ] [I] 2 ] K eq2 ]K ] [I ] [HI] 2[]HI 2 eq22 [H22]] [I 22]] [H 69 0.020 0.020 [HI ] 0.17 M 60 Disturbing and Re-Establishing Equilibrium • Once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same. • However, if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-established. • The new concentrations will be different, but the equilibrium constant will be the same. Unless you change the temperature. 61 Additional aspects of Free Energy • Even though a reaction has a negative G it may occur too slowly to be observed (i.e. combustion). • Thermodynamics gives us the direction of a spontaneous process, it does not give us the rate of the process. • A nonspontaneous process can be driven if coupled with a spontaneous process – this is very important in life processes (i.e., respiration to form ATP), and can be used in industrial processes, such as smelting. • To calculate K values, use ΔG° = -RT ln Keq. This refers to the ΔG difference of the standard states of compounds, before equilibrium is attained. Calculating Keq from ΔG values Let’s calculate Keq from the following reaction, which we previously studied in the Equilibrium chapter: N2O4(g) 2NO2(g) ΔGf° 98.3 2(51.8) ΔG = 5.3 kJ ΔG = -RT ln K 5300 = -8.314(407) ln K ln K = -1.57 R = 8.314 J/mol-K K = 0.208 -- very close to the experimental value Let’s take a look at the dissolution of NH4Cl. NH4Cl(s) NH4+(aq) + Cl-(aq) ΔHf° ΔS° -314.4 0.0946 -132.5 -167.2 ΔH = +14.7 0.1135 0.0565 ΔS = +0.0754 ΔG = ΔH – T ΔS = 14.7 – (298)(0.0754) = 14.7 – 22.47 = -7.77 kJ The reaction is endothermic, but is spontaneous! Hence, ammonium chloride is excellent for cold packs. Le Châtelier’s Principle • Le Châtelier’s principle guides us in predicting the effect on the position of equilibrium when conditions change. • “When a chemical system at equilibrium is disturbed, the system shifts in a direction that will minimize the disturbance.” 65 The Effect of Concentration Changes on Equilibrium • Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found. That has the same Keq. • Removing a product will increase the amounts of the other products and decrease the amounts of the reactants. You can use to this to drive a reaction to completion! • Remember: Adding more of a solid or liquid does not change its concentration and, therefore, has no effect on the equilibrium. 66 The Effect of Concentration Changes on Equilibrium When NO2 is added, some of it combines to make more N2O4. 67 The Effect of Concentration Changes on Equilibrium When N2O4 is added, some of it decomposes to make more NO2. 68 Practice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems: • 2 CO2(g) 2 CO(g) + O2(g) • BaSO4(s) Ba2+(aq) + SO42-(aq) • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) 69 Practice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems, Continued: • 2 CO2(g) 2 CO(g) + O2(g) Shift right, removing some of the added CO2 and increasing the concentrations of CO and O2. • BaSO4(s) Ba2+(aq) + SO42-(aq) Shift left, removing some of the added Ba2+ and reducing the concentration of SO42-. • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) Shift right, removing some of the added CO2 and decreasing the O2, 70 while increasing the concentration of CO2. Effect of Volume Change on Equilibrium • For solids, liquids, or solutions, changing the size of the container has no effect on the concentration. • Changing the volume of a container changes the concentration of a gas. Same number of moles, but different number of liters, resulting in a different molarity. 71 Effect of Volume Change on Equilibrium • Decreasing the size of the container increases the concentration of all the gases in the container. This increases their partial pressures. • If their partial pressures increase, then the total pressure in the container will increase. • According to Le Châtelier’s principle, the equilibrium should shift to remove that pressure. • The way to reduce the pressure is to reduce the number of molecules in the container. • When the volume decreases, the equilibrium shifts to the side with fewer molecules. 72 The Effect of Volume Change on Equilibrium When Sincethe there pressure are more is decreased gas molecules by increasing on the reactants volume, the side position of theof reaction, equilibrium whenshifts the toward pressure the is side increased with thethe greater position number of of equilibrium molecules—the shifts toward reactant the products. side. 73 The Effect of Temperature Changes on Equilibrium • Exothermic reactions release energy and endothermic reactions absorb energy. • If we write heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s principle to predict the effect of temperature changes. However, heat is not matter and not written in a proper equation. 76 The Effect of Temperature Changes on Equilibrium for Exothermic Reactions • For an exothermic reaction, heat is a product. • Increasing the temperature is like adding heat. • According to Le Châtelier’s principle, the equilibrium will shift away from the added heat. • The concentrations of C and D will decrease and the concentrations of A and B will increase. • The value of Keq will decrease. • How will decreasing the temperature effect the system? aA + bB cC + dD + heat C D K eq a b A B c d 77 The Effect of Temperature Changes on Equilibrium for Endothermic Reactions • For an endothermic reaction, heat is a reactant. • Increasing the temperature is like adding heat. • According to Le Châtelier’s principle, the equilibrium will shift away from the added heat. • The concentrations of C and D will increase and the concentrations of A and B will decrease. • The value of Keq will increase. • How will decreasing the temperature effect the system? Heat + aA + bB cC + dD Cc Dd K eq Aa Bb 78 The Effect of Temperature Changes on Equilibrium 79 Practice—Predict the Effect on the Equilibrium When the Temperature Is Reduced. • Heat + 2 CO2(g) 2 CO(g) + O2(g) • BaSO4(s) Ba2+(aq) + SO42-(aq) (endothermic) • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) (exothermic) 80 Practice—Predict the Effect on the Equilibrium When the Temperature Is Reduced • Heat + 2 CO2(g) 2 CO(g) + O2(g) Shift left, reducing the value of Keq. • Heat + BaSO4(s) Ba2+(aq) + SO42-(aq) Shift left, reducing the value of Keq. • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) + Heat Shift right, increasing the value of Keq. 81 Solubility and Solubility Product • Even “insoluble” salts dissolve somewhat in water. Insoluble = less than 0.1 g per 100 g H2O. • The solubility of insoluble salts is described in terms of equilibrium between undissolved solid and aqueous ions produced. AnYm(s) n A+(aq) + m Y-(aq) • Equilibrium constant for this process is called the solubility product constant, Ksp. Ksp = [A+]n[Y-]m • If there is undissolved solid in equilibrium with the solution, the solution is saturated. • Larger Ksp = more soluble. For salts that produce the same number of ions. 82 Example—Determine the Ksp of PbBr2 if its Solubility Is 1.44 x 10-2 M. PbBr2(s) Pb2+(aq) + 2 Br–(aq) init -0 0 equil -0.0144 0.0288 Ksp = [Pb2+][Br–]2 = (0.0144)(0.0288)2 = 1.19 x 10-5 83 Example3.5— Calculating Molar Solubility from Ksp 84 Example 3.5: • Calculate the molar solubility of BaSO4. Ksp = 1.07 x 10-10 at 25 °C 85 Example 3.5: Calculate the molar solubility of BaSO4. Ksp = 1.07 x 10-10 at 25 °C Information: Given: Ksp = 1.07 x 10-10 Find: [BaSO4], M = [Ba2+] = [SO42-] Equation: Ksp = [Ba2+][SO42-] Solution Map: Ksp → [Ba2+] • Apply the solution map: Ksp Ba 2 Ba 2 1.07 10 10 Ba 2 1.03 10-5 Ba 2 1.07 10-10 SO4 2 Ba 2 2 2 Ba 90 Combustion as Redox 2 H2(g) + O2(g) 2 H2O(g) 92 Redox without Combustion 2 Na(s) + Cl2(g) 2 NaCl(s) 2 Na 2 Na+ + 2 e Tro, Chemistry: A Molecular Approach 93 Cl2 + 2 e 2 Cl Reactions of Metals with Nonmetals • consider the following reactions: 4 Na(s) + O2(g) → 2 Na2O(s) 2 Na(s) + Cl2(g) → 2 NaCl(s) • the reaction involves a metal reacting with a nonmetal • in addition, both reactions involve the conversion of free elements into ions 4 Na(s) + O2(g) → 2 Na+2O– (s) 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) 94 Oxidation and Reduction • in order to convert a free element into an ion, the atoms must gain or lose electrons of course, if one atom loses electrons, another must accept them • reactions where electrons are transferred from one atom to another are redox reactions • atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced Ger Na+Cl–(s) 2 Na(s) + Cl2(g) → 2 Na → Na+ + 1 e– oxidation Cl2 + 2 e– → 2 Cl– reduction Leo 95 Electron Bookkeeping • for reactions that are not metal + nonmetal, or do not involve O2, we need a method for determining how the electrons are transferred • chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction even though they look like them, oxidation states are not ion charges! oxidation states are imaginary charges assigned based on a set of rules ion charges are real, measurable charges 96 Rules for Assigning Oxidation States • rules are in order of priority 1. free elements have an oxidation state = 0 Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g) 2. monatomic ions have an oxidation state equal to their charge Na = +1 and Cl = -1 in NaCl 3. (a) the sum of the oxidation states of all the atoms in a compound is 0 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0 97 Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1 4. (a) Group I metals have an oxidation state of +1 in all their compounds Na = +1 in NaCl 4. (b) Group II metals have an oxidation state of +2 in all their compounds Mg = +2 in MgCl2 98 Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation states according to the table below nonmetals higher on the table take priority Nonmetal Oxidation State Example F -1 CF4 H +1 CH4 O -2 CO2 Group 7A -1 CCl4 Group 6A -2 CS2 Group 5A -3 NH3 99 Practice – Assign an Oxidation State to Each Element in the following • Br2 • K+ • LiF • CO2 • SO42- • Na2O2 100 Practice – Assign an Oxidation State to Each Element in the following • Br2 Br = 0, (Rule 1) • K+ K = +1, (Rule 2) • LiF Li = +1, (Rule 4a) & F = -1, (Rule 5) • CO2 O = -2, (Rule 5) & C = +4, (Rule 3a) • SO42- O = -2, (Rule 5) & S = +6, (Rule 3b) • Na2O2 Na = +1, (Rule 4a) & O = -1, (Rule 3a) 101 Oxidation and Reduction Another Definition • oxidation occurs when an atom’s oxidation state increases during a reaction • reduction occurs when an atom’s oxidation state decreases during a reaction CH4 + 2 O2 → CO2 + 2 H2O -4 +1 0 +4 –2 +1 -2 oxidation reduction 102 Oxidation–Reduction • oxidation and reduction must occur simultaneously if an atom loses electrons another atom must take them • the reactant that reduces an element in another reactant is called the reducing agent the reducing agent contains the element that is oxidized • the reactant that oxidizes an element in another reactant is called the oxidizing agent the oxidizing agent contains the element that is reduced 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl2 is the oxidizing agent 103 Identify the Oxidizing and Reducing Agents in Each of the Following 3 H2S + 2 NO3– + 2 H+ 3 S + 2 NO + 4 H2O MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O 104 Identify the Oxidizing and Reducing Agents in Each of the Following red ag ox ag +1 -2 +5 -2 3 H2S + 2 NO3– + 2 H+ 3 S + 2 NO + 4 H2O +1 0 +2 -2 +1 -2 oxidation reduction ox ag red ag +4 -2 +1 -1 MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O +2 -1 0 +1 -2 oxidation reduction 105 Oxidation–Reduction Reactions • We say that the element that loses electrons in the reaction is oxidized. • And the substance that gains electrons in the reaction is reduced. • You cannot have one without the other. • In combustion, the O atoms in O2 are reduced, and the non-O atoms in the other material are oxidized. 106 Combustion as Redox • In the following reaction: 2 Mg(s) + O2(g) 2 MgO(s) • The magnesium atoms are oxidized. Mg0 Mg2+ + 2 e • The oxygen atoms are reduced. O0 + 2 e O2 107 Combustion as Redox, Continued • Even though the following reaction does not involve ion formation, electrons are still transferred. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) • The carbon atoms are oxidized. C4 C+4 + 8 e These are not charges, they are called oxidation numbers, but they help us see the electron transfer. • The oxygen atoms are reduced. O0 + 2 e O2 108 Bonding Theories • explain how and why atoms attach together • explain why some combinations of atoms are stable and others are not why is water H2O, not HO or H3O • one of the simplest bonding theories was developed by G.N. Lewis and is called Lewis Theory • Lewis Theory emphasizes valence electrons to explain bonding • using Lewis Theory, we can draw models – called Lewis structures – that allow us to predict many properties of molecules aka Electron Dot Structures such as molecular shape, size, polarity 109 Why Do Atoms Bond? • processes are spontaneous if they result in a system with lower potential energy • chemical bonds form because they lower the potential energy between the charged particles that compose atoms • the potential energy between charged particles is directly proportional to the product of the charges • the potential energy between charged particles is inversely proportional to the distance between the charges 110 Potential Energy Between Charged Particles 1 q1 q2 E potential 4 0 r • 0 is a constant = 8.85 x 10-12 C2/J∙m • for charges with the same sign, Epotential is + and the magnitude gets less positive as the particles get farther apart • for charges with the opposite signs, Epotential is and the magnitude gets more negative as the particles get closer together • remember: the more negative the potential energy, the more stable the system becomes 111 Potential Energy Between Charged Particles The attraction repulsion between like-charged particles opposite-charged increasesincreases particles as the as particles the particles get closer get closer together. To Bringing bring them closer lowers requiresthe the addition potential energy of more of the energy. system. 112 Bonding • a chemical bond forms when the potential energy of the bonded atoms is less than the potential energy of the separate atoms • have to consider following interactions: nucleus-to-nucleus repulsion electron-to-electron repulsion nucleus-to-electron attraction 113 Types of Bonds Types of Atoms metals to nonmetals nonmetals to nonmetals metal to metal Type of Bond Ionic Covalent Metallic 114 Bond Characteristic electrons transferred electrons shared electrons pooled Electronegativity • measure of the pull an atom has on bonding electrons • increases across period (left to right) and • decreases down group (top to bottom) fluorine is the most electronegative element francium is the least electronegative element • the larger the difference in electronegativity, the more polar the bond negative end toward more electronegative atom 115 Electronegativity Scale 116 Electronegativity and Bond Polarity • If difference in electronegativity between bonded atoms is 0, the bond is pure covalent equal sharing • If difference in electronegativity between bonded atoms is 0.1 to 0.4, the bond is nonpolar covalent • If difference in electronegativity between bonded atoms 0.5 to 1.9, the bond is polar covalent • If difference in electronegativity between bonded atoms larger than or equal to 2.0, the bond is ionic 4% 0 0.4 Percent Ionic Character 51% 2.0 Electronegativity 117Difference “100%” 4.0 Bond Polarity ENCl = 3.0 3.0 - 3.0 = 0 Pure Covalent Tro, Chemistry: A Molecular Approach ENCl = 3.0 ENH = 2.1 3.0 – 2.1 = 0.9 Polar Covalent 118 ENCl = 3.0 ENNa = 1.0 3.0 – 0.9 = 2.1 Ionic 119 Soaps are useful for cleaning because soap molecules have both a hydrophilic end, which dissolves in water, as well as a hydrophobic end, which is able to dissolve nonpolar grease molecules. Although grease will normally adhere to skin or clothing, the soap molecules can form micelles which surround the grease particles and allow them to be dissolved in water. Tro's Introductory Chemistry, Chapter 15 120 Applied to a soiled surface, soapy water effectively holds particles in colloidal suspension so it can be rinsed off with clean water. The hydrophobic portion (made up of a long hydrocarbon chain) dissolves dirt and oils, while the ionic end dissolves in water. Therefore, it allows water to remove normallyinsoluble matter by emulsification. Sometimes the absence of oxygen in cold and humid environment causes corpses to naturally accumulate a soap-like coating, adipocere, as covering the Soap Lady on exhibit in the Mutter Museum. 121 Properties of Acids • Sour taste. • Change color of vegetable dyes. • React with “active” metals, not noble metals. I.e., Al, Zn, Fe, but not Cu, Ag or Au. Zn + 2 HCl ZnCl2 + H2 Corrosive. • React with carbonates, producing CO2. Marble, baking soda, chalk, limestone. CaCO3 + 2 HCl CaCl2 + CO2 + H2O • React with bases to form ionic salts. And often water. 122 Common Acids Chemical name Formula Old name Strength Nitric acid HNO3 Aqua fortis Strong Sulfuric acid H2SO4 Vitriolic acid Strong Hydrochloric acid HCl Muriatic acid Strong Phosphoric acid H3PO4 Moderate Chloric acid HClO3 Moderate Acetic acid HC2H3O2 Hydrofluoric acid HF Carbonic acid H2CO3 Boric acid H3BO3 Vinegar Weak Weak Soda water Weak Weak 123 Properties of Bases • • • • A.k.a. alkalis. Taste bitter. Feel slippery. Change color of vegetable dyes. Different color than acid. Litmus = blue. • React with acids to form ionic salts. And often water. Neutralization. 124 Common Bases Chemical name Sodium hydroxide Potassium hydroxide Calcium hydroxide Magnesium hydroxide Ammonium hydroxide Formula Strength KOH Common name Lye, caustic soda Caustic potash Ca(OH)2 Slaked lime Strong Mg(OH)2 Milk of magnesia Weak NH4OH, Ammonia water, {NH3(aq)} aqueous ammonia Weak NaOH Strong Strong 125 Acid–Base Reactions • Also called neutralization reactions because the acid and base neutralize each other’s properties. • In the reaction of an acid with a base, the H+1 from the acid combines with the OH-1 from the base to make water. • The cation from the base combines with the anion from the acid to make the salt. acid + base salt + water 2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l) • The net ionic equation for an acid-base reaction often is: H+1(aq) + OH-1(aq) H2O(l) As long as the salt that forms is soluble in water. 126 Process for Predicting the Products of an Acid–Base Reaction 1. Determine what ions each aqueous reactant has. 2. Exchange ions. (+) ion from one reactant with (-) ion from the other. H+ combines with OH− to make water. 3. Balance charges of combined ions to get formula of the salt. 4. Balance the equation. Count atoms. 5. Determine solubility of the salt. Use the solubility rules. If the salt is insoluble or slightly soluble, it will precipitate. 127