The molecular nature of matter and change bài giảng môn học general chemistry ii

Lecture PowerPoint Chapter 16 Chemistry Kinetics: Rates and Mechanisms of Chemical Reactions The Molecular Nature of Matter and Change Martin S Silberberg 16-1 Copyright  The McGraw-Hill Companies, Inc Permission required for reproduction or display 16-2 Figure 16.1 A faster reaction (top) and a slower reaction (bottom) Kinetics: Rates and Mechanisms of Chemical Reactions 16.1 Focusing on Reaction Rate 16.2 Expressing the Reaction Rate 16.3 The Rate Law and Its Components 16.4 Integrated Rate Laws: Concentration Changes over Time 16.5 Theories of Chemical Kinetics 16.6 Catalysis: Speeding Up a Reaction 16-3 Figure 16.2 16-4 The wide range of reaction rates Factors That Influence Reaction Rate • Particles must collide in order to react • The higher the concentration of reactants, the greater the reaction rate – A higher concentration of reactant particles allows a greater number of collisions • The physical state of the reactants influences reaction rate – Substances must mix in order for particles to collide • The higher the temperature, the greater the reaction rate – At higher temperatures particles have more energy and therefore collide more often and more effectively 16-5 16-6 Figure 16.3 The effect of surface area on reaction rate A hot steel nail glows feebly when placed in O2 Figure 16.4 Sufficient collision energy is required for a reaction to occur The same mass of steel wool bursts into flame 16-7 16-8 Table 16.1 Expressing the Reaction Rate Concentration of O3 at Various Times in its Reaction with C2H4 at 303 K Reaction rate is measured in terms of the changes in concentrations of reactants or products per unit time Time (s) For the general reaction A → B, we measure the concentration of A at t1 and at t2: Rate = change in concentration of A change in time =- conc A2 - conc A1 t2 - t1 =- [A] t Square brackets indicate a concentration in moles per liter The negative sign is used because the concentration of A is decreasing This gives the rate a positive value 16-9 Figure 16.5 Concentration of O3 (mol/L) 0.0 3.20x10-5 10.0 2.42x10-5 20.0 1.95x10-5 30.0 1.63x10-5 40.0 1.40x10-5 50.0 1.23x10-5 60.0 1.10x10-5 rate = =-  [C2H4] t  [O3] t 16-10 Three types of reaction rates for the reaction of O3 and C2H4 Figure 16.6A Plots of [reactant] and [product] vs time C2H4 + O3 → C2H4O + O2 [O2] increases just as fast as [C2H4] decreases Rate = - = 16-11 [C2H4] t [C2H4O] t =- = [O3] t [O2] t 16-12 Figure 16.6B Plots of [reactant] and [product] vs time In general, for the reaction H2 + I2 → 2HI [HI] increases twice as fast as [H2] decreases Rate = - Rate = [H2] =- t [IH] t = -2 [I2] t [H2] t aA [HI] = t = -2 + bB → cC + dD where a, b, c, and d are the coefficients for the balanced equation, the rate is expressed as: [I2] rate = - t a [A] [B] = [C] =b t c t t = [D] d t The expression for the rate of a reaction and its numerical value depend on which substance serves as the reference 16-13 16-14 Sample Problem 16.1 Expressing Rate in Terms of Changes in Concentration with Time PROBLEM: Hydrogen gas has a nonpolluting combustion product (water vapor) It is used as a fuel abord the space shuttle and in earthbound cars with prototype engines: 2H2(g) + O2(g) → 2H2O(g) The Rate Law For any general reaction occurring at a fixed temperature aA (b) When [O2] is decreasing at 0.23 mol/L·s, at what rate is [H2O] increasing? PLAN: We choose O2 as the reference because its coefficient is For every molecule of O2 that disappears, two molecules of H2 disappear, so the rate of [O2] decrease is ½ the rate of [H2] decrease Similarly, the rate at which [O2] decreases is ½ the rate at which [H2O] increases 16-15 + bB → cC + dD Rate = k[A]m[B]n (a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time The term k is the rate constant, which is specific for a given reaction at a given temperature The exponents m and n are reaction orders and are determined by experiment The values of m and n are not necessarily related in any way to the coefficients a and b 16-16 Reaction Orders Figure 16.7 Plots of reactant concentration, [A], vs time for first-, second-, and zero-order reactions A reaction has an individual order “with respect to” or “in” each reactant For the simple reaction A → products: If the rate doubles when [A] doubles, the rate depends on [A]1 and the reaction is first order with respect to A If the rate quadruples when [A] doubles, the rate depends on [A]2 and the reaction is second order with respect to [A] If the rate does not change when [A] doubles, the rate does not depend on [A], and the reaction is zero order with respect to A 16-17 16-18 Figure 16.8 Plots of rate vs reactant concentration, [A], for first-, second-, and zero-order reactions Individual and Overall Reaction Orders For the reaction 2NO(g) + 2H2(g) → N2(g) + 2H2O(g): The rate law is rate = k[NO]2[H2] The reaction is second order with respect to NO, first order with respect to H2 and third order overall Note that the reaction is first order with respect to H2 even though the coefficient for H2 in the balanced equation is Reaction orders must be determined from experimental data and cannot be deduced from the balanced equation 16-19 16-20 Sample Problem 16.2 Determining Reaction Orders from Rate Laws PROBLEM: For each of the following reactions, use the give rate law to determine the reaction order with respect to each reactant and the overall order k[NO]2[O (a) 2NO(g) + O2(g) → 2NO2(g); rate = 2] (b) CH3CHO(g) → CH4(g) + CO(g); rate = k[CH3CHO]3/2 + (c) H2O2(aq) + 3I (aq) + 2H (aq) →I3 (aq) + 2H2O(l); rate = k[H2O2][I-] PLAN: We inspect the exponents in the rate law, not the coefficients of the balanced equation, to find the individual orders We add the individual orders to get the overall reaction order Determining Reaction Orders For the general reaction A + 2B → C + D, the rate law will have the form Rate = k[A]m[B]n To determine the values of m and n, we run a series of experiments in which one reactant concentration changes while the other is kept constant, and we measure the effect on the initial rate in each case SOLUTION: 16-21 16-22 Table 16.2 Initial Rates for the Reaction between A and B Experiment Initial Rate (mol/L·s) Initial [A] (mol/L) Initial [B] (mol/L) 1.75x10-3 2.50x10-2 3.00x10-2 3.50x10-3 5.00x10-2 3.00x10-2 3.50x10-3 2.50x10-2 6.00x10-2 7.00x10-3 5.00x10-2 6.00x10-2 Finding m, the order with respect to A: We compare experiments and 2, where [B] is kept constant but [A] doubles: Rate Rate k[A] m [B] n = k[A] m 16-23 m = [A] m [A]1 = [A]2 m [A]1 m 3.50x10-3 mol/L·s 1.75x10-3mol/L·s [B] is kept constant for experiments and 2, while [A] is doubled Then [A] is kept constant while [B] is doubled [B] n1 = 5.00x10-2 mol/L 2.50x10-2 mol/L Dividing, we get 2.00 = (2.00)m so m = 16-24 Table 16.3 Finding n, the order with respect to B: O2(g) + 2NO(g) → 2NO2(g) We compare experiments and 1, where [A] is kept constant but [B] doubles: Rate Rate k[A] m [B] n = 3 k[A] m [B] n1 n = [B] [B]1 Experiment Initial Rate (mol/L·s) [O2] [NO] 3.21x10-3 1.10x10-2 1.30x10-2 m 3.50x10-3 mol/L·s = 1.75x10-3mol/L·s Rate = k[O2]m[NO]n Initial Reactant Concentrations (mol/L) [B]3 n = n [B]1 Initial Rates for the Reaction between O2 and NO 6.00x10-2 mol/L 6.40x10-3 2.20x10-2 1.30x10-2 3.00x10-2 12.48x10-3 1.10x10-2 2.60x10-2 9.60x10-3 3.30x10-2 1.30x10-2 28.8x10-3 1.10x10-2 3.90x10-2 mol/L Dividing, we get 2.00 = (2.00)n so n = 16-25 16-26 Sample Problem 16.3 Determining Reaction Orders from Rate Data PROBLEM: Many gaseous reactions occur in a car engine and exhaust system One of these is NO2(g) + CO(g) → NO(g) + CO2(g) rate = k[NO2]m[CO]n Use the following data to determine the individual and overall reaction orders: Experiment Initial Rate (mol/L·s) Sample Problem 16.3 PLAN: We need to solve the general rate law for m and for n and then add those orders to get the overall order We proceed by taking the ratio of the rate laws for two experiments in which only the reactant in question changes concentration SOLUTION: Initial [NO2] Initial [CO] (mol/L) (mol/L) 0.0050 0.10 0.10 0.080 0.40 0.10 0.0050 0.10 0.20 16-27 16-28 Sample Problem 16.4 Determining Reaction Orders from Molecular Scenes PROBLEM: At a particular temperature and volume, two gases, A (red) and B (blue), react The following molecular scenes represent starting mixtures for four experiments: Table 16.4 Overall Reaction Units of k Order (t in seconds) Expt no: Initial rate (mol/L·s) 0.50x10-4 1.0x10-4 2.0x10-4 ? (a) What is the reaction order with respect to A? With respect to B? The overall order? (b) Write the rate law for the reaction (c) Predict the initial rate of experiment PLAN: We find the individual reaction orders by seeing how a change in each reactant changes the rate Instead of using concentrations we count the number of particles 16-29 Units of the Rate Constant k for Several Overall Reaction Orders mol/L·s (or mol L-1s-1) 1/s (or s-1) L/mol·s (or L mol-1s-1) L2/mol2·s (or L2 mol-2s-1) General formula: L Units of k = order-1 mol unit of t The value of k is easily determined from experimental rate data The units of k depend on the overall reaction order 16-30 Integrated Rate Laws Figure 16.9 Information sequence to determine the kinetic parameters of a reaction An integrated rate law includes time as a variable Series of plots of concentration vs time First-order rate equation: rate = - [A] = k [A] Determine slope of tangent at t0 for each plot Initial rates Compare initial rates when [A] changes and [B] is held constant (and vice versa) Reaction orders Substitute initial rates, orders, and concentrations into rate = k[A]m[B]n, and solve for k Rate constant (k) and actual rate law ln t Second-order rate equation: [A] rate = = k [A]2 t Zero-order rate equation: rate = - [A] = k [A]0 [A]0 [A]t = - kt 1 = kt [A]t [A]0 [A]t - [A]0 = - kt t 16-31 16-32 Sample Problem 16.5 Determining the Reactant Concentration after a Given Time Figure 16.10A Graphical method for finding the reaction order from the integrated rate law PROBLEM: At 1000oC, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87 s-1, to two molecules of ethylene (C2H4) (a) If the initial C4H8 concentration is 2.00 M, what is the concentration after 0.010 s? (b) What fraction of C4H8 has decomposed in this time? First-order reaction integrated rate law ln [A]0 = - kt [A]t PLAN: We must find the concentration of cyclobutane at time t, [C4H8]t The problem tells us the reaction is first-order, so we use the integrated first-order rate law: ln [C4H8 ]0 [C4H8 ]t straight-line form ln[A]t = -kt + ln[A]0 = - kt A plot of ln [A] vs time gives a straight line for a first-order reaction 16-33 16-34 Figure 16.10B Graphical method for finding the reaction order from the integrated rate law Figure 16.10C Graphical method for finding the reaction order from the integrated rate law Second-order reaction Zero-order reaction integrated rate law [A]t - [A]0 integrated rate law [A]t - [A]0 = - kt = kt straight-line form 1 = kt + [A]t [A]0 A plot of 16-35 vs time gives a straight line for a second-order reaction [A] straight-line form [A]t = - kt + [A]0 A plot of [A] vs time gives a straight line for a first-order reaction 16-36 Figure 16.11 Graphical determination of the reaction order for the decomposition of N2O5 Reaction Half-life The half-life (t1/2) for a reaction is the time taken for the concentration of a reactant to drop to half its initial value For a first-order reaction, t1/2 does not depend on the starting concentration ln 0.693 t1/2 = = k k The concentration data is used to construct three different plots Since the plot of ln [N2O5] vs time gives a straight line, the reaction is first order 16-37 Figure 16.12 The half-life for a first-order reaction is a constant Radioactive decay is a first-order process The half-life for a radioactive nucleus is a useful indicator of its stability 16-38 A plot of [N2O5] vs time for three reaction half-lives for a first-order process t1/2 = ln k = 0.693 k Sample Problem 16.6 Using Molecular Scenes to Find Quantities at Various Times PROBLEM: Substance A (green) decomposes to two other substances, B (blue) and C (yellow), in a first-order gaseous reaction The molecular scenes below show a portion of the reaction mixture at two different times: (a) Draw a similar molecular scene of the reaction mixture at t = 60.0 s (b) Find the rate constant of the reaction (c) If the total pressure (Ptotal) of the mixture is 5.00 atm at 90.0 s, what is the partial pressure of substance B (PB)? 16-39 Sample Problem 16.7 16-40 Determining the Half-Life of a First-Order Reaction PROBLEM: Cyclopropane is the smallest cyclic hydrocarbon Because its 60° bond angles allow poor orbital overlap, its bonds are weak As a result, it is thermally unstable and rearranges to propene at 1000°C via the following firstorder reaction: The rate constant is 9.2 s-1, (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value? PLAN: The reaction is first order, so we find t1/2 using the half-life equation for a first order reaction Once we know t1/2 we can calculate the time taken for the concentration to drop to 0.25 of its initial value 16-41 Half-life Equations For a first-order reaction, t1/2 does not depend on the initial concentration For a second-order reaction, t1/2 is inversely proportional to the initial concentration: t1/2 = k[A]0 (second order process; rate = k[A]2) For a zero-order reaction, t1/2 is directly proportional to the initial concentration: t1/2 = [A]0 2k0 (zero order process; rate = k) 16-42 Table 16.5 An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions Collision Theory and Concentration Zero Order First Order Second Order Rate law rate = k rate = k[A] rate = k[A]2 Units for k mol/L·s [A]0 2k 1/s ln k Integrated rate law in straight-line form [A]t = -kt + [A]0 ln[A]t = -kt + ln[A]0 L/mol·s k[A]0 = kt + [A] Plot for straight line [A]t vs t ln[A]t vs t Slope, y intercept -k, [A]0 -k, ln[A]0 Half-life t The basic principle of collision theory is that particles must collide in order to react [A]0 vs t [A]t k, [A]0 An increase in the concentration of a reactant leads to a larger number of collisions, hence increasing reaction rate The number of collisions depends on the product of the numbers of reactant particles, not their sum Concentrations are multiplied in the rate law, not added 16-43 16-44 Temperature and the Rate Constant Figure 16.13 The number of possible collisions is the product, not the sum, of reactant concentrations Temperature has a dramatic effect on reaction rate For many reactions, an increase of 10°C will double or triple the rate add another collisions Experimental data shows that k increases exponentially as T increases This is expressed in the Arrhenius equation: collisions add another k = rate constant A = frequency factor Ea = activation energy k = Ae-Ea/RT Higher T larger k increased rate collisions 16-45 16-46 Figure 16.14 Increase of the rate constant with temperature for the hydrolysis of an ester Activation Energy In order to be effective, collisions between particles must exceed a certain energy threshold Expt [Ester] [H2O] T (K) 0.100 0.100 0.100 0.100 0.200 0.200 0.200 0.200 288 298 308 318 Rate k (mol/L·s) (L/mol·s) 1.04x10-3 0.0521 2.20x10-3 0101 3.68x10-3 0.184 6.64x10-3 0.332 When particles collide effectively, they reach an activated state The energy difference between the reactants and the activated state is the activation energy (Ea) for the reaction The lower the activation energy, the faster the reaction Smaller Ea larger f larger k increased rate Reaction rate and k increase exponentially as T increases 16-47 16-48 Figure 16.15 Energy-level diagram for a reaction Temperature and Collision Energy An increase in temperature causes an increase in the kinetic energy of the particles This leads to more frequent collisions and reaction rate increases At a higher temperature, the fraction of collisions with sufficient energy equal to or greater than Ea increases Reaction rate therefore increases Collisions must occur with sufficient energy to reach an activated state This particular reaction is reversible and is exothermic in the forward direction 16-49 16-50 Figure 16.16 The effect of temperature on the distribution of collision energies Molecular Structure and Reaction Rate For a collision between particles to be effective, it must have both sufficient energy and the appropriate relative orientation between the reacting particles The term A in the Arrhenius equation is the frequency factor for the reaction k = Ae -Ea/RT A = pZ p = orientation probability factor Z = collision frequency The term p is specific for each reaction and is related to the structural complexity of the reactants 16-51 16-52 Figure 16.23 Catalysis: Speeding up a Reaction Reaction energy diagram for a catalyzed (green) and uncatalyzed (red) process A catalyst is a substance that increases the reaction rate without itself being consumed in the reaction In general, a catalyst provides an alternative reaction pathway that has a lower total activation energy than the uncatalyzed reaction A catalyst will speed up both the forward and the reverse reactions A catalyst does not affect either H or the overall yield for a reaction 16-53 16-54 Figure 16.24 The catalyzed decomposition of H2O2 A homogeneous catalyst is in the same phase as the reaction mixture A small amount of NaBr is added to a solution of H2O2 Chemical Connections Figure B16.1 Satellite images of the increasing size of the Antarctic ozone hole (purple) Oxygen gas forms quickly as Br-(aq) catalyzes the H2O2 decomposition; the intermediate Br2 turns the solution orange 16-55 16-56 Chemical Connections Figure B16.2 Reaction energy diagram for breakdown of O2 by Cl atoms (Not drawn to scale.) 16-57 10 03-May-20 Calculating DG° DG°rxn can be calculated using the Gibbs equation: DG°sys = DH°sys - T DS°sys Sample Problem 20.4 Calculating DG°rxn from Enthalpy and Entropy Values PROBLEM: Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated +5 +7 -1 Δ 4KClO3(s) 3KClO4(s) + KCl(s) DG°rxn can also be calculated using values for the standard free energy of formation of the components DG°rxn = SmDG°products - SnDG°reactants 20-43 20-44 Sample Problem 20.5 Calculating DG°rxn from DG°f Values PROBLEM: Use DG°f values to calculate DG°rxn for the reaction in Sample Problem 20.4: 4KClO3(s) Δ 3KClO4(s) + KCl(s) DG° and Useful Work DG is the maximum useful work done by a system during a spontaneous process at constant T and P In practice, the maximum work is never done Free energy not used for work is lost to the surroundings as heat DG is the minimum work that must be done to a system to make a nonspontaneous process occur at constant T and P A reaction at equilibrium (DGsys = 0) can no longer any work 20-45 Figure 20.14 20-46 An expanding gas lifting a weight Effect of Temperature on Reaction Spontaneity DGsys = DHsys - T DSsys Reaction is spontaneous at all temperatures If DH < and DS > DG < for all T Reaction is nonspontaneous at all temperatures If DH > and DS < DG > for all T 20-47 20-48 03-May-20 Effect of Temperature on Reaction Spontaneity Table 20.1 Reaction Spontaneity and the Signs of DH, DS, and DG Reaction becomes spontaneous as T increases If DH > and DS > DG becomes more negative as T increases DH DS -TDS DG – + – – Spontaneous at all T + – + + Nonspontaneous at all T + + – + or – Spontaneous at higher T; – – + + or – Spontaneous at lower T; nonspontaneous at lower T nonspontaneous at higher T Reaction becomes spontaneous as T decreases If DH < and DS < DG becomes more negative as T decreases 20-49 Description 20-50 Sample Problem 20.6 Using Molecular Scenes to Determine the Signs of DH, DS, and DG PROBLEM: The following scenes represent a familiar phase change for water (blue spheres) Sample Problem 20.7 Determining the Effect of Temperature on ΔG PROBLEM: A key step in the production of sulfuric acid is the oxidation of SO2(g) to SO3(g): 2SO2(g) + O2(g) → 2SO3(g) At 298 K, DG = -141.6 kJ; DH = -198.4 kJ; and DS = -187.9 J/K (a) What are the signs of DH and DS for this process? Explain (b) Is the process spontaneous at all T, no T, low T, or high T? Explain 20-51 (a) Use the data to decide if this reaction is spontaneous at 25°C, and predict how DG will change with increasing T (b) Assuming DH and DS are constant with increasing T, is the reaction spontaneous at 900.° C? 20-52 Figure 20.15 The effect of temperature on reaction spontaneity The sign of DG switches at DH T= ΔS Sample Problem 20.8 Finding the Temperature at Which a Reaction Becomes Spontaneous PROBLEM: At 25°C (298 K), the reduction of copper(I) oxide is nonspontaneous DG = 8.9 kJ) Calculate the temperature at which the reaction becomes spontaneous PLAN: We need to calculate the temperature at which DG crosses over from a positive to a negative value We set DG equal to zero, and solve for T, using the values for DH and DS from the text SOLUTION: T= DH DS = 58.1 kJ = 352 K 0.165 kJ/K At any temperature above 352 K (= 79°C), the reaction becomes spontaneous 20-53 20-54 03-May-20 Chemical Connections Figure B20.1 The coupling of a nonspontaneous reaction to the hydrolysis of ATP Chemical Connections Figure B20.2 The cycling of metabolic free energy A spontaneous reaction can be coupled to a nonspontaneous reaction so that the spontaneous process provides the free energy required to drive the nonspontaneous process The coupled processes must be physically connected 20-55 20-56 Chemical Connections DG, Equilibrium, and Reaction Direction Figure B20.3 ATP is a high-energy molecule A reaction proceeds spontaneously to the right if Q < K; Q < so ln Q < and DG < K K A reaction proceeds spontaneously to the left if Q > K; Q > so ln Q > and DG > K K A reaction is at equilibrium if Q = K; Q = so ln Q = and DG = K K When ATP is hydrolyzed to ADP, the decrease in charge repulsion (A) and the increase in resonance stabilization (B) causes a large amount of energy to be released 20-57 20-58 DG, Q, and K ΔG and the Equlibrium Constant DG = RT ln Q = RT lnQ – RT lnK K For standard state conditions, Q = and DG° = -RT lnK If Q and K are very different, DG has a very large value (positive or negative) The reaction releases or absorbs a large amount of free energy A small change in DG° causes a large change in K, due to their logarithmic relationship If Q and K are nearly the same, DG has a very small value (positive or negative) The reaction releases or absorbs very little free energy As DG° becomes more negative, K becomes larger As DG° becomes more positive, K becomes smaller To calculate DG for any conditions: DG = DG° + RT lnQ 20-59 20-60 10 03-May-20 Sample Problem 20.9 Using Molecular Scenes to Find DG for a Reaction at Nonstandard Conditions PROBLEM: These molecular scenes represent three mixtures in which A2 (black) and B2 (green) are forming AB Each molecule represents 0.10 atm The equation is Table 20.2 The Relationship Between DG° and K at 298 K DG° (kJ) K 100 3x10-18 50 2x10-9 10 2x10-2 7x10-1 -1 1.5 -10 5x101 -50 6x108 -100 3x1017 -200 1x1035 Significance no forward reaction; {Essentially reverse reaction goes to completion and reverse reactions proceed {Forward to same extent A2(g) + B2(g) REVERSE REACTION 9x10-36 FORWARD REACTION 200 reaction goes to completion; {Forward essentially no reverse reaction 20-61 Sample Problem 20.10 2AB(g) DGo = -3.4 kJ/mol (a) If mixture is at equilibrium, calculate K (b) Which mixture has the most negative DG, and which has the most positive? (c) Is the reaction spontaneous at the standard state, that is PA2 = PB2 = PAB = 1.0 atm? 20-62 Calculating DG at Nonstandard Conditions Figure 20.16 Free energy and the extent of reaction PROBLEM: The oxidation of SO2(g) is too slow at 298 K to be useful in the manufacture of sulfuric acid, so the reaction is run at high T 2SO2(g) + O2(g) → 2SO3(g) (a) Calculate K at 298 K and at 973 K, (DG°298 = -141.6 kJ/mol of reaction as written; using DH° and DS°values at 973 K, DG°973 = -12.12 kJ/mol for reaction as written.) (b) Two containers are filled with 0.500 atm of SO2, 0.0100 atm of O2, and 0.100 atm of SO3; one is kept at 25°C and the other at 700.°C In which direction, if any, will the reaction proceed to reach equilibrium at each temperature? (c) Calculate DG for the system in part (b) at each temperature 20-63 Each reaction proceeds spontaneously (green curved arrows) from reactants or products to the equilibrium mixture, at which point DG = After that, the reaction is nonspontaneous in either direction (red curved arrows) Free energy reaches a minimum at equilibrium 20-64 11 Lecture PowerPoint Chapter 21 Chemistry Electrochemistry: The Molecular Nature of Matter and Change Chemical Change and Electrical Work Martin S Silberberg 21-1 Copyright  The McGraw-Hill Companies, Inc Permission required for reproduction or display 21-2 Electrochemistry: Chemical Change and Electrical Work Key Points About Redox Reactions 21.1 Half-Reactions and Electrochemical Cells •Oxidation (electron loss) always accompanies reduction (electron gain) 21.2 Voltaic Cells: Using Spontaneous Reactions to Generate Electrical Energy 21.3 Cell Potential: Output of a Voltaic Cell •The oxidizing agent is reduced, and the reducing agent is oxidized 21.4 Free Energy and Electrical Work 21.5 Electrochemical Processes in Batteries •The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent 21.6 Corrosion: A Case of Environmental Electrochemistry 21.7 Electrolytic Cells: Using Electrical Energy to Drive a Nonspontaneous Reaction 21-3 21-4 Half-Reaction Method for Balancing Redox Reactions Figure 21.1 A summary of redox terminology Zn(s) + 2H+(aq) Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions Zn2+(aq) + H2(g) •Each reaction is balanced for mass (atoms) and charge OXIDATION One reactant loses electrons Zn loses electrons Reducing agent is oxidized Zn is the reducing agent and becomes oxidized Oxidation number increases REDUCTION Other reactant gains electrons Oxidizing agent is reduced •One or both are multiplied by some integer to make the number of electrons gained and lost equal •The half-reactions are then recombined to give the balanced redox equation The oxidation number of Zn increases from x to +2 Advantages: •The separation of half-reactions reflects actual physical separations in electrochemical cells •The half-reactions are easier to balance especially if they involve acid or base •It is usually not necessary to assign oxidation numbers to those species not undergoing change Hydrogen ion gains electrons Hydrogen ion is the oxidizing agent and becomes reduced Oxidation number decreases The oxidation number of H decreases from +1 to 21-5 21-6 Balancing Redox Equations in Acidic Conditions Write the skeletons of the oxidation and reduction half-reactions Balance all elements other than O and H Balance the oxygen atoms by adding H2O molecules on the side of the arrow where O atoms are needed Balance the hydrogen atoms by adding H+ ions on the side of the arrow where H atoms are needed Balance the charge by adding electrons, e- If the number of electrons lost in the oxidation half-reaction is not equal to the number of electrons gained in the reduction halfreaction, multiply one or both of the half- reactions by a number that will make the number of electrons gained equal to the number lost Add the half-reactions as if they were mathematical equations Check to make sure that the atoms and the charge are balanced Balancing Redox Equations in Basic Conditions Steps #1-8 Begin by balancing the equation as if it were in acid solution If you have H+ ions in your equation at the end of these steps, proceed to Step #9 Otherwise, skip to Steps 9-12 Add enough OH- ions to each side to cancel the H+ ions Be sure to add the OH- ions to both sides to keep the charge and atoms balanced • 10 Combine the H+ ions and OH- ions that are on the same side of the equation to form water 11 Cancel or combine the H2O molecules 12 Check to make sure that the atoms and the charge balance If they balance, you are done If they not balance, re-check your work in Steps 1-11 21-8 Balancing Redox Reactions in Acidic Solution Balancing Redox Reactions in Acidic Solution Cr2O72-(aq) + I-(aq) 6e- + 14H+(aq) + Cr2O72- Divide the reaction into half-reactions - +6 -1 Cr2O72-(aq) + I-(aq) I- Cr3+ I2 Cr3+ I- Determine the O.N.s for the species undergoing redox Cr2O72- continued Cr3+(aq) + I2(aq) I2 + 7H2O(l) + 2e- Cr(+6) is the oxidizing agent and I(-1) is the reducing agent +3 Cr3+(aq) + I2(aq) Multiply each half-reaction by an integer, if necessary - Cr is going from +6 to +3 I- I2 + 2e- X3 I is going from -1 to Balance atoms and charges in each half-reaction - Add the half-reactions together 6e- + 14H+ + Cr2O72- 14H+(aq) + Cr2O72- Cr3+ + 7H2O(l) Cr3+ + 7H2O(l) I2 + 6e- Add 6e - to left net: +6 net: +12 6e- + 14H+(aq) + Cr2O72- Cr3+ I14H+(aq) + Cr2O72-(aq) + I-(aq) + 7H2O(l) 2Cr3+(aq) + 3I2(s) + 7H2O(l) Do a final check on atoms and charges 21-9 21-10 Balancing Redox Reactions in Basic Solution Figure 21.2 The redox reaction between dichromate ion and iodide ion Balance the reaction in acid and then add OH- so as to neutralize the H+ ions 14H+(aq) + Cr2O72-(aq) + I-(aq) + 14OH-(aq) 14H2O + Cr2O72- + I- Cr2O72- 2Cr3+(aq) + 3I2(s) + 7H2O(l) + 14OH-(aq) 2Cr3+ + 3I2 + 7H2O + 14OHCr3+ + I2 Reconcile the number of water molecules 7H2O + Cr2O72- + I- I- 2Cr3+ + 3I2 + 14OH- Do a final check on atoms and charges 21-11 21-12 Sample Problem 21.1: PROBLEM: Balancing Redox Reactions by the Half-Reaction Method Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in redox titrations It reacts in basic solution with the oxalate ion to form carbonate ion and solid mangaese dioxide Balance the skeleton ionic reaction that occurs between NaMnO4 and Na2C2O4 in basic solution: MnO4-(aq) + C2O42-(aq) PLAN: Sample Problem 21.1: 4H+ + MnO4- +3e- MnO2+ 2H2O C2O42- + 2H2O 4H+ + MnO4- +3e- MnO2+ 2H2O C2O42- + 2H2O MnO2(s) + CO32-(aq) 8H+ + 2MnO4- +6e- SOLUTION: 4H+ + MnO4- MnO2 +4 MnO2 C2O42+3 C2O42- CO32+4 CO32- MnO2 + 2H2O C2O42- + 2H2O +3e- +2e- 2CO32- + 4H+ + 2e2CO32- + 4H+ + 2eX3 X2 Proceed in acidic solution and then neutralize with base MnO4+7 MnO4- Balancing Redox Reactions by the Half-Reaction Method continued: 2CO32- + 4H+ 2MnO2+ 4H2O 6CO32- + 12H+ + 6e- 8H+ + 2MnO4- +6e- 2MnO2+ 4H2O 3C2O42- + 6H2O 6CO32- + 12H+ + 6e- 2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l) 2MnO2(s) + 6CO32-(aq) + 4H+(aq) + 4OH2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq) 21-13 3C2O42- + 6H2O + 4OH2MnO2(s) + 6CO32-(aq) + 2H2O(l) 21-14 The spontaneous reaction between zinc and copper(II) ion Figure 21.4 Figure 21.3 General characteristics of voltaic and electrolytic cells VOLTAIC CELL System Energy does is released work on from its spontaneous surroundings redox reaction X(s) Y(s) Oxidation half-reaction X X+ + e- ELECTROLYTIC CELL Surroundings(power Energy is absorbed tosupply) drive a nonspontaneous redox reaction work on system(cell) A(s) B(s) Oxidation half-reaction AA + e- Reduction half-reaction Y++ e- Y Reduction half-reaction B++ eB Overall (cell) reaction X + Y+ X+ + Y DG < Overall (cell) reaction A- + B+ A + B DG > 21-15 Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) 21-16 Figure 21.5 A voltaic cell based on the zinc-copper reaction Notation for a Voltaic Cell components of anode compartment components of cathode compartment (oxidation half-cell) (reduction half-cell) phase of lower phase of higher oxidation state oxidation state phase of higher oxidation state phase of lower oxidation state phase boundary between half-cells Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s) Examples: Zn(s) Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s) graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite Reduction half-reaction Cu2+(aq) + 2eCu(s) inert electrode Overall (cell) reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) 21-17 21-18 Figure 21.6 A voltaic cell using inactive electrodes Oxidation half-reaction 2I-(aq) I2(s) + 2e- Reduction half-reaction MnO4-(aq) + 8H+(aq) + 5eMn2+(aq) + 4H2O(l) Sample Problem 21.2: PROBLEM: PLAN: Diagramming Voltaic Cells Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge Measurement indicates that the Cr electrode is negative relative to the Ag electrode Identify the oxidation and reduction reactions and write each halfreaction Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction) Voltmeter e- SOLUTION: salt bridge Oxidation half-reaction Cr(s) Cr3+(aq) + 3e- Cr Reduction half-reaction Ag+(aq) + eAg(s) Cr3+ Ag + Overall (cell) reaction Cr(s) + Ag+(aq) Cr3+(aq) + Ag(s) Overall (cell) reaction 2MnO4-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l) 21-19 Ag K+ NO3- Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s) 21-20 Why Does a Voltaic Cell Work? Table 21.1 Voltages of Some Voltaic Cells The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit Voltaic Cell Voltage (V) Ecell > for a spontaneous reaction Volt (V) = Joule (J)/ Coulomb (C) 21-21 Figure 21.7 Common alkaline battery 1.5 Lead-acid car battery (6 cells = 12V) 2.0 Calculator battery (mercury) 1.3 Electric eel (~5000 cells in 6-ft eel = 750V) 0.15 Nerve of giant squid (across cell membrane) 0.070 21-22 Determining an unknown E0half-cell with the standard reference (hydrogen) electrode Sample Problem 21.3: PROBLEM: Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- Calculating an Unknown E0half-cell from E0cell A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) E0cell = 1.83V Calculate E0bromine given E0zinc = -0.76V PLAN: The reaction is spontaneous as written since the E0cell is (+) Zinc is being oxidized and is the anode Therefore the E0bromine can be found using E0cell = E0cathode - E0anode SOLUTION: anode: Zn(s) Zn2+(aq) + 2e E0Zn as Zn2+(aq) + 2eOverall (cell) reaction Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l) 21-23 Reduction half-reaction 2H3O+(aq) + 2eH2(g) + 2H2O(l) - E = +0.76 Zn(s) is -0.76V E0cell = E0cathode - E0anode = 1.83 = E0bromine - (-0.76) E0bromine = 1.86 - 0.76 = 1.07V 21-24 Table 21.2 Selected Standard Electrode Potentials (298K) •By convention, electrode potentials are written as reductions E0(V) Half-Reaction •When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell Reverse the sign of the potential +2.87 +1.36 +1.23 +0.96 +0.80 +0.77 +0.40 +0.34 0.00 -0.23 -0.44 -0.83 -2.71 -3.05 F2(g) + 2e2F-(aq) Cl2(g) + 2e2Cl-(aq) + MnO2(g) + 4H (aq) + 2eMn2+(aq) + 2H2O(l) + NO3 (aq) + 4H (aq) + 3eNO(g) + 2H2O(l) Ag+(aq) + eAg(s) 3+ 2+ Fe (g) + e Fe (aq) O2(g) + 2H2O(l) + 4e4OH-(aq) 2+ Cu (aq) + 2e Cu(s) 2H+(aq) + 2eH2(g) N2(g) + 5H+(aq) + 4eN2H5+(aq) 2+ Fe (aq) + 2e Fe(s) 2H2O(l) + 2eH2(g) + 2OH-(aq) Na+(aq) + eNa(s) Li+(aq) + eLi(s) •The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E0cell •When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents Example: 21-25 Cu2+(aq) + Zn2+(aq) stronger oxidizing agent + weaker oxidizing agent Cu(s) weaker reducing agent 21-26 Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength PROBLEM: (a) Combine the following three half-reactions into three spontaneous, balanced equations (A, B, and C), and calculate E0cell for each Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (2 of 4) SOLUTION: (1) NO3-(aq) + 4H+(aq) + 3e(2) N2(g) + 5H+(aq) + 4e(3) MnO2(s) +4H+(aq) + E0 = 0.96V NO(g) + 2H2O(l) N2H5+(aq) 2e- Mn2+(aq) (1) E0 = -0.23V + 2H2O(l) E0 In ranking the strengths, compare the combinations in terms of NO3-(aq) + 4H+(aq) (2) N2H5+(aq) = 1.23V Put the equations together in varying combinations so as to produce (+) E0cell for the combination Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E0 Balance the number of electrons gained and lost without changing the E0 (1) NO3-(aq) + 4H+(aq) + 3e- Rev (2) N2H5+(aq) (a) (b) Rank the relative strengths of the oxidizing and reducing agents: PLAN: Zn(s) stronger reducing agent (A) + 3e- (1) NO(g) + 2H2O(l) (3) MnO2(s) +4H+(aq) + 2e- X4 X3 + 4H+(aq) + 3e- E0 = -0.96V Mn2+(aq) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- (B) 2NO(g) + 3MnO2(s) + 4H+(aq) E0 = +0.23V E0cell = 1.19V 4NO(g) + 3N2(g) + 8H2O(l) NO3-(aq) (3) MnO2(s) +4H+(aq) + 2e- E0cell 21-27 NO(g) + 2H2O(l) N2(g) + 5H+(aq) + 4e- 4NO3-(aq) + 3N2H5+(aq) + H+(aq) Rev (1) NO(g) + 2H2O(l) NO(g) + 2H2O(l) E0 = 0.96V N2(g) + 5H+(aq) + 4e- Mn2+(aq) + 2H2O(l) E0 = 1.23V X2 E0cell = 0.27V X3 2NO3-(aq) + 3Mn3+(aq) + 2H2O(l) 21-28 Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (3 of 4) Rev (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- (3) MnO2(s) +4H+(aq) + 2e(2) N2H5+(aq) E0 = +0.23V Mn2+(aq) + 2H2O(l) E0 = 1.23V (C) N2H5+(aq) + 2MnO2(s) + 3H+(aq) Mn2+(aq) + 2H2O(l) A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of E0cell = 1.46V N2(g) + 5H+(aq) + 4e- (3) MnO2(s) +4H+(aq) + 2e- Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (4 of 4) Oxidizing agents: MnO2 > NO3- > N2 X2 Reducing agents: N2H5+ > NO > Mn2+ N2(g) + 2Mn2+(aq) + 4H2O(l) (b) Ranking oxidizing and reducing agents within each equation: (A): oxidizing agents: NO3- > N2 reducing agents: N2H5+ > NO (B): oxidizing agents: MnO2 > NO3- reducing agents: NO > Mn2+ (C): oxidizing agents: MnO2 > N2 reducing agents: N2H5+ > Mn2+ 21-29 21-30 Figure 21.8 The reaction of calcium in water Relative Reactivities (Activities) of Metals Oxidation half-reaction Ca(s) Ca2+(aq) + 2e- Reduction half-reaction 2H2O(l) + 2eH2(g) + 2OH-(aq) Metals that can displace H2 from acid Metals that cannot displace H2 from acid Metals that can displace H2 from water Metals that can displace other metals from solution Overall (cell) reaction Ca(s) + 2H2O(l) Ca2+(aq) + H2(g) + 2OH-(aq) 21-31 21-32 Free Energy and Electrical Work DG a -Ecell DG = wmax = charge x (-Ecell) K E0cell 1 >0 at equilibrium >0 E0cell •When Q = and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell •When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell 0.0592 Ecell = E0cell - log Q n Overall (cell) reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) 21-37 Figure 21.11 21-38 A concentration cell based on the Cu/Cu2+ half-reaction Sample Problem 21.7: PROBLEM: Calculating the Potential of a Concentration Cell A concentration cell consists of two Ag/Ag+ half-cells In half-cell A, electrode A dips into 0.0100M AgNO3; in half-cell B, electrode B dips into 4.0x10-4M AgNO3 What is the cell potential at 298K? Which electrode has a positive charge? E0cell will be zero since the half-cell potentials are equal Ecell is calculated from the Nernst equation with half-cell A (higher [Ag+]) having Ag+ being reduced and plating out, and in half-cell B Ag(s) will be oxidized to Ag+ PLAN: SOLUTION: Ag+(aq, 0.010M) half-cell A Ecell = E0cell Oxidation half-reaction Cu(s) Cu2+(aq, 0.1M) + 2e- Reduction half-reaction Cu2+(aq, 1.0M) + 2eCu(s) log Ag+(aq, 4.0x10-4M) half-cell B [Ag+]dilute [Ag+]concentrated Ecell = V -0.0592 log 4.0x10-2 = 0.0828V Half-cell A is the cathode and has the positive electrode Overall (cell) reaction Cu2+(aq,1.0M) Cu2+(aq, 0.1M) 21-39 Figure 21.12 0.0592V 21-40 The laboratory measurement of pH Table 21.3 Some Ions Measured with Ion-Specific Electrodes Pt Glass electrode Reference (calomel) electrode Hg AgCl on Ag on Pt 1M HCl Thin glass membrane 21-41 Paste of Hg2Cl2 in Hg KCl solution Species Detected Typical Sample NH3/NH4+ Industrial wastewater, seawater CO2/HCO3- Blood, groundwater F- Drinking water, urine, soil, industrial stack gases Br- Grain, plant tissue I- Milk, pharmaceuticals NO3- Soil, fertilizer, drinking water K+ Blood serum, soil, wine H+ Laboratory solutions, soil, natural waters Porous ceramic plugs 21-42 Figure 21.13 The corrosion of iron Figure 21.14 21-43 Enhanced corrosion at sea 21-44 Figure 21.15 The effect of metal-metal contact on the corrosion of iron faster corrosion The use of sacrificial anodes to prevent iron corrosion cathodic protection 21-45 Figure 21.17 Figure 21.16 21-46 The tin-copper reaction as the basis of a voltaic and an electrolytic cell Figure 21.18 The processes occurring during the discharge and recharge of a lead-acid battery VOLTAIC(discharge) voltaic cell Oxidation half-reaction Sn(s) Sn2+(aq) + 2eReduction half-reaction Cu2+(aq) + 2eCu(s) Overall (cell) reaction Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s) 21-47 electrolytic cell Oxidation half-reaction Cu(s) Cu2+(aq) + 2eReduction half-reaction 2+ Sn (aq) + 2e Sn(s) Overall (cell) reaction Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s) ELECTROLYTIC(recharge) 21-48 Sample Problem 21.8: Table 21.4 Comparison of Voltaic and Electrolytic Cells Electrode PROBLEM: Cell Type DG Ecell Name Process Voltaic 0 Anode Oxidation - Voltaic 0 Cathode Reduction + Electrolytic >0 0