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Handouts Transport numbers to obtain values of mobility u and of ionic molar conductivities λ for single, individual ions t±: transport, transference or migration numbers t+ for a cation; t- for an anion t+ is the fraction of current, I+, carried by the cation total current: I = I+ + I-: I +_ t+_ = I+ + I - t± means t+ or t-, and I± means the corresponding I+ or I-: I+ ; = I - t+ = t- I++ I - I++ I - _I = I + + I - ; thus t+ + t - = 1 and 0 t+_ 1 For a general electrolyte z + | |- + | |- Aa+ Bbz- _ a Az+ (aq)+ b B z- (aq) the electro-neutrality condition must hold: a z+ = b | z- | in dissociation equilibrium for n initial mol of a weak electrolyte we will have in solution (1-α)n undissociated molecules naα cations and nbα anions transport numbers can be measured for each ion separately Now, what is their relation to mobilities and ionic molar conductivities? F is the charge on 1 mol of charge, so a mol of a cation with charge z+ have a charge of Q+ = aFz+ (for 1 mol AaBb) and Q- = bFz- (for 1 mol AaBb) So the electricity flowing through a given area A in unit time is aFz+u+ for 1 mol AaBb in solution and the negative ion current is bFz-u-, thus aF z+ u+ a z + u+ t+ = = aF z+ u+ + bF | z - | u - a z+ u+ + b | z - | u - _ bF | z - | u - b | z- | u- t- = = aF z+ u+ + bF | z - | u - a z+ u+ + b | z - | u - and thus t+ + t- = 1. Since az+ = b|z-|, we can replace b|z-| in the denominator of t+ with az+, and az+ in the denominator of t- with b|z-|: a z + u+ u+ t+ = = a z + u+ + a z+ u - u + + u - b | z- | u- u- t- = = b | z - | u+ + b | z - | u - u + + u - With the short hand writing ( Aa Bb ) ; + ( Az++ ) ; - ( B|z-|- ) we know from before + + = F z + u + ; u+ = F z+ - - = F | z- | u- ; u- = F | z- | = a + + b - = F(a z+ u+ + b | z - | u - ) and therefore + u+ F z+ t+ = = u+ + u - + + - F z+ F | z - | - u- F | z- | t- = = u+ + u - - + - F z+ F | z - | Now we can cancel F and since az+ = b|z-| we can replace b 1 a 1 a 1 b 1 z+ = | z- | ; = ; | z - | = z+ ; = a z+ b | z - | b | z - | a z+ and thus + z+ + = a + t+ = = + + b - + - a + + b - b + z+ a z+ a - | z- | - b - t- = = = a + + - a + + b - a + + - b | z- | | z- | b Since (see before) = a + + b - a + b - thus t + = ; t- = For any strong electrolyte, t+ and t- can be measured for different values of c+ = ac and c- = bc and then extrapolated to t+o and t-o. Since t+o and t-o can be measured individually, we can split Λo into o ; o= o o o = t+ - o + t- a b For NaCl e.g. a = b = 1 and z+ = |z-| = 1 but for MgO e.g. a = b = 1 but z+ = |z-| = 2 Methods to measure t+ and t- Hittorf method After the anode and cathode solutions are connected, a small current is sent through the Hittorf cell for a period of time. Then the connection is closed, and the anode and cathode solutions after electrolysis can be collected separately and undergo chemical analysis, which yields the concentration changes. For simplicity let us assume M+A- electrolytes (can be done for MaAb ones also, but one has to care then for a and b) cation current fraction: t+ anion current fraction: t- After passing of 1 mol electrons through the cell, F charges have passed through thus Ft+ charges are transported to the - electrode (cathode) by M+ ions and Ft- charges are transported to the + electrode (anode) by A- ions and because 1 mol of charge has passed: 1 mol of M+ was discharged at the cathode and 1 mol of A- was discharged at the anode Cathode solution By passage of 1 mol of electrons = F charges t- mol of A- ions are lost by transport to the anode 1 mol M+ ions are discharged and thus lost t+ mol M+ ions are transported into the cathode solution: t+ mol M+ ions are gained = -t+ mol M+ ions are lost together (1 - t+) mol M+ ions are lost = t- mol M+ ions are lost thus t- mol MA are lost in the cathode solution Anode solution By passage of 1 mol of electrons = F charges t+ mol of M+ ions are lost by transport to the cathode 1 mol A- ions are discharged and thus lost t- mol A- ions are transported into the anode solution: t- mol A- ions are gained = -t- mol A- ions are lost together (1 - t-) mol A- ions are lost = t+ mol A- ions are lost thus t+ mol MA are lost in the anode solution: amount MA lost in anode solution t+ = amount MA lost in cathode solution t - t+ 1 - t- = = 1 - t+ t- Chemistry 311, Section 02, Lecture01 ö Instructor: W. Fِ rner Office: 4/147-3 Phone: 3553 email: forner@kfupm.edu.sa Lectures: Build. 5 - 201; SMW 11:00 - 11:50 am Office hours: SMW 9:00 - 10:50 in case I am not in office, check right outside Building 4, maybe I am only 5 minutes out for smoking You can also make extra appointments for meetings by email or phone Quizzes: in almost all Review lectures, others will be announced about 1 week before Unexcused absence in a quiz: 0 out of 10, no makeup Homework is assigned but will not be collected. However, if you solve the homework you can get good grades in Quizzes (will be homework problems only) Quizzes (out of 100) will be averaged to 75 each unexcused absence (excuses from student affairs must be in 1 week after the absence): -1 point from classwork; 9 unexcused absences: DN Major dates, grading of the course, reading assignments: see syllabus Sections which are not in the syllabus are not covered in class and are not part of exams Overview Electrochemistry Chapter 7 : Solutions of electrolytes (electrolysis) Chapter 8 : Electrochemical cells (batteries) 1. Major Exam Chemical Kinetics Chapter 9 : Elementary (one-step) reactions Chapter 10 : Kinetics and mechanisms of composite reactions, e.g.: Step (1) : A+BC Step (2) : C+DE Overall (C cancels) : A+B+DE 2. Major Exam Surface Chemistry Chapter 18 : Adsorption and adsorption isotherms, heterogeneous catalysis Transport Properties Chapter 19 : Viscosity of gases and liquids and diffusion Final Exam: Chapters 7,8,9,10,18,19 (comprehensive) ELECTROCHEMISTRY (Solutions of electrolytes) Objectives general introduction, Faraday's law of electrolysis, Molar conductivity We are concerned with 1. Properties, like conductivity or dissociation of electrolytes in solutions Electrolyte: a compound that dissociates into ions when it is dissolved, e.g. in water or other solvents 2. Processes that happen at electrodes when immersed in electrolyte solutions Informations can be obtained by the investigation of electrical effects in solutions Measurement of conductivity versus concentration yields: extent of ionization (e.g. in water) association of ions in solutions (movement of ions bound close together) movement of ions in the solvent (mostly water) Basic concepts for electricity Electrical units Electrostatic force between 2 charges (unit As) Q1 and Q2 at distance r: Q1 Q 2 F= 4 o r 2 true in vacuum only εo: permittivity or dielectric constant of the vacuum: 8.854 x 10-12 C2J-1m-1 1 C = 1 As; 1 J = 1 Nm = 1 kgm2/s2 = 1 VAs Thus 1 C2J-1m-1 = 1 A2s2/(VAsm) = 1 As/(Vm) force not in vacuum but in a medium with relative dielectric constant ε (no unit): Q1 Q 2 F= 4 o r 2 r = o : relative permittivity Electric field: force/unit charge electric field of charge Q at distance r: Q E= 4 o r 2 unit: 1 N/C = 1 (J/m)/C = (1 VAs)/(Asm) = 1 V/m electric potential φ at distance r from charge Q (if the field depends only on distance and not on angles) is defined such that d E=- thus Edr = - d dr Integration: Q dr d = = - Edr = - 4 o r 2 Q = 4 o r The unit is 1 V Faraday's law of electrolysis 1. law: the mass of the element produced (deposited) at an electrode is proportional to the amount of electricity (charge) passed through the solution, where Q = It; unit: 1 As = 1 C (Coulomb) 2. law: the mass of the element produced (deposited) at an electrode is proportional to the equivalent weight of the element M+ + e- M: equivalent weight = molar weight M2+ + 2e- M: equivalent weight = 1/2 molar weight charge of 1 electron: 1.602 x 10-19 C charge of 1 mol of electrons: 1.602 x 10-19 C x NA (Avogadros's number) = = 1.602 x 10-19 C x 6.022 x 1023 mol-1 = 96472 C/mol = F (Faraday's constant) Ag+ + e- Ag H+ + e- 1/2 H2 Thus a charge of 96472 C produces 1 mol Ag, but 1/2 mol H2 Electrolysis: the negative electrode attracts cations and is called cathode the positive electrode attracts anions and is called anode Batteries: the negative electrode produces electrons and is called anode the positive electrode takes up electrons and is called cathode A solution of gold (III) nitrate, Au(NO3)3 is electrolized using a current of 0.0250 A until 1.20 g Au (molar mass MM = 197.0 g/mol) is deposited at the cathode. 1. What is the amount of electricity (charge) passed through the solution? Au3+ + 3e- Au equivalent to 1/3 Au3+ + e- 1/3 Au 1 mol e- yields 1/3 mol Au 96472 C of charge yield 1/3 mol Au moles of Au deposited: (1.20 g Au)/(197.0 g/mol) = 6.091 x 10-3 mol Au 1 mol e 6.091 10-3 mol Au = 18.27 10-3 mol e 1 mol Au 3 Multiply by Faraday's constant to get the charge: C Q = 96472 18.27 10-3 mol = 1.76 103 C (As) mol 2. What is the time needed for that? Q = It and thus the time is t = Q/I: 1.76 103 As t= = 7.04 104 s 0.0250 A = 1.17 103 min = 19.6 h 3. The anode reaction is 2 H2O O2 + 4H+ + 4e- 2 oxygen atoms are oxidized from oxidation number -2 in water to 0 in oxygen, and thus 4e- are released. Divided by 4 the equation for 1 e- is: 1/2 H2O 1/4 O2 + H+ + e- How much (volume) of O2 gas is obtained at STP (0oC, 1 atm) ? 1 mol e- 1/4 mol O2 and thus we have 1 mol O 2 18.27 10-3 mol e- 4 mol e- = 4.57 10-3 mol O 2 The molar volume of O2 at STP can be obtained from the ideal gas equation PVm = nRT at P = 1 atm, T = 273.15 K and n= 1 mol and is Vm = 22.4 L/mol (1 L = 1 dm3) and thus L V O2 = 22.4 4.57 10-3 mol O2 = 0.102 L mol O2 = 0.102 dm3 Molar Conductivity The nature of solutions can be obtained from conductivity measurements: non-electrolytes: no dissociation into ions, no charge carriers, low conductivity electrolytes: dissociation into ions when dissolved, high conductivity two types of electrolytes: strong electrolytes: 100% of the salt is dissociated into ions when dissolved, e.g. NaCl: NaCl Na+(aq) + Cl-(aq) in the solution (no matter if all dissolves or not) there are no "NaCl molecules", only ions weak electrolytes: only partially dissociated and there are also electrolyte molecules in the solution not only ions, e.g. acetic acid CH3COOH (HAc). dissociation equilibrium: CH3COOH(aq) <=> CH3COO-(aq) + H+(aq) or HAc(aq) <=> Ac-(aq) + H+(aq) in solution all 3 species are present measurement of conductivity vs concentration of the solute gives informations Ohm's law: resistance R is proportional to the voltage V and inversly proportional to the current strength I (proportional to 1/I): R = V/I unit of R: 1 Ohm = 1 Ω = 1 V/A (1 Volt/Ampere) conductance G = 1/R; unit 1 S (Siemens) = 1 Ω-1 When the current flows through a cell of area A (area of the electrodes) and length l (distance between the electrodes), the conductivity κ is defined as A l G = ; = G l A the conductivity is thus the conductance of a unit cube For solutions the conductivity is also called the electrolytic conductivity unit: 1 Sm-1 = 1 Ω-1m-1 more common is the use of cm instead of m: 1 Scm-1 = 1 Ω-1cm-1 S S S 1 = 1 -2 = 100 cm 10 m m Kohlrausch: definition of the equivalent conductivity or also called molar conductivity Λ: = c where c is the molarity of the electrolyte example: the electrolytic conductivity of 0.1 M (0.1 mol/L = 0.1 mol/dm3) acetic acid is 5.3 x 10-4 Ω-1cm-1, what is the molar conductivity? 5.3 10-4 -1 cm-1 = = c 0.1 mol dm-3 _ 1 cm -1 -1 = 5.3 10 -3 mol ( 10+1 cm )-3 1 cm -1 -1 = 5.3 10 -3 mol 10-3 cm-3 = 5.3 -1 2 2 cm S cm = 5.3 mol mol Λ(c): in all cases Λ slightly decreases when c is increased: more electrolyte = larger c and smaller molar conductivity G(c), κ(c): in all cases G and κ increase when c is increased: more electrolyte = larger c = more ions = larger conductance G and larger conductivity κ Lecture 02 The electrolytic conductivity at 20oC of 1.00 x 10-2 M aqueous solution of acetic acid (HAc) was found to be 1.60 x 10-4 S cm-1. Some Λo values are Λo(HCl) = 426.2 S cm2 mol-1, Λo(NaAc) = 91.0 S cm2 mol-1 and Λo(NaCl) = 126.5 S cm2 mol-1. (a) Calculate the degree of ionization of acetic acid. Solution: c = 1.00 10-2 M = 1.00 10-2 mol(10 cm )-3 = 1.00 10-5 mol cm-3 1.60 10-4 S cm-1 (HAc)= = -5 -3 = 16.0 S cm2 mol-1 c 10 mol cm _ (HAc)= (HCl)+ (NaAc)- (NaCl) o o o o = (426.2+ 91.0 - 126.5) S cm2 mol-1 = 390.7 S cm2 mol-1 16.0 = = = 0.04095= 0.0410 o 390.7 (b) Calculate the equilibrium constant of acetic acid with units in the above solution. Solution: 2 = 1.00 -2 M 0.040952 Ka= c 10 (1 - ) 1 - 0.04095 = 1.75 10-5 M (c) What is the van't Hoff factor and the osmotic pressure of this solution? Solution: v is the theoretical number of particles per dissolved molecule in the solution, if all molecules of HAc would dissociate (this is not really the case, so theoretical number). Since HAc, if all molecules would dissociate, would yield H+ and Ac-, and thus v = 2 ions per molecule. When the degree of dissociation is α, then in solution we have for each dissolved molecule (1 - α) undissociated molecules and vα ions. Thus the van't Hoff factor is: i = 1 - + v = 1 - 0.0410 + 2 0.0410 = 1.0410 The concentration is in mol/L and must be changed to mol/m3: mol mol c = 1.00 10- 2 = 1.00 10- 2 dm 3 ( 10-1 m )3 mol = 10.0 3 m and the osmotic pressure Π is = icRT mol J = 1.0410 10.0 3 8.315 293.15 K m K mol kg = 2.54 104 = 2.54 104 Pa = 0.254 bar m s2 = 0.251 atm units: N kg 1 Pa = 1 2 =1 m m s2 1 = 10-5 bar = atm 1.01325 10 5 Lecture 034 Alternating Currents (AC) Concept of ionic atmosphere moving together with the ions: molar conductivity Λ depends on the frequency of an alternating potential If the frequency is large enough (very fast change of direction of move) i.e. so large, that the oscillation time is smaller (faster) than the relaxation time of the ionic atmosphere: then the ions are stationary and the atmosphere stays symmetric The relaxation and electrophoretic effects, which cause Λ to decrease with increasing concentration c, vanish at large AC frequencies Thus at large AC frequencies Λ increases with increasing c (more ions = charge carriers at higher c) Wien effect large potentials of about 20000 V/cm or larger result in a high speed of the ions Thus the ions move so fast, that the ionic atmosphere cannot follow them Thus at very large potentials Λ also increases with increasing c, because the effects from the ionic atmosphere vanish when the ions are faster than the atmospheres Dissociation fields weak electrolytes in large fields (potentials): Λ becomes large and weak electrolytes behave like strong electrolytes because the large potentials force the dissolved, undissociated molecules in solutions of weak electrolytes to dissociate. lecture 04: linearized plots, independent migration of ions: mobility and ionic molar conductivity of ions Question: which kinds of plots must be linear in case of strong and weak electrolytes? Determination if an electrolyte is strong or weak from linear plots strong electrolytes molar conductivity as function of concentration: = o - a c Thus if the electrolyte is a strong one then a plot of Λ versus c must be linear (Correlation coefficient r2 close to 1) the intercept of the straight line gives Λo which is a very good value for strong electrolytes, the slope is -a (the constant in the formula) weak electrolytes Ostwald's dilution law gives c 2 K= 1- The degree of dissociation (fraction of dissociated molecules) is o and thus c 2 c 2 c 2 2 2 K = o = o = o ( o - ) o - 1 1- o o Multiplication by the total denominator Λo(Λo - Λ): K o ( o - ) = c 2 K o - K o = c 2 2 Division by KΛo2Λ: 1 1 c - = o K o2 1 1 1 = + c o K o 2 Thus if the elctrolyte is a weak one, then a plot of 1/Λ versus (cΛ) = κ must give a straight line (Correlation coefficient r2 close to 1) with intercept 1 1 intercept = ; thus o = o intercept But Λo values for weak electrolytes from such plots are not very good. To obtain correct Λo values in this case the independent migration of ions law must be used But using intercepts from such plots and the slope gives good K values: 2 1 intercept slope = = K o 2 K 2 intercept K= slope In general: derivations given in labs can be asked in exams!! Lecture 05 Hittorf Cell nA: number of moles of MA lost in the anode compartment nC: number of moles of MA lost in the cathode compartment nT: total number of moles of MA deposited at the electrodes nT = charge passed/Faraday's constant, if M+, and A- charge passed = current strength (I) x time (t) nT = nA + nC ratio of the transport numbers: t+/t- = nA/nC since t- = 1 - t+: t+ n = A ; t+ nC = (1 - t+ ) n A 1 - t + nC nA n t+ = = A n A + nC nT and therefore nA n t+ = ; t- = C It/F It/F Moving boundary method for example M+ (can be H+) or A- M'+ and A'- must be slower than M+ and A- the indicator solution must change colour when M+ and A- arrive in it for example an acid/base indicator if M+ is H+ and A- is OH- A distinct moving boundary is obtained between M'A and MA on one side and between MA and MA' on the other side because M'+ is slower than M+ and cannot overtake it and because A'- is slower than A- and cannot overtake it moved distances from start (a, b): aa' proportional to u+ and bb' proportional to u- to make the boundaries visible indicators must be added that change colour when M+ or A- are there aa bb t+ = ; t- = aa + bb aa + bb Often seen nonsense lab results: t+ = 360 which is nonsense, because 0 t+ 1 must be and to make t+ + t- = 1, t- = -359 which is also nonsense, because 0 t- 1 must also be mistake because the students forget that they read their current in mA and not in A Lab: 1 boundary in acid solution + acid base indicator at the cathode a neutral, yellow CdCl2 solution is produced, but the H+ (acid in the solution) ions move faster to the cathode, and the indicator changes colour when the H+ ions leave the - in the beginning - acidic solution below boundary: yellow, neutral CdCl2 solution (H+ ions moved out) above boundary: red (methylorange) acidic solutions (H+ still there) Q = It charge transported by the H+ ions, F Faraday's constant, c known concentration of the acid solution, A diameter of the tube, aa' distance travelled by the boundary in time t: Q aa = t+ FcA Lecture 07 Consider an electrolyte AaBb Then for electroneutrality we must have az+ = b|z-| and for ativity coefficients a+b a b y+_ = y+ y- (a + b) log10 y+_ = a log10 y+ + b log10 y- Now we insert the DHLL: log10 yi = - zi2 B I to obtain (a + b) log 10 y+_ = - (a z+ + b z -2 )B I 2 a 2 z+ b2 z -2 2 =- a + b B I electroneutrality: (b|z-|)2 = (az+)2: 2 1 1 (a + b) log 10 y+_ = - a 2 z+ + B I a b a+b = - a 2 z+ 2 B I ab a 2 log 10 y+_ = - z+ B I b electroneutrality: b|z-| = az+: a/b = |z-|/z+ and thus log10 y+_ = - z+ | z- | B I and in water at 25oC: I log10 y+_ = - 0.51 z+ | z- | mol L-1 ionic strength I for 1 M solutions of different electrolytes: uni-univalent: NaCl Na+(aq) + Cl-(aq) 1 1 ci z i = (1 M (1 ) + 1 M (-1 ) ) = 1 M 2 2 2 I= 2 i 2 summation always over ALL ions in a solution uni-bivalent: Na2SO4 2 Na+(aq) + SO42-(aq) 1 1 I= ci z i 2 2 i 2 = (2 1 M (1 )2 + 1 M (-2 )2 ) = 3 M uni-trivalent: Na3PO4 3 Na+(aq) + PO43-(aq) 1 1 I= ci z i 2 2 i 2 = (3 1 M (1 )2 + 1 M (-3 )2 ) = 6 M bi-trivalent: Mg3(PO4)2 3 Mg2+(aq) + 2 PO43-(aq) 1 1 I= ci z i 2 2 i 2 = (3 1 M (2 )2 + 2 M (-3 )2 ) = 15 M Lecture 08 Ionic equilibria Objectives: activity coefficients from equilibrium constants and solubility products Equilibrium is obtained very rapidly in ionic solutions example: HAc <=> H+ + Ac- HAc: CH3COOH; Ac-: CH3COO- activity: a = cy activity coefficient is written as english lower case y or as greek lower case gamma (γ) The dissociation constant Ka of acetic acid is practically a H + a Ac- [ H + ] [ Ac ] y+ y - - Ka= = a HAc [HAc] yu yu: activity coefficient of the undissociated acid about equal to 1 because the molecule has no charge (not exactly 1 because of the dipole moment) the activity coefficients of cations (y+) and of anions (y-) are not 1, because of the charges and their interactions mean activity coefficient y± for AB compunds: y± = (y+y-)1/2; (y±)2 = y+y- for AaBb compounds: (y±)a+b = (y+)a(y-)b Therefore + - [ H ][ Ac ] Ka= y+ _ 2 [HAc] taking logarithm basis 10 gives u c +c - log 10 K u = log 10 H Ac a + 2 log 10 y+_ c HAc eq u c H + c Ac- Kc = u c HAc eq superscript u: units taken out subscript eq: equilibrium superscript o: standard state: 25oC, 1 atm, all concentrations equal to 1 M (not equilibrium). so superscript o at an equilibrium constant means only 25oC, 1 atm Thus u c +c - log 10 H Ac = log 10 K u - 2 log 10 y a c HAc eq = - pK a - 2 log 10 y+_ Ostwald's dilution law: c H + c Ac- ; = 2 =c c HAc eq 1- o Thus u c 2 log 10 = log 10 K u - 2 log 10 y 1- a eq On the left and right hand side the same unit must be taken out (mol/L) DHLL: u c 2 log 10 1- = log 10 K u + 2 z+ | z - | B I a eq B = 0.51 dm3/2mol1/2 in water at 25oC 1 I= ci zi2 ; c H + = c , c Ac- = c 2 i _ c HAc = c(1 - ) (no ion) I = c 12 + c 12 = c 1 2 if there are no other ions present, which in case must be added to I. Thus one has to measure Λ at different concentrations c and obtain Λo. Then one can calculate ionic strength for each c value and u 2 c o u c 2 1 - = log 10 log 10 1- o Now, at small c values only a plot of this quantity versus square root of the ionic strength I1/2 will give a straight line where c is small enough that DHLL is obeyed. The straight line at small I (c) has the slope 2B when z+|z-| = 1 as in our case and in water at 25oC: 2B = 1.02 dm3/2 mol1/2 at 25oC and 1 atm, Ku is also written as Ko (without units) When I and thus c goes to 0, then the equation tells u c 2 log10 1- = log10 K o Because at c = 0 we have I = 0 and according to DHLL y± = 1 and thus log10y± = 0 At larger I (larger c) where DHLL is not obeyed (no straight line) the difference between the measured points and log10Ko is equal to -2log10y± and the mean activity coefficient can be obtained at all c values, since u c 2 - 2 log 10 y+_ = log 10 1- - log 10 K o values of y+ or y- cannot be obtained, only the mean value by plotting u c 2 f = log10 1- versus I When I is very small, the plot can also be done as f versus m (molality) Solubility products The solubility product Ks is a constant in saturated (as much electrolyte is dissolved as possible) solutions of electrolytes. Example: AgCl <=> Ag+(aq) + Cl-(aq) The largest amount of AgCl that can be dissolved is the solubility s: s = [AgCl]max = [Ag+]max = [Cl-]max and K s = a Ag+ aCl - = [ Ag ] [ Cl - ] max y+ y- + max = s s y+_ = s 2 y+_ since y+_ = y+ y- 2 2 2 Example: PbCl2 <=> Pb2+(aq) + 2 Cl-(aq) The largest amount of PbCl2 that can be dissolved is the solubility s: s = [PbCl2]max = [Pb2+]max = 1/2 [Cl-]max and thus [Cl-]max = 2s K s = a Pb2+ aCl - = [ Pb ] ([ Cl - ] max )2 y+ y -2 2 2+ max = s (2s )2 y+_ = 4 s3 y+_ since y+_ = y+ y -2 3 3 3 Adding of an inert electrolyte An electrolyte inert to AgCl is one that has no ions that are also in AgCl NaBr is inert to AgCl since it has no Ag+ or Cl- ions NaCl is non-inert to AgCl because it has Cl- ions as also AgCl then [Cl-] from NaCl is much larger than [Cl-]max from AgCl and K s = a Ag+ aCl - = [ Ag ] [ Cl - ] y+ y - + max = s (s + [ Cl - ] NaCl ) y+_ s[ Cl - ] NaCl y+_ 2 2 since [ Cl - ] NaCl much larger than s adding of an inert electrolyte brings about no chemical change, but it changes y± because it changes I: no inert NaBr: 1 1 I = ([ Ag+ ] max z+ + [ Cl - ] max z -2 ) = (s 12 + s 12 ) = s 2 2 2 inert NaBr present: 1 I = ([ Ag + ] m ax12 + [ Cl - ] m ax12 + [ Na+ ] 12 + [ Br - ] 12 ) 2 1 = (2s + 2[NaBr]) = s + [NaBr] 2 At low I DHLL (at higher I the CI correction must be added) is valid DHLL: log10 y+_ = - z+ | z- | B I Thus, when I increases because of adding an electrolyte, then the activity coefficients drops (becomes smaller) because of the - sign, when I is so small that DHLL holds at high I log10y± increases when I increases, can even become positive, because the CI correction (positive) must be added low I y± decreases with increasing I, but Ks must be a constant: 2 2 K s = s y = constant+_ When Ks must be the same, then when y± decreases, s must increase thus more AgCl goes into solution than into pure water when NaBr is added to it: salting in effect large I y± increases with increasing I, but Ks must be a constant: 2 2 K s = s y = constant +_ When Ks must be the same, then when y± increases, s must decrease thus less AgCl goes into solution than into pure water when NaBr is added to it: salting out effect Example: Ks(BaSO4) = 9.2 x 10-11 M2 (mol2/L2 = mol2/dm6) What is y± in a solution (in water at 25oC) that is saturated in BaSO4 and contains also 0.05 M KNO3 and 0.05 M KCl, salting in or salting out effect? assume that DHLL applies In the solution we can neglect the Ba2+ and SO42- concentrations because they are much smaller than those of KNO3 and KCl: 1 I = ([ K + ] KNO 3 12 + [ NO -3 ] 12 + 2 + [ K + ] KCl 12 + [ Cl - ] 12 ) 1 = (0.05+ 0.05 + 0.05 + 0.05) M = 0.1 M 2 DHLL: L mol log 10 y+_ = - z+ | z - | B I = - 2 2 0.51 0.1 mol L = - 0.645 y+_ = 10-0.645 = 0.226 note that for BaSO4 z+ = +2 and z- = -2, thus |z-| = +2 The solubility product is 2 2 Ks= s y +_ and thus Ks s= = 4.24 10-5 M y+_ In pure water s is so small that I is almost 0 and thus y± is almost 1: K s = s thus s= K s = 9.6 10 M 2 -6 In the KNO3/KCl solution more BaSO4 is dissolved than in pure water: salting in effect If for a salt Ks is larger, i.e. 10-3 or so, the above is only the first step in an iteration: use the solubility to calculate a new I, a new y and a new solubility. Repeat until there is no more change salting out happens only when I is so large that the CI correction becomes important and z+ | z - | B I + CI log10 y+_ = - 1 + aB I Consider a simple AB electrolyte: AB(s) <=> A+(aq) + B-(aq), then K s = a a+ a B- = [ A ][ B ] y+ y - = [ A ][ B ] y+_ + - + - 2 log 10 K o = log 10 ([ A+ ][ B- ] ) + 2 log 10 y+_ u s _ log 10 ([A ] ][ B- ] ) = log 10 K o - 2 log 10 y + u s In a saturated solution that contains only AB and inert electrolytes we have s = [ A+ (aq) ] m ax= [ B- (aq) ] m ax= [AB(aq) ] m ax and thus log 10 ( su ) = log 10 K o - 2 log 10 y+_ 2 s 1 g = log 10 su = log 10 K o - log 10 y s 2 1 DHLL : g = log 10 su = log 10 K o + B I s 2 Plot as before g = log10su versus I1/2 then at low I (DHLL) a straight line with slope B (0.51 L1/2mol- 1/2 in water at 25oC) is obtained and its intercept is 0.5 x log10Kso. since 1 1 - log10 y+_ = g - log10 K o = log10 su - log10 K o s s 2 2 Lecture 45: Final Exam from May, 1999 1. The solubility product of BaSO4 is 9.2 x 10-11 mol2dm-6. (a) Calculate the mean activity coefficient of the Ba2+ and SO42- ions in a solution in water at 25oC, that is 0.05 M in NaNO3 and 0.05 M in KCl, assuming the DHLL to apply. Solution DHLL: log 10 +_ = - 0.51 z+ | z - | I M -1 ; z+ | z - |= + 2 | -2 |= 4 ionic strength : I= 1 2 0.05 M 12 ( Na+ ) + 0.05M (-1 )2 ( NO -3 ) + + 0.05M 12 ( K + ) + 0.05M (-1 )2 ( Cl - ) 1 = 4 0.05 M = 0.10 M 2 Since the concentrations of Ba2+ and SO42- ions are about the square root of the solubility product, and therefore about 10-5 to 10-4 M, they can be neglected as compared to 0.05 M. Then log10 +_ = - 0.51 4 0.1 = - 0.645 +_ = 10-0.645 = 0.226 (b) What is the solubility of BaSO4 in that solution, and Solution The constant solubility product implies that the concentrations of Ba2+ and SO42- in there are actually each one equal to the solubility of BaSO4 (saturated solution): K sp = aBa2+ aSO4- = cBa2+ + cSO4- - = s2 +_ 2 2 2 Note: the above is true only for AB electrolytes. K sp 9.2 10-11 M 2 s= = = 4.24 10-5 M +_ 0.226 (c) in pure water? Solution: Since in pure water the ionic strength contains only the terms for the Ba2+ and for the SO42- ions, which are very small due to the small Ksp, we can approximate I by 0: I 0 _ log10 +_ 0 _ +_ 1 _ K sp = s 2 +_ s 2 _ s = K sp = 9.59 10-6 M 2 The fact that more BaSO4 dissolves in NaNO3/KCl solution than in pure water is called "salting in" effect. 2. The cell Pt(s), H2 (1 bar) | HCl (m) | AgCl(s) | Ag(s) gives the following emf values at 25oC: m/(mol kg-1) E/V (E + (2RT/F) ln mu )/V __________________________________________ 3.215 x 10-3 0.5200 4.488 x 10-3 0.5039 5.619 x 10-3 0.4933 -3 7.311 x 10 0.4812 __________________________________________ (a) What is the cell reaction? Solution right side: reduction AgCl(s) + e- Ag(s) + Cl-(aq) left side : oxidation H2(g) 2H+(aq) + 2e- The cell reaction is the sum of the two with the electrons cancelling out. That can be done by 2 x reduction + oxidation or by reduction + 1/2 x oxidation: 1 H 2 (g)+ AgCl(s) Ag(s)+ HCl(aq) ; z = 1 2 ln a H + aCl - = RT o u Nernst : E = E - zF RT u RT =E - o ln ( c H + cCl - ) - ln ( + - ) = F F RT u RT =E - o 2 ln ( c HCl ) - ln ( +_ )2 = F F 2RT 2RT = Eo - ln mu - ln +_ F F 2RT 2RT _ E+ ln mu = E o - ln +_ F F Note: superscript u at the molality m of HCl is not a power, but means "remove the unit before taking the ln". Further note, that [H+]=[Cl-]=[HCl]=m at all times. Since E is positive, the reaction goes in the direction as derived above (if it would be reversed, E would be negative). It can be shown that the expression for the emf of this cell over a range of concentrations of HCl is given by 2RT o 2RT E+ u ln m = E - ln +_ F F (b) Complete the table above and determine Eo on the supplied sheet of graph paper with the axis already marked (plot y versus mu!). Linear regression is also acceptable. Solution: Trick: we define 2RT y E+ ln mu and x mu F A plot of y versus x2 could give us a straight line y=ax2+b with slope a and intercept b, when such plots are done for small concentrations as here. But when x is small then ln(1+x) = x NOT lnx = x. Since DHLL applies for small concentrations, we can replace ln by log10 with the help of 2.303 and insert the DHLL. Also a slope of 2RT/F is not important. Thus 2RT y = Eo - ln +_ ; DHLL : log 10 +_ = - z+ | z - | B I F _HCL : z+ | z - |= 1 ; ln +_ = 2.303 log 10 +_ 2RTB _ y = Eo + 2.303 I ; HCl : I = m F 2RTB _ y = Eo + 2.303 m F Then a plot of y vs I1/2 is a straight line where DHLL applies. Since with HCl I=0.5(c(H+)+c(Cl-))=m, thus here a plot of y vs m1/2 must give a straight line. At the intercept mu is 0. Thus (here: molality not molarity, thus NOT B=0.51M-1/2) 2RTB y( mu = 0) = b = E o + 2.303 m = E o _ b = E o F Completed table (2RT/F = 5.1389 x 10-2 V): m (10-3 mol/kg) E(V) y = E + (2RT/F)ln mu (V) 3.215 0.5200 0.2250 4.488 0.5039 0.2261 5.619 0.4933 0.2270 7.311 0.4812 0.2284 m (10-3 mol/kg) ln mu x = (mu)1/2 3.215 -5.7399 0.056701 4.488 -5.4063 0.066993 5.619 -5.1816 0.074960 7.311 -4.9184 0.085504 A plot y vs mu, assuming x1/2=x for small x, is not that good, cause a Taylor series of x1/2 around x=0 is not possible. In linear regression we have N xi y i - ( xi )( y i ) a= i i i ; D= N x -(x 2 i i 2 ) D i i ( xi2 )( yi ) - ( xi )( xi y i ) i i i i b= D where the number of (xi, yi) points N=4. From the table we get: Σyi = 0.9065 V Σxi = 0.28416 Σxi2 = 0.020633 Σxiyi = 0.064450 V (note that x=mu and thus has no units) D = 4 0.020633- (0.28416 )2 = 1.7851 10-3 4 0.064450V - 0.28416 0.9065V a= = 0.1171V 1.7851 10-3 0.020633 0.9065V - 0.28416 0.064450V b= = 1.7851 10-3 = 0.2183V = E o So, our final regression function is y = 0.1171V mu + 0.2183V (b) Calculate the activity coefficient for a 6.000 x 10-3 m solution of HCl. Solution: First we need y(0.006 m) for that concentration: y(0.006 m) = 0.1171V 0.006 + 0.2183 V = 0.2274 V Then using the right hand side of the above given equation we obtain: 2RT y = Eo - ln +_ F _ y(0.006 m) = 0.2183V - 5.1389 10- 2 V ln +_ (0.2183- 0.2274)V _ ln +_ = = - 0.1771 5.1389 10- 2 V _ +_ = e-0.1771 = 0.8377 3. A solution of A is mixed with an equal volume of a solution of B containing the same number of moles (initial concentrations ao and bo are equal) and the reaction A+BC occurs. At the end of 1 h, A is 75% reacted. How much of A (in % of the initial concentration ao) will be left unreacted at the end of 2 h, if (a) the reaction is first order in A and zero order in B. a t d[A] d[A] = - k[A] _ = - k dt dt ao [A] 0 a ln = - kt ao after 1 h : a = 25% a o = 0.25 ao 75% of ao has reacted, i.e. 25% of ao remains unreacted: ln 0.25 ln 0.25 = - k 1 h _ k = - = 1.386 h-1 1h a a(2 h) -1 2h = e- kt _ = e-1.386 h = 0.0625 ao ao _ a = 0.0625 ao = 6.25% ao unreacted after 2 h (b) the reaction is both first order in A and first order in B. d[A] = - k[A][B]= - k[A ] 2 cause a o = bo thus a = b at all times dt a t d[A] 1 1 1 1 [A ] 2 = - k dt _ - ( - ) = - kt ; kt = - 0 a ao a ao ao 1 1 1 3 _ k 1h= - _ k= (4 - 1) = 0.25 a o a o hao hao 3 1 1 6 1 1 1 7 _ 2 h= - _ = - _ = hao a ao ao a ao a ao 1 _ a = a o = 0.143 a o = 14.3% a o 7 4. Acetaldehyde decomposes thermally and the main products are methane and carbon monoxide. A likely mechanism is: CH3CHO CH3 + CHO (rate constant k1) (1) CH3 + CH3CHO CH4 + CH3CO (rate constant k2) (2) CH3CO CO + CH3 (rate constant k3) (3) CH3 + CH3 C2H6 (rate constant k4) (4) Derive the expression for the rate of formation of CO: d[CO] k1 3/2 = k2 [ CH 3 CHO ] dt 2 k4 (To simplify the steady state treatment, further reactions of the radical CHO have been omitted and their rate equations may be ignored) Solution: d[CO] v= = k 3 [ CH 3 CO] dt CH3CO is an intermediate with unknown, small and constant concentration. It must be obtained from the steady state approximation for the two intermediates CH3 and CH3CO in terms of reactant and product concentrations and rate constants. The reactant is CH3CHO, main products are CH4 and CO, C2H6 is a by-product, CHO is also an intermediate radical, but its reactions are neglected: d[ CH 3 ] (A) : = 0 = k 1 [ CH 3 CHO] - k 2 [ CH 3 ][ CH 3 CHO] - dt - 2 k 4 [ CH 3 ] 2 + k 3 [ CH 3 CO] d[ CH 3 CO] (B) : = 0 = k 2 [ CH 3 ][ CH 3 CHO] - k 3 [ CH 3 CO] dt Note that, looking at reaction (4) alone, we get the following rate term: 1 v4 = vCH 3 ,4 = - k 4 [ CH 3 ] 2 2 _ vCH 3 ,4 = - 2 k 4 [ CH 3 ] 2 Note, that the terms at k2 and k3 are with opposite signs in both equations. Thus when we add the equations we get 0 on the left hand side and those terms cancel out: (A)+ (B) : 0 = k 1 [ CH 3 CHO] - 2 k 4 [ CH 3 ] 2 k1 _ [ CH 3 ] = [ CH 3 CHO] 2 k4 k1 into (B) : k 2 [ CH 3 CHO [ CH 3 CHO] - k 3 [ CH 3 CO] = 0 2 k4 k 2 k1 _ [ CH 3 CO] = [ CH 3 CHO ] 3/2 k3 2 k4 k1 v = k 3 [ CH 3 CO] = k 2 [ CH 3 CHO ] 3/2 2 k4 5.(A) The density of liquid mercury at 273K is 13.6 g cm-3 and the surface tension is 0.47 N m-1. If the contact angle is 140o, calculate the capillary depression in a tube of 1 mm diameter. Solution: rhg = 2 cos 2 cos h= rg 2 0.47 kg s - 2 cos( 140o ) = 0.5 10-3 m 13.6 ( 10-3 kg)(10- 2 m )-3 9.81 m s - 2 = - 1.08 10- 2 m = - 10.8 mm (B) If a molecule dissociates on being adsorbed, the process is referred to as dissociative adsorption. (a) Derive the Langmuir adsorption isotherm for dissociative adsorption: K[ A2 ] = 1 + K[ A2 ] Solution: Reaction of 1 A2 with 2 surface sites: AA | | | | A2 + -S-S- -S-S- Thus: adsorption : va = k a [ A2 ](1 - )2 fraction of free sites squared _desorption : v d = k d 2 fraction of occupied sites squared In equilibrium the rates are equal: v a = v d _ k a [ A2 ](1 - ) = k d 2 2 2 k k = a [ A2 ] = K[ A2 ] ; K = a (1 - )2 k d kd = K[ A2 ] ; = K[ A2 ] - K[ A2 ] 1- K[ A2 ] _(1+ K[ A2 ] ) = K[ A2 ] ; = 1 + K[ A2 ] (b) What is the expression for the rate of a reaction, assuming that the mechanism of the reaction is unimolecular. Solution: unimolecular reactions on surfaces after dissociation: k K[ A2 ] v = k = 1 + K[ A2 ] (c) Sketch in the following graph the variation of the rate with [A2]1/2. Solution: (d) What would be the order of the reaction, when [A2] is (i) low Solution: 1 K[ A2 ] » 1 _ v = k K[ A2 ] : order 2 (ii) high Solution: k K[ A2 ] K[ A2 ] « 1 _ v = = k : order 0 K[ A2 ] 6. (a) In a normal adult at rest the average speed of flow of blood through the aorta is 0.33 m s-1. The radius of the aorta is 9 mm and the viscosity of blood at body temperature , 37oC, is about 4.0 x 10-3 kg m-1 s-1. (i) Calculate the rate of flow of blood through the aorta. Solution: area : A = r 2 = 3.1416 (9 10-3 )2 m2 = 2.54 10-4 m2 rate of volume flow= average speed area A _ 3 m -5 m dV = 0.33 2.54 10 m = 8.40 10 -4 2 = s s dt (ii) Calculate the pressure drop along 0.5 m length of the aorta. Solution: Pouisseuille equation: dV r 4 P 8l dV = ; P = 4 dt 8l r dt _ kg 8 4.0 10-3 0.5 m 3 ms -5 m P = 8.4 10 3.1416 (9 10 ) m -3 4 4 s 1 Torr = 65.2 Pa = 0.49 Torr 133.322 Pa (b) The diffusion coefficient for glucose in water is 6.81 x 10-10 m2 s-1 at 25oC. The viscosity of water at 25oC is 8.937 x 10-4 kg m-1 s-1 and the density of glucose is 1.55 g cm-3. Assume the glucose molecule to be spherical and to obey Stokes's law. (i) Estimate the radius of a glucose molecule. Solution: kBT k T S tokess law : D = _ r= B 6 r 6D J 298.15 K 1.381 10- 23 r= K 2 -10 m kg 6 3.1416 6.81 10 8.937 10-4 s ms Jm = 0.359 10-9 = 0.359 nm kg m2 2 s (ii) Estimate the molar mass of glucose. Solution: The radius of a spherical molecule gives the volume of 1 molecule: 4 4 V = r 3 = 3.1416 (3.59 10-9 )3 m3 = 1.938 10-28 m3 3 3 The density and the volume give the mass of 1 molecule: -3 10 kg m = V = 1.55 -2 3 1.938 10- 28 m3 = 3.00 10- 25 kg ( 10 m ) The mass of 1 molecule and Avogadro's number give the molar mass MM: MM = m N A = 3.00 10-22 g 6.022 1023 mol -1 = 180.7 g mol -1 Transport numbers to obtain values of mobility u and of ionic molar conductivities λ for single, individual ions t±: transport, transference or migration numbers t+ for a cation; t- for an anion t+ is the fraction of current, I+, carried by the cation total current: I = I+ + I-: I +_ t+_ = I+ + I - t± means t+ or t-, and I± means the corresponding I+ or I-: I+ ; = I - t+ = t- I++ I - I++ I - _I = I + + I - ; thus t+ + t - = 1 and 0 t+_ 1 For a general electrolyte z++ |z-|- A a B b _ a Az++ (aq)+ b B|z-|- (aq) the electro-neutrality condition must hold: a z+ = b | z- | in dissociation equilibrium for n initial mol of a weak electrolyte we will have in solution (1-α)n undissociated molecules naα cations and nbα anions transport numbers can be measured for each ion separately Now, what is their relation to mobilities and ionic molar conductivities? F is the charge on 1 mol of charge, so a mol of a cation with charge z+ have a charge of Q+ = aFz+ (for 1 mol AaBb) and Q- = bFz- (for 1 mol AaBb) So the electricity flowing through a given area A in unit time is aFz+u+ for 1 mol AaBb in solution and the negative ion current is bFz-u-, thus aF z+ u+ a z + u+ t+ = = aF z+ u+ + bF | z - | u - a z+ u+ + b | z - | u - _ bF | z - | u - b | z- | u- t- = = aF z+ u+ + bF | z - | u - a z+ u+ + b | z - | u - and thus t+ + t- = 1. Since az+ = b|z-|, we can replace b|z-| in the denominator of t+ with az+, and az+ in the denominator of t- with b|z-|: a z + u+ u+ t+ = = a z + u+ + a z+ u - u + + u - b | z- | u- u- t- = = b | z - | u+ + b | z - | u - u + + u - With the short hand writing ( Aa Bb ) ; + ( Az++ ) ; - ( B|z-|- ) we know from before + + = F z + u + ; u+ = F z+ - - = F | z- | u- ; u- = F | z- | = a + + b - = F(a z+ u+ + b | z - | u - ) and therefore + u+ F z+ t+ = = u+ + u - + + - F z+ F | z - | - u- F | z- | t- = = u+ + u - - + - F z+ F | z - | Now we can cancel F and since az+ = b|z-| we can replace b 1 a 1 a 1 b 1 z+ = | z- | ; = ; | z - | = z+ ; = a z+ b | z - | b | z - | a z+ and thus + z+ + = a + t+ = = + + b - b a + + b - + + - z+ a z+ a - | z- | - b - t- = = = a + + - a + + b - a + + - b | z- | | z- | b Since (see before) = a + + b - a + b thus t + = ; t- = - For any strong electrolyte, t+ and t- can be measured for different values of c+ = ac and c- = bc and then extrapolated to t+o and t-o. Since t+o and t-o can be measured individually, we can split Λo into = t+ ; o = t o o o o + o - - a b For NaCl e.g. a = b = 1 and z+ = |z-| = 1 but for MgO e.g. a = b = 1 but z+ = |z-| = 2 Methods to measure t+ and t- Hittorf method After the anode and cathode solutions are connected, a small current is sent through the Hittorf cell for a period of time. Then the connection is closed, and the anode and cathode solutions after electrolysis can be collected separately and undergo chemical analysis, which yields the concentration changes. For simplicity let us assume M+A- electrolytes (can be done for MaAb ones also, but one has to care then for a and b) cation current fraction: t+ anion current fraction: t- After passing of 1 mol electrons through the cell, F charges have passed through thus Ft+ charges are transported to the - electrode (cathode) by M+ ions and Ft- charges are transported to the + electrode (anode) by A- ions and because 1 mol of charge has passed: 1 mol of M+ was discharged at the cathode and 1 mol of A- was discharged at the anode Cathode solution By passage of 1 mol of electrons = F charges t- mol of A- ions are lost by transport to the anode 1 mol M+ ions are discharged and thus lost t+ mol M+ ions are transported into the cathode solution: t+ mol M+ ions are gained = -t+ mol M+ ions are lost together (1 - t+) mol M+ ions are lost = t- mol M+ ions are lost thus t- mol MA are lost in the cathode solution Anode solution By passage of 1 mol of electrons = F charges t+ mol of M+ ions are lost by transport to the cathode 1 mol A- ions are discharged and thus lost t- mol A- ions are transported into the anode solution: t- mol A- ions are gained = -t- mol A- ions are lost together (1 - t-) mol A- ions are lost = t+ mol A- ions are lost thus t+ mol MA are lost in the anode solution: amount MA lost in anode solution t+ = amount MA lost in cathode solution t - t+ 1 - t- = = 1 - t+ t-

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