AP Chemistry: General & Solubility Equilibrium

Transcription

1 AP Chemistry: General & Solubility Equilibrium AgCl Ag Cl = = [ x][ x] = x sp 10 2 x = [Ag ] = [Cl ] = molar solubility of AgCl = René McCormick AP IS A REGISTERED TRADEMAR OF THE COLLEGE BOARD, WHICH WAS NOT INVOLVED IN THE PRODUCTION OF, AND DOES NOT ENDORSE, THIS PRODUCT

2 INFORMATION IN THE TABLE BELOW AND IN THE TABLES ON PAGES 3-5 MAY BE USEFUL IN ANSWERING THE QUESTIONS IN THIS SECTION OF THE EXAMINATION. DO NOT DETACH FROM BOO. 1 H Li Na Rb Cs Fr (223) 4 Be Mg Ca PERIODIC TABLE OF THE ELEMENTS Sr Ba Ra Sc Y *La Ac Ti Zr Hf V Nb Ta Cr Mo W Mn Tc (98) 75 Re Fe Ru Os Co Rh Ir Rf Db Sg Bh Hs Mt (261) (262) (263) (262) (265) (266) 28 Ni Pd Pt (269) 29 Cu Ag Au (272) 30 Zn Cd Hg (277) 5 B Al Ga In Tl C Si Ge Sn Pb N P As Sb Not yet named Bi O S Se Te Po (209) 9 F Cl Br I At (210) *Lanthanide Series Actinide Series Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu (145) Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) 2 He Ne Ar r Xe Rn (222) 2 Page 15

3 General Equilibrium These terms may be used in a question testing your understanding of EQUILIBRIUM initial concentration/pressure; original concentration/pressure; placed in a container; at equilibrium; LeChatlier s principle (reaction shift direction); ; c ; p ; equilibrium concentration; percent dissociation; equilibrium expression; law of mass action; change in concentration ey Formulas or Relationships Equilibrium Expression (AA Law of Mass Action) coefficients coefficient [Products] (Products P atmospheres) c = [Reactants] coefficients p = Either setup for Q (Reactants P ) coefficient p = c (RT) Δn atmospheres Be sure to leave out any solids and liquids and realize that changing their amounts can in no way affect the position of the equilibrium since they are not variables that are included in the equilibrium expression. Manipulating for different versions of the same reaction: reverse rxn = 1/; double coefficients = 2 ; ½ the coefficients = 1/2 ey Concepts Interpreting : If is >1 then reaction favors products b/c ratio is top heavy If is <1 then reaction favors reactants b/c ratio is bottom heavy Think about the reaction as running a race did the reaction make it over half way to the finish line before it stalled out or did it just barely make it off the starting line before it stalled out? LeChâtelier s Principle: If a system at equilibrium is stressed the system will shift direction until equilibrium is re-established. This will result in different concentration/pressure values for the individual members of the reaction but the VALUE of will not change without temperature. Connections to Other Chapters Thermodynamics: ΔG = ΔH TΔS o Electrochem: Ecell = E table logq n The easiest question you can be asked is What is the value of ΔG or E cell when the system is at equilibrium? Answer: ZERO Potential Pitfalls and Unit Warnings No units on LeChatlier tricky questions: Only temperature will change the value of ; no shifting can affect value of Amounts of solids and liquids that are not in the expression can not change position or size of Adding an inert gas (He, Ar, Ne ) to a gaseous equilibrium has NO EFFECT since it is not represented in the expression. In general equilibrium don t forget that these equations may have 1:2 ratios [i.e. (2x) 2 terms] are possible don t get lazy after acids and bases only involved 1:1 coefficients. Neglecting the x will be O. You should never have to use the quadratic equation on the AP exam. Page 1

4 Figures and Graphs You May Need to Interpret As time goes by, the concentration of reactants decreases and the concentration of products increases. Slope of curve is determined by stoichiometric coefficients. When the rates of increase and decrease are the same (i.e. level out) equilibrium is established. Equilibrium Concept Map 2008 René McCormick & Mary Payton. All rights reserved. 2 Page 2

5 2003 AP CHEMISTY FREE-RESPONSE QUESTIONS (Form B) CHEMISTRY Section II (Total time 90 minutes) Part A Time 40 minutes YOU MAY USE YOUR CALCULATOR FOR PART A. CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if you do not. Attention should be paid to significant figures. Be sure to write all your answers to the questions on the lined pages following each question in the booklet with the goldenrod cover. Do NOT write your answers on the lavender insert. Answer Question 1 below. The Section II score weighting for this question is 20 percent. 2 HI(g) H 2 (g) I 2 (g) 1. After a 1.0 mole sample of HI(g) is placed into an evacuated 1.0 L container at 700., the reaction represented above occurs. The concentration of HI(g) as a function of time is shown below. (a) Write the expression for the equilibrium constant, c, for the reaction. (b) What is [HI] at equilibrium? Copyright 2003 by College Entrance Examination Board. All rights reserved. Available to AP professionals at apcentral.collegeboard.com and to students and parents at 6 Page 3

6 2003 AP CHEMISTY FREE-RESPONSE QUESTIONS (Form B) (c) Determine the equilibrium concentrations of H 2 (g) and I 2 (g). (d) On the graph above, make a sketch that shows how the concentration of H 2 (g) changes as a function of time. (e) Calculate the value of the following equilibrium constants at (i) c (ii) p (f) At 1,000, the value of c for the reaction is In an experiment, 0.75 mole of HI(g), 0.10 mole of H 2 (g), and 0.50 mole of I 2 (g) are placed in a 1.0 L container and allowed to reach equilibrium at 1,000. Determine whether the equilibrium concentration of HI(g) will be greater than, equal to, or less than the initial concentration of HI(g). Justify your answer. Copyright 2003 by College Entrance Examination Board. All rights reserved. Available to AP professionals at apcentral.collegeboard.com and to students and parents at 7 Page 4

7 2008 AP CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) CHEMISTRY Section II (Total time 95 minutes) Part A Time 55 minutes YOU MAY USE YOUR CALCULATOR FOR PART A. CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if you do not. Attention should be paid to significant figures. Be sure to write all your answers to the questions on the lined pages following each question in the goldenrod booklet. Do NOT write your answers on the lavender insert. Answer Questions 1, 2, and 3. The Section II score weighting for each question is 20 percent. 1. Answer the following questions regarding the decomposition of arsenic pentafluoride, AsF 5 (g). (a) A 55.8 g sample of AsF 5 (g) is introduced into an evacuated 10.5 L container at 105 C. (i) What is the initial molar concentration of AsF 5 (g) in the container? (ii) What is the initial pressure, in atmospheres, of the AsF 5 (g) in the container? At 105 C, AsF 5 (g) decomposes into AsF 3 (g) and F 2 (g) according to the following chemical equation. AsF 5 (g) AsF 3 (g) F 2 (g) (b) In terms of molar concentrations, write the equilibrium-constant expression for the decomposition of AsF 5 (g). (c) When equilibrium is established, 27.7 percent of the original number of moles of AsF 5 (g) has decomposed. (i) Calculate the molar concentration of AsF 5 (g) at equilibrium. (ii) Using molar concentrations, calculate the value of the equilibrium constant, eq, at 105 C. (d) Calculate the mole fraction of F 2 (g) in the container at equilibrium The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents). -6- Page 5

8 Solubility Equilibrium You may be faced with a sp problem if it contains these words or phrases Solubility, molar solubility, sp, precipitation ey formulas and relationships For a poorly soluble salt, XY the reaction and sp equation are: XY (s) X 1 (aq) Y 1 (aq) and sp = [X 1 ][Y 1 ] For a poorly soluble salt XY 2 the reaction and sp equation are: XY 2(s) X 2 (aq) 2Y 1 (aq) and sp = [X 2 ][Y 1 ] 2 Q sp vs. sp can be used to determine whether precipitation will occur: (Remember Q is set up just like using the values you know at the time) If Q > sp then ppt will occur and solid will be present at the bottom If Q = sp the solution is saturated and will begin to show ppt at bottom If Q < sp the solution does not yet have enough ions to make a ppt (no soild) ey Concepts The sp value is telling you one of two things: (1) the amount of ions that will dissolve as a poorly soluble salt gets wet (2) the minimum concentration of ions necessary in a solution for precipitation to begin Therefore there are three types of problems you may encounter with sp : (1) Calculate sp -- calculating the sp from given concentrations. (a) Plug in numbers given to sp equation (b) Apply coefficient ratio if needed to arrive at additional concentration values. (2) Calculate x -- determine the solubility [concentration of dissolved ions] from a sp value (a) without any common ions mentioned use x and 2x type variables in [ ]. (b) if common ion is present, use its concentration directly in [ ] and x for the other. **Neglecting the x portion of the common ion is acceptable. (3) Determine when conditions are right for precipitation within a solution (a) How much must be added to start precipitation? --- Solve for [x] in sp equation. (b) Which ions of a mixture will precipitate first? ---The smaller the sp, the earlier it will precipitate. ---Think time-line : [0 M] sp = sp = sp = sp = ppt#1 starts ppt#2 starts Potential Pitfalls and Unit Warnings Students often get stuck knowing when to use x and 2x and when to apply the exponent. If either [X] and [Y] are given as numbers, use the number in the appropriate [ ] and remember to square/cube it if necessary. USE x for the unknown concentration. If [X] and [Y] are both unknown to you THEN apply the coefficient ratio and use x and 2x as necessary in the appropriate [ ]. Remember to square/cube them if necessary by René McCormick and Mary Payton. All rights reserved. 1 Page 6

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10 2004 AP CHEMISTRY FREE-RESPONSE QUESTIONS CHEMISTRY Section II (Total time 90 minutes) Part A Time 40 minutes YOU MAY USE YOUR CALCULATOR FOR PART A. CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if you do not. Attention should be paid to significant figures. Be sure to write all your answers to the questions on the lined pages following each question in the booklet with the pink cover. Do NOT write your answers on the green insert. Answer Question 1 below. The Section II score weighting for this question is 20 percent. 1. Answer the following questions relating to the solubilities of two silver compounds, Ag 2 CrO 4 and Ag 3 PO 4. Silver chromate dissociates in water according to the equation shown below. Ag 2 CrO 4 (s) 2 Ag (aq) CrO 4 2 (aq) sp = at 25 C (a) Write the equilibrium-constant expression for the dissolving of Ag 2 CrO 4 (s). (b) Calculate the concentration, in mol L 1, of Ag (aq) in a saturated solution of Ag 2 CrO 4 at 25 C. (c) Calculate the maximum mass, in grams, of Ag 2 CrO 4 that can dissolve in 100. ml of water at 25 C. (d) A mol sample of solid AgNO 3 is added to a 1.00 L saturated solution of Ag 2 CrO 4. Assuming no volume change, does [CrO 4 2 ] increase, decrease, or remain the same? Justify your answer. In a saturated solution of Ag 3 PO 4 at 25 C, the concentration of Ag (aq) is M. The equilibriumconstant expression for the dissolving of Ag 3 PO 4 (s) in water is shown below. sp = [Ag ] 3 [PO 4 3 ] (e) Write the balanced equation for the dissolving of Ag 3 PO 4 in water. (f) Calculate the value of sp for Ag 3 PO 4 at 25 C. (g) A 1.00 L sample of saturated Ag 3 PO 4 solution is allowed to evaporate at 25 C to a final volume of 500. ml. What is [Ag ] in the solution? Justify your answer. Copyright 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6 Page 8

11 STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25 C Half-reaction E ( V ) F 2 ( g) 2e Æ 2F Co e - 2 Æ Co Au 3 e - Æ Au( s ) 1.50 Cl 2 ( g) 2 e - Æ 2Cl O 2 ( g) 4H 4e Æ 2H2O( l ) Hg 2e - Æ Hg Hg 2 e - Æ Hg( l ) 0.85 Br 2 ( l) 2 e - Æ 2Br - Ag e - Æ Ag( s ) 0.80 Hg2 2 e - Æ 2Hg( l ) Fe e - Æ Fe 0.77 I 2 ( s) 2e Æ 2I Cu e - Æ Cu( s ) 0.52 Cu 2 e - Æ Cu( s ) 0.34 Cu e - Æ Cu Sn 2 e - Æ Sn 0.15 S( s) 2 H 2 e - Æ H2S( g ) H 2e - Æ H 2 ( g ) 0.00 Pb 2 e - Æ Pb( s ) 0.13 Sn 2 e - Æ Sn( s ) 0.14 Ni 2 e - Æ Ni( s ) 0.25 Co 2 e - Æ Co( s ) 0.28 Cd 2 e - Æ Cd( s ) Cr e - Æ Cr 0.41 Fe 2 e - Æ Fe( s ) Cr 3 e - Æ Cr( s ) 0.74 Zn 2 e - Æ Zn( s ) H2O( l) 2e - Æ H 2 ( g) 2OH 0.83 Mn 2 e - Æ Mn( s ) Al 3 e - Æ Al( s ) 1.66 Be 2 e - Æ Be( s ) 1.70 Mg 2 e - Æ Mg( s ) 2.37 Na e - Æ Na( s ) 2.71 Ca 2 e - Æ Ca( s ) 2.87 Sr 2 e - Æ Sr( s ) 2.89 Ba 2 e - Æ Ba( s ) 2.90 Rb e - Æ Rb( s ) 2.92 e - Æ ( s ) 2.92 Cs e - Æ Cs( s ) 2.92 Li e - Æ Li( s ) Page 9

12 ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS ATOMIC STRUCTURE E = hv c = lv l = h p = mu mu En = joule 2 n EQUILIBRIUM [ H ][ A ] a = [ HA] b w [ OH ][ HB ] = [ B] 14 = [ OH ][ H ] = C = a b ph = log [ H ], poh = log [ OH ] 14 = ph poh = log [ A ph p ] a [ HA] = log [ HB poh p ] b [ B] p = log, p = log a a b b = ( RT), p c Dn where Dn = moles product gas moles reactant gas THERMOCHEMISTRY/INETICS DS = S products - S reactants DH = DH products - DH reactants DG = DG products - DG reactants DG = DH - TDS =- RT ln = RT log =-n E DG = DG RT ln Q = DG RT log Q q = mcdt C p ln A ln k DH = DT t    f f - ln A = -kt = kt A A t 0 0 e j Ea = - 1 ln A R T    f f E = energy u = velocity v = frequency n = principal quantum number l = wavelength m = mass p = momentum Speed of light, c = m s Planck s constant, h = J s Boltzmann s constant, k = J Avogadro s number = mol Electron charge, e = coulomb 1 electron volt per atom = 96.5 kj mol Equilibrium Constants a b w p c (weak acid) (weak base) (water) (gas pressure) (molar concentrations) S = standard entropy H = standard enthalpy G = standard free energy E = standard reduction potential T = temperature n = moles m = mass q = heat c = specific heat capacity C E p a = molar heat capacity at constant pressure = activation energy k = rate constant A = frequency factor Faraday's constant, = 96, 500 coulombs per mole of electrons -1-1 Gas constant, R = 831. J mol = L atm mol -1-1 = 831. volt coulomb mol Page 10