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AP® Chemistry
                                2009 Free-Response Questions
                                           Form B




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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.




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STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25∞C

                                           Half-reaction                         E ∞(V)
                                   -              Æ              -
        F2 ( g) + 2 e                                       2F                     2.87
             3+            -                      Æ              2+
        Co         +e                                       Co                     1.82
                               -                  Æ
        Au    3+
                      + 3e                                  Au(s)                  1.50
                                   -
        Cl2 (g ) + 2 e                            Æ         2 Cl -                 1.36
        O2 (g) + 4 H + + 4 e -                    Æ         2 H 2 O(l )            1.23
                                   -
        Br2 (l ) + 2 e                            Æ         2 Br -                 1.07
        2 Hg2+ + 2 e -                            Æ         Hg22+                  0.92
        Hg2+ + 2 e -                              Æ         Hg(l )                 0.85
                           -                      Æ
        Ag + e+
                                                            Ag(s)                  0.80
                                   -              Æ
        Hg2      2+
                       + 2e                                 2 Hg(l )               0.79
                           -                      Æ
        Fe   3+
                   +e                                       Fe 2+                  0.77
        I 2 (s) + 2 e -                           Æ         2 I-                   0.53
        Cu+ + e -                                 Æ         Cu(s)                  0.52
                               -                  Æ
        Cu   2+
                      + 2e                                  Cu(s)                  0.34
                           -                      Æ
        Cu   2+
                      +e                                    Cu+                    0.15
        Sn 4+ + 2 e -                             Æ         Sn 2+                  0.15
        S(s) + 2 H + + 2 e -                      Æ         H 2S(g )               0.14
                               -                  Æ
        2H + 2e
              +
                                                            H2 (g)                 0.00
                               -                  Æ
        Pb   2+
                   + 2e                                     Pb(s)                – 0.13
                               -                  Æ
        Sn   2+
                   + 2e                                     Sn(s)                – 0.14
                               -                  Æ
        Ni   2+
                   + 2e                                     Ni(s)                – 0.25
                               -                  Æ
        Co   2+
                      + 2e                                  Co(s)                – 0.28
                               -                  Æ
        Cd   2+
                      + 2e                                  Cd(s)                – 0.40
                           -                      Æ
        Cr   3+
                   +e                                       Cr 2+                – 0.41
        Fe 2+ + 2 e -                             Æ         Fe(s)                – 0.44
                               -                  Æ
        Cr   3+
                   + 3e                                     Cr(s)                – 0.74
                               -                  Æ
        Zn   2+
                   + 2e                                     Zn(s)                – 0.76
                                       -
        2 H 2 O(l ) + 2 e                         Æ         H 2 ( g ) + 2 OH -   – 0.83
        Mn 2+ + 2 e -                             Æ         Mn(s)                – 1.18
                               -                  Æ
        Al   3+
                   + 3e                                     Al(s)                – 1.66
                               -                  Æ
        Be   2+
                   + 2e                                     Be(s)                – 1.70
                                   -              Æ
        Mg       2+
                      + 2e                                  Mg(s)                – 2.37
                           -                      Æ
        Na + e
             +
                                                            Na(s)                – 2.71
                               -                  Æ
        Ca   2+
                   + 2e                                     Ca(s)                – 2.87
                               -                  Æ
        Sr   2+
                  + 2e                                      Sr(s)                – 2.89
                               -                  Æ
        Ba   2+
                   + 2e                                     Ba(s)                – 2.90
                       -                          Æ
        Rb + e
             +
                                                            Rb(s)                – 2.92
                      -                           Æ
        K +e
          +
                                                            K(s)                 – 2.92
                       -                          Æ
        Cs + e
             +
                                                            Cs(s)                – 2.92
                       -                          Æ
             +
        Li + e                                              Li(s)                – 3.05




                                                      -3-                          GO ON TO THE NEXT PAGE.
ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS
ATOMIC STRUCTURE                                               E   =   energy            u = velocity
           E = hv         c = lv                               v   =   frequency         n = principal quantum number
                h                                              l   =   wavelength        m = mass
           l=             p = mu
               mu                                              p   =   momentum
               -2.178 ¥ 10 -18
          En =                 joule
                     n2                                                  Speed of light, c = 3.0 ¥ 108 m s-1

EQUILIBRIUM                                                        Planck’s constant, h = 6.63 ¥ 10 -34 J s
          [H + ][A - ]                                       Boltzmann’s constant, k = 1.38 ¥ 10 -23 J K -1
    Ka =
             [HA]
          [OH - ][HB+ ]                                            Avogadro’s number = 6.022 ¥ 1023 mol -1
    Kb =
                [B]
                                                                       Electron charge, e = -1.602 ¥ 10 -19 coulomb
    K w = [OH ][H + ] = 1.0 ¥ 10 -14 @ 25 C
               -

        = K a ¥ Kb                                            1 electron volt per atom = 96.5 kJ mol -1
       pH = - log [H + ], pOH = - log[OH - ]
       14 = pH + pOH                                                   Equilibrium Constants
                               [A - ]                                  K a (weak acid)
       pH = pK a + log
                               [HA]                                    K b (weak base)
                          [HB+ ]                                       K w (water)
     pOH = pK b + log
                           [B]                                         K p (gas pressure)
      pK a = - log K a , pK b = - log K b                              K c (molar concentrations)
                          Dn
       K p = K c ( RT )        ,
where D n = moles product gas - moles reactant gas                     S = standard entropy
                                                                    H = standard enthalpy
THERMOCHEMISTRY/KINETICS                                            G = standard free energy
 DS =Â S products -Â S reactants                                    E     =   standard reduction potential
DH = Â DHf products -Â DH f reactants                                T    =   temperature
                                                                     n    =   moles
DG =      Â DGf products -Â DGf reactants                            m    =   mass
                                                                     q    =   heat
DG = DH - T D S                                                       c   =   specific heat capacity
   = - RT ln K = -2.303 RT log K                                    Cp    =   molar heat capacity at constant pressure
      = -n E                                                           Ea = activation energy
                                                                        k = rate constant
 DG = DG + RT ln Q = DG + 2.303 RT log Q
                                                                        A = frequency factor
  q = mcDT
      DH                                                                               = 96,500 coulombs per mole
 Cp =                                                      Faraday’s constant,
      DT
                                                                                         of electrons
 ln [A ] t - ln [A]0 = - kt                                        Gas constant, R = 8.31 J mol -1 K -1
        1
             -
                 1
                     = kt                                                              = 0.0821 L atm mol -1 K -1
     [A] t [A]0
                                                                                       = 62.4 L torr mol -1 K -1

 ln k =
           R T()
          - Ea 1
                 + ln A
                                                                                       = 8.31 volt coulomb mol -1 K -1




                                                     -4-                               GO ON TO THE NEXT PAGE.
GASES, LIQUIDS, AND SOLUTIONS

                    PV = nRT                                                     P    =   pressure
Ê                                                                                V    =   volume
     n aˆ
       2
Á P + 2 ˜ (V - nb) = nRT
Ë                                                                                T    =   temperature
     V ¯
                                                                                 n    =   number of moles
                                                        moles A
                     PA = Ptotal ¥ X A , where X A =                             D    =   density
                                                       total moles
                   Ptotal = PA + PB + PC + ...                                   m    =   mass
                            m                                                    u    =   velocity
                       n=
                            M
                  K = C + 273
              PV1 P2V2                                                         urms   =   root-mean-square speed
               1
                    =                                                           KE    =   kinetic energy
               T1      T2
                      m                                                           r   =   rate of effusion
                 D=                                                              M    =   molar mass
                      V
                        3kT     3RT                                               p   =   osmotic pressure
              urms =        =
                         m       M                                                i   =   van’t Hoff factor
                      1 2                                                             =
  KE per molecule = mu                                                          Kf        molal freezing-point depression constant
                      2
                      3                                                         Kb    =   molal boiling-point elevation constant
     KE per mole = RT                                                            A    =   absorbance
                      2
                 r1     M2                                                       a    =   molar absorptivity
                    =
                 r2     M1                                                       b    =   path length
      molarity, M = moles solute per liter solution                              c    =   concentration
         molality = moles solute per kilogram solvent                            Q    =   reaction quotient
              DT f = iK f ¥ molality                                             I    =   current (amperes)
                    DTb = iK b ¥ molality                                        q    =   charge (coulombs)
                      p = iMRT                                                    t   =   time (seconds)
                      A = abc                                                   E = standard reduction potential
                                                                                 K = equilibrium constant


OXIDATION-REDUCTION; ELECTROCHEMISTRY
                                                                                   Gas constant, R = 8.31 J mol -1 K -1
                                                                                                       = 0.0821 L atm mol -1 K -1
           [C] c [D] d                                                                                 = 62.4 L torr mol -1 K -1
    Q=         a      b
                          , where a A + b B Æ c C + d D
           [A] [B]                                                                                     = 8.31 volt coulomb mol -1 K -1
     I =
           q                                                               Boltzmann’s constant, k = 1.38 ¥ 10 -23 J K -1
           t
                                                                                          K f for H2 O = 1.86 K kg mol -1
                    RT                0.0592
  Ecell = Ecell -      ln Q = Ecell -        log Q @ 25 C                                 K b for H2 O = 0.512 K kg mol -1
                    n                    n
                                                                                                 1 atm = 760 mm Hg
            nE
log K =                                                                                                = 760 torr
           0.0592
                                                                                              STP = 0.00 C and 1.0 atm
                                                                             Faraday’s constant, = 96,500 coulombs per mole
                                                                                                     of electrons




                                                                     -5-                         GO ON TO THE NEXT PAGE.
2009 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 this 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. A pure 14.85 g sample of the weak base ethylamine, C2H5NH2 , is dissolved in enough distilled water to make
   500. mL of solution.

    (a) Calculate the molar concentration of the C2H5NH2 in the solution.

    The aqueous ethylamine reacts with water according to the equation below.
                                                      Æ
                                 C2H5NH2(aq) + H2O(l) ¨ C2H5NH3+(aq) + OH-(aq)

    (b) Write the equilibrium-constant expression for the reaction between C2H5NH2(aq) and water.

    (c) Of C2H5NH2(aq) and C2H5NH3+(aq), which is present in the solution at the higher concentration at
        equilibrium? Justify your answer.

    (d) A different solution is made by mixing 500. mL of 0.500 M C2H5NH2 with 500. mL of 0.200 M HCl.
        Assume that volumes are additive. The pH of the resulting solution is found to be 10.93.

          (i) Calculate the concentration of OH-(aq) in the solution.
         (ii) Write the net-ionic equation that represents the reaction that occurs when the C2H5NH2 solution is
              mixed with the HCl solution.
         (iii) Calculate the molar concentration of the C2H5NH3+(aq) that is formed in the reaction.
         (iv) Calculate the value of Kb for C2H5NH2 .




                                            © 2009 The College Board. All rights reserved.
                                    Visit the College Board on the Web: www.collegeboard.com.

                                                              -6-                           GO ON TO THE NEXT PAGE.
2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)


                                                   H+ Æ
                                       S2O32-(aq) æææ SO32-(aq) + S(s)

2. A student performed an experiment to investigate the decomposition of sodium thiosulfate, Na2S2O3 , in acidic
   solution, as represented by the equation above. In each trial the student mixed a different concentration of
   sodium thiosulfate with hydrochloric acid at constant temperature and determined the rate of disappearance of
   S2O32−(aq). Data from five trials are given below in the table on the left and are plotted in the graph on the right.




   (a) Identify the independent variable in the experiment.

   (b) Determine the order of the reaction with respect to S2O32-. Justify your answer by using the information
       above.

   (c) Determine the value of the rate constant, k, for the reaction. Include units in your answer. Show how you
       arrived at your answer.

   (d) In another trial the student mixed 0.10 M Na2S2O3 with hydrochloric acid. Calculate the amount of time it
       would take for the concentration of S2O32- to drop to 0.020 M .

   (e) On the graph above, sketch the line that shows the results that would be expected if the student repeated the
       five trials at a temperature lower than that during the first set of trials.




                                          © 2009 The College Board. All rights reserved.
                                  Visit the College Board on the Web: www.collegeboard.com.

                                                            -7-                           GO ON TO THE NEXT PAGE.
2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)


                                           2 H2O2(aq) → 2 H2O(l) + O2(g)

3. The mass of an aqueous solution of H2O2 is 6.951 g. The H2O2 in the solution decomposes completely
   according to the reaction represented above. The O2(g) produced is collected in an inverted graduated tube
   over water at 23.4°C and has a volume of 182.4 mL when the water levels inside and outside of the tube are
   the same. The atmospheric pressure in the lab is 762.6 torr, and the equilibrium vapor pressure of water at
   23.4°C is 21.6 torr.

   (a) Calculate the partial pressure, in torr, of O2(g) in the gas-collection tube.

   (b) Calculate the number of moles of O2(g) produced in the reaction.

   (c) Calculate the mass, in grams, of H2O2 that decomposed.

   (d) Calculate the percent of H2O2 , by mass, in the original 6.951 g aqueous sample.

   (e) Write the oxidation number of the oxygen atoms in H2O2 and the oxidation number of the oxygen atoms in
       O2 in the appropriate cells in the table below.

                                                          Oxidation Number
                                            Substance
                                                          of Oxygen Atoms
                                              H2O2
                                                O2


   (f) Write the balanced oxidation half-reaction for the reaction.




                                                     STOP
            If you finish before time is called, you may check your work on this part only.
                  Do not turn to the other part of the test until you are told to do so.



                                          © 2009 The College Board. All rights reserved.
                                  Visit the College Board on the Web: www.collegeboard.com.

                                                            -8-
2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)


                                                     CHEMISTRY
                                            Part B
                                        Time— 40 minutes
                             NO CALCULATORS MAY BE USED FOR PART B.


Answer Question 4 below. The Section II score weighting for this question is 10 percent.

4. For each of the following three reactions, write a balanced equation in part (i) and answer the question in
   part (ii). In part (i), coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous
   unless otherwise indicated. Represent substances in solutions as ions if the substances are extensively ionized.
   Omit formulas for any ions or molecules that are unchanged by the reaction. You may use the empty space at the
   bottom of the next page for scratch work, but only equations that are written in the answer boxes provided will
   be graded.




    (a) A barium nitrate solution and a potassium fluoride solution are combined and a precipitate forms.

          (i) Balanced equation:




         (ii) If equimolar amounts of barium nitrate and potassium fluoride are combined, which reactant, if any,
              is the limiting reactant? Explain.
        ______________________________________________________________________________________
        ______________________________________________________________________________________
        ______________________________________________________________________________________




                                           © 2009 The College Board. All rights reserved.
                                   Visit the College Board on the Web: www.collegeboard.com.

                                                             -9-                           GO ON TO THE NEXT PAGE.
2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)


(b) A piece of cadmium metal is oxidized by adding it to a solution of copper(II) chloride.

      (i) Balanced equation:




     (ii) List two visible changes that would occur in the reaction container as the reaction is proceeding.
    ______________________________________________________________________________________
    ______________________________________________________________________________________
    ______________________________________________________________________________________

(c) A hydrolysis reaction occurs when solid sodium sulfide is added to distilled water.

      (i) Balanced equation:




     (ii) Indicate whether the pH of the resulting solution is less than 7, equal to 7, or greater than 7. Explain.
    ______________________________________________________________________________________
    ______________________________________________________________________________________
    ______________________________________________________________________________________




                                       © 2009 The College Board. All rights reserved.
                               Visit the College Board on the Web: www.collegeboard.com.

                                                        -10-                           GO ON TO THE NEXT PAGE.
2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)


Answer Question 5 and Question 6. The Section II score weighting for these questions is 15 percent each.

Your responses to these questions will be graded on the basis of the accuracy and relevance of the information cited.
Explanations should be clear and well organized. Examples and equations may be included in your responses where
appropriate. Specific answers are preferable to broad, diffuse responses.

5. Answer the following questions about nitrogen, hydrogen, and ammonia.

    (a) In the boxes below, draw the complete Lewis electron-dot diagrams for N2 and NH3 .




    (b) Calculate the standard free-energy change, ΔG°, that occurs when 12.0 g of H2(g) reacts with excess
        N2(g) at 298 K according to the reaction represented below.
                                          Æ
                          N2(g) + 3 H2(g) ¨ 2 NH3(g)                     DG298 = -34 kJ mol-1

    (c) Given that ΔH 298 for the reaction is −92.2 kJ mol-1, which is larger, the total bond dissociation energy
        of the reactants or the total bond dissociation energy of the products? Explain.

    (d) The value of the standard entropy change, ΔS298 , for the reaction is −199 J mol-1K-1. Explain why the
        value of ΔS298 is negative.

    (e) Assume that ΔH° and ΔS ° for the reaction are independent of temperature.
         (i) Explain why there is a temperature above 298 K at which the algebraic sign of the value of ΔG°
             changes.
         (ii) Theoretically, the best yields of ammonia should be achieved at low temperatures and high pressures.
              Explain.




                                          © 2009 The College Board. All rights reserved.
                                  Visit the College Board on the Web: www.collegeboard.com.

                                                           -11-                           GO ON TO THE NEXT PAGE.
2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)


6. Answer the following questions about electrochemical cells.




   It is observed that when silver metal is placed in aqueous thallium(I) fluoride, TlF, no reaction occurs. When the
   switch is closed in the cell represented above, the voltage reading is +1.14 V.

   (a) Write the reduction half-reaction that occurs in the cell.

   (b) Write the equation for the overall reaction that occurs in the cell.

   (c) Identify the anode in the cell. Justify your answer.

   (d) On the diagram above, use an arrow to clearly indicate the direction of electron flow as the cell operates.
   (e) Calculate the value of the standard reduction potential for the Tl+/Tl half-reaction.

   The standard reduction potential, E°, of the reaction Pt2+ + 2 e− → Pt is 1.20 V.

   (f) Assume that electrodes of pure Pt, Ag, and Ni are available as well as 1.00 M solutions of their salts.
       Three different electrochemical cells can be constructed using these materials. Identify the two metals that
       when used to make an electrochemical cell would produce the cell with the largest voltage. Explain how you
       arrived at your answer.

   (g) Predict whether Pt metal will react when it is placed in 1.00 M AgNO3(aq). Justify your answer.




                                                          STOP

                                                    END OF EXAM



                                          © 2009 The College Board. All rights reserved.
                                  Visit the College Board on the Web: www.collegeboard.com.

                                                           -12-

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Chemistry AP Free-Response Questions Form B 2009

  • 1. AP® Chemistry 2009 Free-Response Questions Form B The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 5,600 schools, colleges, universities and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools and 3,800 colleges through major programs and services in college readiness, college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT® and the Advanced Placement Program® (AP®). The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities and concerns. © 2009 The College Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central, SAT, and the acorn logo are registered trademarks of the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. Permission to use copyrighted College Board materials may be requested online at: www.collegeboard.com/inquiry/cbpermit.html. Visit the College Board on the Web: www.collegeboard.com. AP Central is the official online home for the AP Program: apcentral.collegeboard.com.
  • 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. -2- GO ON TO THE NEXT PAGE.
  • 3. STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25∞C Half-reaction E ∞(V) - Æ - F2 ( g) + 2 e 2F 2.87 3+ - Æ 2+ Co +e Co 1.82 - Æ Au 3+ + 3e Au(s) 1.50 - Cl2 (g ) + 2 e Æ 2 Cl - 1.36 O2 (g) + 4 H + + 4 e - Æ 2 H 2 O(l ) 1.23 - Br2 (l ) + 2 e Æ 2 Br - 1.07 2 Hg2+ + 2 e - Æ Hg22+ 0.92 Hg2+ + 2 e - Æ Hg(l ) 0.85 - Æ Ag + e+ Ag(s) 0.80 - Æ Hg2 2+ + 2e 2 Hg(l ) 0.79 - Æ Fe 3+ +e Fe 2+ 0.77 I 2 (s) + 2 e - Æ 2 I- 0.53 Cu+ + e - Æ Cu(s) 0.52 - Æ Cu 2+ + 2e Cu(s) 0.34 - Æ Cu 2+ +e Cu+ 0.15 Sn 4+ + 2 e - Æ Sn 2+ 0.15 S(s) + 2 H + + 2 e - Æ H 2S(g ) 0.14 - Æ 2H + 2e + H2 (g) 0.00 - Æ Pb 2+ + 2e Pb(s) – 0.13 - Æ Sn 2+ + 2e Sn(s) – 0.14 - Æ Ni 2+ + 2e Ni(s) – 0.25 - Æ Co 2+ + 2e Co(s) – 0.28 - Æ Cd 2+ + 2e Cd(s) – 0.40 - Æ Cr 3+ +e Cr 2+ – 0.41 Fe 2+ + 2 e - Æ Fe(s) – 0.44 - Æ Cr 3+ + 3e Cr(s) – 0.74 - Æ Zn 2+ + 2e Zn(s) – 0.76 - 2 H 2 O(l ) + 2 e Æ H 2 ( g ) + 2 OH - – 0.83 Mn 2+ + 2 e - Æ Mn(s) – 1.18 - Æ Al 3+ + 3e Al(s) – 1.66 - Æ Be 2+ + 2e Be(s) – 1.70 - Æ Mg 2+ + 2e Mg(s) – 2.37 - Æ Na + e + Na(s) – 2.71 - Æ Ca 2+ + 2e Ca(s) – 2.87 - Æ Sr 2+ + 2e Sr(s) – 2.89 - Æ Ba 2+ + 2e Ba(s) – 2.90 - Æ Rb + e + Rb(s) – 2.92 - Æ K +e + K(s) – 2.92 - Æ Cs + e + Cs(s) – 2.92 - Æ + Li + e Li(s) – 3.05 -3- GO ON TO THE NEXT PAGE.
  • 4. ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS ATOMIC STRUCTURE E = energy u = velocity E = hv c = lv v = frequency n = principal quantum number h l = wavelength m = mass l= p = mu mu p = momentum -2.178 ¥ 10 -18 En = joule n2 Speed of light, c = 3.0 ¥ 108 m s-1 EQUILIBRIUM Planck’s constant, h = 6.63 ¥ 10 -34 J s [H + ][A - ] Boltzmann’s constant, k = 1.38 ¥ 10 -23 J K -1 Ka = [HA] [OH - ][HB+ ] Avogadro’s number = 6.022 ¥ 1023 mol -1 Kb = [B] Electron charge, e = -1.602 ¥ 10 -19 coulomb K w = [OH ][H + ] = 1.0 ¥ 10 -14 @ 25 C - = K a ¥ Kb 1 electron volt per atom = 96.5 kJ mol -1 pH = - log [H + ], pOH = - log[OH - ] 14 = pH + pOH Equilibrium Constants [A - ] K a (weak acid) pH = pK a + log [HA] K b (weak base) [HB+ ] K w (water) pOH = pK b + log [B] K p (gas pressure) pK a = - log K a , pK b = - log K b K c (molar concentrations) Dn K p = K c ( RT ) , where D n = moles product gas - moles reactant gas S = standard entropy H = standard enthalpy THERMOCHEMISTRY/KINETICS G = standard free energy DS =Â S products -Â S reactants E = standard reduction potential DH = Â DHf products -Â DH f reactants T = temperature n = moles DG = Â DGf products -Â DGf reactants m = mass q = heat DG = DH - T D S c = specific heat capacity = - RT ln K = -2.303 RT log K Cp = molar heat capacity at constant pressure = -n E Ea = activation energy k = rate constant DG = DG + RT ln Q = DG + 2.303 RT log Q A = frequency factor q = mcDT DH = 96,500 coulombs per mole Cp = Faraday’s constant, DT of electrons ln [A ] t - ln [A]0 = - kt Gas constant, R = 8.31 J mol -1 K -1 1 - 1 = kt = 0.0821 L atm mol -1 K -1 [A] t [A]0 = 62.4 L torr mol -1 K -1 ln k = R T() - Ea 1 + ln A = 8.31 volt coulomb mol -1 K -1 -4- GO ON TO THE NEXT PAGE.
  • 5. GASES, LIQUIDS, AND SOLUTIONS PV = nRT P = pressure Ê V = volume n aˆ 2 Á P + 2 ˜ (V - nb) = nRT Ë T = temperature V ¯ n = number of moles moles A PA = Ptotal ¥ X A , where X A = D = density total moles Ptotal = PA + PB + PC + ... m = mass m u = velocity n= M K = C + 273 PV1 P2V2 urms = root-mean-square speed 1 = KE = kinetic energy T1 T2 m r = rate of effusion D= M = molar mass V 3kT 3RT p = osmotic pressure urms = = m M i = van’t Hoff factor 1 2 = KE per molecule = mu Kf molal freezing-point depression constant 2 3 Kb = molal boiling-point elevation constant KE per mole = RT A = absorbance 2 r1 M2 a = molar absorptivity = r2 M1 b = path length molarity, M = moles solute per liter solution c = concentration molality = moles solute per kilogram solvent Q = reaction quotient DT f = iK f ¥ molality I = current (amperes) DTb = iK b ¥ molality q = charge (coulombs) p = iMRT t = time (seconds) A = abc E = standard reduction potential K = equilibrium constant OXIDATION-REDUCTION; ELECTROCHEMISTRY Gas constant, R = 8.31 J mol -1 K -1 = 0.0821 L atm mol -1 K -1 [C] c [D] d = 62.4 L torr mol -1 K -1 Q= a b , where a A + b B Æ c C + d D [A] [B] = 8.31 volt coulomb mol -1 K -1 I = q Boltzmann’s constant, k = 1.38 ¥ 10 -23 J K -1 t K f for H2 O = 1.86 K kg mol -1 RT 0.0592 Ecell = Ecell - ln Q = Ecell - log Q @ 25 C K b for H2 O = 0.512 K kg mol -1 n n 1 atm = 760 mm Hg nE log K = = 760 torr 0.0592 STP = 0.00 C and 1.0 atm Faraday’s constant, = 96,500 coulombs per mole of electrons -5- GO ON TO THE NEXT PAGE.
  • 6. 2009 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 this 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. A pure 14.85 g sample of the weak base ethylamine, C2H5NH2 , is dissolved in enough distilled water to make 500. mL of solution. (a) Calculate the molar concentration of the C2H5NH2 in the solution. The aqueous ethylamine reacts with water according to the equation below. Æ C2H5NH2(aq) + H2O(l) ¨ C2H5NH3+(aq) + OH-(aq) (b) Write the equilibrium-constant expression for the reaction between C2H5NH2(aq) and water. (c) Of C2H5NH2(aq) and C2H5NH3+(aq), which is present in the solution at the higher concentration at equilibrium? Justify your answer. (d) A different solution is made by mixing 500. mL of 0.500 M C2H5NH2 with 500. mL of 0.200 M HCl. Assume that volumes are additive. The pH of the resulting solution is found to be 10.93. (i) Calculate the concentration of OH-(aq) in the solution. (ii) Write the net-ionic equation that represents the reaction that occurs when the C2H5NH2 solution is mixed with the HCl solution. (iii) Calculate the molar concentration of the C2H5NH3+(aq) that is formed in the reaction. (iv) Calculate the value of Kb for C2H5NH2 . © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. -6- GO ON TO THE NEXT PAGE.
  • 7. 2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) H+ Æ S2O32-(aq) æææ SO32-(aq) + S(s) 2. A student performed an experiment to investigate the decomposition of sodium thiosulfate, Na2S2O3 , in acidic solution, as represented by the equation above. In each trial the student mixed a different concentration of sodium thiosulfate with hydrochloric acid at constant temperature and determined the rate of disappearance of S2O32−(aq). Data from five trials are given below in the table on the left and are plotted in the graph on the right. (a) Identify the independent variable in the experiment. (b) Determine the order of the reaction with respect to S2O32-. Justify your answer by using the information above. (c) Determine the value of the rate constant, k, for the reaction. Include units in your answer. Show how you arrived at your answer. (d) In another trial the student mixed 0.10 M Na2S2O3 with hydrochloric acid. Calculate the amount of time it would take for the concentration of S2O32- to drop to 0.020 M . (e) On the graph above, sketch the line that shows the results that would be expected if the student repeated the five trials at a temperature lower than that during the first set of trials. © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. -7- GO ON TO THE NEXT PAGE.
  • 8. 2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) 2 H2O2(aq) → 2 H2O(l) + O2(g) 3. The mass of an aqueous solution of H2O2 is 6.951 g. The H2O2 in the solution decomposes completely according to the reaction represented above. The O2(g) produced is collected in an inverted graduated tube over water at 23.4°C and has a volume of 182.4 mL when the water levels inside and outside of the tube are the same. The atmospheric pressure in the lab is 762.6 torr, and the equilibrium vapor pressure of water at 23.4°C is 21.6 torr. (a) Calculate the partial pressure, in torr, of O2(g) in the gas-collection tube. (b) Calculate the number of moles of O2(g) produced in the reaction. (c) Calculate the mass, in grams, of H2O2 that decomposed. (d) Calculate the percent of H2O2 , by mass, in the original 6.951 g aqueous sample. (e) Write the oxidation number of the oxygen atoms in H2O2 and the oxidation number of the oxygen atoms in O2 in the appropriate cells in the table below. Oxidation Number Substance of Oxygen Atoms H2O2 O2 (f) Write the balanced oxidation half-reaction for the reaction. STOP If you finish before time is called, you may check your work on this part only. Do not turn to the other part of the test until you are told to do so. © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. -8-
  • 9. 2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) CHEMISTRY Part B Time— 40 minutes NO CALCULATORS MAY BE USED FOR PART B. Answer Question 4 below. The Section II score weighting for this question is 10 percent. 4. For each of the following three reactions, write a balanced equation in part (i) and answer the question in part (ii). In part (i), coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solutions as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You may use the empty space at the bottom of the next page for scratch work, but only equations that are written in the answer boxes provided will be graded. (a) A barium nitrate solution and a potassium fluoride solution are combined and a precipitate forms. (i) Balanced equation: (ii) If equimolar amounts of barium nitrate and potassium fluoride are combined, which reactant, if any, is the limiting reactant? Explain. ______________________________________________________________________________________ ______________________________________________________________________________________ ______________________________________________________________________________________ © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. -9- GO ON TO THE NEXT PAGE.
  • 10. 2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) (b) A piece of cadmium metal is oxidized by adding it to a solution of copper(II) chloride. (i) Balanced equation: (ii) List two visible changes that would occur in the reaction container as the reaction is proceeding. ______________________________________________________________________________________ ______________________________________________________________________________________ ______________________________________________________________________________________ (c) A hydrolysis reaction occurs when solid sodium sulfide is added to distilled water. (i) Balanced equation: (ii) Indicate whether the pH of the resulting solution is less than 7, equal to 7, or greater than 7. Explain. ______________________________________________________________________________________ ______________________________________________________________________________________ ______________________________________________________________________________________ © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. -10- GO ON TO THE NEXT PAGE.
  • 11. 2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) Answer Question 5 and Question 6. The Section II score weighting for these questions is 15 percent each. Your responses to these questions will be graded on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized. Examples and equations may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses. 5. Answer the following questions about nitrogen, hydrogen, and ammonia. (a) In the boxes below, draw the complete Lewis electron-dot diagrams for N2 and NH3 . (b) Calculate the standard free-energy change, ΔG°, that occurs when 12.0 g of H2(g) reacts with excess N2(g) at 298 K according to the reaction represented below. Æ N2(g) + 3 H2(g) ¨ 2 NH3(g) DG298 = -34 kJ mol-1 (c) Given that ΔH 298 for the reaction is −92.2 kJ mol-1, which is larger, the total bond dissociation energy of the reactants or the total bond dissociation energy of the products? Explain. (d) The value of the standard entropy change, ΔS298 , for the reaction is −199 J mol-1K-1. Explain why the value of ΔS298 is negative. (e) Assume that ΔH° and ΔS ° for the reaction are independent of temperature. (i) Explain why there is a temperature above 298 K at which the algebraic sign of the value of ΔG° changes. (ii) Theoretically, the best yields of ammonia should be achieved at low temperatures and high pressures. Explain. © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. -11- GO ON TO THE NEXT PAGE.
  • 12. 2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B) 6. Answer the following questions about electrochemical cells. It is observed that when silver metal is placed in aqueous thallium(I) fluoride, TlF, no reaction occurs. When the switch is closed in the cell represented above, the voltage reading is +1.14 V. (a) Write the reduction half-reaction that occurs in the cell. (b) Write the equation for the overall reaction that occurs in the cell. (c) Identify the anode in the cell. Justify your answer. (d) On the diagram above, use an arrow to clearly indicate the direction of electron flow as the cell operates. (e) Calculate the value of the standard reduction potential for the Tl+/Tl half-reaction. The standard reduction potential, E°, of the reaction Pt2+ + 2 e− → Pt is 1.20 V. (f) Assume that electrodes of pure Pt, Ag, and Ni are available as well as 1.00 M solutions of their salts. Three different electrochemical cells can be constructed using these materials. Identify the two metals that when used to make an electrochemical cell would produce the cell with the largest voltage. Explain how you arrived at your answer. (g) Predict whether Pt metal will react when it is placed in 1.00 M AgNO3(aq). Justify your answer. STOP END OF EXAM © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. -12-