Chemistry 115

Midterm Exam

 

May 3, 2001

 

Name ______________________________

1. Matching: Place the letter corresponding to the best answer in front of each of the following definitions. (10)

_R__ The equilibrium constant for the solubility of a salt, i.e. saturated solution of a salt.

_K__ The equilibrium constant for the formation of a complex ion.

_P__ A molecule or anion that donates a pair of electrons to bind with a metal ion

_T__ The study of the laws that govern the energy and entropy changes of physical and chemical events.

_C _ The energy needed to rupture all the bonds in one mole of a substance in the gas state to produce atoms

in the gas state.

_I__ The thermodynamic function that describes the degree of disorder or randomness of a system

_L__ The thermodynamic function that describes the spontaneity for change of a system.

_J__ One mole of electrons

_D__ The electrode at which reduction occurs

_F__ The SI unit of electrical charge; the amount of charge that passes a fixed point of a wire when a

current of 1 A flows for 1 s

A. ampere

F. coulomb

K. formation constant

P. ligand

B. anode

G. electrolysis

L. Gibbs free energy

R. solubility product

C. atomization energy

H. enthalpy

M. instability constant

S. spontaneous change

D. cathode

I. entropy

N. internal energy

T. thermodynamics

E. complex

J. Faraday

O. ion product

 

 

  1. Multiple Choice: Place the letter corresponding to the best answer in front of each of the following. (28)

_A__ A large excess of MgF2(s) is maintained in contact with 1.00 L of pure water to produce a saturated

solution of MgF2. When an additional 1.00 L of pure water is added to the mixture and equilibrium

reestablished, compared to its value in the original saturated solution, [Mg2+] will be

    1. the same C. half as large
    2. twice as large D. some unknown fraction of the original [Mg2+]

Must maintain the same Ksp from the solid dissolving

_B__ Compared to its solubility in pure water, the solubility of NiCO3 in 1.0 M Na2CO3 is expected to be

A. greater B. less C. the same D. cannot predict

Common ion reduces solubility since less of the solid will be able to dissolve

_A__ Compared to its solubility in pure water, the solubility of NiCO3 in 6.0 M HCl is expected to be

A. greater B. less C. the same D. cannot predict

Acid increases solubility because as HCO3- and H2CO3 are formed more solid must dissolve

_A__ Compared to its solubility in pure water, the solubility of NiCO3 in 6.0 M NH3 is expected to be

A. greater B. less C. the same D. cannot predict

As Ni(NH3)62+ is formed, NiCO3 (or Ni(OH)2 will dissolve to replace the Ni2+

_A__ If solid Na2CO3 is added to a solution that is 0.10 M in both Ni2+ and Cu2+, the cation that would

precipitate first is CuCO3 has the smaller Ksp

    1. Cu2+ C. Both will start to precipitate at the same time

B. Ni2+ D. Neither will precipitate

 

_B__ The best way to ensure the complete precipitation from a saturated aqueous solution of H2S of some

metal cation, M2+, as its sulfide, MS(s) is to

    1. add an acid C. heat the solution
    2. raise the pH D. increase [H2S] in the solution

As is done to separate Cation Groups II and III in the qual scheme

_C__ The process for which entropy (of the system) decreases is

    1. a block of ice melting C. precipitation of CaCO3 from "hard" water
    2. CO2 being released from a soft drink D. mixing a can of juice concentrate with water

Formation of a solid from ions increases the order of the system

_B__ The condition that is necessary for a spontaneous reaction is that

    1. DSsys is positive C. DSsurr is positive
    2. DSuniv is positive D. DS should be negative for a spontanteous reaction

_B__ If a diatomic gas, such as Cl2, is dissociated at constant temperature, the entropy

A. decreases B. increases C. is constant D. cannot predict without numbers

Forming two particles from one, which increases the disorder.

_B__ If for a particular chemical reaction DHo = + 37 kJ and DSo = + 63 J/K, the reaction will proceed

spontaneously at

A. all temperatures B. high temperature only C. low temperature only D. no temperature

DG = DH - TDS must be negative which DS will cause at higher temperatures

_D__ The condition that is necessary for a spontaneous reaction is that for the system

A. DGo is positive B. DGo is negative C. DG is positive D. DG is negative

DGo is related to the size of the equilibrium constant at standard conditions.

_B__ At constant temperature and pressure, the condition for equilibrium is

A. DGo = 0 B. DG = 0 C. DGo = 1 D. DG = 1

Again it is the DG not the DGo which is related to the change rather than the K

_B__ The reaction rates of many spontaneous reactions are actually very slow. The best explanation among

the following is

    1. Such reactions are endothermic C. The entropy change is negative
    2. The activation energy of the reaction is large D. The free energy change is positive

Speed of the reaction is related to the kinetics, not the thermodynamics.

_C__ If it is necessary to employ electric current (electrolysis) to carry out a chemical reaction, then for that

reaction

A. DH is positive B. DS is positive C. DG is positive D. DH = DS

Electrolysis is used to carry out a nonspontaneous reaction

PROBLEMS: Watch significant figures and give units.

  1. If calcium ions are to be removed from a city water supply by the addition of soda ash, Na2CO3 , what final concentration of CO32- is required to reduce [Ca2+] in the water to 1.0 x 10-5 M? (5)
  2. Ca2+ + CO32- = CaCO3(s) but Ksp reaction is in opposite direction

    Ksp = 2,8 x 10-9 = [Ca2+][CO32-] = [1.0 x 10-5][CO32-]

    [CO32-] = 2.8 x 10-9 / 1.0 x 10-5 = 2.8 x 10-4 M

     [CO32-] = |_2.8 x 10-4 M _

     

  3. Calculate the solubility of Ni(OH)2 in = Ni2+ + 2 OH-; Ksp = 2.0 x 10-15 (10)

a. distilled water

Ksp = 2.0 x 10-15 = [Ni2+][OH-]2 = [x][2x]2 = 4x3

X = (2.0 x 10-15 / 4)1/3 = 7.94 x 10-6M

  sol = _7.9 x 10-6 M_____

    1. solution buffered at pH = 11.00; pOH = 3.00; [OH-] = 1.0 x 10-3 M

Ksp = 2.0 x 10-15 = [Ni2+][OH-]2 = [Ni2+][1.0 x 10-3]2

[Ni2+] = 2.0 x 10-15 / 1.0 x 10-6 = 2.0 x 10-9 M

sol = _2.0 x 10-9 M___

 

  1. A saturated H2S solution is to be used to separate the cations in a solution that is 0.10 M in ZnCl2 and 0.20 M in MnCl2. What pH will provide the maximum separation of the two ions? Ksp MnS = 3 x 10-11, Kspa = 3 x 1010; Ksp ZnS = 2 x 10-22; Kspa = 2 x 10-1. (8)
  2. Zn2+ is more insoluble and will precipitate first; work with Mn2+ to keep it in solution.

    MnS(s) + 2 H+ = Mn2+ + H2S; Kspa = 3 x 1010 = [Mn2+][H2S] / [H+]2 = [0.20][0,10] / [H+]2

    [H+] = [(0.20)(0.10) / 3 x 1010]1/2 = [6.667 x 10-13]1/2 = 8.16 x 10-7 M

    pH = 6.088

      pH= |_6.09_____

     

  3. Potassium thiocyanate, KSCN, is often used to determine the presence of Fe3+ ions in solution by the formation of the red FeSCN2+ ion. What is the concentration of Fe3+ when 0.50 L each of 0.0015 M Fe(NO3)3 and 0.20 M KSCN are mixed? (Kf FeSCN2+ = 8.9 x 102) large; mainly products (5)
  4. Fe3+ +

    SCN- =

    FeSCN2+

    500moL x .0015M = 0.75 mmol

    500mL x .20M = 100 mmol

    0 mmol

    -0.75 mmol

    -0.75 mmol

    +0.75 mmol

    0 +x

    99.25 mmol/1000 mL +x

    0.75mmol/1000mL -x

    Kf = [FeSCN2+] /[Fe3+][SCN-] = 890 = [0.00075 x] /[x][0.09925 +x] ~ [0.00075] /[x][0.09925]

    X ~ [0.00075] /[0.09925][890] = 8.5 x 10-6 M

      [Fe+3] = |_8.5 x 10-6 M____

     

  5. Use bond energies to estimate the heat of reaction at 298 K for ICl3(g) + 2 H2(g) à HI(g) + 3 HCl(g). BE [I-Cl]= 209 kJ. (6)
  6. 3 I-Cl + 2 H-H à H-I + 3 H-Cl

    DHrxn = S Bonds broken - S Bonds formed

    = 3(209) +2(435) [297 + 3(431)] = 1497 1590 = -93 kJ

    DHrxn = |_-93 kJ__

     

  7. The boiling point of chloroform (CHCl3), is 61.7oC. The enthalpy of vaporization is 31.4kJ/mol. Calculate the entropy of vaporization T = 61.7 x 273.2 = 334.9 K (5)
  8. DG = DH - TDS = 0 since the system is at equilibrium at a phase change

    DH = TDS

    DS = DH / T = 31.4 kJ x 103 J/kJ / 334.9 K = 93.76 J/K

     DS = |_93.8 J/K_

     

  9. Calculate DGo at 25oC for the reaction BaSO4(s) = Ba2+(aq) + SO42-(aq). The values of DGof at 25oC are (in kJ/mol): BaSO4(s), -1353; Ba2+(aq), -561; SO42-, -742. What is the value of the solubility-product constant, Ksp, for this reaction at 25oC? (8)
  10. DGo = SDGoproducts - SDGoreactants = -561 742 [-1353] = 50 kJ

    DG = DGo + RTlnK = 0 at equilibrium

    lnK = - DGo / RT = - 50 kJ x 103J/kJ / [8.314 J/K x 298K] = -20.18

    K = invln(-20.18) = 1.72 x 10-9

     Ksp = |_1.7 x 10-9_

     

  11. Consider the reaction 2 CO(g) + O2(g) à 2 CO2(g) (18)
  1. Is this reaction expected to spontaneous at only high, |only low|, all, or no temperatures? (Circle the correct answer) and then explain briefly.
  2. DGo = DHo - TDSo and since there is less gas in the products, DSo is negative so that the reaction is not spontaneous when the temperature is "high" It is determined below that the reaction is exothermic, which means it will be spontaneous as long as DHo is numerically larger than the TDSo term

     

  3. Briefly give what is the limiting condition would be for a spontaneous reaction.
  4. DG must be negative for a spontaneous reaction; when it is equal to 0, it would be the point at which it is switching from negative to positive, or nonspontaneous.

     

  5. Make the necessary thermodynamic calculations to estimate the temperature range over which the reaction is spontaneous.
  6. DHo = SDHoproducts - SDHoreactants = 2(-393.5) [2(-110.5) + 0] = -566.0 kJ

    DSo = SDSoproducts - SDSoreactants = 2(213.6) [2(197.6) + 205.1] = 427.2 600.3 = -173.1 J/K

     

    DHo = |_-566.0 kJ

    DSo = |_-173.1 J/K

    DGo = 0 = DHo - TDSo

    DHo = TDSo

    T = DHo / DSo = -566.0 kJ x 103 J/kJ / (-173.1 J/K) = 3269.7 K

    Reaction is spontaneous to approximately 3270 K

    Temp = |_3270 K_

     

  7. In light of your answers above, how do you explain the fact that carbon monoxide from auto exhaust is a major health hazard if a catalytic convertor is not used to change it to CO2

The calculations show that CO is more stable towards oxidation as the temperature is raised. Thus it is more stable at the higher temperatures of combustion, even if adequate oxygen is present for oxidation to CO2

 

  1. What mass of platinum could be plated onto a ring from the electrolysis of a platinum(II) salt at a 0.100 ampere current for 30.0 sec? (5)

0.100amp x 30.0 sec x (1 mol e-/96500ampsec) x (1 mol Pt/2 mol e-) x 195.1 g/mol Pt = 0.00303 g

Mass = |_0.00303 g___