Chemistry 212 Exam 3-Key	Name:
May 1, 2000		please print

 

Instructions: Read the whole examination before starting to work on individual problems.  Work the easiest problems first.  Budget your time to insure that you finish the examination.  Don’t waste time on problems that you find hard to finish until you have worked the others.  Remember the use of programmable calculators is forbidden and would be considered a serious honor code violation.

 

Answers:

 

1.  a	7.  b	13.  a
2.  c	8.  c	14.  d
3.  d	9.  a	15.  b
4.  b	10.  b	16.  c
5.  d	11.  c
6.  c	12.  b

Solutions:

1.  Which reaction below would result in a net decrease in entropy?

a.  2KCl(s) + 3O2(g) « 2KClO3(s)	c.  AgBr(s) « Ag+(aq) + Cl-(aq)
b.  CaCO3(s) « CaO(s) + CO2(g)	d.  NH4NO3(s) « N2O(g) + 2H2O(g)

Solution 1: Look for the reaction that has a smaller number of moles of gases on the product side.

2.  What is the standard cell voltage at 25.0°C for the reverse of the reaction below?

Pb²⁺(aq) + Cd(s) « Pb(s) + Cd²⁺(aq)        K = 1.356x10⁹

a.  +0.270 V	c.  -0.270 V
b.  +0.117 V	d.  -0.117 V

Solution 2:

The reverse reaction would develop  -0.270 V.

3.  Which of the answers below is the correct representation according to the shorthand method of the following galvanic cell?

2Fe³⁺(aq) + 3Ni(s) « 2Fe(s) + 3Ni²⁺(aq)

a.  Fe3+(aq)| Fe(s) || Ni(s)| Ni2+(aq)	c.  Ni(s)|Ni2+(aq) || Fe(s)| Fe3+(aq)
b.  Fe3+(aq)| Ni(s) || Fe(s)| Ni2+(aq)	d.  Ni(s)| Ni2+(aq) || Fe3+(aq)| Fe(s)

Solution 3: Nickel metal is first since it is the anode.  Iron metal is last since it is the cathode!

4.   What is the solubility of silver sulfide in water, if it dissolves according to the reaction below?

Ag₂S(s) « 2Ag⁺(aq) + S^2-(aq)     K_sp = 6.31x10^-51

a.  6.31x10-51	c.  9.28x10-18
b.  1.16x10-17	d.  1.28x10-17

Solution 4:

5.  Using Boltzmann’s formula, calculate the entropy of 1.563x10¹⁰ molecules of a compound if they are randomly distributed in one of two ways.

a.  DS = 4.16x10-24 J/K	c.  DS = 6.50x10-14 J/K
b.  DS = 9.01x1010 J/K	d.  DS = 1.50x10-13 J/K

Solution 5: determine the number of moles and then use the Boltzmann relationship.

 

6.  What is the free energy change at 25.0°C for the reaction below, if the standard entropies of reactants and products are S^o(C(gr)) = 5.74 J/mol×K, S^o(H₂(g)) = 130.68 J/mol×K and S^o(C₆H₆(l)) = 173.30 J/mol×K and the heat of formation of C₆H₆(l) is 49.00 kJ/mol.

6C(gr) + 3H₂(g) « C₆H₆(l)

a.  DGo = -1.24x102 kJ	c.  DGo =  1.24x102 kJ
b.  DGo = -26.448 kJ	d.  DGo = 26.448 kJ

Solution 6: DG = DH - TDS.  You are given DH; calculate DS from the given entropies of reactants and products.

DS = 173.3 - (3×130.68 + 6×5.74) = -253.3 J/K

DG = 49.00 - 0.29815×(-253.3) = 124.4 kJ

7.  What is the molar concentration of  when an excess of solid mercury (I) chloride is added to 0.186 M NaCl so that some solid remains?  The solubility reaction is: Hg₂Cl₂(s) « (aq) + 2Cl^-(aq)      K_sp = 1.288x10^-18

a.  4.81x10-7 M	c.  1.79x10-19 M
b.  3.73x10-17 M	d.  3.53x10-20 M

Solution:

8.  The free energy of formation for C₆H₆(g) is 129.72 kJ/mol.  What is the equilibrium constant at 25.0°C for its formation?

a.  4.60x10-53	c.  1.88x10-23
b.  2.18x10+52	d.  5.32x10+22

Solution:  Remember the relationship between free energy change and the equilibrium constant.

9.   What is the cell voltage for the following cell?

Sn²⁺(aq) + Mg(s) « Sn(s) + Mg²⁺(aq)       DG^o = -430.3 kJ

a.  +2.23 V	c.  -2.23 V
b.  +0.14 V	d.  -0.14 V

Solution: 

10.  What is the standard cell potential for the following cell?  Will the reaction be spontaneous?

Ni(s) + Mg2+(aq) «	Ni2+(aq) + Mg(s)
Ni2+(aq) + 2e- «	Ni(s)	Eo = -0.260 V
Mg2+(aq) + 2e- «	Mg(s)	Eo = -2.370 V

 

a.  +2.110 V, spontaneous	c.  +4.220 V, non-spontaneous
b.  -2.110 V, non-spontaneous	d.  -4.220 V, spontaneous

Solution: .  Negative voltage means that the reaction is non-spontaneous.

11.  What must be the reaction quotient, Q, if the free energy is -243.56 kJ at 25.0°C?

2C₂H₂(g) + H₂(g) « C₄H₆(g)         DG^o = -216.79 kJ

a.  0.26	c.  2.03x10-5
b.  2.82x10-5	d.  1.32x10-25

Solution:

 

12.  What is the molar concentration of Al³⁺(aq) if the cell potential was 0.872 V and the molar concentration of Cr³⁺(aq) is 8.18x10^-5 M?

Cr³⁺(aq) + Al(s) « Cr(s) + Al³⁺(aq)                       V

a.  2.70x10-2 M	c.  9.73x10-5 M
b.  2.26x10-2 M	d.  9.40x10-4 M

Solution: Use the Nernst equation. 

13.  Calculate  for C₈H₁₀(g) at 25.0°C from the information below, if (CH₄(g)) is -50.79 kJ/mol.

8CH₄(g) « C₈H₁₀(g) + 11H₂(g)      DG^o = 534.56 kJ/mol

a.  +128.0 kJ	c.  +16.303 kJ
b.  -16.030 kJ	d.  -128.0 kJ

Solution:

14.  What is the total entropy change at 25.0°C for the reaction below, if the standard entropies of reactants and products are S^o(C(gr)) = 5.74 J/mol×K, S^o(H₂(g)) = 130.68 J/mol×K, and S^o(C₁₀H₁₀(l)) = 368.99 J/mol×K and the heat of formation of C₈H₁₀(g) is 29.80 kJ/mol.

8C(gr) + 5H₂(g) « C₈H₁₀(g)

 

a.  -330 J/K	c.  +430 J/K
b.  +330 J/K	d.  -430 J/K

Solution: DS = 368.99 J/K - (5×130.68 J/K + 8×5.74 J/K) = -330 J/K.

 

15.  Which of the combinations below will not produce any precipitate?  The concentrations shown are after mixing, but before reaction.

a.  3.37x10-7 M Pb(NO3)2 + 1.35x10-4 M Na2CrO4; Ksp(PbCrO4) = 2.818x10-13

b.  1.72x10-7 M AgNO3 + 1.96x10-15 M NaBr; Ksp(AgBr) = 5.495x10-13

c.  3.47x10-20 M Bi(NO3)2 + 6.96x10-14 M Na2S; Ksp(Bi2S3) = 1.000x10-97

d.  1.34x10-16 M CuNO3 + 1.87x10-14 M Na2S; Ksp(Cu2S) = 1.995x10-47

Solution:  Determine Q_sp for each.  If Q_sp < K_sp, there will be no precipitate.  By inspection one can see that answer b meets this criterion.

16.  How many grams of tin metal will be plated out in an electrolysis system operated at 6.050 A for 251.8 seconds from a solution containing Sn²⁺(aq)

a.  6.651x10-5 g	c.  0.937 g
b.  0.838 g	d.  2.561x10-2 g

Solution: