chapter 22 chemistry of the main group...

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CHAPTER 22

CHEMISTRY OF THE MAIN GROUP ELEMENTS II: GROUPS 18, 17, 16, 15 AND HYDROGEN

PRACTICE EXAMPLES 1A (D) This question involves calculating E for the reduction half-reaction:

ClO3(aq) + 3 H2O(l) + 6 e Cl(aq) + 6 OH(aq)

Here we will consider just one of the several approaches available to solve this problem. The four half-reactions (and their associated E values) that are used in this method to come up with the “missing E value” are given below. (Note: the E for the first reaction was determined in Example 22-1)

1) ClO3(aq) + H2O(l) + 2 e ClO2

(aq) + 2 OH(aq) E = 0.295 V G1 = 2FE

2) ClO2(aq) + H2O(l) + 2 e OCl(aq) + 2 OH(aq) E = 0.681 V G2 = 2FE

3) OCl(aq) + H2O(l) + 1 e 1/2 Cl2(aq) + 2 OH(aq) E = 0.421 V G3 = 1FE

4) 1/2 Cl2(aq) + 1 e Cl(aq) E = 1.358 V G4 = 1FE

Although the reactions themselves may be added to obtain the desired equation, the E for this equation is not the sum of the E values for the above four reactions. The E value for the desired equation is actually the weighted average of the E values for reactions (1) to (4). It can be calculated by summing up the free energy changes for the four reactions. (The standard voltages for half-reactions of the same type are not additive. The G values for these reactions can, however, be summed together.)

When (1), (2), (3) and (4) are added together we obtain

1) ClO3(aq) + H2O(l) + 2 e ClO2

(aq) + 2 OH(aq) E1 = 0.295 V

+ 2) ClO2(aq) + H2O(l) + 2 e OCl(aq) + 2 OH(aq) E2 = 0.681 V

+ 3) OCl(aq) + H2O(l) + 1 e 1/2 Cl2(aq) + 2 OH(aq) E3 = 0.421 V

+ 4) 1/2 Cl2(aq) + 1 e Cl(aq) E4 = 1.358 V

ClO3(aq) + 3 H2O(l) + 6 e Cl(aq) + 6 OH(aq) E5 = ?

and G5 = G1 + G2 + G3 + G4

so, 6FE5 = 2F(0.295 V) + 2F(0.681V) + 1F(0.421 V) + 1F(1.358 V)

Hence, E5 = 6F-

V) 1F(1.358- V) 1F(0.421- 2F(0.681V)- V) 2F(0.295- = 0.622 V

Chapter 22: Chemistry of the Main- Group Elements II: Groups 18, 17, 16, 15, and Hydrogen

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1B. (D) This question involves calculating E for the reduction half-reaction:

2 ClO3(aq) + 12H+(aq) + 10 e Cl2(aq) + 6 H2O(l)

The three half-reactions (and their associated E values) are given below:

(1) ClO3(aq) + 3H+(aq) + 2 e HClO2(aq) + H2O(l) E = 1.181 V

(2) HClO2(aq) + 2 H+(aq) + 2 e HClO(aq) + H2O(l) E = 1.645 V

(3) 2 HClO (aq) + 2 H+(aq) + 2 e Cl2(aq) + 2 H2O(l) E = 1.611 V (remember that G= nFE)

Although the reactions themselves can be added to obtain the desired equation, the E for this equation is not the sum of the E values for the above three reactions. The E for the desired equation is actually the weighted average of the E values for reactions (1) to (3). It can be obtained by summing up the free energy changes for the three reactions. (For reactions of the same type, standard voltages are not additive; G values are additive, however.) When (1) and (2) (each multiplied by two) are added to (3) we obtain:

2(1) 2 ClO3(aq) + 6 H+(aq) + 4 e 2 HClO2(aq) + 2 H2O(l) E1 = 1.181 V

2(2) 2 HClO2(aq) + 4 H+(aq) + 4 e 2 HClO(aq) + 2 H2O(l) E2 = 1.645 V (3) 2 HClO (aq) + 2 H+(aq) + 2 e Cl2(aq) + 2 H2O(l) E3 = 1.611 V

2 ClO3(aq) + 12 H+(aq) + 10 e Cl2(aq) + 6 H2O(l) E4 = ?

and G4 = G1 + G2 + G3

so, 10FE4 = 4F(1.181 V) + 4F(1.645 V) + 2F(1.611 V)

Hence, E4 = F10-

V) 2F(1.611- V) 4F(1.645- V) 4F(1.181- = 1.453 V

2A (M) The dissociation reaction is the reverse of the formation reaction, and thus G for

the dissociation reaction is the negative of fG

1 12 2 f2 2HF g H g + F (g) = = 273.2 kJ/molG G

We know that

3 1 1P P= ln +273.2 10 J/mol = 8.3145 J mol K 298 K lnG RT K K

3 1

110 48 48P p1 1

273.2 10 J molln = = 110 = e = 1.7 10 2 10

8.3145 J mol K 298 KK K

Virtually no dissociation of HF(g) into its elements occurs.


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2B (M) The dissociation reaction with all integer coefficients is twice the reverse of the formation reaction.

2 2 p2HCl g H g + Cl g = 2 95.30 kJ / mol = +190.6 kJ / mol = lnG RT K

ln K

p=G

RT=

190.6 103 J / mol

8.3145 J mol1 K1 298 K= 76.9 K

p= e76.9 = 4 1034

We assume an initial HCl(g) pressure of P atm, and calculate the final pressure of Cl2(g) and H2(g), x atm.

Reaction: 2 HCl(g) H2(g) + Cl2(g)

Initial: P atm 0 atm 0 atm Changes: –2x atm +x atm +x atm Equil: (P – 2x) atm x atm x atm

22 2 34 17

p 22

{H g } {Cl g }= = = = 4 10 2 10

{HCl g } 2 22

P P x x x xK

P P x P xP x

17 17= 2 10 2 2 10x P x P

17 152% decomposition = 100% = 2 2 10 100% 4 10 % decomposed

x

P

INTEGRATIVE EXAMPLE A (M) The half-reactions are: NO2

- +2OH- NO3- +H2O+2e- Eo 0.04V

{NO2- +H2O+e- NO+2OH-} 2 Eo 0.46V

_________________________________________________________________________________________________________

3NO2- H2O NO3

+2NO+2OH Ecello 0.42V

Eo for the first reaction (oxidation) can be calculated by using the approach in Example 22-1. The calculated value is very small, and so the Eo for the disproportionation reaction is negative. The disproportionation of NO2

- to NO3- and NO is therefore not

spontaneous. B (M) The half-reactions are: HNO2 +H2O NO3

- +3H+ +2e- Eo 0.934V

{HNO2 +H+ +e- NO+H2O} 2 Eo 0.996V ________________________________________________________________________________________________________________

3HNO2 NO3- +H+ +H2O+2NO Ecell

o 0.062V Eo for the first reaction (oxidation) can be calculated by using the approach in Example

22-1. Ecello for the reaction is positive and therefore disproportionation of HNO2 to NO3

- and NO is spontaneous under standard conditions.


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EXERCISES Periodic Trends in Bonding and Acid-Base Character of Oxides 1. (E) LiF, BeF2, BF3, CF4, NF3, OF2. LiF is an ionic compound, BeF2 is a network covalent compound, and the others are molecular covalent compounds. 2. (E) NaF, MgF2, AlF3, SiF4, PF5, SF6, and ClF5. NaF and MgF2 are ionic compounds, AlF3 is a network covalent compound, and the others are molecular covalent compounds. 3. (E) The metallic character of the elements increases as we move down a group, and so too does the basic character of the oxides. Bi2O3 is most basic, P4O6 is least basic (and is actually acidic), and Sb4O6 is amphoteric. 4. (M) The metallic character of the elements increases as we move down a group, and so too does the basic character of the oxides. Stated another way, the acidic character of the oxides decreases as we move down a group. On this basis, we expect SeO2 to be the acidic oxide and TeO2 to be the amphoteric oxide. The Noble Gasses 5. (M) First we use the ideal gas law to determine the amount in moles of argon.

21 1

145 atm 55 L3.25 10 mol Ar

0.08206 L atm mol K 299 K

PVn

RT

2 522.414 L Ar STP 100.000 L airL air = 3.25 10 mol Ar = 7.8 10 L air

1 mol Ar 0.934 L Ar

6. (M) First we compute the volume occupied by 5.00 g of He at STP, and then the

volume of natural gas. From Chapter 6 we saw that 1 mol of gas at STP occupied 22.414 L.

2

1 mol He 22.414 L He 100 L airnatural gas volume = 5.00 g He

4.003 g He 1 mol He 8 L He

natural gas volume = 3.5 10 L air at STP 300 L

7. (M) (a) (b) (c)

O Xe O

O O Xe O

O

O

Xe

F

F

FF

F+


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(a) The Lewis structure has three ligands and one lone pair on Xe. 3XeO has a

trigonal pyramidal shape.

(b) The Lewis structure has four ligands on Xe. 4XeO has a tetrahedral shape.

(c) There are five ligands and one lone pair on Xe in +5XeF . Its shape is square

pyramidal.

8. (M) We use VSEPR theory to predict the shapes of the species involved. (a) 2 2O XeF has a total of 2 6 +8 + 2 7 = 34 valence electrons = 17 pairs.

(b) 3 2O XeF has a total of 3 6 +8 + 2 7 = 40 valence electrons = 20 pairs.

(c) 4OXeF has a total of 6 +8 + 4 7 = 42 valence electrons = 21pairs.

We draw a plausible Lewis structure for each species below. (a) (b) (c) (a) There are four ligands and one lone pair on Xe in 2 2O XeF . Its shape is a see-saw.

(b) There are five ligands and no lone pairs on Xe in 3 2O XeF . Its shape is trigonal

bipyramidal.

(c) The Lewis structure has five ligands and one lone pair on Xe. 4OXeF has a square pyramidal shape.

9. (E) 3 XeF4(aq) + 6 H2O(l) 2 Xe(g) + 3/2 O2(g) + 12 HF(g) + XeO3(s) 10. (E) 2 XeF6(aq) + 16 OH(aq) Xe(g) + XeO6

4(aq) + O2(g) + 8 H2O(l) + 12 F(aq) 11. (E) For these noble gases, the bond energy of the noble gas fluorine is too small to offset the energy required to break the F—F bond. 12. (E) The bond energy of the Xe=O bond is too small to offset the energy required to break the O=O bond.

F Xe F

O

FFO Xe O

O

FFF Xe F

O

O


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The Halogens 13. (M) Iodide ion is slowly oxidized to iodine, which is yellow-brown in aqueous

solution, by oxygen in the air.

Oxidation: { 22I aq I aq + 2e } 2 = 0.535VE

Reduction: +2 2O g + 4 H aq + 4e 2 H O(l) = +1.229 VE

Net: +2 2 2 cell4 I aq + O g + 4 H aq 2 I aq + 2 H O(l) = +0.694 VE

Possibly followed by: 2 3I aq + I aq I aq

14. (E) 2

6 5 4 6 4 3 2MnF + 2SbF MnF + 2SbF ; 2 MnF 2 MnF + F g

15. (M) Displacement reactions involve one element displacing another element from

solution. The element that dissolves in the solution is more “active” than the element supplanted from solution. Within the halogen group, the activity decreases from top to bottom. Thus, each halogen is able to displace the members of the group below it, but not those above it. For instance, molecular bromine can oxidize aqueous iodide ion but molecular iodine is incapable of oxidizing bromide ion:

Br2(aq) + 2 I(aq) 2 Br(aq) + I2(aq) however, I2(aq) + 2 Br(aq) NO RXN

The only halogen with sufficient oxidizing power to displace O2(g) from water is F2(g):

2 H2O(l) 4 H+ + O2(g) + 4 e E1/2ox = 1.229 V {F2(g) + 2 e 2 F(aq)} 2 E1/2red = 2.866 V

2 F2(g) + 2 H2O(l) 4 H+ + O2(g) + 4 F(aq) Ecell = 1.637 V

The large positive standard reduction potential for this reaction indicates that the reaction will occur spontaneously, with products being strongly preferred under standard state conditions. None of the halogens reacts with water to form H2(g). In order to displace molecular hydrogen from water, one must add a strong reducing agent, such as sodium metal.

16. (M) We first list the relevant values of halogen properties.

F Cl Br I (a) Covalent radius (pm): 71 99 114 133 (b) Ionic radius(pm): 133 181 196 220 (c) First ionization energy (kJ/mol): 1681 1251 1140 1008 (d) Electron affinity(kJ/mol): –328.0 –349.0 –324.6 –295.2 (e) Electronegativity: 4.0 3.0 2.8 2.5 (f) Standard reduction potential: +2.886 V +1.358 V +1.065 V +0.535 V

We can do a reasonably good job of predicting the properties of astatine by simply looking at the difference between Br and I and assuming that the same difference exists between I and At.


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(a) Covalent radius: 152 pm (b) Ionic radius: 244pm

(c) 1st ionization energy: 876kJ/mol (d) Electron affinity: 265kJ/mol

(e) Electronegativity: 2.2 (f) E 0.005V to 0.010V

Depending on the technique that you use, you will arrive at different answers. For example, 260 kJ/mol is a reasonable estimation of the electron affinity by the following reasoning. The difference between the value for Cl and Br is 24 kJ/mol, the difference between the values for Br and I is 29 kJ/mol, so the difference between I and At should be about 35 kJ/mol.

17. (M) (a)

mass F2

= 1 km3 1000 m

1 km

100 cm

1 m

3

1.03 g

1 cm3

1 lb

454 g

1 ton

2000 lb

1 g F

1 ton

37.996 g F2

18.998 g F

= 2 109 g F2

= 2 106 kg F2

(b) Bromine can be extracted by displacing it from solution with 2Cl g . Since there

is no chemical oxidizing agent that is stronger than F2(g), this method of displacement would not work for F2(g). Even if there were a chemical oxidizing agent stronger than F2(g), it would displace 2O (g) before it displaced F2(g).

Obtaining F2(g) would require electrolysis of one of its molten salts, obtained from seawater evaporate.

18. (E)

3 4 223

23 4 22

1 mol 3 Ca PO CaF

1009 g

1000 g 2 mol FMass F = 1.00 10 kg rock

1 kg 1 mol 3 Ca PO CaF

19.00 g F 1 kg= 37.7 kg fluorine

1 mol F 1000 g

19. (M) In order for the disproportionation reaction to occur under standard conditions, the

E for the overall reaction must be greater than zero. To answer this question, we must refer to the Latimer diagrams provided in Figure 22-4 and the answer to Practice Example 22-1B.

(i) Reduction half reaction (acidic solution) Cl2(aq) + 2 e 2 Cl(aq) E1/2 red = 1.358 V

(ii) Oxidation half reaction (acidic solution) Cl2(aq) + 6 H2O(l) 2 ClO3

(aq) + 12 H+(aq) + 10 e E1/2 ox = 1.453 V

Combining (i) 5 with (ii) 1, we obtain the desired disproportionation reaction: 6 Cl2(aq) + 6 H2O(l) 2 ClO3

(aq) + 12 H+(aq) + 10 Cl(aq) Ecell = 0.095 V

Since the final cell voltage is negative, the disproportionation reaction will not occur


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spontaneously under standard conditions. Alternatively, we can calculate Keq by using lnKeq = G/RT and G = nFE. This method gives a Keq = 8.6 1017. Clearly, the reaction will not go to completion under standard conditions.

20. (D) To find out whether the reaction will go to completion, we must first calculate the

standard potential for the disproportionation of hypochlorous acid (HOCl). As was the case for Exercise 15, we will use information contained in Fig. 22.4 to answer this question.

(i) 2 HOCl(aq) + 2 H2O(l) 2 HClO2(aq) + 4 H+(aq) + 4 e -E(i) = 1.645 V) (oxidation)

(ii) 2 HOCl(aq) + 2 H+(aq) + 2 e Cl2(aq) + 2 H2O(l) E(ii) = +1.611 V

(iii) Cl2(aq) + 2 e 2 Cl(aq) E(iii) =+1.358 V

Next we need to determine the potential of the following half reaction using the relationship between standard free energies and standard potential

(iv) 2 HOCl(aq) + 2 H+(aq) + 4 e 2 Cl(aq) + 2 H2O(l) E(iv)

We can determine this using half reaction (ii) and(iii) since adding (ii) and (iii) give (iv). Hence,

4F E(iv) = 2F(1.611 V) + 2F(1.358 V) So, E(iv) = 1.485 V

Therefore, by adding equation (i) to equation (iv), we can obtain the standard cell voltage for the disproportionation of HOCl. We divide by 2 to give the target

(i)2 HOCl(aq) + H2O(l) HClO2(aq) + 2 H+(aq) + 2 e -E(i) = 1.645 V)

(iv)2 HOCl(aq) + H+(aq) + 2 e Cl(aq) + H2O(l) E1/2 red = 1.485 V

2 HOCl(aq) HClO2(aq) + Cl(aq) + H+(aq) Ecell = 0.160 V (disproportionation reaction of HOCl(aq))

Since the standard cell potential is negative, the equilibrium will favor the reactants and thus the reaction will not go to completion as written (i.e., reactants will predominate at equilibrium and thus the reaction will be far from complete). Alternatively, we can calculate Keq from the reaction by using lnKeq = G/RT and G = nFE. This gives Keq = 3.9 106. Clearly, the reaction will not go to completion because Keq is very small (reactants strongly predominate at equilibrium, not products).

21. (M) First we must draw the Lewis structure for all of the species listed. Following this, we will deduce their electron-group geometries and molecular shapes following the VSEPR approach.

(a) BrF3: 28 valence electrons

VSEPR class: AX3E2

Thus BrF3 is T-shaped FBr

F

F


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(b)

IF5

42 valence electrons VSEPR class: AX5E Thus IF5 is square pyramidal

FI

F

FF

F

(c) Cl3IF 36 valence electrons

VSEPR class: AX4E2

Thus Cl3IF is square planar

FI

Cl

ClCl

22. (M) To answer this question we need to apply the VSEPR method to each species:

(i) ClF2+ (total number of valence electrons = 20 e)

Molecular shape: AX2E2 angular (<109.5 due to lone pairs on Cl atom)

F FCl

(ii) IBrF(total number of valence electrons = 22 e) Molecular shape: AX2E3 linear (~180 )

I FBr

(iii) OCl2 (total number of valence electrons = 20 e) Molecular shape: AX2E2 angular (<109.5 due to lone pairs on O or Cl atom) Note: ClOCl preferred over ClClO because all atoms have a formal charge of zero.

Cl ClO

or

Cl OCl

(iv) ClF3 (total number of valence electrons = 28 e) Molecular shape: AX3E2 T-shaped molecule

FCl

F

F

(v) SF4(total number of valence electrons = 34 e) Molecular shape: AX4E See-saw molecular geometry (or distorted tetrahedron) F

F

F S

F The VSEPR treatment of each species indicates only IBrF has a linear structure. Since (i) and (iii) belong to the same VSEPR class (AX2E2), it is not unreasonable to assume that they have the same “bent structure”.

23. (M) (a) The half-reactions are: Oxidation: 3I- I3

- +2e-

Reduction: I2 (aq)+2e- 2I-(aq) _________________________________________________________________________________________________________

Overall: I2 (aq)+I-(aq) I3- (aq)

From the known value of the equilibrium constant at 25 oC, we can calculate Go and consequently Eo :


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Go RT ln K nFEo

8.314JK-1mol-1 298.15K ln K 2 96485Cmol-1 Eo

Eo 8.314JK-1mol-1 298.15K ln 7.7 102

2 96485Cmol-1 0.085V

(b) Set up the ICE table and solve for the equilibrium concentrations: I2 (aq) + I- (aq) É I3

- (aq) 0.0010M 0.0050M 0 0.0010-x 0.0050-x x

K 7.7 102 x

(0.0010 x) (0.0050 x)

x

5.0 106 0.0060x x2

7.7 102 (5.0 106 0.0060x x2 ) x

7.7 102 x2 5.62x 3.85 103 0

Solving the quadratic equation gives x=7.6510-3 M. 24. (M) (a) Cl3

has the same structure as I3 , which is given in Figure 22.7

(b) Set up the ICE table and solve for the equilibrium concentrations: Cl3

- (aq) É Cl- (aq) + Cl2 (aq) 0 0.0010M 0.0010M x 0.0010-x 0.0010-x

K 5.5

(0.0010 x) (0.0010 x)

x

0.00102

x

1.0 106

x

x 1.0 106

5.5 1.8 107 M

Oxygen

25. (E)(a) 2HgO(s) 2Hg(l) O (g)2 (b) 2KClO (s) 2KClO (s) O (g)4 3 2

26. (E) (a) +

3 2 2 2O g + 2 I aq + 2 H aq I aq + H O(l) + O g

(b) S s + H2O l + 3 O3 g H2SO4 aq + 3 O2 g (c) 4 3

3 2 26 62 Fe CN aq + O g + H O(l) 2 Fe CN aq + O g + 2 OH aq

27. (M) We first write the formulas of the four substances: N O Al O P O CO2 4 2 3 4 6 2, , , . The

one constant in all these substances is oxygen. If we compare amounts of substance with the same amount (in moles) of oxygen, the one with the smallest mass of the other element will have the highest percent oxygen.

3 mol N O2 4 contains 12 mol O and 6 mol N: 6 14.0 = 84.0 g N 4 mol Al O2 3 contains 12 mol O and 8 mol Al: 8 27.0 = 216 g Al 2 mol P O4 6 contains 12 mol O and 8 mol P: 8 31.0 = 248 g P


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6 mol CO2 contains 12 mol O and 6 mol C: 6 12.0 = 72.0 g C

Thus, of the oxides listed, CO2 contains the largest percent oxygen by mass.

28. (M) 2NH NO (s) 2N (g) O (g) + 4H O(g)4 3 2 2 2400 C

2H O (l) 2H O(l) O (g)2 2 2 2

2KClO (s) 2KCl(s) O (g)3 2 3

(a) All three reactions have two moles of reactant, and the decomposition of

potassium chlorate produces three moles of O2 g , compared to one mole of

O2 g for each of the others. The potassium chlorate decomposition produces

the most oxygen per mole of reactant.

(b) If each reaction produced the same amount of O2 g , then the one with the

lightest mass of reactant would produce the most oxygen per gram of reactant. When the first two reactions are compared, it is clear that 2 mol H O2 2 have less mass than 2 mol NH NO4 3 . (Notice that each mole of NH NO4 3 contains the amounts of elements—2 mol H and 2 mol O—present in each mole H O2 2 plus some more.) So now we need to compare hydrogen peroxide with potassium chlorate. Notice that 6 2 2 H O produces the same amount of O2 g as 2 3 KClO .

6 mol H O2 2 has a mass of 34.01 6 = 204.1 g , while 2 mol KClO3 has a mass

of 122.2 2 = 244.2 g . Thus, H O2 2 produces the most O2 g per gram of

reactant. 29. (M) Recall that fraction by volume and fraction by pressure are numerically equal.

Additionally, one atmosphere pressure is equivalent to 760 mmHg. We combine these two facts.

533 6

0.04 mmHg of O{O } = 760 mmHg = 3 10 mmHg

10 mmHg of atmosphereP

30. (M)

12 3

23

33

1 mol O 0.08206 L atm5 10 molecules 220 K

molecule mol K6.022 10moleP

1 L1 cm

1000 cm

nRT

V

7 4760 mmHg= 1.5 10 atm = 1.1 10 mmHg

1 atmP


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31. (M) The electrolysis reaction is 2H2O(l) electrolysis 2H

2(g) O

2(g) . In this reaction, 2

moles of H2 g are produced for each mole of O2 g . By the law of combining

volumes, we would expect the volume of hydrogen to be twice the volume of oxygen produced. (Actually the volumes are not exactly in the ratio of 2:1 because of the different solubilities of oxygen and hydrogen in water.)

32. (M) The electrolysis reaction is 2H2O(l) electrolysis 2H

2(g) O

2(g) . In this reaction, 2

moles of water are decomposed to produce each mole of O2 g . We use the data in the

problem to determine the amount of O2 g produced, which we then convert to the

mass of H O2 needed.

mass H2O(l)=

736.7 mmHg 1 atm

760 mmHg

22.83 mL 1 L

1000 mL

0.08206 L atm

mol K (25.0 273.2)K

2 mol H2O

1 mol O2

22

2

18.02 g H O= 0.03259 g H O decomposed

1 mol H O

33. (D) Since the pKa for H2O2 had been provided to us, we can find the solution pH simply

by solving an I.C.E. table for the hydrolysis of a 3.0 % H2O2 solution (by mass). Of course, in order to use this method, the mass percent must first be converted to molarity. We must assume that the density of the solution is 1.0 g mL1.

[H2O2] = 2 23.0 g H O

100 g solution

solutionmL1

solutiong1 2 2

2 2

1 mol H O

34.015 g H O

L1

mL1000= 0.88 M

The pKa for H2O2 is 11.75. The Ka for H2O2 is therefore 1011.75 or 1.8 1012. By comparison with pure water, which has a Ka of 1.8 1016 at 25 C, one can see that H2O2 is indeed a stronger acid than water but the difference in acidity between the two is not that great. Consequently, we cannot ignore the contribution of protons of pure water when we work out the pH of the solution at equilibrium.

Reaction: H2O2(aq) + H2O(l) -12

aK = 1.8 10 H3O+(aq) + HO2

(aq)

Initial: 0.88 M 1.0 107 M 0 M Change: x M + x M +x M

Equilibrium: (0.88x) M (1.0 107 + x)M x M (~0.88 M)

So, 1.8 1012 = 7( 1.0 10 )

~0.88

x x x2 + 1.0 107x 1.58 1012 = 0

x = 7 14 121.0 x 10 1.0 10 4(1.58 10 )

2


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The root that makes sense in this context is x = 1.2 106 M. Thus, the [H3O

+] = 1.2 106 M + 1.0 107 M = 1.3 106 M Consequently, the pH for the 3.0 % H2O2 solution (by mass) should be 5.89 (i.e., the solution is weakly acidic)

34. (E) Oxide ion reacts essentially quantitatively with water to form hydroxide ion.

+2 2Li O s + H O l 2 Li aq + 2OH aq

We first calculate OH and then the solution’s pH.

2 2

2 2

0.050 g Li O 1000 mL 1 mol Li O 2 mol OHOH = = 0.00446 M

750.0 mL 1 L 29.88 g Li O 1 mol Li O

pOH = log 0.00446 = 2.350 pH = 14.000 pOH = 14.000 2.350 = 11.650

35. (E) 3 O = O 2 O O O

Bonds broken: 3 (O=O) = 3 498 kJ/mol = 1494 kJ/mol

Bonds formed: 4 (O O)

o = +285 kJ/mol = bonds broken bonds formed = 1494 kJ/mol 4 (O O)H

4 (O O) = 1494 kJ/mol 285 kJ/mol = 1209 kJ/mol

O O = 1209 kJ/mol + 4 = 302 kJ/mol

36. (E) (all gaseous species)

kJ498 +142

O==O +O—O kJmolAverage bond energy = = = 3202 2 mol

37. (E) (a) 2H S, while polar, forms only weak hydrogen bonds. 2H O forms much

stronger hydrogen bonds, leading to a higher boiling point.

(b) All electrons are paired in 3O , producing a diamagnetic molecule and thus a

liquid at room temperature.

38. (M) (a)

O O O O O H O O H

The O—O bond in 2O is a double bond, which should be short (121 pm). That in

3O is a “one-and-a-half” bond, of intermediate length (128 pm). And that in

2 2H O is a longer (148 pm) single bond.

O OO23 O O


1108

(b) The +2O ion has one less electron than does the 2O molecule. Based on Lewis

structures,

one would predict that +2O would have a bond order of 1.5, thereby making its

OO bond weaker than the double bond of 2O , and therefore longer, in contradiction to the experimental evidence. The molecular orbital picture of the two species suggests the opposite. According to this model, O2 has a bond order of 2.0 while that for O2

+ is 2.5. Thus MO theory predicts a shorter bond for O2+.

1s 1s 2s 2s 2p 2p 2p 2p

2O

bond order = no. bonding electrons no. antibonding electrons 2

bond order = 10 6 + 2 = 2.0

1s 1s 2s 2s 2p 2p 2p 2p

+2O

bond order = no. bonding electrons no. antibonding electrons 2 = 10 5 2 = 2.5

39. (M) Reactions that have Keq values greater than 1000 are considered to be essentially

quantitative (i.e., they go virtually 100% to completion). So to answer this question we need only calculate the equilibrium constant for each reaction using the equation Ecell = (0.0257/n)lnKeq.

(a) H2O2(aq) + 2 H+(aq) + 2 e 2 H2O(l) E1/2red = +1.763 V

2 I(aq) I2(s) + 2 e E1/2ox = 0.535 V

H2O2(aq) + 2 H+(aq) +2 I(aq) I2(s) + 2 H2O(l) E1/2cell = +1.228 V(n = 2 e)

lnKeq = 1.228 V 2

0.0257 V

= 95.56 Keq = 3.2 1041

Therefore the reaction goes to completion (or very nearly so).

(b) O2(g) + 2 H2O(l) +4 e 4 OH(aq) E1/2red = +0.401 V

4 Cl(aq) 2 Cl2(g) + 4 e E1/2ox = 1.358 V

O2(g) + 2 H2O(l) +4 Cl(aq) 2 Cl2(g) + 4 OH(aq) E1/2cell = 0.957 V (n = 4 e)

lnKeq = -0.957 V 4

0.0257 V

= 148.95 Keq = 2.1 1065

O O O O


1109

The extremely small value of Keq indicates that reactants are strongly preferred and thus, the reaction does not even come close to going to completion.

(c) O3(g) + 2 H+(aq) + 2 e O2(g) + H2O(l) E1/2red = +2.075 V

Pb2+(aq) + 2 H2O(l) PbO2(s) + 4H+(aq) + 2 e E1/2ox = 1.455 V

O3(g) + Pb2+(aq) + H2O(l) PbO2(s) + 2 H+(aq) + O2(g) Ecell = 0.620 V

lnKeq = 0.62 V 2

0.0257 V

= 48.25 Keq = 9.0 1020

Therefore the reaction goes to completion (or very nearly so).

(d) HO2(aq) + H2O(l) + 2 e 3 OH(aq) E1/2red = +0.878 V

2 Br(aq) Br2(l) + 2 e E1/2ox = 1.065 V

HO2(aq) + H2O(l) + 2 Br(aq) Br2(s) + 3 OH(aq) E1/2cell = 0.187 V

lnKeq = -0.187 V 2

0.0257 V

= 14.55 Keq = 4.8 107

The extremely small value of Keq indicates the reaction heavily favors reactants at equilibrium and thus, the reaction does not even come close to going to completion.

40. (M) (a) HgO(s) Hg(l) + 1/2 O2(g)

(b) KClO4(s) KCl(s) + 2 O2(g)

(c) Hg(NO3)2(s) Hg(l) + 2 NO2(g) + O2(g) or Hg(NO3)2(s) Hg(NO2)2(g) + O2(g)

(d) H2O2(l) H2O(l) + 1/2 O2(g) 41. (M) The reaction is 2 KClO3(s)

2MnO (s) 2 KCl(s) + 3 O2(g).

1.0 gKClO3 1 mol

122.549 g

3 molO2

2 molKClO3

32 gO2

1molO2

0.391gO2

pV nRT V nRT

p

0.01224 8.314JK 1mol1 298.15K

101kPa 1atm

101.325kPa

30L

42. (M) The reaction is 2 HgO(s) 2 Hg(l) + O2(g).

Depending on the temperature


1110

1.0 g HgO 1 mol

216.589 g HgO

1 molO2

2 mol HgO 2.31103mol O2

pO2 (756 23.76) 752mmHg

pV nRT V nRT

p

2.31103 8.314JK 1mol1 298.15K

752mmHg 1atm

760mmHg

5.8L

Sulfur 43. (E) (a) ZnS, zinc sulfide (b) 3KHSO , potassium hydrogen sulfite

(c) 2 2 3K S O , potassium thiosulfate (d) 4SF , sulfur tetrafluoride

44. (E) (a) 4 2CaSO 2H O, calcium sulfate dehydrate (b) 2H S aq , hydrosulfuric acid

(c) 4NaHSO , sodium hydrogen sulfate (d) 2 2 7H S O aq , disulfuric acid

45. (M) (a) 2 2FeS s + 2 HCl aq FeCl aq + H S aq MnS(s), ZnS(s), etc. also are

possible.

(b) 3 2 2 2CaSO s + 2 HCl aq CaCl aq + H O(l) +SO g

(c) Oxidation: 2 +2 2 4SO aq + 2 H O(l) SO aq + 4H aq + 2 e

Reduction: + 2+2 2MnO s + 4 H aq + 2 e Mn aq + 2 H O(l)

Net: 222 2 4SO aq + MnO s Mn aq +SO aq

(d) Oxidation: 2 +2 3 2 2S O aq + H O(l) 2SO g + 2H aq + 4e

Reduction: 2 +2 3 2S O aq + 6H aq + 4e 2S s + 3H O(l)

Net: 2 +2 3 2 2S O aq + 2H aq S s +SO g + H O(l)

46. (M) (a) 2 2S s + O g SO g

2 22 Na s + 2 H O(l) 2 NaOH aq + H g

2 2 3 22 NaOH aq +SO g Na SO aq + H O(l)

(b) 2 2S s + O g SO g

2 +2 2 2 4 cellSO g + 2H O(l) + Cl g SO aq + 2Cl aq + 4H aq = +1.19VE

2 22 Na s + 2 H O(l) 2 NaOH aq + H g

2+4 2 4 22 NaOH aq + 2H aq +SO aq Na SO aq + 2H O(l)


1111

(c) 2 2S s + O g SO g

2 22 Na s + 2H O(l) 2 NaOH aq + H g

2 2 3 22 NaOH aq +SO g Na SO aq + H O(l)

2 3 2 3Na SO aq + S s NaS O aq

(Thus, one must boil the reactants in an alkaline solution.) 47. (M) The decomposition of thiosulfate ion is more highly favored in an acidic solution. If

the white solid is Na2SO4, there will be no reaction with strong acids such as HCl. By contrast, if the white solid is Na2S2O3, SO2(g) will be liberated and a pale yellow precipitate of S(s,rhombic) will form upon addition of HCl(aq).

2 +2 3 2 2S O aq + 2H aq S s +SO g + H O(l)

Consequently, the solid can be identified by adding a strong mineral acid such as HCl(aq). 48. (M) Sulfites are easily oxidized to sulfates by, for example, 2O in the atmosphere. On the

other hand, there are no oxidizing agents naturally available in reasonable concentrations

that can oxidize 24SO to a higher oxidation state, such as in 2

2 8S O . Similarly, the

atmosphere of Earth is an oxidizing one, reducing agents are not present to reduce 24SO

to a species with a lower oxidation state, except in localized areas.

49. (D) +Na aq will not hydrolyze, being only very weakly polarizing. But 4HSO aq

will ionize further, 22 = 1.1 10K for 4HSO aq . We set up the situation, and solve

the quadratic equation to obtain +3H O .

4 4 44

4 4

12.5 g NaHSO 1000 mL 1 mol NaHSO 1 mol HSOHSO = = 0.416 M

250.0 mL soln 1 L soln 120.1g NaHSO 1 mol NaHSO

Reaction: 2 +4 2 4 3HSO aq + H O(l) SO aq + H O aq

Initial: 0.416M 0M 0M

Changes: M + M + Mx x x

Equil: 0.416 M M Mx x x

Noting the moderate value of the equilibrium constant (K = 0.011), the full quadratic equation must be solved (i.e. no assumption can be made here).

+ 2 23 4 2

2

4

2

H O SO= = 0.011 = = 0.0046 0.011

0.416HSO

0 = + 0.011 0.0046

xK x x

x

x x

2 4 2+

3

4 0.011 1.2 10 +1.8 10= = = 0.062 = H O

2 2

b b acx

a

pH = log 0.062 = 1.21


1112

50. (M) Oxidation: 2 2 +3 2 4{SO aq + H O(l) SO aq + 2H aq + 2e } 5

Reduction: + 2+4 2{MnO aq +8H aq + 5e Mn aq + 4H O(l)} 2

Net: 2 2+ 2+

3 4 4 25 SO aq + 2 MnO aq + 6H aq 5SO aq + 2 Mn aq + 3 H O(l)

23 2 34

2 3 24 3

2 32 3

2 3

5 mol SO 1 mol Na SO1 L 0.0510 mol MnOmass Na SO = 26.50 mL

1000 mL 1 L soln 2 mol MnO 1 mol SO

126.0 g Na SO= 0.426 g Na SO

1 mol Na SO

51. (M) The question is concerned with assaying for the mass percent of copper in an ore. The

assay in this instance involves the quantitative determination of the amount of metal in an ore by chemical analysis. The titration for copper in the sample does not occur directly, but rather indirectly via the number of moles of I3

(aq) produced from the reaction of Cu2+ with I:

2 Cu2+ + 5 I(aq) 2 CuI(s) + I3(aq)

The number of moles of I3(aq) produced is determined by titrating the iodide-treated sample

with sodium thiosulfate. The balanced oxidation reaction that forms the basis for the titration is: I3

(aq) + 2 S2O32(aq) 3 I(aq) + S4O6

2(aq) The stoichiometric ratio is one I3

(aq) reacting with two S2O32(aq) in this titration.

The number of moles of I3 formed

= 0.01212 L S2O32(aq)

2

32

2

32

OS L 1

OSmoles0.1000

2

32

3

OS mol 2

Imol1= 6.060 104 moles I3

Therefore, the number of moles of Cu2+ released when the sample is dissolved is

= 6.060 104 moles I3

2

3

2 mol Cu

1 mol I

= 1.212 103 moles of Cu2+

Consequently, the mass percent for copper in the ore is

1.212 103 moles of Cu2+ 2

2

63.546 g Cu

1 mol Cu

oreCuofg1.100

1 100% = 7.002%

52. (M) First you must realize that pressure and temperature are irrelevant to answering this question. The reaction is Pb2+(aq) + H2S(aq) 2 H+(aq) + PbS(s)

This means that there is a one to one relationship between the moles of PbS and H2S. If we realize that 1 m3 is 1000 L (40 times larger than the sample used), then we can

calculate the mass of PbS in 1 m3, namely, 40 × 0.535 g PbS = 21.4 g PbS.

The mass of sulfur can be calculated using stoichiometry.

mass of S = 21.4 g PbS PbSg239.3

PbS mol 1

1 mol S

1 mol PbS×

32.066 g S

1 mol S= 2.87 g S (per 1 m3)


1113

Alternatively, knowing that the mass percent of S in PbS is 13.40 %. We can calculate the

mass of sulfur as: 21.4 g PbS × 13.40 g S

100 g PbS= 2.87 g S (per 1 m3 natural gas).

53. (E) (a) +4; (b) +5; (c) -2 and (d) +4 54. (E) (a) +1; (b) +2; (c) +2 and (d) +2.5 Nitrogen Family

55. (M) (a) The Haber-Bosch process is the principal artificial method of fixing

atmospheric nitrogen. 2 2 3N g + 3 H g 2 NH g

(b) The first step of the Ostwald process:

3 2 2850 C, Pt4NH (g) 5 O (g) 4NO(g) 6H O(g)

(c) The second and third steps of the process: 2 NO g + O2 g 2 NO2 g

3 NO2 g + H2O l 2 HNO3 aq + NO g

56. (M) (a) 4 3 2 2 2400°C2 NH NO (s) 2 N (g) + O (g) + 4 H O(g)

The equation was balanced by inspection, by realizing first that each mole of NH4NO3 s would produce 1 mol N2 g and 2 mol H2O l , with 1 mol O [or

1/2 mol O2 g ] remaining.

(b) NaNO3 s NaNO2 s +1 / 2 O2 g or 2 NaNO3 s 2 NaNO2 s + O2 g

(c) Pb NO3 2

s PbO s + 2 NO2 g +1 / 2 O2 g 2 Pb NO3

2s 2 PbO s + 4 NO2 g + O2 g

57. (D) Begin with the chemical formulas of the species involved:

Na2CO3 aq + O2 g + NO g NaNO2 aq

Oxygen is reduced and nitrogen (in NO) is oxidized. We use the ion-electron method.

Two couples: 2 2NO g NO aq and O products

Balance oxygens: H O NO NO O H O2 2 2 2+ 2

Balance hydrogens: H O NO NO H H O H O2 2+ +

2 2+ + 2 4 + 2

Balance charge: H O NO NO H e e H O H O2 2+ +

2 2+ + 2 + 4 + 4 + 2

Combine: 4 + 4 + 4 + 2 + 4 +82+

2 2 2+ H O NO H O H O NO H

Simplify: 2 + 4 + 4 + 42 2 2+ H O NO O NO H

Add spectator ions: 4 + 2 4 + 2+3

2 +3

2 Na CO Na CO

2 3 2 2 2 2 22 Na CO + 2 H O + 4 NO + O 4 NaNO + 2 H O + 2 CO


1114

simplify: 2 Na2CO3 aq + 4 NO g + O2 g 4 NaNO2 aq + 2 CO2 g

58. (E) mass % HNO

mol HNO g HNO

1 mol HNO

1 L soln mL

1 L g

1 mL soln

HNO3

3

333

1563 01

1000 141100% 67%

.

.

59. (E) 75 109 gal 15 miles

1 gal

5 g

1 mile

1 kg

1000 g= 6 109 kg of nitrogen oxides released.

60. (E) 2 NH3(g) + 3 NO(g) 5/2 N2(g) + 3 H2O(g) (or, dividing by 2):

NH3(g) + 3/2NO(g) 5/4N2(g) + 3/2H2O(g)

Hrxn = Hf products - Hf reactants

Hrxn = [5/4 mol(0 kJ mol-1) + 3/2 mol(-241.8 kJ mol-1)] -[3/2 mol(90.25 kJ mol-1) + 1 mol(-46.11 kJ mol-1)] = -452 kJ

61. (M) (a) 2 2 42 NO g N O g

(b) i) HNO2(aq) + N2H5+(aq) HN3(aq) + 2H2O(l) + H+(aq)

ii) HN3(aq) + HNO2(aq) N2(g) + H2O(l) + N2O(g)

(c) H3PO4(aq) + 2 NH3(aq) (NH4)2HPO4(aq)

62. (M) (a) +3 3 23 Ag s + 4 H aq + 4 NO aq 3 AgNO aq + NO g + 2 H O(l)

(b) 3 2 2 2 2 22CH NNH l + 4 O g 2 CO g + 4 H O l + N g

(c) NaH2PO4(s) + 2Na2HPO4(s) 2H2O(l) + Na5P3O10(s) 63. (M)

(a) (CH3)2NNH2: Each molecule has a total of 26 valence electrons.

N N H

HC

CH

H

H

H

H H

(b) ClNO2: Each molecule has a total of 24 valence electrons.

N OCl

O


1115

(c) H3PO3 Each molecule has a total of 26 valence electrons.

P OO

O

HH

H 64. (M) Hyponitrous acid is a weak diprotic acid while nitramide has an amide group, 2NH

OH N N O H

H

H N O N O O

OH

H N N

1 2 3

Both Lewis structures for nitramide are plausible based on the information supplied. To choose between them would require further information about #2 and #3, such as whether the molecule contains a nitrogen-nitrogen bond. Experimental evidence indicates the structure adopted is the one with the nitrogen-nitrogen bond, i.e. structure #3.

65. (E) (a) 24HPO , hydrogen phosphate ion

(b) 2 2 7Ca P O , calcium pyrophosphate or calcium diphosphate

(c) 6 4 13H P O , tetrapolyphosphoric acid

66. (E) (a) 2HONH , hydroxylamine

(b) 4CaHPO , calcium hydrogen phosphate

(c) 3Li N , lithium nitride

67. (M) (i) 2 H+(aq) + N2O4(aq) + 2 e 2 HNO2(aq) E1 = +1.065 V (ii) 2 HNO2(aq) + 2 H+(aq) + 2 e 2 NO(aq) + 2 H2O(l) E2 = +0.996 V

(iii) N2O4(aq) + 4 H+(aq) + 4 e 2 NO(aq) + 2 H2O(l) E3 = ? V

Recall that G = nFE and that G values, not standard voltages are additive for reactions in which the number of electrons do not cancel out. So, 4FE3 = 2F(1.065 V) + 2F(0.996 V) E3 = 1.031 V (4 sig figs)

68. (M) (i) 2 NO3(aq) + 2 H2O(l) + 2 e N2O4(aq) + 4 OH(aq) E1 = 0.86 V

(ii) N2O4(aq) + 2 e 2 NO2(aq) E2 = +0.87 V

(iii) 2 NO3(aq) + 2 H2O(l) + 4 e 2 NO2

(aq) + 4 OH(aq) E3 = ? V

Recall that G = nFE and that G values, not standard voltages, are additive for reactions in which the numbers of electrons do not cancel out.

So, 4FE3 = 2F(0.86 V) + 2F(0.87 V) 4FE3 = 1.72F 1.74 F = 0.02F (1 sig fig) E3 = 0.005 V (1 sig fig)


1116

69. (M) (a)The nitrogen atom cannot bond to five fluorine atoms because, as a second-row element, it cannot accommodate more than four electron pairs.

(b) the NF3 molecule is trigonal pyramidal. The lone pair on the N atom causes the F—N—F bond angle to decrease from the ideal tetrahedral bond angle of 109° to 102.5°.

70. (M) On the basis of electronegativities, the N—H and N—F bonds are both quite polar. However, the polarities of the bonds are opposite. In NH3, the H atoms are slightly positive and in NF3, the N atom is slightly positive. The lone pair on N presumably enhances the bond dipoles in NH3 and diminishes the bond dipoles in NF3. Consequently, the NH3 is very polar and NF3 is less so.

Hydrogen 71. (M) The four reactions of interest are: (Note: H combustion = Hf products - Hf

reactants)

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) H combustion = -890.3 kJ (Molar mass CH4 = 16.0428 g mol-1)

C2H6(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(l) H combustion = -1559.7 kJ (Molar mass C2H6 = 30.070 g mol-1)

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) H combustion = -2219.9 kJ (Molar mass C3H8 = 44.097 g mol-1)

C4H10(g)+13/2 O2(g) 4 CO2(g) + 5 H2O(l) H combustion = -2877.4 kJ (Molar mass C4H10 = 58.123 g mol-1)

72. (M) Using the answers obtained in question 57, the per gram energy release is: CH4(g) -55.5 kJ; C2H6(g) -51.9 kJ; C3H8(g) -50.3 kJ; C4H10(g) -49.5 kJ

(a) C4H10(g) evolves the most energy on a per mole basis (-2877.4 kJ). (b) CH4(g) evolves the most energy on a per gram basis (-55.5 kJ). (c) CH4(g) is the most desirable alkane from the standpoint of emission, producing

the least quantity of CO2(g) per mole and/or per gram of fuel burned (as well per kJ of energy produced).

73. (E) (a) 2 Al s + 6 HCl aq 2 AlCl3 aq + 3 H2 g (b) 3 CO g + 7 H2 g C3H8 g + 3 H2O g (c) MnO (s) 2H (g) Mn(s) 2H O(g)2 2 2 74. (M) (a) 2H O(l) O (g) 2H (g)2

electricity2 2

(b) 2 HI aq + Zn s ZnI2 aq + H2 g Any moderately active metal is

suitable. (c) Mg s + H2SO4 aq MgSO4 aq + H2 g Any strong acid is suitable.


1117

(d) CO g + H2O g CO2 g + H2 g

75. (M) 2 2 22CaH s + 2 H O(l) Ca OH aq + 2 H g

2 22Ca s + 2 H O(l) Ca OH aq + H g

2 22 Na s + 2 H O(l) 2 NaOH aq + H g

(a) The reaction that produces the largest volume of H2 g per liter of water also

produces the largest amount of H2(g) per mole of water used. All three reactions use two moles of water and the reaction with CaH2 s produces the most H2(g).

(b) We can compare three reactions that produce the same amount of hydrogen; the one that requires the smallest mass of solid produces the greatest amount of H2 per gram of solid. The amount of hydrogen we will choose, to simplify matters, is 2 moles, which means that we compare 1 mol CaH2 (42.09 g) with 2 mol Ca (80.16 g) and with 4 mol Na (91.96 g). Clearly CaH2 produces the greatest amount of H2 per gram of solid.

76. (M) The balanced equation is C17H33COOH l + H2 g C17H35COOH s . One

mole of oleic acid requires one mole of H2 g .

V nRT

P

1.00 mol 0.08206L atm

mol K 298 K

752 mm Hg 1 atm

760 mmHg

0.95 L calculated

1.0 L produced 23.5 L H

2(g)

77. (M) Greatest mass percent hydrogen: The atmosphere is mostly N2(g) and O2(g) with only a trace of hydrogen containing gas

molecules. Seawater is H2O(l), natural gas is CH4(g) and ammonia is NH3(g). Each of these compounds has one non-hydrogen atom, and the non-hydrogen atoms have approximately the same mass (~14 + 2 g mol-1). Since CH4 has the highest hydrogen atom to non-hydrogen atom ratio, this molecule has the greatest mass percent hydrogen.

78. (M) The reaction is CaH2 s + 2 H2O l Ca OH 2 aq + 2 H2 g

First we calculate the amount of H2 g needed and then the mass of CaH2 s required.

mol Hmm Hg

atm760 mmHg

L

L atm mol K K mol H2 1 2

PV

RT

7221

235

0 08206 2732 19 79 29

1. ( . . ).

mass CaH mol H mol CaH

mol H

g CaH

mol CaH g CaH2 2

2

2

2

22= 9.29

1

2

42.09

1= 196


1118

79. (M) NH2− has 8 valence electrons (and is isoelectronic with H2O). The first four MO’s

are fully occupied. Because the 2px orbital on N is strongly bonding in the bent

configuration, as shown in Figure 22-27, the energy of the NH2− will be much lower

for the bent configuration. (The same argument applies to H2O.) On the basis of

molecular orbital theory, we expect that NH2− will be bent.

80. (M) NH2+ has 6 valence electrons and is isoelectronic with CH2. On this basis, we

anticipate that structure of NH2+ will be similar to that of CH2: a slightly bent

molecule with two unpaired electrons.

INTEGRATIVE AND ADVANCED EXERCISES

81. (M) In step (1), oxygen is converted to H2O by the reaction H2(g) + 1

2O2(g) →

H2O(l). Step (2) ensures that unreacted hydrogen from step (1) is also converted to H2O(l). The dehydrated zeolite has a very strong affinity for water molecules and thus, H2O(l) is removed from the gas mixture.

82. (M) The AsF6− is octahedral. The F—AsF bond angles are 90°. The most important

Lewis structure for the HCNKrF+ ion is shown below. If we assume that the lone pairs on Kr occupy equatorial positions (to minimize lone pair repulsions), then the HCNKrF+ ion is linear:

+1H—C N —Kr:—F:

83. (E) The electrolysis reaction is 2 2H O(l) H (g) O (g)2

electroysis2 2 . In this reaction, 2

moles of water are decomposed to produce one mole of oxygen and two moles of hydrogen. We use the data in the exercise to determine the volumes of oxygen and hydrogen produced.

n(H2O)

m

M

17.3g

18.00g/mol 0.96mol

V (O2)=

nRT

p 0.96molH

2O

1molO2

2molH2O

0.08206Latm/molK 298.15K

755mmHg 1atm

760mmHg

11.8L

V (H2)=

nRT

p 0.96molH

2O

2molO2

2molH2O

0.08206Latm/molK 298.15K

755mmHg 1atm

760mmHg

23.6L


1119

84. (M) NaNO2 decolorizes acidic solution of KMnO4 according to the following balanced chemical equation:

2KMnO4 + 3H2SO4 + 5NaNO2 5NaNO3 + K2SO4 + 2MnSO4 + 3H2O NaNO3 does not react with acidic solution of KMnO4. 85. (D) First we balance the equation, then determine the number of millimoles of NH3(g)

that are produced, and finally find the [NO3–] in the original solution.

The skeleton half-equations: 2-3 3 4NO (aq) NH (g) Zn(s) Zn(OH) (aq)

3 3 2

3 3 2

Balance O's and H's: NO (aq) NH (g) 3 H O(l)

NO (aq) 9 H (aq) NH (g) 3 H O(l)

3 2 3Balance charge and add OH (aq) NO (aq) 6 H O(l) 8 e NH (g) 9 OH (aq)

2-4Add OH (aq)'s and then electrons: Zn(s) 4 OH (aq) Zn(OH) (aq) 2 e

2-4

- -3 2 3

3

Oxidation: { Zn(s) 4 OH (aq) Zn(OH) (aq) 2 e } 4

Reduction: NO (aq) + 6 H O(l) + 8 e NH (g) + 9 OH (aq)

Net: NO (aq) + 4 Zn(s) +

- 2-2 4 36 H O(l) + 7 OH (aq) Zn(OH) (aq) + NH (g)

The titration reactions are the following.

2 3 4HCl(aq) NaOH(aq) NaCl(aq) H O(l) NH (aq) HCl(aq) NH Cl(aq)

0.1000 mmol NaOH 1 mmol HClmmol excess HCl 32.10 mL 3.210 mmol HCl

1 mL soln 1 mmol NaOH

0.1500 mmol HClmmol HCl at start 50.00 mL

1 mL s

33 3

3

33

7.500 mmol HCloln

1 mmol NHmmol NH produced (7.500 3.210) mmol HCl 4.290 mmol NH

1 mmol HCl

1 mmol NO4.290 mmol NH

3 1 mmol NH[NO ] 0.1716 M

25.00 mL soln

Notice that it was not necessary to balance the equation, since NO3– and NH3 are the only

nitrogen-containing species involved, and thus they must be in a one-to-one molar ratio. 86. (M) First we compute the root-mean-square speed of O at 1500 K.

m/s105.1kg/mol 0.0160

1500Kmol J 314.833rms

311

K

M

RTu

Kinetic Energy = 2

21KE mu 3 2 20

23

0.0160 kg/mol0.5 (1.5 10 m/s) 3.0 10 J/atom

6.022 10 atoms/mol


1120

87. (M) Reduction: { 2 NO3- (aq) + 10 H+ (aq) + 8 e- N

2O(g) + 5 H

2O(l) } 7

Oxidation:{C6H

11O(aq)+3H

2O(l) HOOC(CH

2)

4COOH(aq)+7H+ (aq)+7e-} 8

Net: 14 NO3- (aq) + 14 H+ (aq) + 8 C

6H

11O(aq) 7 N

2O(g) + 8 HOOC(CH

2)

4COOH(aq) + 11 H

2O(l)

88. (D) The Lewis structures for the two compounds are drawn below. The four bonded

groups in XeO4 and no lone pairs on the central atom give it a tetrahedral shape in which there are four oxygen atoms on the periphery of the molecule. These rounded tetrahedral should not stick well together. In fact, each oxygen is slightly negatively charged and they should repel each other. Thus, we expect weak intermolecular forces between XeO4 molecules. In XeO3, on the other hand, the three bonds and one lone pair produces a trigonal pyramidal molecule in which the central Xe (carrying a slight positive charge), is exposed to other molecules. There thus can be strong dipole-dipole forces leading to strong intermolecular forces and a relatively high boiling point vis-à-vis XeO4.

Xe OO

O

Xe

O

O O

O

4+ 3+

Note: You can also expand the octets to reduce formal charges. There are a number of resonance forms, however, Lewis structures(octets) are shown.

89. (E) The N center in the ammonium ion is the reducing agent, while Cl in the perchlorate anion

is the oxidizing agent. g)(O 2g)(ClO(g)H 4g)(Ns)(ClONH 2 222244 90. (E)Hf = Bonds broken in reactants – Bonds formed in products (a) ½ Cl2(g) + ½ F2(g) Cl—F(g) Hf = ½ (159 kJ) + ½ (243 kJ) 251 kJ = -50 kJ (b) ½ O2(g) + F2(g) F—O—F(g) Hf = ½ (498 kJ) + (159 kJ) 2(213 kJ) = -18 kJ (c) ½ O2(g) + Cl2(g) Cl—O—Cl(g) Hf = ½ (498 kJ) + (243 kJ) 2(205 kJ) = +82 kJ (d) ½ N2(g) + 3/2 F2(g) NF3(g) Hf = ½ (946 kJ) + 1.5 (159 kJ) 3(280 kJ) = -128.5 kJ 91. (E) The electrode reaction is: aq)(F 2e 2g)(F 2

G o 2 Gfo[F (aq)] G

fo[F

2(g)] 2( 278.8 kJ/mol) (0 kJ/mol)

G o 557.6 kJ/mol 2nFE o

V 890.2C/mol 485,962

J/mol 106.557 3o

E

This value compares favorably with the value of +2.866 V in Appendix D. 92. (M) Each simple cubic unit cell has one Po atom at each of its eight corners, but each corner

is shared among eight unit cells. Thus, there is a total of one Po atom per unit cell. The edge of that unit cell is 335 pm. From this information we obtain the density of polonium.


1121

3

3

12

23

g/cm 23.9

m 1

cm 100

pm 10

m 1pm 335

Po mol 1

Po g 209

atoms Po 106.022

Po mol 1atom Po 1

density

93. (M) First we write the molecular orbital diagram of each of the species. From each

molecular orbital diagram we determine the number of bonding electrons (b) and the number of antibonding electrons (*) and thus the bond order [(number of bonding electrons — number of antibonding electrons) 2 ]. Species with high bond order have strong bonds, which are short. Those with low bond order have weak bonds, which are long.

Number of e- b

1s *1s b

2s *2s b

2 p b2 p *

2 p *2 p Antibonding

e- Bond order

O2+ (2 × 8) – 1 =

15

5 2.5

O2 (2 × 8) = 16

6 2

O2– (2 × 8) + 1 =

17 7 1.5

O22– (2 × 8) + 2 =

18 8 1.0

We have only listed the number of antibonding electrons above because, for each species, the number of bonding electrons is the same, namely 10. Recall that the bond order is determined as follows: bond order = (number of bonding electrons — number of antibonding electrons) 2

(a) In order of increasing bond distance: O2

+ < O2 < O2– < O2

2–

(b) In order of increasing bond strength: O22– < O2

– < O2 < O2+

94. (M)(a)

Cl

CF Cl

Cl

Cl

CF

Cl

Cl

As shown in the Lewis structures above, the net result of this reaction is breaking a C—

Cl bond. From Table 10-3, the energy of this bond is 339 kJ/mol, and this must be the energy of the photons involved in the reaction.


1122

(b) E = h or = E/h on a molecular basis. Thus, we have the following.

nm 353m 1

nm10m1053.3

s/1050.8

m/s 1000.3

s1050.8mol/106.022sJ 10626.6

J/mol 10339

97

14

8

114

2334

3

c

This radiation is in the near ultraviolet region of the electromagnetic spectrum. 95. (M) (a) % P indicates the number of grams of P per 100 g of material, while % P4O10

indicates the number of grams of P4O10 per 100 g of material.

P g 436.0P mol 1

P g 974.30

OP mol 1

P mol 4

OP g 89.283

OP mol 1OP g 1.000P mass

104104

104104

Thus, multiplying the mass of P4O10 by 0.436 will give the mass (or mass percent) of P. %BPL indicates the number of grams of Ca3(PO4)2 per 100 g of material.

4 103 4 2 4 10

4 10 4 10

3 4 2 3 4 23 4 2

3 4 2

1 mol P O 4 mol P mass Ca (PO ) 1.000 g P O

283.89 g P O 1 mol P O

1 mol Ca (PO ) 310.18 g Ca (PO ) 2.185 g Ca (PO )

2 mol P 1 mol Ca (PO )

Thus, multiplying the mass of P4O10 (283.88) by 2.185 will give the mass (or mass %) of BPL. (b) A %BPL greater than 100% means that the material has a larger %P than does pure Ca3(PO4)2.

(c) 3 4 2 2

3 4 2 2 3 4 2 2

1 mol 3Ca (PO ) CaF6 mol P 30.974 g P%P 100% 18.43% P

1mol 3Ca (PO ) CaF 1008.6 g 3Ca (PO ) CaF 1 mol P

BPL %4.92OP %3.42185.2O%P2.185BPL%

OP %3.42436.0

43.18

0.436

P%O%P

104104

104104

96. (M) The chemical equation is 2Cl (g) 2 Cl(g) . From Appendix D, o [Cl(g)] 121.7 kJ/molfH , 243.4 kJ/2 mol Cl or 243.4 kJ/mol Cl2.

.K J 107.3223.1165.2)(2g)](Cl[Cl(g)][2 12

ooorxn

SSS

We assume the values of oH and oS are unchanged from 298 K to 1000 K, and we calculate the value of oG at 1000 K.

Go H o TS o 243.4 kJ (1000 K 0.1073 kJ K1) 136.1 kJ Go RT ln Kp

ln KpGo

RT

136.1103 J/mol

8.314 J mol1 K1 1000 K 16.37 K

p e16.37 7.8 108

Since the value of Kp is so small, we assume the initial Cl2(g) pressure is 1.00 atm.


1123

2Reaction: Cl (g) 2Cl(g)

Initial: 1.00 atm 0 atm

Changes: atm 2 atm

Equil: (1.00 ) 2 atm

x x

x x

2

2 2 28Cl

Cl

84

(2 ) 47.8 10

1.00 1.00

7.8 101.4 10

4

p

P x xK

P x

x

Our assumption, that x << 1.00 atm, obviously is valid. x is the degree of dissociation. The % dissociation now can be calculated. % dissociation = x × 100% = 1.4 × 10–4 × 100% = 0.014 %

97. (E) The peroxo prefix of peroxonitrous acid hints at the presence of a —O—O—

linkage in the molecule. nitric acid H O N O

O

peroxonitrous acid H O O N O

98. (M) The Lewis structures of these two compounds are similar. If we designate CH3

as Me and SiH3 as Sl, the two Lewis structures are Me N Me

Me

Sl N Sl

Sl

The pyramidal structure of N(CH3)3 indicates a tetrahedral electron pair geometry, with

sp3 hybridization for N in N(CH3)3 and the lone pair of electrons residing in an sp3

hybrid orbital. The N—C bond involves the overlap of )(C—)N( 33 spsp . The C—H

bonds are )H(1—)C( 3 ssp . The planar arrangement of N(SiH3)3 indicates that the nitrogen atom is sp2 hybridized; the lone pair is in a 2p atomic orbital on the central N atom, perpendicular to the SiN plane in the center of the molecule. If we assume that the Si atom is sp3 hybridized (to account for the four bonds each Si atom forms), the N—Si bond involves the overlap between )Si(—)N( 32 spsp . Thus, the Si—H bonds are

)H(1—)Si( 3 ssp . 99. (M) pH = 3.5 means [H+] = 10–3.5 = 3 × 10–4 M. This is quite a dilute acidic solution, and we

expect H2SO4 to be completely ionized under these circumstances.

43 2 4 2 4

2 4 2 42 4

1 mol H SO 98.1 g H SO3 10 mol H mass H SO 1.00 10 L 15 g H SO

1 L 1 mol H SO2 mol H

22

2

2

2

2

2

2

26

23

33

2

Cl g 37needed Cl g 100

usedCl g 115

Cl mol 1

Cl g 9.70

Br mol 1

Cl mol 1

Br g 8.159

Br mol 1

seawater g 10

Br g 70

cm 1

g 03.1

L 1

cm 1000L 1000.1Cl mass


1124

100.(M)

E Eo 0.0592

nlog

10Q ;

assuming E 0 for a process that is no longer spontaneous:

log10

Q nEo

0.0592

(16)(0.065)

0.0592 17.6 P

SO 2 110-6

Q 1017.6 3.9 1017 (P

SO 2)8

18(H )16

(110-6 )8

18(H )16

Solving for [H ] yields a value of 8 105 M, which corresponds to a pH of 4.1

Thus, the solution is still acidic.

101. (M) In the process of forming XeF2 and XeCl2, either a Cl—Cl bond or a F—F

bond is broken. We note that the F—F bond is much weaker than the Cl—Cl bond and thus much less energy is required to break it. Since greater stability indicates that the resulting product is of lower energy, the greater stability for XeF2 compared to XeCl2 can be partially explained by the need to expend less energy to break the bonds of the reactants. (Another reason, of course, is that the shorter Xe—F bond is stronger, and hence more stable, than the somewhat longer Xe—Cl bond.)

102. (M) e 2aq)(H 2HF(aq) 4g)(XeOOH 3g)(XeF :Oxidations 324

32 2 2

4

34 2 3 2

{2 H O(l) O (g) 4 H (aq) 4 e }

Reduction: {XeF (g) 4 e Xe(g) 4 F (aq)} 2

Net: 3 XeF (g) 6 H O(l) 2 XeO (g) 12 HF(aq)

2 O (g) 2 Xe(g)

The fact that O2 is also produced indicates that there are two oxidation half-reactions. The production ratio of Xe and XeO3 indicates the amount by which the other two half-reactions must be multiplied before they are added. Then the half-reaction for the production of O2 is multiplied by

23 to balance charge.

103. (M) S(s) + 2 H+(aq) + 2 e- H2S(g) E = 0.174 V S(s) + 2 H+(aq) + 2 e- H2S(aq) E = 0.144 V

The difference is that one results in an aqueous solution being formed, while the other gives a gaseous product. They can both be correct because the product has a different phase (the products have different Gf values, hence one would expect different values).

104. (M) The initial concentration of Cl2(aq)= 0.090M70.9g/mol

6.4g


1125

-4cK 4.4 10

2 2 Cl (aq) H O(l) HOCl(aq) H (aq) Cl (aq)

initial 0.090 M 0 M 0 M 0 M

equil. (0.090- ) M x

2

3-4

2

M M M

{where is the molar solubility of Cl in water}

[HOCl(aq)][ H (aq)] [Cl (aq)]4.4 10

[Cl (aq)] 0.090-

By using successive approximations, we find

x x x

x

xK

x

x

2

0.030 M, so:

[HOCl(aq)] [H (aq)] [Cl (aq)] 0.030 M and [Cl (aq)] 0.090 0.030 0.060 M

105. (M)

- + + o2 2 3

+ + - + o3 2 5 2

+ + - + o42 5

2e N (g) 4 H (aq) + 2H O(l) 2 NH OH (aq) E = -1.87 V

2 NH OH (aq) + H (aq) + 2e N H (aq) + 2 H O(l) E = 1.42 V

N H (aq) + 3 H (aq) + 2e 2 NH (aq) E = 1.2

+ - + o

42

75 V

N (g) + 8 H (aq) + 6e 2 NH (aq) E = 0.275 V (see below)

o o o o o o o o1 2 3 tot tot 1 2 3

o o oo 1 1 2 2 3 3

tottot

otot

G total = G + G + G = -n FE = -nFE -nFE -nFE

n E n E n EE =

n

E = [2(-1.87) +2(1.42) + 2(1.275)]/6 = 0.275 V

+ - o3 2

+ - + o42

moving on...

2 HN (aq) 3 N ( ) 2H (aq) 2 e E 3.09 V (oxidation)

3 N (g) 24 H (aq) 18e 6 NH (aq) E 0.275 V (top equation multiplied by 3)

g

+ - +

432 HN (aq) 22 H (aq) 16 e 6 NH (aq) [ ]sum

+ - + o43

o o o o o ototal 1 2 tot 1 2

ototal

[divide by 2]

HN (aq) 11 H (aq) 8 e 3 NH (aq) E 0.70 V

since

G = G + G = -nFE = -nFE -nFE

E = [2(3.09) +18(0.275) ]/16 = 0.70 V

106. (E) Ignoring distortions arising from different bond lengths and different bond orders,

we can say the following:

(a) in FClO2, the Cl atom is bonded to three atoms and a lone pair of electrons. The

electron group geometry is tetrahedral. The molecular structure is ammonia-like

(trigonal pyramidal).


1126

(b) In FClO3, the Cl atom is bonded to four atoms and has no lone pairs. The electron

group geometry is tetrahedral. The molecular structure is methane-like (tetrahedral).

(c) In F3ClO, the Cl atom is bonded to four atoms and has one lone pair. The electron

group geometry is trigonal bipyramidal with the lone pair in an equatorial position.

The molecular structure is SF4-like (seesaw).

107. (E) Both molecules are V shaped, with a lone pair of electrons on the central atom.

For O3, the structure is a hybrid of two equivalent structures. In each of these

structures, the central atom has a formal charge of +1. The oxygen–oxygen bond

order is between 1 and 2. Although many resonance structures can be drawn for SO2,

in the most important structure, the formal charge on the S atom is zero and the

sulfur–oxygen bonds are double bonds.

108. (M)

Sb

F

F

F

F

F

F

HN N N N N

N N N N N

N N N N N

109. (M) ofH = [

3

2(498) − 3 (36)] = 639 kJ mol−1

The formation reaction is very endothermic and thus energetically unfavorable. This

result supports the observation that Xe(g) does not react directly with O2(g) to form


1127

XeO3(g). Because the reaction converts 2.5 moles of gas into 1 mole of gas, we

expect ofS < 0; thus, the reaction is also entropically unfavorable.

FEATURE PROBLEMS 110. (D) The goal here is to demonstrate that the three reactions result in the decomposition

of water as the net reaction: Net: 2 2 +2 2 2 H O H O First balance each equation.

(1) 3 + 4 + +2 2 3 4 2 FeCl H O Fe O HCl H Balance by inspection. Notice that there are 3 Fe and 4 O on the right-hand side. Then balance Cl.

3 + 4 + 6 +2 2 3 4 2 FeCl H O Fe O HCl H

(2) Fe O HCl Cl FeCl H O O3 4 2 3 2 2+ + + + Try the half-equation method.

Cl e Cl2 + 2 2 But realize that Fe O Fe O FeO3 4 2 3= .

2 2 + + 63+2 FeO Fe O e Now combine the two half-equations.

2 + 3 2 + + 623+

2 FeO Cl Fe O Cl

Add in 2 +12 4 + 62 3+ 3+

2 Fe O H Fe H O

2 + 3 +12 6 + + 6 + 63 4 2+ 3+

2 2 Fe O Cl H Fe O Cl H O

And 12 Cl spectators: 2 + 3 +12 6 + + 63 4 2 3 2 2 Fe O Cl HCl FeCl O H O

(3) FeCl FeCl Cl3 2 2+ by inspection 2 2 +3 2 2 FeCl FeCl Cl

One strategy is to consider each of the three equations and the net equation. Only equation (1) produces hydrogen. Thus, we must run it twice. Only equation (2) produces oxygen. Since only one mole of O2(g) is needed, we only have to run it once. Equation (3) can balance out the Cl2 required by equation (2), but we have to run it three times to cancel all the Cl2(g).

2 2 3 4 22 1 6 FeCl (s) + 8 H O(l) 2 Fe O (s) +12 HCl(l) + 2 H (g)

3 4 2 3 2 21 2 2 Fe O (s) + 3 Cl (g) +12 HCl(g) 6 FeCl (s) + O (g) + 6 H O(l)

3 2 23 3 6 FeCl (s) 6 FeCl (s) + 3 Cl (g)

Net: 2 2 22 H O(l) 2 H (g) + O (g)

111. (D) We begin by calculating the standard voltages for the two steps in the decomposition

mechanism. Step 1 involves the reduction of Fe3+ and the oxidation of H2O2. The two half-reactions that constitute this step are:

(i) Fe3+(aq) + e Fe2+(aq) E1,red = 0.771 V

(ii) H2O2(aq) O2(g) + 2 H+(aq) + 2 e E2,ox = 0.695 V


1128

The balanced reaction is obtained by combining reaction (i), multiplied by two, with reaction (ii).

2(i) 2(Fe3+(aq) + e Fe2+(aq))

(ii) H2O2(aq) O2(g) + 2 H+(aq) + 2 e

(iii) 2 Fe3+(aq) + H2O2(aq) 2 Fe2+(aq) +O2(g) + 2 H+(aq),

for which Ecell = Ei(red) + Eii(ox) = 0.771 V + (0.695 V) = 0.076 V

Since the overall cell potential is positive, this step is spontaneous. The next step involves oxidation of Fe2+(aq) back to Fe3+(aq), (i.e. the reverse of reaction (i) and the reduction of H2O2(aq) to H2O(l) in acidic solution, for which the half-reaction is

(iv) H2O2(aq) + 2 H+(aq) + 2 e 2 H2O(l) Eiv(red) = 1.763 V

Combining (iv) with two times the reverse of (i) gives the overall reaction for the second step:

(i) {Fe2+(aq) Fe3+(aq) + e} 2

iv) H2O2(aq) + 2 H+(aq) + 2 e 2 H2O(l)

(v) 2 Fe2+(aq) + H2O2(aq) + 2 H+(aq) 2 Fe3+(aq) + 2 H2O(l)

Thus the overall cell potential for the second step in the mechanism, via equation (v) is

Ecell = Ei(red) + Eiv(red) = 0.771 V + (1.763 V) = 0.992 V.

Since the overall standard cell potential is positive, like step 1, this reaction is spontaneous.

The overall reaction arising from the combination of these two steps is:

Step 1 2 Fe3+(aq) + H2O2(aq) 2 Fe2+ + (aq) + O2(g) + 2 H+(aq)

Step 2 2 Fe2+(aq) + H2O2(aq) + 2 H+(aq) 2 Fe3+(aq) + 2 H2O(l)

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

The overall potential, Eoverall = Estep1+ Estep2 = 0.076 V + 0.992 V = 1.068 V Therefore, the reaction is spontaneous at standard conditions.

To determine the minimum and maximum E values necessary for the catalyst, we need to consider each step separately.

In step 1, if E(1) is less than 0.695 V, the overall voltage for the first step will be negative and hence non-spontaneous. In step 2, if the oxidation half-reaction has a potential that is more negative than 1.763 V, the overall potential for this step will be negative, and hence non-spontaneous. Consequently, E(1) must fall between 0.695 V and 1.763 V in order for both steps to be spontaneous. On this basis we find that

(a) Cu2+(aq) + 2 e Cu(s) E1/2red = 0.337 V cannot catalyze the reaction.


1129

(b) Br2(l) + 2 e 2 Br(aq) E1/2red = 1.065 V may catalyze the reaction.

(c) Al3+(aq) + 3 e Al(s) E1/2red = 1.676 V cannot catalyze the reaction.

(d) Au3+(aq) + 2 e Au+(s) E1/2red = 1.36 V may catalyze the reaction.

In the reaction of hydrogen peroxide with iodic acid in acidic solution, the relevant half-reactions are:

2 IO3(aq) + 12 H+(aq) + 10 e I2(s) + 6 H2O(l) E1/2red = 1.20 V

H2O2(aq) O2(g) + 2 H+(aq) + 2e E1/2ox= 0.695 V

Thus, both steps in the decomposition of H2O2, as described above, are spontaneous if IO3

is used as the catalyst. As the iodate gets reduced to I2, the I2 forms a highly colored complex with the starch, resulting in the appearance of a deep blue color solution. Some of the H2O2 will be simultaneously oxidized to O2(g) by the iodic acid When sufficient I2(s) accumulates, the reduction of H2O2 by I2 begins to take place and the deep blue color fades as I2 is consumed in the reaction. Additional iodate is formed concurrently, and this goes on to oxidize the H2O2(aq), thereby causing the cycle to repeat itself. Each cycle results in some H2O2 being depleted. Thus, the oscillations of color can continue until the H2O2 has been largely consumed.

112. (D) ClO3- ClO2 HClO2

(?) (?)

1.1181 V

In order to add ClO2 to the Latimer diagram drawn above, we must calculate the voltages denoted by (?) and (??) . The equation associated with the reduction potential (?) is

(i) 2 H+(aq) + ClO3(aq) + 1 e ClO2(g) + H2O(l)

The standard voltage for this half-reaction is given in Appendix D:

To finish up this problem, we just need to calculate the standard voltage (??) for the half-reaction (ii):

(ii) H+(aq) + ClO2(aq) + 1 e HClO2(g) Eo = (??)

To obtain the voltage for reaction (ii), we need to subtract reaction (i) from reaction (iii) below, which has been taken from Figure 22-2:

(iii) 3 H+(aq) + ClO3(aq) + 2 e HClO2(g) + H2O(l) Eo = 1.181 V

Thus,

(iii) 3 H+(aq) + ClO3(aq) + 2 e HClO2(g) + H2O(l) Eo(iii) = 1.181 V

1 (i) 2 H+(aq) + ClO3(aq) + 1 e ClO2(g) + H2O(l) Eo(i) = 1.175 V

Net (ii) H+(aq) + ClO2(g) + 1 e HClO2(g) Eo(ii) = (??)

Since reactions (i) and (iii) are both reduction half reactions, we cannot simply subtract the potential for (i) from the potential for (iii). Instead, we are forced to obtain the voltage for


1130

(ii) via the free energy changes for the three half reactions. Thus,

G(ii) = G(iii) G(i) = 1FEo(ii) = 2F(1.181 V) + 1F(1.175 V)

Dividing both sides by –F gives

Eo(ii) = 2(1.181 V) – 1.175 V So, Eo(??) = 1.187 V. 113. (D) (a) As before, we can organize a solution around the balanced chemical equation.

Equation: 2I aq 2 4I CCl

Initial: 31.33 10 M 0 M Initial: 0.0133 mmol 0 mmol Changes: mmolx + mmolx

Equil: 0.0133 mmol mmolx x

2 4c

2

[I (CCl )] 10.0 mL85.5

0.0133[I (aq)]10.0 mL

x

Kx

1.137 85.5x x 86.5 1.137x 72 4

1.13= = 0.01315 mmol I in CCl

86.5x

2I aq = 0.0133 0.01315 mmol = 0.00015 mmol

222 2 2

2

253.8 mg Imass I = 0.00015 mmol = 3.8 10 mg I = 0.038 mg I

1 mmol I remaining

(b) We have the same set-up, except that the initial amount 2I aq = 0.00015mmol .

2 4c

2

[I (CCl )]= 85.5

[I (aq)] 0.00015

xK

x

0.0128 85.5x x 86.5 0.0128x

2 4

0.0128= = 0.0001480 mmol = I CCl ,

86.5x

62 2I aq = 0.00015 0.0001480 = 2.0 10 mmol I

6 422 2 2 2

2

253.8 mg Imass I = 2.0 10 mmol in H O = 5.1 10 mg I = 0.00051 mg I

1 mol I

(c) If twice the volume of 4CCl were used for the initial extraction, the equilibrium concentrations would have been different from those in part (a).

Equation: 2I aq 2 4I CCl

Initial: 31.33 10 M 0 M Initial: 0.0133 mmol 0 mmol Changes: mmolx + mmolx

Equil: 0.0133 mmol mmolx x

2 4c

2

[I (CCl )] 20.0 mL= 85.50.0133[I (aq)]10.0 mL

x

Kx

2 1.137 85.5x x 86.0 1.137x 22 4

1.137= = 1.32 10 mmo I CCl

86.0x


1131

3

2 22 2

2

1.33 10 mmol I 253.8 mg Itotal mass I = 10.0 mL soln = 3.376 mg I

1 mL 1 mol I

2 22 4 2

2

253.8 mg Imass I in CCl =1.322 10 mmol = 3.356 mg I

1 mol I

2mass I remaining in water = 3.376 mg 3.356 mg = 0.020 mg Thus, two smaller volume extractions are much more efficient than one large

volume extraction. 114. (D) (a) The pyroanions, a series of structurally analogous anions with the general

formula X2O7n, are known for Si, P and S. The Lewis structures for these three

anions are drawn below: (Note: Every member of the series has 56 valence e)

X O

O

O

O

X

O

O

O

X = Si : Si2O76-

X = P : P2O74-

X = S : S2O72-

2+

Based upon a VSEPR approach, we would predict tetrahedral geometry for each X atom (i.e. Si, P+, S2+) and a bent geometry for each bridging oxygen atom Therefore, a maximum of five atoms in each pyroanion can lie in a plane:

O

XO O

XO

O

OO

Atoms labeled 1-5 are all in the same plane. (b) The related “mononuclear” acids, which contain just one third-row element are

H4SiO4, H3PO3 and H2SO4. A series of pyroacids can be produced by strongly heating the “mononuclear” acids in the absence of air. In each case, the reaction proceeds via loss of water:

2 H4SiO4 H6Si2O7(l) + H2O(l) 2 H3PO3 H4P2O7(l) + H2O(l) 2 H2SO4 H2S2O7(l) + H2O(l) (c) The highest oxidation state that Cl can achieve is VII (+7); consequently, the

chlorine containing compound that is analogous to the mononuclear acids mentioned earlier is HClO4. The strong heating of perchloric acid in the absence of air, should, in principle, afford Cl2O7, which is the neutral chlorine analogue of the pyroanions. Thus, Cl2O7 is the anhydride of perchloric acid.

2 HClO4 Cl2O7(l) + H2O(l)


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115. (D) (a) The bonding in the XeF2 molecule can be explained quite simply in terms of

a 3center, 4 electron bond that spans all three atoms in the molecule. The bonding in this molecular orbital description involves the filled 5pz orbital of Xe and the half-filled 2pz orbitals of the two F-atoms. The linear combination of these three atomic orbitals affords one bonding, one non-bonding and one anti-bonding orbital, as depicted below:

F(2pz)Xe(5pz)F(2pz)

Antibonding MO

Bonding MO

Non-bonding MO

+

+

+ +

+ +

+

+ +

++ +

+

(b) If one assumes that the order of energies for molecular orbitals is: bonding MO < non-bonding MO < antibonding MO, the following molecular orbital representation of the bonding can be sketched.

Thus, a single bonding pair of electrons is responsible for holding all three atoms together. The non-bonding pair of electrons is localized primarily on the two Fatoms. This suggests that the bond possesses substantial ionic character. Bond order is defined as one-half the difference between the number of bonding electrons and the number of antibonding electrons. In the case of XeF2, the overall bond order is

therefore: 1.0e2

e0e2

.

(i.e., each XeF bond has a bond order of 0.5)

Antibonding

Bonding

Non-bonding

5pz 2pz 2pz

XeAtomicOrbital

2 xFAtomicOrbitals

XeF2

MolecularOrbitals

(c) By invoking a molecular orbital description based upon three-center bonding for the

three FXeF units in XeF6, one obtains an octahedral structure in which there are six identical XeF bonds. The extra nonbonding pair of electrons is delocalized over the entire structure in this scheme. In spite of its manifest appeal, this explanation of the bonding in XeF6 is incorrect, or at best a stretch, because the actual structure for this molecule is a distorted octahedron. A more accurate description of the stereochemistry adopted by XeF6 is provided by VSEPR theory. In this approach, the


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shape of the molecule is determined by repulsions between bonding and non-bonding electrons in the valence shell of the central Xe atom.

Accordingly, in XeF6, six bonding pairs and one lone pair of electrons surround the Xe. Having the repulsions of 7 pairs of electrons to cope with, the XeF6 molecule adopts a distorted structure that approaches either a monocapped octahedron or a pentagonal bipyramidal arrangement of electron pairs, depending on how one chooses to view the structure. In either case, these two shapes are much closer to the true shape for XeF6 than that predicted by the molecular orbital treatment involving three, three-center bonds. By contrast, the molecular orbital description involving three-center bonds gives far better results when applied to XeF4. In this instance, with two three-center FXeF bonds in the structure, molecular orbital theory predicts that XeF4 should adopt a square planar structure.

The result here is quite satisfactory because XeF4 does in fact exhibit square planar geometry. It is worth noting, however, that a square planar shape for XeF4 is also predicted by VSEPR theory. Despite the fact that the molecular orbital method has made some inroads as of late, VSEPR is still the best approach available for rationalizing the molecular geometries of noble gas compounds.

116. (D) (a) In the phase diagram sketched in the problem, extend the liquid-vapor

equilibrium curve (the vapor pressure curve) to lower temperatures (supercooled liquid region). Extend the S-vapor equilibrium curve (sublimation curve) to the temperature at which it intersects the extended vapor pressure curve. This should come at 113 °C. Draw a line from this point of intersection to the “peak” of the S phase region. Erase the three lines that bound the S region in the original sketch. The remaining three lines would represent the equilibria, S-V, S-L, and L-V, producing the phase diagram if S (rhombic) were the only solid form of sulfur.

(b) If rhombic sulfur is heated rapidly, the transition to monoclinic sulfur at 95.3 °C

might not occur. (Solid-state transitions are often very slow.) In this case, rhombic sulfur would melt at 113 °C, as described in the modified phase diagram in part (a). If the molten rhombic sulfur is further heated and then cooled, the liquid very likely will freeze at the equilibrium temperature of 119 °C, producing monoclinic, not rhombic, sulfur.


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SELF-ASSESSMENT EXERCISES 117. (E) (a) The halogens can combine with each other to form interhalogens and

polyhalide ions. (b) Polyphosphates are anionic phosphate polymers linked between hydroxyl groups and hydrogen atoms. (c) The halogens react with each other to form interhalogen compounds. The general formula of most interhalogen compounds is XYn, where n = 1, 3, 5 or 7 (X is the less electronegative of the two halogens). (d) Disproportionation is the chemical reaction in which the same element is spontaneously oxidized and reduced.

118. (E) (a) The Frasch process is a method to extract sulfur from underground deposits.

(b) The water-gas shift reaction is a chemical reaction in which carbon monoxide reacts with water vapor to form carbon dioxide and hydrogen: CO + H2O → CO2 + H2. It is a very important industrial reaction. (c) Eutrophication is an increase in the concentration of chemical nutrients in an ecosystem to an extent that increases the primary productivity of the ecosystem. (d) Electrode potential diagrams are used to simplify actual electrochemical cells.

119. (E) (a) An acid anhydride is an organic compound that has two acyl groups bound to

the same oxygen atom. Acid salt is an anionic compound. (b) Azide functionality is -N3. A nitride is a compound of nitrogen with a less electronegative element where nitrogen has an oxidation state of −3. (c) Elemental phosphorus exists in two major forms - white phosphorus and red phosphorus, they are structurally different. (d) Both terms are similar and represent a class of compounds in which hydrogen is bound to a very electropositive element (such as alkali earth metals).

120. (E) (b) 121. (E) (d) 122. (E) (c) 123. (E) (a) 124. (E) (d) 125. (E) (b) 126. (E) (a) and (b) 127. (M) (a) Cl2 (g)+2NaOH(aq) NaCl(aq)+NaOCl(aq)+H2O(l) (b) 2NaI(s)+2H2SO4 (concd aq) Na2SO4 (a)+2H2O(g)+SO2 (g)+I2 (g) (c) Cl2 (g)+2KI3(aq) 2KCl(aq)+3I2 (s) (d) 3NaBr(s)+H3PO4 (concd aq) Na3PO4 +3HBr(g)


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(e) 5HSO3- (aq)+2MnO4

- (aq)+H+(aq) 5SO42- (aq)+2Mn2+(aq)+3H2O(l)

128. (M) (a) 2KClO3(s) 2KCl(s)+3O2 (g) a catalyst such as MnO2 is required,

electrolysis of H2O is an alternate method. (b) 3Cu(s)+8H+(aq)+2NO3

- (aq) 3Cu2+(aq)+4H2O(l)+2NO(g) , difficult to control reaction so that NO(g) is the only reduction product.

(c) Zn(s)+2H+(aq) Zn2+(aq)+H2 (g) , a number of other metals can be used. (d) NH4Cl(aq)+NaOH(aq) NaCl(aq)+H2O(l)+NH3(g) , other ammonium salts

and other bases work as well. (e) CaCO3(s)+2HCl(aq) CaCl2 (aq)+H2O(l)+CO2 (g) , other carbonates and

acids can be used. 129. (M) (a) LiH(s)+H2O(l) LiOH(aq)+H2 (g) (b) C(s)+H2O(g) CO(g)+H2 (g) (c) 3NO2 (g)+H2O(l) 2HNO3(aq)+NO(g)

130. (M) 3Cl2 (g)+I-(aq)+3H2O(l) 6Cl-(aq)+IO3

- (aq)+6H+(aq)

Cl2 (g)+2Br-(aq) 2Cl- (aq)+Br2 (aq)

131. (M) One conversion pathway is ton lb kg g H2SO4 g Smg SL

seawaterm3 kg3 seawater, leading to about 5.0 kg3 of seawater. 132. (M) (a) AgAt, (b) sodium perxenate, (c) MgPo, (d) tellurous acid, (e) K2SeSO3, (f)

potassium perastatate 133. (D) Convert the Eo values to Go values for the reduction of H2SeO3 to Se and from

Se to H2Se. Add those two Go values to obtain Go for the reduction of H2SeO3 to H2Se. Convert this Go value to Eo , which proves to be 0.38 V.

134. (M) (a) SO3, (b) SO2, (c) Cl2O7, (d) I2O5 135. (M) (a) false, (b) true, (c) true.

chapter 22 chemistry of the main group...

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