Exam 3 Worksheet Answers Exam 3 Worksheet Answers – Chemistry 102 Chapter 6 – Energy Relationships in Chemical Reactions 1. Heat a. –125 J b. –110 J c. –93.1 J d. –1.02 kJ e. –95.2 J f. –174 J g. –31.6 J 2. Specific heat a. 0.509 b. 0.449 c. 0.38 d. 4.18 e. 0.388 f. 0.710 g. 0.129 3. Heat and final temperature a. 50.9 ºC b. 45.6 ºC c. 37.6 ºC d. 85.2 ºC e. 38.7 ºC f. 61.9 ºC g. –64.4 ºC 4. Heat and final temperature a. 49.1 ºC b. 54.4 ºC c. 62.4 ºC d. 14.8 ºC e. 61.3 ºC f. 38.1 ºC g. 164.4 ºC 5. Heat transfer from a hotter body to water a. 49.4 ºC b. 46.6 ºC 1
Exam 3 Worksheet Answers c.
d.
e.
f.
43.4 ºC 43.8 ºC 58.7 ºC 31.3 ºC 6. 32.0ºC 7. Enthalpy of solution a. ΔHsoln = –55.29 kJ∙mol–1; b. ΔHsoln = +20.07 kJ∙mol–1; c. ΔHsoln = +4.14 kJ∙mol–1; d. ΔHsoln = –2.33 kJ∙mol–1; Tf = 38.2 ºC Tf = 19.8 ºC Tf = 21.4 ºC Tf = 22.8 ºC 8. Enthalpy of combustion a. ΔHcomb(exp) = –2800 kJ∙mol–1; ΔHcomb(calc) = –2537.3 kJ∙mol–1; error = 10.4% b. ΔHcomb(exp) = –5000 kJ∙mol–1; ΔHcomb(calc) = –5160.1 kJ∙mol–1; error = 3.10% c. ΔHcomb(exp) = –380 kJ∙mol–1; ΔHcomb(calc) = –393.5 kJ∙mol–1; error = 3.43% 9. Enthalpy of combustion a. 278 L b. 349 L c. 156 L d. 178 L e. 168 L f. 109 L 10. Enthalpy of reaction a. –57.2 kJ b. 178 kJ c. –126 kJ d. –336.6 kJ e. –1133.6 kJ exothermic endothermic exothermic exothermic ∆Hf = –134.5 kJ∙mol–1 (CHCl3(l)) exothermic 11. Enthalpy of reaction – Hess’s Law a. – 135 kJ exothermic b. – 108 kJ exothermic c. – 127.8 kJ exothermic d. – 78.6 kJ exothermic e. – 53.0 kJ exothermic 12. Enthalpy of formation a. – 463 kJ⋅mol−1 exothermic b. 14.4 kJ⋅mol−1 endothermic −1
c. – 155.1 kJ⋅mol exothermic 2
Exam 3 Worksheet Answers d. – 188 kJ⋅mol−1 13. Enthalpy of formation a. – 277.6 kJ⋅mol−1 b. – 527.5 kJ⋅mol−1 c. – 363.3 kJ⋅mol−1 d. – 947.7 kJ⋅mol−1 exothermic exothermic exothermic exothermic exothermic 14. Formation reactions a. 6C ( s ) + 3H 2 ( g ) + 1 O 2 ( g ) → C6 H 5OH ( l ) 2
b. C ( s ) + O2 ( g ) → CO2 ( g ) c. C ( s ) + 1 O 2 ( g ) → CO ( g ) 2
d. 8C ( s ) + 9H 2 ( g ) → C8 H18 ( l ) e. N 2 ( g ) + 1 O 2 ( g ) → N 2 O ( g ) 2
f. N 2 ( g ) + 2H 2 ( g ) → N 2 H 4 ( g ) Chapter 7 – Electronic Structure of Atoms 1. Balmer series ΔE = –3.03×10–19 J ν = 4.57×1014 Hz a. (n1 = 3) b. (n1 = 4) ΔE = –4.09×10–19 J ν = 6.17×1014 Hz c. (n1 = 5) ΔE = –4.58×10–19 J ν = 6.91×1014 Hz d. (n1 = 6) ΔE = –4.84×10–19 J ν = 7.31×1014 Hz 2. Lyman series ΔE = –1.64×10–18 J ν = 2.47×1015 Hz a. (n1 = 2) ΔE = –1.94×10–18 J ν = 2.92×1015 Hz b. (n1 = 3) c. (n1 = 4) ΔE = –2.04×10–18 J ν = 3.08×1015 Hz ΔE = –2.09×10–18 J ν = 3.16×1015 Hz d. (n1 = 5) 3. Quantum numbers a. (4, 2, ± 2, ± ½); (4, 2, ± 1, ± ½); (4, 2, 0, ± ½) b. (5, 1, ± 1, ± ½); (5, 1, 0, ± ½) c. (3, 0, 0, ± ½); (3, 1, ± 1, ± ½); (3, 1, 0, ± ½) (3, 2, ± 2, ± ½); (3, 2, ± 1, ± ½); (3, 2, 0, ± ½) d. (4, 0, 0, ± ½); (4, 1, ± 1, ± ½); (4, 1, 0, ± ½) (4, 2, ± 2, ± ½); (4, 2, ± 1, ± ½); (4, 2, 0, ± ½) (4, 3, ± 3, ± ½); (4, 3, ± 2, ± ½); (4, 3, ± 1, ± ½); (4, 3, 0, ± ½) e. (2, 0, 0, ± ½) f. (7, 3, ± 3, ± ½); (7, 3, ± 2, ± ½); (7, 3, ± 1, ± ½); (7, 3, 0, ± ½) 3
λ = 656 nm λ = 486 nm λ = 434 nm λ = 410 nm λ = 121 nm λ = 103 nm λ = 97.2 nm λ = 94.9 nm Exam 3 Worksheet Answers 4. v < vi = xi < x < viii < iii < iv < ix < i < ii = vii < xii 5. Quantum number a. F b. G c. E d. F e. G f. G g. E h. F i. F j. F k. E l. G Chapter 8 – The Periodic Table 1. Electronic configurations ↑↓
, diamagnetic a. [Ar] 4s2 4s
↑ ↑
b. [Ar] 4s2 3d2 , paramagnetic 3d
↑↓ ↑↓ ↑↓
c. [Ar] 4s0 3d0 , diamagnetic 3p
↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
, paramagnetic 4s
3d
↑↓ ↑ ↑ ↑ ↑
[Ar] 4s0 3d6 , paramagnetic 3d
↑ ↑ ↑
[Ar] 4s0 3d3 , paramagnetic 3d
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
[Ar] 4s0 3d10 , diamagnetic 3d
↑ ↑
[Xe] 6s2 4f14 5d10 6p2 , paramagnetic 6p
d. [Ar] 4s1 3d10 e.
f.
g.
h.
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
, diamagnetic 5d
↑↓ ↑ ↑
, paramagnetic 3p
i.
[Xe] 6s2 4f14 5d10 j.
[Ne] 3s2 3p4 4
Exam 3 Worksheet Answers k. [Ne] 3s2 3p6 l.
[Xe] 6s2 4f11 m. [Rn] 7s2 5f7 n. [Kr] 5s0 4d8 o. [Kr] 5s0 4d10 ↑↓ ↑↓ ↑↓
, diamagnetic 3p
↑↓ ↑↓ ↑↓ ↑↓ ↑
4f
↑
↑
↑
↑
↑
↑
↑
↑ ↑
5f
↑ ↑
, paramagnetic , paramagnetic ↑↓ ↑↓ ↑
, paramagnetic 4d
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
, diamagnetic 4d
2. Periodic Trends Largest radii Highest IE a. Si N b. Ge P +
c. Na Na d. Br‐ Br e. Se Cl 3. Elements are placed in the periodic table based on their physical and chemical properties. Elements within a group have more properties in common than elements within the same period. The metallic character of elements increases down a group (from top to bottom) and decreases across a period (from left to right). In the group 4a elements, carbon is a nonmetal, silicon and germanium are metalloids and tin and lead are metals. Tin and lead will react with acid to form hydrogen and a 2+ cation. These metals can also exist in a 4+ oxidation state. For silicon and carbon, the 4+ oxidation state is the most stable. (Think about CO2 versus CO.) Lead(II) is more stable than lead(IV). In period 3, the metallic character decreases from left to right, sodium, magnesium and aluminum are metals, silicon is a metalloid and phosphorus, sulfur, chlorine and argon are nonmetals. The periodicity of the oxides of these elements in water are basic (on the left – sodium, etc) to amphoteric (aluminum) to acidic (silicon continuing to the right – not including argon). Chapter 9 – Chemical Bonding I: The Covalent Bond 1. a. I b. IC c. C d. IC e. C f. N g. IC 5
Exam 3 Worksheet Answers h.
i.
j.
k.
l.
m.
n.
o.
p.
q.
I I C N C C N IN C I 2. For the Born‐Haber cycle: the standard molar enthalpy of formation (of the solid compound), the enthalpy of sublimation of the metal, the ionization energy(s) of the metal, the electron affinity of the nonmetal, the bond enthalpy of the nonmetal are needed to determine the lattice energy for the compound. a. LiCl(s) → Li+(g) + Cl–(g) b. MgCl2(s) → Mg2+(g) + 2Cl–(g) c. KF(s) → K+(g) + F–(g) d. CaF2(s) → Ca2+(g) + 2F–(g) 3. 6
Exam 3 Worksheet Answers 4. Lewis dot structures – lone pairs are terminal atoms are not shown for clarity here but must be included on a quiz or exam. S
a.
O , bent, sp
O
:O :
2
, polar C
H , trigonal planar, sp2, polar :
H
:
b. H
O
C
c. H
d. H
H
C – tetrahedral, sp3, polar; O – bent, sp3, polar N : , linear, sp, polar H
:
C
:N
O:
N
:
e.
, linear, sp, polar :O :
S:
f.
Cl
Cl , trigonal pyramidal, sp3, polar F
C
F
Cl
Cl
g.
H
C
h.
, tetrahedral, sp3, polar C
H , linear, sp, non‐polar H
H
C
i.
C
O
O
N
N
j.
H , trigonal planar, sp2, non‐polar H
O , trigonal planar, sp
O
2
, nonpolar H
O
O
C
C
O
O
k.
Cl
H
C
C
l.
, trigonal planar, sp2, nonpolar H
Cl
H
, trigonal planar, sp2, polar O
C
m.
HO
OH , trigonal planar, sp
2
, polar 7
Exam 3 Worksheet Answers 5. S
O:
:
:
:
:
S
:O
O:
:O
-1
:
:N
O
:
N
+1
:
N
-1
:
:
:
O:
:
:
N
+1
:
:N
:
:
:
these are equivalent structures but based on formal charges, the structure given in 4a is the best. -2
N
+1
O:
+1
, first structure is most valid since O is more electronegative than N. :
0
:O :
S:
Cl
:O :
S: +1
0
Cl
Cl
O
Cl
, first structure is more valid O
N
O
-1
N
O
, equivalent to the structure given in 4j 6. All breaking of bonds is endothermic and all making of bonds is exothermic. a. Break 1 mol of C–O, O–H, 3 mol of C–H, and 3/2 mol of O=O; make 2 mol of C=O and 4 mol of O–H b. Break 1 mol of C=C, 2 mol of C–H, and 5/2 mol of O=O; make 4 mol of C=O and 2 mol of O–H c. Break 1 mol of N≡N, and 3 mol of H–H; make 6 mol of N–H d. Break 1 mol of C=O and ½ mol of O=O; make 2 mol of C=O e. Break 1 mol of F–F and H–H; make 2 mol H–F f. Break 1 mol of Cl–Cl and H–H; make 2 mol H–Cl g. Break 2 mol of N–O and N=O; make 2 mol of N–O, N=O, and 1 mol of N–N 7. a. –637 kJ; exothermic b. –1229 kJ; exothermic c. –107 kJ; exothermic d. Need CO e. –543 kJ; exothermic f. –185 kJ; exothermic g. –193 kJ; exothermic 8. a. –638.4 kJ b. –1255.4 kJ c. –92.6 kJ d. –283 kJ e. –543.2 kJ f. –184.6 kJ g. –58.04 kJ 8
Exam 3 Worksheet Answers 9. The differences observed between 7 and 8 can generally be attributed to the methods required for approximating bond enthalpies for non‐diatomic molecules. The universal application of one bond enthalpy regardless of the environment of the bond (for example, C–C bond energy within molecules varies depending on the molecule) can lead to error. 10. (Only one shown, although the idea is the same for all) 11. This reaction was purposefully chosen. Clearly, this proposed “intermediate” state is ridiculous because this reaction occurs through the formation of a N–N bond. The idea of bond enthalpies and much better proposed intermediates will be discussed next semester in kinetics. 9