Chapter 6 - Solutions
1. a. CH3(CH2)10COOH (s) + 17 O2 (g)  12 CO2 (g) + 12 H2O (l)
Hrxn = 12(393.5 kJ) + 12(285.8 kJ) – (774.6 kJ) = 7377 kJ
C12H22O11 (s) + 12 O2 (g)  12 CO2 (g) + 11 H2O (l)
Hrxn = 12 (393.5 kJ) + 11(285.8 kJ) – (2225.8 kJ) = 5640 kJ
15.0 g CH 3 (CH 2 )10 COOH 
1 mol CH 3 (CH 2 )10 COOH
 7377 kJ

200.32 g CH 3 (CH 2 )10 COOH 1 mol CH 3 (CH 2 )10 COOH
Hrxn = 552.4 kJ
m (C12H22O11) =  552.4 kJ 
1 mol C12 H 22 O11
342.3 g C12 H 22 O11

- 5640 kJ C12 H 22 O11 1 mol C12 H 22 O11
m (C12H22O11) = 33.5 g C12H22O11
b. For the same amount of energy, more grams of carbohydrates are needed than
grams of fat molecules.
3. a. SO2 has 9 degrees of freedom: (3)(Number of atoms).
O
S
O
O
S
O
The VSEPR electron arrangement is AX2E; therefore, SO2 is nonlinear resulting
in SO2 having 3 rotational degrees of freedom.
Overall, SO2 has 3 translational degrees of freedom, 3 rotational degrees of
freedom, and 3 vibrational degrees of freedom.
High-temperature limit of Cv means all modes are accessible.
1
1
Cv = 3   R + 3   R + 3R = 6R = 6(8.3145 J·mol-1·K-1) = 49.9 J·mol-1·K-1
2
2
b.
Vib 1:  =
h vib
kB
=
(6.626 10 34 J·s)(3.5 1013 Hz)
1.38 10  23 J·K - 1
= 1680 K
Vib 2:  =
h vib
Vib 3:  =
h vib
kB
kB
=
=
(6.626 10 34 J·s)(4.11013 Hz)
1.38 10  23 J·K - 1
(6.626 10 34 J·s)(1.6 1013 Hz)
1.38 10  23 J·K - 1
= 1969 K
= 768 K
Therefore, at 1000 K, 1 vibrational mode is accessible.
1
1
Cv = 3   R + 3   R + R = 4R = 4(8.3145 J·mol-1·K-1) = 33.3 J·mol-1·K-1
2
 2
c. From part b, no vibrational modes are accessible at 298 K.
1
Cv = 3   R + 3
2
1
  R = 3R = 3(8.3145 J·mol-1·K-1) = 24.9 J·mol-1·K-1
 2
5. Zn (s) + 2 HCl (aq)  Zn2+ (aq) + H2 (g) + 2 Cl (aq)
Hrxn = 2(167.16 kJ) + (153.89 kJ)  2(167.16 kJ) = 153.89 kJ
n (Zn) = 8.5 g Zn 
1 mol Zn
= 0.13 mol Zn
65.39 g Zn
n (HCl) = 0.800 L HCl 
0.500 mol HCl
= 0.400 mol HCl
1 L HCl
Zn is limiting reagent.
q = 8.5 g Zn 
- 153.89 kJ
1 mol Zn

= 20.0 kJ
1 mol Zn
65.39 g Zn
q = nCT
n (soln) = 800.0 ml soln  (1.00 g·cm-3) 
1 mol H O
2
18.02 g H O
= 44.4 mol soln
2
T =


20.0 kJ


 75.92 J·mol -1·K -1


44.4 mol soln 




 10 3 J 
 = 5.93 K

 1 kJ 


Tf (HCl soln) = 298 K + 5.93 K = 304 K
7. Energy to heat 350 g of water: T = 75 oC = 75 K
q = nCT = 350 g(4.184 J·g -1·K -1)75K  1.10105 J
What is energy of one photon?
E = h =
hc

=
 6.626 10  34 J·s  2.998 108 m·s - 1 


 3



  10 mm 
 4.4110  23 J·photon - 1

4.5 mm
m 


# photons =
J
J·photon -1

1.10 105 J
4.4110  23 J·photon -1
 2.49 10 27 photons
9. a. maleic acid (C4H4O4) (s) + 3 O2 (g)  4 CO2 (g) + 2 H2O (g)
H = 1355.2 kJ
4 CO2 (g) + 2 H2O (g)  fumaric acid (C4H4O4) (s) + 3 O2 (g)
H = +1334.7 kJ
net rxn: maleic acid  fumaric acid
H = 1355.2 kJ + 1334.7 kJ = 20.5 kJ
b. Fumaric acid has a lower standard enthalpy of formation because the
isomerization reaction in which maleic acid forms fumaric acid is exothermic;
that is, energy is released in forming fumaric acid from maleic acid. Hence it
must have taken less energy to form fumaric acid from the elements.
c. U is more negative than H for the combustion reactions.
H = U + PV = U + nRT
For this reaction, more moles of gas are produced in the products than there were
in the reactants. At constant T and P (conditions under which this reaction was
run), greater moles of product are produced. Hence nRT > 0. Hcombustion < 0.
Because U = H  nRT, U has to be negative and more negative than H.