Chapter 6 - Solutions

1. a. CH3(CH2)10COOH (s) + 17 O2 (g) ® 12 CO2 (g) + 12 H2O (l)

DHrxn = 12(-393.5 kJ) + 12(-285.8 kJ) – (-774.6 kJ) = -7377 kJ

C12H22O11 (s) + 12 O2 (g) ® 12 CO2 (g) + 11 H2O (l)

DHrxn = 12 (-393.5 kJ) + 11(-285.8 kJ) – (-2225.8 kJ) = -5640 kJ

DHrxn = -552.4 kJ

m (C12H22O11) =

m (C12H22O11) = 33.5 g

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).

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.

Cv = 3R + 3R + 3R = 6R = 6(8.3145 J·mol-1·K-1) = 49.9 J·mol-1·K-1

b.

Vib 1: Q = = = 1680 K

Vib 2: Q = = = 1969 K

Vib 3: Q = = = 768 K

Therefore, at 1000 K, 1 vibrational mode is accessible.

Cv = 3 R + 3 R + R = 4R = 4(8.3145 J·mol-1·K-1) = 33.3 J·mol-1·K-1

c.  From part b, no vibrational modes are accessible at 298 K.

Cv = 3 R + 3 R = 3R = 3(8.3145 J·mol-1·K-1) = 24.9 J·mol-1·K-1

5.  Zn (s) + 2 HCl (aq) ® Zn2+ (aq) + H2 (g) + 2 Cl- (aq)

DHrxn = 2(-167.16 kJ) + (-153.89 kJ) - 2(-167.16 kJ) = -153.89 kJ

n (Zn) = 8.5 g Zn ´ = 0.13 mol Zn

n (HCl) = 0.800 L HCl ´ = 0.400 mol HCl

Zn is limiting reagent.

q = 8.5 g Zn ´ ´ = -20.0 kJ

q = nCDT

n (soln) = 800.0 ml soln ´ (1.00 g·cm-3) ´ = 44.4 mol soln

DT = ´ = 5.93 K

Tf (HCl soln) = 298 K + 5.93 K = 304 K

7.  Energy to heat 350 g of water: DT = 75 oC = 75 K

q = nCDT =

What is energy of one photon?

E = hn = =

# photons =

9. a. maleic acid (C4H4O4) (s) + 3 O2 (g) ® 4 CO2 (g) + 2 H2O (g)

DH = -1355.2 kJ

4 CO2 (g) + 2 H2O (g) ® fumaric acid (C4H4O4) (s) + 3 O2 (g)

DH = +1334.7 kJ

net rxn: maleic acid ® fumaric acid DH = -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. DU is more negative than DH for the combustion reactions.

DH = DU + PDV = DU + DnRT

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 DnRT > 0. DHcombustion < 0. Because DU = DH - DnRT, DU has to be negative and more negative than DH.