Chemistry 65.100 A and V 1996/1997

Final Exam Answers

PART A.

1. 	DGo is the standard free energy change for the reaction 

	Eo is the standard cell potential (calculated at 25oC, 1 atm with all reactants and products in their standard states).

F is the Faraday constant (96,500 C mol^-1, not necessary to quote value)

n is the number of electrons transferred in the balanced reaction

2. The intermolecular forces in hexane are dispersion forces only. These are weak compared to the H-bonds between water molecules. Thus, at a given temperature, it is more likely that hexane molecules will have sufficient energy to overcome the dispersion forces and escape into the gas phase.

3. The fundamental reason for all colligative properties is that the vapor pressure of a solvent is lowered upon addition of a solute.

4. A stronger acid has a weaker conjugate base. Since hypochlorous acid is the weaker acid, this implies that it does not dissociate as much in aqueous solution. In other words, its conjugate base (OCl^-_(aq)) is more strongly associated with the H⁺. In other words, OCl^- is a stronger base.

An alternate answer is simply that K_b = K_w/K_a. Since K_a for acetic acid is larger, K_b for its conjugate base (CH₃COO^-) is smaller.

5. A unit cell is the repeating unit in a solid material that can be stacked together in 3 dimensions to form the overall solid.

6. At a higher temperature, (1) there are more collisions per unit time and (2) the collisions are more energetic.

7. There is in fact a change in mass during a chemical reaction, but the change is so small that we can not measure it. Use of E=mc² could demonstrate this.

8. Alkanes contain only single bonds between carbon atoms. The carbons and the groups attached to them are thus free to rotate. In an alkene, the double prevents this rotation, and the groups attached to the carbons can therefore be held on one side or the other of the molecule.

PART B.

1. 150 kg/h x 1000 g/kg x (1 mol product / 108 g product) x (2 mol e^- / mol product)

x 96000 C/mol e^- = 2.67 x 10⁸ C/h

= 7.41 x 10⁴ C/s = 74100 A.

2. ln (p₂/p₁) = (DH_vap/R) (1/T₁ - 1/T₂)

or, ln p₂ = lnp₁ + (DH_vap/R) (1/T₁ - 1/T₂)

here, T₁ = 56.5^oC = 329.5 K

p₁ = 760 Torr (i.e. the normal boiling point)

T₂ = 25.0^oC = 298.0 K

thus, ln p₂ = ln(760) + (32000 J mol^-1/8.314 J K^-1 mol^-1) (1/329.5 K - 1/298.0 K)

= 5.40

thus, p₂ = exp(5.40) = 221.1 Torr

3. DT = k_f m

k_f = 1.86^oC kg mol^-1

m = 1.0 g/L x (1 mol / 9.0 x 10⁴g) x (1 L / 1 kg) = 1.11 x 10^-5 mol kg^-1

thus, DT = 1.86^oC kg mol^-1 (1.11 x 10^-5 mol kg^-1) = 2.07 x 10^{-5 o}C

OR, the freezing point of the solution = -2.07 x 10^{-5 o}C

P = MRT

M = 1.0 g/L x (1 mol / 9.0 x 10⁴ g) = 1.11 x 10^-5 mol L^-1

thus, P = 1.11 x 10^-5 mol L^-1 (0.082 L atm K^-1 mol^-1) (298 K)

= 2.71 x 10^-4 atm

The osmotic pressure is much easier to measure in this case than the very small freezing point depression.

4a. This is a simple application of the Henderson Hasselbach equation,

pH = pK_a + log ([base]/[acid])

here, pK_a is that of the NH₄⁺_(aq). This i calculated from the given value of K_b for NH₃ (the conjugate base of NH₄⁺):

K_a = K_w / K_b = 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10

thus, pK_a = -log₁₀(K_a) = -log₁₀(5.56 x 10^-10) = 9.26

[base] = [NH₃] = 0.50 M

[acid] = [NH₄⁺] = 0.50 M

thus, pH = 9.26 + log (0.50/0.50)

= 9.26

4b. After addition of 0.020 mol HCl_(aq), 0.020 mol of the NH₃ will have been consumed according to: NH_3(aq) + HCl_(aq) ® NH₄Cl_(aq) + H₂O_(l)

thus, now [NH₃] = 0.50 - 0.020 = 0.48 M

and, [NH₄⁺] = 0.50 + 0.020 = 0.52 M

thus, pH = 9.26 + log(0.48/0.52)

9.23

5a. If the edge length is 495 pm, the diagonal length of one face is l:

l² = 495² + 495² = 490050 pm²

or, l = 700.0 pm

this diagonal is 4 times the radius of one atom. Thus, r = 700/4 = 175.0 pm

5b. The FCC unit cell contains 4 atoms. Thus, the mass of a unit cell = 4 (mass of one Pb atom)

= 4 (207.2 g/mol) / 6.023 x 10²³ atoms/mol = 1.38x 10^-21 g

The unit cell volume = d³

d = 495 pm = 495 x 10^-12 m = 495 x 10^-10 cm

thus, volume = (495 x 10^-10 cm)³ = 1.21 x 10^-22 cm³

density = mass/volume

= 1.38 x 10^-21g / 1.21 x 10^-22 cm³

= 11.3 g/cm³

6a. The differential rate law will be of the form rate = k[NO]^a[Cl₂]^b. To find a and b, examine the data in the table. If [NO] is held constant at 0.10 and [Cl₂] is doubled, the rate doubles from 0.18 to 0.36. Thus, the reaction must be first order in [Cl₂] (i.e. b=1). If now [NO] is doubled and [Cl₂] is held constant, the rate quadruples from 0.36 to 1.44. Thus, the reaction must be second order in [NO] (i.e. a=2).

The differential rate law is thus: rate = k[NO]²[Cl₂]

6b. The rate constat can be found by rearranging the differential rate law: k = rate / ([NO]²[Cl₂])

We can use any of the three sets of data to find k. Using the first set,

k = 0.18 mol L^-1 min^-1 / ((0.10 mol L^-1)²(0.10 mol L^-1))

= 180 mol^-2 L² min^-1

6c. The reaction likely occurs as written, since the stoichiometric constants are the same as the orders for both NO and Cl₂.

7a. ³²P₁₅ ® ⁰e + ?

To balance the charge is this reaction, the product nuclide must have a nuclear charge of 16. It must therefore be an isotope of element #16 (i.e. sulfur). To balance the number of nucleons, the product must have A = 32. Thus, the product must be ³²S₁₆.

7b. k = 0.693/t_1/2 (for a first order process such as radioactive decay)

= 0.693 / 14.3 d = 0.0485 day^-1

7c. N = N_oexp(-kt)

= 175.0 mg exp(-0.0485 day^-1 (35.0 day))

= 32.0 mg

7d. activity = rate of decay = kN

here, N = number of radioactive nuclei = 0.175 g x (1 mol/32 g/mol) x 6.023 x 10²³

= 3.29 x 10²¹ nuclei

thus, activity = kN = 0.0485 day^-1 x 3.29 x 10²¹ nuclei

= 1.60 x 10²⁰ day^-1

/24/3600 = 1.85 x 10¹⁵ s^-1

/3.7 x 10⁷ = 5.00 x 10⁷ mCi

8a. (i) para-dimethylbenzene

(ii) 4-chloro-2-heptene

(iii) tert-butylcyclohexane

(iv) 1,1,1-trichlorobutane

8b. (i) CH₂CH₂ + Br₂ ® CH₂BrCH₂Br (1,2-dibromoethane)

(ii) benzene + iodine ® iodobenzene

(iii) phenylamine + HCl ® phenylamine hydrochloride

(iv) acetic acid + 1-propanol ® propyl acetate

(v) acetic acid + diethylamine ® N,N-diethylacetamide