Chemistry 65.100 A and V Final Exam
April, 2001
ANSWERS

Part A.

  1. DGo is the standard free energy change (i.e. at 25oCand 1 atm pressure, with all species in their standard states), DHo is likewise the standard enthalpy change, T is the absolute temperature, and DSo is the standard entropy change for the reaction.
  2. Referring to the table of standard reduction potentials, Ni is more easily reduced than iron, hence Ni is less easily oxidized. Al is more easily oxidized than Fe. Therefore, Al could cathodically protect Fe, whereas Ni could not.
  3. The rate of a reaction tells us nothing about its standard free energy change!
  4. If a plot of 1/[NO2] versus time gives a straight line, the reaction must be second order with respect to [NO2]. Since NO2 is the only reactant, the reaction must be second order overall.
  5. Making a buffer using a weak acid requires the addition of a soluble salt of the weak acid. Thus, any soluble salt of the nitrite ion could be used, such as NaNO2, KNO2, etc.
  6. The presence of the solute lowers the vapor pressure of the water. To get the vapor pressure back up to 1 atm (i.e. to boil the solution) thus requires that it be heated to a higher temperature.
  7. In theory, either isotope could be used. However, if we were to use 14C to date an object this old, so many half-lives of this isotope would have elapsed that measurement of the remaining 14C would be practically impossible. The 40K is a much better choice.
  8. Aldehyde, ketone, carboxylic acid, ester, amide all contain the C=O group.

 Part B.

1a. The activation energy comes from rearranging the Arrhenius equation:

ln (k) = -Ea/RT

Thus, a plot of ln(k) vs. 1/T gives a straight line having slope = -Ea/R. Using the given data, the slope of the line is:

slope = (ln(4.9 x 10-3) - ln(3.5 x 10-5)) / (1/338 K- 1/298 K) = -12440 K

Thus, Ea = -R (slope) = -8.314 J K-1 mol-1 (-12440 K) = 103400 J mol-1 = 103.4 kJ mol-1

1b. The simplest way to do this is to calculate the value of A, and use it in the Arrhenius equation to find k at 300 K:

A = k / (exp(-Ea/RT)

= 4.9 x 10-3 s-1/ exp(-103400/(8.314x338)) = 4.68 x 1013 s-1

Thus at 300 K, k = A exp(-Ea/RT) = 4.68 x 1013 s-1 exp(-103400/(8.314 x 360)) = 0.0464 s-1

1c. The half life of a first order reaction is t1/2 = -ln(2)/k = 0.693 / k
= 0.693 / 4.9 x 10-3 s-1
= 141 s

1d. For a first order process, [A] = [A]o exp (-kt). Since the pressure is low, we can use partial pressures in place of concentrations:
p(N2O5)1000 s = 0.10 atm exp (-4.9 x 10-3 s-1 x 1000 s)
= 7.45 x 10-4 atm

2(i). Using the principles that total number of nucleons and total charge are conserved, the reactions are balanced as follows:

(a) 24095Am + 42He ® 24397Bk + 10n

(b) 23892U + 126C ® 24498Cf + 6 10n

(c) 24998Cf + 105B ® 257103Lw + 2 10n

(ii) For the reaction 21H + 31H ® 42He + 10n, the change in mass is:

Dm = m(42He) + m(10n) - m(21H) - m(31H)

= 4.00260 + 1.00866 - 2.01410 - 3.01605 = -0.01889 amu

x 1.66 x 10-27 kg/amu = -3.14 x 10-29 kg

Per mole, this loss of mass is 3.14 x 10-29 kg x 6.023 x 1023 mol-1 = 1.89 x 10-5 kg/mol

The energy associated with this much mass is DE:

DE = Dmc2
= 1.89 x 10-5 kg/mol (3.00 x 108 m s-1)2
= 1.70 x 1012 J mol-1
= 1.70 x 109 kJ mol-1

 

 3(i)a. 1,3-dichloropentane
3(i)b. methyl-isobutyl ketone
3(i)c. diphenyl amine
3(i)d. 5-methyloctanoic acid

3(ii)a. Oxidation of this primary alcohol (butanol) yields the corresponding aldehyde, butanal:

 

3(ii)b. Oxidation of this secondary alcohol (2-butanol) yields the corresponding ketone, 2-butanone (or methyl ethyl ketone):

3(ii)c. Acid plus alcohol yields an ester, in this case butyl pentanoate:

3(ii)d. Acid plus amine produces an amide, in this case N-ethyl acetamide:

 

Part C.

4. The two half reactions are:

PbO2(s) + 3 H+(aq) + HSO4-(aq) + 2 e- ® PbSO4(s) + 2 H2O(l) Eo = +1.63 V
Pb(s) + HSO4-(aq) ® PbSO4(s) + 2 e- Eo = +0.30 V

The overall reaction is thus:

Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4-(aq) ® 2 PbSO4(s) + 2 H2O(l) Eo = +1.93 V

(a) For this overall reaction, the reaction quotient Q is:

Q = 1/([H+(aq)]2[HSO4-(aq)]2)
= 1/((4.5)2(4.5)2)
= 0.00244

Thus, E = Eo - RT/nF ln(Q)
= 1.93 - 8.314(298)/2(96487) ln(0.00244)
= 2.01 V

(b) If the cell potential has decreased to 1.80 V (which is less than the standard cell potential), we expect that the concentration of sulfuric acid must have decreased, since it is a reactant. Solving the Nernst equation for Q, we have:

ln(Q) = (Eo - E)/(RT/nF)
= (1.93 - 1.80)/((8.314x298)/(2x96487))
= 10.12

Thus, Q = exp(10.12) = 24970
But, Q = 1/([H+(aq)]2[HSO4-(aq)]2)
=1/(x2x2)
=1/x4

Thus, x4 = 1/Q
= 1/24970
= 4.00 x 10-5

Thus, x = [H+(aq)] = [HSO4-(aq)] = [H2SO4(aq)] = 0.079 M

5. In a face centred cubic unit cell, there are 4 atoms in total. In a face-centred cubic unit cell, the diagonal of one face = 4r. Thus, (4r)2 = l2 + l2

Or, 16 r2 = 2 l2, or r2 = l2/8
or r = l/81/2
= 4.08 x 10-8/2.83
= 1.44 x 10-8 cm
= 144 pm

The volume of the unit cell is the cube of the length of one side:

V = l3
= (392 x 10-12 m)3
= 6.02 x 10-29 m3
= 6.02 x 10-23 cm3

Since we know the density, we can calculate the mass of one unit cell:

munit cell = rV
= 21.45 g cm-3 x 6.02 x 10-23 cm3
= 1.29 x 10-21 g

Since this is a face centred cubic unit cell, we know it contains 4 atoms in total. The mass of one atom is therefore:

massatom = 1.29 x 10-21 g / 4
= 3.23 x 10-22 g

One mole of atoms therefore has a mass of:

3.23 x 10-22 g x 6.02 x 1023 mol-1
= 194.1 g/mol

The metal with the closest atomic weight is Platinum, Pt.

 

6. The first dissociation is: H3AsO4(aq) + H2O(l) ¾ H2AsO4(aq) + H3O+(aq)
For this dissociation, [H2AsO4(aq)] [H3O+(aq)] / [H3AsO4(aq)] = Ka1 = 0.006

Starting with 1 M H3AsO4(aq), the equilibrium expression will be:

x2/(1-x) = 0.006

Solving using the quadratic formula (the acid is too strong not to use it), we obtain:

[H3AsO4(aq)] = 1-x = 0.925 M
[H3O+(aq)] = x = 0.0745 M
[H2AsO4-(aq)] = x = 0.0745 M
pH = -log10[H3O+(aq)] = 1.13

The second dissociation is H2AsO4-(aq) + H2O(l) ¾ HAsO4-2(aq) + H3O+(aq)
For this dissociation, [HAsO4-2(aq)] [H3O+(aq)] / [H2AsO4-(aq)] = Ka2 = 1.1 x 10-7

Here, note that [H3O+(aq)] = [H2AsO4-(aq)] = 0.0745 M from the first step. Thus,

[HAsO4-2(aq)] = Ka2 = 1.1 x 10-7 M

The third dissociation is HAsO4-2(aq) + H2O(l) ¾ AsO4-3(aq) + H3O+(aq)
For this dissociation, [AsO4-3(aq)] [H3O+(aq)] / [HAsO4-2(aq)] = Ka3 = 3.0 x 10-12
Rearranging, [AsO4-3(aq)] = Ka3 [HAsO4-2(aq)]/ [H3O+(aq)]
= 3.0 x 10-12 (1.1 x 10-7)/0.0745 = 4.43 x 10-18 M


7. DTb = Kbm

or, m = DTb / Kb = 2.63oC / 40.0 oC kg mol-1 = 0.06575 mol kg-1

The molality of the reserpine is m = moles reserpine / kg camphor
Thus, moles reserpine = m x kg camphor
= 0.06575 mol kg-1(0.025 kg)
= 0.001643 mol

Thus, the molecular weight of the compound = g/mol
= 1.00 g / 0.001643 mol
= 608 g/mol


8. (a) PV = nRT,

thus P = nRT/V = 0.500 mol(0.082 L atm K-1 mol-1)(298 K)/1.00 L
= 12.22 atm

(b) P = nRT/(V-nb) - a(n/V)2
= 0.500 mol(0.082 L atm K-1 mol-1)(298 K)/(1.00 L - 0.500 mol(0.0391 L mol-1)
- 1.39 atm L2 mol-2 (0.500 mol/1.00 L)2
= 12.46 - 0.35
= 12.11 atm

(c) The pressure calculated using the van der Waals equation is smaller than that using the ideal gas law. This is because the van der Waals equation takes into account the size of the N2 molecules and the fact that they do interact, albeit weakly, with one another.

(d) NH3 is a polar molecule. We would expect a greater difference between pressures calculated using the ideal and the van der Waals equation than in the case of nitrogen because the ammonia molecules would interact to a greater extent, lowering the pressure quite a bit below that predicted by the ideal gas law.

 

9. The Clausius-Clapeyron equation relates vapor pressure to temperature:

ln(p2/p1) = (DHvap/R) [1/T1 - 1/T2]

Solving for T2, we have: T2 = [1/T1 - R/DHvap ln(p2/p1)]-1

Note that p1 is the vapor pressure of the liquid at its normal boiling point, i.e. p1 = 1 atm = 760 Torr

thus, T2 = [1/(273.1+56.5) - 8.314/32000 ln(630/760)]-1
= 324.4 K
= 51.3oC

 

10. When lead iodide dissolves in water, the following equilibrium is established:

PbI2(s) ¾ Pb+2(aq) + 2 I-(aq)

for which Ksp = [Pb+2(aq)][I-(aq)]2 = 1.4 x 10-8
(a) In water, [Pb+2(aq)] = x and [I-(aq)] = 2x
Thus, x(2x)2 = 1.4 x 10-8
4x3 = 1.4 x 10-8
x = 0.00152

Thus, the solubility of lead iodide in water is 1.52 x 10-3 M.

(b) In 0.10 M Pb(NO3)2(aq), the concentration of lead ions is much greater, since this is a soluble salt of lead. Thus, [Pb+2(aq)] = 0.10 M.

But, [I-(aq)]2 = Ksp/[Pb+2(aq)]
Thus, [I-(aq)] = (Ksp/[Pb+2(aq)])1/2
= (1.4 x 10-8 / 0.10)1/2
= 3.74 x 10-4 M

And the solubility of lead iodide = 1/2 [I-(aq)] = 1.87 x 10-4 M

(c) In 0.10 M NaI, the solubility will also be supressed because of the common ion, I-:

But, [Pb-+2(aq)] = Ksp/[I-(aq)]2
= 1.4 x 10-8 / (0.10)2
= 1.4 x 10-6 M
Thus, the solubility of PbI2 = 1.4 x 10-6 M.

11(a) DGo = -RT ln(Kc)
= -8.314 J K-1 mol-1 (298K) ln(9.1 x 10-6)
= 28760 J mol-1
= 28.8 kJ mol-1

(b) The reaction quotient under these conditions is:

Q = [Fe+2]2[Hg+2]2/([Fe+3]2[Hg2+2])
= 1.002 x 1.002 / (1.002 x 1.00)
= 1.00

Since Q>Kc, the reaction will have to shift to the left (i.e. back to the reactants) until Q = Kc to attain equilibrium.

(c) DG = DGo - RT ln(Q)
= 28800 J mol-1 - 8.314 J K-1 mol-1 (298) ln(0.012 x 0.0252 / (0.202 x 0.010))
= 50510 J mol-1

Here, Q<Kc, so the reaction will have to proceed to the right to achieve equilibrium.