Chemistry 65

Chemistry 65.100 A and V Final Exam
April 23, 1999
ANSWERS

Part A.

E is the cell potential

E^o is the standard cell potential
R is the gas constant
T is the absolute temperature
n is the number of moles of electrons transferred in the balanced REDOX reaction
F is the Faraday constant
Q is the reaction quotient

Cl^- is an ion common to SrCl₂ and NaCl. Adding a common DECREASES the solubility of sparingly soluble salts such as SrCl₂.
Diamond has uses only sp³ orbitals for bonding, implying complete covalency, i.e. the electrons are localized between two C atoms, and are not mobile. In graphite, there is sp² hybridization, and the extra electron from each C is contained in pi-orbitals. These electrons are mobile (delocalized), and thus can carry an electric current.
According to Arrhenius, k = A exp(-E_a/RT). If E_a is positive, k will increase with temperature.
The equilibrium constant varies only with temperature. Adding a reactant will not change the equilibrium constant.
Aldehyde, ketone, carboxylic acid, ester, amide all contain the C=O group.
The first ionization of an acid involves removing the proton from a singly charged species (e.g. removing the H⁺ from the HCO₃^- ion). The second therefore involves removing a proton from a doubly charges species (e.g. removing H⁺ from the CO₃^-2 ion), and electrostatically, this requires more energy.
Going across a period, we are adding electrons to the same shell. They thus have essentially the same energies. However, we are also adding protons to the nucleus, which causes a greater attraction of the electrons to the nucleus, drawing the orbital close to the nucleus. Another way of stating this is that the effective nuclear charge is greater for elements at the right end of a period, or that the shielding is smaller at the right.

Part B.

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

ln (k) = -E_a/RT

Thus, a plot of ln(k) vs. 1/T gives a straight line having slope = -E_a/R. Using the given data, the slope of the line is:
slope = (ln(4.5 x 10³) - ln(1.00 x 10⁴)) / (1/274 K- 1/283 K) = -6880 K
Thus, E_a = -R (slope) = -8.314 J K^-1 mol^-1 (-6880 K) = 57200 J mol^-1 = 57.2 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(-E_a/RT)

= 4.5 x 10³ s^-1/ exp(-57200/(8.314 x 274)) = 3.61 x 10¹⁴ s^-1

Thus at 300 K, k = A exp(-E_a/RT) = 3.61 x 10¹⁴ s^-1 exp(-57200/(8.314 x 300)) = 3.96 x 10⁴ s^-1

1c. The half life of a first order reaction is t_1/2 = -ln(2)/k = 0.693 / k

= 0.693 / 4.5 x 10³ s^-1
= 1.54 x 10^-4 s

1d. For a first order process, [A] = [A]_o exp (-kt)
= 1.00 mol L^-1 exp (-4.5 x 10³ s^-1 x 2 x 10^-4 s)
= 0.41 mol L^-1

2. All radioactive decay is first order. Thus, N/N_o = e^-kt, or ln (N/N_o) = -kt
k = 0.693 / t_1/2 = 0.693 / 5730 y = 1.21 x 10^-4 y^-1

Activity (= rate of radioactive decay) is proportional to N, the number of ¹⁴C atoms remaining. Thus, ln(N/N_o) = ln(11 / 14) = ln(0.786) = -0.241

Thus, t = -(ln(N/N_o)/k = -(-0.241)/1.21 x 10^-4 y^-1 = 1993 y

3(i)a. 1-ethyl-2-chlorobenzene
3(i)b. 2-methyl-2-butanol
3(i)c. methyl pentanoate OR pentanoic acid, methyl ester

3(ii)a. This is an addition reaction, where the hydrogen gets added to the C atoms at either end of the double bond forming 2-methylpentane, or

3(ii)b. This reaction forms an ether by elimination of water to yield dipropyl ether, or:

3(ii)c. Acid plus alcohol produces an ester, in this case ethyl acetate, or:

Part C.

4a. Oxidation reaction:
I^-_(aq) à I_2(s)
2 I^-_(aq) à I_2(s)
2 I^-_(aq) à I_2(s) + 2 e^-

Reduction reaction:
Fe⁺³_(aq) à Fe⁺²_(aq)
Fe⁺³_(aq) + e^- à Fe⁺²_(aq)
2 x reduction reaction plus oxidation reaction =

2 I^-_(aq) + 2 Fe⁺³_(aq) à I_2(s) + 2 Fe⁺²

4b. E^o = E^o_ox + E^o_red = -0.54V + 0.77 V = +0.23 V = +0.23 J/C

DG^o = -nFE^o
= -2 mol e^- (96487 C mol^-1)(0.23 J/C) = -44380 J/mol - -44.4 kJ/mol
Since DG^o < 0, the reaction is spontaneous

4c. For this, we use the Nernst equation, E = E^o - RT/nF ln(Q)

Here, E^o = +0.23 V
R = 8.314 J K^-1 mol^-1
T = 50^oC = 323 K
n = 2 mol e^-
F = 96487 C/mol e^-
Q = [Fe⁺²]²/([Fe⁺³]²[I^-]² = .01²/((.01²)(.01)²) = 1 x 10⁴

Thus, E = 0.23 - 8.314(323)/(2(96487)) ln(1 x 10⁴) = 0.10 V

5. In a face centred cubic unit cell, there are 4 atoms in total. Thus, the unit cell mass = 4 x mass of one gold atom = 4(197 g/mol)/6.02 x 10²³ mol^-1 = 1.31 x 10^-21 g.

Since density = r = m/V, then V = m/r = 1.31 x 10^-21 g / (19.32 g cm^-3)
= 6.78 x 10^-23 cm³

Since this is a cubic unit cell, the cell volume = (edge length)³. Thus, the edge length = l = (6.78 x 10^-23 cm³)^1/3 = 4.08 x 10^-8 cm.

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/8^1/2
= 4.08 x 10^-8/2.83
= 1.44 x 10^-8 cm
= 144 pm

6. This is an application for the Henderson-Hasselbalch equation, pH = pK_a + log([base]/[acid])

Here, the acid is H₂PO₄^-_(aq) and the base is HPO₄^-2_(aq).
Also, pK_a = -log₁₀(K_a) = -log₁₀(6.31 x 10^-8) = 7.20
Thus, 7.40 = 7.20 + log([HPO₄^-2]/[H₂PO₄^-])
or, log([HPO₄^-2]/[H₂PO₄^-]) = 0.20
or, ([HPO₄^-2]/[H₂PO₄^-]) = 1.58
or, [HPO₄^-2] = 1.58[H₂PO₄^-]
But, we also know that ([H₂PO₄^-] + [HPO₄^-2]) = 0.0200 M
Thus, [H₂PO₄^-] + 1.58[H₂PO₄^-] = 0.0200 M
2.58[H₂PO₄^-] = 0.0200 M
[H₂PO₄^-] = 0.0078 M
Thus, [HPO₄^-2] = 1.58[H₂PO₄^-]
= 1.58(0.0078) = 0.0122

7a. DT_b = K_bm
or, m = DT_b / K_b = (80.31 - 80.10)^oC / 2.53 ^oC kg mol^-1 = 0.0830 mol kg^-1
The molality of the methyl salicylate is m = moles methyl salicylate / kg solvent
Thus, moles methyl salicylate = m(kg solvent)
= 0.083 mol kg^-1(0.099 kg)
= 0.00822 mol
Thus, the molecular weight of the compound = g/mol
= 1.25 g / 0.00822 mol
= 152 g/mol

	NH₃	N₂	H₂
initial	1.8	0	0
change	-2x	+x	+3x
equilibrium	1.8-2x	x	3x

Thus at equilibrium, [N₂][H₂]³/[NH₃]² = x(3x)³/(1.8 - 2x)² = K_eq = 6.3
or, 27x⁴/(1.8-2x)² = 6.3
Taking the square root of both sides, 5.20x²/(1.8-2x) = 2.51
or, 5.20 x² + 5.02x - 4.52 = 0
i.e. a quadratic of the form ax² + bx + c = 0, where a=5.20, b=5.02 and c=-4.52
Using the quadratic formula, x = 0.57 or -1.53
Thus, x = 0.57
[NH₃] = 1.8 - 2x = 0.66
[N₂] = x = 0.57
[H₂] = 3x = 1.71

Check: x(3x)³/(1.8 - 2x)² = 6.5

8b. The total concentration of gas is [NH₃] + [N₂] + [H₂] = .66+.57+1.71
= 2.94 mol/L
x 2.00 L = 5.88 mol gas in total
p = nRT/V = 5.88 mol(0.082 L atm K^-1 mol^-1)(450+273 K)/2.00 L
= 174 atm

9. ln(p₂/p₁) = DH_vap/R {1/T₁ - 1/T₂} or DH_vap = R ln(p₂/p₁) / (1/T₁ - 1/T₂)

here, p₁ = 400
T₁ = 18^oC = 291 K
p₂ = 760 (i.e. the boiling point)
T₂ = 35^oC = 308 K
R = 8.314 J K^-1 mol^-1
Thus, DH_vap = 8.314 ln(760/400) / (1/291 - 1/308) = 28130 J/mol = 28.1 kJ/mol

10. Fe(OH)_3(s) ¾ Fe⁺³_(aq) + 3 OH^-_(aq)
K_sp = [Fe⁺³_(aq)][OH^-_(aq)]³ = 2.6 x 10^-39
The solubility of the compound is how much Fe(OH)₃ dissolves and is the same as [Fe⁺³_(aq)]

However, [Fe⁺³_(aq)] = K_sp/[OH^-_(aq)]³

a. If pH = 7.00, [OH^-] = 10^-7 M. Thus, [Fe⁺³] = 2.6 x 10^-39 / (10^-7)³

= 2.6 x 10^-18 mol L^-1

b. If pH = 5.00, [OH^-] = 10^-9 M. Thus, [Fe⁺³] = 2.6 x 10^-39 / (10^-9)³

= 2.6 x 10^-12 mol L^-1

c. If pH = 11.00, [OH^-] = 10^-3 M. Thus, [Fe⁺³] = 2.6 x 10^-39 / (10^-3)³

= 2.6 x 10^-30 mol L^-1

11a. If [NO] is held constant at 0.10 mol L^-1 and [Cl₂] is doubled, the rate is also doubled. Thus, rate is proportional to [Cl₂]. Also, if [Cl₂] is held constant at 0.20, and [NO] is doubled, the rate increases four fold, i.e. rate is proportional to [NO]². The rate equation is therefore rate = k[NO]²[Cl₂]

11b. Rearranging the above, k = rate / ([NO]²[Cl₂]). Using the first experiment (we could use any of them),
k = 0.18 mol L^-1 min^-1 / (0.10 mol L^-1)²(0.10 mol L^-1)
= 180 mol^-2 L^-1 min^-1

11c. Since the orders of both reactants are the same as their stoichiometric coefficients in the overall reactions, it is possible that the reaction occurs in one step.