CHEMISTRY 115

CHEMISTRY 115	HOMEWORK 4	NAME ______________________
DUE MAY 2, 2001		COMPUTER HOMEWORK: 21-Thermo: ALL

1. Calculate the melting point of aluminum, Al(s)à Al (l) given that DH = +10.0 kJ/mol and DS = +9.50 J/molK.

DG = 0 = DH – TDS; DH = TDS;

T = DH / DS = 10.0 kJ x 10³ J/kJ / 9.50 J/K = 1052.6 K = 779^oC

m.p. = |_1053 K = 779^oC____

2. Consider the reaction H2O(l) = H+(aq) + OH^-(aq). Given DGfo for H+(aq) = 0 and for OH^-(aq) = -157.2 kJ/mol, calculate DG_f^o H₂O(l) = -237.2 kJ

a. DGo at 25oC.
DG_rxn^o = SDG_f^o products - SDG_f^o reactants = (-157.2) – (-237.2) kJ = 80.0 kJ

DG^o= |_80.0 kJ________

b. Kw at 25 oC.
DG = 0 = DG^o + RT lnK

lnK = - DG^o / RT = -80.0 kJ x 10³J/kJ /( 8.314 J/K x 298K) = -32.

taking inv ln both sides, K = 9.5 x 10^-15

Kw = |_9.5 x 10^-15_______

c. Calculate the value of free energy DG^o₇₇₃ for the above reaction at 500 ^oC .
DG^o₇₇₃ = DH^o₂₉₈ – TDS^o₂₉₈;

DH^o₂₉₈ = SDH_f^o products - SDH_f^o reactants = 0 + (-230) – (-285.9) = 55.9 kJ

DS^o₂₉₈ = SDS^o products - SDS^o reactants = 0 + (-10.75) – 69.96 = -80.71 J/K

DG^o₇₇₃ = 55.9 kJ – 773K(-80.79 J/K) x 10^-3kJ/J = 55.9 + 62.39 = 118.3 kJ

DG^o₇₇₃= |_118 kJ________
This answer does not make much sense, perhaps partially because it is so far above the boiling point of water. Normally would expect the entropy change to be positive with the formation of the more disordered ions but perhaps a very small H⁺creates enough greater order.

3. The pollutant gas hydrogen sulfide is removed from natural gas by the reaction:

2H2S(g) + SO2(g) -> 3S(s) + 2 H2O(g).

a. What is the equilibrium constant for this reaction at 25oC?
DG_rxn^o = SDG_f^o products - SDG_f^o reactants = 3((0) + 2(-228.6) –[ 2(-33.6) + (-300) kJ =

-457.2 + 67.2 + 300 kJ = -457.2 + 367.2 kJ = -90 kJ

DG = 0 = DG^o + RT lnK

lnK = - DG^o / RT = -(-90. kJ )x 10³J/kJ /( 8.314 J/K x 298K) = +36.33

taking inv ln both sides, K = 5.97 x 10¹⁵

K = |_6.0 x 10¹⁵______

Estimate the value of the equilibrium constant for this reaction at 100oC.
DH^o₂₉₈ = SDH_f^o products - SDH_f^o reactants = 3(0) + 2(-241.8) – [2(-20.6) + (-297)] = -145.4 kJ

DS^o₂₉₈ = SDS^o products - SDS^o reactants = 3(31.8) + 2(188.7) – [ 2(206) + 248] = -187.2 J/K

DG^o₇₃₃ = DH^o₂₉₈ – TDS^o₂₉₈ = -145.4 kJ –373K (-.1872 kJ/K) = -145.5 + 69.83 = -75.57 kJ

DG = 0 = DG^o + RT lnK

lnK = - DG^o / RT = -(-75.6kJ )x 10³J/kJ /( 8.314 J/K x 373K) = +24.37

taking inv ln both sides, K = 3.83 x 10¹⁰

K = |_3.8 x 10¹⁰ ______
Because the reaction is exothermic with a negative entropy change, the reaction becomes less favorable as temperature is increased and the K becomes smaller.

5. How much energy would be released during the formation of the bonds in one mole of acetone

O

||

( CH₃-C–CH₃ ) molecules.
-[6 (C-H) + 2 (C-C) + C=O] = -[6(412) + 2(348) + 743] = -3911 kJ

The sign is actually negative because energy is released when bonds are formed; the phrasing of the question could permit a + sign since it asking for the energy released.

Note that this number is nothing like the heat of formation of acetone, which would be a liquid formed from solid C and diatomic hydrogen and oxygen molecules rather than the gaseous atoms that are assumed for bond energies.

E = |_(-)3911 kJ_______

The standard heat of formation of ethylene, C₂H₄(g) is +52.284 kJ/ mol. Calculate the C=C bond energy in this molecule. 2C(s) + 2 H₂(g) = C₂H₄(g) with DH^o_f= 52.284 kJ/mol

Bond energy reaction is 2C(g) + 4 H(g) = C₂H₄(g) thus need to modify the heat of formation by reversing and

C₂H₄(g) = 2C(s) + 2 H₂(g), DH^o= -52.284 kJ/mol

2C(s) = 2C(g), DH_rxn= 2(716.67 kJ) kJ = 1433.34 kJ

2 H₂(g) = 4 H(g), DH_rxn= 4(217.89 kJ) kJ = 871.56 kJ

Thus for the reaction C₂H₄(g) = 2C(g) + 4 H(g) DH^o= -52.284 kJ + 2304.9 = +2252.6 KJ

Or for the atomization reaction: [4(C- H) + C=C] = 4(412) + [C=C]

= (+2252.6 kJ) = 1648 kJ + [C=C]

[C=C] = 2252.6 – 1648 = 604.6 kJ = 605 kJ

Bond energy = |_605 kJ____

7. How many moles of electrons are required to produce 5.00 g of Mg from molten MgCl₂ = Mg²⁺ + 2 Cl^-

Mg²⁺ + 2 e^- = Mg

5.00 g Mg x 1 mol Mg / 24.31 g x 2 mol e^- / mol Mg²⁺ = 0.4114 mol

mol = |_0.411 mol ______

8. A Cu2+(aq) solution is electrolyzed using a current of 0.500 A. Calculate the mass of Cu that would be plated out after 500 min. Cu²⁺ + 2 e^- = Cu

0.500 A x 500 min x 60 sec/min = 15000 amp sec = 15000 C

15000 C x [1 mol e^- / 96500 C] x [1 mol Cu / 2mol e^-] x [63.55 g / mol Cu] = 4.939 g Cu

mass = |_4.94 g_____

9. How many hours would it take to produce 35.0 g of lead from PbSO₄ during the charging of a lead storage battery using a current of 1.50 A?

The half reaction is PbSO₄ + 2e^- à Pb + SO₄^2-
35.0 g Pb x [1 mol Pb / 207.2 g] x [2 mol e^- / 1 mol Pb] x [96500 amp sec / mole e^-] x 1/1.50 amp =

21734 sec = 362 min = 6.04 hr

Hours = |_6.04 hr_____