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BOOK PROBLEMS CHAPTER 14
23. b CH4 at 300K will have more thermal motion, and therefore higher entropy, than CH4 at 250 K.
c. The drink. The ice cube is a much more ordered structure.
d. Let’s see. Don’t they both have S = 0 according to the 3rd Law, assuming that they are perfect
x’tals?
25.a.
H2(g) + 1/2 O2(g) H2O(l)
ΔSo = Σ nSof products - Σ nSof reactants = Sof [H2O(l)] – { Sof [O2(g)] + 1/2Sof[H2(g)]}
= +70.0 J/K mol – (130.6 J/K mole + 1/2(205 J/K mole) = -163.1 J/K mole
Going from less ordered to more ordered
28 a. C(s, graphite)  C(s, diamond)
ΔGo = Σ nΔGof products - Σ nΔGof reactants = ΔGof [C(s, diamond)] – { ΔGof [C(s, graphite) ] }
= 2.9 kL/mol – 0.0 kJ/mol = +2.9 kJ/mol
b. C(s, diamond)  C(s, graphite)
ΔGo = Σ nΔGof products - Σ nΔGof reactants = ΔGof [C(s, graphite)] –( ΔGof [C(s, diamond) ] )
= 0.0 kJ/mol – 2.9 kL/mol) = - 2.9 kJ/mol
Diamonds AREN’T FOREVER!
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35 d. CH3CH2OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
ΔHo = 2 ΔHfo [CO2 (g)] + 3 ΔHfo [H2O (l)] - {ΔHfo [CH3CH2OH(l)] + 3 ΔHfo [O2 (g)] }
= 2(-393.5 kJ/mol) + 3(-285.8 kJ/mol) – {-277.7 kJ/mol + 3(0.0 kJ/mol)} = -1366.7 kJ/mol
ΔSo = Σ nSof products - Σ nSof reactants = 2 Sof [CO2(g) ] + 3 Sof [H2O(l)] – {( Sof [CH3CH2OH(l)] + 3Sof[O2(g)]}
= 2(213.6 J/K mol) + 3(70.0 J/K mol) – { 160.7 J/K mol + 205.0 J/K mol } = -138.5 J/ K mol
ΔGo = 2 ΔGfo [CO2 (g)] + 3 ΔGfo [H2O (l)] - {ΔGfo [CH3CH2OH(l)] + 3 ΔGfo [O2 (g)] }
= 2(-394.4 kJ/mol) + 3(-273.2 kJ/mol) – {-174.9 kJ/mol + 3(0.0 kJ/mol)} = -1325.5 kJ/mol
OR: ΔGo = ΔHo -T ΔSo.
At 25o C = 298.15 K
ΔGo = -1366.7 kJ/mol – 298.15 K(-138.5 J/ K mol)(1 kJ/1000J) = 1325.5 kJ/mol
40 a. NaHCO3(S)
 NaOH(s) + CO2(g)
At 25oC (298 K) :
ΔGo = ΔGfo [CO2 (g)] + ΔGfo [NaOH (s)] - {ΔGfo [NaHCO3(s)]}
= (-379.5 kJ/mol) + (-394.4 kJ/mol) – {-851.0 kJ/mol} = +77.1 kJ/mol
Since ΔGo is +, the reaction is NOT spontaneous at 25oC
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41 a. CaCO3(s)  CaO(s) + CO2(g)
Kp = PCO2
To get the value of Kp, get ΔGo:
At 25oC (298 K) :
ΔGo = ΔGfo [CaO (s)] + ΔGfo [CO2 (g)] - {ΔGfo [CaCO3(s)]}
= (-603.3 kJ/mol) + (-394.4 kJ/mol) – {-1128.8 kJ/mol} = +131.1 kJ/mol
ΔGo is +, therefore the reaction is NOT SPONTANEOUS at 25o C (298.15 K)
ΔGo = -RTlnKp
lnKp =- ΔGo /RT = 131.1 kJ/mole(1000 J/kJ)/[(8.3145 J/K mol)(298.15 K) = -52.885
Kp = e- ΔGo /RT = e-52.885 = 1.08 x 10-23. Pretty teeny.
They want to compare Kp with Q, the reaction quotient. Q is just the concentrations of reactant and
product, regardless of whether they are in equilibrium or not.
Q = Pco2 = 1 atm.
Q >>> K and we are FAR from equilibrium.
b. What temperature must we be at so that Pco2 = 1 atm and we are at equilibrium?
ΔGo = ΔHo -T ΔSo , and at eq, ΔGo = -RTlnKp
-RTlnKp = ΔHo -TΔSo
-RTlnKp + TΔSo = ΔHo
T(-RlnKp + ΔSo ) = ΔHo
T = ΔHo/( ΔSo - RlnKp)
Now all we have to do is get ΔHo and ΔSo
ΔHo = ΔHfo [CaO (s)] + ΔHfo [CO2 (g)] - {ΔHfo [CaCO3(s)]}
= (-634.9 kJ/mol) + (-393.5kJ/mol) – {-1206.9kJ/mol} = +178.5 kJ/mol = 178500 J/mol
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ΔSo = Sfo [CaO (s)] + Sfo [CO2 (g)] - {Sfo [CaCO3(s)]}
= (38.1 J/K mol) + (213.6 J/K mol) – {92.9 J/K mol} = +158.8 J/K mol
Kp (new)= 1 atm
T = ΔHo/( ΔSo - RlnKp) = 178500 J/mol/(158.8 J/K mol – 8.3145 j/K mol x ln(1 atm)
= 178500 J/mol/(158.8 J/K mol – 8.3145 j/K mol x 0) = 1124 K (851 oC)
43. CH3COOH(aq) 
H+(aq) + CH3COO -(aq) . Ka = 1.8 x 10-5
At eq, ΔGo = -RTlnKa = -(8.3145 J/K mol)(298.15 K)ln(1.8 x 10-5) = 27000 J/mol
ΔGo is +, this rx is NOT spontaneous at 25oC. HAc is a WEAK acid.
For a lark, consider
HCl
(aq)
 H+(aq) + Cl -(aq) . Ka  1.0 x 106
At eq, ΔGo = -RTlnKa = -(8.3145 J/K mol)(298.15 K)ln(1.0 x 106) = - 34288 J/mol
ΔGo is -, this rx IS spontaneous at 25oC. HCl is a STRONG acid.