Advanced Placement Examination in Chemistry

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The Advanced Placement Examination in Chemistry

Part II - Free Response Questions & Answers
1970 to 2007

Thermodynamics

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Advanced Placement Examination in Chemistry. Questions copyright© 1970-2007 by the College Entrance Examination Board, Princeton, NJ 08541. Reprinted with permission. All rights reserved. apcentral.collegeboard.com. This material may not be mass distributed, electronically or otherwise. This publication and any copies made from it may not be resold.

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Consider the first ionization of sulfurous acid:
H2SO3(aq) ( H+(aq) + HSO3-(aq)
Certain related thermodynamic data are provided below:
H2SO3(aq)H+(aq)HSO3-(aq)
---------------------------------------------
Hf( kcal/mole-145.50-151.9
S° cal/mole K56026
(a)Calculate the value of ΔG° at 25°C for the ionization reaction. (b)Calculate the value of K at 25°C for the ionization reaction. (c)Account for the signs of ΔS° and ΔH° for the ionization reaction in terms of the molecules and ions present. Answer:

(a)[pic]
= [-159.9] - [-145.5] kcal = -14.4 kcal
[pic]
= (26 - 56) cal = -30 cal/K
ΔG° = ΔH - TΔS = -14400 - (298)(-30) cal
= -5.46 kcal
(b)K = e-ΔG/RT = e-(-5460/(1.9872)(298)) = 10100
(c)

1971
Given the following data for graphite and diamond at 298K.
S°(diamond) =0.58 cal/mole deg
S°(graphite) =1.37 cal/mole deg
ΔHf° CO2(from graphite) =-94.48 kilocalories/mole
ΔHf° CO2(from diamond) =-94.03 kilocalories/mole
Consider the change: C(graphite) = C(diamond) at 298K and 1 atmosphere. (a)What are the values of ΔS° and ΔH° for the conversion of graphite to diamond. (b)Perform a calculation to show whether it is thermodynamically feasible to produce diamond from graphite at 298K and 1 atmosphere. (c)For the reaction, calculate the equilibrium constant Keq at 298K Answer:

(a)ΔS° = S°(dia.) - S°(graph.) = (0.58 - 1.37) cal/K = -0.79 cal/K CO2 ( C(dia.) + O2ΔH = + 94.03 kcal/mol
C(graph.) + O2 → CO2ΔH = - 94.48 kcal/mol
C(graph.) → C(dia.)ΔH = -0.45 kcal/mol
(b)(G° = ΔH° - TΔS° = -450 - (298)(-0.79) cal
= -223.52 cal/mol; a ΔG° < 0 indicates feasible conditions (c)Keq = e-ΔG/RT = e-(-223.52/(1.9872)(298)) = -0.686

1972
Br2 + 2 Fe2+(aq) → 2 Br-(aq) + 2 Fe3+(aq)
For the reaction above, the following data are available:
2 Br-(aq) → Br2(l) + 2e-E° =-1.07 volts
Fe2+(aq) → Fe3+(aq) + e-E° =-0.77 volts
S°, cal/mole °C
Br2(l)58.6Fe2+(aq)-27.1
Br-(aq)19.6Fe3+(aq)-70.1
(a) Determine ΔS°
(b) Determine ΔG°
(c) Determine ΔH°
Answer:
(a)[pic]
= [(19.6)(2)+(-70.1)(2)]-[58.6+(-27.1)(2)] cal
= -105.4 cal = -441 J/K
(b)E°cell = [+1.07 + (-0.77)] v = 0.30 v
ΔG°=-n(E°=-(2)(96500)(0.30v)=-57900 J/mol
(c)ΔH° = ΔG° + TΔS° = 57900 + 298(-441) J
= -73.5 kJ/mol

1974
WO3(s) + 3 H2(g) → W(s) + 3 H2O(g)
Tungsten is obtained commercially by the reduction of WO3 with hydrogen according to the equation above. The following data...
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