The Advanced Placement Examination in
Chemistry
Part II - Free Response Questions & Answers
1970 to 2008
Thermodynamics
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Advanced Placement Examination in Chemistry. Questions
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Questions With ANSWERS
1979 B
Hf
S
Compound (kilocalories/mole) (calories/mole K)
H2O(l)
-68.3
16.7
CO2(g)
-94.1
51.1
O2(g)
0.0
49.0
C3H8
?
64.5
When 1.000 gram of propane gas, C3H8, is burned at 25C and 1.00 atmosphere, H2O(l) and
CO2(g) are formed with the evolution of 12.03 kilocalories.
(a) Write a balanced equation for the combustion reaction.
(b) Calculate the molar enthalpy of combustion, Hcomb, of propane.
(c) Calculate the standard molar enthalpy of formation, Hf, of propane gas.
(d) Calculate the entropy change, Scomb, for the reaction and account for the sign Scomb.
Answer:
(a) C3H8 + 5 O2(g)  3 CO2(g) + 4 H2O(l)
(b)
12.03 kcal 44.10 g
H comb. 

 530.5 kcal mol
1.000 g
1 mol
This answer should be -529.3 kcal/mol




(c) H comb.  3 H f CO 2  4 H f H 2 O  H f C 3 H 8
-529.3 kcal = [3(-94.1) + 4(-68.3) - X] kcal
Hf. = -26.2 kcal/mol
(d) S
= [3(51.1) + 4(16.7)] - [64.5 + 5(49.0)]
comb. [ 3 S _CO 2  4 S _H 2 O ] [ S_C 3 H 8  5 S _O 2 ]
= -89.4 cal/mol.K
Entropy decreases due to loss of highly disordered gaseous species upon combustion.
1981 D
PCl5(g)  PCl3(g) + Cl2(g)
For the reaction above, H = +22.1 kilocalories per mole at 25C
(a) Does the tendency of reactions to proceed to a state of minimum energy favor the formation
of the products of this reaction? Explain
Answer:
(a) No, since reaction is endothermic, the products must be at higher energy than the reactants.
OR
ln KP = -H/RT + constant; if H>0, ln KP is less than if H<0. OR
G = H - TS. Low free energy (G0) is not favored by H>0.
1984 B
Substance
Standard Heat of
Formation, Hf,
in kJ mol-1
Absolute
Entropy, S,
in J mol-1 K-1
------------------
-----------------------------
------------------------
C(s)
0.00
5.69
CO2(g)
-393.5
213.6
H2(g)
0.00
130.6
H2O(l)
-285.85
69.91
O2(g)
0.00
205.0
C3H7COOH(l)
?
226.3
The enthalpy change for the combustion of butyric acid at 25C, Hcomb, is -2,183.5 kilojoules
per mole. The combustion reaction is
C3H7COOH(l) + 5 O2(g)  4 CO2(g) + 4 H2O(l)
(a) From the above data, calculate the standard heat of formation, Hf, for butyric acid.
(b) Write a correctly balanced equation for the formation of butyric acid from its elements.
(c) Calculate the standard entropy change, Sf, for the formation of butyric acid at 25C. The
entropy change, S, for the combustion reaction above is -117.1 J K-1 at 25C.
(d) Calculate the standard free energy of formation, Gf, for butyric acid at 25C.
Answer:
_
(a) H H_
f (products ) Hf (reactants ) = [4(393.5) + 4(285.85) - 2183.5] kJ = -533.8 kJ
(b) 4 C(s) + 4 H2(g) + O2(g)  C3H7COOH(l)
(c) Sf (butyric acid) = S(butyric acid) - [4 S(C) + 4 S(H2) + S(O2)] = 226.3 -[4(5.69) + 4(130.6) +
205] = -523.9 J/K
(d) Gf = Hf - TSf = 533.8 - (298)(-0.5239) kJ = -377.7 kJ
1985 D
(a) When liquid water is introduced into an evacuated vessel at 25C, some of the water
vaporizes. Predict how the enthalpy, entropy, free energy, and temperature change in the
system during this process. Explain the basis for each of your predictions.
(b) When a large amount of ammonium chloride is added to water at 25C, some of it dissolves
and the temperature of the system decreases. Predict how the enthalpy, entropy, and free
energy change in the system during this process. Explain the basis for each of your
predictions.
(c) If the temperature of the aqueous ammonium chloride system in part (b) were to be
increased to 30C, predict how the solubility of the ammonium chloride would be affected.
Explain the basis for each of your predictions.
Answer:
(a) H>0 since heat is required to change liquid water to vapor
S>0 since randomness increases when a liquid changes to vapor.
G<0 since the evaporation takes place in this situation.
T<0 since the more rapidly moving molecules leave the liquid first. The liquid remaining
is cooler.
(b) H>0. The system after dissolving has a lower temperature and so the change is
endothermic.
S>0, since the solution is less ordered than the separate substances are.
G<0. The solution occurred and so is spontaneous.
(c) Solubility increases. The added heat available pushes the endothermic process toward more
dissolving.
1988 B
Enthalpy of
Combustion, H
Absolute
Entropy, S
Substance
(kiloJoules/mol)
(Joules/mol-K)
C(s)
-393.5
5.740
H2(g)
-285.8
130.6
C2H5OH(l)
-1366.7
160.7
H2O(l)
-69.91
(a) Write a separate, balanced chemical equation for the combustion of each of the following:
C(s), H2(g), and C2H5OH(l). Consider the only products to be CO2 and/or H2O(l).
(b) In principle, ethanol can be prepared by the following reaction:
2 C(s) + 2 H2(g) + H2O(l)  C2H5OH(l)
Calculate the standard enthalpy change, H, for the preparation of ethanol, as shown in the
reaction above.
(c) Calculate the standard entropy change, S, for the reaction given in part (b).
(d) Calculate the value of the equilibrium constant at 25C for the reaction represented by the
equation in part (b).
Answer:
(a) C + O2  CO2
2 H2 + O2  2 H2O
C2H5OH + 3 O2  2 CO2 + 3 H2O
(b) 2 C + 2 O2  2 CO2
H = 2(-393.5) = -787.0 kJ
2 H2 + O2  2 H2O
H = 2(-285.8) = -571.6 kJ
2 CO2 + 3 H2O  C2H5OH + 3 O2
H = -(-1366.7) = +1366.7 kJ
2 C + 2 H2 + H2O  C2H5OH
H = +8.1 kJ
OR
Hcomb. C(s) = Hf CO2(g)
Hcomb. H2(g) = Hf H2O(l)
C2H5OH + 3 O2  2 CO2 + 3 H2O
H = -1366.7 kJ
_
H H_
f (products ) Hf (reactants ) = [2(-393.5) + 3(-258.8)] - [Hf
C2H5OH
+ 0] kJ = -
277.7 kJ/mol
2 C + 2 H2 + H2O  C2H5OH
_
H H_
f (products ) Hf (reactants ) = [-277.7] - [0 + 0 + (-285.8)] kJ = +8.1 kJ
(c)  S   S(products) S(reactants)
= [160.7] - [11.5 + 261.2 + 69.9] J/mol.K = -181.9 J/mol.K

(d) G = H - TS = 8100 -(298)(-181.9) J = 62300 J
Keq = e-G/RT = e-(62300/(8.3143)(298)) = 1.210-11
1988 D
An experiment is to be performed to determine the standard molar enthalpy of neutralization of a
strong acid by a strong base. Standard school laboratory equipment and a supply of standardized
1.00 molar HCl and standardized 1.00 molar NaOH are available.
(a) What equipment would be needed?
(b) What measurements should be taken?
(c) Without performing calculations, describe how the resulting data should be used to obtain
the standard molar enthalpy of neutralization.
(d) When a class of students performed this experiment, the average of the results was -55.0
kilojoules per mole. The accepted value for the standard molar enthalpy of neutralization of
a strong acid by a strong base is -57.7 kilojoules per mole. Propose two likely sources of
experimental error that could account for the result obtained by the class.
Answer:
(a) Equipment needed includes a thermometer, and a container for the reaction, preferably a
container that serves as a calorimeter, and volumetric glassware (graduated cylinder, pipet,
etc.).
(b) Measurements include the difference in temperatures between just before the start of the
reaction and the completion of the reaction, and amounts (volume, moles) of the acid and
the base.
(c) Determination of head (evolved or absorbed): The sum of the volumes (or masses) of the
two solutions, and change in temperature and the specific heat of water are multiplied
together to determine the heat of solution for the sample used. q = (m)(cp)(T).
Division of the calculated heat of neutralization by moles of water produced, or moles of H+,
or moles of OH-, or moles of limiting reagent.
(d) Experimental errors: heat loss to the calorimeter wall, to air, to the thermometer; incomplete
transfer of acid or base from graduated cylinder; spattering of some of the acid or base so
that incomplete mixing occurred, … Experimenter errors: dirty glassware, spilled solution,
misread volume or temperature, …
1990 B
Substance
C2H4Cl2(g)
C2H5Cl(g)
HCl(g)
Cl2(g)
Standard Free Energies of
Formation at 298 K
Gf 298 K, kJ mol-1
-80.3
-60.5
-95.3
0
Average Bond Dissociation
Energies at 298 K
Bond
Energy, kJ mol-1
C-H
414
C-C
347
C-Cl
377
Cl-Cl
243
H-Cl
431
The tables above contain information for determining thermodynamic properties of the reaction
below.
C2H5Cl(g) + Cl2(g)  C2H4Cl2(g) + HCl(g)
(a) Calculate the H for the reaction above, using the table of average bond dissociation
energies.
Answer:
(a) H = energy of bonds broken - energy of bonds formed
C2H5Cl + Cl2  C2H4Cl2 + HCl
H = (2794 + 243) - (2757 + 431) kJ = -151 kJ
OR
CH + Cl-Cl  C-Cl + HCl (representing the changes)
H = (414) + 243) - (377 + 431) = -151 kJ
1995 B
Propane, C3H8, is a hydrocarbon that is commonly used as fuel for cooking.
(a) Write a balanced equation for the complete combustion of propane gas, which yields CO2(g)
and H2O(l).
(b) Calculate the volume of air at 30C and 1.00 atmosphere that is needed to burn completely
10.0 grams of propane. Assume that air is 21.0 percent O2 by volume.
(c) The heat of combustion of propane is -2,220.1 kJ/mol. Calculate the heat of formation, Hf,
of propane given that Hf of H2O(l) = -285.3 kJ/mol and Hf of CO2(g) = -393.5 kJ/mol.
(d) Assuming that all of the heat evolved in burning 30.0 grams of propane is transferred to 8.00
kilograms of water (specific heat = 4.18 J/g.K), calculate the increase in temperature of
water.
Answer:
(a) C3H8 + 5 O2  3 CO2 + 4 H2O
(b) 10.0 g C3H8  1 mol C3H8/44.0 g  5 mol O2/1 mol C3H8) = 1.14 mol O2
VO
(c)
2

o
nRT
P

1.14 mol 0 . 0821
L a t m
m ol K
303 K 
1 . 00 atm
o
o
o
= 28.3 L O2 ; f(28.3 L,21.0%) = 135 L of air
o
H comb  H f (CO )  H f (H O ) H f (C H )  H f (O ) 
2
2
3 8
2
-2220.1 = [3(-393.5) + 4(-285.3)] - [X+ 0]
X = Hcomb) = -101.6 kJ/mol
(d) q = 30.0 g C3H8  1 mol/44.0 g  2220.1 kJ/1 mol = 1514 kJ
q = (m)(Cp)(T)
1514 kJ = (8.00 kg)(4.184 J/g.K)(T)
T = 45.2
1996 B
C2H2(g) + 2 H2(g)  C2H6(g)
Information about the substances involved in the reaction represented above is summarized in
the following tables.
Substance
S (J/molK)
Hf (kJ/mol)
C2H2(g)
200.9
226.7
H2(g)
130.7
0
C2H6(g)
----
-84.7
Bond
Bond Energy (kJ/mol)
C-C
347
C=C
611
C-H
414
H-H
436
(a) If the value of the standard entropy change, S, for the reaction is -232.7 joules per
moleKelvin, calculate the standard molar entropy, S, of C2H6 gas.
(b) Calculate the value of the standard free-energy change, G, for the reaction. What does the
sign of G indicate about the reaction above?
(c) Calculate the value of the equilibrium constant, K, for the reaction at 298 K.
(d) Calculate the value of the CC bond energy in C2H2 in kilojoules per mole.
Answer:
(a) -232.7 J/K = S(C2H6) - [2(130.7) + 200.9] J/K
S(C2H6) = 229.6 J/K
(b) H =  Hâ
-  Hâ
= -84.7 kJ - [226.7 + 2(0)] kJ = -311.4 kJ
G = H - TS = -311.4 kJ - (298K)(-0.2327 kJ/K) = -242.1 kJ
A G < 0 (a negative G) indicates a spontaneous forward reaction.
(c) Keq = e-G/RT = e-(-242100/(8.314)(298)) = 2.741042
(d) H = bond energy of products - bond energy of reactants
-311.4 kJ = [(2)(436) + Ecc + (2)(414)] - [347 + (6)(414)] kJ
E = 820 kJ
(products)
(reactants)
2001 B
2 NO(g) + O2(g)  2 NO2(g) H°= -114.1 kJ, S°= -146.5 J K-1
The reaction represented above is one that contributes significantly to the formation of
photochemical smog.
(a) Calculate the quantity of heat released when 73.1 g of NO(g) is converted to NO2(g).
(d) Use the data in the table below to calculate the bond energy, in kJ mol-1, of the nitrogenoxygen bond in NO2 . Assume that the bonds in the NO2 molecule are equivalent (i.e., they
have the same energy).
Bond Energy
(kJ mol-1)
Nitrogen-oxygen bond in NO
607
Oxygen-oxygen bond in O2
495
Nitrogen-oxygen bond in
NO2
?
Answer:
1 mol NO 114.1 kJ
(a) 73.1 g  30.007 g  2 mol NO = 139 kJ
(d) 2 NO(g) + O2(g)  2 NO2(g) + 114.1 kJ
H = enthalpy of bonds broken – enthalpy of bonds formed
-114.1 = [(607)(2) + 495] - 2X
X = 912 kJ / 2 N=O bonds
456 kJ = bond energy for N=O bond
2002 D Required (repeated in lab procedures)
A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction
represented above. The student combines equal volumes of 1.0 M HCl and 1.0 M NaOH in an
open polystyrene cup calorimeter. The heat released by the reaction is determined by using the
equation q = mc∆T.
Assume the following.
• Both solutions are at the same temperature before they are combined.
• The densities of all the solutions are the same as that of water.
• Any heat lost to the calorimeter or to the air is negligible.
• The specific heat capacity of the combined solutions is the same as that of water.
(a) Give appropriate units for each of the terms in the equation q = mc∆T.
(b) List the measurements that must be made in order to obtain the value of q.
(c) Explain how to calculate each of the following.
(i) The number of moles of water formed during the experiment
(ii) The value of the molar enthalpy of neutralization, ∆Hneut, for the reaction between
HCl(aq) and NaOH(aq)
(d) The student repeats the experiment with the same equal volumes as before, but this time
uses 2.0 M HCl and 2.0 M NaOH.
(i) Indicate whether the value of q increases, decreases, or stays the same when compared
to the first experiment. Justify your prediction.
(ii) Indicate whether the value of the molar enthalpy of neutralization, ∆Hneut, increases, decreases, or stays the same when compared to the first experiment. Justify your
prediction.
(e) Suppose that a significant amount of heat were lost to the air during the experiment. What
effect would this have on the calculated value of the molar enthalpy of neutralization,
∆Hneut? Justify your answer.
Answer:
(a) q in J, m in grams, C in J/g˚C, T in ˚C
(b) mass or volume of each solution
starting temperature of each reagent
ending temperature of mixture
(c) (i) both are 1 M acid and base and react on a 1:1 basis
1 mol HCl
1 mol H+
volume  1000 mL  1 mol HCl = mol of H+
H+ + OH–  H2O
joules released
(ii) mol H O produced
2
(d) (i) increases. Twice as much water is produced so it is twice the energy released in the same
volume of solution
twice energy
(ii) same. twice mol water = same result
(e) smaller. heat lost to the air gives a smaller amount of temperature change in the solution,
which leads to a smaller measured heat release