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AP Chemistry 6: Thermodynamics
A.
Enthalpy (H): Bond Energy (5.3 to 5.5, 8.8)
1. chemical reactions typically involve breaking bonds
between reactant atoms and forming new bonds
2. breaking bonds takes energy  chemical system gains
bond energy; surroundings lose energy (heat, etc.)
3. forming bonds releases energy  chemical system
loses energy, surroundings gain energy
4. change in energy called “change in enthalpy”—H
a. when energy required to break bonds > energy
released to form new bonds, +H (endothermic)
1. products at a higher energy state than
reactants (weaker bonds)
2. surroundings lose energy (cool down)
b. when energy required to break bonds < energy
released to form new bonds, –H (exothermic)
1. products at a lower energy state than
reactants (stronger bonds)
2. surroundings gain energy (heat up)
5. thermochemical equation
a. chemical equation with H
1. listed to the right of equation
2. included as reactant (endothermic) or product
(exothermic)
b. H can be used in dimensional analysis process
6. H from calorimetry
a. reactants are put in an insulated container filled
with water, where heat is exchanged between
reactants and water, but no heat is lost
b. by conservation of energy: Hreaction = –Qwater
1. Q = mcT for simple coffee cup calorimeter—
aqueous reactions
a. m = mass of water
b. c = specific heat of water (4.18 J/g•K)
c. T = change in temperature (Tf – Ti)
temperature can stay in oC, since
1 oC = 1 K (don't add 273 to ToC!)
2. Q = (C + mc)T for “bomb" calorimeter
a. C = “bomb constant” accounts for all
non-water components that change
temperature
b. all other letters are the same as the
simple calorimeter
7. H using bond energy (B.E.) data
Bond Energies in (kJ/mol)
Single
Multiple
H C N O S
F Cl Br
I C=C 614
H 436 413 391 463 339 567 431 366 299 C=N 615
C
348 293 358 259 485 328 276 240 C=O 799
N
163 201
272 200 243
N=N 418
O
146
190 203
243 N=O 607
S
266 327 253 218
O=O 495
F
155 253 237
O=S 523
Cl
242 218 208 CC 839
Br
193 175 CN 891
151 CO 1072
I
NN 941
S=S 418
a. energy needed to break a bond (i.e. C–H) in a
diatomic, gaseous molecule, which contains the
bond type
1. is approximately the same for any molecule
2. affected by molecular bonding  only works
for gaseous species
3. positive value (+ B.E.) for breaking bonds
b. forming bonds (– B.E.)
c. H = B.E.reactants – B.E.products
Name __________________________
B.
C.
D.
Entropy (S): Disorder (19.2)
1. atoms/molecules have inherent disorder depending on
a. number of atoms—more internal motion = disorder
b. spacing of molecules—farther apart = disorder
c. speed of molecules—faster = disorder
2. predict increase in disorder for physical changes (+S)
a. spread out: evaporation, diffusion and effusion
(solution: spread out solute and solvent (+S), but
bond solute-solvent (-S)  ?, but usually +S)
b. motion: melting and boiling
3. predict increase in disorder for chemical changes (+S):
moles gaseous products > moles gaseous reactants
Thermodynamic Data (5.6 to 5.7, 19.4)
So (kJ/mol•K)
Species
Hfo (kJ/mol)
Al
0.0
+0.0283
Al3+
-531.0
-0.3217
Al2O3
-1675.7
+0.0509
1. standard heat of formation (Hfo) data
a. Ho for the formation of one mole of compound
from its elements at standard temperature (25oC)
Al: Al(s)  Al(s)  no reaction
Al3+: Al(s)  Al3+ + 3 eAl2O3: 2 Al(s) + 3/2 O2(g)  Al2O3(s)
b. Hfo for elements in natural state = 0.0
c. more negative = more stable (harder to decompose)
2. standard entropy (So) data
a. amount of disorder compared to H+ (simplest form
of matter), which is zero by definition
b. listed in J/mol•K on AP exam, so you will have to
convert to kJ/mol•K for most calculations
3. calculations using the thermodynamic data chart
a. altering Hfo
1. opposite sign for the reverse reaction
C + 2 Cl2  CCl4 = –139.4 kJ
 CCl4  C + 2 Cl2 = +139.4 kJ
2. multiply by number of moles (coefficient)
1 mole CCl4= –139.4 kJ
 2 mole CCl4 = –278.8 kJ
b. calculate H for a reaction using Hfo
1. Hess’s Law: H for a multi-step reaction equals
the sum of H for each step
-(-74.8)
CH4(g)  C + 2 H2
-393.5
C + O2  CO2(g)
2(-241.8)
+ 2 H2 + O2  2 H2O(g)
CH4(g) + 2 O2  CO2(g) + 2 H2O(g) -802.3
2. H Ho = Hfoproducts – Hforeactants
c. calculate S for a reaction using So

S So = Soproducts – Soreactants
Gibbs Free Energy (G): Overall Energy State (19.5 to 19.6)
1. combination of enthalpy and entropy: G = H + TS
2. for a chemical or physical change: Go = Ho – ToSo
a. To = 298 K
b. where T  298 K:G  Ho – TSo
3. determining if a process is spontaneous (G < 0)
a. lower potential energy (-H)—chemical reactions
b. greater disorder (+S)—physical changes
c. depends on temperature
1. threshold temperature (Tthreshold)
2. occurs when G = 0  Tthreshold = Ho/So
d. summary chart
H
S
Spontaneous Process (G <0)
for temperatures above Tthreshold
+
+
+
–
at no temperatures
–
+
at all temperatures
for temperatures below Tthreshold
–
–
d.
Temperature (oC)
Heat of Reaction Lab—Use calorimetry to determine H
for a series of reactions, compare the results with
thermodynamic data, and combine the results to verify
Hess' law.
Heat about 75 mL of water to about 70oC. Place a
Styrofoam cup in a 250-mL beaker. Add 50.0 mL cold tap
water to the cup. Record the temperature TC. Measure out
50.0 mL of the hot water and place in a second Styrofoam
cup. Record the temperature TH. Pour the hot water into the
cold water, cover the cup, insert the thermometer in the
hole, and mix gently. Record the temperature every 20
seconds for 3 minutes. Discard the water.
a. (1) Record the temperatures.
TC
TH
time (s) 20
40 60 80 100 120 140 160 180
To C
(2) Graph the temperature vs. time data. Draw a best
fit straight line (use a ruler).
20
60
100
140
180
Time (s)
(3) Use the y-intercept to determine Tmix.
Tmix (y-intercept)
b. Calculate the following.
(1) Average of the hot and cold temperatures.
Tav = (TH – TC)/2
(2) Heat lost from the water.
QL = mc(Tav – Tmix)
(3) C.
C = QL/(Tmix – TC)
Place a Styrofoam cup in a 250 mL beaker. Add 50.0 mL of
3.00 M NaOH. Record the temperature To. Pour 50.0 mL of
3.00 M HCl into the NaOH, cover, insert the thermometer,
and mix gently. Record the temperature every 20 seconds
for 3 minutes. Discard the mixture.
c. (1) Record the temperatures.
To
time (s) 20
40 60 80 100 120 140 160 180
ToC
(2) Graph the temperature vs. time data. Draw a best
fit straight line (use a ruler).
Temperature (oC)
1.
20
60
100
140
Time (s)
(3) Use the y-intercept to determine Tmix.
Tmix (y-intercept)
180
Calculate Hreaction per mole of reactant based on the
calorimetry data.
T (K)
Qwater (kJ)
Hreaction/mole
e.
Calculate Hreaction per mole of reactant based on Hfo.
OH-(aq) + H+(aq)  H2O(l)
H
%
Place a Styrofoam cup in a 250 mL beaker. Add 50.0 mL of
3.00 M NH4Cl. Record the temperature To. Pour 50.0 mL of
3.00 M NaOH into the NH4Cl, cover, insert the thermometer,
and mix gently. Record the temperature every 20 seconds
for 3 minutes. Discard the mixture.
f. (1) Record the temperatures.
To
time (s) 20
40 60 80 100 120 140 160 180
ToC
(2) Graph the temperature vs. time data. Draw a best
fit straight line (use a ruler).
Temperature (oC)
Experiments
20
100
140
Time (s)
(3) Use the y-intercept to determine Tmix.
Tmix (y-intercept)
g.
60
180
Calculate Hreaction per mole of reactant based on the
calorimetry data.
T (K)
Qwater (kJ)
Hreaction/mole
h.
Calculate Hreaction per mole of reactant based on Hfo.
NH4+(aq) + OH-(aq)  NH3(aq) + H2O(l)
H
%
Place a Styrofoam cup in a 250 mL beaker. Add 50.0 mL of
3.00 M NH3. Record the temperature To. Pour 50.0 mL of
3.00 M HCl into the NH3, cover, insert the thermometer, and
mix gently. Record the temperature every 20 seconds for 3
minutes. Discard the mixture.
i. (1) Record the temperatures.
To
time (s) 20
40 60 80 100 120 140 160 180
ToC
4.
Temperature (oC)
(2) Graph the temperature vs. time data. Draw a best
fit straight line (use a ruler).
20
100
140
Time (s)
(3) Use the y-intercept to determine Tmix.
Tmix (y-intercept)
j.
60
b.
the number of moles of MgSO4 dissolved.
c.
H (in kJ) for the dissolving of one mole of MgSO4.
180
Calculate Hreaction per mole of reactant based on the
calorimetry data.
5.
H2(g) + F2(g)  2 HF(g)
Estimate H for the reaction using the bond energy values.
6.
C2H2(g) + 2 H2(g)  C2H6(g)
Estimate H for the reaction using the bond energy values.
7.
A bomb calorimeter with a constant of 921 J/oC contains
1,000 g of water. The combustion of 1.00 g of ethene
(C2H4) increases the temperature 9.3oC. Determine
a. Qwater for the combustion process.
T (K)
Qwater (kJ)
Hreaction/mole
k.
12.8 g of MgSO4 is dissolved in 250. g of H2O in a coffee
cup calorimeter. The temperature of the solution increases
from 23.8oC to 33.1oC. Determine
a. Qwater for the solution process.
Calculate Hreaction per mole of reactant based on Hfo.
NH3(aq) + H+(aq)  NH4+(aq)
H
b.
the number of moles of ethene reacted.
c.
H (in kJ) for the combustion of one mole of C2H4.
d.
Write the equation for the combustion of ethene (C2H4).
e.
Calculate H using bond energies.
%
l.
Show that the chemical equations and H from
part (d) – part (g) = part (j).
Practice Problems
1.
a.
b.
2.
8.
How many grams of iron are needed to generate
1.00 x 104 kJ of heat?
CaSO4(s) + CO2(g)  CaCO3(s) + SO3(g)
H = 224 kJ
a. How much heat is absorbed when 10.0 g CaSO4 react.
b.
3.
A. Enthalpy
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
H = -1640 kJ
How much heat is released to produce 10.0 g Fe2O3?
9.
B. Entropy
Predict whether S > 0, S < 0 or S 0.
>0
 0 < 0
Melting ice at 0oC
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Distilling alcohol-water mixture
C. Thermodynamic Data
2 Na2O2(s) + 4 HCl(g)  4 NaCl(s) + 2 H2O(l) + O2(g)
Determine H from the thermochemical reactions below.
2 Na2O2(s) + 2 H2O(l) 
s) + O2(g) H1 = -126 kJ
NaOH(s) + HCl(g) 
s) + H2O(l)
H2 = -179 kJ
How much CaCO3 is produced when 500. kJ is absorbed?
10.
C2H2(g) + 5 N2O(g)  2 CO2(g) + H2O(g) + 5 N2(g)
Determine H from the thermochemical equations below.
2 C2H2(g) + 5 O2(g) 
2(g) + 2 H2O(g) H1 = -2512 kJ
N2(g) + ½ O2(g) 
H2 = 104 kJ
2O(g)
11.
NO(g) + O(g)  NO2(g)
Determine H for the above reaction using the following
thermochemical equations.
NO(g) + O3(g)  NO2(g) + O2(g)
H1 = -198.9 kJ
O3(g)  3/2 O2(g)
H2 = -142.3 kJ
O2(g)  2 O(g)
H3 = 495.0 kJ
When 1.51 g of NH4Cl are dissolved in 100. g of water the
temperature drops 1.00oC. Determine
a. Qwater for the solution process.
b.
the number of moles of NH4Cl dissolved.
c.
H (in kJ) for the dissolving of one mole of NH4Cl.
12. a.
b.
Write the equation for the combustion of methanol,
CH3OH(l). (other reactants and products are gaseous).
13.
C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
Calculate Ho.
b.
Calculate So.
c.
Calculate G at 20oC.
d.
At which temperature (if any) will the reaction be
spontaneous?
Calculate H using Hfo values.
1.00 g of methanol is burned in a bomb calorimeter that
contains 1200 g of water. The temperature increases 3.4 K.
c. Calculate the heat generated by the combustion reaction.
d.
a.
17.
Calculate the calorimeter constant of the bomb.
Ca(s) + SO3(g) + 2 H2O(l)  CaSO3•2 H2O(s)
H = -795 kJ and S = -0.2535 kJ/K for the reaction.
a. Calculate Hfo for CaSO3•2 H2O.
18. When H2SO4(l) is dissolved in water, the temperature of
the mixture increases. Predict the sign of H, S and G
for this process (justify your answer).
+/–
Justification
H
S
b.
Calculate So for CaSO3•2 H2O.
G
19.
D. Gibbs Free Energy
14. Consider the reaction at 25oC:
Cu(s) + 4 H+(aq) + 2 NO3-(aq)  Cu2+(aq) + 2 NO2(g) + 2 H2O(l).
a. Calculate Ho using Hfo values.
15.
b.
Calculate So using So values.
c.
Calculate Go.
NH4NO3(s)  NH4+(aq) + NO3-(aq)
Determine the following for the above reaction.
a. Is the reaction exothermic or endothermic?
b.
Is there an increase or decrease in entropy?
c.
Is the reaction spontaneous at 25oC?
a.
2 SO2(g) + O2(g)  2 SO3(g)
Calculate Ho.
b.
Calculate So.
c.
Calculate G at 400 K.
d.
Determine the temperature range where the reaction
is spontaneous.
16.
C2H5OH(l)  C2H5OH(g)
Calculate the boiling point (threshold temperature) given
the information: H = 37.95 kJ and S = 0.1078 kJ/K.
Summary
Change in Enthalpy (H)
Chemical reactions typically involve breaking some bonds
between reactant atoms and forming new bonds. Breaking
bonds absorbs energy, therefore the chemical system gains
bond energy and the surroundings lose energy, typically in the
form of heat. In contrast, forming bonds releases energy;
resulting in lose of energy by the chemical system and a gain in
energy by the surroundings (also in the form of heat).
When energy required to break bonds is greater than the
energy released to form new bonds, then products are at a
higher energy state than reactants (making the product bonds
weaker than the reactant bonds) and energy of the system
increases (+H), which is described as endothermic because
the surroundings typically lose heat energy and cool down.
Alternatively, when energy required to break bonds is less than
the energy released to form new bonds, then products are at a
lower energy state than reactants (making the product bonds
stronger than the reactant bonds) and energy of the system
decreases, –H, which is described as exothermic because the
surroundings typically gain heat energy and warm up. The
change in enthalpy, H, is listed to the right of a balanced
chemical equation. H can be treated in the same way as a
coefficient when using dimensional analysis.
The amount of heat transferred between the system and the
surroundings is measured experimentally by calorimetry. A
calorimeter measures the temperature change accompanying a
process. The temperature change of a calorimeter depends on
its heat capacity, the amount of heat required to raise its
temperature by 1 K. The heat capacity for one mole of a pure
substance is called its molar heat capacity; for one gram of the
substance, we use the term specific heat. Water has a very high
specific heat, c = 4.18 J/g•K. The exchange of heat, q, with the
surroundings is the product of the surrounding medium's specific
heat (c), mass (m), and change in temperature (T), such that
q = mcT. If a Bomb calorimeter is used, then the bomb
constant (C) is in the equation: q = (C + mc)T.
Bond energy, B.E., measures the energy needed to break a
covalent bond in a diatomic, gaseous molecule. The bond
energy is approximately the same for any gaseous molecule.
Change in enthalpy is estimated by adding the bond energies of
all bonds that are broken and subtracting the bond energies of
all bonds formed: H = B.E.react – B.E.prod.
Change in Entropy (S)
All chemical systems have an inherent amount of disorder
because of the complexity of the atomic arrangement within
molecules, the spacing of molecules with respect to each other;
and the overall motion of the system. Increases in complexity,
spacing and overall motion result in increased disorder as
measured by change in entropy, S. A positive S for physical
changes can be predicted based on whether the molecules
spread out. Evaporation, diffusion and effusion have +S
values. Dissolving is more complicated because spreading out
solute and solvent increases disorder, but formation of hydration
bonds between solute and solvent decreases disorder, therefore
it is impossible to predict the sign for S (although most
dissolving is +S. All chemical reactions that result in more
moles of gas products compared to reactants have a +S.
Thermodynamic Data
The standard enthalpy of formation, Hfo, of a substance is
the enthalpy change for the reaction in which one mole of
substance is formed from its constituent elements under
standard conditions of 1 atm pressure and 25oC (298 K). For any
element in its most stable state under standard conditions, Hfo
= 0 kJ/mol. Most compounds have negative values of Hfo.
Large negative Hfo indicate a strong bond and stable
compound. The standard entropy So is based H+ having So = 0
kJ/mol•K (although the AP exam often lists the values in
J/mol•K). The thermodynamic data chart lists the Hfo and So for
common substances.
Hfo applies to situations involving more than one mole,
where Hfo is multiplied by the number of moles, and involving
decomposition, where H = -Hfo. An important use of Hfo and
So is calculate H and S for a wide variety of reactions under
laboratory conditions, where H  Ho = Hfoprod – Hforeact and
S  So = Soprod – Soreact.
H depends only on the initial and final states of the
system. Thus, the enthalpy change of a process is the same
whether the process is carried out in one step or in a series of
steps. Hess's law states that if a reaction is carried out in a
series of steps, H for the reaction will be equal to the sum of
the enthalpy changes for the steps. We can therefore calculate
H for any process, as long as we can write the process as a
series of steps for which H is known.
Change in Free Energy (G)
The Gibbs free energy (or just free energy) G combines
enthalpy and entropy. For processes that occur at constant
temperature, G = H – TS. The sign of G relates to the
spontaneity of the process. When G is negative, the process is
spontaneous. When G is positive, the process is
nonspontaneous; the reverse process is spontaneous. At
equilibrium the process is reversible and G = 0 kJ/mol.
The values of H and S generally do not vary much with
temperature. As a consequence, the dependence of G with
temperature is governed mainly by the value of T in the
expression G = H –TS. The threshold temperature,
T = H/S, is when a reaction goes from spontaneous 
nonspontaneous. This only occurs when H and S are both
positive or both negative. When are both positive, the reaction
is spontaneous at all temperatures above the threshold. When
are both negative, the reaction is spontaneous at all
temperatures below the threshold. When H and S have
opposite signs, then the reaction is spontaneous at all
temperatures (-H and +S) or spontaneous at no temperature
(+H and –S).
Practice Multiple Choice
Briefly explain why the answer is correct in the space provided.
1.
I2(g) + 3 Cl2(g)  2 ICl3(g)
According to the data in the table below, what is the value
of Ho , in kJ, for the reaction represented above?
Bond
CI–CI
I–I
I–Cl
Bond Energy (kJ/mole)
150
240
210
(A) - 870
(B) - 390
(C) +180
(D) + 450
2.
C2H4(g) + 3 O2(g)  2 CO2(g) + 2 H2O(g)
For the reaction, H is -1,300 kJ. What is the value of H,
in kJ, if the combustion produced liquid water rather than
water vapor? (H for H2O(l)  H2O(g) is 45 kJ/mol)
(A) -1,300 (B) -1,210 (C) -1,345 (D) -1,390
3.
CH4 (g) + 2 O2(g)  CO2(g) + 2 H2O(l) Ho = -900 kJ
What is the standard heat of formation of CH4, in kJ/mol,
as calculated from the data below?
(HfoH2O = -300 kJ/mol, HfoCO2 = -400 kJ/mol)
(A) -200
(B) -100
(C) 100
(D) 200
4.
H2(g) + ½ O2(g)  H2O(l)
2 Na(s) + ½ O2(g)  Na2O(s)
Na(s) + ½ O2(g) + ½ H2(g)  NaOH(s)
What is H for the reaction below?
Na2O(s) + H2O(l)  2 NaOH(s)
(A) x + y + z
(B) x + y – z
(C) x + y - 2z
(D) 2z - x - y
5.
Which is true when ice melts at its normal melting point?
(A) H < 0, S > 0, G = 0 (B) H < 0, S < 0, G > 0
(C) H > 0, S < 0, G < 0 (D) H > 0, S > 0, G = 0
6.
Which of the following reactions has the largest positive
value of S per mole of Cl2?
(A) H2(g) + Cl2(g)  2 HCl(g)
(B) Cl2(g) + O2(g)  Cl2O(g)
(C) Mg(s) + Cl2(g)  MgCl2(s)
(D) 2 NH4Cl(s)  4 H2(g) + Cl2(g)
7.
Ice is added to hot water in an insulated container, which is
then sealed. What has happened to the total energy and
the total entropy when the system reaches equilibrium?
(A) Energy and entropy remain constant
(B) Energy remains constant, entropy decreases
(C) Energy remains constant, entropy increases
(D) Energy decreases, entropy increases
8.
N2(g) + 3 H2(g)  2 NH3(g)
The above reaction is thermodynamically spontaneous at
298 K, but becomes nonspontaneous at higher
temperatures. Which of the following is true at 298 K?
(A) G, H, and S are all positive.
(B) G, H, and S are all negative.
(C) G and H are negative, but S is positive.
(D) G and S are negative, but H is positive.
Ho = x
Ho = y
Ho = z
9.
3 C2H2(g)  C6H6(g)
What is the standard enthalpy change, Ho, for the
reaction represented above?
(HfoC2H2 is 230 kJ•mol-1; HfoC6H6 is 80 kJ•mol-1)
(A) -610 kJ (B) 150 kJ (C) -770 kJ (D) 610 kJ
10. When solutions of NH4SCN and Ba(OH)2 are mixed in a
closed container, the temperature drops and a gas is
produced. Which of the following indicates the correct
signs for G, H, and S for the process?
(A) –G –H –S (B) –G +H –S
(C) –G +H +S (D) +G –H +S
11.
c.

d.
3.
b. Determine So for the reaction above using the table
CO(g)
CO2(g)
O2(g)
Substance
So (J/mol•K)
197.7
213.7
205.1

c.
b.
Calculate
for the combustion reaction at
25oC.

c.
Determine Go for the above reaction at 298 K.

4.
The dissolving of AgNO3(s) in water is represented by the
equation: AgNO3(s)  Ag+(aq) + NO3-(aq)
a. Is G positive, negative, or zero? Justify your answer.

Consider the combustion of butanoic acid at 25oC:
HC4H7CO2(l) + 5 O2(g)  4 CO2(g) + 4 H2O(l
Ho= -2,183.5 kJ
o
So (kJ/mol•K)
Substance
Hf (kJ/mol)
CO2(g)
-393.5
0.2136
H2O(l)
-285.8
0.0699
O2(g)
0.0
0.2050
C3H7COOH(l)
?
0.2263
a. Calculate Hfo, for butanoic acid.
So
The combustion of carbon monoxide is represented by the
equation: CO(g) + ½O2(g) 
2(g)
a. Determine Ho for the reaction above using the values.
C(s) + ½ O2(g)  CO(g)
Ho298 = -110.5 kJ•mol-1
C(s) + O2(g)  CO2(g)
Ho298 = -393.5 kJ•mol-1

Practice Free Response
1.
Calculate the F–F bond energy using the information
above and the bond energies
(NN = 946 kJ/mol, N–F = 272 kJ/mol).

X(s)  X(l)
Which of the following is true for any substance undergoing
the process represented above at its normal melting point?
(A) S < 0
(B) H = 0
(C) H = TG
(D) H = TS
12. For a reaction, Ho = -150 kg/mol and So = -50 J/mol•K.
Which statement is true about this reaction?
(A) It is spontaneous at high temperature only.
(B) It is spontaneous at low temperature only.
(C) It is spontaneous at all temperatures.
(D) It is non-spontaneous at all temperatures.
Calculate the heat released when 0.256 mol of NF3(g)
is formed from N2(g) and F2(g) at 1.00 atm and 298 K.
b.
The solution cools when AgNO3(s) is dissolved. Is H
positive, negative or zero? Justify your answer.

c.
Is S positive, negative, or zero? Justify your answer.

5.
Consider the thermochemical equation:
C2H6(g) + 7/2 O2(g)  2 CO2(g) + 3 H2O(l) H = -1559.7 kJ
a. Calculate H for the thermochemical equations:
2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(l)
2 CO2(g) + 3 H2O(l) C2H6(g) + 7/2 O2(g) 
b. The heat of vaporization of H2O(l) is +44.0 kJ/mol.
Calculate H for the equation:
C2H6(g) + 7/2 O2(g)  2 CO2(g) + 3 H2O(g)
Calculate Go for the combustion reaction at 25oC.

c.
The heat of formation of CO2(g) and H2O(l) are -393.5
kJ/mol and -285.8 kJ/mol. Calculate Hfo of C2H6(g).
Consider the synthesis reaction:
N2(g) + 3 F2(g)  2 NF3(g)
(Ho298 = -264 kJ mol-1, So298 = -278 J K-1 mol-1)
a. Calculate Go298 for the reaction.
d.
How much heat is evolved when 1.00 g of C2H6(g) is
burned to give CO2(g) + H2O(l) in an open container?

e.
What is the bomb constant C if the change in
temperature is 13.13oC when 1.00 g of C2H6 reacts in
the bomb calorimeter that contains 250 g H2O?
d.
What is the spontaneous temperature range?

2.
b.

For what temperature range is the reaction spontaneous?