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AP Chemistry 7: 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. H Ho = Hfoproducts – Hforeactants
2. 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
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
–
–
1.
2.
Enthalpy
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
H = -1640 kJ
8.
a. How much heat is released to produce 10.0 g Fe2O3?
10.0 g Fe2O3 x 1 mol x -1640 kJ = -51.3 kJ
160 g
2 mol
b. How many grams of iron are needed to generate
1.00 x 104 kJ of heat?
-1.00 x 104 kJ x 4 mol x 55.8 g = 1360 g
-1640 kJ 1mol
9.
CaSO4(s) + CO2(g)  CaCO3(s) + SO3(g)
H = 224 kJ
a. How much heat is absorbed when 10.0 g CaSO4 react.
10.0 g CaSO4 x 1 mol x 224 kJ = 16.5 kJ
136 g 1 mol
b. How much CaCO3 is produced when 500. kJ is absorbed?
Qwater = mcT = (100 g)(4.18 J/g•K)(-1.00 K) = -418 J
H = ½H1 + -5H2 = ½(-2512 kJ) + -5(104 kJ) = -1776 kJ
the number of moles of NH4Cl dissolved.
1.51 g NH4Cl x 1 mol/53.5 g = 0.0282 mol
c.
H (in kJ) for the dissolving of one mole of NH4Cl.
0.418 kJ/0.0282 mol = 14.8 kJ/mol
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.
Qwater = (250 g)(4.18 J/g•K)(33.1 – 23.8)K = 9720 J
b.
the number of moles of MgSO4 dissolved.
12.8 g MgSO4 x 1 mol/120. g = 0.107 mol
c.
H (in kJ) for the dissolving of one mole of MgSO4.
-9.72 kJ/0.107 mol = -90.8 kJ
5.
6.
7.
H = H1 + 4H2 = -126 kJ + 4(-179 kJ) = -842 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.
4.
>0
 0 < 0
x
Melting ice at 0oC
x
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
x
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
x
Distilling alcohol-water mixture
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
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)
500 kJ x 1 mol/224 kJ x 100 g/1 mol = 223 g
3.
Entropy
Predict whether S > 0, S < 0 or S 0.
H2(g) + F2(g)  2 HF(g)
Estimate H for the reaction using the bond energy values.
H = (H–H) + (F–F) – 2(F–H)
H = 436 + 155 – 2(567) = -543 kJ
C2H2(g) + 2 H2(g)  C2H6(g)
Estimate H for the reaction using the bond energy values.
H = 2(C–H) + 1(CC) + 2(H–H) – 6(C–H) – 1(C–C)
H = 2(413) + 839 + 2(436) – 6(413) – 348 = -289 kJ
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.
Q = (C + mc)T
Q = [921 + (1000 g)(4.18 J/g•K)](9.3 K) = 4.8 x 104 J
b. the number of moles of ethene reacted.
1.00 g C2H4 x 1 mol/28.0 g = 0.0357 mol
c.
H (in kJ) for the combustion of one mole of C2H4.
-48 kJ/0.0357 mol = - 1300 kJ
d.
Write the equation for the combustion of ethene (C2H4).
C2H4(g) + 3 O2(g)  2 CO2(g) + 2 H2O(g)
e. Calculate H using bond energies.
H = 4(C–H) + 1(C=C) + 3(O=O) – 4(C=O) – 4(O–H)
H = 4(413) + 614 + 3(495) – 4(799) – 4(463) = -1297 kJ/mol
10.
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
H = H1 + -H2 + -½H3
H = -198.9 kJ + 142.3 kJ + -½(495.0 kJ) = -304.1 kJ
12. a. Write the equation for the combustion of methanol,
CH3OH(l). (other reactants and products are gaseous).
2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(g)
b. Calculate H using Hfo values.
H = 2 HfoCO2 + 4 HfoH2O – 2 HfoCH3OH – 3 HfoO2
H = 2(-393.5) + 4(-241.8) – 2(-238.7) – 3(0.0) = -1276.8 kJ
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.
1.00 g CH3OH x 1 mol x 1276.8 kJ = 20.0 kJ
32.0 g
2 mol
d. Calculate the calorimeter constant of the bomb.
Q = (C + mc)T
20000 J = [C + (1200 g)(4.18 J/g•K)](3.4 K)  C = 866 J/K
13.
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.
H = HfoCaSO3•2 H2O – HfoCa – HfoSO3 – 2 HfoH2O
-795 = HfoCaSO3•2 H2O – 0 – (-395.7) – 2(-285.8)
HfoCaSO3•2 H2O = -1762 kJ
b. Calculate So for CaSO3•2 H2O.
So = SoCaSO3•2 H2O – SoCa – SoSO3 – 2SoH2O
-0.2535 = SoCaSO3•2 H2O – 0.0414 – 0.2567 – 2(0.0699)
SoCaSO3•2 H2O = 0.1844 kJ/K
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.
H = HfoCu2+ + 2 HfoNO2 + 2 HfoH2O – HfoCu – 4 HfoH+ – 2 HfoNO3
H = 64.8 + 2(33.2) + 2(-285.8) – 0 – 4(0) – 2(-205.0) = -30.4 kJ
b. Calculate So using So values.
So = SoCu2+ + 2 SoNO2 + 2 SoH2O – SoCu – 4 SoH+ – 2 SoNO3So = -.0996 + 2(.2400) + 2(.0699) – .0332 – 2(.1464) = 0.1942 kJ/K
c. Calculate Go.
Go = H –TS
Go = -30.4 kJ – (298 K)(0.1942 kJ/K) = -88.3 kJ
11.
17.
18.
19.
Heat of Reaction Lab
20. 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
ToC
(2) Graph the temperature vs. time data. Draw a best
fit straight line (use a ruler).
Temperature (oC)
16.
NH4NO3(s)  NH4+(aq) + NO3-(aq)
Determine the following for the above reaction.
a. Is the reaction exothermic or endothermic?
H = HfoNH4+ + HfoNO3- – HfoNH4NO3
H = (-132.5) + (-205.0) – (-365.6) = 28.1 kJ  endo
b. Is there an increase or decrease in entropy?
S = SoNH4+ + SoNO3- – SoNH4NO3
S = 0.1134 + 0.1464 – 0.1511 = 0.1087 kJ/K  increase
c. Is the reaction spontaneous at 25oC?
Go = H –TS = 28.1 kJ – (298 K)( 0.1087kJ/K)
Go = -4.3 kJ  spontaneous
2 SO2(g) + O2(g)  2 SO3(g)
a. Calculate Ho.
Ho = 2 HfoSO3 – 2 HfoSO2 – HfoO2
Ho = 2(-395.7) – 2(-296.8) – 0 = -197.8 kJ
b. Calculate So.
So = 2 SoSO3 – 2 SoSO2 – SoO2
So = 2(0.2567) – 2(0.2481) – 0.2050 = -0.1878 kJ/K
c. Calculate G at 400 K.
G = Ho –TSo
G = -197.8 – (400)(-0.1878) = -122.7 kJ
d. Determine the temperature range where the reaction
is spontaneous.
Tthreshold = Ho/So = (-197.8)/(-0.1878) = 1053 K
 All temperatures below 1053 K
C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
a. Calculate Ho.
Ho = 2 HfoCO2 + 3 HfoH2O – HfoC2H5OH – 3 HfoO2
Ho = 2(-393.5) + 3(-285.8) – (-277.6) – 3(0) = -1366.8 kJ
b. Calculate So.
So = 2 SoCO2 + 3 SoH2O – SoC2H5OH – 3 SoO2
So = 2(.2136) + 3(.0699) – (.1607) – 3(.2050) = -.1388 kJ/K
c. Calculate G at 20oC.
G = Ho –TSo = -1366.8 – (293)(-0.1388)
G = -1326.1 kJ
d. At which temperature (if any) will the reaction be
spontaneous?
T = Ho/So = -1366.8/-0.1388 = 9847 K
 All temperatures below 9847 K
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
–
Temperature increases
H
S +/– Can't predict about dissolving
–
dissolving occured
G
C2H5OH(l)  C2H5OH(g)
Calculate the boiling point (threshold temperature) given
the information: H = 37.95 kJ and S = 0.1078 kJ/K.
T = Hvap/Svap = 37.95/0.1078 = 352 K (79oC)
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).
20
60
Temperature (oC)
15.
100
140
180
Time (s)
(3) Use the y-intercept to determine Tmix.
Tmix (y-intercept)
d. Calculate Hreaction per mole of reactant based on the
calorimetry data.
T (K)
Qwater (kJ)
Hreaction/mole
e.
H
%
Calculate Hreaction per mole of reactant based on Hfo.
OH-(aq) + H+(aq)  H2O(l)
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).
Calculate Hreaction per mole of reactant based on the
calorimetry data.
T (K)
Qwater (kJ)
Hreaction/mole
k.
Calculate Hreaction per mole of reactant based on Hfo.
NH3(aq) + H+(aq)  NH4+(aq)
H
Temperature (oC)
%
Show that the chemical equations and H from
part (d) – part (g) = part (j).
OH-(aq) + H+(aq)  H2O(l)
-55.8 kJ
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
3.3 kJ
NH3(aq) + H+(aq)  NH4+(aq)
-52.5 kJ
l.
20
100
140
180
Time (s)
(3) Use the y-intercept to determine Tmix.
Tmix (y-intercept)
g. Calculate Hreaction per mole of reactant based on the
calorimetry data.
60
T (K)
Qwater (kJ)
Hreaction/mole
h.
j.
Calculate Hreaction per mole of reactant based on Hfo.
NH4+(aq) + OH-(aq)  NH3(aq) + H2O(l)
H
%
Temperature (oC)
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
(2) Graph the temperature vs. time data. Draw a best
fit straight line (use a ruler).
20
60
100
140
Time (s)
(3) Use the y-intercept to determine Tmix.
Tmix (y-intercept)
180
Practice Quiz
Multiple Choice (no calculator)
Briefly explain why the answer is correct in the space provided.
1
2
3
4
5
6
7
8
9
10 11 12
B
D
B
D
D
D
C
B
A
C
D
B
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
H = (I-I) + 3(Cl-Cl) – 6(I-Cl)
H = (150) + 3(240) – 6(210) = -390 kJ
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
H = Hcomb – 2HH2O(l)
H = -1300 kJ – 2(45 kJ) = -1390 kJ
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
H = HfoCO2 +2 HfoH2O – HfoCH4
-900 kJ = -400 kJ + 2(-300 kJ) – HfoCH4HfoCH4 = -100 kJ
4.
H2(g) + ½ O2(g)  H2O(l)
Ho = x
2 Na(s) + ½ O2(g)  Na2O(s)
Ho = y
Na(s) + ½ O2(g) + ½ H2(g)  NaOH(s)
Ho = z
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
the first and second reactions are reversed and the
third reaction is doubled H = -x + -y + 2z
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
Bonds are broken  +H, liquids more disordered
than solids  +S, threshold temperature 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)
Large positive S for reactions that produce many
more moles of gas compared to reactants.
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
The total energy constant because insulation. Entropy
increases because melted ice has more disorder.
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.
Nonspontaneous at high temperature then H and S
are negative. Spontaneous at 298 K G is negative.
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
H = HfoC6H6 – 3HfoC2H2
H = (80 kJ) – 3(230 kJ) = -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
Temperature drop means +H. The gas product
means +S. The reaction occurs –G.
11.
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
Threshold temperature, G = 0.
G = H – TS = 0  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.
Ho and So are negative  range of spontaneous
temperatures are all temperatures below the threshold.
Free Response (calculator)
1. 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
2.
a. Calculate Hfo, for butanoic acid.
Ho = 4HfoCO2 + 4HfoH2O – HfoC3H7COOH – 5HfoO2
-2183.5 = 4(-393.5) + 4(-285.8) – HfoC3H7COOH – 5(0)
HfoC3H7COOH = -533.7 kJ/mol
b. Calculate So for the combustion reaction at 25oC.
So = 4SoH2O + 4SoCO2 – SoC3H7COOH -5SoO2
So = 4(0.0699) + 4(0.2136) – (0.2263) – 5(0.2050)
So = -0.1173 kJ/K
c. Calculate Go for the combustion reaction at 25oC.
Go = Ho –TSo
Go = -2183.5 kJ – (298 K)(-0.1173 kJ/K)
Go = -2148.5 kJ
d. What is the spontaneous temperature range?
T = Ho/So = -2183.5 kJ/-0.1173 kJ/K = 19,000 K
 < 19,000 K
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.
Go = Ho –TSo = -264 kJ – (298 K)(-0.278 kJ/K)
Go = -181 kJ
b. For what temperature range is the reaction spontaneous?
T = Ho/So = -264 kJ/-0.278 kJ/K = 950 K and below
c.
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.
0.256 mol NF3 x -264 kJ/2 mol NF3 = -33.8 kJ
d.
3.
4.
Calculate the F–F bond energy using the information
above and the bond energies
(NN = 946 kJ/mol, N–F = 272 kJ/mol).
Ho = (NN) + 3(F–F) – 6(N–F)
-264 kJ = (946 kJ) + 3(F–F) – 6(272 kJ)
 (F–F) = 141 kJ/mol
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
CO(g)  ½ O2(g) + C(s)
Ho = 110.5 kJ
C(s) + O2(g)  CO2(g)
Ho = -393.5 kJ
CO(g) + ½ O2(g)  CO2(g)
Ho = -283.0 kJ
o
b. Determine S for the reaction above using the table
CO(g)
CO2(g)
O2(g)
Substance
So (J/mol•K)
197.7
213.7
205.1
So = SoCO2 – SoCO - ½ SoO2
So = 213.7– 197.7– ½(205.1) = -86.5 J/K
c. Determine Go for the above reaction at 298 K.
Go = Ho - TSo
Go = (-283 kJ) - (298 K)(-.0865 kJ/K) = -257.2 kJ
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.
G is negative because AgNO3 is soluble in water, 
the solution process is spontaneous and G < 0.
b. The solution cools when AgNO3(s) is dissolved. Is H
positive, negative or zero? Justify your answer.
H is positive because heat is absorbed during
dissolving.
c. Is S positive, negative, or zero? Justify your answer.
S is positive because G = H – TS and G < 0 and
H > 0  S must be > 0.