1
THERMODYNAMICS
I.
TERMS AND DEFINITIO
NS
A. Review of Definitions
1. Thermodynamics = Study of the exchange of heat, energy and work between a
system
and its
surroundings
.
a. System = That part of universe of interest. (reaction vessel, etc.)
b. Surroundings = Rest of universe.
c. Some types of systems.
1) Open systems = Mass can be exchanged in addition to heat, energy and work.
These are important to engineers in flow systems
2) Closed systems = No mass can be exchanged, only heat
, energy and work. These
are usually encountered in the laboratory
3) Isolated system = No mass, heat or work can be exchanged. The universe is an
example of an isolated system.
d. Exchange takes place at boundary between system and surroundin
gs during a
change in
state
of system.
1) State = That condition in which all variables are fixed and unvarying.
2) When one or more of these variables are changed, the system changes state.
3) Examples:
Temperature Changes
Gas at V
1
, P
1
, T
1

> Gas at V
2
,P
2
, T
2
( PV = nRT is the equation of state for an ideal gas )
Phase changes ( solid

> liquid; liquid

> gas: etc. )
Reactants

> Products
2. Can control some variables or con
ditions during a change.
a. Isothermal Change = One at constant temperature (supply or take away just enough
heat so that the temperature remains constant).
b. Adiabatic Change = No heat is exchanged during change, that is , the system
is insulat
ed from its surroundings.
c. Isobaric Change = One at constant pressure.
2
d. Normal laboratory conditions are isothermal and isobaric ones.
e. Reversible change = an idealized change carried out very slowly by allowing one
restraining variable to b
e infinitesimally different from it’s equilibrium value; the
system is essentially at equilibrium at all times during the change.
B. State Functions ( or State Variables ).
1. Variables (properties) whose value depend only on the state of the system.
a. Define the state.
b. Value does not depend on the past history of the system.
c. Examples: State functions

T, P, V, Energy.
Not State functions

work(W), heat change(q)
2. Thermodynamics is concerned with h
ow the state variables change during a change of
state. It views these changes in the light of three laws.
Recall that in science a law is simply a summary of experience.
C. Work ( W )
1. Sense of W.
W is positive when work is done on the system
by the surroundings.
When the system does work, W is negative; the system has to expend energy.
2. Some types of work.
a. Mechanical work

exert a force through a distance.
W =
–
( force )x( distance ) or W =
–
f
(
Δ
x)
If force is not con
stant, W =
–
∫
f
d
x
b. Work of expansion of a gas under constant pressure. W =
–
P∆V
3. Dimension conversion.
a. Because of the different formulas for calculating work, a number of different
units for work can be obtained. They should be c
onverted to the usual SI
unit of joules (kg m
2
/s
2
). This can be easily done using the Gas Constant, R, which has
a general dimension of
.
3
Example
. A gas is expanded from a volume of 5.00 L to 9.00 L under a
co
nstant pressure of 2.00 atm. Calculate the work done on expansion in joules.
W =
–
P(V
2

V
1
) P = 2.00 atm V
2
= 9.00 L V
1
= 5.00 L
W =
–
(2.00 atm )( 9.00 L

5.00 L ) =
–
8.00 L atm
What is the work in joules?
W =
–
(
8.00 L atm )
=
–
811 J
D. Heat (q).
1. Sense of q.
a. q is positive if heat energy is absorbed by the system (flows into the system)
b. q is negative if heat is evolved by the system (flows out of the system).
2. The valu
e of q can be measured experimentally in
calorimetry
.
II. THE FIRST LAW OF THERMODYNAMICS
A. Statement of the first law.
1. Law of Conservation of Energy. The total energy of the universe (or any isolated
system ) is constant. Energy can neither be c
reated nor destroyed but can be
converted from one form to another.
2. First Law in symbols.
For any change in the state of a system
Δ
E = q + W
a. E = Internal Energy = energy intrinsically possessed by a system
(due to mass, structu
re, temperature, etc.)
1)
E is State Function.
In a change from some
Initial State
Final State
Δ
E = change in E = E
final

E
initial
In general, the sense of
Δ
is final

initial
2) If
Δ
E is + , then E
final
> E
initial
, the int
ernal energy increases.
If
Δ
E is

, then E
final
< E
initial
, the internal energy decreases.
3) Do not know the absolute values of E but are only interested in its changes.
b. W is the work done on the system and q is the heat absorbed by
the system.
Neither W nor q are state functions.
4
3. Changes at constant volume and only PV work possible.
a. At constant volume,
Δ
V = 0,
∴
W = 0
b.
Δ
E = q
v
(subscript v means constant volume)
A measurement of q
v
gives
Δ
E for the change. Th
is can be measured in a constant
volume (bomb) calorimeter
4.
Changes at constant pressure and only PV work.
a. At constant pressure, q
P
= ∆H, H = Enthalpy = E + PV
b. If
Δ
H is + , H
final
> H
initial
, enthalpy increases; heat is absorbed.
If
Δ
H is

, H
final
< H
initial
, enthalpy decreases; heat is evolved.
c.
Exothermic
change = one in which
Δ
H is
–
, heat is liberated.
Endothermic
change = one in which
Δ
H is + , heat is absorbed.
d. Chemical Reactions.
1) For a reaction,
Δ
H =
Σ
H
pr
oducts

Σ
H
reactants
2) The absolute values of H's are not known, however, we can develop a set
of
Enthalpies of Formation
to use in calculating
Δ
H's for reactions
5. Heats of Reactions.
a. Standard Enthalpies of Formation,
Δ
H
f
.
Δ
H
f
=
Δ
H when 1 mole of a compound is prepared from its elements under
standard conditions.
b. Standard Conditions.
1) All gases at partial pressures of 1 atm.
2) All solutes at an effective concentration of 1 Molar.
3) Al
l liquids and solids in their most stable form.
4) Note that no temperature is stipulated for standard conditions. Most data
are reported at 25
C.
5) A superscript zero (
) indicates standard conditions.
c. Use of
Δ
H
f
Calculate
ΔΗ
's for reactions.
ΔΗ
=
∑ΔΗ
f
's (products)

∑Δ
H
f
's (reactants)
Δ
H
f
’s have been measured for most common substances and can be found in
Tables. They are usually listed for 25 C.
5
III. SPONTANEOUS CHANGES
A. Entropy and the Second Law.
1.
Second Law can be stated in several ways. One statement is that “it is impossible
for an isolated system to spontaneously order itself”.
a. This does not depend on energy considerations. The First Law deals only with
the conservation of energy
not on the likelihood of the energy change actually
occurring in the direction contemplated.
b. For example we know from experience that when a 500g block of Cu is dropped, the
spontaneous energy changes are:
Potential Energy Kin
etic Energy Thermal Energy.
As far as the First Law is concerned, the process could just as well work backwards.
That is, the Cu block could have absorbed energy, jumped up its original height. We
know from experience that this will not spon
taneously happen.
c. When a cylinder of compressed gas is opened in a room, the compressed gas
escapes from the cylinder into the air. Why wouldn't the air in the
room “escape" into the cylinder? In other words, why do gases spontaneously
go from regions of high pressure to regions of low pressure?
d.
In both cases, the nonspontaneous direction requires that the molecules
involved would, at the same instant in time, all decide to go in one particular
direction. Although this di
rection could be energetically feasible, the
probability (or chance) of this happening is so small that we can be assured that
it will not occur.
e.
In general, the more ways an isolated system can distribute its energy, the less likely
it i
s that the system will change out of that state to one that is more energy restrictive.
The Second Law requires that any spontaneous change be in the direction of
increased energy distribution possibilities.
2. Entropy, S.
a. Measures the dispersal o
f energy as a function of temperature. The dimensions are in
J/K or J/mol K (dS = dq
rev
/T) . The more ways a system has the distribute its energy,
the higher will be its entropy. For example in going from solid

> liquid

> gas
entropy increases.
6
b. A perfectly ordered crystal of a substance at 0 K, is most restrictive in the ways its
energy can be distributed and would therefore have zero entropy (S = 0).
This is a statement of the
Third Law of Thermodynamics
and allows absolute
entrop
ies to be determined. The entropies of substances depend on their molecular
complexity, their phase and the temperature. The molar entropies for most common
substances have been measured under standard conditions and are found in standard
thermodynam
ic tables. Some values are listed below for 25
C.
Substance
S
(J/mol K)
H
2
O
(s)
47.91
H
2
O
(l)
69.91
H
2
O
(g)
188.83
CH
4(g)
186.26
C
2
H
6(g)
229.60
C
3
H
8(g)
237.4
C
4
H
10(g)
310.1
1) S
increase
s as you go from solid

> liquid

> gas
2) S
increases as the complexity of the molecule increases.
3) S
increases as temperature increases. For H
2
O
(l)
at 50
C S
J / mol
Κ
and at
C, S
= 86.8 J / mol K.
2. Standa
rd Entropy Changes.
a. Entropy is a state function, that is, for any change in state, the change in entropy,
Δ
S =
S
final
–
S
initial
b. For phase changes,
Δ
S
can be obtained from the S
's of the pure phases.
Example
. For water
Δ
S
fus
= S
li
quid)
–
S
(solid) = 69.6 J / mol K
–
47.6 J / mol K = 22.0 J / mol K
Δ
S
vap
= S
(gas)
–
S
(liquid) = 188.8 J / mol K
–
69.6 J / mol K = 118.9 J / mol K
These are the values at 25
C. The values at the normal melting and boiling points will
be s
lightly different.
c. For chemical reactions,
Δ
S
=
Σ
S
(products)
–
Σ
S
(reactants).
B. Spontaneous Reactions.
1. For spontaneous chemical reactions the total entropy must increase. This means that
7
Δ
S
tot
=
Δ
S
sys
+
Δ
S
surr
> 0
a.
Δ
S
sys
ca
n be determined from the S
's of the reactants and products. However,
Δ
S
surr
can not be easily obtained.
b. If heat is absorbed by the surroundings, its entropy should increase. Therefore,
at constant temperature,
Δ
S
surr
q/T. At constant pres
sure, q
p
=
Δ
H.
c. Therefore, at constant temperature and pressure
Δ
S
surr
=
But ∆H
surr
=
–
∆H
sys
∴Δ
S
surr
=
–
∴Δ
S
tot
=
Δ
S
sys
–
> 0.
∴
T
Δ
S
tot
= T
Δ
S
sys
–
Δ
H
sys
> 0
We now h
ave a set of parameters we can use to predict spontaneity at constant T and P.
2. Free Energy ( Gibbs Free Energy or Gibbs Energy ), G
a. G = H
–
TS
Since H, S, and T are state functions, G is also a state function.
b. At constant temperature an
d pressure,
Δ
G =
Δ
H
–
T
Δ
S
Note T
Δ
S
tot
= T
Δ
S
sys
–
ΔΗ
sys
=
–
Δ
G
sys
∴
at constant T and P
Δ
G
sys
is negative for all spontaneous changes
Δ
G
sys
= 0 at equilibrium.
c. Note if
Δ
G
sys
> 0, the change is not spontaneous. The reverse direction is
spo
ntaneous direction.
3. Consider the phase change H
2
O(s)

> H
2
O(l ) at 1 atm pressure. Given that
Δ
H
(fus) = 6010 J / mol and
Δ
S
(fus) = 22.0 J / mol K. Assuming that
Δ
H
and
Δ
S
are independent of temperature, calculate
Δ
G
(fus) at

10.0
C,
0.0
C, and 10.0
C
t,
C
T, K
ΔΗ
(fus)
T
Δ
S
(fus)
Δ
G
(fus)=∆H°(fus)

T∆S°(fus)

10.0 263.2 6010 5790 +220 Nonspontaneous
0.0 273.2 6
010 6010 0 Equilibrium
10.0 283.2 6010 6230
–
220 Spontaneous
8
Note that T
fp
=
In the same way T
bp
=
C.
∆G°’s for chemical Reactions. Two ways to obtain.
1. From ∆H°’s and ∆S°’s.
Given the following data at 25
C
Substance
ΔΗ
f
(kJ / mol)
S
J / mol K )
SO
3
(g)
–
395.72
256.76
SO
2
(g)
–
296.83
248.22
NO(g)
90.25
210.
76
N
2
O(g)
82.05
219.85
For the reaction N
2
O(g) + SO
3
(g)

> 2NO(g) + SO
2
(g)
Calculate ∆H°, ∆S° and ∆G° at 25 °C.
a.
Δ
H
Δ
H
=
Δ
H
f
(SO
2
) + 2
Δ
H
f
(NO)
–
Δ
H
f
(N
2
O)
–
Δ
H
f
(SO
3
)
Δ
H
=
–
296.83 kJ + 2( 90.25 kJ)
–
( 82.05
kJ)
–
(
–
395.72 kJ)
Δ
H
= 197.34 kJ/mol
b.
Δ
S
Δ
S
= S
(SO
2
) + 2S
(NO)

S
(N
2
O)

S
(SO
3
)
Δ
S
J / K + 2(210.76 J / K)
–
219.85 J / K
–
256.76 J / K = 193.13 J /mol K
c.
Δ
G
at 25
C.
Δ
G
=
ΔΗ
–
T
Δ
S
Ν
ote
Δ
H
is in k
J/mol and
Δ
S
is in J / mol K
Δ
G
= 197,340 J/mol
–
(298 K )(193.13 J /mol K) = 139,787 J/mol = 139.77 kJ/mol
The dimension kJ/mol (or J/mol) means per mol of the balanced equation.
Note that the reaction is nonspontaneous under standard condit
ions. That is,
It is nonspontaneous at 25 °C when
P
N
2
O =
P
NO
=
P
SO
3
=
P
SO
2
= 1.0 atm.
2. From standard Free Energies of Formation (
Δ
G
f
’s)
a.
Δ
G
f
=
Δ
G
when one mole of a compound is prepared from its elements.
c.
Δ
G
f
's have been calculated
at 25
C for many compounds and are listed with
Δ
H
f
's and S
’s. As with
Δ
H
f
's,
Δ
G
f
's of elements = 0
d. Given the following
Δ
G
f
's at 25
C, calculate
Δ
G
for the reaction
9
N
2
O
(g)
+ SO
3(g)

> 2NO
(g)
+ SO
2(g)
.
Substance
Δ
G
f
(
kJ / mol )
SO
3
(g)
–
371.06
SO
2
(g)
–
300.19
NO(g)
86.55
N
2
O(g)
104.20
Δ
G
= 2
Δ
G
f
(NO) +
Δ
G
f
(SO
2
)
–
Δ
G
f
(SO
3
)
–
Δ
G
f
(N
2
O)
= 2(86.55 kJ / mol) + (
–
300.19 kJ / mol )
–
(
–
371.06 kJ / mol )
–
104.20 kJ / mol
= 1
39.77 kJ / mol which is the same answer obtained earlier from
Δ
H
f
's and S
's.
D.
Δ
G Under Nonstandard Conditions.
1. Chemical Potential ( Free Energy per mole)
a.
The Chemical Potential of a component , i , of a mixture,
i
= Change in G
i
with
moles
of
i
G
i
=
i
n
i
i
in J / mol.
b. The value of
depends on the nature of the substance, its concentration, and
temperature. In general
i
=
i
+ RT ln a
i
where a
i
is the activity of the component = effective concentratio
n.
Will assume:
1) For gases, activity is the partial pressure in atm. a
i
=
P
i
2) For solutes, activity is the Molarity a
i
= M
i
3) For pure liquids and solids, a
i
= 1
2. Consider the gas phase reaction cC + bB

> dD + eE
Δ
G = G(products)

G(reactants) = d
D
+ e
E
–
c
C
–
b
B
Since
C
=
C
+ RT ln
a
C
,
etc.
Δ
G = d
D
+ e
E
−
c
C
−
b
B
dRTln
a
D
+ eRTln
a
E
–
cRTln
a
C
–
bRTln
a
B
but
Δ
G
= d
D
+ e
E
–
c
C
–
b
B
∴Δ
G =
Δ
G
+ RTln
a
D
d
+ RTln
a
E
e
–
RTln
a
B
b
–
RTlna
C
c
10
∴
If all reactants and products are gases, such as, cC(g) + bB(g)

> dD(g) + eE(g), and
using partial pressures (in atm) for activities, you get
a. ∆G for is usually written in the
form
Q
is called the
Reaction Quotient
(or the
Mass Action Expression
).
For gaseous reactions where the partial pressures are used for activities, a subscript P is
attached to Q (Q
P
). The subscript P indicates that part
ial pressures are used to express
the activities of gases.
b
.
For the reaction Zn(
s
) + 2 H
+

> Zn
2+
+ H
2
(
g
),
c.
Q measures the extent of the reaction. At the start of the reaction when no products
are pre
sent Q = 0 and
Δ
G

∞
that is, a large negative number and any reaction
would be spontaneous. As the reaction proceeds, Q
increases and
Δ
G becomes less
negative. Will reach a point where
Δ
G = 0. At this point equilibrium is established
between
reactants and products and the reaction is complete.
d. At equilibrium Q
reaches a constant value. This value is called the
Equilibrium
Constant
and is given the symbol K. For gaseous reactions, Q is Q
P
and K is K
P
.
The subscript p indicates t
hat gas concentrations are expressed in partials pressures
in atm.
At equilibrium:
Δ
G = 0
Q = K
∆G = 0 = ∆G° + RT ln K or ln K =
–
∴
K = e
11
∴
the values of the equilibrium constants for reactions can be calculated
from tabulated thermodynamic data.
4. Q vs. K
Since ∆G°
=
–
RTlnK, the ∆G equation could be written as
∆G =
–
RTlnK + RTlnQ = RTln
If Q < K, then ∆G is
–
and the reaction is spontaneous as written
If Q > K, then ∆G is + and the reaction is not spontaneous as written, the reverse
r
eaction is spontaneous.
3. Example.
Given the following data at 25 °C
Substance
Δ
H
f
°
(kJ/mol)
S°(J.mol K)
NO(
g
)
90.25
210.76
N
2
O(
g
)
82.05
219.85
NO
2
(
g
)
33.18
240.06
For the reaction. N
2
O(
g
) +
NO
2
(
g
)

> 3NO(
g
),
a. calculate ∆H°, ∆S° and ∆G° at 25 °C.
∆H° = 3∆H
f
°(NO)

∆H
f
°(N
2
O)
–
∆H
f
°(NO
2
) = 3(90.25)
–
82.05
–
33.18 = 155.52 kJ/mol
∆S° = 3S°(NO)
–
S°(N
2
O)
–
S°(NO
2
) = 3(210.76)
–
219.85
–
240.06 = 172.37 J/mol K
or 0.17237 kJ/
mol K
∆G° = ∆H°
–
T∆S° = 155.52 kJ/mol
–
(298 K)(0.17237 kJ/mol K) = 104.15 kJ/mol
b. calculate ∆G when P
= 1.0x10
–
3
atm, P
= 500 atm and P
= 3.0x10
4
atm.
Q
P
=
=
6.7x10
–
17
∆G = ∆G° + RTlnQ
P
= 104.15 kJ/mol + (8.314x10
–
3
kJ/mol K)(298 K)ln(6.7x10
–
17
)
∆G = 11.9 kJ/mol The reaction is still not spontaneous
c. calculate K
P
at 25 °C.
12
lnK
P
=
–
=
–
=
–
42.039
K
P
=
e
–
42.039
= 5.5x10
–
19
Q
P
must be less than 5.5x10

19
for the reaction to be spontaneous at 25 °C.
E. Variation of
Δ
G
with Temperature. Temperature Dependence of K.
1. For small temperature changes,
Δ
H
and
Δ
S
do not change much and can be
co
nsidered as being temperature independent. However,
Δ
G
changes
with temperature.
a. The temperature dependence of
Δ
G
can be used to calculate the variation of
K
p
with temperature.
ln K =
–
=
–
=
–
b. This means that a plot of ln K
p
vs
. 1/T will be a straight line whose slope =
–
Δ
H
/R.
Note: If
Δ
H
is negative (the reaction is exothermic), then K
p
decreases as T increases.
If
Δ
H
is positive (the reaction is endothermic), then K
p
increases as T increases.
2. Plots.
C
B
ln K
P
A
1/T
Of the three reactions, A is exothermic, and B and C are endothermic. Of the two
e
ndothermic reactions, reaction B has the higher
Δ
H
.
13
3. For calculations
:
Subtract the two equations to eliminate
Δ
S
R
This equation can be used to estimate
Δ
H
if K
p
is known at two temperatures
or to estimate K
p
at some temperature when
Δ
H
is known and a K
p
at some other
temperature.
4. Example.
For the reaction N
2
O
(g)
+ NO
2(g)

> 3NO
(g)
Δ
H
= 155.52 kJ / mol,
Δ
S
= 173.37 J / mol K, and K
p
at 25
C = 6.24x10

19
.
Calculate K
p
at 500
C.
Two ways to work thi
s.
a. Calculate
Δ
G
at 500
C. Note K = 500 + 273 = 773K
Δ
G
773
= 155,520 J / mol
–
(773
K
)(173.37 J / mol
K
) = 21,505 J / mol
∴
ln K
p
=
=
–
21,505
J
/
mol
) / (8.314
J
/
mol K
)(773
K
) =
–
3.46
ln K
p
= e
–
3
.46
= 3.1x10
–
2
b. Use
K
p1
= 5.5x10

19
T
1
= 298K; K
p2
= ? T
2
= 773K
14
K
p2
/ 5.5x10

19
= e
38.572
= 5.64x10
16
K
p2
= ( 5.5x10

19
)( 5.64x10
16
) = 3.1x10

2
5.
Variation of ∆G°
with temperature
Sign of Change in ∆G° (=∆H°
–
T∆S°)
Example
∆H°
∆S°
with Temperature
(∆H° & ∆S°)
+
+
∆G° is + at low temperatures, but
H
2
O(
s
)

> H
2
O(
l
)
becomes
–
at high temperatures
∆H°= 6.01 kJ/mol, ∆S°= 22.0 J
/mol K
+
–
∆G° is + at any temperature
3 O
2
(
g
)

> 2 O
3
(
g
)
∆H°= 286 kJ/mol, ∆S°=
–
137 J/mol K
–
+
∆G° is
–
at all temperatures
2 H
2
O
2
(
l
)

> 2 H
2
O(
l
) + O
2
(
g
)
∆H°=
–
196 kJ/mol, ∆S°= 125 J/mol K
–
–
∆G° is
–
at low temper
ature, but
Na
+
(
aq
) + Cl
–
(
aq
)

> NaCl(
s
)
+ at high temperatures
∆H°=
–
3.9 kJ/mol, ∆S°=
–
43.2 J/mol K
F. SUMMARY
Using the thermodynamic tables one can:
a. Calculate the heat change at constant pressure (
Δ
H
).
b. Determine if a re
action is spontaneous under standard conditions at 25 °C.
c. Determine if a reaction is spontaneous under any set of concentrations at 25
°C.
d. Calculate the equilibrium constant of the reaction at 25 °C.
e. Calculate ∆G° at temperatures other than
25 °C.
f. Determine if a reaction is spontaneous under standard conditions at any
temperature.
g. Determine if a reaction is spontaneous under any set of concentrations at any
temperature.
h.
Estimate the equilibrium constant of the react
ion at any temperature.
15
THERMODYNAMICS PROBLEM SET
CHEM 1304
Use the table of thermodynamic data given at the end of this problem set and in the
appendix 3 of your textbook as necessary for working these problems.
1.
Without referring to the thermodyna
mic tables, indicate the sign of ∆S° for the
following reactions.
a.
3O
2
(g)

> 2O
3
(g)
b.
2H
2
O
2
(
l
)

> 2H
2
O (
l
) + O
2
(g)
S°(H
2
O
2
(l)) = 110.0 J/mol
c.
2KCl (s) + 4 O
2
(g)

> 2KClO
4
(s)
d.
Pb
2+
(aq) + 2 Cl
–
(aq)

> PbCl
2
(s)
e.
Mg(OH)
2
(s)

> MgO (s) + H
2
O (
l
)
2. Calculate ∆G° for each of the reactionsin question 1 at 25° C. Under standard
conditions, which of these reactions will be spontaneous?
3.
For which of the reactions in question1, will K increase as th
e temperature
increases?
4..
Calculate ∆S°, ∆H°, and ∆G° for the reaction in which ethane C
2
H
6
is burned in
oxygen to form H
2
O (
l
) and CO
2
(g) at 25° C.
5.
For the reaction CO (g) + 2H
2
(g)

> CH
3
OH (
l
) calculate the values of
∆G° and K a
t 25° C.
6.
Using the thermodynamic data in your text, and the fact that ∆H
f
°(CS
2
(g)) = 115.3
kJ/mol and S°(CS
2
(g)) = 237.8 J/mol K, calculate the normal boiling point of CS
2
(
l
).
7.
For the reaction PCl
3
(g) + Cl
2
(g)

> PCl
5
(g) at 25° C,
calculate ∆H°, ∆S°,
∆G°, and K.
8.
For the reaction N
2
H
4
(
l
) + 2NO
2
(g)

> 2N
2
O (g) + 2H
2
O (g) calculate:
a. ∆G°, ∆H°, and ∆S° at 25° C.
b. ∆G at 25° C when P
N
2
O
= 1.0 x 10

4
atm, P
H
2
O
= 2.0 x 10

2
atm, P
NO
2
= 1000 atm
c. K at 500 ° C
16
9.
Calculate the equilibrium vapor pressure of N
2
H
4
(
l
) at 25 °C and at 50 °C.
10.
Calculate the vapor pressure of I
2
(s) at 25°C.
Thermodynamic Data at 25 °C
Substance
∆H°
f
(kJ/mol)
S°(J/mol K)
PCl
3
(g)

287.0
311.7
PCl
5
(g)

374.9
364.5
I
2
(g)
62.4
260.6
N
2
H
4
(l)
50.6
121.2
N
2
H
4
(g)
95.4
238.4
H
2
O
2
(l)

187.8
110.0
Thermodynamics Problem Set
Answers
1. ∆S is
–
: a, c, d. ∆S is +: b, e
2.
∆H°(kJ/mol)
∆S°(J/mol

K)
∆G°(kJ/mol)
a. 3O
2
(g)

> 2O
3
(g)
284.4

139.8
326.1
b. 2H
2
O
2
(
l
)

> 2H
2
O (
l
) + O
2
(g)
–
196.4
124.8
–
233.6
c. 2KCl (s) + 4 O
2
(g)

> 2KClO
4
(s)
4.8
–
683.4
208.4
d. Pb
2+
(aq) + 2 Cl
–
(aq)

> PbCl
2
(s)
–
26.4
2.1
–
27.0
e. Mg(OH)
2
(s)


> MgO (s) + H
2
O (
l
)
37.1
33.6
27.1
reactions b and d will be spontaneous
3.
a, c
and
e
4.
2 C
2
H
6
+ 7 O
2

> 4 CO
2
+ 6 H
2
O(l)
∆S° =

621.2 J/K, ∆H° =

3119 kJ, and ∆G° =

2935 kJ
5.
∆G° =

166.3
–
(
–
137.3) =
–
29.0 kJ
ln K = 2.90x10
4
/(8.31)(298) = 11.71 ; K = 1.21x10
5
=
1
P
co
P
H
2
2
6.
CS
2
(l)

> CS
2
(g) ∆H° = 27.4 kJ, ∆S° = 86.8 J/K, T = 315.7 K o
r 42.5° C
7.
∆H° =

87.9 kJ ∆S° =

170.2 J/K
∆G° =

37.2 kJ K = 3.3x10
6
8.
a. ∆G°, ∆H°, and ∆S° at 25° C.
∆H° =

439.0 kJ, ∆S° = 215.2 J/K, ∆G° =

503.2 kJ
17
b. ∆G at 25° C when P
N
2
O
= 1.0 x 10

4
atm,
P
H
2
O
= 2.0 x 10

2
atm, P
NO
2
= 1000 atm
∆G =

602.4 kJ
c. K at 500 ° C
K = 8.0x10
40
9.
14.3 Torr, 57.9 Torr
10.
∆G°
= 19.34 kJ/mol; P = 0.31 Torr
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