Chemistry 360
Dr. Jean M. Standard
Fall 2016
Name _____________KEY________________
Exam 2 Solutions
#∂ H &
( for a gas obeying the equation of state
1.) (14 points) Determine %
$ ∂ P 'T
Z = €
PVm
= 1 + BP + CP 2 ,
RT
where B and C are constants.
Since a partial derivative of H is requested, it is helpful to start with the fundamental equation for H,
dH = T dS + V dP.
Take the partial derivative with respect to V (at constant T) of both sides of the equation,
!∂ H $
!∂ S $
#
& = T #
& + V .
" ∂ P %T
" ∂ P %T
The partial derivative on the right side of the equation cannot be determined from the equation of state.
However, one of the Maxwell relations can be used,
!∂ S $
!∂ V $
#
& = − #
& .
" ∂ P %T
" ∂ T %P
Substituting yields
!∂ H $
!∂ V $
#
& = −T #
& + V .
" ∂ P %T
" ∂ T %P
Now, the partial derivative on the right side can be calculated from the equation of state. Since the equation of
PV
state is Z = m = 1 + BP + CP 2 , solving for Vm gives
RT
PVm
= 1 + BP + CP 2
RT
RT
Vm = 1 + BP + CP 2
P
RT
or Vm = + BRT + CRTP .
P
(
)
Now, converting to V instead of Vm leads to the expression
V = €
nRT
+ nBRT + nCRTP .
P
2
1.
Continued
Taking the partial derivative,
!∂ V $
nR
#
& = + nBR + nCRP .
P
" ∂ T %P
Substituting into the expression for the partial derivative of H gives the result,
!∂ H $
!∂ V $
#
& = − T #
& + V
" ∂ P %T
" ∂ T %p
! nR
$
= −T # + nBR + nCRP & + V
"P
%
! nRT
$
= − #
+ nBRT + nCRTP & + V
" P
%
= −V + V
!∂ H $
#
& = 0.
" ∂ P %T
3
ΔH !f
2.) (14 points) The standard molar enthalpy of formation of Fe2O3 (s) is
ΔH !f
standard molar enthalpy of formation of SO2 (g) is
= −824.2 kJ/mol , and the
= −296.8 kJ/mol (both at 298 K). Use this
information, along with the standard molar enthalpy change of the following reaction at 298 K,
€
ΔH R! = −1655 kJ/mol,
2 FeS2 (s) + 112 O 2 (g) → Fe
€ 2O3 (s) + 4 SO 2 (g)
to determine the standard molar enthalpy change of the reaction shown below at 298 K:
Fe (s) + 2 S (s)
→
FeS 2 (s) .
[Note: The standard state of iron is Fe(s) and the standard state of sulfur is S(s) at 298 K.]
€
The formation reaction for Fe2O3 is
2 Fe (s) +
3O
2 2
(g )
→ Fe 2O3 (s) .
The formation reaction for SO2 is
S (s) + O 2 (g) → SO 2 (g) .
The desired reaction can be constructed from the reactions given in the problem in the following way:
2
1
2
1
2
[
[ S (s)
[2 Fe (s)
+ O 2 ( g) → SO 2 ( g)
+
3
O
2 2
→ Fe 2O 3 (s)]
( g)
Fe 2O 3 (s) + 4 SO 2 (g)
Fe (s)
+ 2 S (s)
]
→
2 FeS 2 (s) +
11
2
O 2 (g)
]
FeS 2 (s)
→
From the combination of reactions above, the enthalpy of reaction can be calculated from Hess' Law,
€
ΔH !f ( FeS 2 ) = 2ΔH !f (SO 2 ) + 12 ΔH !f ( Fe 2O 3 ) − 12 ΔH r! ,
!
where ΔH r corresponds to the standard molar enthalpy of the reaction between FeS2 and O2 given in the
€
problem. Substituting,
ΔH !f ( FeS 2 ) = 2ΔH !f (SO 2 ) +
€
1
2
ΔH !f ( Fe 2O 3 ) −
= 2(−296.8 kJ/mol) +
ΔH !f
€
( FeS 2 )
= − 178.2 kJ/mol.
1
2
1
2
ΔH r! ,
(−824.2 kJ/mol)
−
1
2
(−1655 kJ/mol)
4
3.) (14 points) Explain why we are able to obtain absolute molar entropies of substances; that is, what is the
theoretical foundation that allows the determination of absolute molar entropies? Discuss any trends
that may be observed in the magnitudes of the absolute molar entropies of the following substances and
the reasoning behind your answer: He (g), O2 (g), CH4 (g), C6H6 (g).
The opportunity for an absolute entropy scale is provided by the Third Law of Thermodynamics, which sets the
absolute entropy of a perfect crystalline substance as 0 at 0 K.
The absolute entropies tabulated above are all for gases of varying structural complexity. As the substances
vary from a monatomic gas (He) to a diatomic gas (O2) to polyatomic gases (CH4 and C6H6), the entropy is
expected to increase. This is a result of the increase in the number of degrees of freedom of the substances. As
we saw in our discussion of the Equipartition Theorem, the number of degrees of freedom increases as the
complexity of the molecule increases. For the complex molecules with more degrees of freedom, there are
more ways of distributing the available energy among the degrees of freedom, which increases the disorder in
the system. Thus, more complex substances have higher entropies than less complex substances.
5
4.) (15 points) True/false, short answer, multiple choice.
a.) True or False : In the statistical definition of entropy, S = kB lnW , the quantity W corresponds to the
energy.
b.) True or False : The Debye equation gives the form of the constant pressure molar heat capacity of a
solid at low temperature and has the form C !p,m
(T ) = aT 2 .
c.) Short answer
_____Adiabatic (or bomb)____ calorimetry is carried out at constant volume, whereas
_____Solution______ calorimetry is carried out at constant pressure.
d.) Short answer
For two chemical reactions that combine to give a third reaction (the overall reaction), the enthalpy change
of the overall reaction (3) is given as ΔH 3 = ΔH1 + ΔH 2 , which is an example of the application of
_________Hess' Law________ .
€
e.) Multiple Choice: Which of the following corresponds to the correct relationship?
!∂ G $
1) T = #
& .
" ∂ P %S
"∂ U %
2) T = − $
' .
# ∂ V &S
"∂ A %
3) T = − $
' .
# ∂ V &S
4)
!∂ H $
T =#
& .
" ∂ S %P
6
5.) (14 points) Using the values in the table below reported at 298 K, determine
hydrogenation of acetylene to ethane,
ΔH R!
at 700 K for the
C2 H 2 ( g) + 2 H 2 (g) → C2 H 6 (g) .
Assume that C p,m is independent of temperature for all species in the reaction.
€
C2H2 (g)
C2H6 (g)
ΔH !f
C p,m
(kJ/mol)
(Jmol–1K–1)
226.73
€
–84.68
H2 (g)
43.93
€
52.63
0
28.82
The first step is to determine the enthalpy of reaction at 298 K:
ΔH R! = ΔH !f (C2 H 6 ) − ΔH !f (C2 H 2 ) − 2ΔH !f ( H 2 )
= −84.68 kJ/mol − ( 226.73 kJ/mol) − 2 ( 0 )
ΔH R! = −311.41 kJ/mol.
The temperature dependence of the molar enthalpy of reaction is given by
ΔH R! (T2 ) = ΔH R! (T1 ) + ΔC p,m ⋅ (T2 − T1 ).
This equation assumes that the molar heat capacities are independent of temperature. In the equation, ΔC p,m is
the difference in molar heat capacities between products and reactants with the appropriate signed
stoichiometric coefficients included,
€
∑ν C
ΔC p,m =
i
p,m
(i )
.
i
For this reaction, the heat capacity difference is
€
ΔC p,m = ∑ν iC p,m (i )
i
= C p,m (C2 H 6 ) − C p,m (C2 H 2 ) − 2C p,m (O 2 )
= 52.63 − 43.93 − 2 ( 28.82 ) Jmol−1K −1
ΔC p,m = −48.94 Jmol−1K −1.
Substituting, the molar enthalpy of reaction at 700 K is
ΔH R! ( 700 K ) = ΔH R! ( 298 K ) + ΔC p,m ⋅ ( 700 − 298 K )
⎛ 1 kJ ⎞
= −311.41 kJ/mol + −48.94 Jmol−1K −1 ( 700 − 298 K ) ⎜
⎟
⎝ 1000 J ⎠
(
= −311.41 −19.67 kJ/mol
ΔH R! ( 700 K ) = −331.08 kJ/mol .
)
7
6.) (14 points) Calculate the entropy change ΔS for the following transformation involving one mole of
H2O,
€ H 2O (s, 200 K) → H 2O (ℓ, 300 K) .
You may need to use the following data:
€
!
−1
T fus = 273 K ; ΔH fus = 6.01 kJ mol
!
−1
T vap = 373 K ; ΔH vap = 40.65 kJ mol
€
€
€
C p,m (s) = 36.2 J mol−1K −1 ; C p,m ( ℓ) = 75.3 J mol−1K −1 ; C p,m ( g) = 33.6 J mol−1K −1
€
€
The process may be broken up into three steps: (1) heating the solid from 200 to 273 K; (2) the S-L phase
€
€
transition at 273 K; and (3) heating the liquid from 273 to 300 K. The molar entropy change can then be
written as a sum of these three steps,
ΔSm = ∫
C p,m ( s)
T fus
T
T1
dT + ΔH !fus
T fus
+ ∫
T2
C p,m ( ℓ)
T fus
T
dT .
Assuming that the heat capacities are independent of temperature (they are listed in the problem as constants),
the expression becomes
ΔSm = ∫
T fus
C p,m ( s)
T
T1
= C p,m ( s) ∫
T fus
T1
dT + ΔH !fus
T fus
+ ∫
T2
C p,m ( ℓ)
T fus
T
dT
ΔH !fus
T2 1
1
dT + + C p,m ( ℓ) ∫
dT
T fus T
T
T fus
⎛T ⎞
⎛T ⎞
ΔH !fus
ΔSm = C p,m ( s) ln ⎜ fus ⎟ + + C p,m ( ℓ) ln ⎜⎜ 2 ⎟⎟ .
T fus
⎝ T1 ⎠
⎝ T fus ⎠
Substituting,
⎛T ⎞
⎛T ⎞
ΔH !fus
ΔSm = C p,m ( s) ln ⎜ fus ⎟ +
+ C p,m ( ℓ) ln ⎜⎜ 2 ⎟⎟
T fus
⎝ T1 ⎠
⎝ T fus ⎠
⎛ 273K ⎞
(6010 J/mol)
= 36.2 J mol−1K −1 ln ⎜
⎟ +
273K
⎝ 200 K ⎠
(
)
+
( 75.3J mol
−1
⎛ 300 K ⎞
K −1 ln ⎜
⎟
⎝ 273K ⎠
)
= 11.3 + 22.0 + 7.1 J mol−1K −1
ΔSm = 40.4 J mol−1K −1 .
−1
So, for one mole of H2O, the entropy change is ΔS = 40.4 J K .
€
8
7.) (15 points) True/false, short answer, multiple choice.
a.) True or False: For the reaction 2 CO (g) + O2 (g) → 2 CO2 (g), the entropy change is expected to be
negative.
b.) True or False : The standard state is defined to be the pure form of an element at 1 bar and 25°C.
c.) Short answer
The ___Second Law of Thermodynamics____ states that entropy increases for a spontaneous process in
an isolated system.
d.) Short answer
The _______Helmholtz Energy A_________ is defined as U − TS .
e.) Multiple Choice: At 298 K, ΔGR! = –158 kJ/mol for a certain chemical reaction. In addition, this reaction
has ΔH R! > 0 . Which of the following is a reasonable expectation for ΔGR! at 500 K?
1) ΔGR! = –195 kJ/mol .
2) ΔGR! = –158 kJ/mol .
3) ΔGR! = –141 kJ/mol .
4) ΔGR! = +158 kJ/mol .