Download Chemistry 520 - Problem Set 6

Chemistry 520 - Problem Set 6
Due Wednesday February 17, 1999
1. Chang, problem 6.14 (page 6.37)
The dierences come from the dierences in thedenitions of the zero in which is given
by the third law of thermodynamics and f and f where the values are zero for
elements in their standard states at 0 K.
If instead of comparing f = ,394 359 kJ mol,1 to
S
H
G
f ,
H
TS
G
:
= ,393 509 , 298 0 21374 = ,457 20 kJ mol,1
:
:
:
we use,
f = f , f G
H
T
S
where
f = (CO2 ) , (C) , (O2 )
= 213 74 , 5 74 , 205 138 = 2 862 J mol,1 K,1
S
S
S
:
S
:
:
:
:
Using this value for f ,
S
f , f = ,393 509 , 298 0 002862 = ,394 36
H
T
S
:
:
:
which is in good agreement with f 2. Chang, problem 6.15 (page 6.38)
For a reaction to be spontaneous, r must be smaller than zero. Therefore, the minimum
value of entropy necessary for the reaction to occur at 72 C = 345 K when r = 19 kJ
is
G
G
H
r , r
r
H
T
S
<
S
>
0
r
H
T
19 000 = 55 J mol,1 K,1
345
>
3. Chang, problem 6.16 (page 6.38)
NO! All that tells us is if a reaction is thermodynamically favored. It does not tell us
how long the reaction will take.
G
1
4. The thermodynamic data for transitions from ordered, helical conformations of polypeptides and polinucleotides to disordered states gives entropy and enthalpy changes that are
positive, as expected. Heat is absorbed in these transitions and entropy (or disorder) increases. However, when the synthetic polypeptide, polybenzyl-L-glutamate undergoes a
transition from an ordered helix to a disordered coil in an ethylene dichloride-dichloroacetic
acid solvent at 39C and 1 atm, r = ,4 0 kJ (per mol amide) and r = ,12 J/K
(per mol amide).
H
:
S
(a) Give a possible explanation for the experimental result that heat is released and
entropy is decreased for this transition. Does increasing the temperature favor the
helix-coil transition?
Both of these observations can be explained by the interactions of the two forms
of the polypeptide with the solvent. If more solvent molecules are bound to the
coiled form than the helix then the reaction will be exothermic (products are at
lower internal energy than the reactants) and the products are more ordered. Since
r 0, increasing the temperature will make r less negative, making the helixcoil transition less favorable.
(b) Is the reaction spontaneous at 39 C? What thermodynamic criterion did you use to
reach this conclusion?
Since r
r at this temperature, r , r
0 and the reaction is
spontaneous at 39C.
(c) At what temperature ( C) will the helix-coil reaction be reversible? This temperature
is often called the \melting" temperature for the helix. Assume that 0 and 0
are independent of temperature. [60 C]
S <
G
H
< T
S
H
T
S
<
H
r , r
H
T
S
= 0
r = ,4000 = 333 K = 60 C
= r
,12
S
H
T
S
(d) Can a reaction occur in an isolated system which leads to a decrease in the entropy
of the system? If it can, give an example; if it cannot, state why not.
No. For an isolated system, = 0 which means that surr = 0 and univ =
+ surr 0 which violates the second law of thermodynamics!
q
S
S
S
S
<
5. An electrochemical battery is used to provide 1 milliwatt of power for a (small) light. The
chemical reaction in the battery is:
1
3
2 N2 (g) + 2 H2 (g) ,! NH3 (g)
(a) What is the free energy change for the reaction at 25C, 1 atm?
(1)
r = f (NH3 ) , 12 f (N2 ) , 23 f (H2 )
= f (NH3 ) = ,16 45 kJ mol,1
G
G
G
G
G
:
(b) Calculate the change in Gibbs free energy for this reaction at 50C, 1 atm. State any
assumptions made in the calculation.
If we assume that the enthalpy and the entropy are both independent of temperature,
r (323) = r (298) , (323 , 298) r
G
G
where
2
S
r
(NH3 ) , 12 (N2 ) , 32 (H2 )
= 192 45 , 21 191 61 , 23 130 684 = ,99 38 J mol,1 K,1
=
S
S
S
S
:
:
:
:
Therefore,
r (323) = ,16 45 , 25 (,0 09938) = ,13 97 kJ
G
:
:
:
(c) The limiting reactant in the battery is 100 g of H2 . Calculate the maximum length
of time (sec) the light can operate at 25C. [5 44 108 s]
From the answer to part (a), the maximum amount of work that can be done by the
battery is 16 450 J/ mol NH3 . For 100 g of H2 ,
1 mol H g H 100 g H = 544 000 J
= 16 450 J mol NH3 23 mol NH3 mol H2 2 016
2
2
2
:
w
=
=
=
:
Since the batter is operating at 1 mwatt (or 10,3 J/sec), the battery can run for
5 44 108 s or 17 years!
:
6. The earth's atmosphere behaves as if it is approximately isentropic { the molar entropy of
air is independent of the altitude up to about 10 km. It is well known that pressure and
temperature vary with altitude.
Calculate the temperature of the atmosphere 10 km above the earth where = 210 Torr.
The temperature and pressure at the surface of the earth (0 km) are 25C and 760 Torr,
respectively. You may assume air behaves as an ideal gas with P = 72 . You may ignore
gravitational inuences. [-66.6C]
To solve this problem, we use the relationship derived in class, taking = 0 for an
isoetropic system.
P
nR
C
S
S
=
P
C
0 = 72
ln
72
f
Ti
T
f
T
nR
= ln
=
i
T
ln
f
Ti
T
ln
P
f
Ti
T
,
nR
,
ln
nR
ln
f
i
P
P
f
Pi
P
f
i
P
27
f
i
P
P
= 298 210
760
27
= 206 K = ,66 6 C
:
7. At 298 K the standard enthalpy of combustion of sucrose is -5645 kJ/mol and the standard
Gibbs free energy of the reaction is -5797 kJ/mol. Estimate the additional non-expansion
work that may be obtained by raising the temperature to blood temperature, 37C.
(298) + 1 , 1 r (310) = 310 r 298
310 298 r
G
G
H
= ,5803 kJ mol,1
Therefore, the maximum amount of non-PV work is 5803 kJ per mole of sucrose, and the
additional non-PV work is 5803-5797 = 6 kJ.
3