Download Solutions for Chapter 8 End-of-Chapter Problems

Solutions for Chapter 8 End-of-Chapter Problems
Problem 8.1.
(a) Ice cream melting will decrease the organization of the system. Rather than being in
identifiable scoops of ice cream perched securely on a cone, there now will be liquid ice cream
dripping on the outside of the cone, onto your hands, and possibly onto your clothes as well.
(b) Students entering a classroom with fixed seating are about to undergo an increase in
organization. The chairs are permanently arranged so that a specified order will be maintained.
(c) A shrub forming spring flowers will be increasing the organization of the system as these
new structures are formed. Complex molecules are being arranged into even more complex
cells, which in turn are organized into new units of the shrub.
(d) Obtaining pure water from seawater requires an increase in the organization of the system.
Starting with the salt solution, water molecules are separated from the solution, leaving behind a
more concentrated brine solution (or even the previously dissolved salts as crystalline solids).
(e) Shuffling a new deck of cards decreases the organization of the deck as the ordered
sequence of suits and values is mixed up.
(f) Raking leaves into a single pile definitely represents an increase in organization, at least
until the next puff of wind comes along.
Problem 8.2.
A water-soluble dye spreads out when it is mixed with water because the mixed state is more
probable (see Section 8.3 for why it is more probable). We know from everyday life that the
reverse process never happens, (because the unmixed final state would be less probable).
Problem 8.3.
(a) It is unlikely that liquid food coloring added to cookie dough will spontaneously disperse
through the semisolid cookie dough in a finite amount of time, because the molecules within the
dough are not free to translate from one part to another. If you do enough mixing and stirring, it
will be possible to turn the entire sample of dough the desired green color, but this external
input of mechanical mixing is necessary.
(b) Drops of green food coloring are more easily mixed with a liquid than with cookie dough.
The mixing starts as the drops are added to the liquid, and molecular motion in the liquid will
enable the process to continue because all the molecules in the system are in constant
translational motion without any external influence (although stirring will disperse the dye
molecules more quickly).
Problem 8.4.
If an intravenous infusion having a higher concentration than the solution within the red blood
cells was given, water from the blood cell would move out of the blood cell causing crenation
(shriveling) of the cells.
Problem 8.5.
(a) Swimming in a freshwater lake makes your body water-logged and increases your need to
urinate, because osmosis occurs to transfer water through your semipermeable cell membranes
to your more concentrated body solutions from the fresh water.
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Chapter 8
(b) Cucumbers wrinkle, lose water and shrink when placed in brine (concentrated salt water
solution), because water from the less concentrated liquid inside the cucumber is transferred by
osmosis through the semipermeable cell membranes into the higher concentration brine
solution.
(c) Prunes and raisins swell in water, because water is transferred by osmosis through the
semipermeable cell membranes of the prune or raisin skin into the more concentrated solution
within the dried fruit.
(d) You get thirsty and your skin wrinkles when you swim in the ocean, because the salty ocean
water is more concentrated than your body fluids and water is transferred by osmosis through
the semipermeable cell membranes in your skin into the ocean. The loss of water from the cells
near the surface of the skin causes them to change shape and size, resulting in the wrinkles that
disappear when their normal complement of water is restored. Your body reacts to the need to
replace the lost water by causing you to be thirsty.
(e) Salt and sugar are used to preserve many foods, because the bacteria, which cause the food
to spoil, die when water is transferred out of the liquid in the cells by osmosis through their
semipermeable cell membranes.
(f) Many saltwater fish die when placed in fresh water and vice versa, because osmosis through
the semipermeable membranes of their skin causes the solute concentrations in the fish’s body
tissue to decrease (saltwater fish in fresh water) or decrease (freshwater fish in salt water).
Problem 8.6.
When a dye solution mixes with water in an unstirred container, the mixing occurs because:
(i) the mixed state has a higher number of distinguishable arrangements than the unmixed state.
This is true, given our model that says the more arrangements, the higher the probability that the
state will be observed.
(ii) the mixed state has a lower heat content than the unmixed state. This is false, because
mixing like this takes place without any transfer of energy in or out of the system.
(iii) the mixed state is more probable than the unmixed state. This is true, given our model that
says the more arrangements, the higher the probability that the state will be observed.
Problem 8.7.
(a) The probability that a broken eggshell will reform into an unbroken eggshell is 0. A broken
eggshell has never been observed reforming itself into an unbroken egg. It might be possible to
painstakingly glue all of the pieces back into a whole shape, but this would require considerable
work (and skill) on your part and would not actually recreate the original shell (which had no
glue).
(b) The probability that a drop of food coloring will disperse throughout a cup of water at room
temperature is 1. A drop of food coloring, given enough time, can disperse throughout a cup of
water at room temperature. The molecules of water and of dye are constantly in motion, helping
to achieve the complete mixture.
(c) The probability of drawing the ace of spades from a shuffled deck of cards is 1/52 or 0.019.
There is only one ace of spades in a deck of 52 cards.
(d) The probability of finding all of the pepperoni slices on one half of a pizza surface should be
close to 0. Unless a pizza has been ordered to be only half pepperoni, the consumer expects that
a pepperoni pizza will have the slices even distributed over the entire pizza. There is the
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possibility of human error in placing the pepperoni, but such an initial error would likely be
detected by the worker or the supervisor in charge.
(e) The probability of a large oak tree falling to the ground and then returning to the upright
position is 0. The large oak tree fell to the ground because of disease, severe weather, or
intentional cutting. A large, heavy tree, responding to gravitational force, has never been
observed to spontaneously right itself. It is true that smaller trees, responding to circular winds
associated with hurricanes or typhoons, may bend first one way and then back the other way.
The probability of such trees successfully righting themselves and remaining healthy is low.
(f) The probability of recovering six pairs of clean socks from the dryer after washing six pairs
of dirty socks should equal 1, given that the person doing the washing is careful and looks for
all the socks that may be hidden in other clothes or those that are dropped somewhere in the
process. However, in the experience of this author, the probability hovers mysteriously around
0.
Problem 8.8.
Examples from daily life that illustrate the statement, “If a system can exist in more than one
observable state, any changes will be in direction toward the state that is most probable,” might
include, water at room temperature which exists in a liquid state rather than solid (ice) or
gaseous (steam). Therefore the steam evolved during boiling water condenses in room
temperature, and a cube of ice melts taken out from freezer melts while left at room
temperature.
(a) The most probable state for water at –5 ° C will be ice, because this temperature is below the
freezing point of water.
(b) The most probable state for water at 130 ° C will be steam (gas), because this temperature is
above the boiling point of water.
(c) The most probable state for sugar crystals in a glass of hot water will be dissolved in the
water, because sugar is a soluble compound and, given enough time, will dissolve completely in
water (as long as the saturated concentration is not attained).
(d) The most probable state for a drop of perfume in a room will be to evaporate and spread
through the room. Perfume is made to be volatile (so it will enter the gas phase and make it to
the olfactory sensors in our nose), so it will evaporate and then the perfume molecules will mix
with the air (like dye molecules mixing into water, except faster, because the gas molecules are
further apart and travel further between collisions than molecules in a liquid).
(e) The most probable state for iron filings in a magnetic field is to line up along the magnetic
field lines as illustrated in Chapter 4, Section 4.3, Figure 4.11.
Problem 8.9.
(a) There are ten possible arrangements for two identical objects in five labeled boxes (one
object per box), as shown in this table.
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Arrangement Box 1
1
X
2
X
3
X
4
X
5
6
7
8
9
10
Box 2
X
Box 3
Box 4
Box 5
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
(b) This is the same number of arrangements that we found in Investigate This 8.4 for placing
three identical objects in five boxes. To see why this is so, compare the arrangements you
tabulated in Investigate This 8.4 with this set of arrangements. For every arrangement in which
there are three objects, there is a corresponding arrangement here in which the occupied and
unoccupied boxes have swapped identities. Three occupied boxes and two unoccupied boxes
have the same number of arrangements as two occupied boxes and three unoccupied boxes. The
reason is that “emptiness” can be arranged just like identical objects, so it does not matter
whether we focus on arranging objects in boxes or “emptiness” in boxes. The result is the same
in terms of number of arrangements. (See Problem 8.95 for another system like this.)
Problem 8.10.
(a) There are 15 different arrangements for two dye molecules mixing into six cells (as in
Figure 8.5). These are the possibilities (space is conserved by picturing the layers side by side).
Arrangement
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
4
Cell 1
X
X
Cell 2
X
X
Cell 3
Cell 4
Cell 5
Cell 6
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
ACS Chemistry Chapter 8 suggested solutions
X
X
X
X
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(b) Two dye molecule mixing into the top two layers of three cells produce 15 different
arrangements (three with both molecules in the top layer, three with both molecules in the next
layer, and nine with one molecule in each layer). In Figure 8.5, three dye molecules were mixed
into the top two layers, producing 20 different relationships. There are more possible
arrangements when there are more molecules. (Note that this result is different from the
relationship of Problem 8.9 to Investigate This 8.4. Here, the two dye molecules do not occupy
all of the spaces unoccupied by dye molecules in Figure 8.5.)
(c) These results suggest that the more particles that mix into a given volume, the greater the
number of arrangements. This is true to a point. In these countable systems, when the number of
solute molecules exceeds the number of solvent molecules, the number of arrangements begins
to decrease. When this happens, you can think of the solute and solvent molecules exchanging
roles, since the “solvent” is usually taken as the species in excess. In real systems, such as
aqueous solutions, this role reversal can usually only occur for solutes that are miscible with the
water, solutions of methanol or ethanol, for example.
Problem 8.11.
In Figure 8.6, as the number of boxes per molecule increases from 2 to 20, the number of
possible arrangements for three solute molecules in the cells increases far more than ten fold.
Figure 8.6 shows that for 20 boxes per molecule, the value of W is a little more than 34,000. For
two boxes per molecule (the case shown in Figure 8.5), the value of W is 20. The approximate
ratio (factor of increase) is 1,700.
Problem 8.12.
(a) The number of distinguishable arrangements, W, of n identical objects among N labeled
boxes for N = 6 and n = 3, is:
N!
6!
6 ⋅5 ⋅ 4 ⋅3⋅ 2 ⋅1
6 ⋅5 ⋅ 4
6 ⋅5 ⋅ 4
=
=
=
=
= 5·4 = 20
W=
(N − n)!n! (6 − 3)!3! (3⋅ 2 ⋅1)(3⋅ 2 ⋅1) (3⋅ 2 ⋅1) (3⋅ 2 ⋅1)
This result is the same as the number of arrangements shown in Figure 8.5.
(b) These are the numbers of distinguishable arrangements for the mixtures represented in
Figure 8.6. In cases where N becomes quite a bit larger than n, the first step in calculating the
ratio of factorials is to cancel (N – n)! from the numerator and denominator. Then cancel n! in
the denominator with appropriate terms in the numerator before doing the multiplications. These
results are the same as those given in Figure 8.6.
1 1
3!
=
= = 1 (because 0! ≡ 1)
N = 3, n = 3: W =
(3 − 3)!3! 0! 1
N = 6, n = 3: W = 20 [See part (a).]
9 ⋅8 ⋅7 ⋅ 6! 9 ⋅8 ⋅7
9!
N = 9, n = 3: W =
=
=
= 3·4·7 = 84
(9 − 3)!3!
6!3!
3⋅ 2 ⋅1
12 ⋅11⋅10
12!
N = 12, n = 3:
W=
=
= 2·11·10 = 220
(12 − 3)!3!
3 ⋅2 ⋅1
15 ⋅14 ⋅13
15!
N = 15, n = 3:
W=
=
= 5·7·13 = 455
(15 − 3)!3!
3 ⋅2 ⋅1
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N = 30, n = 3:
N = 45, n = 3:
N = 60, n = 3:
Chapter 8
30 ⋅ 29⋅ 28
30!
=
= 5·29·28 = 4,060
(30 − 3)!3!
3⋅ 2 ⋅1
45 ⋅ 44 ⋅43
45!
W=
=
= 15·22·43 = 14,190
(45 − 3)!3!
3 ⋅2 ⋅1
60 ⋅ 59⋅ 58
60!
W=
=
= 10·59·58 = 34,220
(60 − 3)!3!
3⋅ 2 ⋅1
W=
(c) For a mixture of 5 objects in 20 boxes and a mixture of 10 objects in 20 boxes, we find hat
the numbers of distinguishable arrangements are:
20 ⋅19⋅18 ⋅17 ⋅16
20!
N = 20, n = 5: W =
=
= 1·19·3·17·16 = 15,504
(20 − 5)!5!
5 ⋅4 ⋅ 3⋅ 2 ⋅1
20!
20!
N = 20, n = 10: W =
=
= 184,756 (used calculator)
(20 − 10)!10! 10!10!
The conclusion in Problem 8.10(c) is reinforced by these results: the more solute molecules in a
given volume, the greater the number of arrangements. The same restriction applies here. To see
the affect on W when n > N/2, calculate W for N = 20 and n = 9 and 11. Is the result at all
surprising? Why? Can you explain why you get this result? What other choices of n would give
similar results? (See Problem 8.95 for another system like this.)
Problem 8.13.
In order to increase the concentration of sucrose in a plant cell (normally 0.5%) surrounded by a
semipermeable membrane through which water can pass but sucrose cannot, place the cell in a
solution with concentration of sucrose greater than 0.5% [process (ii)]. Some of the water inside
the cell will move by osmosis through the semipermeable membrane into the solution that is
more concentrated in sucrose. Since water leaves the cell, the concentration of the sucrose in the
cell would increase. Placing the cell in pure water [process (i)] will result in water from the
outside passing into the cell by osmosis, thus diluting the concentration of sucrose in the cell.
Placing the cell in a sucrose solution with a concentration less than 0.5% [process (iii)] will
result in water from this lower concentration solution to pass into the cell by osmosis, thus
diluting the sugar concentration. In processes (ii) and (iii), osmosis will continue until the
concentration of sucrose inside and outside the cell is the same (as long as the cell and its
membrane stay intact). For case (i), water will continue to enter the cell until it either ruptures or
the pressure inside rises enough to prevent further osmotic flow (see Section 8.13). Plant cells
have a strong wall surrounding their membrane and are quite resistant to rupture.
Problem 8.14.
(a) Starting with the initial arrangement of solute and solvent molecules shown in Figure 8.7(a),
the number of molecular arrangements will increase after osmosis takes place. The maximum
number of molecular arrangements results from spreading the solute through the largest volume
of solution possible, and the volume of solution is increasing because of movement of solvent
molecules into the solution (osmosis).
(b) This diagram shows what will happen if six of the nine solvent molecules in Figure 8.7(a)
pass from the solvent into the solution. The solid dots indicate solute molecules.
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initial [Figure 8.7(a)]
final
The number of possible distinguishable arrangements, Wsoln, of the three solute molecules, n =
3, in twelve cells (boxes), N = 15, is:
15 ⋅14 ⋅13
N!
15!
Wsoln =
=
=
= 5·7·13 = 455
( N − n )!n! (15 − 3)!3!
3 ⋅2 ⋅1
The number of possible distinguishable arrangements for the (three) solvent molecules on the
left in the final state is unity: Wsolv = 1. Therefore, the number of possible distinguishable
arrangements for the system in its final state is, Wsyst = Wsoln·Wsolv = 455·1 = 455. This value can
be compared to 84 for the initial system [from Figure 8.7(a)] and 220 [from Figure 8.7(b)] when
three of the water molecules have moved through the semipermeable membrane. Continuation
of the osmotic flow of solvent is favored.
(c) The process that is taking place in this system is osmosis, not reverse osmosis. In (forward)
osmosis, solvent molecules move through a semipermeable membrane from a less concentrated
solution (or pure solvent) to a more concentrated solution, which dilutes the more concentrated
solution. The net effect is to increase the number of molecular arrangements for the system,
which provides the direction for the change. (Reverse osmosis is introduced in Problems 8.84,
8.85, and 8.86.)
Problem 8.15.
When red blood cells, with a concentration of solute particles of about 2%, are placed in a 3%
sucrose solution [solution (iii)], water will move out of the red blood cells by osmosis, causing
the cells to shrivel (crenate). If the red blood cells are placed in either distilled water [solution
(iv)] or 1% sucrose solution [solution (i)], the concentration of solute outside the cells is lower
than inside, so water will pass into the cells by osmosis and the cells will swell. If the red blood
cells are placed in a 2% sucrose solution [solution (ii)], there will be no net movement of water
into or out of the cells, because the concentration of particles is the same on both sides of the
membrane. Water will still pass through the membrane, but the same amount will pass into the
cell as passes out.
Problem 8.16.
(a) If energy is transferred from a warmer object in contact with an identical, but cooler, object,
the number of energy arrangements after the transfer is larger than the number before the
transfer. Initially, the number of possible arrangements of energy quanta in the warmer object is
greater than the number in the cooler object. As thermal energy is transferred, the number of
arrangements in the warmer object will decrease and the number of arrangements in the cooler
object will increase. At any point in this process, the product of the number of energy
arrangements in the warm and cool object will be larger than any previous to this point. When
transfer has occurred until the two identical objects have the same number of energy quanta, the
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product of the number of energy arrangements in each (which will be identical at this point) will
reach a maximum. Any net transfer in either direction will result in a lower number of energy
arrangements for the system. At this point, the two objects will have the same temperature.
(b) The two objects are specified as being identical so their masses must be the same and they
must be made of the same material. When net energy transfer stops, the final temperature will
be halfway between the initial warmer and the cooler temperatures (or, looked at another way,
the final temperature will be the average of the initial temperatures).
Problem 8.17.
(a) The observable property that differs between two identical four-atom solids, one having
four quanta of energy and the other having ten, is the temperature. The four-atom solid with ten
energy quanta has a higher temperature than the four-atom solid with four energy quanta.
(b) To determine whether transfer of three quanta of energy from the solid with ten quanta to
the solid with four quanta of energy is likely, we need to compare the total number of
distinguishable arrangements of the energy in the initial and final states. We can use the data in
Figure 8.9 to find the number of energy arrangements, W, in each of the solids under each
condition. Values taken from the graph [calculated with the equation in Problem 8.19.] are:
number of quanta
4
7
10
W
35
120
286
The total number of arrangements in a system is the product of the number of arrangements
possible in each independent component of the system. For the initial state:
Wtot = W10 quanta × W4 quanta = 286 × 35 = 10,010
For the final state, after the transfer of two quanta:
Wtot = W7 quanta × W7 quanta = 120 × 120 = 14,400
There are about 1.4 times as many ways to arrange the energy quanta when both solids have
seven quanta compared to when one solid has four and the other has ten. The larger the number
of possible arrangements, the more likely it is that a given condition will exist, so the proposed
energy transfer is favored.
(c) In this example, the number of ways to arrange the energy quanta when the transfer has
occurred is about 1.4 times as many as the original number of possible arrangements. In Worked
Example 8.8, in which two quanta were transferred from a four-atom solid initially having eight
quanta to one with four quanta, the final state has about 1.2 times as many possible arrangement
as the initial state. In both cases, the number of possible arrangements increases when energy is
transferred from a warmer to a cooler object and both transfers are likely to occur, as we know
from experience.
Problem 8.18.
Thermal energy moves from a hot body to a cold body, because the number of distinguishable
arrangements for the energy quanta is higher in the final state [statement (iii)]. Energy is
conserved in this transfer (and no work is done), so the amount of thermal energy is constant.
Thus, statements (i) and (ii) are incorrect. If the number of arrangements were higher in the
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initial state, as statement (iv) says, then no transfer of energy would occur, since spontaneous
changes occur in a direction that increases the number of distinguishable arrangements.
Problem 8.19.
(a) The number of distinguishable arrangements, W, of n identical quanta among N atoms for
N = 4 and n = 2, is:
5!
5 ⋅4
( N + n − 1)! (4 + 2 − 1)!
W=
=
=
=
= 10
( N − 1)!n!
(4 − 1)!2!
3!2! 2 ⋅1
This is the same number of arrangements you found in Investigate This 8.7. Since the
toothpicks-in-candies system models the quanta-in-atoms system, we expect the results to be the
same.
(b) These are the numbers of distinguishable arrangements for different numbers of identical
quanta in a four atom solid. In cases where n becomes quite a bit larger than N, the first step in
calculating the ratio of factorials is to cancel n! from the numerator and denominator. Then
cancel (N – 1)! in the denominator with appropriate terms in the numerator before doing the
multiplications. These results are the same as those given in Figure 8.9.
(4 + 0 − 1)!
N = 4, n = 0: W =
= 1 (because 0! ≡ 1)
(4 − 1)!0!
4!
(4 + 1 − 1)!
N = 4, n = 1: W =
=
=4
(4 − 1)!1!
3!1!
N = 4, n = 2: W = 10 (See part (a).)
6!
6⋅5⋅4
(4 + 3 − 1)!
=
=
= 20
(4 − 1)!3!
3!3!
3 ⋅ 2 ⋅1
7!
(4 + 4 − 1)!
7⋅6⋅5
N = 4, n = 4: W =
=
=
= 35
3!4!
(4 − 1)!4!
3 ⋅ 2 ⋅1
8!
(4 + 5 − 1)!
8⋅7⋅6
N = 4, n = 5: W =
=
=
= 56
(4 − 1)!5!
3!5!
3⋅ 2 ⋅1
9!
(4 + 6 − 1)!
9⋅8⋅7
N = 4, n = 6: W =
=
=
= 3·4·7= 84
3!6!
(4 − 1)!6!
3⋅ 2 ⋅1
(4 + 7 − 1)! 10! 10 ⋅ 9⋅ 8
N = 4, n = 7: W =
=
=
= 10·3·4 = 120
3!7!
3 ⋅2 ⋅1
(4 − 1)!7!
(4 + 8 − 1)! 11! 11⋅10 ⋅9
N = 4, n = 8: W =
=
=
= 11·5·3 = 165
3!8!
3⋅ 2 ⋅1
(4 − 1)!8!
(4 + 9 − 1)! 12! 12 ⋅11⋅10
N = 4, n = 9: W =
=
=
= 2·11·10 = 220
3!9!
3 ⋅2 ⋅1
(4 − 1)!9!
13!
13 ⋅12 ⋅11
(4 + 10 − 1)!
N = 4, n = 10:
W=
=
=
= 13·2·11 = 286
3!10!
3⋅ 2 ⋅1
(4 − 1)!10!
N = 4, n = 3: W =
(c) The initial state of the system described in this part is two 8-atom solids with 10 and 16
quanta, respectively. The number of distinguishable arrangements of energy in each solid is:
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(8 + 10 − 1)! 17 ⋅16 ⋅15⋅14 ⋅13⋅12⋅11
=
= 17·4·1·2·13·1·11 = 19,448
7 ⋅ 6⋅ 5⋅ 4 ⋅ 3⋅ 2⋅1
(8 − 1)!10!
(8 + 16 − 1)! 23 ⋅22 ⋅ 21⋅20 ⋅19 ⋅18 ⋅17
W16 =
=
= 23·11·1·1·19·3·17 = 245,157
7 ⋅6 ⋅ 5⋅ 4 ⋅3 ⋅2 ⋅1
(8 − 1)!16!
W10 =
[It is easier for these calculations to use the factorial function on your calculator, if you have
this function available.] The overall number of distinguishable arrangements of energy is:
Wtotal = W10·W16 = 4.77 × 109.
After the proposed energy transfer, the two solids have 14 and 12 quanta, respectively. The
number of distinguishable arrangements of energy in each solid is:
(8 + 14 − 1)! 21⋅ 20 ⋅19 ⋅18⋅17⋅16 ⋅15
W14 =
=
= 1·1·19·3·17·8·15 = 116,280
7 ⋅6 ⋅ 5⋅ 4 ⋅ 3⋅ 2⋅1
(8 − 1)!14!
(8 + 12 − 1)! 19 ⋅18⋅17⋅16 ⋅15 ⋅14⋅13
W12 =
=
= 19·1·17·2·3·2·13 = 50,388
7⋅ 6⋅ 5 ⋅4 ⋅ 3⋅ 2 ⋅1
(8 − 1)!12!
The overall number of distinguishable arrangements is:
Wtotal = W14·W12 = 5.86 × 109
The total number of arrangements has increased in this change, but the energy transfer ends up
with the solids at different temperatures and the originally cooler one is now warmer. We know
from experience that this is not a likely process, so why has the total number of arrangements
increased? We need to consider the process of energy transfers in more detail. Note that the
number of arrangements just calculated would be the number for the system, if two quanta of
energy had been transferred from the warmer (16 quanta) solid to the cooler (10 quanta) solid.
This would be a transfer from a warmer to a cooler solid with the final temperatures still not
equal, but the warmer body would still be the warmer. This process would be likely, since the
final number of energy arrangements is larger than the initially, and it is in accord with our
experience that warm solids transfer thermal energy to cooler solids. What we need to find is the
number of arrangements available when just enough quanta (three) have been transferred to
bring both solids to the same temperature (that is, with the same number of quanta, 13). In this
case, the number of distinguishable arrangements of each solid is:
(8 + 13 − 1)! 20 ⋅19 ⋅18 ⋅17 ⋅16 ⋅15 ⋅14
W13 =
=
= 1·19·3·17·16·5·1 = 77,520
(8 − 1)!13!
7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
The overall number of distinguishable arrangements is:
Wtotal = W13·W13 = 6.01 × 109
As you see, this 13:13 system of two identical solids with the 26 quanta divided equally
between them has more possible energy arrangements than either the 16:10 or 14:12 cases
previously calculated. This equal distribution is the most Δprobable for this system. If we
imagine transferring one quantum from one of these solids to the other, we will end up with a
14:12 system with a lower number of energy arrangements, so the process will not be favored
and explains why the originally proposed four-quantum transfer will not occur (the transfer will
not “overshoot” the system with the solids at the same temperature).
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Problem 8.20.
The second law of thermodynamics states that the total entropy of an isolated system increases
in any spontaneous change. Or Observed changes take place in a direction that increases the net
entropy of the universe (system and its surroundings). The mathematical statement of the law is:
ΔSnet = kln
(
Wfinal
)
Winitial > 0 for spontaneous processes
Where ΔSnet is the net entropy change for the system and surroundings, W is the number of
distinguishable states for the system and surroundings, and k is a positive constant, the
Boltzmann constant.
Problem 8.21.
The way we have defined it, positional entropy is a measure of the number of different
distinguishable arrangements of the particles in a system. Thermal entropy is a measure of the
number of different distinguishable arrangements of energy quanta among the particles in a
system. [In fact, positional entropy is also a measure of energy arrangements and, for a gas,
counts the number of ways the particles in the system can be distributed among the translational
energy levels available to them. The qualitative results we get from particle arrangements in
space mirror those we get from the correct treatment and are easier to visualize. See Problem
8.96.]
Problem 8.22.
If a change takes place that involves only thermal energy transfers (with no change in positional
arrangements), then the net entropy change for the process will be just the thermal entropy
change and it is true that the process will take place in a direction that increases thermal (net)
entropy. (An example is the transfer of energy between a warm solid and a cool solid.) The
same argument is true for positional entropy and processes that involve only changes in
positional arrangements with no thermal energy transfers. (An example is the mixing of a
soluble dye with water.) In the general case, both positional and thermal entropy changes must
be accounted for in determining the net entropy change for a process and hence the direction
that will be observed.
Problem 8.23.
We can determine absolute entropies for compounds and elements because there is a reference
point for entropies – absolute zero. This is the temperature at which all substances are solids. In
perfect crystals, there is no disorder and all particles are in their lowest energy state. Therefore,
the number of distinguishable arrangements is one, that is, W = 1, and the entropy at this
temperature is zero, S = klnW. Entropies at other temperatures can be calculated relative to S at
0 K. There is no such reference point for enthalpy and so we can only calculate changes in
enthalpy, ΔH, from one state to another.
Problem 8.24.
(a) The entropy of a mole of solid mercury is higher than a mole of mercury vapor. This
statement is false. Gas phases have higher positional entropy because the volume available to
each particle (atom of mercury, in this case) is so much higher than in the condensed phases.
One way to correct the statement is to say: The entropy of a mole of solid mercury is lower than
a mole of mercury vapor.
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(b) If the net entropy increases during a process, the process is spontaneous. This statement is
true. Spontaneous changes take place in the direction that increases net entropy.
(c) The entropy of a system depends on the pathway it took to its present state. This statement is
false. Entropy is a state function that is independent of pathway, so the correct statement might
be: The entropy of a system does not depend on the pathway it took to its present state.
(d) The more positive its net entropy change, the faster the process. This statement is false.
Positive change of net entropy indicates that a change is favorable, but it does not indicate the
rate of the process. A correct statement could be: The net entropy change for a process, is not
related to the speed of the process.
(e) The entropy of one liter of a 0.1 M aqueous solution of NaCl is lower than the entropy of the
undissolved 0.1 mole of NaCl and one liter of water. This statement is false. The entropy of the
mixture is greater than the entropy for the separated components because more molecular
arrangements exist for the system with the soluble compound dissolved in the solvent.
(f) The entropy is larger for a system of two identical blocks of copper in which the temperature
of each block is 60 ° C than for the same blocks of copper if one has a temperature of 20 ° C and
the other is at 120 ° C. This statement is true. There are more energy arrangement and greater
entropy for systems in which two identical objects have the same temperature than when one
object is warmer than the other. See Problem 8.19(c).
(g) There is no reference point for measuring entropy. This statement is false. The reference
point for measuring entropy is the state of the substances at absolute zero, 0 K, at which
temperature the entropy of a perfectly ordered substance is zero. (Absolute zero has never been
experimentally achieved, but experiments have come within a billionth of a degree of 0 K.) A
correct statement could be: The reference point for measuring the entropy of a substance is its
state at 0 K, where its entropy is zero, for a perfectly ordered crystal.
Problem 8.25.
(a) The positional entropy change for precipitation of BaSO4 upon mixing Ba(NO3)2(aq) and
H2SO4(aq), will be positive, because the Ba2+(aq) and SO42–(aq) have lower positional entropy
than the same ions in the crystal lattice. The reason the aquated ions have low positional entropy
is because many water molecules are oriented around these highly charged ions, making the
entropy of the solution low due to the loss of positional entropy of the water molecules.
(b) The positional entropy change for cooling of ammonia gas from room temperature to –
50 ° C, which results in NH3(g) →NH3(l), will be negative because ammonia gas has higher
positional entropy than ammonia in the condensed phase.
(c) The positional entropy change for formation of ammonia in the reaction, N2(g) + 3H2(g) →
2NH3(g), will be negative because there are fewer moles of gas in the product than in the
reactants. This means that some positional entropy is lost in this process, because fewer moles
of gas have lower entropy.
(d) The positional entropy change for decomposition of CaCO3 in the reaction, CaCO3(s) →
CaO(s) + CO2(g), will be positive because a solid is converted into a solid and gas. Some
positional entropy is gained in this process, because gases have higher positional entropy than
solids.
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(e) The change in positional entropy for formation of NO in the reaction, N2(g) + O2(g) →
2NO(g), is hard to predict in this case, because of the same number of moles of gas is involved
in the reactants and in product.
Problem 8.26.
[This problem would be better placed in Section 8.14 where molecular organization is discussed
in terms of the source of the net entropy that “drives” organization on Earth.]
The second law of thermodynamics states that the net entropy change (accounting for the
system and surroundings) for a spontaneous process is positive. If a change in a system, such as
evolving more complex organisms from simpler ones, reduces the number of possible
distinguishable arrangements of the components, then the change in the system can only be
spontaneous if it is accompanied by an increase in the number of arrangements of its
surroundings. In the case of evolution (or even the formation of the first most primitive cell),
energy is required to bring the components together in just the right way (to form the correct
bonding arrangements, for example). In almost all cases, this energy is supplied by
electromagnetic radiation from the sun that is captured in various ways on earth and used to
bring about the evolutionary changes. The nuclear processes in the sun transfer a prodigious
amount of energy to its surroundings, so there are enormous increases in their entropy. Thus, the
entropy increase of the sun and its surroundings are much more than enough to “drive” the
relatively small evolutionary changes that reduce the entropy of the evolving system.
Problem 8.27.
Directly applying the principles of thermodynamics to workspace objects is not a good idea. In
thermodynamics, we assume that the energy of the system is sufficient that all possible
arrangements can be reached from one another. For example, gas molecules are in constant
motion, so the system can easily move from one arrangement to another. On the other hand, if
books are arranged on a shelf, we know that (barring some kind of high energy intervention
such as an earthquake) they are going to stay arranged on the shelf. They lack the kinetic energy
(motion) to make a transition to other arrangements. So, thermodynamics is not responsible for
clutter in a friend's workspace.
Problem 8.28.
H3C
H2
C
C
H2
O
C
OH
H2C CH2
O
O
H2C CH2
(a) In the isomeric pair butanoic acid,
, and dioxane,
, the butanoic acid has
the higher entropy under the same pressure and temperature conditions. There are more
positional arrangements for butanoic acid due to bond rotations than the constrained ring of
dioxane. “Floppy” molecules will almost always have higher entropies than isomers that are less
floppy, because of rings or other more compact structures. This is because there are more low
energy motions, mainly rotations of one part of the molecule relative to the rest, for the floppier
molecule and thus more distinguishable ways to distribute the same amount of energy.
H3C
H2
C
C
H2
H2
C
CH3
CH3
H3C C
CH3
H3C
(b) In the isomeric pair pentane,
, and 2,2-dimethylpropane,
, the pentane
has the higher entropy under the same pressure and temperature conditions. Again, as in part
(a), there are more positional arrangements for pentane due to bond rotations. Pentane is a
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floppier molecule than the more compact 2,2-dimethylpropane, as you can see if you build
models of the two molecules.
H2C CH2
C
H2
H2C CH2
C
H2
(c) In the isomeric pair cyclopropane,
, and propene,
, the propene has the higher
entropy under the same pressure and temperature conditions. Again, as in parts (a) and (b), there
are more positional arrangements for propene due to bond rotations. Propene is a floppier
molecule than the more rigid cyclopropane.
Problem 8.29.
The statement, “observed phase changes always take place in a direction that increases
entropy,” is incomplete and can be misleading. It is the net entropy that points the direction of
observed changes, including phase changes. Without this qualifier, there is sometimes confusion
when only thermal or only positional entropy is considered. For example, this chapter started
with the observation that complex, highly-ordered snowflakes can form from water vapor or
liquid droplets. This apparently spontaneous ordering does not increase the positional entropy
for the water molecules involved in the change. As the snowflakes form, there is a decrease in
positional entropy of the system, which does not favor the phase change. However, there is an
increase in thermal energy of the surroundings that does favor the phase change. A positive
value for net entropy change, which must consider both system and surroundings, does indicate
the correct direction for observed phase changes.
Problem 8.30.
(a) In the pair H2O(s) and H2O(l), the H2O(l) has the higher entropy, because a molecule in the
liquid has a larger volume to move in (the whole volume of liquid instead of a tiny localized
volume around its position in the crystal).
(b) In the pair CaCl2 (aq) and CaCl2(s), the CaCl2(s) has the higher entropy. This result is
difficult to predict, because the entropy of dissolution of salts of multiply-charged and singlycharged ions depends sensitively on the interactions of specific ions with the solvent water.
Multiply-charged ions have a greater orienting effect on the water molecules in their vicinity
than do singly-charged ions. For salts of multiply-charged cations and anions, the effect is
always to make the entropy of the aquated ions less than the entropy of the ions in the crystal
[see Problem 8.25(a)]. For the case here, we need to look at the entropy values in Appendix B to
determine which state has the higher entropy. For the solid, the entropy is 104.6 J·K–1·mol–1, and
for the sum of the aquated ions (Cl– taken twice), the entropy is 59.9 J·K–1·mol–1. (Calcium
chloride is quite soluble in water, because the enthalpy of solution is large and negative, so the
entropy change of the thermal surroundings is positive enough to compensate for the loss of
entropy upon dissolution.)
(c) In the pair 5.0 g H2O(l) at 1 ° C and 5.0 g H2O(l) at 60 ° C, the 5.0 g H2O(l) at 60 ° C has the
higher entropy, because the system has more thermal energy to distribute among the same
number of molecules, so there will be more distinguishable arrangements of energy quanta in
the system and, thus, higher entropy.
(d) In the pair H2O(g) and H2O(l), the H2O(g) has the higher entropy, because the molecules in
the gas have a larger volume to move in (the whole volume of the container instead of just the
volume occupied by the liquid).
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(e) In the pair CO2 (g) and CO2(s), dry ice, the CO2 (g) has the higher entropy, because a
molecule in the gas has a larger volume to move in (the whole volume of the container instead
of a tiny localized volume around its position in the crystal).
Problem 8.31.
When water freezes, the entropy change of the system is negative, ΔSwater < 0. The thermal
energy change when the water freezes (and many hydrogen bonds form) is negative, ΔHwater < 0,
and the thermal energy change in the surroundings is positive, ΔHsurround = –ΔHwater. The entropy
ΔH surround
change of the surroundings is ΔSsurround =
T , where T is the kelvin temperature of the
surroundings. If the thermal surroundings are at a temperature less than 273 K, ΔSsurround +
ΔSwater = ΔSnet > 0 and the process is spontaneous. Thus, water freezes, if the temperature of
surroundings is below 273 K.
Problem 8.32.
(a) When solid CO2 (dry ice) changes to carbon dioxide gas, the positional entropy increases. A
molecule in the gas has a larger volume to move in (the whole volume of the container instead
of a tiny localized volume around its position in the crystal) and, therefore, more distinguishable
arrangements possible than in the solid.
(b) When liquid ethanol (C2H5OH) freezes, the positional entropy decreases. There are fewer
possible arrangements of ethanol molecules in the solid state than when in the liquid state,
because a molecule in the liquid has a larger volume in which to move (the entire volume of the
liquid instead of a tiny localized volume around its position in the crystal).
(c) When mothballs made of naphthalene (C10H8) sublime, the positional entropy increases.
There are many more possible arrangements of molecules of naphthalene in the gas phase than
in the solid state, because a molecule in the gas has a larger volume to move in (the whole
volume of the container instead of a tiny localized volume around its position in the crystal).
(d) When plant materials burn to form carbon dioxide and water, the positional entropy
increases. There are many more possible arrangements of molecules in the gases that result
from combustion of the plant material, because they are free to move about in the entire volume
of their container instead of being constrained to a small volume within the molecules in the
plant.
Problem 8.33.
Consider an isolated system where a balloon containing steam, H2O(g), is immersed into a
thermally isolated container of liquid water at 10.0 ° C.
(a) In an isolated system where a balloon containing steam, H2O(g), is immersed into a
thermally isolated container of liquid water at 10.0 ° C, the steam will condense to liquid water.
(b) In this system, the entropy of the water molecules in the balloon will decrease as the water
vapor cools and changes from a gas to a liquid and then the liquid decreases in temperature until
it is the same temperature as the water outside the balloon. Both the positional and thermal
entropy of the water will decrease as the gas condenses and cools. The positional entropy
decreases because the gas changes to a liquid and thermal entropy of the water in the balloon
decreases because thermal energy is transferred to the surrounding water and there are fewer
quanta to distribute in the water in the balloon.
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(c) The entropy of the water outside the balloon will increase. Thermal energy is transferred to
the water outside the balloon, so its thermal entropy will increase.
(d) The net entropy change for this process in the isolated system will be positive. Our
experience tells us that steam in thermal contact with cool water will condense, so this must be
the direction of the observable change in the system and net entropy increases in spontaneous
(observable) processes.
Problem 8.34.
(a) This diagram, modeled after Figure 8.11, is for melting of H2S(s) at temperatures above the
normal melting point, 187 K. [Note that the diagram is qualitatively identical to Figure 8.11,
which was for the melting of H2O(s) at temperatures above 273 K. This is to be expected for the
melting of any solid above its normal melting point.]
As we see, melting increases positional entropy (blue arrow). Melting requires energy from the
surroundings, so the process is endothermic and the enthalpy change for the system is positive.
The surroundings lose an equivalent amount of energy and we can calculate the thermal entropy
change of the surroundings as shown. Loss of thermal energy from the surroundings decreases
their thermal entropy (red arrow). Since the temperature, T, is above the normal (equilibrium)
melting point of the solid, the thermal entropy change is smaller in magnitude than the
positional entropy change and their sum, the net entropy change (green arrow) is positive.
(b) This diagram is for H2S(l) freezing to H2S(s) at 187 K, the normal melting/freezing point.
Compare this diagram to the one in part (a).
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Freezing decreases positional entropy (blue arrow). Freezing releases energy to the
surroundings, so the process is exothermic and the enthalpy change for the system is negative.
The surroundings gain an equivalent amount of energy and we can calculate the thermal entropy
change of the surroundings as shown. Gain of thermal energy by the surroundings increases
their thermal entropy (red arrow). Since the temperature, T, is the normal (equilibrium) freezing
point of the liquid, the thermal entropy change is identical in magnitude to the positional
entropy change and their sum, the net entropy change is zero, so no net entropy change arrow is
shown.
(c) The phase change for water changing from solid to liquid takes place at 0 ° C (273 K), while
that for H2S at –86 ° C (187 K). The difference can be explained by hydrogen bonding. This
intermolecular force is a major factor in water, but is quite weak in H2S, where the larger size of
the sulfur atom and the smaller electronegativity difference are unfavorable for hydrogen
bonding. The hydrogen bonding in H2O makes the energy required to disrupt the interactions in
the solid larger than that required in H2S. Since less energy is required by the H2S, enough
energy is available at lower temperature and its melting point is lower.
Problem 8.35.
The order of increasing entropy of vaporization is methane < ammonia < water. There are a
ΔHl →g
couple of ways to reason out this ordering. One is to use the relationship ΔSl→g =
T ,
which is valid at the boiling temperature of the liquid where the net entropy change for the
process is zero, because the two phases are in equilibrium. Water, with its extensive network of
hydrogen bonds, has a higher enthalpy of vaporization (energy required to break the hydrogen
bonds among the molecules and get them into the gas phase) than either ammonia or methane.
Since the enthalpy of vaporization is highest for water, its entropy of vaporization is highest.
Ammonia is also hydrogen bonded in the liquid, but, as we discussed in Chapter 1, Section 1.7,
Consider This 1.26, the hydrogen bonding cannot be as extensive as in water, because each
molecule has only a single nonbonding pair of electrons with which to hydrogen bond. Because
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it has some hydrogen bonding, the enthalpy of vaporization of ammonia will be higher than that
of methane, which has no hydrogen bonding capacity and whose molecules are held together in
the liquid phase only by dispersion forces, which are relatively weak for this small molecule.
Thus, the entropy of vaporization for ammonia will be higher than that for methane.
Problem 8.36.
(a) The complete combustion of ethanol, C2H5OH(l), changes 3 moles of gas (and 1 mole of the
condensed liquid phase) into 2 moles of gas (and 3 moles of condensed liquid phase):
C2H5OH(l) + 3O2 (g) →2CO2(g) + 3H2O(l)
For this reaction, ΔH° = –1367 kJ and ΔS° = –139 J.K–1. The decrease in number of moles of
gas means that there is a decrease in positional entropy for the reaction. You might argue that 4
moles of reactants have changed to 5 moles of products, which might be expected to increase
positional entropy. However, the entropies of gas phase substances are a good deal larger than
the entropy of condensed phases, so arguments based on changes in gaseous moles usually are a
better predictor of the sign of the entropy change. The fact that the experimental value for the
entropy change is negative indicates that the reduction in number of moles of gas is the
controlling factor in the entropy change for the system.
(b) The overall reaction is highly exothermic, releasing 1367 kJ. The sum of the bond enthalpies
for formation of the product bonds must be substantially greater than the sum of the bond
enthalpies for the bonds that must be broken in the reactants. The C=O bonds in CO2 are
particularly strong (because of the delocalization of the pi bonding electrons in the molecule),
which, together with the rather weak O=O bond in O2, helps to account for the large
exothermicity of this and other combustion reactions.
(c) Considering what we said in part (a), this reaction is not favored by positional entropy
change, which is negative. There is, however, a large release of energy to the surroundings,
which means a large positive thermal entropy change for the reaction, so the net entropy change
is positive and quite large:
ΔS
o
net
= ΔS
o
system
o
ΔH system
−
T
⎛ –1367 × 10 3 J ⎞
–1
⎟⎠ = +4449 J·K
298 K
o
ΔSnet
= –138 J ⋅ K –1 − ⎜
⎝
The large positive value means that the combustion is highly probable. Remember that it also is
an exothermic reaction. Ethanol is used in many types of automotive fuels as an octane booster
now that lead has been eliminated from gasoline in the US.
Problem 8.37.
(a) To calculate ΔHol→g and ΔSol→g for the phase transition, Br2(l) →Br2(g), we need the standard
enthalpies of formation and standard entropies for Br2(l) and Br2(g) (from Appendix B):
ΔH° l→g = ΔH° f[Br2(g)] – ΔH° f[Br2(l)] = [30.91 kJ.mol–1] – [0 kJ.mol–1] = 30.91
kJ.mol–1
ΔS° l→g = S° [Br2(g)] – S° [Br2(l)] = [245.46 J·K–1·mol–1] – [152.23 J·K–1·mol–1]
= 93.23 J·K–1·mol–1
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⎛ ΔH o l→ g ⎞
(b) We can use the relation, ΔS° net = ΔS° l→g – ⎜
T ⎟⎠ , to find the net entropy change at
⎝
different temperatures (assuming that ΔS° and ΔH° are independent of temperature):
(i) 273 K (0 ° C):
3
–1
ΔS° net = (93.23 J·K–1·mol–1) – ⎛ (30.91 × 10 J ⋅ mol ) 273 K ⎞
⎝
⎠
–1
–1
= –19.99 J·K ·mol
3
–1
(ii) 313 K (40 ° C): ΔS° net = (93.23 J·K–1·mol–1) – ⎛ (30.91 × 10 J ⋅ mol ) 313 K ⎞
⎝
⎠
–1
–1
= –5.52 J·K ·mol
3
–1
(iii) 343 K (70 ° C): ΔS° net = (93.23 J·K–1·mol–1) – ⎛ (30.91 × 10 J ⋅ mol ) 343 K ⎞
⎝
⎠
–1
–1
= 3.11 J·K ·mol
(c) The results from part (b) show that the vaporization becomes spontaneous between 40 ° C
and 70 ° C. This means that at some temperature in this range, ΔS° net = 0, and the liquid and gas
ΔH o l→g
are in equilibrium at one atmosphere pressure. At this temperature, ΔS° l→g =
Tbp , which
we can solve to find T, the boiling temperature of bromine:
Tbp =
ΔH o l→g
ΔS o l→g
3
–1
= (30.91 × 10 J ⋅ mol )
(93.23 J ⋅ K –1 ⋅ mol–1 )
= 331.5 K (58.5
° C)
[The handbook value for the boiling point is 58.78 ° C, so this approach gives excellent results
in this case.]
Problem 8.38.
For the phase change, H2O(l) →H2O(g), ΔH° l→g is positive (energy must be put into the system
to change liquid to gaseous water). The enthalpy change will contribute to a negative net
⎛ ΔH o l→g ⎞
entropy change for the vaporization, because ΔS° net = ΔS° l→g – ⎜
T ⎟⎠ . However, for this
⎝
change, ΔS° l→g is also positive (a gas has higher entropy than its liquid), so as the temperature
ΔH o l→g
rises (which makes the
T term smaller) the net entropy change, ΔS° net, will ultimately
become dominated by the ΔS° l→g term.
Problem 8.39.
(a) The reasoning for this solution is identical to that for Problem 8.38. To calculate ΔHol→g and
ΔSol→g for the phase transition, RCHO(l) →RCHO(g), where RCHO is CH3CHO (ethanal), we
need the standard enthalpies of formation and standard entropies for RCHO(l) and RCHO(g)
(from Appendix B):
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ΔHol→g = ΔHof[RCHO(g)] – ΔHof[RCHO(l)] = –166.36 kJ.mol–1 – (–192.30 kJ.mol–1)
= 25.94 kJ.mol–1
ΔSol→g = So[RCHO(g)] – So[RCHO(l)] = 264.22 J.K–1.mol–1 – 160.2 J.K–1.mol–1
= 104.0 J.K–1.mol–1
(b) At some temperature, ΔS° net = 0, for this phase change and the liquid and gas are in
ΔH o l→g
equilibrium at one atmosphere pressure. At this temperature, ΔS° l→g =
Tbp , which we
can solve to find T, the boiling temperature of bromine:
Tbp =
ΔH o l→g
ΔS
3
–1
= (25.94 × 10 J ⋅ mol )
o
l→g
(104.0 J ⋅ K –1 ⋅ mol–1 )
= 249.4 K (–23.8
° C)
[The handbook value for the boiling point is 20.8 ° C, which is far from this calculated result.
There is likely to be a problem with the thermodynamic values in Appendix B. An internet
search for thermodynamic data relevant to this problem suggests that the enthalpy of
vaporization is probably correct, so the problem probably lies with the entropies. However, the
values in Appendix B are confirmed (within ±1%) by thermodynamics data tables available on
the Web. However, one reference in a Google search
(http://64.233.187.104/search?q=cache:fT1JmaIGvyIJ:web.chem.ucla.edu/~luceigh/BA
L/BAL_EXAMS/CHEM14B/14BF99FINANSW.pdf+%22standard+entropy%22+%2Ba
cetaldehyde&hl=en),
gave a value of 250.3 J.K–1.mol–1 for the standard entropy of gaseous ethanal. If this value is
used in this problem, we get ΔSol→g ≈ 90 J.K–1.mol–1 and Tbp ≈ 288 K (15.2 ° C), which is a
good deal closer to the experimental value. If ΔSol→g ≈ 88 J.K–1.mol–1, the calculated boiling
point would agree with the experimental value. Students should note that not all data in tables
are necessarily correct.]
Problem 8.40.
(a) To calculate ΔHol→g for the phase transition, CH3OH(l) →CH3OH(g), we need the standard
enthalpies of formation for CH3OH(l) and CH3OH(g) (from Appendix B):
ΔH° l→g = ΔH° f[CH3OH(g)] – ΔH° f[CH3OH(l)] = (–200.66 kJ.mol–1) – (–238.86 kJ.mol–
1
)
= 38.20 kJ.mol–1
(b) Using the value for ΔH° l→g from part (a), we can use the boiling point of methanol, 65.0 ° C
(338.2 K), to find:
ΔS° l→g =
ΔH o l→g
Tbp
3
–1
= ⎛ (38.20 × 10 J ⋅ mol ) 338.2 K ⎞ = 113.0 J. K–1.mol–1
⎝
⎠
To use this relationship, we assume that the standard enthalpy and entropy changes are not
temperature dependent.
(c) Using the standard entropies of formation for CH3OH(l) and CH3OH(g) (from Appendix B),
we find:
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Entropy and Molecular Organization
ΔSol→g = So[CH3OH(g)] – So[CH3OH(l)] = (239.81 J. K–1.mol–1) – (126.8 J. K–1.mol–1)
= 113.0 J. K–1.mol–1
There is good agreement between this value and the calculated result from part (b).
Problem 8.41.
(a) The value ΔG° = –1325 kJ.mol–1, for the complete combustion of ethanol, C2H5OH(l),
predicts that this reaction will take place at standard pressure and 298 K, because the free
energy change is < 0 for this temperature and pressure.
(b) If this reaction were to come to equilibrium, the Gibbs free energy change, ΔG = ΔG° , at
standard pressure, would be zero and we can use this relationship (together with the values for
ΔH° and ΔS° from Problem 8.26) to find the temperature where this would occur:
0 = ΔG = ΔG° = ΔH° – TΔS°
0 = (–1367 × 103 J) – T(–139 J·K–1);
∴T ≈ 9800 K
(c) In order to use the relationship in part (b), we assume that ΔH° and ΔS° are constant over
the temperature range from 273 K to 9800 K. This is definitely not a valid assumption for such a
large temperature range. Long before this high temperature is reached, the molecules in this
reaction will begin to disintegrate into smaller fragments and atoms, or even ions and electrons.
One lesson from this sort of result is that reactions with such large enthalpy changes are
unlikely to come to equilibrium at standard pressure under any temperature conditions.
Problem 8.42.
To find out whether the reaction, 2H2O2(g) →2H2O(g) + O2(g), is likely to occur at 298 K (that
is, whether we expect H2O2 (g) to be stable or likely to decompose at 298 K), we can combine
the enthalpy and entropy values, ΔH = –106 kJ and ΔS = 58 J·K–1, to get ΔG.
ΔG = ΔH – TΔS = (–106 × 103 J) – (298 K)(58 J·K–1) = –123 × 103 J = –123 kJ
The decomposition of H2O2 (g) has a high negative free energy change, so at this temperature
H2O2(g) is not stable, the decomposition is spontaneous (although not rapid—see Chapter 11).
Problem 8.43.
(a) The free energy change for a reaction that is nonspontaneous at 300 K must be positive at
this temperature, ΔG > 0. If the reaction has a positive entropy change of 130 J·K–1, the
contribution of the –TΔS term to ΔG is negative. Therefore, the enthalpy change for the reaction
must be positive, ΔH > 0, in order for ΔG > 0. This reaction is endothermic since ΔH is positive.
(b) The minimum value of ΔH that will make this reaction nonspontaneous is the value that will
make ΔG = 0:
ΔG = 0 = ΔH – TΔS = ΔH – (300 K)(130 J·K–1)
ΔH = 39.0 kJ
For ΔH ≥ 39.0 kJ, ΔG ≥ 0.
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Chapter 8
Problem 8.44.
Because there are no absolute reference states for enthalpies and free energies, we define the
reference state as the elements in their most stable form at the standard state and 298 K. Thus,
the standard enthalpies and free energies of formation, ΔH0f and ΔGof , are defined as zero to
provide a reference for other standard enthalpies and free energies of formation. There is an
absolute reference state for entropies, the entropy of the substance at absolute zero, 0 K, which,
for a perfectly ordered crystal is S0 = 0. The subscript “0” on the symbol indicates the value at
0 K. The entropies of all substances are referred to this absolute value.
Problem 8.45.
The answers to this problem are based on the relationship
ΔG = ΔH – TΔS, the criterion for spontaneity, ΔG < 0, and their
relationships summarized in this graphic.
(a) Processes that fall in quadrant A are always spontaneous. A
negative ΔH and positive ΔS always combine to give ΔG < 0.
(b) Processes that fall in quadrant D are never spontaneous. A
positive ΔH and negative ΔS always combine to give ΔG > 0.
(c) Processes that fall in quadrants B and C could possibly be spontaneous, depending upon the
temperature at which the processes are carried out. In quadrant B, the positive ΔH always
disfavors spontaneity. However, the –TΔS factor (with positive ΔS) is always negative and, at
sufficiently high temperature could outweigh ΔH, that is, |–TΔS | > |ΔH | and ΔG < 0.
Conversely, in quadrant C, the negative ΔH always favors spontaneity, but the –TΔS factor
(with negative ΔS) is always positive. At sufficiently low temperature the enthalpy effect could
outweigh the entropy effect, that is, |ΔH | > |–TΔS | and ΔG < 0.
(d) Quadrant C represents the freezing of a liquid. In this process, both ΔS and ΔH are negative
and, as we showed in part (c), at sufficiently low temperature the enthalpy effect could outweigh
the entropy effect, that is, |ΔH | > |–TΔS |, ΔG < 0, and the liquid freezes.
(e) Quadrant B represents the melting of a solid. In this process, both ΔS and ΔH are positive
and, as we showed in part (c), at sufficiently high temperature the entropy effect could outweigh
the enthalpy effect, that is, |–TΔS | > |ΔH |, ΔG < 0, and the solid melts.
Problem 8.46.
At a certain temperature, ΔG for the reaction CO2(g) ⇔ C(s) + O2(g), is found to be 42 kJ.
(i) “The system is at equilibrium” is an incorrect statement, because ΔG = 0 for a system at
equilibrium.
(ii) “The process is not possible” is a correct statement (under these conditions), because ΔG > 0
for the change means the reaction is nonspontaneous, that is, will not be observed.
(iii) “CO2 will be formed spontaneously” is a correct statement, because the reverse reaction,
formation of CO2, will have ΔG = –42 kJ and reactions are spontaneous if ΔG < 0 for the
change.
(iv) “CO2 will decompose spontaneously” is an incorrect statement, because reactions are only
spontaneous if ΔG < 0 for the change.
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Problem 8.47.
(a) Considering the oxidation of methane, CH4(g) + 2O2(g) →CO2(g) + 2H2O(g), we predict that
the change in entropy will be close to zero because 3 molecules of gaseous reactants produce 3
molecules of gaseous products.
(b) We are asked to find the standard entropy change for the reaction, using the standard
enthalpies and free energies of formation, ΔH° f and ΔG° f, of the reactants and products. We
will first calculate ΔH° and ΔG° for the reaction and then use the ΔG° = ΔH° – TΔS°
relationship rearranged to solve for ΔS° :
ΔH° = [(1 mol)ΔH° f(C02) + (2 mol)ΔH° f (H2O)] – (1 mol)ΔH° f(CH4)
= [–393.51 kJ + 2·(–241.82kJ)] – (–74.81kJ) = –802.34 kJ
ΔG° = [(1 mol)ΔG° f(C02) + (2 mol)ΔG° f (H2O)] – (1 mol)ΔG° f(CH4)
= [–394.36 kJ + 2·(–228.57 kJ)] – (–50.72kJ) = –800.78 kJ
(–802.34 × 10 3 J) – (–800.78 × 10 3 J)
ΔH o – ΔG o
=
= –5.23 J·K–1
298 K
T
(c) The calculated change in entropy is small, just as we predicted in part (a). [Using the
standard entropies in Appendix B to calculate ΔS° gives –5.14 J·K–1. This value is, identical,
within the uncertainties of the data, to the value calculated in part (b), which indicates that the
data in Appendix B are internally consistent.]
ΔS° =
Problem 8.48.
A large negative ΔG indicates that the reaction, 2H2(g) + O2(g) →2H2O(l), is spontaneous, but it
does not indicate anything about the speed of the reaction. Although the reaction is
spontaneous, it is very slow, so we will not observe any water formed after mixing oxygen and
hydrogen at room temperature.
Problem 8.49.
(a) For the reaction, C(graphite) + 2H2(g) →CH4(g), ΔH° = −74.8 kJ and ΔS° = −80.8 J·K-1. This
table and plot show ΔH° , TΔS° , and ΔG° as a function of temperature for the reaction under
standard conditions, assuming that ΔH° and ΔS° are constant and do not vary with temperature.
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Entropy and Molecular Organization
T(K)
200
300
400
500
600
700
800
900
1000
1100
1200
1300
ΔH° (kJ)
-74.8
-74.8
-74.8
-74.8
-74.8
-74.8
-74.8
-74.8
-74.8
-74.8
-74.8
-74.8
Chapter 8
ΔS° (kJ·K–1) TΔS° (kJ)
–0.0808
–16.2
–0.0808
–24.2
–0.0808
–32.3
–0.0808
–40.4
–0.0808
–48.5
–0.0808
–56.6
–0.0808
–64.6
–0.0808
–72.7
–0.0808
–80.8
–0.0808
–88.9
–0.0808
–97.0
–0.0808
–105
ΔG° (kJ)
–58.6
–50.6
–42.5
–34.4
–26.3
–18.2
–10.2
–2.08
6.00
14.1
22.2
30.2
(b) From the plot we find the intersection of the ΔH° and TΔS° lines, that is, where
ΔG° = ΔH° – TΔS° = 0. This is at about 925 K and corresponds to T = ΔH° /ΔS° .
(c) Since the data extrapolation required to estimate the equilibrium temperature is more than
600 degrees (298 K to over 900 K), the assumptions that ΔS° and ΔH° are constant,
independent of T, almost certainly are not valid. The temperature estimated in part (b) is likely
to be off from the experimental value.
Problem 8.50.
Liquid and gaseous water are in equilibrium at 100 ° C and one atmosphere, 101.3 kPa,
pressure. Under these conditions ΔG = 0 kJ for the phase change. At one bar, 100 kPa, pressure,
the phases are not quite in equilibrium at 100 ° C. The slightly lower vapor pressure of water
will be in equilibrium with the liquid at a temperature slightly below 100 ° C. If the temperature
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ACS Chemistry Chapter 8 suggested solutions
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Entropy and Molecular Organization
is increased (holding the pressure constant at one bar), vaporization will become spontaneous,
that is, ΔG < 0 kJ, but we have no information from the problem or background in the textbook
to calculate the value for ΔG (which will be quite close to zero) under these conditions.
Problem 8.51.
The challenge in this problem is to reconcile two viewpoints about the direction of change: (1)
observed changes occur in the direction that increases the net entropy of the universe, ΔSnet, the
sum of positional and thermal entropy changes for the process and (2) observed changes occur
in the direction that minimizes the energy of the system undergoing change and maximizes its
entropy. The thermal entropy change for a process is a measure of the entropy change in the
thermal surroundings, ΔSthermal = –ΔH/T, where ΔH is the enthalpy change in the system
undergoing change. If ΔH is negative (an exothermic process), then ΔSthermal is positive and acts
to increase the net entropy of the world, ΔSnet. The positional entropy change, ΔSpositional, is the
entropy change in the system. From the point of view of the system alone, maximizing its
entropy means maximizing ΔSpositional, which is in the direction to make ΔSnet large and positive.
Minimizing the energy of the system means losing as much energy as possible to the
surroundings, an exothermic process. This exothermicity makes ΔSthermal positive and large,
which, again, is in the direction to make ΔSnet large and positive. Thus, the two viewpoints come
down to the same thing. The problem with the energy-entropy viewpoint is that it obscures the
underlying basis for the change, the increase in net entropy, and makes it seem that entropy and
energy effects are somehow in competition, which is not the case.
Problem 8.52.
(a) The relationship among ΔG, ΔH, and TΔS for a system is.
ΔG = ΔH –TΔS
For the combustion reaction, CH3COCH3(g)+ 4O2(g) →3CO2(g)) + 3H2O(l), at 298 K, we are
given the values ΔG = –1784 kJ and ΔH = –1854 kJ. Substituting into the above equation gives:
–1784 kJ = –1854 kJ – TΔS
The difference between the ΔG and ΔH terms is only 70 kJ, which is the value of TΔS. The TΔS
term is small relative to the values of ΔG and ΔH. The sign of ΔS must be negative for the two
sides of the equation to be equal in value. A negative value for ΔS makes sense, because the
reaction involves five moles of gaseous reactants giving only three moles of gaseous products.
The reduction in moles of gas in the system predicts a decrease in entropy for the reaction.
(b) A small value for TΔS means a small value for ΔS, which we equate with the positional
entropy change, ΔSposition for the system. Since ΔS < 0 [from part (a)], the contribution of the
positional entropy change to the net entropy change, ΔSnet, for the reaction will be small and
negative. Conversely, the thermal entropy change, ΔSthermal = – ΔH T , will be a large positive
contribution to ΔSnet. Thermal energy released to the surroundings provides the driving force in
this and most combustion reactions.
(c) Rearranging the equation in part (a), gives us ΔS for this reaction:
(–1854 × 103 J) – (–1784 × 10 3 J)
ΔH – ΔG
ΔS =
=
= –235 J·K–1
298 K
T
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Chapter 8
Problem 8.53.
(a) For the reaction, 4Ag(s) + O2(g) →2Ag2O(s), to be spontaneous, we need ΔG < 0. Under
constant atmospheric pressure conditions, we can use the standard enthalpy and entropy
changes, ΔH° = –61.1 kJ and ΔS° = –0.132 kJ.K–1, to calculate ΔG = ΔG° , at 25 ° C (298 K):
ΔG = –61.1 kJ – (298 K)(–0.132 kJ.K–1) = –21.8 kJ
The reaction is spontaneous at 25 ° C.
(b) For the reaction at 273 ° C (546 K), we calculate:
ΔG = –61.1 kJ – (546 K)(–0.132 kJ.K–1) = 11.0 kJ
The reaction is not spontaneous at 273 ° C. This calculation is based on the assumption that the
enthalpy and entropy changes are constant, independent of temperature. This assumption is
probably not strictly valid, but the variation over about a 250 ° C range is likely not to be so
large as to make our conclusion wrong. Somewhere in this temperature range, the reaction is
probably in equilibrium at standard pressure.
Problem 8.54.
These thermodynamic data are for the species
present in the gas-phase dimerization reaction
of methanoic acid (dotted lines represent
hydrogen bonds):
2
O
H C
O H
O
H O
H C
C H
O H
O
Species
HC(O)OH(g)
[HC(O)OH]2(g)
ΔHof
kJ·mol–1
–362.63
–785.34
So
J·K–1·mol–1
251.0
347.7
(a) We can use the data in the table to calculate ΔH° , ΔS° , and ΔG° for the dimerization
reaction and determine whether it is spontaneous at standard pressure and 298 K:
ΔH° = (1 mol)ΔH° f(dimer) – (2 mol)ΔH° f(acid) = (–785.34 kJ) – 2·(–362.63 kJ)
= –60.08 kJ
ΔS° = (1 mol)S° (dimer) – (2 mol)S° (acid) = (347.7 J·K–1) – 2·(251.0 J·K–1)
= –154.3 J·K–1
ΔG° = ΔH° – TΔS° = –60.08 kJ – (298 K)(–0.1543 kJ·K–1) = –14.10 kJ
The negative value of ΔG° indicates a spontaneous dimerization reaction at standard pressure
and 298 K.
(b) The ΔH° calculated value of –60.08 kJ is essentially the energy released when two moles of
hydrogen bonds are formed during the dimerization. Therefore, the enthalpy change per mole of
hydrogen bonds would be half of this value, –30.04 kJ.
(c) Unlike enthalpy values, entropy values cannot be averaged, because the probable steps in the
reaction have quite different entropy changes. The entropy change for two molecules of
methanoic acid finding each other to form the first hydrogen bond is quite large, because two
particles free to move about anywhere in the system become one particle. The solutions to
Problems 8.52 and 8.54 suggest that a little more than 100 J·K–1 are lost for each mole of gas
that is lost going from reactants to products. If this is the case, then more than two-thirds of the
entropy change in this reaction is due to the loss of particles in the dimerization. The dimer with
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Entropy and Molecular Organization
only one hydrogen bond between the acid monomers will be more floppy (see Problem 8.28)
than the cyclic structure formed when the second hydrogen bond is made. Thus, there is a
further loss of entropy in this second step of the reaction, but the change is a good deal smaller
than for the first interaction. Since the value of free energy change depends on the value of
entropy change, we cannot estimate the free energy change per hydrogen bond formed, in the
same way we did for the enthalpy.
Problem 8.55.
We use data from Appendix B to calculate ΔH° , ΔS° , and ΔG° for these dissolution reactions:
KCl(s) ⇔ K+(aq) + Cl–(aq)
ΔH° = [(1 mol)ΔH° f(K+(aq)) + (1 mol)ΔH° f(Cl–(aq))] – [(1 mol)ΔH° f(KCl(s)]
= [(–252.38 kJ) +(–167.16 kJ)] – [–436.75 kJ] = 17.21 kJ
ΔS° = [(1 mol)S° (K+(aq)) + (1 mol)S° (Cl–(aq))] – [(1 mol)S° (KCl(s)]
= [(102.5 J·K–1) +(56.5 J·K–1)] – [82.59 J·K–1] = 76.4 J·K–1
ΔG° = [(1 mol)ΔG° f(K+(aq)) + (1 mol)ΔG° f(Cl–(aq))] – [(1 mol)ΔG° f(KCl(s)]
= [(–283.27 kJ) +(–131.23 kJ)] – [–409.14 kJ] = –5.36 kJ
AgCl(s) ⇔ Ag+(aq) + Cl–(aq)
ΔH° = [(1 mol)ΔH° f(Ag+(aq)) + (1 mol)ΔH° f(Cl–(aq))] – [(1 mol)ΔH° f(AgCl(s)]
= [(105.58 kJ) +(–167.16 kJ)] – [–127.07 kJ] = 65.49 kJ
ΔS° = [(1 mol)S° (Ag+(aq)) + (1 mol)S° (Cl–(aq))] – [(1 mol)S° (AgCl(s)]
= [(72.68 J·K–1) +(56.5 J·K–1)] – [96.2 J·K–1] = 33.0 J·K–1
ΔG° = [(1 mol)ΔG° f(Ag+(aq)) + (1 mol)ΔG° f(Cl–(aq))] – [(1 mol)ΔG° f(AgCl(s)]
= [(77.11 kJ) +(–131.23 kJ)] – [–109.79 kJ] = 55.67 kJ
(a) The calculated results support the solubilities discussed in Chapter 2. The negative ΔG° for
dissolving KCl(s) indicates that it will dissolve in water and our solubility rules suggest that
ionic compounds of alkali metal cations and halide anions (each with a single charge) should be
soluble. In fact, we observed that potassium salts are quite soluble in water. The positive ΔG°
for dissolving AgCl(s) indicates that it will not be spontaneous, even though it is also an ionic
compound with a single charge on each ion. However, we pointed out in Chapter 2 that AgCl(s)
(and the other silver halides) is an insoluble salt, so the positive ΔG° is not really a surprise.
(b) The solubility of KCl(s) should increase as the temperature is increased. With increasing
temperatures, the value of TΔS° will grow larger, which will make ΔG° (= ΔH° – TΔS° ) more
negative.
(c) The solubility of AgCl(s) should also increase as the temperature is increased, because ΔS°
is positive, as in part (b). However, the enthalpy is so large and positive that increasing the
temperature even to the boiling point of water will still yield a large positive value for ΔG° and
the dissolution will not occur to an appreciable extent.
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Entropy and Molecular Organization
Chapter 8
Problem 8.56.
(a) We can use data from Appendix B to find ΔGo at 298 K for the conversion of ethyne to
benzene, 3C2H2(g) ⇔ C6H6(l):
ΔG° = (1 mol)ΔG° f(C6H6(l)) – (3 mol)[ΔG° f(C2H3(g))] = (124.3 kJ) – 3·(209.20 kJ)
= –503.30 kJ
Under standard conditions, ΔG = ΔG° , and the reaction is spontaneous in the direction written,
that is to produce benzene from ethyne.
(b) In the reaction converting gaseous ethyne to liquid benzene, the number of moles of gas
decreases from three to zero, so the standard entropy change for this reaction, ΔS° , should be
relatively large and negative [more negative than –300 J·K–1 according to the reasoning in the
solution to Problem 8.54(c)]. The –TΔS° term in the expression, ΔG° = ΔH° = TΔS° , should be
positive and relatively large, so there will be a temperature above which the decrease in entropy
will dominate and ΔG = ΔG° > 0 for the ethyne-to-benzene conversion at standard pressure.
Under these conditions, the decomposition of benzene to ethyne, the reverse reaction, will be
spontaneous. We can check the entropy values in Appendix B to see whether our reasoning is
correct. For the ethyne-to-benzene conversion:
ΔS° = (1 mol)S° (C6H6(l)) – (3 mol)[S° (C2H3(g))] = (173.3 J·K–1) – 3·(200.94 J·K–1)
= –429.52 J·K–1
This result confirms our reasoning that the standard entropy change for the reaction is large and
negative. If we assume that the standard enthalpy and entropy changes are independent of
temperature, we can get an estimate of the temperature at which ΔG = ΔG° = 0 and the ethyneto-benzene conversion changes from being spontaneous to nonspontaneous. This temperature
will probably be fairly high, so it makes more sense to do the calculations for the reaction
producing gaseous benzene, 3C2H2(g) ⇔ C6H6(g):
ΔH° = (1 mol)ΔH° f(C6H6(g)) – (3 mol)[ΔH° f(C2H3(g))] = (82.9 kJ) – 3·(226.73 kJ)
= –597.3 kJ
ΔS° = (1 mol)S° (C6H6(g)) – (3 mol)[S° (C2H3(g))] = (269.31 J·K–1) – 3·(200.94 J·K–1)
= –333.51 J·K–1
o
TΔG = 0 = ΔH
ΔS
o
3
= (–597.3 × 10 J)
(–333.51 J ⋅ K –1 )
≈ 1800 K
Note that the standard entropy change for the gas phase reaction is not as negative as that for the
reaction giving liquid benzene. This reflects the fact that the decrease in moles of gas is from
three to one, so the entropy decrease is not as large.
Problem 8.57.
(a) We can use data from Appendix B to calculate ΔH° , ΔS° , and ΔG° at 298 K for the
reaction: N2H4(l) + H2O2(l) →N2(g) + 2H2O(l). The relevant thermodynamic quantities are:
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Chapter 8
Entropy and Molecular Organization
compound
ΔH° f, kJ·mol–1
S° , J·K–1·mol–1
ΔG° f, kJ·mol–1
N2H4(l)
H2O2(l)
N2(g)
H2O(l)
50.63
–187.78
0
–285.83
121.21
109.6
191.61
69.91
149.34
–120.35
0
–237.13
ΔH° = {(1 mol)ΔH° f[N2(g)] + (2 mol)ΔH° f[H2O(l)]}
– {(1 mol)ΔH° f[N2H4(l)] + (1 mol)ΔH° f[H2O2(l)]}
= (0 kJ) + 2·(–285.83 kJ) – (50.63 kJ) – (–187.78 kJ) = – 434.51 kJ
ΔS° = {(1 mol)S° [N2(g)] + (2 mol)S° [H2O(l)]}
– {(1 mol)S° [N2H4(l)] + (1 mol)S° [H2O2(l)]}
= (191.61 J·K–1) + 2·(69.91 J·K–1) – (121.21 J·K–1) – (109.6 J·K–1) = 100.6 J·K–1
ΔG° = {(1 mol)ΔG° f[N2(g)] + (2 mol)ΔG° f[H2O(l)]}
– {(1 mol)ΔG° f[N2H4(l)] + (1 mol)ΔG° f[H2O2(l)]}
= (0 kJ) + 2·(–237.13 kJ) – (149.34 kJ) – (–120.35 kJ) = –503.25 kJ
(b) The standard Gibbs free energy change for this reaction is large and negative. Under
standard conditions the reaction is spontaneous with a large “driving force.” Since the enthalpy
change for the reaction is large and negative and the entropy change is large and positive, the
reaction should be spontaneous under essentially all conditions (see Problem 8.45), unless either
the enthalpy or entropy is highly dependent on changes in conditions, which is not likely.
(c) As the temperature increases, the –TΔS term will get larger and more negative, making ΔG
larger and more negative. Increasing the temperature makes the reaction even more favorable.
At higher temperatures, the hydrazine, hydrogen peroxide, and water will vaporize and this will
require some thermal energy, but there is a great deal released in the reaction, so this will not
change the sign of ΔG.
Problem 8.58.
(a) Use the data from Appendix B to calculate ΔH° , ΔS° , and ΔG° for the reaction that converts
pyruvic acid to ethanal and carbon dioxide, CH3COC(O)OH(l) ⇔ CH3CHO(l) + CO2(g):
ΔH° = {(1 mol)ΔH° f[CH3CHO(l)] + (1 mol)ΔH° f[CO2(g)]}
– {(1 mol)ΔH° f[CH3COC(O)OH(l)]}
= {(–192.30 kJ) + (–393.51 kJ)} – (–584.5 kJ) = –1.3 kJ
ΔS° = {(1 mol)S° [CH3CHO(l)] + (1 mol)S° [CO2(g)]} – {(1 mol)S° [CH3COC(O)OH(l)]}
= {(160.2 J·K–1) + (213.74 J·K–1)} – (179.5 J·K–1) = 194.4 J·K–1
ΔG° = {(1 mol)ΔG° f[CH3CHO(l)] + (1 mol)ΔG° f[H2O(l)]}
– {(1 mol)ΔG° f[CH3COC(O)OH(l)]}
= {(–128.12 kJ) + (–394.36 kJ)} – (–463.38 kJ) = –59.10 kJ
To check the internal consistency of these values, calculate ΔH° – TΔS° and compare the result
to ΔG° :
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Chapter 8
ΔH° – TΔS° = (–1.3 kJ) – (298 K)(0.1944 kJ·K–1) = –59.2 kJ
This value is identical (within the uncertainties of the data values) to that calculated from
standard free energies of formation.
(b) The reaction is spontaneous under standard conditions, since ΔG = ΔG° < 0.
(c) Compressing a gas decreases its entropy (less volume to spread out in), so the entropy of the
CO2 would decrease, if its pressure is raised to 100 atm. This change would decrease the
positive entropy change for the reaction.
(d) Since the positive entropy change decreases, if the pressure of the CO2 is increased, the free
energy change will be less negative at the higher pressure, that is, there will be a lower driving
force for the reaction and the reaction will be less likely to proceed as written (although it will
still be spontaneous under these conditions).
(e) Le Chatelier’s principle says that a system at equilibrium will react to a disturbance in a way
that minimizes the effect of the disturbance. If the stress on the system in this case is an increase
in the concentration (molecules per liter) of a product (increased pressure of CO2), the reaction
will go in reverse to relieve the stress and use up some of the products. The free energy change
is in this same direction—less tendency to form products [see part (d)], so the prediction based
on Le Chatelier’s principle is consistent with the thermodynamic analysis.
Problem 8.59.
(a) We can use data from Appendix B to calculate ΔH° , ΔS° , and ΔG° at 298 K for the water
gas reaction: C(s) + H2O(g) →CO(g) + H2(g). The relevant thermodynamic quantities are:
compound
C(s)
H2O(g)
CO(g)
H2(g)
ΔHof, kJ·mol–1
0
–241.82
–110.53
0
So, J·K–1·mol–1
5.740
188.83
197.67
130.68
ΔGof, kJ·mol–1
0
–228.57
–137.17
0
ΔH° = {(1 mol)ΔH° f[CO(g)] + (1 mol)ΔH° f[H2(g)]}
– {(1 mol)ΔH° f[C(s))] + (1 mol)ΔH° f[H2O(g)]}
= (–110.53 kJ) + (0 kJ) – (0 kJ) – (–241.82 kJ) = – 131.29 kJ
ΔS° = {(1 mol)S° [CO(g] + (1 mol)S° [H2(g)]}
– {(1 mol)S° [C(s)] + (1 mol)S° [H2O(g)]}
= (197.67 J·K–1) + (130.68 J·K–1) – (5.740 J·K–1) – (188.83 J·K–1) = 133.78 J·K–1
ΔG° = {(1 mol)ΔG° f[CO(g] + (1 mol)ΔG° f[H2(g)]}
– {(1 mol)ΔG° f[C(s)] + (1 mol)ΔG° f[H2O(g)]}
= (–137.17 kJ) + (0 kJ) – (0 kJ) – (–228.57 kJ) = –91.40 kJ
Under standard conditions at 298 K, the reaction is not spontaneous: ΔG = ΔGo > 0.
(b) Since the entropy change for the reaction is large and positive. increasing the temperature
will make –TΔS a larger negative value, which, at high enough temperature, should make ΔG <
0 and make the reaction spontaneous.
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Entropy and Molecular Organization
(c) There is an increase in number of moles of gas in the water gas reaction. If we disturb the
system by increasing the gas pressure, Le Chatelier’s principle predicts that the system will
respond by using up some of the gas, that is, by going in reverse toward reactants. This is the
unfavorable direction for the reaction, so we should decrease the gas pressure to favor products.
We also know that the entropy of gases increases as the volume available per molecule
increases, so decreasing the pressure (by increasing the volume) will increase the entropy of the
gases. Since there are more product than reactant gases, the net change in entropy for this
change will be positive, so this thermodynamic argument also suggests lowering the pressure to
favor the reaction as written.
(d) Parts (b) and (c) suggest raising the temperature and lowering the pressure to favor the
reaction. This is choice (iii) among the reaction conditions suggested for consideration.
Problem 8.60.
[NOTE: The structure in the problem should probably be labeled as a typical fat molecule.]
The polar ends of the fat molecules (the right-hand ends in the structure shown) in waxed paper
will interact with the polar cellulose molecules (by H bonding) that make up paper and “cover”
them with a hydrophobic layer of the nonpolar ends of the fat molecules sticking out from the
cellulose polymeric chains. Water molecules will contact this hydrophobic layer and not be able
to interact with the polar cellulose. The strong interaction of water molecules with one another
(H bonding) will hold the water together in beads (droplets) on the hydrophobic surface.
Problem 8.61.
(a) The positional entropy change for methane-water clathrate formation involves a combination
of factors. Forming a mixture increases the positional entropy. However, restricting a gas,
methane, from its usual freedom of motion decreases the positional entropy. Furthermore,
formation of the ordered cages of water molecules around the methane molecules decreases the
positional entropy of the water. The net effect is a decrease in positional entropy, which is why
the solubility of methane in water is low.
(b) At lower temperatures, the movement of water molecules is less vigorous and it is easier for
them to form the ice-like cages required for clathrate formation. Also, the solution of methane is
water is slightly exothermic. At lower temperatures the positive thermal entropy change in the
surroundings will be greater than at higher temperatures, thus favoring the solution of methane
and formation of the clathrate.
(c) At higher pressures, the entropy of the gas decreases because its volume decreases and there
is less room for molecules to move around. The difference in entropy between the gas and the
gas molecules dissolved in water becomes less, so the decrease in entropy for methane
dissolving becomes smaller. Overall, this means that the positional entropy change for
dissolving and clathrate formation becomes a bit more favorable and clathrate formation more
favored.
Problem 8.62.
Benjamin Franklin found that a teaspoon of oil poured on the surface of a calm pond spread out
to form an oil patch about 1/2-acre in area. If Franklin’s oil patch is a continuous film of oil one
molecule thick, then we can equate the volume of one teaspoon of oil with the volume of the
one-molecule thick layer. If we further assume that oil molecules are a simple shape, like cubes,
that stack together as tightly as possible in both the three dimensional and two dimensional case,
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we can figure out the size of the molecule (cube). A teaspoon holds about 5 mL of liquid or
about 5 × 10–6 m3. An acre is about 4000 m2; a half acre is about 2 × 103 m2. If the side of the
cubic oil molecule is d meters long, the volume of the oil film is (2 × 103 m2)(d m) = 2d × 103
m3. Setting the two volumes equal and solving for d, gives d = 2.5 × 10–9 m = 2.5 nm = 2500
pm. Carbon-carbon bond lengths are about 150 pm and oil or fat molecules contain at least 50
carbons, so a molecule similar in shape to a phospholipid could be folded about in such a way as
to have a diameter of about 2500 pm. Franklin’s observation more than 200 years ago (before
Dalton proposed his atomic model for matter) is in accord with what we now know about
molecular structure.
Problem 8.63.
For ambiphilic compounds with the same polar head but non-polar tails of different lengths, the
lengths of the tails can affect the ability of the compounds to form micelles in water. Micelles
can form when the size of the polar heads is large enough to cover the roughly spherical volume
occupied by the hydrophobic tails tucked inside the “shell” of polar heads, as shown in Figure
8.20(c) and reproduced here:
If the nonpolar tails are too long, the polar heads are not large enough to form a shell that can
completely enclose them, as shown in this drawing:
As the tails get longer, micelles are harder to form and the ambiphiles become less soluble or
perhaps form other, more extended structures more like bilayers.
Problem 8.64.
(a) Suppose you add a little bit of an ambiphilic compound that can’t form micelles in water
because its non-polar tail is too long to a system of liquid water with a less dense non-polar
liquid floating on it. The molecules of the ambiphile will collect at the interface between the two
phases with their nonpolar tails in the upper nonpolar liquid and their polar heads in the water:
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ACS Chemistry Chapter 8 suggested solutions
Chapter 8
Entropy and Molecular Organization
(b) Suppose you add more of the ambiphilic compound to the system in part (a). Since the
ambiphile cannot form a micelle in water, it will preferentially dissolve in the nonpolar solvent
and form a “reversed” micelle with the nonpolar tails on the outside and the polar heads inside:
Problem 8.65.
Experiments to test the feasibility of extracting solutes from the water by getting them to
dissolve in the reversed micellin a non-polar liquid are being carried out in several laboratories.
The polar head probably has to have some hydrogen bonding properties, in order to compete
with water for solutes that are hydrogen bonded with water. Ions that are hydrated but do not
hydrogen bond to water might be extracted by polar heads that can chelate the ions, in order to
compete with hydration by the water molecules.
[NOTE: Among other systems, experiments are being done with reversed micelles and liquid
CO2 as the non-polar liquid to see if such extractions from water can be done without generating
a lot of waste solvents that could pollute the environment. See Chemistry & Engineering News,
September 29, 1997, page 6, for photos demonstrating the extraction.]
Problem 8.66.
(a) In order to calculate how many phospholipid molecules are required to form the cell
membrane of hepatocytes, cubical cells with edge lengths of about 15 × 10–6 m, we have to
make an assumption about the area of the surface of the cell membrane occupied by a
phospholipid molecule. You could make a model of the glycerol molecule and the next few
atoms bonded to it in the phospholipid shown in Figure 8.22 and then measure the size with the
scale that comes with the model set. The measurement is probably different in different
dimensions, but we will assume that the “shoulders” of each molecule occupy a square that is
about 500 pm (500 × 10–12 m) on a side. The area of this square is 2.5 × 10–19 m2. The area of
one surface of the cell is (15 × 10–6 m)2 = 2.3 × 10–10 m2 and there are six sides, so the outside
area of the cell surface is 1.4 × 10–9 m2. The membrane is a bilayer and we will assume that the
area of both the inner and outer surfaces is the same, so the total area that must be covered is
2.8 × 10–9 m2. The number of phospholipid molecules required to cover this area is about:
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Entropy and Molecular Organization
(2.8 × 10 –9 m 2 )
(2.5 × 10 –19 m 2 )
Chapter 8
= 1 × 1010.
(b) The number of moles represented by 1 × 1010 phospholipid molecules is:
(1 × 1010 molecules)
(6 × 10 23 molecules ⋅ mol –1 )
≈ 2 × 10–14 mol
To determine the mass of this many moles, we need the molar mass of a phospholipid. There are
a great variety of phospholipids with different molar masses. Use the model shown in Figure
8.22 to get an idea of the molar mass. The molecule shown has the formula, C44H86O8NP, with a
molar mass of 8 × 102 g. Thus, the mass of phospholipid in the hepatocyte membrane is about
(8 × 102 g·mol–1)(2 × 10–14 mol) = 2 × 10–11 g·= 20 pg (picogram). For comparison purposes,
the mass of the cell, assuming a density of 1 g·mL–1 = 1 × 106 g·m–3, is
(1 × 106 g·m–3)(15 × 10–6 m)3 = 3 × 10–9 g = 3000 pg
This calculation suggests that the outer membrane phospholipids make up a little less than 1%
of the mass of the cell. There are many internal structures in a cell that are also enclosed in
phospholipid bilayer membranes and these add to the mass of phospholipid just calculate.
Problem 8.67.
Mayonnaise is basically a mixture of oil and water that does not separate (at least, it is not
supposed to separate). This sort of mixture is called an emulsion. In order to prevent separation,
egg yolk is added to the mixture. The phospholipids in the added yolk form micelles that hold
the oil in their interiors (like a detergent micelle) and keep the mixture from separating.
Problem 8.68.
The more uniformly the phospholipid molecules can stack together, side by side, in a bilayer
membrane, the less freedom each has to move about. Good stacking leads to a “stiffer”
membrane in which molecules are limited in their motion. To make the membrane more flexible
or fluid, the organism needs to disrupt the uniform stacking. Note from Figure 8.22 that double
bonds (unsaturation) in the fatty acid chain makes the chain kinked, so it cannot pack as
uniformly with its neighbors. Thus, a cold-water fish would probably have more unsaturated
fatty acids in its phospholipids, in order to increase the fluidity of its membranes even as the
temperature tends to stiffen them. The properties of fats and oils provide a familiar analogy.
Fats and oils are esters of glycerol with three fatty acids bonded to the alcohol groups. The
difference between a fat and an oil is that a fat is a solid at room temperature and an oil is liquid.
The cause of this difference is a much higher proportion of unsaturated fatty acid chains in oils.
The molecules of oil can’t stack together as easily and must be taken to a lower temperature
before they can solidify.
Problem 8.69.
Aren’t icebergs made of salt water like the ocean water? Icebergs are actually pieces broken off
from glaciers that have reached the edge of the sea. Glaciers are made from rain and snow fall
which contain only small amounts of dissolved solids, so icebergs are not salty. Even if icebergs
were made by freezing ocean water, they would contain only small amounts of salt, because
pure solid solvent (water) crystallizes from the solution, not a solid containing sodium ions,
chloride ions, and water molecules.
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ACS Chemistry Chapter 8 suggested solutions
Chapter 8
Entropy and Molecular Organization
Problem 8.70.
The entropy of a solution (a mixture of solvent and soluble solute particles) is greater than the
entropy of the pure solvent, because the positional entropy in the mixture is greater. The entropy
of a pure liquid solvent is greater than the entropy of its solid, because the particles in the liquid
have the entire volume of the liquid available to them, but those in the solid are stuck in one
place in the crystal lattice. Thus, we have: Ssolution > Ssolvent > Ssolid.
Problem 8.71.
Colligative properties depend on entropy differences between two solutions or solution and pure
solvent. These differences depend on the number of arrangements of the particles and not the
interactions between particles, so colligative properties depend only on the number of solute
particles and not their chemical properties.
Problem 8.72.
Freezing point lowering is a colligative property, so its magnitude depends on the number of
particles (ions or molecules) in a given volume of a solution. Assuming that the solutions in this
problem are aqueous and that each of the ionic salts dissociates completely in solution, we get 2,
3, and 4 moles of ions per liter, respectively, from compounds (i) NaCl, (ii) CaBr2, and (iii)
Al(NO3)3. The aluminum nitrate solution will have the lowest freezing point.
Problem 8.73.
Most antifreeze products available today contain ethylene glycol, HOCH2CH2OH
(1,2-ethanediol), which is a relatively nonvolatile liquid that is miscible with water. Dissolving
antifreeze in water lowers the freezing point of the solution, so the liquid is protected from
freezing in cold weather. If the coolant solution freezes in your radiator or engine, it can cause a
great deal of damage, because it expands upon freezing and can break an engine block. A
second great advantage of ethylene glycol is that it increases the boiling point of its aqueous
solutions, so your engine is less likely to boil over in the summer. Thus, this antifreeze might
also be called “antiboil.”
Problem 8.74.
Each of these 1.5 m solutions has the same molality, but not the same number of particles in
solution. MgCl2 dissolves to give three moles of particles per mole dissolved (a mole of
Mg2+(aq) cations and two moles of Cl–(aq) anions). NaCl dissolves to give two moles of
particles per mole dissolved (a mole of Na+(aq) cations and a mole of Cl–(aq) anions). Glucose
does not ionize or dissociate, so it dissolves to give one mole of solute per mole dissolved. Since
freezing point lowering is a colligative property that depends on the number of solute particles
in the solution, 1.5 m MgCl2 would give the greatest freezing point lowering.
Problem 8.75.
The freezing point of a solution made by dissolving 5.00 g of sucrose, C12H22O11, in 100.0 g of
water can be calculated from the freezing point lowering expression, ΔTfp = kfp·m. The freezing
point lowering constant for water is –1.86 ° C·m–1. The molality of the solution is the number of
moles of sucrose per kilogram of solvent:
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Entropy and Molecular Organization
Chapter 8
5.00 g sucrose (in 100.0 g water) = (5.00 g
⎛ 1 mol sucrose ⎞ ⎛
⎞
1
sucrose) ⎜
⎟
⎜
⎝ 342.3 g ⎠ ⎝ 0.1000 kg water ⎟⎠
= 0.146 m
ΔTfp = kfp·m = (–1.86 ° C·m–1)(0.146 m) = –0.272 ° C
Tfp(solution) = Tfp(solvent) + ΔTfp = (0 ° C) + (–0.272 ° C) = –0.272 ° C
Problem 8.76.
(a) The molality of a solution of 10.00 g of n-decane, CH3(CH2)8CH3, dissolved in 100.00 g of
benzene is:
⎛ 1 mol decane ⎞ ⎛
⎞
1
= 0.7452 m
10.00 g decane = (10.00 g decane) ⎜
⎟
⎜
⎝ 134.2 g ⎠ ⎝ 0.10000 kg benzene ⎟⎠
(b) The freezing point lowering of this solution is:
ΔTfp = kfp·m = (–5.12 ° C·m–1)(0.7452 m) = –3.82 ° C
(c) The freezing point of the solution is:
Tfp(solution) = Tfp(solvent) + ΔTfp = (5.50 ° C) + (–3.82 ° C) = 1.68 ° C
Problem 8.77.
(i) A 0.30 m solution in benzene has a freezing point lowering and freezing point of:
ΔTfp = kfp·m = (–5.12 ° C·m–1)(0.30 m) = –1.5 ° C
Tfp(solution) = Tfp(solvent) + ΔTfp = (5.50 ° C) + (–1.5 ° C) = 4.0 ° C
(ii) A 0.10 m solution in cyclohexane has a freezing point lowering and freezing point of:
ΔTfp = kfp·m = (–20.2 ° C·m–1)(0.10 m) = –2.0 ° C
Tfp(solution) = Tfp(solvent) + ΔTfp = (6.5 ° C) + (–2.0 ° C) = 4.5 ° C
The benzene solution will freeze at a lower temperature.
Problem 8.78.
(a) An aqueous solution of 1.0 m sodium chloride is predicted to have a freezing point of –
3.72 ° C, because the salt dissolves to give two ions. Thus, the molality of particles in the
solution is 2.0 m and the predicted freezing point lowering and freezing point are:
ΔTfp = kfp·m = (–1.86 ° C·m–1)(2.0 m) = –3.72 ° C (data uncertainly gives –3.7 ° C)
Tfp(solution) = Tfp(solvent) + ΔTfp = (0 ° C) + (–3.72 ° C) = –3.72 ° C
(b) The actual freezing point of this solution is −3.53 ° C, because the ions tend to associate
(stick together) in solution and not act completely independently. The actual molality of
particles is not quite twice the molality of dissolved salt. You can get an estimate of the
effective molality of particles from the experimental freezing point lowering:
ΔTfp
(–3.53 o C)
m=
=
= 1.9 m
kfp
(–1.86 o C ⋅ m –1 )
36
ACS Chemistry Chapter 8 suggested solutions
Chapter 8
Entropy and Molecular Organization
Problem 8.79.
A solution made by dissolving 5.00 g of sodium sulfate, Na2SO4, in 100.0 g of water has three
times as many moles of particles as moles of Na2SO4 dissolved, because each mole of the solid
dissolved ionizes to give two moles of Na+(aq) cations and one mole of SO42–(aq) anions. The
molality, m, of particles in the solution is:
5.00 g Na2SO4 = (5.00 g
⎞ ⎛ 3 mol particles ⎞
⎛ 1 mol Na 2SO 4 ⎞ ⎛
1
Na2SO4) ⎜
⎟
⎝
142.0 g ⎠ ⎜⎝ 0.1000 kg H 2O ⎟⎠ ⎜⎝ 1 mol Na 2 SO 4 ⎟⎠
= 1.056 m
The freezing point lowering and freezing point of this solution are:
ΔTfp = kfp·m = (–1.86 ° C·m–1)(1.056 m) = –1.96 ° C
Tfp(solution) = Tfp(solvent) + ΔTfp = (0 ° C) + (–1.96 ° C) = –1.96 ° C
[If the data in Problem 8.78 are correct, it is likely that the actual freezing point of a solution
this concentrated will be somewhat higher (closer to zero) than this calculation predicts.]
Problem 8.80.
The molality, freezing point lowering, and freezing point of a solution of 1.00 g of sorbital,
C6H14O6, in 100.0 g of water are:
⎞
⎛ 1 mol Na 2 SO 4 ⎞ ⎛
1
= 0.0549 m
1.00 g C6H14O6 = (1.00 g C6H14O6) ⎜
⎟
⎜
⎝
⎠ ⎝ 0.1000 kg H 2 O ⎟⎠
182 g
ΔTfp = kfp·m = (–1.86 ° C·m–1)(0.0549 m) = –0.102 ° C
Tfp(solution) = Tfp(solvent) + ΔTfp = (0 ° C) + (–0.102 ° C) = –0.102 ° C
Problem 8.81.
To calculate the molality of a solution of 3.5 milligrams of hemoglobin (molar mass = 64,000 g)
dissolved in 5.0 mL of water, we assume that 5.0 mL of water is 5.0 g of water. The molality,
freezing point lowering, and freezing point of this solution are:
⎞
⎛ 1 mol hemoglobin ⎞ ⎛
1
0.0035 g hemoglobin = (0.0035 g hemoglobin) ⎜
⎟
⎜
⎝
⎠ ⎝ 0.0050 kg H 2 O ⎟⎠
64,000 g
= 1.09 × 10–5 m
ΔTfp = kfp·m = (–1.86 ° C·m–1)(1.09 × 10–5 m) = –2.0 × 10–5 ° C
Tfp(solution) = Tfp(solvent) + ΔTfp = (0 ° C) + (–2.0 × 10–5 ° C) = –2.0 × 10–5 ° C
This change is too small to measure, so freezing point depression is not a good method for
determining molecular weights of biomolecules, such as proteins.
Problem 8.82.
The relationship between osmotic pressure, Π, and the concentration, c (the molarity), of
molecules or ions in a solution is: Π = cRT. The average osmotic pressure of blood serum is 7.7
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Entropy and Molecular Organization
Chapter 8
atm at 25 ° C. Therefore, to find the molarity of particles in the serum, we can rearrange the
osmotic pressure relationship to solve for M at Π = 7.7 atm:
(7.7 atm)
Π
c=
=
= 0.31 M
(0.08206 L ⋅ atm ⋅ K –1 ⋅ mol–1 )(298 K)
RT
The concentration of glucose solution which is isotonic with blood serum (has the same osmotic
pressure) is 0.31 M. The concentration of NaCl which is isotonic with blood is 0.16 M, because
NaCl ionizes in solution to form two particles, Na+ cations and Cl– anions.
Problem 8.83.
Suppose 0.150 g of a newly isolated protein dissolved in 25.0 mL of water has an osmotic
pressure of 0.00342 atm at 277 K. The molar mass of the solute can be determined from the
osmotic pressure that gives the concentration of the solute, c (see the solution to Problem 8.82).
The defining equation for the concentration can then be rearranged to calculate the molar mass
from the value of the concentration and the mass of solute, m, dissolved in a known volume, V,
of solvent:
m
m ⋅ R ⋅T
(0.150 g)(0.08206 L ⋅ atm ⋅ K –1 ⋅ mol –1 )(277 K)
molar mass =
=
=
(0.00342 atm)(0.0250 L)
c ⋅V
Π ⋅V
= 3.99 × 104 g·mol–1
Problem 8.84.
The osmotic pressure, Π, is the pressure required to decrease the entropy of a solution to match
that of the pure water (or a more dilute solution) on the other side of a semipermeable
membrane, so there is no net flow of water through the membrane. If a pressure greater than Π
is applied to the solution, the entropy of the solution will decrease still further and be less than
the entropy of the pure water. Under these circumstances, the net entropy will increase, if water
molecules pass from solution into the pure liquid. Thus, water will flow from the solution into
the pure water. Molecular level pictures can show the volume available to the water molecules
decreasing as the pressure is increased and the solution is compressed a tiny bit. If the pressure
is high enough the compression is large enough to make the entropy even lower than for pure
water.
Problem 8.85.
In reverse osmosis, the flow of water is reversed from the direction we associate with regular
osmosis. The flow is from solution to pure liquid, rather than the other way around. Purification
(desalting) of salt (ocean) water to produce fresh, water for irrigation and drinking is one
practical use of reverse osmosis that is being used in arid parts of the world near oceans or seas,
such as the Arabian peninsula.
[Another method of desalting, flash evaporation with energy supplied by power plants by waste
heat from power plants, was discussed as a coupled process in the solution to Problem 7.47.
Both these methods are in practical use on very large scales, especially in the middle east.
Reverse osmosis is more energy efficient, but flash evaporation competes when the heat energy
is available from some source that has another primary use, such as power production.]
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ACS Chemistry Chapter 8 suggested solutions
Chapter 8
Entropy and Molecular Organization
Problem 8.86.
Reverse osmosis (see Problems 8.84 and 8.85) will begin when the applied pressure on the
solution just exceeds the osmotic pressure of the solution. Thus, to solve this problem about
desalting seawater, we need to find the osmotic pressure, Π, of the seawater solution. The
concentrations are given in ppm instead of molarity, so we have to convert them to moles per
liter. We will assume that the ppm are parts per million by mass. For example, 18,000 ppm of
Cl– is 18,000 g of chloride ion in one million grams of solution. We will further assume that one
million grams of solution is one million mL (or 1000 L) of solution. The number of moles of
each ion in 1000 L is:
ion
Cl–
Na+
Mg2+
SO42–
K+
concentration, ppm
18000
10500
1200
870
379
TOTAL
mol·(1000 L)–1
510
460
50
9
10
1039
The table shows that the total number of moles of ions in 1000 L of the seawater is 1039. Thus,
the total ionic molarity of the seawater is 1.04 M. This is the value we need to use to get Π:
Π = cRT = (1.04 M)(8.314 × 103 Pa·L·mol–1·K–1)(298 K) = 2.58 × 106 Pa ≈ 25 atm
A pressure of above 25 atm has to be exerted on the seawater to get reverse osmosis to occur.
This is a high pressure, but not extraordinarily high, so the process is quite feasible. The major
technical problem is design of “membranes” that can withstand such pressures, but this is partly
solved by constructing them as cylinders with the seawater and high pressure on the outside and
fresh water flowing out on the inside. A cylinder can withstand a great deal more pressure
(uniformly applied) than a flat surface.
Problem 8.87.
The resin coating on the time release plant fertilizer called Osmocote® needs to be
semipermeable to water. There is little water inside the granules to begin with, so water from
the surroundings enters the granules by osmosis and begins to dissolve the fertilizer compounds
to form a highly concentrated solution. Osmosis continues to add water to the granules and their
size increases (they “swell into plump capsules”). As the granules swell, it is likely that the resin
coating finally begins to develop tiny pores through which the contents can ooze and
“continuously release nutrients for approximately 9 months.” The product name seems to
combine the ideas of a coating (“cote”) that has osmotic (“osmo”) properties.
Problem 8.88.
The only statement that applies correctly to the growing plant is (ii). The net entropy of the
system (plant and its precursor molecules such as carbon dioxide and water) and surroundings
(everything else, especially energy from the sun as well as the enormous entropy increase the
sun undergoes in producing and spreading the energy into the solar system) increases. For
example, photosynthesis occurs to produce more complex molecules from simple ones and uses
the energy of electromagnetic radiation from the sun to drive the reaction. As we have just
shown, statement (i) is incorrect. Statement (iii) is incorrect, because it would not provide a
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Chapter 8
mechanism for a positive net entropy change for the system plus surroundings. Statement (iv)
suggests that there is something about the complexity of biological systems that makes the
second law not applicable. No system, simple or complex has yet been found to which the
second law does not apply. That is why it is called a “law.” No exceptions have been found.
Problem 8.89.
(a) A fertilized chicken egg in an incubator (a constant temperature and pressure environment)
is an open system. Since gases from the atmosphere can enter the egg, they can act as reactants
for reactions within the egg.
(b) The same molecules, such as ATP, that are coupled to synthetic reactions in other living
things are at work in the egg. They are produced in the same way as in other organisms by
oxidation of food molecules (mostly the fat in the yolk) by atmospheric oxygen. The oxidations
all have high net positive entropy changes and the hydrolysis of the ATP formed in these
processes also has a net positive entropy which can be “harnessed” by coupling the hydrolysis
to the synthetic reactions required to form the chick. The net entropy change in the processes is
positive and consistent with the second law of thermodynamics.
(c) As metabolism occurs in the egg, energy not “captured” in the form of potential energy of
chemical bonds (in ATP, for example) is lost from the egg to the surroundings as thermal
energy. (If an egg were insulated, so it did not lose thermal energy to its surroundings, it would
get warmer.)
(d) Free energy must decrease during the chick development process. The overall process is
spontaneous, so ΔG < 0.
Problem 8.90.
(a) Since the polymer →product reaction is endothermic, we know that ΔH > 0 for the system.
Also, since the reaction is spontaneous, ΔG < 0 for the system. In order for ΔG to be negative, if
ΔH is positive, we must have ΔS > 0, so that –TΔS < 0.
(b) The product must have a higher entropy than the reactant, since ΔS > 0. One possibility is
for the polymer to be breaking down to smaller pieces, thus giving more freedom of motion to
the parts that made up the polymer. This process would require the net breaking of bonds, which
would require energy, which is consistent with the observed endothermic reaction. Thus, the
most likely reaction is some sort of breakdown of the polymer.
Problem 8.91.
The data table for this problem is reproduced here for easy reference.
40
ACS Chemistry Chapter 8 suggested solutions
Chapter 8
Entropy and Molecular Organization
Enthalpies of formation and entropies for selected gas-phase hydrocarbons.
name
formula
condensed structural formula
ethene
propene
cyclopropane
1-butene
cyclobutane
1-pentene
2-methyl-1-butene
3-methyl-1-butene
2-methyl-2-butene
trans-2-pentene
cyclopentane
1-hexene
cyclohexane
1-heptene
1-octene
1-nonene
1-decene
1-undecene
1-dodecene
C2H4
C3H6
C3H6
C4H8
C4H8
C5H10
C5H10
C5H10
C5H10
C5H10
C5H10
C6H12
C6H12
C7H14
C8H16
C9H18
C10H20
C11H22
C12H24
H2C=CH2
H2C=CHCH3
C’s bonded in an equilateral triangle
H2C=CHCH2CH3
C’s bonded in a non-planar square
H2C=CHCH2CH2CH3
H2C=C(CH3)CH2CH3
H2C=CHCH(CH3)2
H3CCH=C(CH3)2
H3CCH=CHCH2CH3
C’s bonded in a non-planar pentagon
H2C=CH(CH2)3CH3
C’s bonded in a non-planar hexagon
H2C=CH(CH2)4CH3
H2C=CH(CH2)5CH3
H2C=CH(CH2)6CH3
H2C=CH(CH2)7CH3
H2C=CH(CH2)8CH3
H2C=CH(CH2)9CH3
ΔHfo
kJ mol–1
52.28
20.41
53.30
1.17
26.65
–20.92
–36.62
–28.95
–42.55
–31.76
–77.24
–41.7
–123.1
–62.16
–82.93
–103.5
–124.1
–144.8
–165.4
So
J K mol–1
–1
219.8
266.9
237.4
307.4
265.4
347.6
342.0
333.5
338.5
342.3
292.9
386.0
298.2
424.4
462.8
501.2
539.6
578.1
616.5
(a) The difference in bonding between an alkene and the cycloalkane with the same number of
carbons is the bonding between carbons. (The number of C–H bonds is the same in both
isomers.) The cycloalkane has as many C–C single bonds as there are carbons in the ring. The
alkene has a C=C double bond that replaces two of the C–C single bonds. The difference in
enthalpy between the two isomers is just the difference between a C=C double bond enthalpy
and two C–C single bond enthalpies. Figure 7.9 in Chapter 7 showed this difference graphically.
Two C–C single bonds are stronger by 74 kJ·mol–1 than one C=C double bond. Stronger
bonding should lead to a more negative enthalpy of formation for the cyclic isomers. This
prediction is borne out by the data for the five and six carbon isomers. The enthalpy of
formation of cyclopentane is about 56 kJ·mol–1 more negative than the enthalpy of formation of
1-pentene. For cyclohexane and 1-hexene, the difference is about 81 kJ·mol–1. Just the opposite
is the case for cyclobutane and cyclopropane, which have more positive enthapies of formation
than butene and propene, respectively. When you make molecular models of these eight
molecules, you find that it is hard to force the ball-and-stick models to make triangular and
almost square carbon structures. There is no problem making the five- and six-carbon rings,
although the five-carbon ring is still a bit strained. The double bond in the alkenes is a little
difficult to form, but the rest of the bonding is easy to model. The smaller rings are strained in
the models and they are strained in the real molecules. The strain weakens the bonding between
the carbons and makes them weaker than the average bond energies we used in the calculation.
ACS Chemistry FROG
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Entropy and Molecular Organization
Chapter 8
Thus, the strain makes cyclopropane and cyclobutane less stable, relative to their elements, than
the corresponding alkenes. If the difference between cyclohexane and 1-hexene, 81 kJ·mol–1,
represents the difference for a cyclic molecule that is not strained, then we might suppose that
cyclopentane is strained by about 25 kJ·mol–1 [= (81 kJ·mol–1) – (56 kJ·mol–1)], because it is
only 56 kJ·mol–1 more stable than 1-pentene. By similar reasoning, the strain in cyclopropane is
114 kJ·mol–1 [= (81 kJ·mol–1) – (–33 kJ·mol–1)], because it is 33 kJ·mol–1 less stable than
propene and the strain in cyclobutane is 107 kJ·mol–1 [= (81 kJ·mol–1) – (–26 kJ·mol–1)].
(b)This table lists the differences (fourth column) between the standard entropies for the linear
and cyclic isomers with 3, 4, 5, and 6 carbons.
number of
carbons
3
4
5
6
S° linear isomer
J K–1 mol–1
266.9
307.4
347.6
386.0
S° cyclic
isomer J K–1
mol–1
237.4
265.4
292.9
298.2
ΔS° (linear –
cyclic) J K–1 mol–1
32.5
42.0
54.7
87.8
We would expect the linear isomers to have higher entropies, as they do, because the chains are
free to flop about into many different forms. The ends of the chain are bonded together in the
cyclic isomers: the freedom of bond rotation (floppiness) is lost and the entropy of the cyclic
forms is lower. The longer the chain, the more floppy it is and the more freedom of motion is
lost going to the cyclic isomer. As we see, the entropy loss for the 6-carbon chain is almost
three times as great as for the 3-carbon chain.
(c) The enthalpies of formation get more negative and the entropies more positive as the chain
lengths increase for the linear 1-alkenes. The data are shown in these two plots:
There is an essentially linear decrease in enthalpy of formation, to become more negative by
21 kJ·mol–1 for each added carbon. Similarly, there is a linear increase in the entropy of about
39 J K–1 mol–1 for each carbon added. Each addition of a CH2 unit to the chain seems to have
the same effect. A linear extrapolation of these plots (from the slopes of the lines–the change
per added CH2) gives ΔHf° = –207 kJ·mol–1 and S° = 695 J K–1 mol–1 for 1-tetradecene, C14H28.
The plots look quite linear, so we can be pretty confident that these values for 1-tetradecene are
within about ±1% of their experimental values.
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ACS Chemistry Chapter 8 suggested solutions
Chapter 8
Entropy and Molecular Organization
For the enthalpies, we can also use bond enthalpies and the enthalpy of formation of C(g) to
estimate the change in enthalpy of formation for adding a CH2 to the chain.
ΔH, kJ·mol–1
reaction
net change
C(s) →C(g)
717
vaporize carbon
H2(g)→2H(g)
436
break H–H bond
C(g) + 2H(g) →CH2(g)
–828
make two C–H bonds
Cn alkene(g) + CH2(g) →C(n+1) alkene(g)
–347
make C–C bond
Cn alkene(g) + C(s) +H2(g) →C(n+1) alkene(g)
–22
summation
This calculated value is very close to the experimental value of –21 kJ·mol–1.
(d) An analysis for the cyclic isomers similar to that for the linear isomers is difficult because of
the effect of strain in the cyclic isomers. We don’t have enough data and the smaller rings are
strained [as we calculated in part (a)], so their values have to be disregarded. The analysis on the
basis of bond enthalpies at the end of part (c) is applicable here and predicts about a 22 kJ·mol–1
more negative enthalpy of formation for each additional CH2 in the ring. If we extrapolate this
bond enthalpy increment, we would predict a decrease of 44 kJ·mol–1 from cyclohexane and an
enthalpy of formation of –167 kJ·mol–1 [= (–123 kJ·mol–1) + (–44 kJ·mol–1)] for cyclooctane,
C8H16.
[Unfortunately, this analysis is flawed, because rings with more than six carbons are also
strained. The six-membered ring, cyclohexane, can twist into two conformations (see Problem
5.39) in which the bond angles around all the carbons are the tetrahedral angle exhibited by
linear hydrocarbons. Smaller and larger rings cannot do this, so there is some strain in their
structures. The consequence is that (for gas phase molecules) ΔH° f(cycloheptane) = –
116 kJ·mol–1 and ΔH° f(cyclooctane) = –126 kJ·mol–1. These values are essentially the same as
for cyclohexane, so the amount of strain energy almost exactly compensates for the increase in
bond energy due to the increasing number of CH2 groups in the ring.]
Extrapolation of entropy data for the 3-. 4-, 5-, and 6-membered rings is also complicated by
what appears to be a non-linear increase in entropy as the CH2 groups are added. The increases
in entropy from cyclopropane to cyclobutane and from cyclobutane to cyclopentane are each
about 28 J·K–1·mol–1, but the increase from cyclopentane to cyclohexame is only 5 J·K–1·mol–1.
Thus, we cannot tell from these data whether the increase per CH2 group added begins to
decrease at the 6-membered ring or whether the value for cyclohexane is somehow a problem.
The standard entropies for gaseous cycloheptane and cyclooctane are available in the on-line
National Institute of Standards and Technology (NIST) Chemistry WebBook at
http://webbook.nist.gov/chemistry/, and we can analyze these entropy data together with those
for the smaller rings as shown in this plot:
ACS Chemistry FROG
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Entropy and Molecular Organization
Chapter 8
The line, with a slope of about 27 J·K–1·mol–1 per carbon atom, is the best straight line through
the data points represented by squares. The data point represented by a circle is the standard
entropy of gaseous cyclohexane. It appears that there may be a problem with the value for
cyclohexane, but this, like the others, is the value given in the NIST Chemistry WebBook, so it
is presented on the same basis as the others. Note that, if the standard entropy value for
cyclohexane were about 315 J·K–1·mol–1 (where it would fall on the line on this plot), the
entropy difference between the linear and cyclic C6H12 isomers in part (a) would be about
71 J·K–1·mol–1. This value would be more in line with the increasing differences as the number
of carbons increases in the table in part (a). Perhaps more experiments on the thermodynamics
of cyclohexane are needed to resolve apparent anomalies like these, or discover the molecular
basis for the cyclohexane value.
(e)One way to approach the analysis of the C5H10 alkene isomers is to list the isomers in order
of decreasing entropies to see if any trends in the structures and/or enthalpies of formation are
discernable. We choose to order on the basis of entropy, because entropy is more obviously
related to structure.
S° , J K–1 mol–1
ΔH° f, kJ mol–
1-pentene
347.6
–20.92
H2C
trans-2-pentene
342.3
–31.76
H 3C
isomer
structure
1
CHCH2CH2CH3
C
H
2-methyl-1-butene
2-methyl-2-butene
342.0
338.5
–36.62
–42.55
CH3
H2C
333.5
C
CH2CH3
H3C
C
H3C
3-methyl-1-butene
CH
CH2CH3
C
CH3
H
CH3
–28.95
H 2C
CHCH
CH3
44
ACS Chemistry Chapter 8 suggested solutions
Chapter 8
Entropy and Molecular Organization
The table shows that the more compact the structure, the lower its entropy. The most floppy
structure is 1-pentene. You can twist its model into many different conformations and it has a
higher entropy than any of the others. At the bottom of the list, 2-methyl-2-butene and
3-methyl-1-butene have little flexibility, except the rotation of their methyl groups. The absolute
values of the entropy are high because the molecules are gases and they have many atoms. In
general, the more atoms, the higher the entropy. You saw this “complexity” effect in the graph
for the entropy as a function of number of carbons in parts (c) and (d). Patterns in the enthalpies
of formation are harder to see, but it does seem that the enthalpy of formation (stability)
increases as the number of carbons bonded to the double-bonded carbons increases. The least
negative values are for 1-pentene and 3-methyl-1-butene, the isomers with only one carbon
bonded to the double bond. 2-methyl-2-butene, with three carbons bonded to the double bond
has the most negative enthalpy of formation. [But to confuse things just a bit, the standard
enthalpy of formation of cis-2-pentene is about –28 kJ·mol–1.] A deeper knowledge of bonding
is required to figure out why the stability increases in this way.
Problem 8.92.
The reaction of breaking water down to its elements is the reverse of the reaction forming water
from its elements. The enthalpy change for the breakdown has the same magnitude, but is
opposite in sign to the enthalpy of formation of water. The data in Appendix B give the enthalpy
of formation of water (gas) as –241.82 kJ·mol–1, from which we get +241.82 kJ·mol–1 for the
enthalpy change of the breakdown. This is the net enthalpy associated with breaking four H–O
bonds in water and forming two H–H bonds and one O=O bond. Although the value will vary
somewhat under nonstandard conditions, it will always be positive, that is, energy will always
be required to cause the breakdown. Thus, investors should be wary of an inventor who claims
that this process can be made to occur without an input of energy. Catalysts cannot change the
overall thermodynamics of a reaction.
Problem 8.93.
(a) A solution, a mixture of components, has a higher entropy than pure solvent:
Ssolution > Ssolvent.
(b) Since the enthalpy of vaporization, ΔHvap, for the same amount of solvent from pure solvent
or from a solution are the same, ΔHvap/T will be the same, at the same temperature, for both pure
solvent and a solution.
(c) The entropy change of the surroundings, ΔSsurr, for the vaporization process is –ΔHvap/T.
ΔHvap is positive (vaporization is an endothermic process) and the same for both pure solvent
and a solution [part (b)], so ΔSsurr < 0 and has the same numerical value for vaporization of pure
solvent and a solution. Thus, in order for ΔSnet to be zero, the positional entropy changes for
(solvent →gas) and (solution →gas) must be equal to –ΔSsurr, that is, be positive and of the same
numerical value as ΔSsurr.
(d) We know from part (c) that ΔS(solvent →gas) = ΔS(solution →gas). Since Ssolution > Ssolvent
[from part (a)], we need to have Sgas over solution > Sgas over solvent, in order to have the entropy
differences be the same. The relationships are shown in this diagram:
ACS Chemistry FROG
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Entropy and Molecular Organization
Chapter 8
(e) Since the gas in equilibrium with the solution has the higher entropy, it must have the lower
pressure (which is more volume per molecule for a given amount of gas). Thus, the equilibrium
pressure of gaseous solvent over a solution is lower than over pure solvent. This is consistent
with and explains the lowering of the vapor pressure by a nonvolatile solute.
(f) A beaker of water and a beaker of a 1 M aqueous glucose solution are placed side-by-side
inside a sealed, transparent container held at constant temperature. In this system, the vapor
pressure of water over the pure water is higher than the vapor pressure of water over the sucrose
solution. If the vapor pressure of water in the container gets higher than the equilibrium vapor
pressure over the sucrose solution, water molecules from the vapor will condense into the
solution. This is an example of Le Chatelier’s principle. The disturbance to the equilibrium,
solution ⇔ water vapor, is addition of more water vapor (vaporized from the pure solvent). The
solution equilibrium system will respond by trying to use up some of the extra gaseous water,
that is, some water vapor will condense in the glucose solution. What we will observe over time
is the liquid level in the beaker of pure water decreasing and the liquid level in the beaker of
solution increasing. If this process continues for a long time (diffusion of gases is a slow
process) all of the water from the beaker of pure water will end up in the beaker of glucose
solution, which will be diluted by the process.
Problem 8.94.
This is a schematic representation of the binding of a substrate molecule in the active site on an
enzyme. The bottom diagram shows the interaction of the substrate with a metal ion in solution.
+
+
(a) The free squiggle molecule can move about and twist into whatever shape is allowed by its
bonding. In order to bind to the enzyme, one end of the squiggle has to fold into a particular
shape. There are fewer ways to do this than to twist about randomly, so the positional entropy of
the molecule will decrease as it folds into this particular shape. Also, when two molecules, the
enzyme and the squiggle, that are free to move about independently of one another come
46
ACS Chemistry Chapter 8 suggested solutions
Chapter 8
Entropy and Molecular Organization
together they lose some freedom of movement. This is a further contribution to a decrease in
positional entropy. For the binding process shown, the positional entropy will decrease.
(b) The complex formed between the squiggle molecule and the metal ion holds the squiggle
molecule in the folded form that binds well with the enzyme. The squiggle molecule in the
metal-bound form is not free to twist about randomly, so there will not be a great loss of
positional entropy when it binds with the enzyme. The freedom of movement through the
solution is lost, so the positional entropy will still decrease upon binding, but not as much as in
the absence of the metal ion. We are assuming that the presence of the metal in the folded
squiggle does not interfere with the binding between the squiggle and the enzyme. An example
of a situation like the one pictured here is the binding of ATP to many of the proteins with
which it interacts. Many of these interactions do not take place unless magnesium ion,
Mg2+(aq), is present in the solution to complex with the ATP and hold it in the proper shape for
the interaction. See Chapter 7, Problem 7.48.
Problem 8.95.
(a) There is only one way to arrange 12 identical solvent molecules in the 12 boxes, as in the
lower 12 boxes in Figure 8.4: Wsolvent = 1. Since there are only three boxes and three solute
molecules left to arrange, we also have Wsolute = 1. Hence, Wtotal = Wsolvent·Wsolute = 1·1 = 1.
(b) If there are nine solvent molecules in the lower nine boxes, this leaves three to arrange
among the top six boxes. The number of distinguishable ways to arrange these three identical
solvent molecules among the six boxes is the same as the number of ways we got for arranging
the three identical solute molecules in these six boxes (Figure 8.5), so Wsolvent = 20. Again, once
the solvent molecules are arranged, there is only one way to arrange the solute molecules, so
Wsolute = 1. Hence, Wtotal = Wsolvent·Wsolute = 20·1 = 20. This answer is the same as we got before
(Figure 85) when we focused on arrangements of solute molecules.
(c) If there are six solvent molecules in the lower six boxes, this leaves six to arrange among the
top nine boxes. Using the formula in Problem 8.12, we have:
9!
9 ⋅ 8 ⋅ 7 ⋅ 6! 9 ⋅ 8 ⋅ 7
N = 9, n = 6: W =
=
=
= 3·4·7 = 84
3 ⋅ 2 ⋅1
(9 − 6)!6!
3!6!
Again, once the solvent molecules are arranged, there is only one way to arrange the solute
molecules, so Wsolute = 1. Hence, Wtotal = Wsolvent·Wsolute = 84·1 = 84. Figure 8.6 and the
accompanying text show that there are 84 arrangements of the three solute molecules in the top
nine boxes. We get an identical result when we focus on the solvent instead of the solute.
(d) With 12 solvent molecules arranged among 15 boxes we have:
15 ⋅14 ⋅ 13
15!
N = 15, n = 9:
W=
=
= 5·7·13 = 455
3 ⋅2 ⋅1
(15 − 9)!3!
Again, once the solvent molecules are arranged, there is only one way to arrange the solute
molecules, so Wsolute = 1. Hence, Wtotal = Wsolvent·Wsolute = 455·1 = 455. Figure 8.6 and the
accompanying text show that there are 455 arrangements of the three solute molecules in the15
boxes. We get an identical result when we focus on the solvent instead of the solute.
(e) The above results demonstrate that it makes no difference whether we focus on the solvent
or solute. The number of distinguishable arrangements for the system as a whole is the same for
each step as the mixing occurs. This has to be true, since there has to be a unique number of
distinguishable arrangements for a particular case, no matter how it is calculated.
ACS Chemistry FROG
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Entropy and Molecular Organization
Chapter 8
Problem 8.96.
At the molecular level, all energies are quantized, including the kinetic energy of motion of
molecules in a gas. The energy levels for this motion are so close together that the number of
energy levels available is many times more than the number of molecules. Thus, the probability
that more than one molecule will occupy any energy level is essentially zero: there will be either
one or zero molecules in each energy level. As the volume of the gas increases, the energy
levels get even closer together. Consider these energy levels in a 12-unit range of energy for
three different total volumes of gas, where V1 < V2 < V3:
Our gas consists of three identical molecules to arrange among the available energy levels. A
way to approach these problems is to make a table for each case showing the arrangements,
their energies, their contribution to the total energy, the total energy, and the average energy per
molecule. Here is the table for all three cases with arrangements denoted by showing which
energy levels are occupied by the three molecules in each case.
arrangement
48
arrangement
energy
contribution
to total energy
total
energy
average energy
per molecule
V1
(0, 6, 12)
18 units
18 units
18 units
6 units/molecule
V2
(0, 4, 8)
(0, 4, 12)
(0, 8, 12)
(4, 8, 12)
12 units
16 units
20 units
24 units
3 units
4 units
5 units
6 units
18 units
6 units/molecule
V3
(0, 3, 6)
(0, 3, 9)
(0, 3, 12)
(0, 6, 9)
(0, 6, 12)
(0, 9, 12)
(3, 6, 9)
(3, 6, 12)
(3, 9, 12)
(6, 9, 12)
9 units
12 units
15 units
15 units
18 units
21 units
18 units
21 units
24 units
27 units
0.9 units
1.2 units
1.5 units
1.5 units
1.8 units
2.1 units
1.8 units
2.1 units
2.4 units
2.7 units
18 units
6 units/molecule
ACS Chemistry Chapter 8 suggested solutions
Chapter 8
Entropy and Molecular Organization
(a) The results for the molecules in a volume V1 are shown in the first section of the table. We
sum the energies of the molecules, 0 units + 6 units + 12 units = 18 units, to get the energy of
the single possible arrangement. Since this is the only arrangement possible, it is the only
contributor to the total energy, which must, therefore, be 18 units. There are three molecules in
the system, so the average energy per molecule is (18 units)/(3 molecules) = 6 units/molecule.
(b) The four possible arrangements and the energy of each for the intermediate volume, V2, case
are shown in the second section of the table. Since there are four equally probable arrangements
(equally probable by our fundamental assumption that all distinguishable arrangements are
equally probable), we divide the energy of each one by four to get its contribution to the total
energy and then sum the resulting values to get the total energy, 18 units, and the average
energy per molecule, 6 units/molecule.
(c) For the largest volume case, third section of the table, we find 10 arrangements of the three
molecules among the five energy levels. Therefore, we have to divide the energy of each one by
10 to get its contribution to the total energy of the system. Summing these contributions, we
once again find a total energy, 18 units, and average energy per molecule, 6 units/molecule.
(d) It may seem surprising to find that the total energy and average energy per molecule are the
same in all these cases. This is because we have restricted the molecules to energies in the range
from zero to 12 units and have chosen equally spaced energy levels in all cases. The average
molecular energy will always be in the middle of the energy range (6 units) and the three of
them together will, therefore, have a total energy of 18 units. In a gas consisting of a large
number of molecules, on the order of Avogadro’s number, there is also a range of energy levels
accessible to the molecules at a given temperature. These levels are extremely closely spaced
and we can assume that the spacing is essentially equal. The number of levels increases (and
spacing decreases) as the volume of the gas increases, but the range of energy remains the same,
if the temperature of the gas remains constant. Our results suggest, therefore, that the average
energy per molecule and hence the total energy of the system remains constant as the number of
energy levels in the range increases. This is consistent with the energy of an ideal gas not being
a function of the volume of the gas at constant temperature.
(e) Note that the results are the same for the number of distinguishable arrangements of three
identical molecules among five different energy levels in part (c) and of three identical objects
among five labeled boxes in Investigate This 8.4(a). In energy-quantized molecular systems, the
different (distinguishable) translational energy quantum levels take the place of the boxes in our
spatial representation of distinguishable arrangements of molecules and, for these countable
systems give identical results. In real systems, the distinguishable arrangements of molecules
among quantized translational energy levels is what gives rise to what we have called
“positional entropy.” Thus, entropy is always a measure of the spreading out of energy, either
molecules among translational energy levels or energy quanta spread among the internal
motions of the molecules (or vibrations in a crystal).
ACS Chemistry FROG
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