Download Descriptive Chemistry Assignment 4

Descriptive Chemistry Assignment 5 – Thermodynamics and Allotropes
Read the sections on “Allotropy” and “Allotropes” in your text (pages 464, 475, 871-2, 882-3) and
answer the following:
1.
Define the word allotrope and draw the correct Lewis structures for the two allotropes of oxygen
(use resonance where needed). Based on bond order, which allotrope should be most stable?
Definition: One of two (or more) forms of an element in the same physical state.
molecular oxygen
ozone (resonance forms)
Molecular oxygen (O2) has a double bond (bond order = 2). Ozone (O3) has an average bond
order of 1.5.
Therefore, molecular oxygen should be the more stable form.
2.
Draw structures of the two allotropes of phosphorus (see page 882). Discuss their relative
stability.
The white phosphorus is less stable because the bond angles in the structure ar strained. The
angles are smaller than would normally be expected for an sp3 hybridized P (the expected bond
angle for sp3 is 109.5o, in white phosphorus the angles are only 60o).
In red phosphorus, some of the angle strain is relieved since the bond on one side of each
tetrahedra is used to join tetrahedra together (this allows one face of the terahedra to spread out).
3.
Look up values for ∆Gfº for these two allotropes of carbon: diamond and graphite. Based on
these values explain which is thermodynamically more stable.
∆Gfo for two allotropes: diamond (∆Gfo = +2.86 kJ/mole) and graphite (∆Gfo = 0.00 kJ/mole).
The fact that graphite has ∆Gfo = 0 kJ/mole indicates that it is the most stable form of C at 298 K.
Also, the ∆Gfo = +2.89 kJ/mole for diamond clearly shows that the diamond is a less
thermodynamically stable form.
2
4.
(a) Look up values for ∆Hfº and ∆Sº for the two allotropes of oxygen discussed above. Explain
whether the difference in the relative stability of these allotropes is due to entropy or enthalpy.
∆Hfo and So for the two allotropes of oxygen discussed above (ozone and O2).
ozone (∆Hfo = +142.2 kJ/mole) and oxygen (∆Hfo = 0.00 kJ/mole)
ozone (So = +0.2376 kJ/Kmole) and oxygen (So = +0.2050 kJ/Kmole)
Consider the following reaction which shows the conversion of the less stable allotrope (ozone)
into the more stable allotrope (molecular oxygen). The values of ∆Horxn, ∆Sorxn, and ∆Gorxn are
calculated by using the sum of products minus the sum of reactants process that should be very
familiar to you by now.
2 O3 (g)
<===>
3 O2 (g)
changes
So (kJ/Kmole)
+ 0.2376
+0.2050
∆Sorxn = +0.1398 kJ/K
∆Hfo (kJ/mole)
+ 142.2
+0.0
∆Horxn = -284.4 kJ
∆Gfo (kJ/mole)
+ 163.4
+0.0
∆Gorxn = -326.8 kJ
As you can see from the table above, the reaction is both exothermic (releases heat) and creates
disorder (in the system).
Both terms, the entropy and enthalpy, (∆Sorxn positive and ∆Horxn negative) contribute to a
negative value of ∆Gorxn. Therefore, this process would be spontaneous at standard conditions at
ALL temperatures.
(b) Look up values for ∆Hfº and ∆Sº for the two allotropes of carbon discussed above. Explain
whether the difference in the relative stability of these allotropes is due to entropy or enthalpy.
∆Hfo and So for the two allotropes of carbon discussed above (diamond and graphite).
diamond (∆Hfo = +1.88 kJ/mole) and graphite (∆Hfo = 0.00 kJ/mole)
diamond (So = +0.0024 kJ/Kmole) and graphite (So = +0.0057 kJ/Kmole)
Consider the following reaction which shows the conversion of the less stable allotrope
(diamond) into the more stable allotrope (graphite). The values of ∆Horxn, ∆Sorxn, and ∆Gorxn are
calculated by using the sum of products minus the sum of reactants process that should be very
familiar to you by now.
C (dia)
<===>
C (graph)
changes
So (kJ/Kmole)
+ 0.0024
+0.0057
∆Sorxn = +0.0033 kJ/K
∆Hfo (kJ/mole)
+ 1.88
+0.0
∆Horxn = -1.88 kJ
∆Gfo (kJ/mole)
+ 2.89
+0.0
∆Gorxn = -2.89 kJ
3
As you can see from the table above, the reaction is both exothermic (releases heat) and creates
disorder (in the system).
Both terms, the entropy and enthalpy, (∆Sorxn positive and ∆Horxn negative) contribute to a
negative value of ∆Gorxn. Therefore, this process would be spontaneous at standard conditions at
ALL temperatures.
However, that does not mean that the reaction will occur at an observable rate – in fact,
“diamonds are forever” because the conversion from diamond to graphite is kinetically hindered
by a very large activation energy. This is due to the structural differences (both in type of
bonding and atomic positions in the two solid state forms of this element).
Consider the figures below that show the structure of diamond and graphite.
Diamond has 4 single bonds to
each carbon atom (sp3
hybridization).
Graphite has a sheet-like structure with 3 bonds to each
carbon in the sheet (sp2 hybridized). The sheets are held
together by interaction of the p orbitals in each sheet.
At first glance, you might think that the diamond is more stable. In it each C atom is bonded to
four other C atoms and all the bonds are covalent. However, in the graphite, the average bond
order in the sheet of C atoms is 1 1/3. However, the distance between the sheets is larger and the
attractions weaker. Also, it is worth noting that diamond has a density of 3.51 g/cm3 while
graphite's is only 2.25 g/cm3 (i.e. the diamond is highly order and tightly packed).
Diamond is formed in nature under conditions of high temperature and high pressure – the high
temperature helps to overcome the kinetic barrier to the reaction and the high pressure favors the
formation of the higher density phase (the diamond). Recall the standard changes in Gibbs Free
energy apply only at standard conditions (1 atm) – the pressures under which diamond are
formed are far from standard!
5.
Apply what you learned in the previous question. For the two allotropes of phosphorus, guess
whether the difference in the relative stability of the allotropes is due to entropy or enthalpy.
Find the necessary data to verify your hypothesis via thermodynamic calculations.
4
The energetics should favor the forms of phosphorus with the least bound strain (the red
phosphorus) but the entropy should favor the smaller molecules (the molecules that have fewer
atoms – the white phosphorus). As you can see the answer is not quite as easy as the two cases
described above. Let’s see if our “intuitive answer” is supported by the numbers!
Consider the values of ∆Gfo for two allotropes of phosphorus: Pred (∆Gfo = -11.25 kJ/mole) and
Pwhite (∆Gfo = 0.00 kJ/mole).
The fact that red phosphorus has ∆Gfo = -11.25 kJ/mole indicates that it is the more stable form of
phosphorus at 298 K. Also, the higher ∆Gfo = 0.0 kJ/mole for white phosphorus clearly shows that
the it is a less thermodynamically stable form.
Consider the vaules of ∆Hfo and So for the two allotropes of carbon discussed above.
Pred (∆Hfo = -17.6 kJ/mole) and Pwhite (∆Hfo = 0.00 kJ/mole)
Pred (So = +0.0228 kJ/Kmole) and Pwhite (So = +0.0441 kJ/Kmole)
We can use these numbers to explain if the difference in the relative stability of the allotropes are
due to entropy or enthalpy.
Consider the following reaction which shows the conversion of the less stable allotrope (white
phosphorus) into the more stable allotrope (red phosphorus). The values of ∆Horxn, ∆Sorxn, and
∆Gorxn are calculated by using the sum of products minus the sum of reactants process that
should be very familiar to you by now.
P (white)
<===>
P (red)
changes
So (kJ/Kmole)
+ 0.0441
+0.0228
∆Sorxn = -0.0213 kJ/K
∆Hfo (kJ/mole)
+ 0.0
-17.6
∆Horxn = -17.6 kJ
∆Gfo (kJ/mole)
+ 0.0
-11.25
∆Gorxn = -11.25 kJ
As you can see from the table above, the reaction is exothermic (releases heat) but the entropy
change is negative (in the system). [THIS MATCHES OUR PREDICTION]
Therefore, the reaction is spontaneous at low temperatures due to the energetics – an exothermic
enthalpy term (∆Horxn negative). But to make the forward reaction proceed more quickly, heat
must be applied.
Doing this causes the above reaction becomes nonspontaneous in the forward direction at T >
826 K due to the entropy term (i.e. the reverse reaction is favored at high temperatures due to the
(∆Sorxn negative). [THIS MATCHES OUR PREDICTION]
As it turns out, the white phospohus is very reactive due to its strained bonds. At room
temperature it reacts spontaneously with oxygen in the air to form phosphorus oxides. To protect
it, it is stored under water. The P4 molecules are held together by weak van der Waals forces and
therefore have a low melting point (about 44o) and boiling point (about 280oC). White
phosphorus is also highly toxic.
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Red phosphorus is not toxic and is used in the manufacture of of the special striking surface used
for safety matches (not the "strike anywhere" matches). Due to the much larger molecular size,
red phosphorus has a higher melting point (590oC at 43 atm). Due to the lower angle strain, red
phosphorus is air stable (does not react with oxygen at room temperature, but does at about
250oC and above). Red phosphorus can be made from white phosphorus by heating the white
phosphorus in the absence of air at about 300oC (this temperature is too low to favor the white
phosphorus form thermodynamically but is hot enough to overcome the activation energy for
conversion of white to red phosphorus).