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.

o o kJ o kJ ∆Gf for two allotropes: diamond (∆Gf = +2.86 /mole) and graphite (∆Gf = 0.00 /mole).

o kJ The fact that graphite has ∆Gf = 0 /mole indicates that it is the most stable form of C at 298 K. o kJ Also, the ∆Gf = +2.89 /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.

o o ∆Hf and S for the two allotropes of oxygen discussed above (ozone and O2).

o kJ o kJ ozone (∆Hf = +142.2 /mole) and oxygen (∆Hf = 0.00 /mole) o kJ o kJ ozone (S = +0.2376 /Kmole) and oxygen (S = +0.2050 /Kmole)

Consider the following reaction which shows the conversion of the less stable allotrope (ozone) o o o into the more stable allotrope (molecular oxygen). The values of ∆H rxn, ∆S rxn, and ∆G rxn 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

o kJ o kJ S ( /Kmole) + 0.2376 +0.2050 ∆S rxn = +0.1398 /K

o kJ o ∆Hf ( /mole) + 142.2 +0.0 ∆H rxn = -284.4 kJ

o kJ o ∆Gf ( /mole) + 163.4 +0.0 ∆G rxn = -326.8 kJ

As you can see from the table above, the reaction is both exothermic (releases heat) and creates disorder (in the system). o o Both terms, the entropy and enthalpy, (∆S rxn positive and ∆H rxn negative) contribute to a o negative value of ∆G rxn. 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.

o o ∆Hf and S for the two allotropes of carbon discussed above (diamond and graphite).

o kJ o kJ diamond (∆Hf = +1.88 /mole) and graphite (∆Hf = 0.00 /mole) o kJ o kJ diamond (S = +0.0024 /Kmole) and graphite (S = +0.0057 /Kmole)

Consider the following reaction which shows the conversion of the less stable allotrope o o o (diamond) into the more stable allotrope (graphite). The values of ∆H rxn, ∆S rxn, and ∆G rxn 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

o kJ o kJ S ( /Kmole) + 0.0024 +0.0057 ∆S rxn = +0.0033 /K

o kJ o ∆Hf ( /mole) + 1.88 +0.0 ∆H rxn = -1.88 kJ

o kJ o ∆Gf ( /mole) + 2.89 +0.0 ∆G rxn = -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). o o Both terms, the entropy and enthalpy, (∆S rxn positive and ∆H rxn negative) contribute to a o negative value of ∆G rxn. 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 Graphite has a sheet-like structure with 3 bonds to each each carbon atom (sp3 carbon in the sheet (sp2 hybridized). The sheets are held hybridization). 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 1 order in the sheet of C atoms is 1 /3. However, the distance between the sheets is larger and the g 3 attractions weaker. Also, it is worth noting that diamond has a density of 3.51 /cm while g 3 graphite's is only 2.25 /cm (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!

o o kJ Consider the values of ∆Gf for two allotropes of phosphorus: Pred (∆Gf = -11.25 /mole) and o kJ Pwhite (∆Gf = 0.00 /mole).

o kJ The fact that red phosphorus has ∆Gf = -11.25 /mole indicates that it is the more stable form of o kJ phosphorus at 298 K. Also, the higher ∆Gf = 0.0 /mole for white phosphorus clearly shows that the it is a less thermodynamically stable form.

o o Consider the vaules of ∆Hf and S for the two allotropes of carbon discussed above.

o kJ o kJ Pred (∆Hf = -17.6 /mole) and Pwhite (∆Hf = 0.00 /mole) o kJ o kJ Pred (S = +0.0228 /Kmole) and Pwhite (S = +0.0441 /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 o o phosphorus) into the more stable allotrope (red phosphorus). The values of ∆H rxn, ∆S rxn, and o ∆G rxn 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

o kJ o kJ S ( /Kmole) + 0.0441 +0.0228 ∆S rxn = -0.0213 /K

o kJ o ∆Hf ( /mole) + 0.0 -17.6 ∆H rxn = -17.6 kJ

o kJ o ∆Gf ( /mole) + 0.0 -11.25 ∆G rxn = -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 o enthalpy term (∆H rxn 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 o (∆S rxn 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. 5 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).