IChO 2008 · Theory problem 6

Clathrate hydrates of chlorine and methane

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A feathery, greenish solid precipitate can be observed if chlorine gas is bubbled into water close to its freezing point. Similar precipitates form with other gases such as methane and noble gases. These materials are interesting because vast quantities of the so-called methane-hydrates are supposed to exist in nature (comparable in quantity with other natural gas deposits).

These precipitates all have related structures. The molecules of water just above its freezing point form a hydrogen-bonded structure. The gas molecules stabilize this framework by filling in the rather large cavities in the water structure forming clathrates. The crystals of chlorine and methane clathrates have the same structure. Their main characteristics are dodecahedra formed from 20 water molecules. The unit cell of the crystal can be thought as a body-centered cubic arrangement built from these dodecahedra which are almost spherical objects. The dodecahedra are connected via additional water molecules located on the faces of the unit cell. Two water molecules can be found on each face of the unit cell. The unit cell has an edge dimension of 1.182 nm.

There are two types of cavities in this structure. One is the internal space in the dodecahedra (A). These are somewhat smaller than the other type of voids (B), of which there are 6 for each unit cell.

6.1 How many type A cavities can be found in a unit cell?

6.2 How many water molecules are there in a unit cell?

6.3 If all cavities contain a guest molecule, what is the ratio of the number of water to the number of guest molecules?

6.4 Methane hydrate is formed with the structure in c) at temperatures between 0-10 °C. What is the density of the clathrate?

6.5 The density of chlorine hydrate is \(1.26\ \mathrm{g/cm^3}\). What is the ratio of the number of water and guest molecules in the crystal?

Which cavities are likely to be filled in a perfect chlorine hydrate crystal? Mark one or more.

☐ Some A  ☐ Some B  ☐ All A  ☐ All B

Covalent radii reflect atomic distances when the atoms are covalently bonded. Non-bonded or van der Waals radii give a measure of the atomic size when they are not bonded covalently (modelled as hard spheres).

Atom Covalent radius (pm) Nonbonded radius (pm)
H 37 120
C 77 185
O 73 140
Cl 99 180

6.6 Based on the covalent and non-bonded radii of these atoms estimate lower and upper bounds for the average radii of the cavities where possible. Show your reasoning.

Let us consider the following processes

Figure 1

6.7 What are the signs of the following molar quantities referring to these reactions in the given direction at 4 °C? Mark with a –, 0 or +.

sign
\(\Delta G_{\mathrm m}(1)\)
\(\Delta G_{\mathrm m}(2)\)
\(\Delta H_{\mathrm m}(1)\)
\(\Delta H_{\mathrm m}(2)\)
\(\Delta S_{\mathrm m}(1)\)
\(\Delta S_{\mathrm m}(2)\)
\(\Delta S_{\mathrm m}(2)-\Delta S_{\mathrm m}(1)\)
\(\Delta H_{\mathrm m}(2)-\Delta H_{\mathrm m}(1)\)

Official English text of the problem.

Answers

  • 6.1: \(2\) type A cavities per unit cell.
  • 6.2: \(46\) water molecules per unit cell.
  • 6.3: Water : guest \(=46:8=23:4=5.75:1\).
  • 6.4: \(\boxed{\rho=0.962\ \mathrm{g\,cm^{-3}}}\).
  • 6.5: Water : chlorine \(=46:6=23:3\approx7.67:1\). Select All B only; the A cavities are empty in the intended ideal structure.
  • 6.6: In the spherical fitting approximation, the free-cavity radii satisfy \[ \boxed{234\ \mathrm{pm}\le r_A<279\ \mathrm{pm}},\qquad \boxed{r_B\ge279\ \mathrm{pm}}. \] Guest sizes provide no upper bound for \(r_B\).
  • 6.7: In the order listed in the question: \[ \boxed{+,\;-,\;-,\;-,\;-,\;-,\;-,\;-}. \] These signs assume the hydrate-forming conditions described in the problem.

Solution

The molecular species are water, \(\ce{H2O}\), SMILES O; methane, \(\ce{CH4}\), SMILES C; and chlorine, \(\ce{Cl2}\), SMILES ClCl. Methane is tetrahedral, and chlorine is a linear diatomic molecule. Ice consists of water molecules. SMILES does not specify these substances’ physical phases or the extended hydrogen-bonded cage structure of a clathrate.

6.1

A body-centered cubic arrangement has eight corner positions, each shared by eight unit cells, and one position wholly inside the cell. Each dodecahedron contains one A cavity. Therefore,

\[ N_A=8\left(\frac18\right)+1=\boxed{2}. \]

Here \(N_A\) denotes the number of A cavities, not Avogadro’s constant.

6.2

The two dodecahedra contribute

\[ 2(20)=40 \]

water molecules per unit cell.

There are also two additional water molecules on each of the six faces. Each face is shared by two neighboring cells, so these contribute

\[ 6(2)\left(\frac12\right)=6. \]

Consequently,

\[ N_{\mathrm{water}}=40+6=\boxed{46}. \]

6.3

There are two A cavities and six B cavities:

\[ N_{\mathrm{guest}}=2+6=8. \]

With one guest molecule in every cavity,

\[ \boxed{N_{\mathrm{water}}:N_{\mathrm{guest}} =46:8=23:4=5.75:1}. \]

6.4

For full methane occupancy, one unit cell contains 46 water molecules and eight methane molecules.

Use the standard constants

\[ N_{\mathrm{Av}}=6.02214\times10^{23}\ \mathrm{mol^{-1}}, \] \[ M(\ce{H2O})=18.015\ \mathrm{g\,mol^{-1}}, \qquad M(\ce{CH4})=16.043\ \mathrm{g\,mol^{-1}}. \]

The edge length and unit-cell volume are

\[ a=1.182\ \mathrm{nm}=1.182\times10^{-7}\ \mathrm{cm}, \]

\[ V_{\mathrm{cell}}=a^3 =1.651400568\times10^{-21}\ \mathrm{cm^3}. \]

The mass per cell is

\[ m_{\mathrm{cell}} =\frac{46M(\ce{H2O})+8M(\ce{CH4})}{N_{\mathrm{Av}}} =\frac{957.034\ \mathrm{g\,mol^{-1}}} {6.02214\times10^{23}\ \mathrm{mol^{-1}}} =1.5892\times10^{-21}\ \mathrm g. \]

Thus,

\[ \rho=\frac{m_{\mathrm{cell}}}{V_{\mathrm{cell}}} =0.96233\ \mathrm{g\,cm^{-3}} \approx\boxed{0.962\ \mathrm{g\,cm^{-3}}}. \]

6.5

Let \(n\) be the number of chlorine molecules per unit cell. The water framework still contains 46 water molecules. With

\[ M(\ce{Cl2})=70.90\ \mathrm{g\,mol^{-1}}, \]

the density satisfies

\[ \rho=\frac{46M(\ce{H2O})+nM(\ce{Cl2})} {N_{\mathrm{Av}}a^3}. \]

Therefore,

\[ n= \frac{\rho N_{\mathrm{Av}}a^3-46M(\ce{H2O})} {M(\ce{Cl2})} =5.986\approx6. \]

The small departure from six is consistent with the stated density’s precision: exactly six chlorine molecules would give \(1.261\ \mathrm{g\,cm^{-3}}\), which rounds to \(1.26\ \mathrm{g\,cm^{-3}}\).

The ideal composition therefore has

\[ \boxed{N_{\mathrm{water}}:N_{\ce{Cl2}} =46:6=23:3\approx7.67:1}. \]

There are exactly six larger B cavities. The intended size-selective interpretation is that chlorine occupies these larger cavities, leaving the smaller A cavities empty. Thus the answer-sheet selection is:

☐ Some A  ☐ Some B  ☐ All A  ☑ All B

The density determines the guest count; identifying the occupied cavities additionally uses this size-based interpretation.

6.6

Approximate a cavity by a sphere capable of enclosing its guest molecule.

For methane, the carbon–hydrogen bond length is estimated from the covalent radii:

\[ d_{\ce{C-H}}=77+37=114\ \mathrm{pm}. \]

A hydrogen atom’s outer nonbonded surface extends a further \(120\ \mathrm{pm}\). Hence a sphere centered on the carbon atom and enclosing methane has radius

\[ r_{\mathrm{methane}} =\max\left(185,\;114+120\right)\ \mathrm{pm} =\boxed{234\ \mathrm{pm}}. \]

For chlorine, the estimated bond length is

\[ d_{\ce{Cl-Cl}}=2(99)=198\ \mathrm{pm}. \]

Measured from the bond midpoint, the outermost surface is one half-bond length plus one chlorine nonbonded radius away:

\[ r_{\mathrm{chlorine}} =\frac{198}{2}+180 =\boxed{279\ \mathrm{pm}}. \]

Methane fits into both cavity types. Under the size-exclusion interpretation of 6.5, chlorine fits into B but is too large for A. Therefore the effective free-space radii satisfy

\[ \boxed{234\ \mathrm{pm}\le r_A<279\ \mathrm{pm}}, \]

\[ \boxed{r_B\ge279\ \mathrm{pm}}. \]

No larger guest that fails to enter B is specified, so this comparison supplies no upper bound for \(r_B\). These are approximate spherical fitting bounds, since the molecules and cages are not perfect spheres.

6.7

Process (1) is freezing:

\[ \ce{H2O(l) -> H2O(s)}. \tag{1} \]

Process (2), written with integer coefficients, is methane clathrate formation:

\[ \ce{8 CH4(g) + 46 H2O(l) -> (CH4)8 * 46H2O(s)}. \tag{2, scaled} \]

The solid product denotes the clathrate’s composition. Dividing this equation by 46 gives the normalization used in the question, with

\[ x=\frac{8}{46}=\frac4{23}. \]

Thus both processes are compared per mole of water.

At \(4\,^\circ\mathrm C\), liquid water is stable relative to ordinary ice, so freezing is unfavorable:

\[ \Delta G_{\mathrm m}(1)>0. \]

Freezing releases heat and produces a more ordered phase:

\[ \Delta H_{\mathrm m}(1)<0, \qquad \Delta S_{\mathrm m}(1)<0. \]

Under the hydrate-forming conditions assumed in the question, process (2) is favorable:

\[ \Delta G_{\mathrm m}(2)<0. \]

It orders the water framework and confines methane that was initially gaseous. The intended qualitative entropy comparison is therefore

\[ \Delta S_{\mathrm m}(2)<\Delta S_{\mathrm m}(1)<0. \]

Using

\[ \Delta H_{\mathrm m}=\Delta G_{\mathrm m}+T\Delta S_{\mathrm m}, \]

both terms on the right are negative for process (2), so

\[ \Delta H_{\mathrm m}(2)<0. \]

Furthermore,

\[ \begin{aligned} \Delta H_{\mathrm m}(2)-\Delta H_{\mathrm m}(1) ={}&\bigl[\Delta G_{\mathrm m}(2)-\Delta G_{\mathrm m}(1)\bigr]\\ &+T\bigl[\Delta S_{\mathrm m}(2)-\Delta S_{\mathrm m}(1)\bigr]<0, \end{aligned} \]

because both bracketed differences are negative.

The completed table is:

Quantity Sign
\(\Delta G_{\mathrm m}(1)\) \(+\)
\(\Delta G_{\mathrm m}(2)\) \(-\)
\(\Delta H_{\mathrm m}(1)\) \(-\)
\(\Delta H_{\mathrm m}(2)\) \(-\)
\(\Delta S_{\mathrm m}(1)\) \(-\)
\(\Delta S_{\mathrm m}(2)\) \(-\)
\(\Delta S_{\mathrm m}(2)-\Delta S_{\mathrm m}(1)\) \(-\)
\(\Delta H_{\mathrm m}(2)-\Delta H_{\mathrm m}(1)\) \(-\)

The solution was checked against the official solution.