IChO 2013 · Theory problem 1

Clathrate gun

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The only gun that is able to kill all living people in one shot

On the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. It is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.

Figure 1

Upon decomposition of \(1.00\ \mathrm{g}\) of a methane hydrate with a fixed composition at \(25\ ^\circ\mathrm{C}\) and atmospheric (\(101.3\ \mathrm{kPa}\)) pressure, \(205\ \mathrm{cm^3}\) of methane is released.

1.1 Determine \(n\) (not necessarily integer) in the formula of methane hydrate, \(\ce{CH4}\cdot n\,\ce{H2O}\).

Real methane hydrate has a non-stoichiometric composition close to \(\ce{CH4 * 6H2O}\). At atmospheric pressure, methane hydrate decomposes at \(-81\ ^\circ\mathrm{C}\). However, under high pressures (e.g. on the ocean floor) it is stable at much higher temperatures.

Decomposition of methane hydrate produces gaseous methane and solid or liquid water depending on temperature.

1.2 Write down the equation of decomposition of \(1\ \mathrm{mol}\) of \(\ce{CH4 * 6H2O}\) producing solid water (ice) \(\ce{H2O(s)}\).

The enthalpy of this process equals \(17.47\ \mathrm{kJ\,mol^{-1}}\). Assume that the enthalpies do not depend on temperature and pressure, the volume change upon decomposition of hydrate is equal to the volume of released methane, and methane is an ideal gas.

1.3 At what external pressure does decomposition of methane hydrate into methane and ice take place at \(-5\ ^\circ\mathrm{C}\)?

1.4 What is the minimum possible depth of pure liquid water at which methane hydrates can be stable?

To answer this question, you should first deduce at which minimum temperature methane hydrate can coexist with liquid water. Choose the correct answer.

\[ \square\ 272.9\ \mathrm{K} \qquad \square\ 273.15\ \mathrm{K} \qquad \square\ 273.4\ \mathrm{K} \]

Large methane hydrate stocks on the floor of Baikal lake, the largest freshwater lake in Russia and in the world, have been discovered in July 2009 by the crew of a deep-submergence vehicle «Mir-2». During the ascent from the depth of \(1400\ \mathrm{m}\) methane hydrate samples started to decompose at the depth of \(372\ \mathrm{m}\).

1.5 Determine the temperature in Baikal lake at the depth of \(372\ \mathrm{m}\). The enthalpy of fusion of ice is equal to \(6.01\ \mathrm{kJ\,mol^{-1}}\).

Total amount of methane in hydrates on the Earth is no less than \(5\cdot10^{11}\) tons.

1.6 By how many degrees would the Earth atmosphere heat up, if such amount of methane is burned by reacting with atmospheric oxygen?

The enthalpy of combustion of methane is \(-889\ \mathrm{kJ\,mol^{-1}}\), the total heat capacity of the Earth’s atmosphere is about \(4\cdot10^{21}\ \mathrm{J\,K^{-1}}\).

Official English text of the problem.

Answers

  • 1.1: \(\boxed{n\approx5.74}\).
  • 1.2: \(\boxed{\ce{CH4 * 6H2O(s) -> CH4(g) + 6H2O(s)}}\).
  • 1.3: \(\boxed{P\approx2.25\ \mathrm{MPa}}\), absolute pressure.
  • 1.4: Tick \(272.9\ \mathrm{K}\). The minimum depth is \(\boxed{h_{\min}\approx252\ \mathrm{m}}\), or approximately \(250\ \mathrm{m}\).
  • 1.5: \(\boxed{T\approx277.3\ \mathrm{K}\approx4.2\ ^\circ\mathrm{C}}\).
  • 1.6: \(\boxed{\Delta T\approx7\ \mathrm{K}}\), equivalent to a temperature rise of approximately \(7\ ^\circ\mathrm{C}\).

Solution

The molecular species used below are methane C, water O, dioxygen O=O, and carbon dioxide O=C=O. Methane hydrate is an extended clathrate lattice rather than a discrete molecule. The disconnected SMILES C.O.O.O.O.O.O records its approximate \(\ce{CH4 * 6H2O}\) composition, but does not describe the water cages or hydrogen bonding. SMILES also does not distinguish liquid water from ice.

Use the standard constants and approximations \[ R=8.31446\ \mathrm{J\,mol^{-1}\,K^{-1}},\qquad \rho_{\mathrm{water}}=1.00\times10^3\ \mathrm{kg\,m^{-3}}, \qquad g=9.81\ \mathrm{m\,s^{-2}}, \] with molar masses \(M(\ce{CH4})=16.04\ \mathrm{g\,mol^{-1}}\) and \(M(\ce{H2O})=18.015\ \mathrm{g\,mol^{-1}}\). All pressures below are absolute.

1.1

The amount of methane released is obtained from the ideal-gas equation: \[ \nu_{\ce{CH4}} =\frac{PV}{RT} =\frac{(101.3\times10^3\ \mathrm{Pa})(205\times10^{-6}\ \mathrm{m^3})} {(8.31446\ \mathrm{J\,mol^{-1}\,K^{-1}})(298.15\ \mathrm{K})} =8.3771\times10^{-3}\ \mathrm{mol}. \]

Each formula unit of hydrate contains one methane molecule. Hence its mass satisfies \[ 1.00\ \mathrm{g} =\nu_{\ce{CH4}} \left[M(\ce{CH4})+nM(\ce{H2O})\right]. \]

Therefore, \[ n= \frac{\dfrac{1.00}{8.3771\times10^{-3}}-16.04}{18.015} =5.7359, \] so \[ \boxed{n=5.74}. \]

1.2

One mole of hydrate releases one mole of methane and six moles of water: \[ \boxed{\ce{CH4 * 6H2O(s) -> CH4(g) + 6H2O(s)}}. \]

For this decomposition into methane and ice, \[ \Delta_{\mathrm r}H_{\mathrm{ice}} =+17.47\ \mathrm{kJ\,mol^{-1}}. \]

1.3

Along the decomposition-equilibrium curve, the Clapeyron equation gives \[ \frac{dP}{dT} =\frac{\Delta_{\mathrm r}H}{T\Delta_{\mathrm r}V}. \]

For one mole of hydrate, one mole of methane is released. Under the stated volume approximation, \[ \Delta_{\mathrm r}V=\frac{RT}{P}. \] Thus, \[ \frac{dP}{P} =\frac{\Delta_{\mathrm r}H}{R}\frac{dT}{T^2}. \]

Integrating with constant enthalpy, \[ \boxed{ \ln\frac{P}{P_0} =\frac{\Delta_{\mathrm r}H}{R} \left(\frac1{T_0}-\frac1T\right) }. \]

The known equilibrium point is \[ T_0=273.15-81=192.15\ \mathrm{K}, \qquad P_0=101.3\ \mathrm{kPa}. \]

At \(-5\ ^\circ\mathrm{C}\), \(T=268.15\ \mathrm{K}\), so \[ P=(101.3\ \mathrm{kPa}) \exp\!\left[ \frac{17470}{8.31446} \left(\frac1{192.15}-\frac1{268.15}\right) \right]. \]

Therefore, \[ \boxed{P=2.247\times10^3\ \mathrm{kPa}\approx2.25\ \mathrm{MPa}}. \]

A pressure above this equilibrium value favors the hydrate, which occupies much less volume than its gaseous decomposition product.

1.4

The minimum depth occurs where the hydrate-decomposition boundary meets the ice–liquid-water melting boundary. At lower pressures, the temperatures required to stabilize hydrate would leave the water frozen.

Ice occupies more volume than the same amount of liquid water. Consequently, \[ \Delta V_{\mathrm{fus}}=V_{\mathrm{liquid}}-V_{\mathrm{ice}}<0, \] and the melting-curve slope is \[ \frac{dT_{\mathrm m}}{dP} =\frac{T_{\mathrm m}\Delta V_{\mathrm{fus}}} {\Delta H_{\mathrm{fus}}}<0. \]

Pressure therefore lowers the melting temperature below \(273.15\ \mathrm{K}\). The appropriate choice is \[ \boxed{T_{\mathrm m}=272.9\ \mathrm{K}}. \]

At this boundary, the equilibrium pressure can be calculated using decomposition into ice: \[ P_{\mathrm m} =(101.3\ \mathrm{kPa}) \exp\!\left[ \frac{17470}{8.31446} \left(\frac1{192.15}-\frac1{272.9}\right) \right] =2.5753\times10^6\ \mathrm{Pa}. \]

Hydrostatic pressure includes the atmospheric pressure at the surface: \[ P_{\mathrm m}=P_{\mathrm{atm}}+\rho gh_{\min}. \]

Hence, \[ h_{\min} =\frac{2.5753\times10^6-1.013\times10^5} {(1.00\times10^3)(9.81)} \ \mathrm{m} =252.2\ \mathrm{m}. \]

Thus, \[ \boxed{h_{\min}\approx252\ \mathrm{m}\approx250\ \mathrm{m}}. \]

1.5

At \(372\ \mathrm{m}\), the external pressure is \[ P_{372}=P_{\mathrm{atm}}+\rho gh =101300+(1000)(9.81)(372) =3.75062\times10^6\ \mathrm{Pa}. \]

Here the decomposition products contain liquid water: \[ \ce{CH4 * 6H2O(s) -> CH4(g) + 6H2O(l)}. \]

To obtain its enthalpy, add the melting of six moles of ice to the reaction in 1.2: \[ \ce{6H2O(s) -> 6H2O(l)}. \]

By Hess’s law, per mole of hydrate, \[ \Delta_{\mathrm r}H_{\mathrm{liquid}} =17.47+6(6.01) =\boxed{53.53\ \mathrm{kJ\,mol^{-1}}}. \]

The ice and liquid-water decomposition branches meet at the boundary used in 1.4: \[ T_{\mathrm m}=272.9\ \mathrm{K}, \qquad P_{\mathrm m}=2.5753\times10^6\ \mathrm{Pa}. \]

Integrating along the liquid-water branch, \[ \ln\frac{P_{372}}{P_{\mathrm m}} =\frac{\Delta_{\mathrm r}H_{\mathrm{liquid}}}{R} \left(\frac1{T_{\mathrm m}}-\frac1T\right). \]

Solving for \(T\), \[ T= \left[ \frac1{272.9} -\frac{8.31446}{53530} \ln\!\left(\frac{3.75062\times10^6}{2.5753\times10^6}\right) \right]^{-1} \approx277.32\ \mathrm{K}. \]

Therefore, \[ \boxed{t=T-273.15\approx4.17\ ^\circ\mathrm{C}\approx4.2\ ^\circ\mathrm{C}}. \]

1.6

The balanced combustion equation, with liquid water as product, is \[ \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}. \]

The stated methane mass is \[ m=5\times10^{11}\ \mathrm{tons} =5\times10^{17}\ \mathrm{g}. \]

Its amount is \[ \nu_{\ce{CH4}} =\frac{5\times10^{17}\ \mathrm{g}} {16.04\ \mathrm{g\,mol^{-1}}} =3.1172\times10^{16}\ \mathrm{mol}. \]

The heat released is \[ Q=\nu_{\ce{CH4}}\left|\Delta_{\mathrm c}H\right| =(3.1172\times10^{16})(889\times10^3) =2.7712\times10^{22}\ \mathrm{J}. \]

Assigning this heat to the atmosphere gives \[ \Delta T =\frac{Q}{C_{\mathrm{atm}}} =\frac{2.7712\times10^{22}\ \mathrm{J}} {4\times10^{21}\ \mathrm{J\,K^{-1}}} =6.93\ \mathrm{K}. \]

Given the approximate inventory and heat capacity, \[ \boxed{\Delta T\approx7\ \mathrm{K}}, \] equivalent to a rise of approximately \(7\ ^\circ\mathrm{C}\) for the quoted methane amount.

The solution was checked against the official solution.