Problem 1

The “clathrate gun” scenario: ocean-floor methane hydrates could decompose explosively as oceans warm. Upon decomposition of 1.00 g of a methane hydrate of fixed composition at 25 °C and 101.3 kPa, 205 cm

^3

of methane is released. (a) Determine

n

\ce{CH4 \cdot n H2O}

. (b) Real hydrate is close to

\ce{CH4 \cdot 6 H2O}

, stable at 1 atm down to its decomposition at

-81

°C; write its decomposition to methane and ice,

\Delta H = +17.47

kJ mol

^{-1}

. Assuming

\Delta H

is T- and p-independent, that the volume change equals the released methane volume, and that methane is ideal, find the external pressure at which decomposition occurs at

-5

°C. (c) What is the minimum possible depth of pure liquid water at which methane hydrates are stable? First choose the minimum temperature at which the hydrate can coexist with liquid water: 272.9 K, 273.15 K, or 273.4 K. (d) In Lake Baikal, hydrate samples raised from 1400 m began to decompose at 372 m depth (

\Delta_{fus}H

(ice) = 6.01 kJ mol

^{-1}

); find the water temperature at 372 m. (e) Total methane in hydrates is

\geq 5 \times 10^{11}

t; by how many degrees would Earth's atmosphere heat if it all burned (

\Delta_c H(\ce{CH4}) = -889

kJ mol

^{-1}

, atmospheric heat capacity

4 \times 10^{21}

J K

^{-1}

Step 2 of 5: Dissociation pressure at −5 °C

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Intuition

Higher pressure stabilises the clathrate cage, so warming the hydrate demands a higher equilibrium methane pressure.

\ce{CH4 \cdot 6H2O(s) -> CH4(g) + 6H2O(s)}\ \ \Delta H = +17.47\ \text{kJ mol}^{-1};\quad \ln\dfrac{p}{p_0} = \dfrac{\Delta H}{R}\left(\dfrac{1}{T_0} - \dfrac{1}{T}\right) \Rightarrow p(-5^\circ\mathrm{C}) = 22\ \text{atm} \approx 2.2\ \text{MPa}

Analysis

With $\Delta V \approx V_{\ce{CH4}} = RT/p$ , Clausius–Clapeyron reduces to $d\ln p/dT = \Delta H/(RT^2)$ . Integrating between $T_0 = 192.15$ K, $p_0 = 1$ atm and $T = 268.15$ K gives $p = 22$ atm.