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 3 of 5: Minimum stable depth

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T_{\min} = 272.9\ \text{K}\ \Rightarrow\ p = 2.58\ \text{MPa};\quad h = \dfrac{p - p_{atm}}{\rho(\ce{H2O})\,g} = \dfrac{2.58 - 0.101}{1000 \times 9.8} \approx 250\ \text{m}

Analysis

Hydrate, liquid water and methane coexist lowest at the quadruple-type point where ice can just appear: 272.9 K (the melting point falls slightly with pressure). The dissociation pressure there is 2.58 MPa, equivalent to about 250 m of water column.

Common pitfall. Do not take 273.15 K: pressure lowers the melting point of ice, so liquid water can exist slightly below it while the hydrate stays stable.