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 4 of 5: Baikal depth 372 m

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\Delta H_{liq} = 17.47 + 6 \times 6.01 = 53.53\ \text{kJ mol}^{-1};\quad \dfrac{1}{T} = \dfrac{1}{T_0} + \dfrac{R}{\Delta H}\ln\dfrac{p_0}{p} \Rightarrow T(372\ \text{m}) = 277.3\ \text{K} \approx 4^\circ\mathrm{C}

Analysis

For decomposition into liquid water, Hess adds the fusion of 6 mol ice: $\Delta H = 53.53$ kJ mol $^{-1}$ . From the coexistence point $T_0 = 272.9$ K, $p_0 = 2.58$ MPa and $p(372\,\text{m}) = 1000 \times 9.8 \times 372 + 101000 = 3.75$ MPa, the integrated equation gives $T = 277.3$ K.