Chemistry Labs

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 cm3^3 of methane is released. (a) Determine nn in CHX4 ⋅n HX2O\ce{CH4 \cdot n H2O}. (b) Real hydrate is close to CHX4 ⋅6 HX2O\ce{CH4 \cdot 6 H2O}, stable at 1 atm down to its decomposition at −81-81 °C; write its decomposition to methane and ice, ΔH=+17.47\Delta H = +17.47 kJ mol−1^{-1}. Assuming ΔH\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 −5-5 °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 (ΔfusH\Delta_{fus}H(ice) = 6.01 kJ mol−1^{-1}); find the water temperature at 372 m. (e) Total methane in hydrates is ≥5×1011\geq 5 \times 10^{11} t; by how many degrees would Earth's atmosphere heat if it all burned (ΔcH(CHX4)=−889\Delta_c H(\ce{CH4}) = -889 kJ mol−1^{-1}, atmospheric heat capacity 4×10214 \times 10^{21} J K−1^{-1})?
Step 1 of 5: Hydrate stoichiometry
← PreviousNext →
n(CHX4)=pVRT=101300×205×10−68.314×298.15=8.38×10−3 mol;n=(1.00−8.38×10−3×16)/188.38×10−3≈5.75n(\ce{CH4}) = \dfrac{pV}{RT} = \dfrac{101300 \times 205 \times 10^{-6}}{8.314 \times 298.15} = 8.38 \times 10^{-3}\ \text{mol};\quad n = \dfrac{(1.00 - 8.38 \times 10^{-3} \times 16)/18}{8.38 \times 10^{-3}} \approx 5.75
Analysis

The ideal-gas law gives 8.38×10−38.38 \times 10^{-3} mol CHX4\ce{CH4} per gram; the remaining mass is water, 4.81×10−24.81 \times 10^{-2} mol, so CHX4 ⋅5.75 HX2O\ce{CH4 \cdot 5.75 H2O} (close to the structural ideal CHX4 ⋅6 HX2O\ce{CH4 \cdot 6 H2O}).