[Big-Daddy]

How should I generally try to express ΔV for calculations using the Clausius-Clapeyron equation?
Example problem:

CH₄·6H₂O decomposes to produce solid water, ice. The enthalpy of this process equals 17.47 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. At what external pressure does decomposition of methane hydrate into methane and ice take place at –5°C?

The reaction equation is CH₄·6H₂O (s)  CH₄ (g) + H₂O (s).

The solution notes that

[tex]\frac{dp}{dT} = \frac{\Delta H}{T \cdot \Delta V}[/tex]

Just as I expected. But now ΔV is given as something much more complicated than what I have seen before:

[tex]\Delta V = \frac{RT}{p} + \frac{6 \cdot M_{H2O}}{\rho_{H2O}} - \frac{M_{CH4.6H2O}}{\rho_{CH4.6H2O}}[/tex]

Why on earth is ΔV like this? And that raises the question, for a general reaction, including solid phases (after all H2O and the methane hydrate are both solid here) as well as gaseous phases, how can I write ΔV, this accurately?

[Corribus]

I'm not sure what the point of confusion is here.  The volume change is just the new volume of the (ideal) gas plus the volume of water ice formed, minus the volume of the starting material. In the latter two cases, the molar volume is expressed as a function of the respective densities.  If you put appropriate values in, I suspect you'll find that the first term far exceeds the second and third.

[Big-Daddy]

Ah I see. So volume is being replaced by V=m/ρ for solids and V=nRT/p for gases. Then

[tex]\Delta V = \frac{n_{CH4} \cdot R \cdot T}{p_{CH4}} + \frac{n_{H2O} \cdot M_{r_{H2O}}}{\rho_{H2O}} - \frac{n_{Hydrate} \cdot M_{r_{Hydrate}}}{\rho_{Hydrate}}[/tex]

What's happening next appears to be that we are only considering volume change per mole of the reaction equation (in which case n_CH4=1, n_Hydrate=1, n_H2O=6 as per the coefficients in the reaction). Is this always the correct procedure, i.e. volume change per mole of reaction used, whenever we use the Clausius-Clapeyron equation?