Question

Question: Snow and ice sublime spontaneously when the partial pressure of water vapor is below the equilibrium vapor pressure of ice. At $0°\mathrm{C}$the vapor pressure of ice is 0.0060 atm (the triple-point pressure of water). Taking the enthalpy of sublimation of ice to be$50.0{\mathrm{kJmol}}^{-1}$, calculate the partial pressure of water vapor below which ice will sublime spontaneously at$-15°\mathrm{C}$.

Short answer

The partial pressure of the water vapour is $0.00167\mathrm{atm}$.

Step-by-step solution

Step-by-step solution

Step 1:

The vapour pressure of ice is 0.0060 atm at $0°\mathrm{C}$.

$$\begin{gathered} 0°\mathrm{C}=0+273\mathrm{K} \\ =273\mathrm{K} \end{gathered}$$

Enthalpy of sublimation of ice is$50.0{\mathrm{kJmol}}^{-1}$

$50.0{\mathrm{kJmol}}^{-1}=50\times {10}^3{\mathrm{Jmol}}^{-1}$

The partial pressure of water vapor below which ice will sublime spontaneously at $-15°\mathrm{C}$.

$$\begin{gathered} -15°\mathrm{C}=-15°+273 \\ =258\mathrm{K} \end{gathered}$$

Step 2:

The reaction of sublimation is as follows,

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)\rightleftarrows {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

By using Ven’t Hoffs equation we can determine the partial pressure of the water vapour as follows,

$$\begin{gathered} \ln \left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)=-\frac{\mathrm{\Delta H}}{\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right] \\ \Rightarrow \ln \left(\frac{{\mathrm{P}}_2}{0.0060}\right)=-\frac{50\times {10}^3}{8.315}\left[\frac{1}{258}-\frac{1}{273}\right] \\ \Rightarrow \ln \left(\frac{{\mathrm{P}}_2}{0.0060}\right)=-1.280 \\ \Rightarrow \frac{{\mathrm{P}}_2}{0.0060}={\mathrm{e}}^{-1.280} \\ \Rightarrow \frac{{\mathrm{P}}_2}{0.0060}=0.277 \\ \Rightarrow {\mathrm{P}}_2=0.00167\mathrm{atm} \end{gathered}$$

Hence, The partial pressure of the water vapour is $0.00167\mathrm{atm}$.