# Determining pressure of sublimation of iodine

The pressure of sublimation of iodine can theoretically be determined by considering the two hypothetical steps: the sublimation of iodine at $\pu{298 K}$, followed by an isothermal expansion of our system until it reaches the sublimation pressure.

a) Determine the enthalpy and entropy change for this process.
b) Determine the pressure of sublimation.

Given Data

• Triple point: $\mathrm{12.07~kPa, 113.0~^\circ C}$
• Melting point: $\mathrm{113.7~^\circ C}$
• Boiling point: $\mathrm{184.3~^\circ C}$
• $C_{\mathrm m,p}{\left(\ce{I2 (s)}\right)} = 55.44 \mathrm{~J~K^{–1}~mol^{–1}}$
• $C_{\mathrm m,p}{\left(\ce{I2 (l)}\right)} = 89.38 \mathrm{~J~K^{–1}~mol^{–1}}$
• $C_{\mathrm m,p}{\left(\ce{I2 (g)}\right)} = 36.90 \mathrm{~J~K^{–1}~mol^{–1}}$
• $\Delta_\text{f}H^\circ_\mathrm m{\left(\ce{I2(g)}\right)} = 62.44 \mathrm{~kJ~mol^{-1}}$
• $S^\circ_\mathrm m{\left(\ce{I2(g)}\right)} = 260.6 \mathrm{~J~K^{-1}~mol^{-1}}$
• $S^\circ_\mathrm m{\left(\ce{I2(s)}\right)} = 116.1 \mathrm{~J~K^{-1}~mol^{-1}}$

My Attempt

I have calculated the enthalpy and entropy change for the first step by using Hess's Law. They are $\mathrm{62.44~kJ~mol^{-1}}$ and $\mathrm{144.5~J~K^{-1}~mol^{-1}}$ respectively.

Now for the second step which is the isothermal expansion:

$$\Delta H = -nC_{\mathrm m,p}\Delta T = 0$$

For entropy:

$$\Delta S = nR\ln\left(\frac{V_\mathrm f}{V_\mathrm i}\right)$$

If assume that I have 1 mole of iodine I know what $n$ and $V_\mathrm i$ is but how am I suppose to know what $V_\mathrm f$ is?

Also I am struggling to understand the question. It says isothermal expansion of our system until it reaches the sublimation pressure. How are you suppose to know when it reaches sublimation pressure if it has already sublimed?

• You know that the change in G in going from the solid to the vapor at the equilibrium sublimation pressure is zero, correct? You also know that the free energy of formation of the solid at 298 K and 1 atm is zero, correct? So, all you need to do is determine the free energy of formation of the vapor at the hypothetical state of 298 K and 1 atm, and then determine the pressure change necessary to bring that down to zero. – Chet Miller Oct 11 '15 at 11:59