Question

Solid iodine, \(\mathrm{I}_{2}(s),\) at \(25.0^{\circ} \mathrm{C}\) has an enthalpy of sublimation of \(56.30 \mathrm{kJ} \mathrm{mol}^{-1}\). The \(C_{P, m}\) of the vapor and solid phases at that temperature are 36.9 and \(54.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\), respectively. The sublimation pressure at \(25.0^{\circ} \mathrm{C}\) is 0.30844 Torr. Calculate the sublimation pressure of the solid at the melting point \(\left(113.6^{\circ} \mathrm{C}\right)\) assuming a. that the enthalpy of sublimation does not change with temperature. b. that the enthalpy of sublimation at temperature \(T\) can be calculated from the equation \(\Delta H_{\text {sublimation}}(T)=\) \(\Delta H_{\text {sublimation}}\left(T_{0}\right)+\Delta C_{P}\left(T-T_{0}\right)\) and \(\Delta C_{P}\) does not change with \(T\).

Short answer

The sublimation pressure at the melting point (113.6°C) is approximately 3.563 Torr when the enthalpy of sublimation does not change with temperature and approximately 4.455 Torr when the enthalpy of sublimation is a function of temperature.

Step-by-step solution

Write the Clausius-Clapeyron equation

The Clausius-Clapeyron equation for sublimation is as follows: \(log_{10}\frac{P_2}{P_1}=-\frac{\Delta H_{sublimation}}{2.303R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\) Where \(P_1\) and \(P_2\) are the sublimation pressures at temperatures \(T_1\) and \(T_2\), respectively, and R is the gas constant(8.314 J/mol K). Given: \(P_1 = 0.30844\, Torr\) \(T_1 = 25.0°C = 298.15\, K\) \(T_2 = 113.6°C = 386.75\, K\) \(ΔH_{sublimation} = 56.30 kJ/mol\) We need to find \(P_2\) (sublimation pressure at 113.6°C).

Calculate \(P_2\) assuming enthalpy of sublimation does not change with temperature (part a)

Using the given enthalpy of sublimation, \(ΔH_{sublimation}\), and the given \(P_1\), \(T_1\), and \(T_2\), we can solve for \(P_2\) using the Clausius-Clapeyron equation: \(log_{10}\frac{P_2}{0.30844\, Torr}=-\frac{56.30\times10^3\, J/mol}{2.303\times 8.314\, J/(mol K)}\left(\frac{1}{386.75\, K}-\frac{1}{298.15\, K}\right)\) Now, solve for \(P_2\): \(P_2 = 0.30844\, Torr \times 10^{-\frac{56.30\times10^3\, J/mol}{2.303\times 8.314\, J/(mol K)}\left(\frac{1}{386.75\, K}-\frac{1}{298.15\, K}\right)} \approx 3.563\, Torr\) So, the sublimation pressure at the melting point (113.6°C) when enthalpy of sublimation does not change with temperature is approximately 3.563 Torr.

Calculate the enthalpy of sublimation at temperature T (part b)

We are given the equation for the enthalpy of sublimation at temperature T: \(\Delta H_{\text{sublimation}}(T)=\Delta H_{\text{sublimation}}\left(T_{0}\right)+ \Delta C_{P}\left(T-T_{0}\right)\) Where \(ΔC_P = C_{P,m}(vapor) - C_{P,m}(solid)\) Using the given \(C_{P,m}\) values, we get: \(ΔC_P = 36.9 - 54.4 = -17.5\, J/(K\cdot mol)\) Now, we can calculate the enthalpy of sublimation at 113.6°C: \(\Delta H_{\text{sublimation}}(386.75\,K) = 56.30\, kJ/mol +(-17.5\, J/(K\cdot mol))(386.75\, K -298.15\, K)\) \(\Delta H_{\text{sublimation}}(386.75\,K) \approx 55.079\, kJ/mol\)

Calculate \(P_2\) when enthalpy of sublimation is a function of temperature (part b)

Now, we can use the updated enthalpy of sublimation at 113.6°C in the Clausius-Clapeyron equation to find \(P_2\): \(log_{10}\frac{P_2}{0.30844\, Torr}=-\frac{55.079\times10^3\, J/mol}{2.303\times 8.314\, J/(mol K)}\left(\frac{1}{386.75\, K}-\frac{1}{298.15\, K}\right)\) Now, solve for \(P_2\): \(P_2 = 0.30844\, Torr \times 10^{-\frac{55.079\times10^3\, J/mol}{2.303\times 8.314\, J/(mol K)}\left(\frac{1}{386.75\, K}-\frac{1}{298.15\, K}\right)} \approx 4.455\, Torr\) So, the sublimation pressure at the melting point (113.6°C) when enthalpy of sublimation is a function of temperature is approximately 4.455 Torr.

Key concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is a powerful relationship in thermodynamics used to predict how a substance's vapor pressure changes with temperature. At its core, it provides a way to estimate the enthalpy of phase transitions, such as melting, boiling, or sublimation. When a substance transitions from solid to gas without passing through the liquid phase, as is the case with sublimation, this equation is particularly handy.

The equation is generally presented as:\begin{align*}log_{10}\frac{P_2}{P_1} = -\frac{\Delta H}{2.303R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)d{align*}Where:

• \(P_1\) and \(P_2\) are the pressures at temperatures \(T_1\) and \(T_2\), respectively,
• \(\Delta H\) is the enthalpy change for the phase transition (e.g., sublimation),
• \(R\) is the ideal gas constant.

This equation assumes that the enthalpy change is independent of temperature, which is a reasonable approximation over small temperature ranges. However, over larger temperature ranges or when high precision is required, the temperature dependence of the enthalpy change should be accounted for. The equation can then be integrated to consider how enthalpy changes with temperature. Whether in its simple or integrated form, the Clausius-Clapeyron equation is a fundamental tool for understanding phase transition dynamics.

Sublimation Pressure

Sublimation pressure is the pressure at which a solid transitions directly to a gas at a given temperature, bypassing the liquid phase entirely. It is an essential concept to grasp for substances that undergo this phase transition. The sublimation pressure reflects how readily a substance will sublimate and is influenced by external factors such as temperature.

In the context of the Clausius-Clapeyron equation, sublimation pressure is a critical component. For iodine, for instance, understanding its sublimation pressure at different temperatures can help predict how it behaves under certain conditions, like when heated. By applying the Clausius-Clapeyron equation and knowing the enthalpy of sublimation, one can calculate the sublimation pressure at a new temperature.The calculation involves determining the ratio of the pressure at two different temperatures. It relies heavily on the assumption that the enthalpy of sublimation remains constant. If this assumption does not hold, modifications to the equation are necessary to accurately predict the sublimation pressure. Therefore, it's crucial for students to recognize the assumptions made in these calculations and when they need to consider temperature's effect on the enthalpy of sublimation.

Phase Transition Thermodynamics

Phase transition thermodynamics is a branch of thermodynamics focusing on the energy changes that occur during a phase change, like sublimation, melting, or boiling. Each phase transition involves an enthalpy change, which is essentially the energy required to alter the physical state of a given amount of substance. This change is quantified by specific enthalpies, such as the enthalpy of sublimation for the solid-to-gas transition.

These energy changes are not arbitrary. They are rooted in molecular interactions and changes in disorder, or entropy, of a substance. The enthalpy of sublimation, for instance, represents the energy needed to break intermolecular forces holding the solid together while also providing enough energy for the molecules to disperse and move freely as a gas.

A critical aspect of phase transition thermodynamics relates to the temperature dependence of these enthalpy changes. While some calculations assume a constant enthalpy change regardless of temperature (a sensible approximation over small temperature differences), the reality is that enthalpy can vary with temperature. This nuance was addressed in part b of the exercise, where the enthalpy of sublimation was corrected for temperature changes to offer a more precise calculation of sublimation pressure at a different temperature.Understanding phase transitions from a thermodynamics perspective is fundamental in various fields, especially in materials science, environmental studies, and engineering, where the behavior of substances under different temperatures and pressures is critical.