Question

A reasonable approximation to the vapor pressure of krypton is given by \(\log _{10}(P / \mathrm{Torr})=b-0.05223(a / T)\). For solid krypton, \(a=10,065\) and \(b=7.1770 .\) For liquid krypton, \(a=9377.0\) and \(b=6.92387 .\) Use these formulas to estimate the triple point temperature and pressure and also the enthalpies of vaporization, fusion, and sublimation of krypton.

Short answer

The triple point temperature and pressure for krypton are approximately 116.6 K and 7.47 Torr, respectively. The enthalpies of vaporization, fusion, and sublimation are approximately 0 J/mol, 1595.9 J/mol, and 1595.9 J/mol, respectively.

Step-by-step solution

Solve for the triple point temperature and pressure

To find the triple point, we need to equalize the vapor pressure equations for solid and liquid krypton: \(\log_{10}(P_{solid} / \mathrm{Torr}) = b_{solid} - 0.05223(a_{solid} / T)\) \(\log_{10}(P_{liquid} / \mathrm{Torr}) = b_{liquid} - 0.05223(a_{liquid} / T)\) Since \(P_{solid} = P_{liquid}\) at the triple point, we have: \(b_{solid} - 0.05223(a_{solid} / T) = b_{liquid} - 0.05223(a_{liquid} / T)\) Solving for T: \(T = \frac{a_{liquid}b_{solid} - a_{solid}b_{liquid}}{0.05223(b_{solid} - b_{liquid})}\) Plugging in the given values of \(a\) and \(b\) for solid and liquid krypton: \(T = \frac{(9377)(7.1770) - (10065)(6.92387)}{0.05223(7.1770 - 6.92387)}\) \(T \approx 116.6\: K\) Now that we have the triple point temperature, we can calculate the triple point pressure by plugging T into the vapor pressure equation for either solid or liquid krypton. We'll use the solid krypton equation: \(\log_{10}(P_{triple} / \mathrm{Torr}) = 7.1770 - 0.05223(10065 / 116.6)\) \(P_{triple} \approx 7.47 \: \mathrm{Torr}\)

Calculate the enthalpy of vaporization

Using the vapor pressure equations from Step 1 and the Clausius-Clapeyron equation, we can find the enthalpy of vaporization: \(\Delta H_{vap} = -\frac{T \cdot R \cdot (\log_{10}(P_{liquid}) - \log_{10}(P_{solid}))}{0.05223}\) At the triple point, \(P_{liquid} = P_{solid} = P_{triple}\), so: \(\Delta H_{vap} = -\frac{(116.6 \: K)(8.314 \: J/mol·K) \cdot (\log_{10}(7.47) - \log_{10}(7.47))}{0.05223}\) \(\Delta H_{vap} = 0 \: J/mol\)

Calculate the enthalpy of sublimation

Using the Clausius-Clapeyron equation to find the enthalpy of sublimation: \(\Delta H_{sub} = -\frac{R \cdot (a_{liquid} - a_{solid})}{0.05223}\) \(\Delta H_{sub} = -\frac{(8.314 \: J/mol·K)(9377 - 10065)}{0.05223}\) \(\Delta H_{sub} \approx 1595.9 \: J/mol\)

Calculate the enthalpy of fusion

To find the enthalpy of fusion, we can use the relationship \(\Delta H_{fus} = \Delta H_{sub} - \Delta H_{vap}\): \(\Delta H_{fus} = 1595.9 \: J/mol - 0 \: J/mol = 1595.9 \: J/mol\) So, the enthalpies of vaporization, fusion, and sublimation for krypton are approximately 0 J/mol, 1595.9 J/mol, and 1595.9 J/mol, respectively, and the triple point temperature is about 116.6 K with a pressure of 7.47 Torr.

Key concepts

Vapor Pressure

Vapor pressure is a measure of a liquid's tendency to become a gas. It represents the pressure exerted by the vapor in equilibrium with its liquid or solid form at a given temperature. This concept helps us understand why liquids evaporate and solids sublime. For krypton, the vapor pressure is given by a formula involving temperature and constants specific to the solid or liquid state:

• For solid krypton: \(\log_{10}(P / \mathrm{Torr})=7.1770-0.05223(10065 / T)\)
• For liquid krypton: \(\log_{10}(P / \mathrm{Torr})=6.92387-0.05223(9377 / T)\)

The triple point is where both solid and liquid states have equal vapor pressure, allowing us to equalize these expressions to find the temperature and pressure at this point.

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is essential for understanding phase transitions like evaporation and sublimation. It relates the change in vapor pressure with temperature to the enthalpy of transition, providing a way to calculate enthalpy changes. The equation is:
\[\frac{d(\ln P)}{dT} = \frac{\Delta H_{transition}}{R T^2}\]
This formula connects the vapor pressure slope with temperature to the enthalpies of vaporization or sublimation. It helps in determining these values using known constants, as we see in the exercise where the enthalpies of sublimation and zero vaporization enthalpy were calculated.

Enthalpy of Sublimation

Enthalpy of sublimation is the energy required to convert a solid directly into a gas at a constant temperature and pressure. For krypton, it's computed using the Clausius-Clapeyron form:

\[\Delta H_{sub} = -\frac{R \cdot (a_{liquid} - a_{solid})}{0.05223}\]
Filling in krypton's values gives us approximately 1595.9 J/mol. This enthalpy represents the sum of both the enthalpies of fusion and vaporization, as sublimation combines solid-to-liquid and liquid-to-gas transitions.

Enthalpy of Fusion

The enthalpy of fusion is the energy needed to change a solid into a liquid without altering its temperature. It can also be found from the relationship between the enthalpies of sublimation and vaporization:

\[\Delta H_{fus} = \Delta H_{sub} - \Delta H_{vap}\]
In this exercise, since the enthalpy of vaporization for krypton at the triple point is zero, the enthalpy of fusion equals the enthalpy of sublimation, which is 1595.9 J/mol. This value indicates the energy needed for the phase change from solid to liquid at the triple point.