STEM Problem Detail

Problem Statement

Prompt

provide me a step by step solution for problem 10.

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## Given Data

$$
\begin{array}{|c|c|c|}
\hline
\text{Phase} & T \text{ (K)} & P \text{ (mm Hg)} \\
\hline
\text{Solid} & 260.35 & 4.18 \\
\text{Solid} & 272 & 10 \\
\text{Liquid} & 298.16 & 41.55 \\
\text{Liquid} & 307.0 & 60 \\
\hline
\end{array}
$$

## Step 1: Heat of Sublimation (Solid-Vapor Equilibrium)

*Theory:* The Clausius-Clapeyron equation for solid-vapor equilibrium:
$$\ln P = -\frac{\Delta H_{sub}}{RT} + C_{sub}$$

Using points $$(T_1, P_1) = (260.35, 4.18)$$ and $$(T_2, P_2) = (272, 10)$$:

$$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{sub}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$

$$\ln\left(\frac{10}{4.18}\right) = \frac{\Delta H_{sub}}{R}\left(\frac{1}{260.35} - \frac{1}{272}\right)$$

$$\ln(2.392) = \frac{\Delta H_{sub}}{8.314}\left(3.841 \times 10^{-3} - 3.676 \times 10^{-3}\right)$$

$$0.872 = \frac{\Delta H_{sub}}{8.314}\left(1.65 \times 10^{-4}\right)$$

$$\Delta H_{sub} = \frac{0.872 \times 8.314}{1.65 \times 10^{-4}} = 43950 \text{ J/mol}$$

$$\boxed{\Delta H_{sub} = 44.0 \text{ kJ/mol}}$$

Sublimation line equation (finding $$C_{sub} = 21.73$$ using point 1):
$$\ln P_{sub} = -\frac{5280}{T} + 21.73$$

where $$5280 = \frac{\Delta H_{sub}}{R}$$

## Step 2: Heat of Evaporation (Liquid-Vapor Equilibrium)

*Theory:* Similarly for liquid-vapor equilibrium:
$$\ln P = -\frac{\Delta H_{vap}}{RT} + C_{vap}$$

Using points $$(T_3, P_3) = (298.16, 41.55)$$ and $$(T_4, P_4) = (307.0, 60)$$:

$$\ln\left(\frac{60}{41.55}\right) = \frac{\Delta H_{vap}}{R}\left(\frac{1}{298.16} - \frac{1}{307.0}\right)$$

$$\ln(1.444) = \frac{\Delta H_{vap}}{8.314}\left(3.354 \times 10^{-3} - 3.257 \times 10^{-3}\right)$$

$$0.367 = \frac{\Delta H_{vap}}{8.314}\left(9.7 \times 10^{-5}\right)$$

$$\Delta H_{vap} = \frac{0.367 \times 8.314}{9.7 \times 10^{-5}} = 31460 \text{ J/mol}$$

$$\boxed{\Delta H_{vap} = 31.5 \text{ kJ/mol}}$$

Vaporization line equation ($$C_{vap} = 16.42$$):
$$\ln P_{vap} = -\frac{3784}{T} + 16.42$$

where $$3784 = \frac{\Delta H_{vap}}{R}$$

## Step 3: Triple Point Coordinates

*Theory:* At the triple point, solid, liquid, and vapor coexist. This occurs at the intersection of the sublimation and vaporization curves where $$P_{sub} = P_{vap}$$:

$$-\frac{5280}{T_{tp}} + 21.73 = -\frac{3784}{T_{tp}} + 16.42$$

$$\frac{5280 - 3784}{T_{tp}} = 21.73 - 16.42$$

$$\frac{1496}{T_{tp}} = 5.31$$

$$T_{tp} = \frac{1496}{5.31} = 281.7 \text{ K}$$

$$\boxed{T_{tp} = 282 \text{ K} \text{ or } 9^\circ\text{C}}$$

Calculate $$P_{tp}$$ using vaporization equation:
$$\ln P_{tp} = -\frac{3784}{281.7} + 16.42 = -13.43 + 16.42 = 2.99$$

$$P_{tp} = e^{2.99} = 19.9 \text{ mm Hg}$$

$$\boxed{P_{tp} = 20.0 \text{ mm Hg} \approx 2666 \text{ Pa}}$$

*Verification:* Using sublimation equation gives the same result:
$$\ln P_{tp} = -\frac{5280}{281.7} + 21.73 = -18.74 + 21.73 = 2.99 \checkmark$$

## Step 4: Heat of Fusion at Triple Point

*Theory:* From Hess's law, the enthalpy of sublimation equals the sum of enthalpies of fusion and vaporization:

$$\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$$

Therefore:
$$\Delta H_{fus} = \Delta H_{sub} - \Delta H_{vap}$$

$$\Delta H_{fus} = 44.0 - 31.5 = \boxed{12.5 \text{ kJ/mol}}$$

## Final Results Summary

$$
\begin{array}{|l|c|l|}
\hline
\text{Property} & \text{Value} & \text{Units} \\
\hline
\text{Triple point temperature } T_{tp} & 282 & \text{K (}9^\circ\text{C)} \\
\text{Triple point pressure } P_{tp} & 20 & \text{mm Hg (2.67 kPa)} \\
\text{Heat of sublimation } \Delta H_{sub} & 44.0 & \text{kJ/mol} \\
\text{Heat of evaporation } \Delta H_{vap} & 31.5 & \text{kJ/mol} \\
\text{Heat of fusion } \Delta H_{fus} & 12.5 & \text{kJ/mol} \\
\hline
\end{array}
$$

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Problem #54