6.2 Phase Changes, Heating Curves & Vapor Pressure

6.2 Phase Changes, Heating Curves & Vapor Pressure

Quick Summary: Phase changes are physical transitions between solid, liquid, and gas accompanied by latent heat. During single-phase warming, heat raises molecular kinetic energy ($q = mc\Delta T$); during phase transitions, temperature remains constant ($q = n\Delta H$) as energy increases intermolecular potential energy. In a closed container, liquid and vapor establish dynamic equilibrium at an equilibrium vapor pressure governed solely by temperature and IMF strength. When vapor pressure matches atmospheric pressure, boiling occurs throughout the liquid bulk, modeled by the Clausius-Clapeyron equation.

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1. Energetics and Thermodynamics of Phase Transitions

Phase changes alter physical states without modifying chemical composition. Each transition involves characteristic changes in enthalpy ($\Delta H$) and entropy ($\Delta S$):

• Endothermic Transitions (Heat Absorbed, $\Delta H > 0$):
  • Fusion (Melting): Solid $\to$ Liquid ($\Delta H_{fus} > 0$)
  • Vaporization: Liquid $\to$ Gas ($\Delta H_{vap} > 0$)
  • Sublimation: Solid $\to$ Gas ($\Delta H_{sub} > 0$)
• Exothermic Transitions (Heat Released, $\Delta H < 0$):
  • Freezing: Liquid $\to$ Solid ($\Delta H_{cryst} = -\Delta H_{fus}$)
  • Condensation: Gas $\to$ Liquid ($\Delta H_{cond} = -\Delta H_{vap}$)
  • Deposition: Gas $\to$ Solid ($\Delta H_{dep} = -\Delta H_{sub}$)

Hess's Law and Enthalpy Comparison

Because enthalpy is a state function, Hess's law dictates that the molar enthalpy of sublimation equals the sum of fusion and vaporization enthalpies: $\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$

For virtually all substances, $\Delta H_{vap} \gg \Delta H_{fus}$:

• Melting: Only disrupts long-range lattice order. Molecules remain in dense contact, retaining 80–90% of condensed-phase IMFs.
• Vaporization: Requires completely separating molecules against all intermolecular attractions into an ideal gas.

For water, $\Delta H_{fus} = 6.01\text{ kJ/mol}$ at 0 °C, whereas $\Delta H_{vap} = 40.67\text{ kJ/mol}$ at 100 °C. Condensing steam releases nearly seven times more heat than freezing water, causing destructive burns.

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2. Heating and Cooling Curves: Quantitative Mechanics

A heating curve tracks temperature against heat added at a constant rate, traversing five stages:

1. Stage 1 (Solid Warming): $q_1 = m \cdot c_{solid} \cdot \Delta T$. Temperature climbs as kinetic energy rises.
2. Stage 2 (Melting Plateau): Solid and liquid coexist at melting point $T_f$. Temperature is constant: $q_2 = n \cdot \Delta H_{fus}$.
3. Stage 3 (Liquid Warming): $q_3 = m \cdot c_{liquid} \cdot \Delta T$. Temperature climbs as kinetic energy rises.
4. Stage 4 (Boiling Plateau): Liquid and gas coexist at boiling point $T_b$. Temperature is constant: $q_4 = n \cdot \Delta H_{vap}$. Plateau length is far greater than Stage 2 because $\Delta H_{vap} \gg \Delta H_{fus}$.
5. Stage 5 (Gas Warming): $q_5 = m \cdot c_{gas} \cdot \Delta T$. Temperature climbs as kinetic energy rises.

Energy Distribution and Slopes

• Warming Segments: Added heat raises kinetic energy ($KE_{avg} = \frac{3}{2}RT$). Slope equals $1/(m \cdot c)$. Because liquid water has a higher specific heat ($c_{liq} = 4.184\text{ J/g}\cdot^\circ\text{C}$) than ice ($2.09\text{ J/g}\cdot^\circ\text{C}$) or steam ($2.01\text{ J/g}\cdot^\circ\text{C}$), the liquid warming line rises with half the slope of the solid or gas lines.
• Phase Plateaus: Temperature is invariant ($\Delta T = 0$), so kinetic energy is constant. Absorbed heat converts entirely into intermolecular potential energy by severing IMFs.
• Supercooling: Cooling a liquid without vibration can lower its temperature below $T_f$ without crystallization until nucleation triggers rapid freezing, releasing latent heat and warming the liquid back to $T_f$.

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3. Heating Curve Stage Calculations Table

Heating $36.04\text{ g}$ ($2.000\text{ mol}$) of ice from $-20.0^\circ\text{C}$ to steam at $120.0^\circ\text{C}$ at $1.00\text{ atm}$:

Vaporization represents $81.34\text{ kJ}$ ($73\%$) of the total energy required.

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4. Dynamic Equilibrium and Vapor Pressure

In a closed vessel at constant temperature:

1. Surface molecules escape into the vapor phase at a constant rate ($r_{evap}$).
2. Vapor molecules collide with the liquid surface and condense ($r_{cond}$), increasing with vapor concentration.
3. Dynamic equilibrium is reached when $r_{evap} = r_{cond}$. Macroscopic vapor pressure remains constant.

Invariance of Vapor Pressure & Boiling Criterion

Equilibrium vapor pressure ($P_{vap}$) depends solely on temperature (exponential increase) and IMF strength (stronger IMFs $\implies$ lower $P_{vap}$). It is strictly independent of liquid volume, container volume, and surface area.

Boiling begins when equilibrium vapor pressure matches atmospheric pressure ($P_{vap} = P_{ext}$), allowing vapor bubbles to form throughout the liquid bulk:

• Normal Boiling Point: Temperature where $P_{vap} = 1.000\text{ atm} = 760.0\text{ torr} = 101.325\text{ kPa}$.
• Altitude: At high elevation ($P_{ext} \approx 630\text{ torr}$ in Denver), water boils at $\approx 95^\circ\text{C}$, slowing cooking. In a pressure cooker ($P \approx 2\text{ atm}$), water boils at $\approx 120^\circ\text{C}$, accelerating cooking.

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5. The Clausius-Clapeyron Equation

The relation between vapor pressure and absolute temperature is given by the Clausius-Clapeyron equation: $\ln P = -\left(\frac{\Delta H_{vap}}{R}\right)\left(\frac{1}{T}\right) + C$ Plotting $\ln P$ versus $1/T$ yields a line with $\text{Slope} = -\frac{\Delta H_{vap}}{R}$, where $R = 8.314\text{ J/(mol}\cdot\text{K)}$.

Between two states $(T_1, P_1)$ and $(T_2, P_2)$: $\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Worked Example: Ethanol Vapor Pressure at 25.0 °C

Ethanol boils normally at $78.4^\circ\text{C}$ ($351.55\text{ K}$) at $760.0\text{ torr}$, with $\Delta H_{vap} = 38.56\text{ kJ/mol}$ ($38,560\text{ J/mol}$). Find $P_2$ at $25.0^\circ\text{C}$ ($298.15\text{ K}$): $\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{298.15} - \frac{1}{351.55} = 0.0033540 - 0.0028445 = +0.0005095\text{ K}^{-1}$ $\ln\left(\frac{P_2}{760.0}\right) = -\frac{38,560}{8.314} \times 0.0005095 = -2.363$ $\frac{P_2}{760.0} = e^{-2.363} = 0.09414 \implies P_2 = 760.0 \times 0.09414 = 71.5\text{ torr}$