10.3: Transiciones de fase

Objetivos de aprendizaje

• Definir transiciones de fase y temperaturas de transición de fase
• Explicar la relación entre las temperaturas de transición de fase y las fuerzas de atracción intermoleculares
• Describir los procesos representados por las curvas típicas de calentamiento y enfriamiento, y calcular los flujos de calor y los cambios de entalpía que acompañan a estos procesos

Somos testigos y utilizamos cambios de estado físico, o transiciones de fase, de muchas maneras. Como un ejemplo de importancia global, considere la evaporación, condensación, congelación y fusión del agua. Estos cambios de estado son aspectos esenciales del ciclo hídrico de nuestra tierra así como muchos otros fenómenos naturales y procesos tecnológicos de importancia central para nuestras vidas. En este módulo se exploran los aspectos esenciales de las transiciones de fase.

Vaporización y Condensación

Cuando un líquido se vaporiza en un recipiente cerrado, las moléculas de gas no pueden escapar. A medida que estas moléculas en fase gaseosa se mueven aleatoriamente, ocasionalmente colisionarán con la superficie de la fase condensada, y en algunos casos, estas colisiones darán como resultado que las moléculas vuelvan a entrar en la fase condensada. El cambio de la fase gaseosa al líquido se llama condensación. Cuando la velocidad de condensación se vuelve igual a la velocidad de vaporización, ni la cantidad de líquido ni la cantidad de vapor en el recipiente cambian. Entonces se dice que el vapor en el recipiente está en equilibrio con el líquido. Tenga en cuenta que esta no es una situación estática, ya que las moléculas se intercambian continuamente entre las fases condensada y gaseosa. Tal es un ejemplo de un equilibrio dinámico, el estado de un sistema en el que los procesos recíprocos (por ejemplo, vaporización y condensación) ocurren a tasas iguales. La presión ejercida por el vapor en equilibrio con un líquido en un recipiente cerrado a una temperatura dada se denomina presión de vapor del líquido (o presión de vapor de equilibrio). El área de la superficie del líquido en contacto con un vapor y el tamaño del recipiente no tienen ningún efecto sobre la presión de vapor, aunque sí afectan el tiempo requerido para que se alcance el equilibrio. Podemos medir la presión de vapor de un líquido colocando una muestra en un recipiente cerrado, como el ilustrado en la Figura\(\PageIndex{1}\), y usando un manómetro para medir el aumento de presión que se debe al vapor en equilibrio con la fase condensada.

Figura\(\PageIndex{1}\): En un recipiente cerrado, se alcanza el equilibrio dinámico cuando (a) la velocidad de las moléculas que escapan del líquido para convertirse en gas (b) aumenta y eventualmente (c) es igual a la velocidad de las moléculas de gas que ingresan al líquido. Cuando se alcanza este equilibrio, la presión de vapor del gas es constante, aunque continúan los procesos de vaporización y condensación.

Se muestran tres imágenes y se etiquetan como “a”, “b” y “c”. Cada imagen muestra una bombilla redonda conectada a la derecha a un tubo que es horizontal, luego se dobla verticalmente, se curva, y luego vuelve a ser vertical para hacer una forma de U. Una válvula está ubicada en la porción horizontal del tubo. La imagen a representa un líquido en la bombilla, etiquetado como “Líquido” y flechas orientadas hacia arriba que se alejan de la superficie del líquido. La frase “Las moléculas escapan a la superficie y forman vapor” se escribe debajo del bulbo, y un líquido gris en la porción en forma de U del tubo se muestra a alturas iguales en los lados derecho e izquierdo. La imagen b representa un líquido en el bulbo, etiquetado como “Líquido” y flechas orientadas hacia arriba que se alejan de la superficie del líquido hacia moléculas dibujadas en la porción superior del bulbo. Un líquido gris en la porción en forma de U del tubo se muestra ligeramente más alto en el lado derecho que en el lado izquierdo. La imagen c representa un líquido en el bulbo, etiquetado como “Líquido” y flechas orientadas hacia arriba que se alejan de la superficie del líquido hacia moléculas dibujadas en la porción superior del bulbo. Hay más moléculas presentes en c que en b. La frase “Equilibrio alcanzado, presión de vapor determinada”, se escribe debajo del bulbo y un líquido gris en la porción en forma de U del tubo se muestra más alto en el lado derecho. Se dibuja una línea horizontal nivelada con cada uno de estos niveles de líquido y la distancia entre las líneas se etiqueta con una flecha de doble punta. Esta sección está etiquetada con la frase “Presión de vapor”.

Las identidades químicas de las moléculas en un líquido determinan los tipos (y fortalezas) de atracciones intermoleculares posibles; en consecuencia, diferentes sustancias exhibirán diferentes presiones de vapor de equilibrio. Las fuerzas de atracción intermolecular relativamente fuertes servirán para impedir la vaporización así como favorecer la “recaptura” de moléculas en fase gaseosa cuando chocan con la superficie del líquido, resultando en una presión de vapor relativamente baja. Las débiles atracciones intermoleculares presentan menos barrera a la vaporización y una menor probabilidad de recaptura de gas, produciendo presiones de vapor relativamente altas. El siguiente ejemplo ilustra esta dependencia de la presión de vapor de las fuerzas de atracción intermoleculares.

Ejemplo\(\PageIndex{1}\): Explaining Vapor Pressure in Terms of IMFs

Dadas las fórmulas estructurales mostradas para estos cuatro compuestos, explicar sus presiones de vapor relativas en términos de tipos y extensiones de IMF:

Solución

El éter dietílico tiene un dipolo muy pequeño y la mayoría de sus atractivos intermoleculares son las fuerzas londinenses. Si bien esta molécula es la más grande de las cuatro bajo consideración, sus IMF son las más débiles y, como resultado, sus moléculas escapan más fácilmente del líquido. También tiene la presión de vapor más alta. Debido a su menor tamaño, el etanol exhibe fuerzas de dispersión más débiles que el éter dietílico. Sin embargo, el etanol es capaz de formar enlaces de hidrógeno y, por lo tanto, exhibe IMF globales más fuertes, lo que significa que menos moléculas escapan del líquido a cualquier temperatura dada, y así el etanol tiene una presión de vapor más baja que el éter dietílico. El agua es mucho más pequeña que cualquiera de las sustancias anteriores y exhibe fuerzas de dispersión más débiles, pero su extenso enlace de hidrógeno proporciona atracciones intermoleculares más fuertes, menos moléculas que escapan del líquido y una presión de vapor más baja que para el éter dietílico o el etanol. El etilenglicol tiene dos grupos −OH, por lo que, al igual que el agua, exhibe extensos enlaces de hidrógeno. Es mucho más grande que el agua y por lo tanto experimenta fuerzas más grandes de Londres. Sus IMF generales son las más grandes de estas cuatro sustancias, lo que significa que su velocidad de vaporización será la más lenta y, en consecuencia, su presión de vapor la más baja.

Ejercicio\(\PageIndex{1}\)

At 20 °C, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols:

At 20 °C, the vapor pressures of several alcohols

Compound	methanol CH3OH	ethanol C2H5OH	propanol C3H7OH	butanol C4H9OH
Vapor Pressure at 20 °C	11.9 kPa	5.95 kPa	2.67 kPa	0.56 kPa

Answer

All these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed:

\[P_{methanol} > P_{ethanol} > P_{propanol} > P_{butanol} \nonumber \]

As temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic energies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in Figure \(\PageIndex{2}\). The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure.

Figure \(\PageIndex{2}\): Temperature affects the distribution of kinetic energies for the molecules in a liquid. At the higher temperature, more molecules have the necessary kinetic energy, KE, to escape from the liquid into the gas phase.

A graph is shown where the y-axis is labeled “Number of molecules” and the x-axis is labeled “Kinetic Energy.” Two lines are graphed and a vertical dotted line, labeled “Minimum K E needed to escape,” is drawn halfway across the x-axis. The first line move sharply upward and has a high peak near the left side of the x-axis. It drops just as steeply and ends about 60 percent of the way across the x-axis. This line is labeled “Low T.” A second line, labeled “High T,” begins at the same point as the first, but does not go to such a high point, is wider, and ends slightly further to the right on the x-axis.

Boiling Points

When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth’s atmosphere. The normal boiling point of a liquid is defined as its boiling point when surrounding pressure is equal to 1 atm (101.3 kPa). Figure \(\PageIndex{3}\) shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid’s boiling point on surrounding pressure.

Figure \(\PageIndex{3}\): The boiling points of liquids are the temperatures at which their equilibrium vapor pressures equal the pressure of the surrounding atmosphere. Normal boiling points are those corresponding to a pressure of 1 atm (101.3 kPa.)

A graph is shown where the x-axis is labeled “Temperature ( degree sign, C )” and has values of 200 to 1000 in increments of 200 and the y-axis is labeled “Pressure ( k P a )” and has values of 20 to 120 in increments of 20. A horizontal dotted line extends across the graph at point 780 on the y-axis while three vertical dotted lines extend from points 35, 78, and 100 to meet the horizontal dotted line. Four lines are graphed. The first line, labeled “ethyl ether,” begins at the point “0 , 200” and extends in a slight curve to point “45, 1000” while the second line, labeled “ethanol”, extends from point “0, 20” to point “88, 1000” in a more extreme curve. The third line, labeled “water,” begins at the point “0, 0” and extends in a curve to point “108, 1000” while the fourth line, labeled “ethylene glycol,” extends from point “80, 0” to point “140, 100” in a very shallow curve.

Example \(\PageIndex{2}\): A Boiling Point at Reduced Pressure

A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in Figure \(\PageIndex{3}\) to determine the boiling point of water at this elevation.

Solution

The graph of the vapor pressure of water versus temperature in Figure \(\PageIndex{3}\) indicates that the vapor pressure of water is 68 kPa at about 90 °C. Thus, at about 90 °C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil.

Exercise \(\PageIndex{2}\)

The boiling point of ethyl ether was measured to be 10 °C at a base camp on the slopes of Mount Everest. Use Figure \(\PageIndex{3}\) to determine the approximate atmospheric pressure at the camp.

Answer

Approximately 40 kPa (0.4 atm)

 

The quantitative relation between a substance’s vapor pressure and its temperature is described by the Clausius-Clapeyron equation:

\[P=Ae^{−ΔH_{vap}/RT} \label{10.4.1} \]

where

• \(ΔH_{vap}\) is the enthalpy of vaporization for the liquid,
• \(R\) is the gas constant, and
• \(\ln A\) is a constant whose value depends on the chemical identity of the substance.

Equation \(\ref{10.4.1}\) is often rearranged into logarithmic form to yield the linear equation:

\[\ln P=−\dfrac{ΔH_\ce{vap}}{RT}+\ln A \label{10.4.2} \]

This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the examples and exercises that follow. If at temperature \(T_1\), the vapor pressure is \( P_1\), and at temperature \(T_2\), the vapor pressure is \(T_2\), the corresponding linear equations are:

\[\ln P_1=−\dfrac{ΔH_\ce{vap}}{RT_1}+\ln A \nonumber \]

and

\[\ln P_2=−\dfrac{ΔH_\ce{vap}}{RT_2}+\ln A \label{10.4.3} \]

Since the constant, ln A, is the same, these two equations may be rearranged to isolate \(\ln A\) and then set them equal to one another:

\(\ln P_1+\dfrac{ΔH_\ce{vap}}{RT_1}=\ln P_2+\dfrac{ΔH_\ce{vap}}{RT_2}\label{10.4.5}\)

which can be combined into:

\[\ln \left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R} \left( \dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \label{10.4.6} \]

Example \(\PageIndex{3}\): Estimating Enthalpy of Vaporization

Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 °C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 °C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane.

Solution

The enthalpy of vaporization, \(ΔH_\ce{vap}\), can be determined by using the Clausius-Clapeyron equation (Equation \(\ref{10.4.6}\)):

\[\ln\left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R}\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \nonumber \]

Since we have two vapor pressure-temperature values

• \(T_1 = 34.0^oC = 307.2\, K\)
• \(P_1 = 10.0\, kPa\) and
• \(T_2 = 98.8 ^oC = 372.0\, K\)
• \(P_2 = 100\, kPa\)

we can substitute them into this equation and solve for \(ΔH_{vap}\). Rearranging the Clausius-Clapeyron equation and solving for \(ΔH_{vap}\) yields:

\[ \begin{align*} ΔH_\ce{vap} &= \dfrac{R⋅\ln\left(\dfrac{P_2}{P_1}\right)}{\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right)} \\[4pt] &= \dfrac{(8.3145\:J/mol⋅K)⋅\ln \left(\dfrac{100\: kPa}{10.0\: kPa}\right)}{\left(\dfrac{1}{307.2\:K}−\dfrac{1}{372.0\:K}\right)} \\[4pt] &=33,800\, J/mol =33.8\, kJ/mol \end{align*} \nonumber \]

Note that the pressure can be in any units, so long as they agree for both P values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid.

Exercise \(\PageIndex{3}\)

At 20.0 °C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 °C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol.

Answer

47,782 J/mol = 47.8 kJ/mol

Example \(\PageIndex{4}\): Estimating Temperature (or Vapor Pressure)

For benzene (C₆H₆), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa?

Solution

If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, ΔH_vap, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation (Equation \(\ref{10.4.1}\)) :

\[\ln\left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R}\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \nonumber \]

Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value (\(T_1\) = 80.1 °C = 353.3 K, \(P_1\) = 101.3 kPa, \(ΔH_{vap}\) = 30.8 kJ/mol) and want to find the temperature (\(T_2\)) that corresponds to vapor pressure P₂ = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for \(T_2\). Rearranging the Clausius-Clapeyron equation and solving for \(T_2\) yields:

\[\begin{align*}
T_2 &=\left(\dfrac{−R⋅\ln\left(\dfrac{P_2}{P_1}\right)}{ΔH_\ce{vap}}+\dfrac{1}{T_1}\right)^{−1} \\[4pt] &=\mathrm{\left(\dfrac{−(8.3145\:J/mol⋅K)⋅\ln\left(\dfrac{83.4\:kPa}{101.3\:kPa}\right)}{30,800\: J/mol}+\dfrac{1}{353.3\:K}\right)^{−1}}\\[4pt]
&=\mathrm{346.9\: K\: or\:73.8^\circ C}
\end{align*} \nonumber \]

Exercise \(\PageIndex{4}\)

For acetone \(\ce{(CH3)2CO}\), the normal boiling point is 56.5 °C and the enthalpy of vaporization is 31.3 kJ/mol. What is the vapor pressure of acetone at 25.0 °C?

Answer

30.1 kPa

 

Enthalpy of Vaporization

Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, \(ΔH_{vap}\). For example, the vaporization of water at standard temperature is represented by:

\[\ce{H2O}(l)⟶\ce{H2O}(g)\hspace{20px}ΔH_\ce{vap}=\mathrm{44.01\: kJ/mol} \nonumber \]

As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:

\[\ce{H2O}(g)⟶\ce{H2O}(l)\hspace{20px}ΔH_\ce{con}=−ΔH_\ce{vap}=\mathrm{−44.01\:kJ/mol} \nonumber \]

Example \(\PageIndex{5}\): Using Enthalpy of Vaporization

One way our body is cooled is by evaporation of the water in sweat (Figure \(\PageIndex{4}\)). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at T = 37 °C (normal body temperature); \(ΔH_{vap} = 43.46\, kJ/mol\) at 37 °C.

Figure \(\PageIndex{4}\): Evaporation of sweat helps cool the body. (credit: “Kullez”/Flickr)

Solution We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:

\[\mathrm{1.5\cancel{L}×\dfrac{1000\cancel{g}}{1\cancel{L}}×\dfrac{1\cancel{mol}}{18\cancel{g}}×\dfrac{43.46\:kJ}{1\cancel{mol}}=3.6×10^3\:kJ} \nonumber \]

Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.

Exercise \(\PageIndex{5}\): Boiling Ammonia

How much heat is required to evaporate 100.0 g of liquid ammonia, \(\ce{NH3}\), at its boiling point if its enthalpy of vaporization is 4.8 kJ/mol?

Answer

28 kJ

Melting and Freezing

When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (Figure \(\PageIndex{5}\).

Figure \(\PageIndex{5}\): (a) This beaker of ice has a temperature of −12.0 °C. (b) After 10 minutes the ice has absorbed enough heat from the air to warm to 0 °C. A small amount has melted. (c) Thirty minutes later, the ice has absorbed more heat, but its temperature is still 0 °C. The ice melts without changing its temperature. (d) Only after all the ice has melted does the heat absorbed cause the temperature to increase to 22.2 °C. (credit: modification of work by Mark Ott).

This figure shows four photos each labeled, “a,” “b,” “c,” and, “d.” Each photo shows a beaker with ice and a digital thermometer. The first photo shows ice cubes in the beaker, and the thermometer reads negative 12.0 degrees C. The second photo shows slightly melted ice, and the thermometer reads 0.0 degrees C. The third photo shows more water than ice in the beaker. The thermometer reads 0.0 degrees C. The fourth photo shows the ice completely melted, and the thermometer reads 22.2 degrees C.

If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal processes of melting and freezing occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing).

The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures.

The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, ΔHfus of the substance. The enthalpy of fusion of ice is 6.0 kJ/mol at 0 °C. Fusion (melting) is an endothermic process:

\[\ce{H2O}_{(s)} \rightarrow \ce{H2O}_{(l)} \;\; ΔH_\ce{fus}=\mathrm{6.01\; kJ/mol} \label{10.4.9} \]

The reciprocal process, freezing, is an exothermic process whose enthalpy change is −6.0 kJ/mol at 0 °C:

\[\ce{H_2O}_{(l)} \rightarrow \ce{H_2O}_{(s)}\;\; ΔH_\ce{frz}=−ΔH_\ce{fus}=−6.01\;\mathrm{kJ/mol} \label{10.4.10} \]

Sublimation and Deposition

Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as sublimation. At room temperature and standard pressure, a piece of dry ice (solid CO₂) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure \(\PageIndex{6}\)). The reverse of sublimation is called deposition, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition.

Figure \(\PageIndex{6}\): Sublimation of solid iodine in the bottom of the tube produces a purple gas that subsequently deposits as solid iodine on the colder part of the tube above. (credit: modification of work by Mark Ott)

This figure shows a test tube. In the bottom is a dark substance which breaks up into a purple gas at the top.

Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, ΔHsub, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by:

\[\ce{CO2}(s)⟶\ce{CO2}(g)\hspace{20px}ΔH_\ce{sub}=\mathrm{26.1\: kJ/mol} \nonumber \]

Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation:

\[\ce{CO2}(g)⟶\ce{CO2}(s)\hspace{20px}ΔH_\ce{dep}=−ΔH_\ce{sub}=\mathrm{−26.1\:kJ/mol} \nonumber \]

Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess’s Law.

\[\mathrm{solid⟶liquid}\hspace{20px}ΔH_\ce{fus}\\\underline{\mathrm{liquid⟶gas}\hspace{20px}ΔH_\ce{vap}}\\\mathrm{solid⟶gas}\hspace{20px}ΔH_\ce{sub}=ΔH_\ce{fus}+ΔH_\ce{vap} \nonumber \]

Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in Figure \(\PageIndex{7}\). For example:

Figure \(\PageIndex{7}\): For a given substance, the sum of its enthalpy of fusion and enthalpy of vaporization is approximately equal to its enthalpy of sublimation.

A diagram is shown with a vertical line drawn on the left side and labeled “Energy” and three horizontal lines drawn near the bottom, lower third and top of the diagram. These three lines are labeled, from bottom to top, “Solid,” “Liquid” and “Gas.” Near the middle of the diagram, a vertical, upward-facing arrow is drawn from the solid line to the gas line and labeled “Sublimation, delta sign, H, subscript sub.” To the right of this arrow is a second vertical, upward-facing arrow that is drawn from the solid line to the liquid line and labeled “Fusion, delta sign, H, subscript fus.” Above the second arrow is a third arrow drawn from the liquid line to the gas line and labeled, “Vaporization, delta sign, H, subscript vap.”

 

Heating and Cooling Curves

In the chapter on thermochemistry, the relation between the amount of heat absorbed or related by a substance, q, and its accompanying temperature change, ΔT, was introduced:

\[q=mcΔT \nonumber \]

where m is the mass of the substance and c is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. Figure \(\PageIndex{8}\) shows a typical heating curve.

Figure \(\PageIndex{8}\): A typical heating curve for a substance depicts changes in temperature that result as the substance absorbs increasing amounts of heat. Plateaus in the curve (regions of constant temperature) are exhibited when the substance undergoes phase transitions.

A graph is shown where the x-axis is labeled “Amount of heat added” and the y-axis is labeled “Temperature ( degree sign C )” and has values of negative 10 to 100 in increments of 20. A right-facing horizontal arrow extends from point “0, 0” to the right side of the graph. A line graph begins at the lower left of the graph and moves to point “0” on the y-axis. This segment of the line is labeled “H, subscript 2, O ( s ).” The line then flattens and travels horizontally for a small distance. This segment is labeled “Solid begins to melt” on its left side and “All solid melted” on its right side. The line then goes steeply upward in a linear fashion until it hits point “100” on the y-axis. This segment of the line is labeled “H, subscript 2, O,( l ).” The line then flattens and travels horizontally for a moderate distance. This segment is labeled “Liquid begins to boil” on its left side and “All liquid evaporated” on its right side. The line then rises to a point above “100” on the y-axis. This segment of the line is labeled “H, subscript 2, O ( g ).”

Consider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water’s temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner. This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid). This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress.

Example \(\PageIndex{6}\): Total Heat Needed to Change Temperature and Phase for a Substance

How much heat is required to convert 135 g of ice at −15 °C into water vapor at 120 °C?

Solution

The transition described involves the following steps:

1. Heat ice from −15 °C to 0 °C
2. Melt ice
3. Heat water from 0 °C to 100 °C
4. Boil water
5. Heat steam from 100 °C to 120 °C

The heat needed to change the temperature of a given substance (with no change in phase) is: q = m × c × ΔT (see previous chapter on thermochemistry). The heat needed to induce a given change in phase is given by q = n × ΔH.

Using these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have:

\[\begin{align*}
q_\ce{total}&=(m⋅c⋅ΔT)_\ce{ice}+n⋅ΔH_\ce{fus}+(m⋅c⋅ΔT)_\ce{water}+n⋅ΔH_\ce{vap}+(m⋅c⋅ΔT)_\ce{steam}\\[7pt]
&=\mathrm{(135\: g⋅2.09\: J/g⋅°C⋅15°C)+\left(135⋅\dfrac{1\: mol}{18.02\:g}⋅6.01\: kJ/mol
\right)}\\[7pt]
&\mathrm{+(135\: g⋅4.18\: J/g⋅°C⋅100°C)+\left(135\: g⋅\dfrac{1\: mol}{18.02\:g}⋅40.67\: kJ/mol\right)}\\[7pt]
&\mathrm{+(135\: g⋅1.84\: J/g⋅°C⋅20°C)}\\[7pt]
&=\mathrm{4230\: J+45.0\: kJ+56,500\: J+305\: kJ+4970\: J}
\end{align*} \nonumber \]

Converting the quantities in J to kJ permits them to be summed, yielding the total heat required:

\[\mathrm{=4.23\:kJ+45.0\: kJ+56.5\: kJ+305\: kJ+4.97\: kJ=416\: kJ} \nonumber \]

Exercise \(\PageIndex{6}\)

What is the total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at −30.0 °C?

Answer

40.5 kJ

Summary

Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance.

Key Equations

• \(P=Ae^{−ΔH_\ce{vap}/RT}\)
• \(\ln P=−\dfrac{ΔH_\ce{vap}}{RT}+\ln A\)
• \(\ln\left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R}\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right)\)

Glossary

boiling point

temperature at which the vapor pressure of a liquid equals the pressure of the gas above it

Clausius-Clapeyron equation

mathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance

condensation

change from a gaseous to a liquid state

deposition

change from a gaseous state directly to a solid state

dynamic equilibrium

state of a system in which reciprocal processes are occurring at equal rates

freezing

change from a liquid state to a solid state

freezing point

temperature at which the solid and liquid phases of a substance are in equilibrium; see also melting point

melting

change from a solid state to a liquid state

melting point

temperature at which the solid and liquid phases of a substance are in equilibrium; see also freezing point

normal boiling point

temperature at which a liquid’s vapor pressure equals 1 atm (760 torr)

sublimation

change from solid state directly to gaseous state

vapor pressure

(also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature

vaporization

change from liquid state to gaseous state