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17.11: Heats of Vaporization and Condensation

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  • Natural resources for power generation have traditionally been waterfalls or use of oil, coal, or nuclear power to generate electricity. Research is being carried out to look for other renewable sources to run the generators. Geothermal sites (such as the geyser pictured above) are being considered because of the steam they produce. Capabilities can be estimated by knowing how much steam is released in a given time at a particular site.

    Heat of Vaporization and Condensation

    Energy is absorbed in the process of converting a liquid at its boiling point into a gas. As with the melting point of a solid, the temperature of a boiling liquid remains constant and the input of energy goes into changing the state. The molar heat of vaporization \(\left( \Delta H_\text{vap} \right)\) of a substance is the heat absorbed by one mole of that substance as it is converted from a liquid to a gas. As a gas condenses to a liquid, heat is released. The molar heat of condensation \(\left( \Delta H_\text{cond} \right)\) of a substance is the heat released by one mole of that substance as it is converted from a gas to a liquid. Since vaporization and condensation of a given substance are the exact opposite processes, the numerical value of the molar heat of vaporization is the same as the numerical value of the molar heat of condensation, but opposite in sign. In other words, \(\Delta H_\text{vap} = -\Delta H_\text{cond}\).

    When \(1 \: \text{mol}\) of water at \(100^\text{o} \text{C}\) and \(1 \: \text{atm}\) pressure is converted to \(1 \: \text{mol}\) of water vapor at \(100^\text{o} \text{C}\), \(40.7 \: \text{kJ}\) of heat is absorbed from the surroundings. When \(1 \: \text{mol}\) of water vapor at \(100^\text{o} \text{C}\) condenses to liquid water at \(100^\text{o} \text{C}\), \(40.7 \: \text{kJ}\) of heat is released into the surroundings.

    \[\begin{array}{ll} \ce{H_2O} \left( l \right) \rightarrow \ce{H_2O} \left( g \right) & \Delta H_\text{vap} = 40.7 \: \text{kJ/mol} \\ \ce{H_2O} \left( g \right) \rightarrow \ce{H_2O} \left( l \right) & \Delta H_\text{cond} =-40.7 \: \text{kJ/mol} \end{array}\]

    Other substances have different values for their molar heats of fusion and vaporization and these substances are summarized in the table below.

    Table 17.11.1: Molar Heats of Fusion and Vaporization
    Substance \(\Delta H_\text{fus}\) \(\left( \text{kJ/mol} \right)\) \(\Delta H_\text{vap}\) \(\left( \text{kJ/mol} \right)\)
    Ammonia \(\left( \ce{NH_3} \right)\) 5.65 23.4
    Ethanol \(\left( \ce{C_2H_5OH} \right)\) 4.60 43.5
    Methanol \(\left( \ce{CH_3OH} \right)\) 3.16 35.3
    Oxygen \(\left( \ce{O_2} \right)\) 0.44 6.82
    Water \(\left( \ce{H_2O} \right)\) 6.01 40.7

    Notice that for all substances, the heat of vaporization is substantially higher than the heat of fusion. Much more energy is required to change the state from a liquid to a gas than from a solid to a liquid. This is because of the large separation of the particles in the gas state. The values of the heats of fusion and vaporization are related to the strength of the intermolecular forces. All of the substances in the table above, with the exception of oxygen, are capable of hydrogen bonding. Consequently, the heats of fusion and vaporization of oxygen are far lower than the others.

    Example 17.11.1

    What mass of methanol vapor condenses to a liquid as \(20.0  \: \text{kJ}\) of heat is released?

    Solution:

    Step 1: List the know quantities and plan the problem.

    Known

    • \(\Delta H = 20.0 \: \text{kJ}\)
    • \(\Delta H_\text{cond} = -35.3 \: \text{kJ/mol}\)
    • Molar mass \(\ce{CH_3OH} = 32.05 \: \text{kJ/mol}\)

    Unknown

    • Mass methanol \(= ? \: \text{g}\)

    First the \(\text{kJ}\) of heat released in the condensation is multiplied by the conversion factor \(\left( \frac{1 \: \text{mol}}{-35.3 \: \text{kJ}} \right)\) to find the moles of methanol that condensed. Then, moles are converted to grams.

    Step 2: Solve.

    \[-20.0 \: \text{kJ} \times \frac{1 \: \text{mol} \: \ce{CH_3OH}}{-35.3 \: \text{kJ}} \times \frac{32.05 \: \text{g} \: \ce{CH_3OH}}{1 \: \text{mol} \: \ce{CH_3OH}} = 18.2 \: \text{g} \: \ce{CH_3OH}\]

    Step 3: Think about your result.

    Condensation is an exothermic process, so the enthalpy change is negative. Slightly more than one-half mole of methanol is condensed.

    Summary

    • Molar heats of condensation and vaporization are defined.
    • Examples of calculations involving these parameters are illustrated.

    Contributors

    • CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.