9.3.3: Calculations for Phase Changes

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Learning Objectives
• Calculate the energy change needed for a phase change.

For each phase change of a substance, there is a characteristic quantity of heat needed to perform the phase change per gram (or per mole) of material. The heat of fusion (ΔHfus) is the amount of heat per gram (or per mole) required for a phase change that occurs at the melting point. The heat of vaporization (ΔHvap) is the amount of heat per gram (or per mole) required for a phase change that occurs at the boiling point. If you know the total number of grams or moles of material, you can use the ΔHfus or the ΔHvap to determine the total heat being transferred for melting or solidification using these expressions:

$\text{heat} = n \times ΔH_{fus} \label{Eq1a}$

where $$n$$ is the number of moles and $$ΔH_{fus}$$ is expressed in energy/mole or

$\text{heat} = m \times ΔH_{fus} \label{Eq1b}$

where $$m$$ is the mass in grams and $$ΔH_{fus}$$ is expressed in energy/gram.

For the boiling or condensation, use these expressions:

$\text{heat} = n \times ΔH_{vap} \label{Eq2a}$

where $$n$$ is the number of moles) and $$ΔH_{vap}$$ is expressed in energy/mole or

$\text{heat} = m \times ΔH_{vap} \label{Eq2b}$

where $$m$$ is the mass in grams and $$ΔH_{vap}$$ is expressed in energy/gram.

Remember that a phase change depends on the direction of the heat transfer. If heat transfers in, solids become liquids, and liquids become solids at the melting and boiling points, respectively. If heat transfers out, liquids solidify, and gases condense into liquids.

Some ΔHfus values are listed in Table $$\PageIndex{1}$$; it is assumed that these values are for the melting point of the substance. Note that the unit of ΔHfus is kilojoules per mole, so we need to know the quantity of material to know how much energy is involved. The ΔHfus is always tabulated as a positive number. However, it can be used for both the melting and the freezing processes, minding that melting is always endothermic (so ΔH will be positive), while freezing is always exothermic (so ΔH will be negative).

Table $$\PageIndex{1}$$: Enthalpies of Fusion for Various Substances
Substance (Melting Point) ΔHfus (kJ/mol)
Water (0°C) 6.01
Aluminum (660°C) 10.7
Benzene (5.5°C) 9.95
Ethanol (−114.3°C) 5.02
Mercury (−38.8°C) 2.29
Example $$\PageIndex{1}$$

What is the energy change when 45.7 g of $$\ce{H2O}$$ melt at 0°C?

Solution

The $$ΔH_{fus}$$ of $$\ce{H2O}$$ is 6.01 kJ/mol. However, our quantity is given in units of grams, not moles, so the first step is to convert grams to moles using the molar mass of $$\ce{H_2O}$$, which is 18.0 g/mol. Then we can use $$ΔH_{fus}$$ as a conversion factor. Because the substance is melting, the process is endothermic, so the energy change will have a positive sign.

$45.7\cancel{g\: H_{2}O}\times \frac{1\cancel{mol\: H_{2}O}}{18.0\cancel{g}}\times \frac{6.01kJ}{\cancel{mol}}=15.3\,kJ \nonumber\nonumber$

Without a sign, the number is assumed to be positive.

Exercise $$\PageIndex{1}$$

What is the energy change when 108 g of $$\ce{C6H6}$$ freeze at 5.5°C?

−13.8 kJ

Like the solid/liquid phase change, the liquid/gas phase change involves energy. The amount of energy required to convert a liquid to a gas is called the enthalpy of vaporization (or heat of vaporization), represented as ΔHvap. Some ΔHvap values are listed in Table $$\PageIndex{2}$$; it is assumed that these values are for the normal boiling point temperature of the substance, which is also given in the table. The unit for ΔHvap is also kilojoules per mole, so we need to know the quantity of material to know how much energy is involved. The ΔHvap is also always tabulated as a positive number. It can be used for both the boiling and the condensation processes as long as you keep in mind that boiling is always endothermic (so ΔH will be positive), while condensation is always exothermic (so ΔH will be negative).

Table $$\PageIndex{1}$$: Enthalpies of Vaporization for Various Substances
Substance (Normal Boiling Point) ΔHvap (kJ/mol)
Water (100°C) 40.68
Bromine (59.5°C) 15.4
Benzene (80.1°C) 30.8
Ethanol (78.3°C) 38.6
Mercury (357°C) 59.23
Example $$\PageIndex{2}$$

What is the energy change when 66.7 g of Br2(g) condense to a liquid at 59.5°C?

Solution

The ΔHvap of Br2 is 15.4 kJ/mol. Even though this is a condensation process, we can still use the numerical value of ΔHvap as long as we realize that we must take energy out, so the ΔH value will be negative. To determine the magnitude of the energy change, we must first convert the amount of Br2 to moles. Then we can use ΔHvap as a conversion factor.

$66.7\cancel{g\: Br_{2}}\times \frac{1\cancel{mol\: Br_{2}}}{159.8\cancel{g}}\times \frac{15.4kJ}{\cancel{mol}}=6.43\,kJ \nonumber\nonumber$

Because the process is exothermic, the actual value will be negative: ΔH = −6.43 kJ.

Exercise $$\PageIndex{2}$$

What is the energy change when 822 g of $$\ce{C2H5OH(ℓ)}$$ boil at its normal boiling point of 78.3°C?

689 kJ

As with melting, the energy in boiling goes exclusively to changing the phase of a substance; it does not go into changing the temperature of a substance. So boiling is also an isothermal process. Only when all of a substance has boiled does any additional energy go to changing its temperature.

Sometimes, a table will contain both the heat of fusion and the heat of vaporization. If you are provided such a table as a reference for performing a calculation, make sure that you are selecting the correct constant from the table. Also pay attention to the units. In the first two tables you were provided units of kJ/mol, but in the following table the units are cal/g. Use the following table to answer the questions which follow it.

Table $$\PageIndex{1}$$: Heats of Fusion and Vaporization for Selected Substances
Substance ΔHfus (cal/g) ΔHvap (cal/g)
aluminum (Al) 94.0 2,602
gold (Au) 15.3 409
iron (Fe) 63.2 1,504
water (H2O) 79.9 540
sodium chloride (NaCl) 123.5 691
ethanol (C2H5OH) 45.2 200.3
benzene (C6H6) 30.4 94.1
Example $$\PageIndex{1}$$

How many cal are necessary to melt 55.8 g of ice (solid H2O) at 0°C?

Solution

We can use the relationship between heat and the heat of fusion (Eq. $$\PageIndex{1}$$b) to determine how many joules of heat are needed to melt this ice:

\begin{align*} \text{heat} &= m \times ΔH_{fus} \\[4pt] & = (55.8\: \cancel{g})\left(\dfrac{79.9\: cal}{\cancel{g}}\right)=4,460\: cal} \end{align*}

Exercise $$\PageIndex{1}$$

How many cal are necessary to vaporize 685 g of H2O at 100°C?