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# 17.3: Heat and Changes of State

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Describe the enthalpy change that occurs as a substance changes between the solid and liquid states.
• Describe the enthalpy change that occurs as a substance changes between the liquid and gas states.
• Calculate the enthalpy change involved in the change of state for any amount of a given substance.
• Calculate the enthalpy changes involved as substances dissolve in water.

## Lesson Vocabulary

• molar heat of condensation
• molar heat of fusion
• molar heat of solidification
• molar heat of solution
• molar heat of vaporization

### Recalling Prior Knowledge

• What happens to the temperature of a system during a change of state?
• What information is contained in a heating or cooling curve?

Enthalpy changes accompany physical processes as well as chemical reactions. In this lesson, you will examine the heat absorbed or released during changes of state as well as the enthalpy change that occurs when a solute is dissolved into a solvent.

## Heats of Fusion and Solidification

Suppose you hold an ice cube in your hand. It feels cold because heat energy leaves your hand and enters the ice cube. What happens to the ice cube? It melts. However, as you learned in an earlier chapter, the temperature during a phase change remains constant. So the heat that is being lost by your hand does not raise the temperature of the ice above its melting temperature of 0°C. Rather, all the heat goes into the change of state. Energy is absorbed during the process of changing ice into water. The water that is produced also remains at 0°C until all of the ice has melted.

All solids absorb heat as they melt to become liquids. The gain of heat in this endothermic process goes into changing the state rather than changing the temperature. The molar heat of fusion (ΔHfus) of a substance is the heat absorbed by one mole of that substance as it is converted from a solid to a liquid. Since the melting of any substance absorbs heat, it follows that the freezing of any substance releases heat. The molar heat of solidification (ΔHsolid) of a substance is the heat released by one mole of that substance as it is converted from a liquid to a solid. Since fusion and solidification of a given substance are the exact opposite processes, the numerical value of the molar heat of fusion is the same as the numerical value of the molar heat of solidification, but opposite in sign. In other words, ΔHfus = −ΔHsolid. Shown below (Figure below) are all of the possible changes of state along with the direction of heat flow during each process.

From left to right, heat is absorbed from the surroundings during melting, evaporation, and sublimation. From right to left, heat is released to the surroundings during freezing, condensation, and deposition.

Every substance has a unique value for its molar heat of fusion, depending on the amount of energy required to disrupt the intermolecular forces present in the solid. When 1 mol of ice at 0°C is converted to 1 mol of liquid water at 0°C, 6.01 kJ of heat are absorbed from the surroundings. When 1 mol of water at 0°C freezes to ice at 0°C, 6.01 kJ of heat are released into the surroundings.

H2O(s)H2O(l)   ΔHfus=6.01 kJ/mol\begin{align*}\text{H}_2\text{O}{(s)} \rightarrow \text{H}_2\text{O}{(l)} \ \ \ \Delta \text{H}_{\text{fus}}=6.01 \ \text{kJ/mol}\end{align*}
H2O(l)H2O(s)   ΔHsolid=6.01 kJ/mol\begin{align*}\text{H}_2\text{O}{(l)} \rightarrow \text{H}_2\text{O}{(s)} \ \ \ \Delta \text{H}_{\text{solid}}=-6.01 \ \text{kJ/mol}\end{align*}

The molar heats of fusion and solidification of a given substance can be used to calculate the heat absorbed or released when various amounts are melted or frozen.

Sample Problem 17.4: Heat of Fusion

Calculate the heat absorbed when 31.6 g of ice at 0°C is completely melted.

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

Known

• mass of ice = 31.6 g
• molar mass of H2O = 18.02 g/mol
• molar heat of fusion (H2O) = 6.01 kJ/mol

Unknown

• ΔH = ? J

The mass of ice is first converted to moles. This is then multiplied by the conversion factor of (6.01 kJ/1 mol) in order to find the kJ of heat absorbed.

Step 2: Solve.

31.6 g ice×1 mol ice18.02 g ice×6.01 kJ1 mol ice=10.5 kJ\begin{align*}\mathrm{31.6 \ g \ ice \times \dfrac{1 \ mol \ ice}{18.02 \ g \ ice} \times \dfrac{6.01 \ kJ}{1 \ mol \ ice}=10.5 \ kJ}\end{align*}

The given quantity is a bit less than 2 moles of ice, so just under 12 kJ of heat is absorbed by the melting process.

Practice Problems
1. What mass of ice at 0°C can be melted by the addition of 559 J of heat?
2. What is ΔH when 250. g of water is frozen?

## Heat of Vaporization and Condensation

Energy is also absorbed when a liquid is converted into a gas. As with the melting 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 (ΔHvap) 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 (ΔHcond) 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, ΔHvap = −ΔHcond.

When 1 mol of water at 100°C and 1 atm pressure is converted to 1 mol of water vapor at 100°C, 40.7 kJ of heat are absorbed from the surroundings. When 1 mol of water vapor at 100°C condenses to liquid water at 100°C, 40.7 kJ of heat are released into the surroundings.

H2O(l)H2O(g)   ΔHvap=40.7 kJ/mol\begin{align*}\text{H}_2\text{O}{(l)} \rightarrow \text{H}_2\text{O}{(g)} \ \ \ \Delta \text{H}_{\text{vap}}=40.7 \ \text{kJ/mol}\end{align*}
H2O(g)H2O(l)   ΔHcond=40.7 kJ/mol\begin{align*}\text{H}_2\text{O}{(g)} \rightarrow \text{H}_2\text{O}{(l)} \ \ \ \Delta \text{H}_{\text{cond}}=-40.7 \ \text{kJ/mol}\end{align*}

Molar heats of fusion and vaporization for some other substances are given in the table below (Table below).

Molar Heats of Fusion and Vaporization
Substance ΔHfus (kJ/mol) ΔHvap (kJ/mol)
Ammonia (NH3) 5.65 23.4
Ethanol (C2H5OH) 4.60 43.5
Methanol (CH3OH) 3.16 35.3
Oxygen (O2) 0.44 6.82
Water (H2O) 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 listed above (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.

Sample Problem 17.5: Heat of Vaporization

What mass of methanol vapor condenses to a liquid as 20.0 kJ of heat are released?

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

Known

• ΔH = -20.0 kJ
• ΔHcond = −35.3 kJ/mol
• molar mass of CH3OH = 32.05 g/mol

Unknown

• mass of methanol = ? g

First, the amount of heat released in the condensation is multiplied by the conversion factor of (1 mol/−35.3 kJ) to find the moles of methanol that condensed. Then, moles are converted to grams.

Step 2: Solve.

20.0 kJ×1 mol CH3OH35.3 kJ×32.05 g CH3OH1 mol CH3OH=18.2 g CH3OH\begin{align*}\mathrm{-20.0 \ kJ \times \dfrac{1 \ mol \ CH_3OH}{-35.3 \ kJ} \times \dfrac{32.05 \ g \ CH_3OH}{1 \ mol \ CH_3OH}=18.2 \ g \ CH_3OH}\end{align*}

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

Practice Problem
1. How much heat is absorbed when 1.00 g of liquid ammonia is vaporized at its boiling point?

### Multi-Step Problems with Changes of State

In the chapter States of Matter, you learned about heating curves and how they show the phase changes that a substance undergoes as heat is continuously absorbed.

Heating curve of water. The process of converting ice below its melting point to steam above its boiling point involves five distinct steps. The enthalpy change for each step can be calculated separately.

In the previous lesson, Thermochemical Equations, you learned how to use the specific heat of a substance to calculate the heat absorbed or released as the temperature of the substance changes. It is possible to combine that type of problem with a change of state to solve a problem involving multiple steps. The figure above (Figure above) shows ice at −30°C being converted in a five-step process to gaseous water (steam) at 140°C. It is now possible to calculate the heat absorbed during that entire process. The process and the required calculation is summarized below.

1. Ice is heated from −30°C to 0°C. The heat absorbed is calculated by using the specific heat of ice and the equation ΔH = m × cp × ΔT.
2. Ice is melted at 0°C. The heat absorbed is calculated by multiplying the moles of ice by the molar heat of fusion.
3. Water at 0°C is heated to 100°C. The heat absorbed is calculated by using the specific heat of water and the equation ΔH = m × cp × ΔT.
4. Water is vaporized to steam at 100°C. The heat absorbed is calculated by multiplying the moles of water by the molar heat of vaporization.
5. Steam is heated from 100°C to 140°C. The heat absorbed is calculated by using the specific heat of steam and the equation ΔH = m × cp × ΔT.

Sample Problem 17.6: Multi-Step Problems using a Heating Curve

Calculate the total amount of heat absorbed (in kJ) when 2.00 mol of ice at −30.0°C is converted to steam at 140.0°C. The required specific heats can be found in the table above (Table above).

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

Known

• 2.00 mol ice = 36.04 g ice
• cp (ice) = 2.06 J/g•°C
• cp (water) = 4.18 J/g•°C
• cp (steam) = 1.87 J/g•°C
• ΔHfus = 6.01 kJ/mol
• ΔHvap = 40.7 kJ/mol

Unknown

• ΔHtotal = ? kJ

Follow the five steps outlined in the text. Note that the mass of the water is needed for the calculations that involve the specific heat, while the moles of water is needed for the calculations that involve changes of state. All heat quantities must be in kilojoules so that they can be added together to get a total for the five-step process.

Step 2: Solve.

1. ΔH1=36.04 g×2.06 J/gC×30C×1 kJ1000 J=2.23 kJ\begin{align*}\mathrm{\Delta H_1=36.04 \ g \times 2.06 \ J/g\cdot^{\circ}C \times 30^{\circ}C \times \dfrac{1 \ kJ}{1000 \ J}=2.23 \ kJ}\end{align*}
2. ΔH2=2.00 mol×6.01 kJ1 mol=12.0 kJ\begin{align*}\mathrm{\Delta H_2=2.00 \ mol \times \dfrac{6.01 \ kJ}{1 \ mol}=12.0 \ kJ}\end{align*}
3. ΔH3=36.04 g×4.18 J/gC×100C×1 kJ1000 J=15.1 kJ\begin{align*}\mathrm{\Delta H_3=36.04 \ g \times 4.18 \ J/g\cdot^{\circ}C \times 100^{\circ}C \times \dfrac{1 \ kJ}{1000 \ J}=15.1 \ kJ}\end{align*}
4. ΔH4=2.00 mol×40.7 kJ1 mol=81.4 kJ\begin{align*}\mathrm{\Delta H_4=2.00 \ mol \times \dfrac{40.7 \ kJ}{1 \ mol}=81.4 \ kJ}\end{align*}
5. ΔH3=36.04 g×1.87 J/gC×40C×1 kJ1000 J=2.70 kJ\begin{align*}\mathrm{\Delta H_3=36.04 \ g \times 1.87 \ J/g\cdot^{\circ}C \times 40^{\circ}C \times \dfrac{1 \ kJ}{1000 \ J}=2.70 \ kJ}\end{align*}

ΔHtotal=ΔH1+ΔH2+ΔH3+ΔH4+ΔH5=113.4 kJ\begin{align*}\mathrm{\Delta H_{total}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5=113.4 \ kJ}\end{align*}

The total heat absorbed as the ice at −30°C is heated to steam at 140°C is 113.4 kJ. By far, the largest absorption of heat comes during the vaporization of the liquid water.

Practice Problem
1. Calculate the heat released when 100.0 g of water at 35.0°C is converted to ice at −18.0°C.

## Heat of Solution

Enthalpy changes also occur when a solute undergoes the physical process of dissolving into a solvent. Hot packs and cold packs (Figure below) use this property. Many hot packs use calcium chloride, which releases heat when it dissolves according to the equation below.

CaCl2(s)Ca2+(aq)+2Cl(aq)+82.8 kJ\begin{align*}\text{CaCl}_{2}{(s)} \rightarrow \text{Ca}^{2+}{(aq)}+2\text{Cl}^-{(aq)}+82.8 \ \text{kJ}\end{align*}

The molar heat of solution (ΔHsoln) of a substance is the heat absorbed or released when one mole of the substance is dissolved in water. For calcium chloride, ΔHsoln = −82.8 kJ/mol.

Chemical hot packs and cold packs work because of the heats of solution of the chemicals inside them. When the bag is squeezed, an inner pouch bursts, allowing the chemical to dissolve in water. Heat is released in a hot pack and absorbed by a cold pack.

Many cold packs use ammonium nitrate, which absorbs heat from the surroundings when it dissolves.

NH4NO3(s)+25.7 kJNH+4(aq)+NO3(aq)\begin{align*}\text{NH}_4\text{NO}_{3}{(s)}+25.7 \ \text{kJ} \rightarrow \text{NH}^+_{4}{(aq)}+ \text{NO}^-_{3}{(aq)}\end{align*}

Cold packs are typically used to treat muscle strains and sore joints. The cold pack is activated and applied to the affected area. As the ammonium nitrate dissolves, it absorbs heat from the body and helps limit swelling. For ammonium nitrate, ΔHsoln = 25.7 kJ/mol.

Sample Problem 17.7: Heat of Solution

The molar heat of solution, ΔHsoln, of NaOH is −445.1 kJ/mol. In a certain experiment, 5.00 g of NaOH is completely dissolved in 1.000 L of water at 20.0°C in a foam cup calorimeter. Assuming no heat loss, calculate the final temperature of the water.

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

Known

• mass of NaOH = 5.00 g
• molar mass of NaOH = 40.00 g/mol
• ΔHsoln (NaOH) = −445.1 kJ/mol
• mass H2O = 1.000 kg = 1000. g (assumes density = 1.000 g/mL)
• Tinitial (H2O) = 20.0°C
• cp (H2O) = 4.18 J/g•°C

Unknown

• Tfinal of H2O = ? °C

This is a multiple-step problem: 1) the mass of NaOH is converted to moles; 2) the resulting value is multiplied by the molar heat of solution; 3) the heat released in the dissolving process is used with the specific heat equation and the total mass of the solution to calculate ΔT; 4) Tfinal is determined from ΔT.

Step 2: Solve.

5.00 g NaOH×1 mol NaOH40.00 g NaOH×445.1 kJ1 mol NaOH×1000 J1 kJ=5.56×104 J\begin{align*}\mathrm{5.00 \ g \ NaOH \times \dfrac{1 \ mol \ NaOH}{40.00 \ g \ NaOH} \times \dfrac{-445.1 \ kJ}{1 \ mol \ NaOH} \times \dfrac{1000 \ J}{1 \ kJ}=-5.56 \times 10^4 \ J}\end{align*}
ΔT=qcp×m=5.56×104 J4.18 J/gC×1005 g=13.2C\begin{align*}\mathrm{\Delta T=\dfrac{q}{c_p \times m}=\dfrac{-5.56 \times 10^4 \ J}{4.18 \ J/g\cdot^{\circ}C \times 1005 \ g}=13.2^{\circ}C}\end{align*}
Tfinal=20.0C+13.2C=33.2C\begin{align*}\mathrm{T_{final}=20.0^{\circ}C+13.2^{\circ}C=33.2^{\circ}C}\end{align*}

The dissolving process releases a large amount of heat, which causes the temperature of the solution to rise. Care must be taken when preparing concentrated solutions of sodium hydroxide because of the large amounts of heat released.

Practice Problem
1. Calculate the final temperature of a solution prepared by dissolving 18.40 g of ammonium nitrate into 250. g of water in a foam cup calorimeter. Both the ammonium nitrate and the water are initially at 22.0°C. Assume that the heat capacity of the solution is the same as that of pure water.

## Lesson Summary

• The molar heat of fusion is the heat absorbed when one mole of a solid melts and is numerically equivalent to the molar heat of solidification, the heat released when one mole of the liquid freezes.
• The molar heat of vaporization is the heat absorbed when one mole of a liquid boils and is numerically equivalent to the molar heat of condensation, the heat released when one mole of the gas condenses.
• Conversion factors can be used to calculate the heat absorbed or released during any change of state. Multiple-step problems, such as ice being transformed into steam, can be solved using specific heats along with heats of fusion and vaporization.
• The molar heat of solution is the heat absorbed or released when one mole of a solute dissolves in water.

## Lesson Review Questions

### Reviewing Concepts

1. How does the molar heat of fusion of a substance compare to the molar heat of solidification?
2. How does the molar heat of vaporization of a substance compare to the molar heat of condensation?
3. State the names given to the following changes of state, and classify each as being endothermic or exothermic.
1. Ar(g) → Ar(l)
2. KBr(s) → KBr(l)
3. C4H10(l) → C4H10(g)
4. CO2(s) → CO2(g)
5. Br2(l) → Br2(s)
4. Explain why the temperature of ice at 0°C does not initially rise when heat is added to it.
5. The molar heat of solution of ammonia is −30.50 kJ/mol. What happens to the temperature of water when ammonia is added to it?

### Problems

1. Heats of fusion and vaporization can also be expressed in J/g or kJ/g.
1. Use the molar mass of H2O to convert the molar heat of fusion and the molar heat of vaporization of water to kJ/g.
2. Calculate the energy required to melt 50.0 g of ice at 0°C and to boil 50.0 g of water at 100°C.
2. Calculate the quantity of heat that is absorbed or released (in kJ) during each of the following processes.
1. 655 g of water vapor condenses at 100°C.
2. 3.25 g of CaCl2(s) is dissolved in water.
3. 8.20 kg of water is frozen.
4. 40.0 mL of ethanol is vaporized at 78.3°C (its boiling point). The density of ethanol is 0.789 g/mL.
3. Various systems are each supplied with 9.25 kJ of heat. Calculate the mass of each substance that will undergo the indicated process with this input of heat.
1. melt ice at 0°C
2. vaporize water at 100°C
3. dissolve NH4NO3 in water
4. vaporize water originally at 25.0°C
4. Calculate the total amount of heat (in kJ) absorbed by the process of converting 4.33 g of ice at −60.0°C to steam at 185°C.
5. An ice cube at 0°C was dropped into a foam cup containing 100. g of water at 40.0°C. The final temperature of the water in the cup is 21.2°C. Assuming no heat loss, what was the mass of the ice cube?
6. The molar heat of solution of NaCl is 3.88 kJ/mol. 125 g of NaCl is dissolved in 500. g of water at 20.0°C in a calorimeter. What is the temperature of the solution when the salt has completely dissolved? Assume the specific heat of the solution is the same as that of water.

## Points to Consider

A formation reaction is a reaction in which elements in their standard states are combined to form a compound.

• How can knowledge of heats of formation be used to determine the enthalpy change for any reaction?
• What does it mean to say that heats of reaction are additive?

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