Heat in Heat in Changes of StateChanges of State

Prentice-Hall Chapter 17.3Prentice-Hall Chapter 17.3Dr. YagerDr. Yager

ObjectivesObjectives

ClassifyClassify the enthalpy change that occurs the enthalpy change that occurs when a substance melts, freezes, boils, when a substance melts, freezes, boils, condenses, or dissolves.condenses, or dissolves.

SolveSolve for the enthalpy change that occurs for the enthalpy change that occurs when a substance melts, freezes, boils, when a substance melts, freezes, boils, condenses, or dissolves.condenses, or dissolves.

Enthalpy changes accompany changes in state.Enthalpy changes accompany changes in state.

Heats of Fusion and SolidificationHeats of Fusion and Solidification

The quantity of heat absorbed by a melting solid is the same as the quantity of heat released when the liquid solidifies.

The The molar heat of fusionmolar heat of fusion (∆(∆HHfusfus)) is the heat is the heat

absorbed by one mole of a solid substance as absorbed by one mole of a solid substance as it melts to a liquid at a constant temperature. it melts to a liquid at a constant temperature.

The The molar heat of solidificationmolar heat of solidification (∆(∆HHsolidsolid)) is is

the heat lost when one mole of a liquid the heat lost when one mole of a liquid solidifies at a constant temperature.solidifies at a constant temperature.

Heats of Fusion and SolidificationHeats of Fusion and Solidification

The quantity of heat absorbed by a melting solid is the same as the quantity of heat released when the liquid solidifies.

∆Hfus =

–∆Hsolid

How many kilojoules of heat are required to melt a 10.0-g popsicle at 0oC? Assume the popsicle has the same molar mass and heat of fusion as water.

moles of water = 10.0 g/(18 g/mol) = 0.56 mol

∆Hfus = 6.01 kJ/mol (melting is endothermic)

∆H = (0.56 mol)(6.01 kJ/mol)

= 3.34 kJ

How many kilojoules of heat are required to melt a 10.0-g popsicle at -20oC? Assume the popsicle has the same molar mass and heat of fusion as water.

moles of water = 10.0 g/(18mg/mol) = 0.56mol

∆Hfus = 6.01 kJ/mol (melting is endothermic)

First raise the temperature to 0oC, then melt:

∆Htotal = ∆Hraise the temp + ∆Hfus

= (10.0 g)[(2.1 J/(g·oC)](20oC) + (.56mol) x

(6.01 kJ/mol)

∆Htotal = 0.42 kJ + 3.34 kJ = 3.76 kJ

Enthalpy changes accompany changes in state.Enthalpy changes accompany changes in state.

Heats of Vaporization Heats of Vaporization and Condensationand Condensation

The quantity of heat absorbed by a vaporizing The quantity of heat absorbed by a vaporizing liquid is the same as the quantity of heat liquid is the same as the quantity of heat released when the vapor condenses.released when the vapor condenses.

The amount of heat necessary to vaporize one The amount of heat necessary to vaporize one mole of a given liquid is called its mole of a given liquid is called its molar heat of molar heat of vaporizationvaporization (∆(∆HHvapvap))..

The amount of heat released when 1 mol of The amount of heat released when 1 mol of vapor condenses at the normal boiling point is vapor condenses at the normal boiling point is called its called its molar heat of condensationmolar heat of condensation (∆(∆HHcondcond))..

Heats of Vaporization Heats of Vaporization and Condensationand Condensation

The quantity of heat absorbed by a vaporizing liquid is the same as the quantity of heat released when the vapor condenses.

∆Hvap = –∆Hcond

How many kilojoules of heat are absorbed when 0.46 g of chloroethane (C2H5Cl, bp 12.3oC) vaporizes at its normal boiling point? The molar heat of vaporization of chloroethane is 26.4 kJ/mol.

moles of C2H5Cl = 0.46 g/(64.5 g/mol)

= 7.1 x 10-3 mol

∆H = (26.4 kJ/mol)) x (7.1 x 10-3 mol) = 0.19 kJ

Heat of SolutionHeat of Solution

During the formation of a solution, heat is During the formation of a solution, heat is either released or absorbed. either released or absorbed.

The enthalpy change caused by dissolution of The enthalpy change caused by dissolution of one mole of substance is theone mole of substance is the molar heat of molar heat of solution (∆solution (∆HHsolnsoln))..

When ammonium nitrate crystals and water mix When ammonium nitrate crystals and water mix inside the cold pack, heat is absorbed as the inside the cold pack, heat is absorbed as the crystals dissolve.crystals dissolve.

How many moles of NH4NO3(s) must be dissolved in water so that 88.0 kJ of heat is absorbed from the water?

∆Hsol = 25.7 kJ/mol

moles of NH4NO3 = (88.0 kJ)/(25.7 kJ/mol)

= 3.42 mol

1. The molar heat of condensation of a 1. The molar heat of condensation of a substance is the same, in magnitude, substance is the same, in magnitude, as its molar heat ofas its molar heat of

a.a. formation.formation.

b.b. fusion.fusion.

c.c. solidification.solidification.

d.d. vaporization.vaporization.

1. The molar heat of condensation of a 1. The molar heat of condensation of a substance is the same, in magnitude, substance is the same, in magnitude, as its molar heat ofas its molar heat of

a.a. formation.formation.

b.b. fusion.fusion.

c.c. solidification.solidification.

d.d. vaporization.vaporization.

2. The heat of condensation of ethanol 2. The heat of condensation of ethanol (C(C22HH55OH) is OH) is 43.5 kJ/mol. As 43.5 kJ/mol. As

CC22HH55OH condenses, the temperature OH condenses, the temperature

of the surroundingsof the surroundings

• stays the same.stays the same.

• may increase or decrease.may increase or decrease.

• increases.increases.

• decreases.decreases.

2. The heat of condensation of ethanol 2. The heat of condensation of ethanol (C(C22HH55OH) is OH) is 43.5 kJ/mol. As 43.5 kJ/mol. As

CC22HH55OH condenses, the temperature OH condenses, the temperature

of the surroundingsof the surroundings

• stays the same.stays the same.

• may increase or decrease.may increase or decrease.

• increases.increases.

• decreases.decreases.

3. 3. Calculate the amount of heat absorbed Calculate the amount of heat absorbed to liquefy 15.0 g of methanol (CHto liquefy 15.0 g of methanol (CH33OH) at OH) at

its melting point. The molar heat of its melting point. The molar heat of fusion for methanol is 3.16 kJ/mol.fusion for methanol is 3.16 kJ/mol.

a.a. 1.48 kJ1.48 kJ

b.b. 47.4 kJ47.4 kJ

c.c. 1.52 1.52 10 1033 kJ kJ

d.d. 4.75 kJ4.75 kJ

3. 3. Calculate the amount of heat absorbed Calculate the amount of heat absorbed to liquefy 15.0 g of methanol (CHto liquefy 15.0 g of methanol (CH33OH) at OH) at

its melting point. The molar heat of its melting point. The molar heat of fusion for methanol is 3.16 kJ/mol.fusion for methanol is 3.16 kJ/mol.

a.a. 1.48 kJ1.48 kJ

b.b. 47.4 kJ47.4 kJ

c.c. 1.52 1.52 10 1033 kJ kJ

d.d. 4.75 kJ4.75 kJ

4.4. How much heat (in kJ) is absorbed How much heat (in kJ) is absorbed when 50 g of NHwhen 50 g of NH44NONO33((ss), 0.510 ), 0.510

moles, are dissolved in water? moles, are dissolved in water?

HHsoln soln = 25.7 kJ/mol= 25.7 kJ/mol

4.4. 12.85 kJ12.85 kJ

5.5. 13.1 kJ13.1 kJ

6.6. 25.7 kJ25.7 kJ

7.7. 1285 kJ1285 kJ

4.4. How much heat (in kJ) is absorbed How much heat (in kJ) is absorbed when 50 g of NHwhen 50 g of NH44NONO33((ss), 0.510 ), 0.510

moles, are dissolved in water? moles, are dissolved in water?

HHsoln soln = 25.7 kJ/mol= 25.7 kJ/mol

4.4. 12.85 kJ12.85 kJ

5.5. 13.1 kJ13.1 kJ

6.6. 25.7 kJ25.7 kJ

7.7. 1285 kJ1285 kJ