thermochemistry energy in state changes. copyright © pearson education, inc., or its affiliates....

Thermochemistry

Energy in State Changes

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Heats of Fusion and Solidification

•All solids absorb heat as they melt to become liquids.

• The gain of heat causes a change of state instead of a change in temperature.

• The temperature of the substance undergoing the change remains constant.

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Heats of Fusion and Solidification• The heat absorbed by one mole of a solid

substance as it melts to a liquid at constant temperature is the molar heat of fusion (ΔHfus).

• The molar heat of solidification (ΔHsolid) is the heat lost when one mole of a liquid substance solidifies at a constant temperature.

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Heats of Fusion and Solidification

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

ΔHfus = –ΔHsolid

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Heats of Fusion and Solidification• The melting of 1 mol of ice at 0°C to 1 mol

of liquid water at 0°C requires the absorption of 6.01 kJ of heat.

• The conversion of 1 mol of liquid water at 0°C to 1 mol of ice at 0°C releases 6.01 kJ of heat.

ΔHfus = 6.01 kJ/mol

ΔHsolid = –6.01 kJ/mol

H2O(s) → H2O(l)

H2O(l) → H2O(s)

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•How many grams of ice at 0°C will melt if 2.25 kJ of heat are added?

Using the Heat of Fusion in Phase-Change Calculations

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.

KNOWNS UNKNOWN

Initial and final temperature are 0°CΔHfus = 6.01 kJ/mol

ΔH = 2.25 kJ

mice = ? g

• Find the number of moles of ice that can be melted by the addition of 2.25 kJ of heat.

• Convert moles of ice to grams of ice.

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•Calculate the amount of heat absorbed to liquefy 15.6 g of methanol (CH4O) at its melting point. The molar heat of fusion for methanol is 3.16 kJ/mol.

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•Calculate the amount of heat absorbed to liquefy 15.6 g of methanol (CH4O) at its melting point. The molar heat of fusion for methanol is 3.16 kJ/mol.

ΔH = 15.6 g CH4O

= 1.54 kJ32.05 g CH4O

1 mol 3.16 kJ

1 mol

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Heats of Vaporization and Condensation

•A liquid that absorbs heat at its boiling point becomes a vapor.

• The amount of heat required to vaporize one mole of a given liquid at a constant temperature is called its molar heat of vaporization (ΔHvap).

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•This table lists the molar heats of vaporization for several substances at their normal boiling point.

Heats of Physical Change

Substance ΔHfus (kJ/mol) ΔHvap (kJ/mol)

Ammonia (NH3) 5.66 23.3

Ethanol (C2H6O) 4.93 38.6

Hydrogen (H2) 0.12 0.90

Methanol (CH4O) 3.22 35.2

Oxygen (O2) 0.44 6.82

Water (H2O) 6.01 40.7

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Heats of Vaporization and Condensation

•Condensation is the exact opposite of vaporization.

• When a vapor condenses, heat is released.

• The molar heat of condensation (ΔHcond) is the amount of heat released when one mole of vapor condenses at its normal boiling point.

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Heats of Vaporization and Condensation•The quantity of heat absorbed by a vaporizing liquid is exactly the same as the quantity of heat released when the vapor condenses.

ΔHvap = –ΔHcond

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•A heating curve graphically describes the enthalpy changes that take place during phase changes.

Remember: The temperature of a substance remains constant during a change of state.

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•How much heat (in kJ) is absorbed when 24.8 g H2O(l) at 100°C and 101.3 kPa is converted to H2O(g) at 100°C?

Using the Heat of Vaporization in Phase-Change Calculations

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Multiply the mass of water in grams by the conversion factors.

ΔH = 24.8 g H2O(l)

= 64.21 kJ

18.0 g H2O(l)

1 mol H2O(l)

1 mol H2O(l)

40.7 kJ

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Heat of Solution

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

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Heat of Solution

• The enthalpy change caused by the dissolution of one mole of substance is the molar heat of solution (ΔHsoln).

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

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CaCl2(s) → Ca2+(aq) + 2Cl–(aq)

ΔHsoln = –82.8 kJ/mol

Heat of Solution

•A practical application of an exothermic dissolution process is a hot pack.• In a hot pack, calcium chloride, CaCl2(s),

mixes with water, producing heat.

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Heat of Solution•The dissolution of ammonium nitrate, NH4NO3(s), is an example of an endothermic process.

• The cold pack shown here contains solid ammonium nitrate crystals and water.

• Once the solute dissolves, the pack becomes cold.

• The solution process absorbs energy from the surroundings.

NH4NO3(s) → NH4+(aq) + NO3

–(aq)

ΔHsoln = 25.7 kJ/mol

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•How much heat (in kJ) is released when 2.50 mol NaOH(s) is dissolved in water?

Calculating the Enthalpy Change in Solution Formation

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Multiply the number of moles by the conversion factor.

ΔH = 2.50 mol NaOH(s)

= –111 kJ 1 mol NaOH(s)

–44.5 kJ

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How much heat (in kJ) is absorbed when 50.0 g of NH4NO3(s) are dissolved in water if Hsoln = 25.7 kJ/mol?

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How much heat (in kJ) is absorbed when 50.0 g of NH4NO3(s) are dissolved in water if Hsoln = 25.7 kJ/mol?

ΔH = 50.0 g NH4NO3

= 16.1 kJ80.04 g NH4NO3

1 mol 25.7 kJ

1 mol

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Key Concepts

•The quantity of heat absorbed by a melting solid is exactly the same as the quantity of heat released when the liquid solidifies; that is, ΔHfus = –ΔHsolid.

•The quantity of heat absorbed by a vaporizing liquid is exactly the same as the quantity of heat released when the vapor condenses; that is, ΔHvap = –ΔHcond.

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

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Glossary Terms

• molar heat of fusion (ΔHfus): the amount of heat absorbed by one mole of a solid substance as it melts to a liquid at a constant temperature

• molar heat of solidification (ΔHsolid): the amount of heat lost by one mole of a liquid as it solidifies at a constant temperature

• molar heat of vaporization (ΔHvap): the amount of heat absorbed by one mole of a liquid as it vaporizes at a constant temperature

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Glossary Terms

• molar heat of condensation (ΔHcond): the amount of heat released by one mole of a vapor as it condenses to a liquid at a constant temperature

• molar heat of solution (ΔHsoln): the enthalpy change caused by the dissolution of one mole of a substance