Chapter 17 Thermochemistry 17.3 Heat in Changes of State 17.1 The Flow of Energy 17.2 Measuring and Expressing Enthalpy Changes 17.3 Heat in Changes of State 17.4 Calculating Heats of Reaction Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Why does sweating help cool you off? CHEMISTRY & YOU Why does sweating help cool you off? When your body heats up, you start to sweat. The evaporation of sweat is your body’s way of cooling itself to a normal temperature. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Heats of Fusion and Solidification What is the relationship between molar heat of fusion and molar heat of solidification? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using the Heat of Fusion in Phase-Change Calculations Sample Problem 17.5 Using the Heat of Fusion in Phase-Change Calculations How many grams of ice at 0°C will melt if 2.25 kJ of heat are added? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze List the knowns and the unknown. Sample Problem 17.5 Analyze List the knowns and the unknown. 1 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. KNOWNS UNKNOWN Initial and final temperature are 0°C ΔHfus = 6.01 kJ/mol ΔH = 2.25 kJ mice = ? g Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknown. Sample Problem 17.5 Calculate Solve for the unknown. 2 Start by expressing ΔHfus as a conversion factor. 1 mol H2O(s) 6.01 kJ Use the thermochemical equation H2O(s) + 6.01 kJ → H2O(l). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknown. Sample Problem 17.5 Calculate Solve for the unknown. 2 Express the molar mass of ice as a conversion factor. 18.0 g H2O(s) 1 mol H2O(s) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknown. Sample Problem 17.5 Calculate Solve for the unknown. 2 Multiply the known enthalpy change by the conversion factors. mice = 2.25 kJ   = 6.74 g H2O(s) 1 mol H2O(s) 6.01 kJ 18.0 g H2O(s) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Evaluate Does the result make sense? Sample Problem 17.5 Evaluate Does the result make sense? 3 To melt 1 mol of ice, 6.01 kJ of energy is required. Only about one-third of this amount of heat (roughly 2 kJ) is available. So, only about one-third mol of ice, or 18.0 g/3 = 6 g, should melt. This estimate is close to the calculated answer. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate the amount of heat absorbed to liquefy 15 Calculate the amount of heat absorbed to liquefy 15.0 g of methanol (CH4O) at its melting point. The molar heat of fusion for methanol is 3.16 kJ/mol. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate the amount of heat absorbed to liquefy 15 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 kJ 32.05 g CH4O 1 mol 3.16 kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Heats of Vaporization and Condensation What is the relationship between molar heat of vaporization and molar heat of condensation? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Heats of Physical Change Interpret Data 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 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Explain why the evaporation of sweat off your body helps cool you off. CHEMISTRY & YOU Explain why the evaporation of sweat off your body helps cool you off. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Explain why the evaporation of sweat off your body helps cool you off. CHEMISTRY & YOU Explain why the evaporation of sweat off your body helps cool you off. Energy is required to vaporize (or evaporate) a liquid into a gas. When liquid sweat absorbs energy from your skin, the temperature of your skin decreases. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Interpret Graphs 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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using the Heat of Vaporization in Phase-Change Calculations Sample Problem 17.6 Using the Heat of Vaporization in Phase-Change Calculations 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? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze List the knowns and the unknown. Sample Problem 17.6 Analyze List the knowns and the unknown. 1 First convert grams of water to moles of water. Then find the amount of heat that is absorbed when the liquid is converted to steam. KNOWNS UNKNOWN Initial and final conditions are 100°C and 101.3 kPa Mass of liquid water converted to steam = 24.8 g ΔHvap = 40.7 kJ/mol ΔH = ? kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknown. Sample Problem 17.6 Calculate Solve for the unknown. 2 Start by expressing the molar mass of water as a conversion factor. 18.0 g H2O(l) 1 mol H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknown. Sample Problem 17.6 Calculate Solve for the unknown. 2 Express ΔHvap as a conversion factor. 1 mol H2O(l) 40.7 kJ Use the thermochemical equation H2O(l) + 40.7 kJ → H2O(g). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknown. Sample Problem 17.6 Calculate Solve for the unknown. 2 Multiply the mass of water in grams by the conversion factors. ΔH = 24.8 g H2O(l)   = 56.1 kJ 18.0 g H2O(l) 1 mol H2O(l) 40.7 kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Evaluate Does the result make sense? Sample Problem 17.6 Evaluate Does the result make sense? 3 Knowing that the molar mass of water is 18.0 g/mol, 24.8 g H2O(l) can be estimated to be somewhat less than 1.5 mol H2O. The calculated enthalpy change should be a little less than 1.5 mol  40 kJ/mol = 60 kJ, and it is. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

B. molar heat of vaporization C. molar heat of solidification The molar heat of condensation of a substance is the same, in magnitude, as which of the following? A. molar heat of fusion B. molar heat of vaporization C. molar heat of solidification D. molar heat of formation Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

B. molar heat of vaporization C. molar heat of solidification The molar heat of condensation of a substance is the same, in magnitude, as which of the following? A. molar heat of fusion B. molar heat of vaporization C. molar heat of solidification D. molar heat of formation Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What thermochemical changes can occur when a solution forms? Heat of Solution Heat of Solution What thermochemical changes can occur when a solution forms? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Heat of Solution During the formation of a solution, heat is either released or absorbed. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Heat of Solution During the formation of a solution, heat is either released or absorbed. The enthalpy change caused by the dissolution of one mole of substance is the molar heat of solution (ΔHsoln). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CaCl2(s) → Ca2+(aq) + 2Cl–(aq) 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. CaCl2(s) → Ca2+(aq) + 2Cl–(aq) ΔHsoln = –82.8 kJ/mol Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

NH4NO3(s) → NH4+(aq) + NO3–(aq) 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 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculating the Enthalpy Change in Solution Formation Sample Problem 17.7 Calculating the Enthalpy Change in Solution Formation How much heat (in kJ) is released when 2.50 mol NaOH(s) is dissolved in water? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze List the knowns and the unknown. Sample Problem 17.7 Analyze List the knowns and the unknown. 1 Use the heat of solution for the dissolution of NaOH(s) in water to solve for the amount of heat released (ΔH). KNOWNS ΔHsoln = –44.5 kJ/mol amount of NaOH(s) dissolved = 2.50 mol UNKNOWN ΔH = ? kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

NaOH(s) → Na+(aq) + OH–(aq) + 44.5 kJ/mol. Sample Problem 17.7 Calculate Solve for the unknown. 2 Start by expressing ΔHsoln as a conversion factor. 1 mol NaOH(s) –44.5 kJ Use the thermochemical equation NaOH(s) → Na+(aq) + OH–(aq) + 44.5 kJ/mol. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknown. Sample Problem 17.7 Calculate Solve for the unknown. 2 Multiply the number of moles by the conversion factor. ΔH = 2.50 mol NaOH(s)  = –111 kJ 1 mol NaOH(s) –44.5 kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Evaluate Does the result make sense? Sample Problem 17.7 Evaluate Does the result make sense? 3 ΔH is 2.5 times greater than ΔHsoln, as it should be. Also, ΔH should be negative, as the dissolution of NaOH(s) in water is exothermic. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How much heat (in kJ) is absorbed when 50 How much heat (in kJ) is absorbed when 50.0 g of NH4NO3(s) are dissolved in water if Hsoln = 25.7 kJ/mol? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How much heat (in kJ) is absorbed when 50 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 kJ 80.04 g NH4NO3 1 mol 25.7 kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

END OF 17.3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.