Transcript Slide 1

EXAMPLE 12.1
Using the Heat of Vaporization in Calculations
Calculate the amount of water in grams that can be vaporized at its boiling point with 155 kJ of heat.
Set up the problem in the usual way. You are given
the number of kilojoules of heat energy and asked to
find the mass of water that can be vaporized with the
given amount of energy. The required conversion
factors are the heat of vaporization and the molar
mass of water.
Draw the solution map beginning with the energy in
kilojoules and converting to moles of water and then
to grams of water.
Follow the solution map to solve the problem.
SKILLBUILDER 12.1
Given: 155 kJ
Find: g H2O
Conversion Factors:
ΔHvap = 40.7 kJ/mol (at 100 °C)
18.02 g H2O = 1 mol H2O
Solution Map:
Solution:
Using the Heat of Vaporization in Calculations
Calculate the amount of heat in kilojoules required to vaporize 2.58 kg of water at its boiling point.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.1
Using the Heat of Vaporization in Calculations
Continued
SKILLBUILDER PLUS
A drop of water weighing 0.48 g condenses on the surface of a 55-g block of
aluminum that is initially at 25 °C. If the heat released during condensation
goes only toward heating the metal, what is the final temperature in Celsius of
the metal block? (The specific heat capacity of aluminum is 0.903 J/g °C.)
FOR MORE PRACTICE
Example 12.7; Problems 51, 52, 53, 54, 55, 56.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.2
Using the Heat of Fusion in Calculations
Calculate the amount of ice in grams that, upon melting (at 0 °C), absorbs 237 kJ of heat.
Set up the problem in the usual way. You are given
the number of kJ of heat energy and asked to find the
mass of ice that absorbs the given amount of energy
upon melting. The required conversion factors are the
heat of fusion and the molar mass of water.
Draw the solution map beginning with the energy in
kilojoules and converting to moles of water and then
to grams of water.
Follow the solution map to solve the problem.
SKILLBUILDER 12.2
Given: 237 kJ
Find: g H2O (ice)
Conversion Factors:
ΔHfus = 6.02 kJ/mol
1 mol H2O = 18.02 g H2O
Solution Map:
Solution:
Using the Heat of Fusion in Calculations
Calculate the amount of heat absorbed when a 15.5-g ice cube melts (at 0 °C).
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.2
Using the Heat of Fusion in Calculations
Continued
SKILLBUILDER PLUS
A 5.6-g ice cube (at 0 °C) is placed into 195 g of water initially at 25 °C. If
the heat absorbed for melting the ice comes only from the 195 g of water, what
is the temperature change of the 195 g of water?
FOR MORE PRACTICE
Example 12.8; Problems 59, 60, 61, 62.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.3
Dispersion Forces
Which halogen, Cl2 or I2 has the higher boiling point?
Solution:
The molar mass of Cl2 is 70.90 g/mol and the molar mass of I2 is 253.80 g/mol. Since I2 has the higher molar mass,
it has stronger dispersion forces and therefore the higher boiling point.
SKILLBUILDER 12.5
Dispersion Forces
Which hydrocarbon, CH4 or C2H6, has the higher boiling point?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 67, 68.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.4
Dipole–Dipole Forces
Which of the following molecules have dipole–dipole forces?
(a) CO2
(b) CH2Cl2
(c) CH4
Solution:
A molecule will have dipole–dipole forces if it is polar. To determine whether a molecule is polar, you must:
1. determine whether the molecule contains polar bonds, and
2. determine whether the polar bonds add together to form a net dipole moment (Section 10.8).
(a) Since the electronegativities of carbon and oxygen
are 2.5 and 3.5, respectively (Figure 10.2), CO2 has
polar bonds. The geometry of CO2 is linear.
Consequently, the polar bonds cancel; the molecule is
not polar and does not have dipole–dipole forces.
(b) The electronegativities of C, H, and Cl are 2.5,
2.1, and 3.5, respectively. Consequently, CH2Cl2 has
two polar bonds (C— Cl) and two bonds that are
nearly nonpolar (C — H). The geometry of CH2Cl2 is
tetrahedral. Since the C — Cl bonds and the C — H
bonds are different, they do not cancel, but sum to a
net dipole moment. Therefore the molecule is polar
and has dipole- dipole forces.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.4
Dipole–Dipole Forces
Continued
(c) Since the electronegativities of C and H are 2.5
and 2.1, respectively, the C — H bonds are nearly
nonpolar. In addition, since the geometry of the
molecule is tetrahedral, any slight polarities that the
bonds might have will cancel. CH4 is therefore
nonpolar and does not have dipole–dipole forces.
SKILLBUILDER 12.4
Dipole–Dipole Forces
Which of the following molecules have dipole–dipole forces?
(a) CI4
(b) CH3Cl
(c) HCl
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 63, 64, 65, 66.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.5
Hydrogen Bonding
One of the following compounds is a liquid at room temperature. Which one and why?
Solution:
The three compounds have similar molar masses.
formaldehyde
30.03 g/mol
fluoromethane
34.04 g/mol
hydrogen peroxide
34.02 g/mol
Therefore, the strengths of their dispersion forces are similar. All three compounds are also polar, so they have
dipole–dipole forces. Hydrogen peroxide, however, is the only compound that also contains H bonded directly to F,
O, or N. Therefore it also has hydrogen bonding and is most likely to have the highest boiling point of the three.
Since the problem stated that only one of the compounds was a liquid, we can safely assume that hydrogen peroxide
is the liquid. Note that although fluoromethane contains both H and F, H is not directly bonded to F, so
fluoromethane does not have hydrogen bonding as an intermolecular force. Similarly, although formaldehyde
contains both H and O, H is not directly bonded to O, so formaldehyde does not have hydrogen bonding either.
SKILLBUILDER 12.5
Hydrogen Bonding
Which has the higher boiling point, HF or HCl? Why?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 12.10; Problems 69, 70, 71, 72, 73, 74.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.6
Identifying Types of Crystalline Solids
Identify each of the following solids as molecular, ionic, or atomic.
(a) CaCl2(s)
(b) Co(s)
(c) CS2(s)
Solution:
(a) CaCl2 is an ionic compound (metal and nonmetal) and therefore forms an ionic solid (CaCl 2 melts at 772 °C).
(b) Co is a metal and therefore forms a metallic atomic solid (Co melts at 1768 °C).
(c) CS2 is a molecular compound (nonmetal bonded to a nonmetal) and therefore forms a molecular solid (CS 2 melts
at –110 °C).
SKILLBUILDER 12.6
Identifying Types of Crystalline Solids
Identify each of the following solids as molecular, ionic, or atomic.
(a) NH3(s)
(b) CaO(s)
(c) Kr(s)
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 77, 78, 79, 80.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.7
Using Heat of Vaporization in Calculations
Calculate the amount of heat required to vaporize 84.8 g of water at its boiling point.
Given: 84.8 g H2O
Find: heat (kJ)
Conversion Factors:
ΔHvap = 40.7 kJ/mol (at 100 °C)
1 mol H2O = 18.02 g H2O
Solution Map:
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.8
Using Heat of Fusion in Calculations
Calculate the amount of heat emitted when 12.4 g of water freezes to solid ice.
Given: 12.4 g H2O
Find: heat (kJ)
Conversion Factors:
ΔHfus = 6.02 kJ/mol
1 mol H2O = 18.02 g H2O
Solution Map:
Solution:
The heat emitted is 4.14 kJ.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.9
Determining the Types of Intermolecular Forces in a
Compound
Determine the types of intermolecular forces present in each of the following substances.
(a) N2
(b) CO
(c) NH3
Solution:
(a) N2 is nonpolar and therefore has only dispersion forces.
(b) CO is polar and therefore has dipole–dipole forces (in addition to dispersion forces).
(c) NH3 has hydrogen bonding (in addition to dispersion forces and dipole–dipole forces).
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 12.10
Using Intermolecular Forces to Determine Melting and/or
Boiling Points
Arrange each of the following in order of increasing boiling point.
(a) F2 , Cl2 , Br2
(b) HF, HCl, HBr
Solution:
(a) Since these all have only dispersion forces, and since they are similar substances (all halogens), the strength
of the dispersion force will increase with increasing molar mass. Therefore, the correct order is F 2 < Cl2 < Br2.
(b) Since HF has hydrogen bonding, it has the highest boiling point. Between HCl and HBr, HBr (because of its
higher molar mass) has a higher boiling point. Therefore the correct order is HCl < HBr < HF.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.