13 Liquids and Solids. 2 Chapter Goals 1.Kinetic-Molecular Description of Liquids and Solids 2.Intermolecular Attractions and Phase Changes 分子間引力及相的改變

1313Liquids and

Solids

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Chapter GoalsChapter Goals1.Kinetic-Molecular Description of Liquids and

Solids2.Intermolecular Attractions and Phase Changes

分子間引力及相的改變The Liquid State

3.Viscosity 黏度4.Surface Tension 表面張力5.Capillary Action 毛細現象6.Evaporation 蒸發7.Vapor Pressure 蒸氣壓8.Boiling Points and Distillation 沸點和蒸餾9.Heat Transfer Involving Liquids 液體之熱傳導

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Chapter GoalsChapter GoalsThe Solid State

10.Melting Point 熔點11.Heat Transfer Involving Solids 固體之熱傳導12.Sublimation and the Vapor Pressure of

Solids 昇華及固體之蒸氣壓13.Phase Diagrams (P versus T) 相圖14.Amorphous Solids and Crystalline Solids

非晶固態及結晶固態15.Structures of Crystals 結晶的結構16.Bonding in Solids 固體之鍵結17.Band Theory of Metals 金屬能帶理論

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Kinetic-Molecular Description Kinetic-Molecular Description of Liquids and Solidsof Liquids and Solids

• Solids and liquids are condensed states 壓縮狀態– The atoms, ions, or molecules in solids

and liquids are much closer to one another than in gases.

– Solids and liquids are highly incompressible 不能壓縮的

• Liquids and gases are fluids 流動的 .– They easily flow.

• The intermolecular attractions in liquids and solids are strong. 液體及固體之分子間的引力很強

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• Schematic representation of the three common states of matter.

The process in which a liquid changes to a solid• Solidification• Crystallization• A more specific term• The formation of a very ordered solid material

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• If we compare the strengths of interactionsstrengths of interactions among particles and the degree of orderingdegree of ordering of particles, we see that

Gases< Liquids < Solids• MiscibleMiscible 可溶混的 liquidsliquids are soluble in each

other. – Examples of miscible liquids:

•Water dissolves in alcohol.•A drop of red ink in the water•Gasoline dissolves in motor oil.

Diffusion

The miscibility of two liquids refers to their ability to mix and produced a homogeneous solution 均質溶液

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Diffusion in solids: very slowly

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• Immiscible liquidsImmiscible liquids are insoluble in each other.– Two examples of immiscible liquids:

•Water does not dissolve in oil.•Water does not dissolve in cyclohexane 環已烷

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Intermolecular Attractions and Intermolecular Attractions and Phase ChangesPhase Changes

Covalent bonds共價鍵結

• Intermolecular forces 分子間作用力– The forces between individual particles

(atoms, molecules, ions) of a substance• Intramolecular forces 分子內作用力

– Covalent and ionic bonds within compounds

Intermolecular forces

Intramolecular forces

hydrogen bonding氫鍵

•分子間吸引力是物質中個別粒子 ( 包括原子、分子或離子 ) 間的作用力，此種作用力遠小於分子內作用力，即化合物內原子間所形成的共價鍵與離子鍵結。

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•There are four important intermolecular attractions.–This list is from strongest attraction to the weakest attraction.

1.Ion-ion interactions 離子 - 離子作用力–The force of attraction between two oppositely charged ions is governed by Coulomb’s law 庫侖定律 .

(q+)(q-)d2

q+ and q- are the ion charges.d is the distance between the e ionsF is the force of attraction

F

(q+)(q-)d

E

Energy has the units of forces x distance, F x dThe energy of attraction, E

※ 庫侖定律：在真空中，兩個靜止點電荷之間的相互作用力的大小與兩點電荷電量大小乘積成正比，與距離平方成反比。


• Most ionic bonding is stronghave relatively high melting points• Ionic substances containing multiply charged

ions, such as Al3+, Mg2+, O2-, and S2-, usually have higher melting and boiling points than ionic compounds containing only singly charged ions, such as Na+, K+, F-, Cl-.

• For a series of ions of similar charges, the closer approached ions of smaller ions result in stronger interionic attractive forces and higher melting points (NaF, NaCl, NaBr)

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Intermolecular Intermolecular Attractions and Phase Attractions and Phase

ChangesChanges• Coulomb’s law determines: 1.The melting and boiling points of ionic

compounds.2.The solubility of ionic compounds.

Example 13-1: Arrange the following ionic compounds in the expected order of increasing melting and boiling points.

NaF, CaO, CaF2

Na+F- < Ca2+F2-<Ca2+O2-

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2.Dipole-dipole interactions 偶極 - 偶極作用–在極性分子中 , 有正及負兩極 , 分子帶正電之一端吸引另一分子帶負電的一端 , 即稱之

–使得極性分子間的引力大於非極性分子–偶極─偶極吸引力一般僅是共價鍵及離子鍵強度的百分之一，它們的強度會隨偶極之間距離的增加而變小

– BrF

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3.Hydrogen bonding 氫鍵– Very strong dipole-dipole interaction– Strong hydrogen bonding occurs among polar

covalent molecules containing H (hydrogen bond donor) and one of the three small, highly electronegative elements — F, O or N (hydrogen bond acceptor)

– Consider H2O a very polar molecule.

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Hydrogen bond


ChangesChanges

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ChangesChanges4.London Forces (Dispersion forces; 分散力 )

– They are the weakest of the intermolecular forces.

– This is the only attractive force in nonpolar molecules such as O2, N2 and monatomic species such as the noble gases.

– Without dispersion forces, such substance could not condense to form liquids or solidify to form solids

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ChangesChanges• In a group of Ar atoms the temporary

dipole in one atom induces other atomic dipoles.– The positively charged nucleus– The electron clouds of an atom in

nearby molecules– Dispersion forces are generally

stronger for molecules that are larger– They exist in all substances

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倫敦分散力 (London dispersion forces)

非極性原子或分子間的作用力稱為倫敦分散力，此種作用力是原子或分子在移動時，其電子雲會產生瞬間偶極距 (instantaneous dipole moments) ，此時非極性原子或分子間藉由此種瞬間偶極相互吸引，稱之為倫敦分散力，又可稱之為瞬間偶極距 - 瞬間偶極距作用力。

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分子間作用力的比較倫敦分散力存在於所有分子中，其強度隨分子量的增加而增加，也與分子的形狀有關。除了倫敦分散力，極性分子間存在的另一種作用力稱為偶極 - 偶極作用力。氫鍵則是鍵結在高電負度原子上之氫原子與另一高電負度原子間之作用力，是所有分子間作用力中最強的一種作用力。不過上述不論那種分子間作用力，其強度皆比共價鍵或離子鍵弱很多。

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恆久偶極矩

作用力的種類作用力發生的原因例子

偶極 - 偶極作用力具有偶極矩 ( 偶極矩大小是根據電負度與分子結構來決定 )

H2O 、HCl

氫鍵XHY

當具強極性 XH 鍵與帶孤對電子的 Y 原子同時存在時會發生，而這可說是一種極端的偶極 - 偶極作用力 ( 當X F、 N 、O)

H2O H2O

偶極 - 誘導偶極作用力

極性分子與非極性分子同時存在時會發生 H2OI2

倫敦分散力發生於分子間 I2I2

分子間作用力

分子間作用力的判斷


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Example 13-1: Intermolecular ForcesIdentify the type of intermolecular forces that are

present in a condensed phase sample of each of the following. For each, make a sketch, including a few molecules, that represents the major type of force. (a) water, H2O (b) iodine, I2 (c) nitrogen dioxide, NO2.

(a)water, H2O •polar •H, O hydrogen bonds•The London forces

(b) iodine, I2•Nonpolar the London forces

(c) nitrogen dioxide, NO2•polar •Dipole-dipole interactions•The London forces

Example 13-1: Intermolecular ForcesIdentify the type of intermolecular forces that are

present in a condensed phase sample of each of the following. For each, make a sketch, including a few molecules, that represents the major type of force. (a) water, H2O (b) iodine, I2 (c) nitrogen dioxide, NO2.

(a)water, H2O •polar •H, O hydrogen bonds•The London forces

(b) iodine, I2•Nonpolar the London forces

(c) nitrogen dioxide, NO2•polar •Dipole-dipole interactions•The London forces

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The Liquid StateThe Liquid StateViscosity Viscosity 黏度黏度

• Viscosity is the resistance to flow.– For example: molasses 糖蜜 , syrup 糖漿 or honey.– Oil for your car is bought based on this property.

• 10W30 or 5W30 describes the viscosity of the oil at high and low temperatures.

• Measure by Viscometer. • The stronger of the intermolecular forces of

attraction the more viscosity • Increasing the size and surface area of

molecule genernally results in increased viscosity

• Temperature increase viscosity decrease

Pentane C5H12 戊烷

Dodecane C12H26 十二烷

黏度是液體內部抵抗液體流動的阻力，當液體的黏度愈大，液體就愈不容易流動。液體的黏度大小與液體分子間作用大小有關，分子間作用力愈大，黏度愈大。

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The Liquid StateThe Liquid StateSurface TensionSurface Tension 表面張力表面張力

• Surface tension is a measure of the unequal attractions that occur at the surface of a liquid.

• The molecules at the surface are attracted unevenly.

液體表面分子僅受到液面下方的分子吸引！此現象導致液體表面分子感受到往液內的淨吸引力。因此，液面分子隨時受到一個向下的拉力，此力即稱『表面張力』。當分子間的作用力愈強時，表面張力愈大

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The Liquid StateThe Liquid StateCapillary Action Capillary Action 毛細作用毛細作用

• Capillary action is the ability of a liquid to rise (or fall) in a glass tube or other container 當一根直徑細小的玻璃管或毛細管放在水中時，水會在細管中往上升，此種液態在細管中爬升的現象稱為毛細作用

• Cohesive forces內聚力 are the forces that hold liquids together. 內聚力表示 :液體分子間的作用力

• Adhesive forces 吸附力 are the forces between a liquid and another surface. 吸附力表示 :液體分子與管壁間的附著力– Capillary rise implies that the:

•Adhesive forces > cohesive forces– Capillary fall implies that the:

• Cohesive forces > adhesive forces• The smaller the bore, the higher the liquid climbs

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The Liquid StateThe Liquid State• Capillary action also affects the meniscus of liquids.毛細現

象會影響圓筒內液體的凹凸面

Adhesive forces > cohesive forcesconcave convex

Adhesive forces < cohesive forces水與玻璃管的吸附力大於水分子間的內聚力，因此呈現下凹的液面

汞分子間的內聚力大於汞與玻璃管的吸附力，因此呈現上凸的液面

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The Liquid StateThe Liquid StateEvaporation Evaporation 蒸發蒸發

•Evaporation is the process in which molecules escape from the surface of a liquid and become a gas. 物質從液體表面變為氣體狀態的過程

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The Liquid StateThe Liquid State•Evaporation is temperature dependent.

The rate of evaporation increases as temperature increases

Only the higher-energy molecules can escape from the liquid phase

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The Liquid StateThe Liquid State•Condensation 凝結 ; 冷凝•In a closed containerDynamic equilibrium

vaporliquidevaporation

condensation

•LeChatelier’s Principle 勒沙特列原理•If the vessel were left open to the air, the equilibrium could not be reached.•A liquid can eventually evaporated entirely

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The Liquid StateThe Liquid StateVapor Pressure Vapor Pressure 蒸氣壓蒸氣壓

• Vapor pressure is the pressure exerted by a liquid’s vapor on its surface at equilibrium.

• Vapor Pressure (torr) and boiling point for three liquids at different temperatures.

0oC 20oC 30oC normal boiling point

diethyl ether 乙醚 185 442 647 36oCEthanol 酒精 12 44 74 78oCwater 水 5 18 32 100oC• Vapor pressures of liquids always increase

as temperature increase液體的沸點是液體的表面蒸氣壓等於外界壓力時的溫度

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• Easily vaporized liquids are called volatile liquids 揮發性液體

• Stronger cohesive forces tend to hold molecules in the liquid state– Methanol molecules are strongly linked by

hydrogen bond lower vapor pressure• Dispersion forces increase with increasing

molecular size larger molecules have lower vapor pressure

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• Vapor pressure can be measured with manometers 壓力計

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The Liquid StateThe Liquid StateVapor Pressure as a function of

temperature液體的沸點是液體的表面蒸氣壓等於外界壓力時的溫度

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The Liquid StateThe Liquid StateBoiling Points Boiling Points 沸點沸點 and Distillationand Distillation 蒸餾蒸餾

•The boiling pointboiling point is the temperature at which the liquid’s vapor pressure is equal to the applied pressure (usually atmospheric 大氣壓 ).•The normalnormal boiling pointboiling point 正常沸點 is the boiling point when the pressure is exactly 1 atm (760 torr). 當外界壓力等於一大氣壓時，此溫度稱為正常沸點– If the applied pressure is lower than 1 atom water boil below 100oC.– 物質的沸點高低與分子間作用力有直接的關係。當分子間作用力愈大，沸點愈高。

•DistillationDistillation 蒸餾 is a method we use to separate mixtures of liquids based on their differences in boiling points.

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The Liquid StateThe Liquid StateDistillationDistillation 蒸餾蒸餾

• Different liquids have different vapor pressure and boil at different temperature.

• Distillation is a process in which a mixture or solution is separated into its components on the basis of the differences in boiling points of the components.

• Distillation is another vapor pressure phenomenon.

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The Liquid StateThe Liquid StateDistillationDistillation 蒸餾蒸餾

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The Liquid StateThe Liquid StateHeat Transfer Involving LiquidsHeat Transfer Involving Liquids• Heat must added to a liquid to raise its

temperature• The amount of heat that must be added to

the stated mass of liquid to raise its temperature by one degree

Specific heat (J/goC) 比熱• molar heat capacity (J/moloC) 莫耳熱容量 ( 使

1mol 物質上升 1oC 所需的熱量• Molar heat (enthalpy焓 ) of vaporization 莫耳

汽化熱改變 1 mol 物質由液態形成氣態所需的能量

– 水在 100oC 的莫耳汽化熱為 40.7 kJ/mol

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The Liquid StateThe Liquid State

40.7KJ/mol= 40.7x 1000JKJ

x mol18g

= 2.26x103 J/g

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The Liquid StateThe Liquid StateHeat Transfer Involving Heat Transfer Involving

LiquidsLiquids• Condensation• Heat of condensation 凝結熱 liquid + heat vapor

evaporation

condensation

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The Liquid StateThe Liquid StateExample 13-2: How much heat is released by 2.00 x 102 g of H2O as it cools from 85.0oC to 40.0oC? The specific heat of water is 4.184 J/goC. q=mCT ?J = 2.00 x 102 g x (4.184J/goC) x( 85.0-40oC) ?J = 3.76x104 J = 37.6kJ

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The Liquid StateThe Liquid StateExample 13-3: The molar heat capacity of ethyl alcohol,

C2H5OH, is 113 J/moloC. How much heat is required to raise the T of 125 g of ethyl alcohol from 20.0oC to 30.0oC?

1 mol C2H5OH = 46.0 g ?mol C2H5OH = 125g x 1mol

46g = 2.72 mol C2H5OH

?J = 2.72mol x 113JmoloC

x (30.0-20.0oC)

= 3.07 KJ

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The Liquid StateThe Liquid State• The calculations we have done up to now tell us

the energy changes as long as the substance remains in a single phase.

• Next, we must address the energy associated with phase changes.– For example, solid to liquid or liquid to gas

and the reverse.• Heat of Vaporization is the amount of heat

required to change 1.00 g of a liquid substance to a gas at constant temperature.– Heat of vaporization has units of J/g.

• Heat of Condensation is the reverse of heat of vaporization, phase change from gas to liquid.

+2260J

-2260J1.0g H2O(l) at 100oC 1.0g H2O(g) at 100oC

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The Liquid StateThe Liquid StateMolar heat of vaporization or DHMolar heat of vaporization or DHvapvap

• The DHvap is the amount of heat required to change 1.00 mole of a liquid to a gas at constant temperature.

DHvap has units of J/mol. Molar heat of condensation

• The reverse of molar heat of vaporization is the heat of condensation.

+40.7kJ

-40.7kJ1.0mol H2O(l) at 100oC 1.0mol H2O(g) at 100oC

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The Liquid StateThe Liquid State

加熱／冷卻曲線 (heating-cooling curve)

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The Liquid StateThe Liquid StateExample 13-4: How many joules of energy must be absorbed by 5.00 x 102 g of H2O at 50.0oC to convert it to steam at 120oC? The molar heat of vaporization of water is 40.7 kJ/mol and the molar heat capacities of liquid water and steam are 75.3 J/mol oC and 36.4 J/mol oC, respectively.50oC H2O(l) 100oC H2O(l) 100oC H2O(g) 120oC H2O(g)

?mol H2O = 500g x 1mol18g = 27.8 mol H2O

?J = 27.8mol x 75.3JmoloCx (100.0-50.0oC)= 1.05x105 J

?J = 27.8mol x 40.7x103Jmol

= 11.31x105 J

?J = 27.8mol x 36.4JmoloCX (120.0-100.0oC)= 0.20x105 J

Total J = 1.05x105 + 11.31x105 + 0.2x105

= 12.56 x105 J or 1.26x103 kJ

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The Liquid StateThe Liquid StateExample 13-5: If 45.0 g of steam at 140oC is slowly bubbled into 450 g of water at 50.0oC in an insulated container, can all the steam be condensed??mol steam = 45.0g x

1mol18g = 2.5 mol steam

?mol H2O = 450g x 1mol18g = 25.0 mol H2O

Calculate the amount of heat required to condense the steam2.5mol x(36.4J/moloC)x(140-100oC)+2.5 mol x (40.7kJ/mol)=105.4kJ

Calculate the amount of heat available in the liquid water

25.0mol x (75.3J/moloC)x(100.0-50oC) = 94.1kJAmount of heat to condense all of the steam is 105kJ

Amount of heat that the liquid water can absorb is 94.1kJThus all of the steam cannot be condensed

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The Liquid StateThe Liquid StateExample 13-6: Arrange the following substances in order of

increasing boiling points.C2H6, NH3, Ar, NaCl, AsH3

Ar < C2H6 < AsH3 < NH3 < NaCl nonpolar nonpolar polar very polar ionic London London dipole-dipole H-bonding

ion-ion

Example 13-2: Heat of VaporizationCalculate the amount of heat, in joules, required

to convert 180.0grams of water at 10.0oC to steam at 105.0oC.

Example 13-2: Heat of VaporizationCalculate the amount of heat, in joules, required

to convert 180.0grams of water at 10.0oC to steam at 105.0oC.

10oC H2O(l) 100oC H2O(l) 100oC H2O(g) 105oC H2O(g)

?J = 180.0g x 4.18JgoC x (100.0-10.0oC)= 0.67x105 J

?J = 180.0g x 2.26x103Jg

= 4.07x105 J

?J = 180.0g x 2.03JgoC x (105.0-100.0oC)= 0.01820x105 J

Total J = 0.67x105 + 4.07x105 + 0.0182x105

= 4.76 x105 J

Example 13-3: Heat of VaporizationCompare the amount of “cooling” experienced by an

individual who drinks 400ml of ice water (0.0oC) with the amount of “cooling” experienced by an individual who “sweats out” 400ml of water. Assume that the sweat is essentially pure water and that all of it evaporates. The density of water is very nearly 1.00g/ml at both 0.0oC and 37oC, average body temperature. The heat of vaporization of water is 2.41kJ/g at 37oC.

The amount of heat lost by perspiration =the amount of heat required to vaporize 400g of

water at 37oC

Raising the temperature of 400.0g of water from 0oC to 37oC

Example 13-3: Heat of VaporizationCompare the amount of “cooling” experienced by an

individual who drinks 400ml of ice water (0.0oC) with the amount of “cooling” experienced by an individual who “sweats out” 400ml of water. Assume that the sweat is essentially pure water and that all of it evaporates. The density of water is very nearly 1.00g/ml at both 0.0oC and 37oC, average body temperature. The heat of vaporization of water is 2.41kJ/g at 37oC.

The amount of heat lost by perspiration =the amount of heat required to vaporize 400g of

water at 37oC

Raising the temperature of 400.0g of water from 0oC to 37oC

?J = 400.0g x 4.18JgoC x (37.0-0.0oC)= 6.19x104 J

or 61.9kJ

?J = 400.0ml x 1.0gml x (2.41x103 J/g)= 9.64x105 J

or 964kJ

Evaporating 400.o ml of water at 37oC requires

Sweating removes more heat than drinking ice water

Example 13-5: Boiling Points Versus Intermolecular Forces

Predict the order of increasing boiling points for the following: H2S; H2O; CH4; H2; KBr.

• KBr: is ionic, it boils at the highest temperature

• Hydrogen bond: H2O the next highest temperature

• polar covalent substance: H2S

• CH4; H2 are nonpolar .

CH4 is larger than H2;

so the dispersion forces are stronger in CH4

CH4 boils at a higher temperature than H2

H2 < CH4 < H2S < H2O < KBr

Example 13-5: Boiling Points Versus Intermolecular Forces

Predict the order of increasing boiling points for the following: H2S; H2O; CH4; H2; KBr.

• KBr: is ionic, it boils at the highest temperature

• Hydrogen bond: H2O the next highest temperature

• polar covalent substance: H2S

• CH4; H2 are nonpolar .

CH4 is larger than H2;

so the dispersion forces are stronger in CH4

CH4 boils at a higher temperature than H2

H2 < CH4 < H2S < H2O < KBr

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The Solid StateThe Solid StateNormal Melting Point

• The normal melting pointnormal melting point is the temperature at which the solid melts (liquid and solid in equilibrium) at exactly 1.00 atm of pressure.

• Freezing point melting point• The melting point increases as the strength

of the intermolecular attractions increase.solid liquid

melting

freezing

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The Solid StateThe Solid State• Which requires more energy?

NaCl(s) NaCl(l)

H2O(s) H2O(l) or

Ion-ion interaction

Hydrogen bond

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Heat Transfer Involving Heat Transfer Involving SolidsSolids

Heat of Fusion• Heat of fusion Heat of fusion 熔化熱 is the amount of heat

required to melt one gram of a solid at its melting point at constant temperature.

•Heat of solidificationHeat of solidification 固化熱 is the reverse of the heat of fusion.

1.00g H2O(s) at 0oC+334J-334J

1.00g H2O(l) at 0oC

Heat of fusion Heat of fusion

Heat of solidificationHeat of solidification

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Heat of fusion溶化熱

Specific heat比熱

Specific heat比熱

Heat of vaporization

Specific heat

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Molar heat of fusionMolar heat of fusion 莫耳熔化熱莫耳熔化熱 or or HHfusionfusion

• The molar heat of fusion is the amount of heat required to melt a mole of a substance at its melting point.

• The molar heat of solidification is the reverse of molar heat of fusion

1.00mole H2O(s) at 0oC+6012J

-6012J1.00mole H2O(l) at 0oC

Molar heat of fusion Molar heat of fusion

Molar heat of solidification Molar heat of solidification

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• The heat of fusion depends on the intermlacular forces of attraction in the solid state– Heats of fusion are usually higher for

substances with higher melting points.

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Example 13-7:Calculate the amount of heat required to

convert 150.0 g of ice at -10.0oC to water at 40.0oC. specific heat of ice is 2.09 J/goC-10oC H2O(s) 0oC H2O(s) 0oC H2O(l) 40oC H2O(l)

?J = 150.0g x 2.09JgoC x (10.0oC)= 3.14x103 J

?J = 150.0g x 334Jg = 5.01x104 J

?J = 150.0g x 4.18JgoC x (40.0-0.0oC)= 2.51x104 J

Total J = 3.14x103 + 5.01x104 + 7.83x104

= 7.83x104J

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Example 13-6: Heat of FusionThe molar heat of fusion, Hfus, of Na is

2.6kJ/mol at its melting point, 97.5oC, How much neat must be absorbed by 5.0g of sodium Na at 97.5oC to melt it?

Example 13-6: Heat of FusionThe molar heat of fusion, Hfus, of Na is

2.6kJ/mol at its melting point, 97.5oC, How much neat must be absorbed by 5.0g of sodium Na at 97.5oC to melt it?

?J = 5.0g x 1mol23g x = 0.57 kJ2.6kJ

mol

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Example 13-2: Heat of FusionCalculate the amount of heat that must be

absorbed by 50.0g of ice at -12.0oC to convert to water at 20.0oC.

Example 13-2: Heat of FusionCalculate the amount of heat that must be

absorbed by 50.0g of ice at -12.0oC to convert to water at 20.0oC.

-10oC H2O(s) 0oC H2O(s) 0oC H2O(l) 20oC H2O(l)

?J = 50.0g x 2.09JgoC x (0.0-(-12).0oC)= 1.25x103 J

?J = 50.0g x 334Jg = 16.7x103 J

?J = 50.0g x 4.18JgoC x (20.0-0.0oC)= 4.18x103 J

Total J = 1.25x103 + 16.7x103 + 4.18x103

= 22.1 kJ

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Sublimation and the Sublimation and the Vapor Pressure of SolidsVapor Pressure of Solids

Sublimation 昇華• In the sublimation process the

solid transforms directly to the vapor phase without passing through the liquid phase.

• Solid CO2 or “dry” ice does this well.

• iodine

solid gassublimation

depositionSublimation can be used to purify volatile solids

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Endothermic 吸熱

Exothermic 放熱

Transitions among the three states of matter

凝結蒸發昇華

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Phase Diagrams (P versus T)Phase Diagrams (P versus T) 相相圖圖

• Phase diagrams are a convenient way to display all of the different phase transitions of a substance.

• The equilibrium pressure-temperature (壓力 - 溫度 ) relationships among the different phases of a given pure substance in a closed system ( 密閉系統 ).

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Phase Diagrams (P versus T)Phase Diagrams (P versus T) 相相圖圖

• This is the phase diagram for water.

三相點

ABMelting curveADSublimation curve

Negative slope of line AB•Ice is less dense than liquid•The network of

hydrogen bonding in ice is more extensive than that in a liquid water

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Phase Diagrams (P versus T)Phase Diagrams (P versus T)Compare water’s phase diagram to carbon dioxide’s phase diagram.

三相點

Liquid CO2 cannot exist at atmospheric pressure

Critical point

For CO2: critical point is at 31oC and 73 atmFor H2O: critical point is at 374oC and 218 atm

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To illustrate the use of a phase diagram in determining the physical state or states of a system under different sets of pressures and temperatures

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AmorphousAmorphous 無定形的無定形的 Solids Solids andand

CrystallineCrystalline 結晶狀的結晶狀的 SolidsSolids• Amorphous solids do not have a well

ordered molecular structure.– Examples of amorphous solids include waxes,

glasses, asphalt柏油 .• Crystalline solids have well defined

structures that consist of extended array of repeating units called unit cellsunit cells.– Crystalline solids display X-ray diffraction

patterns which reflect the molecular structure.

– The Bragg equation, detailed in the textbook, describes how an X-ray diffraction pattern can be used to determine the interatomic distances in crystals.

•Unit cell 單位晶胞–The Lengths of its edges (a,b,c)

–The angles between the edges

(,,)–Three-dimensional arrangement

–Lattice point 晶格

立方體四角體斜方晶

菱形六面體六角體單斜體三斜晶系

金紅石岩鹽

瀉利鹽石膏

重鉻酸鉀矽土方解石

All crystal contain regularly repeating arrangements of atoms, molecules or ions

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Structure of CrystalsStructure of Crystals• Unit cells are the smallest repeating unit

of a crystal.– As an analogy, bricks are repeating units for

buildings.• There are seven basic crystal systems.

螢石立方體黃銅礦四角體霰石斜方晶方解石菱形六面體

綠寶石六角體藍銅礦

單斜體薔薇輝石三斜晶系

圖圖 14.1414.14

三種結晶固體的範例。每一例子只顯示結構的一部分，各結構是在三次元以相同的圖形做連續延伸。 (a) 原子固體。在鑽石中每個圓球代表一個碳原子。 (b) 離子固體。在氯化納固體中，每個圓球分別代表 Na ＋和 Cl －離子。 (c) 分子固體。在冰中，每三個圓球單元代表一個 H2O 分子。點線表示極性水分子之間氫鍵。

P.425

表表 14.314.3

P.426

不同種類固體的範例

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Structure of CrystalsStructure of Crystals• We shall look at the three

variations of the cubic crystal system.

• Simple cubic unit cells單立方– The balls represent the

positions of atoms, ions, or molecules in a simple cubic unit cell.

– In a simple cubic unit cell each atom, ion, or molecule at a corner is shared by 8 unit cells

– Thus 1 unit cell contains 8(1/8) = 1 atom, ion, or molecule.

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Structure of CrystalsStructure of Crystals• Body centered cubic

(bcc) 體心立方 has an additional atom, ion, or molecule in the center of the unit cell.

• On a body centered cubic unit cell there are 8 corners + 1 particle in center of cell.– 1 bcc unit cell

• contains 8(1/8) + 1 = 2 particles.

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Structure of CrystalsStructure of Crystals• A face centered cubic

(fcc) 面心立方 unit cell has a cubic unit cell structure with an extra atom, ion, or molecule in each face.

• A face centered cubic unit cell has 8 corners and 6 faces.– 1 fcc unit cell contains

• 8(1/8) + 6(1/2) = 4 particles.

• Isomorphous– refers to crystals having the

same atomic arrangement• Polymorphous

– Refers to the substances that crystallize in more than one crystalline arrangement

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Bonding in SolidsBonding in Solids• Four categories

– Metallic solids 金屬固體 − Ionic solids 離子固體 – Molecular solids 分子固體 − Covalent solids

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Bonding in SolidsBonding in Solids• Molecular Solids have molecules in each of

the positions of the unit cell.– Molecular solids have low melting

points, are volatile, and are electrical insulators.

• Examples of molecular solids include:– water, sugar, carbon dioxide, benzene

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Bonding in SolidsBonding in Solids• Covalent Solids have atoms that are

covalently bonded to one another • Some examples of covalent solids are:

•Diamond, graphite, SiO2 (sand), SiC碳化矽

石墨

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Bonding in SolidsBonding in Solids• Ionic Solids have ions that occupy the

positions in the unit cell.• Examples of ionic solids include:

– CsCl, NaCl, ZnS

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Bonding in SolidsBonding in Solids• Metallic Solids may be thought of as

positively charged nuclei surrounded by a sea of electrons.

• The positive ions occupy the crystal lattice positions.

• Examples of metallic solids include:– Na, Li, Au, Ag, ……..

• 金屬呈現另一種形式的原子固體。• 大部分金屬的鍵結是強但無方向性。• 電子海模型 (electron sea model) ：金屬原子規則排列於價電子海中，此電子海為所有原子所共用並且在金屬結晶中極易運動。

• 因為金屬結晶的本性，其他元素相對容易被加入，這些含有其他元素的金屬物質稱為合金。

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• 合金 (alloy) 的定義為一物質含有多種元素且具有金屬性質。

• 合金一般有兩種型態：– 取代型合金 (substitutional

alloy) ：是某些金屬原子被其他金屬原子所取代，例如黃銅，在銅內約有1/3 的銅原子被鋅原子所取代，

– 間隙型合金 (interstitial alloy) ：金屬原子緊密堆積出結構且其內的空隙 ( 孔洞 ) 被其他較小的原子所佔據。鋼鐵是一種在孔隙內包含了碳原子的鐵結晶。

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Documents

13 Liquids and Solids. 2 Chapter Goals 1.Kinetic-Molecular Description of Liquids and Solids 2.Intermolecular Attractions and Phase Changes 分子間引力及相的改變