1 chapter 15 chemical thermodynamics 化學熱力學. 2 outline heat changes and thermochemistry...

1

Chapter 15Chemical

Thermodynamics 化學熱力學

2

OutlineHeat Changes and Thermochemistry

1. The First Law of Thermodynamics 熱力學第一定律

2. Some Thermodynamic Terms 3. Enthalpy Changes 焓的變化 4. Calorimetry 熱量的測量5. Thermochemical Equations 熱化學方程式6. Standard States and Standard Enthalpy

Changes7. Standard Molar Enthalpies of Formation,

Hfo

8. Hess’s Law9. Bond Energies10.Changes in Internal Energy, E11.Relationship of H and E

3

Outline

Spontaneity of Physical and Chemical Changes

12.The Two Aspects of Spontaneity13.Dispersal of Energy and Matter14.Entropy 熵 , S, and Entropy Change, DS15.The Second Law of Thermodynamics 16.Free Energy Change, DG, and Spontaneity17.The Temperature Dependence of

Spontaneity

4

The First Law of Thermodynamics

熱力學第一定律• Thermodynamics is the study of the

changes in energy and transfers of energy that accompany chemical and physical processes. 所謂的熱力學是指在化學及物理反應中能量的改變及轉移

• In this chapter we will address 3 fundamental questions.1.Will two (or more) substances react when

they are mixed under specified conditions? 當兩個物質在特定狀態混合時，是否會發生反應 ?

2.If they do react, what energy changes and transfers are associated with their reaction? 如果發生反應，那反應中有那些能量的改變及轉移 ?

3.If a reaction occurs, to what extent does it occur?

5


•Exothermic reactions release energy in the form of heat. ( 放熱反應以熱的形式釋出能量 )–For example, the combustion of propane is exothermic. ( 丙烷的燃燒為放熱反應 )• C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) + 2.22x103 kJ

–The combustion of n-butane is also exothermic ( 正丁烷的燃燒亦為放熱反應 )• 2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l) + 5.78x103 kJ

• A process that absorbs energy from its surroundings is called endothermic.– H2O(s) + 6.02kJ H2O(l)

烷類和過量的氧作用會形成二氧化碳和水

Energy is the capacity to do work or to transfer heat.

6


• Exothermic reactions generate specific amounts of heat.( 放熱反應產生熱 )–This is because the

potential energies of the products are lower than the potential energies of the reactants. ( 此乃因產物的位能比反應物未能來得低 )Kinetic energy 動能

v.s.Potential energy 位能

( 包含化學能 )

活化能(activation energy)

7


• There are two basic ideas of importance for thermodynamic systems. ( 熱力學系統有兩個最基本的概念 )

1.Chemical systems tend toward a state of minimum potential energy.( 系統傾向最小位能 )Some examples of this include:

H2O flows downhill. Objects fall when dropped.

The energy change for these two examples is: Epotential = mgh Epotential = mg(h)

8


2.Chemical systems tend toward a state of maximum disorder. ( 系統傾向最大亂度 )• Common examples of this are:

– A mirror shatters when dropped and does not reform.

– It is easy to scramble an egg and difficult to unscramble it.

– Food dye when dropped into water disperses.

9


• This law can be stated as, “The combined amount of matter and energy in the universe is constant.” ( 熱力學第一定律可定義為物質和能量的總和不會改變 )

• The first law is also known as the Law of Conservation of Energy ( 即為能量守恆定律 ).– Energy is neither created nor destroyed in

chemical reactions and physical changes. 在化學及物理反應時，能量既不會憑空消失，也不會憑空產生，只能從一種形式轉化成另一種形式，或者從一個物體轉移到另一個物體，而總量保持不變

10

Some Thermodynamic Terms

• The substances involved in the chemical and physical changes under investigation are called the system. –In chemistry lab, the system is the chemicals inside the beaker.

• The environment around the system is called the surroundings.

–The surroundings are outside the beaker.• The system plus the surroundings is called the universe.

• The set of conditions that specify all of the properties of the system is called the thermodynamic state of a system.

• For example the thermodynamic state could include:

– The number of moles and identity of each substance.

– The physical states of each substance.– The temperature of the system.– The pressure of the system.

11


• The properties of a system that depend only on the state of the system are called state functions.( 狀態函數指一個系統的特質只與狀態有關而與途徑無關 , 通常以大寫表示 )–The value of a state function is independent of pathway.

–An analog to a state function is the energy required to climb a mountain taking two different paths. ( 走兩條不同的路而到達山頂的位能差相同，但路徑不同 )

• E1 = energy at the bottom of the mountain E1 = mgh1

• E2 = energy at the top of the mountain E2 = mgh2

• E = E2-E1 = mgh2 – mgh1 = mg(h)–Notice that the energy change in moving from the top to the bottom is independent of pathway but the work required may not be!

–Some examples of state functions are:• T, P, V, E, H, and S

–Examples of non-state functions are:• n, q, w

12


• In thermodynamics we are often interested in changes in functions.– We will define the change of any function

X as:– X = Xfinal – Xinitial

• If X increases X > 0• If X decreases X < 0.

13

Enthalpy Change 焓變化 ( 反應熱 )

• Chemistry is commonly done in open beakers on a desk top at atmospheric pressure.–Because atmospheric pressure only changes by small increments, this is almost at constant pressure.

• The enthalpy change, H, is the change in heat content at constant pressure. 焓變化是指定壓下熱含量的變化–H = qP

14

Enthalpy Change• Hrxn is the heat of reaction. 反應熱

– This quantity will tell us if the reaction produces or consumes heat. (Hrxn 的量表示反應產生熱或消耗熱 )

– If Hrxn < 0 the reaction is exothermic.

– If Hrxn > 0 the reaction is endothermic.

• Hrxn = Hproducts - Hreactants

– Hrxn = Hsubstances produced - Hsubstances consumed

– Notice that this is Hrxn = Hfinal – Hinitial

15

• A coffee-cup calorimeter ( 熱卡計 ) is used to measure the amount of heat produced (or absorbed) in a reaction at constant P– This is one method to

measure qP for reactions in solution.

Calorimetry 熱量的測量

16

Calorimetry• If an exothermic reaction is performed in a

calorimeter, the heat evolved by the reaction is determined from the temperature rise of the solution.– This requires a two part calculation.

• Amount of heat gained by calorimeter is called the heat capacity of the calorimeter or calorimeter constant.– The value of the calorimeter constant is

determined by adding a specific amount of heat to calorimeter and measuring the temperature rise.

Amount of heatReleased by reaction=

Amount of heatabsorbed by calorimeter+

Amount of heatabsorbed by solution

Heat capacity (C) =Heat absorbed (J)

Increase in temperature (oC)所謂熱容量 (heat capacity) 是指使一個物體上升 1oC 所需的熱量

17

CalorimetryExample 15-1: When 3.425 kJ of heat is added to a

calorimeter containing 50.00 g of water the temperature rises from 24.00oC to 36.54oC. Calculate the heat capacity of the calorimeter in J/oC. The specific heat of water is 4.184 J/g oC.水的溫度變化 T = (36.54 – 24.00)oC =

12.54oC水所吸收的熱 qp = mCT

= (50.00g) (4.184 J/goC) (12.54oC)

= 2623.37 J 熱卡計所吸收的熱量 = 3425J – 2623J

= 802 J熱卡計的熱容量 = 802J / 12.54oC

= 64.0 J/oC

18

Example 15-2: A coffee-cup calorimeter is used to determine the heat of reaction for the acid-base neutralization

CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O()

When we add 25.00 mL of 0.500 M NaOH at 23.000oC to 25.00 mL of 0.600 M CH3COOH already in the calorimeter at the same temperature, the resulting temperature is observed to be 25.947oC. (The heat capacity of the calorimeter : 27.8 J/0C. The specific heat of the mixture is 4.18 J/g0C and the density: 1.02 g/mL.)溫度變化 T= 25.947 – 23.000 = 2.947 oC熱卡計所吸收的熱量 q= (2.947oC)(27.8 J/oC) = 81.9 J混合溶液的重量 = (25.00ml + 25.00ml) x1.02 g/ml = 51.0g溶液所吸收的熱量 q= mCT = (51.0g)(4.18 J/goC)(2.947oC)

= 628 J反應所產生的總熱量為 = 81.9J + 628J = 709.9J CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O()

25.00mlx0.5 M =12.5mmol

25.00mlx0.6 M =15.0mmol

1 1 1

12.5mmol

Hrxn = 709.9J / 0.0125 mol = 56792 J/mol = 5.68kJ/mol

12.5mmol = 0.0125 mol指每莫耳所放出的熱量

19

Thermochemical Equations 熱化學方程式

•Thermochemical equations are a balanced chemical reaction plus the H value for the reaction. ( 熱化學方程式為平衡的化學式再加上反應所需的熱量 H.)– For example, this is a thermochemical

equation.C5H12(g) + 8 O2(g) 5 CO2(g) + 6 H2O(l) + 3523 kJ• 1 mol of C5H12 reacts with 8 mol of O2 to produce

5 mol of CO2, 6 mol of H2O, and releasing 3523 kJ is referred to as one mole of reactions.

• This is an equivalent method of writing thermochemical equations.

1mole 8moles 5moles 6moles

– H < 0 designates an exothermic reaction.

– H > 0 designates an endothermic reaction

C5H12(g) + 8 O2(g) 5 CO2(g) + 6 H2O(l) Horxn= -

3523 kJ

20

Thermochemical EquationsExample 15-3: When 2.61g of dimethyl ether

(CH3OCH3) is burned at constant pressure, 82.5 kJ of heat is given off. Find H for the reaction.

CH3OCH3(l) + 8 O2(g) 2 CO2(g) + 3 H2O(l)

CH3OCH3 的分子量為 46CH3OCH3 的 mole 數 = 2.61g/ 46

= 5.67x10-2 mole此反應為放熱反應Hrxn = - (82.5 kJ/ 5.67x10-2 mol) = -1455 kJ/ mol rxn

21

Thermochemical EquationsExample 15-4: When aluminum metal is exposed

to atmospheric oxygen, it is oxidized to form aluminum oxide. How much heat is released by the complete oxidation of 24.2 g of aluminum at 25oC and 1atm?

4 Al(s) + 3 O2(g) 2 Al2O3(s) H = -3352 kJ/mol rxn

Al 的原子量為 27Al 的 mole 數 = 24.2g/ 27

= 0.896 moleHrxn = - 3352 kJ/mol rxn 表示 1mole 的反應所釋出的熱量而 1mole 的反應需 4mole 的 Al

0.896 mol4 mol Al

x kJ-3352 kJ

= x= -751kJ

22

Standard States and Standard Enthalpy Changes 標準狀態及標準焓變•Thermochemical standard state conditions

–The thermochemical standard T = 298.15 K. (25oC)

–The thermochemical standard P = 1.00 atm.• 例如標準狀態下 (1 大氣壓 , 25oC) ， H2(g), Hg(l), Na(s), C2H5OH(l), CaCO3(s), CO2(g)

•Thermochemical standard states of matter– For pure substances in their liquid or solid

phase the standard state is the pure liquid or solid.

– For gases the standard state is the gas at 1.00 atm of pressure.• For gaseous mixtures the partial pressure must be

1.00 atm.– For aqueous solutions the standard state is

1.00 M concentration.• Standard enthalpy change, Ho

rxn All at standard states ( 只看反應前、反應後，不管反應過程狀態是否改變 )

23

Standard Molar Enthalpies of Formation, 標準莫耳生成焓 ,Hf

o

•The standard molar enthalpy of formation is defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements. ( 在標準狀態下由各元素形成一莫耳物質所需的熱量及稱之 )– The symbol for standard molar enthalpy of

formation ( 簡稱 heat of formation) is Hfo.

•The standard molar enthalpy of formation for MgCl2 is:

Mg(s) + Cl2(g) MgCl2(s) + 64.18 kJHo

fMgCl2 = -641.8 kJ/mol MgCl2(s)

•The standard molar enthalpy of formation for HBr is:

H2(g) + Br2(l) 2HBr(g) Horxn = -72.8 kJ/mol

rxn ½ H2(g) + ½ Br2(l) HBr(g) Ho

rxn = -36.4 kJ/mol

HofHBr = -36.4 kJ/mol HBr(g)

25

Standard Molar Enthalpies of Formation 標準莫耳生成焓 , Hf

o

• Standard molar enthalpies of formation have been determined for many substances and are tabulated in Table 15-1 (Table6.2) and Appendix 4 in the text.

• Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero.Example 15-5: The standard molar enthalpy of formation for phosphoric acid is -1281 kJ/mol. Write the equation for the reaction for which Ho

rxn = -1281 kJ.P in standard state is P4

Phosphoric acid in standard state is H3PO4(s)3/2 H2(g) + 2 O2(g) + 1/4 P4(s) H3PO4(s) + 1281 kJ

HofH3PO4 = -1281 kJ/mol

26

Standard Molar Enthalpies of Formation, Hf

o

Example 15-6: Calculate the enthalpy change for the reaction of one mole of H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC and one atmosphere.

H2(g) + F2(g) 2HF(g) Std. state Std. state Std. state

for this rxn HoHF = 2 Ho

f

查表得知 : Hof = -271 kJ/mol

所以 HoHF =

(2mol)(-271kJ/mol) = -542 kJ

27

Standard Molar Enthalpies of Formation, Hf

o

• Example 15-6: Calculate the enthalpy change for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere. Ho

fAl2O3 = -1676 kJ/mol Al2O3

2 Al(s) + 3/2 O2(g) Al2O3(s) Al mole = 15.0g/ 27 = 0.556 mole0.556 mol2 mol Al

x kJ-1676 kJ

= x= -466kJ

28

Hess’s Law• Hess’s Law of Heat Summation states

that the enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps.– Hess’s Law is true because H is a state

function.赫士定律 (Hess’s law, 又名反應熱加成性定律 ): 若一反應為二個反應式的代數和時 , 其反應熱為此二反應熱的代數和。 Ho

rxn = Hoa + Ho

b + Hoc + …….

29

Hess’s Law• If we know the following Ho’s

[1] 4 FeO(s) + O2(g) 2 Fe2O3(s) Ho=-560 kJ[2] 2 Fe(s) + O2(g) 2 FeO(g) Ho=-544 kJ [3] 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) Ho=-1648 kJ

• For example, we can calculate the Ho for reaction [1] by properly adding (or subtracting) the Ho’s for reactions [2] and [3].

• Notice that reaction [1] has FeO and O2 as reactants and Fe2O3 as a product. (2x[-2]) + [3] = [1]

–Arrange reactions [2] and [3] so that they also have FeO and O2 as reactants and Fe2O3 as a product.

• Each reaction can be doubled, tripled, or multiplied by half, etc.• The Ho values are also doubled, tripled, etc.• If a reaction is reversed the sign of the Ho is changed.2x [-2] 2 (2 FeO(g) 2 Fe(s) + O2(g)) Ho=2x (-(-

544 kJ) ) [3] 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) Ho=-1648

kJ

[1] 4 FeO(s) + O2(g) 2 Fe2O3(s) Ho=-560 kJ

30

Hess’s LawExample 15-7: Given the following equations

and Hovalues

calculate Ho for the reaction below. N2O (g) + NO2(g) 3 NO(g) Ho= ?

[1] 2N2(g) + O2(g) 2 N2O(g) Ho= 164.1 kJ[2] N2(g) + O2(g) 2 NO(g) Ho= 180.5 kJ [3] N2(g) + 2 O2(g) 2NO2(g) Ho= 66.4 kJ

½ [-1] + 3/2 [2] + ½ [-3]

½ [-1] N2O(g) N2(g) + ½ O2(g) Ho= ½ x (-164.1) kJ

3/2 [2] 3/2 N2(g) + 3/2 O2(g) 3 NO(g) Ho= 3/2 x (180.5) kJ

½ [-3] NO2(g) ½ N2(g) + O2(g) Ho= ½ (-66.4) kJ

N2O (g) + NO2(g) 3 NO(g) Ho= (-82.05) + 270.75 + (-33.2)

= 155.5 kJ (endothermic reaction)

31

Hess’s Law

Horxn = nHo

fproducts - Hofreactants

32

Hess’s LawExample 15-8: Calculate the H o

298 for the following reaction. (Appendix 4, A20)

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O (l)

Horxn = [3 Ho

fCO2(g) + 4 HofH2O(l)]-[Ho

fC3H8(g) + 5 HofO2(g)]

Horxn = [3 (-393.5) + 4 (-286)]-[(-104) + 5 (0.0)]

Horxn = -2220.5 kJ

Exothermic reaction

33

Hess’s LawExample 15-9: Given the following

information, calculate Hfo for H2S(g).

2 H2S(g) + 3 O2(g) 2 SO2(g) + 2 H2O (l) Ho298 = -

1124 kJ

Hfo ? 0 -296.8 -285.8 Horxn = [2 Ho

fSO2(g) + 2 HofH2O(l)]-[2 Ho

fH2S(g) + 3 HofO2(g)]

-1124 kJ = [2x(-296.8) + 2 (-285.8)]-[2HofH2S(g) + 5 (0.0)]

2 HofH2S(g) = -41.2 kJ

HofH2S(g) = -20.6 kJ

34

Bond Energies 鍵能• Bond energy is the amount of energy

required to break the bond and separate the atoms in the gas phase. ( 一莫耳氣態原子之化學鍵斷裂時吸收的能量即稱為鍵能 )– To break a bond always requires an

absorption of energy!

A-B(g) + bond energy A(g) + B(g)

H-Cl(g) + 432 kJ/mol H(g) + Cl(g)

36

Bond Energies• Bond energies can be calculated from

otherHo298values

Example 15-10: Calculate the bond energy for hydrogen fluoride, HF. H-F(g) + Bond Energy (BEHF) H(g) + F (g) (atoms NOT ions) or H-F(g) H(g) + F (g) Ho

298= BEHFHo298 = [Ho

fH(g) + HofF(g)]-[ Ho

fHF(g) ]

Ho298 = [218.0 kJ+ 78.99 kJ]-[ -271 kJ]

Ho298 = 568.0 kJ Bond energy for HF

37

Bond EnergiesExample 15-11: Calculate the average N-H bond

energy in ammonia, NH3.

NH3(g) N(g) + 3H (g) Ho298= 3 BEN-H

Ho298 = [Ho

fN(g) + 3 HofH(g)]-[ Ho

fNH3(g) ]

Ho298 = [472.7 kJ+ 3x(218) kJ]-[ -46.11 kJ]

Ho298 = 1173 kJ 3 N-H Bond energy for NH3

Average BEN-H = 1173/3 = 391 kJ/mol N-H bonds

38

Bond Energies• In gas phase reactions Ho values may be

related to bond energies of all species in the reaction. Ho

298 = BEreactants - BEproducts

39

Bond EnergiesExample 15-12: Use the bond energies listed

in Tables 15-2 and 15-3 to estimate the heat of reaction at 25oC for the reaction below.CH4(g) + 2 O2(g) CO2(g) + 2 H2O (g)

Ho298 = [4 BEC-H + 2 BEO=O]-[2 BEC=O+4 BEO-H]

Ho298 = [4x(414 kJ) + 2x(498 kJ)]-[2x(741 kJ) + 4x(464 kJ)]

Ho298 = -686 kJ

40

Changes in Internal Energy, E ( 內能的改變 )

• The internal energy, E, is all of the energy contained within a substance.– This function includes all forms of energy

such as kinetic, potential, gravitational, electromagnetic, etc. ( 包括所有形式的能量如動能、位能、重力、電磁能等 )

– The First Law of Thermodynamics states that the change in internal energy, E, is determined by the heat flow, q, and the work, w.

41

Changes in Internal Energy, EE = Eproducts – Ereactants

E = q+wq > 0 if the heat is absorbed by the systemq < 0 if the heat is absorbed by the surroundingsw > 0 if the surrounding do work on the systemw < 0 if the system does work on the surroundings

42

Changes in Internal Energy, E• E is negative when energy is released by a system undergoing a chemical or physical change. – Energy can be written as a product of the

process.C5H12(l) + 8 O2(g) 5 CO2(g) + 6 H2O(l) + 3.516x103

kJE = - 3.516x103 kJ• E is positive when energy is absorbed by

a system undergoing a chemical or physical change. – Energy can be written as a reactant of the

process.5 CO2(g) + 6 H2O(l) + 3.516x103 kJ C5H12(l) + 8 O2(g)

E = + 3.516x103 kJ

43

Changes in Internal Energy, E

Example 15-13: If 1200 joules of heat are added to a system in energy state E1, and the system does 800 joules of work on the surroundings, what is the :

1. energy change for the system, EsysE = Eproducts – Ereactants = q+wE = 1200 J + (-800J)E sys= +400 J2. energy change of the surroundings, EsurrE surr= -400 J

3. energy of the system in the new state, E2E surr= E2 – E1

E2 = E1 + Esys

= E1 +400J

44


• In most chemical and physical changes, the only kind of work is pressure-volume work.

• Pressure is force per unit area.

P = forcearea = F

d2 • Volume is distance cubed.

• PV is a work term, i.e., the same units are used for energy and work.

V = d3

PV= Fd2

x d3= Fxd ( 所做的功 )

45


A system that absorbs heat and does work.

46


• Using the ideal gas law PV = nRT, we can look at volume changes of ideal gases at constant T and P due to changes in the number of moles of gas present, ngas.

ngas = (number of moles of gaseous products) - (number of moles of gaseous reactants)

PV = nRTP(V)=(ngas)RT

• Work is defined as a force acting through a specified distance.

w= F x d = -PV = -(ngas)RT w = -(ngas)RT at constant T and R

47


When ThenExamples

1. V2 = V1 PV = 0

ngas = 0

• Consequently, there are three possibilities for volume changes:

CO(g) + H2O(g) H2(g) + CO2(g)2 mol gas 2 mol gas

2. V2 > V1 PV > 0

ngas > 0

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)3 mol gas 1 mol gas

3. V2 < V1 PV < 0

ngas < 0

N2(g) + 3 H2(g) 2 NH3(g)

4 mol gas 2 mol gas

48


• Consider the following gas phase reaction at constant pressure at 200oC.

2 NO(g) + O2(g) 2 NO2(g)

3 mol gas 2 mol gasV2 < V1 thus V<0 and PV<0w= -PV >0Works is done on system by surroundings

• Consider the following gas phase reaction at constant pressure at 1000oC.

PCl5(g) PCl3(g) + Cl2(g)

2 mol gas1 mol gasV2 > V1 thus V>0 and PV>0w= -PV <0Works is done by the system on the surroundings

49

Relationship of H and E• The total amount of heat energy that a

system can provide to its surroundings at constant temperature and pressure is given by

H= E + PV – which is the relationship between H and E.H = change in enthalpy of systemE = change in internal energy of systemPV = work done by system

50

Relationship of H and E• At the start of Chapter 15 we defined

H = qP.• Here we define H = E + PV.

– Are these two definitions compatible?• Remember E = q + w.• We have also defined w = -PV .

– Thus E = q + w = q -PV• Consequently, H = q- PV + PV = q

– At constant pressure H = qP.

51

Relationship of H and E• For reactions in which the volume change is

very small or equal to zero.– For small volume changes V 0 and PV 0 H = E + PV H E– For no volume change, H = E

52

Relationship of H and E• Change in enthalpy, H, or heat of reaction

is amount of heat absorbed or released when a reaction occurs at constant pressure.

• The change in energy, E, is the amount of heat absorbed or released when a reaction occurs at constant volume.

• How much do the H and E for a reaction differ?– The difference depends on the amount of work

performed by the system or the surroundings.

53

Relationship of H and E

Bomb calorimeter

54

Relationship of H and EExample 15-14: In Section 15-5, we noted that

Ho = -3523 kJ/mol for the combustion of n-pentane, n-C5H12. Combustion of one mol of n-pentane at constant pressure releases 3523 kJ of heat. What are the values of the work term and E for this reaction?C5H12(l) + 8 O2(g) 5 CO2(g) + 6 H2O(l) Ho

rxn= - 3523

kJ 8 mol gas 5 mol gasHo

rxn= - 3523 kJ/mol T=298Kw= -PV = -(ngas)RTngas = 5-8 mole = -3 molew= -(-3ml)(8.314 J/mol K)(298) = 7433 J = 7.433 kJH = E + PV E= H-PVE = -3523 kJ – (7.433 kJ) = 3516 kJ

55

Spontaneity of Physical and Chemical Changes

• Spontaneous changes happen without any continuing outside influences. – A spontaneous change has a natural

direction. (product-favored)• For example the rusting of iron occurs

spontaneously. ( 鐵生鏽 )– Have you ever seen rust turn into iron

metal without man made interference?• The melting of ice at room temperature

occurs spontaneously.– Will water spontaneously freeze at room

temperature?

56

The Two Aspects of Spontaneity

• Two factors affect the spontaneity of any physical or chemical change:– Exothermic

• An exothermic reaction does not ensure spontaneity.– For example, the freezing of water is

exothermic but spontaneous only below 0oC.

– Increase in the dispersal of energy and matter• An increase in disorder of the system

also does not insure spontaneity.– It is a proper combination of

exothermicity and disorder that determines spontaneity.

57

Entropy, S ( 熵 )• Entropy is a measure of the disorder or randomness of a system. 熵用以測量一個系統的亂度–The greater the energy dispersal in a system, the higher is its entropy

• As with H, entropies have been measured and tabulated in Appendix K as So

298. • When:

– S > 0 disorder increases (which favors spontaneity).

– S < 0 disorder decreases (does not favor spontaneity).

58

Entropy, S

Suniverse = Ssystem + Ssurroundings >0

• In general for a substance in its three states of matter:

Sgas > Sliquid > Ssolid

• Entropy increase (Ssysytem>0), When– Temperature increase– Volume increase– Mixing of substance– Increasing particle number– Molecular size and complexity– Ionic compounds with similar formulas but

different charges

59

Example Without doing a calculation, predict whether the entropy change will be positive or negativea) C2H6(g) +7/2 O2(g) 3H2O(g) + 2 CO2(g)

b) 3C2H2(g) C6H6(l)

c) C6H12O6(s) + 6 O2(g) 6 CO2(g) + 2 H2O(l)

a) S0>0 b) S0<0 c) S0>0

d) Hg(l), Hg(s), Hg(g)

e) C2H6(g), CH4(g) , C3H8(g)

f) CaS(s), CaO(s)

d) Hg(l)< Hg(s) <Hg(g)

e) CH4(g)< C2H6(g)< C3H8(g) f) CaO(s)< CaS(s)

60

Entropy, S• The Third Law of Thermodynamics

states, “The entropy of a pure, perfect, crystalline solid at 0 K is zero.” 熱力學第三定律是指當一個系統趨近於絕對溫度零度時，系統的熵變化率乃零。

• This law permits us to measure the absolute values of the entropy for substances.– To get the actual value of S, cool a

substance to 0 K, or as close as possible, then measure the entropy increase as the substance heats from 0 to higher temperatures.

– Notice that Appendix K has values of S not S.

61

• Entropy changes for reactions can be determined similarly to H for reactions.So

298 = nSoproducts - nSo

reactantsn n

Example 15-15: Calculate the entropy change for the following reaction at 25oC. Use appendix K.2 NO2(g) N2O4(g)

So298 = nSo

products - nSoreactantsn n

= SoN2O4(g) - 2So

NO2(g)

= (304.2 J/mol K) - 2x(240.0 J/mol k)= -175.8 J/mol k

• The negative sign of S indicates that the system is more ordered.

• If the reaction is reversed the sign of S changes.– For the reverse reaction So

298= +0.1758 kJ/mol K • The + sign indicates the system is more

disordered.

62

Entropy, S• Example 15-16: Calculate So

298 for the reaction below. Use appendix K.

3 NO(g) N2O(g) + NO2(g)

So298 = So

N2O(g) + SoNO2(g) – 3 So

NO(g)

= [219.7 + 240.0 – 3x(210.4)] J/mol K

= -172.4 J/mol K (-1.724 kJ/mol K)

• Changes in S are usually quite small compared to E and H.– Notice that S has units of only a fraction of a

kJ while E and H values are much larger numbers of kJ.

63

Example 15-15: Calculation of ΔSo for a phase change

Use the values of standard molar entropies in Appendix K to calculate the entropy change for the vaporation of one mole of bromine at 25oC.

Br2(l) Br2(g)

The values ΔSo from Appendix K:

Example 15-15: Calculation of ΔSo for a phase change

Use the values of standard molar entropies in Appendix K to calculate the entropy change for the vaporation of one mole of bromine at 25oC.

Br2(l) Br2(g)

The values ΔSo from Appendix K:Br2(l) Br2(g)

So, J/mol•k152.2 245.4

ΔSo= ΣnSoproducts - Σ nSo

reactants

= 1(245.4) – 1(152.2)

= 93.2 J/mol•K

64

Example 15-16: Calculation of ΔSorxn

Use the values of standard molar entropies in Appendix K to calculate the entropy change at 25oC and one atmosphere pressure for the reaction . Do you think the reaction is spontaneous?

N2H4(l) + 2H2O2(l) N2(g) + 4H2O(g) ΔHorxn= -642.2

kJ/mol rxn

The values ΔSo from Appendix K: N2H4(l) H2O2(l) N2(g) H2O(g)

Example 15-16: Calculation of ΔSorxn

Use the values of standard molar entropies in Appendix K to calculate the entropy change at 25oC and one atmosphere pressure for the reaction . Do you think the reaction is spontaneous?

N2H4(l) + 2H2O2(l) N2(g) + 4H2O(g) ΔHorxn= -642.2

kJ/mol rxn

The values ΔSo from Appendix K: N2H4(l) H2O2(l) N2(g) H2O(g)

So, J/mol•k121.2 109.6

ΔSorxn= ΣnSo

products - Σ nSoreactants

= [1(191.5)+4(188.7)] – [1(121.2)+2(109.6)]= 605.9 J/mol•K

191.5 188.7

Δ H<0 exothermic reactionΔ S >0Spontanoue reaction

65

The Second Law of Thermodynamics

• The second law of thermodynamics states, “In spontaneous changes the universe tends towards a state of greater disorder.” 熱力學第二定律指自發性的反應傾向最大亂度

• Spontaneous processes have two requirements:1.The free energy change of the system

must be negative. 系統內自由能的改變為負值2.The entropy of universe must increase. 亂

度必增加• Fundamentally, the system must be capable of

doing useful work on surroundings for a spontaneous process to occur.

66

Free Energy Change, G, and Spontaneity

• In the mid 1800’s J. Willard Gibbs determined the relationship of enthalpy, H, and entropy, S, that best describes the maximum useful energy obtainable in the form of work from a process at constant temperature and pressure.– The relationship also describes the

spontaneity of a system.• The relationship is a new state

function, G, the Gibbs Free Energy. G= H - TS (at constant T & P)

67

• The change in the Gibbs Free Energy, G, is a reliable indicator of spontaneity of a physical process or chemical reaction.– G does not tell us how quickly the

process occurs.• Chemical kinetics, the subject of Chapter 16,

indicates the rate of a reaction.• Sign conventions for G.

– G > 0 reaction is nonspontaneous– G = 0 system is at equilibrium– G < 0 reaction is spontaneous

• 化學反應的 DG>0 ，表示反應產生的能量不足以克服阻抗，反應不能夠自然地發生。

• 化學反應的 DG=0 ，表示反應處於平衡狀態，正向與逆向反應的驅力相等。

• 化學反應的 G<0 ，表示該反應釋放的能量足以克服外圍的阻抗，反應會順利地朝產物的方向進行，指示這是一個自發性的反應。DH 越小越有助於 DG<0 ，也就是反應放熱越多越有助於 DG<0 。

DS 越大越有助於 DG<0 ，也就是產物的亂度變得越大越有助於DG<0 。這並不意味，吸熱反應或減小亂度的反應不會自然發生。一個反應會不

會自然發生，完全看 DG是不是小於 0。

68


• Changes in free energy obey the same type of relationship we have described for enthalpy, H, and entropy, S, changes.

Go298 = nGo

products - nGoreactantsn n

69


• Example 15-17: Calculate Go298 for the

reaction in Example 15-8. Use appendix K.C3H8 (g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)

Go298 = nGo

products - nGoreactants

Go298 = [3 Go

fCO2(g) + 4 GofH2O(l)) ]-[Go

fC3H8(g) + 5GofO2(g)]

= {[3x(-394.4) + 4x(-273.3)]-[(-23.49)+5(0)]} kJ/mol= - 2108.9 kJ/mol

Go298 < 0, so the reaction is spontaneous at

standard state conditions.• If the reaction is reversed:

Go298 > 0, and the reaction is

nonspontaneous at standard state conditions.

70

The Temperature Dependence of

Spontaneity• Free energy has the relationship G = H -

TS.• Because 0 ≤ H ≥ 0 and 0 ≤ S ≥ 0, there

are four possibilities for G.

H S G Forward reaction spontaneity

< 0 > 0 < 0 Spontaneous at all T’s.

< 0 < 0T

dependent Spontaneous at low T’s.

> 0 > 0T

dependent Spontaneous at high T’s.

> 0 < 0 > 0Nonspontaneous at all

T’s.

71

The Temperature Dependence of Spontaneity

72

The Temperature Dependence of Spontaneity

Example 15-18: Calculate So298 for the following

reaction. In example 15-8, we found that Ho

298= -2219.9 kJ, and in Example 15-17 we found that Go

298= -2108.5 kJ.C3H8 (g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)

Go = Ho - TSo TSo = Ho- Go

So = Ho - Go T

So =[(-2219.9) – (-2108.50)] kJ

298 K

So = -0.374 kJ/K =-374 J/k

So298 = -374 J/K which indicates that the

disorder of the system decreases .For the reverse reaction,3CO2(g)+ 4H2O(g) C3H8(g) +

5 O2(g)

So298 = +374 J/K which indicates that the

disorder of the system increases .

73

The Temperature Dependence of

SpontaneityExample 15-18: Use thermodynamic data to

estimate the normal boiling point of water. And What is the percent error H2O (l) H2O(g)

Because this is an equilibrium process G=0Thus H = TS and T = H/SAssume H@BP Ho

298

Ho = HoH2O(g) - Ho

H2O(l) Ho = [(-241.8) – (-285.8)]Ho = 44.0 kJ @ 25oC

Assume S@BP Sorxn

Sorxn = So

H2O(g) - SoH2O(l)

Sorxn = [188.7 – 69.91]

Sorxn = 118.8 J/k

= 0.1188 kJ/kT = H/S Ho/So = 44.0 kJ/0.1188 kJ/K = 370 K 370 K – 273 K = 97 oC

% error =(370-373) K 373 K

x100% =-0.8% error

1 chapter 15 chemical thermodynamics 化學熱力學. 2 outline heat changes and thermochemistry...

Documents