1 Energetics 6.1What is Energetics? 6.2Enthalpy Changes Related to Breaking and Forming of Bonds 6.3Standard Enthalpy Changes 6.4Experimental Determination

1

EnergeticsEnergetics

6.16.1 What is Energetics?What is Energetics?

6.26.2 Enthalpy Changes Related to Breaking and Enthalpy Changes Related to Breaking and Forming of BondsForming of Bonds

6.36.3 Standard Enthalpy ChangesStandard Enthalpy Changes

6.46.4 Experimental Determination of Enthalpy Experimental Determination of Enthalpy ChaChanges by Calorimetrynges by Calorimetry

6.56.5 Hess’s LawHess’s Law

6.66.6 Calculations involving Standard Enthalpy Calculations involving Standard Enthalpy Changes of ReactionsChanges of Reactions

66

2

What is energetics?What is energetics?

Energetics is the study of energy changes associated with chemical reactions.Energetics is the study of energy changes associated with chemical reactions.

Thermochemistry is the study of heat changes associated with chemical reactions.Thermochemistry is the study of heat changes associated with chemical reactions.

6.1 What is energetics? (SB p.136)

3

U = kinetic energy + potential energy

Internal Energy (U)

4

Energy

translational rotational vibrational

heat

T (K)Kinetic Energy

Potential EnergyRelative position among particlesBond breaking P.E.

Bond forming P.E.

5

H – H(g) H(g) + H(g) P.E.

H(g) + H(g) H – H(g) P.E.

Na(g) Na+(g) + e P.E.

Bond breaking : -

Bond forming : -

Ionization : -

6

Reaction coordinate

Inte

rnal

energ

y bond breaking

4H(g) + 2O(g)

bond forming

U = U2 – U1 = -(y-x) kJ

2H2(g) + O2(g)U1

2H2O(l)U2

Q.1

7

Internal energy and enthalpyInternal energy and enthalpy6.1 What is energetics? (SB p.137)

H = U + PV

enthalpyInternal energy

8

Internal energy and enthalpyInternal energy and enthalpy

e.g. Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)


qv = U = -473 kJ mol1 qp = H = -470 kJ mol1

Mg Mg

9

qv = qp – 3

= qp – w = qp - PV

Work done against the surroundings

qv = U = -473 kJ mol1 qp = H = -470 kJ mol1

PV (Nm2)(m3)

= NmForce displacement

10


H = U + PV

= U + PV at constant P

qp = qv + PVHeat

change at fixed P

Heat change at

fixed V

Work done

11


qp = qv + PV

On expansion, PV > 0

Work done by the system against the surroundings

System gives out less energy to the surroundings

qp is less negative than qv (less exothermic)

12


qp = qv + PV

On contraction, PV < 0

Work done by the surroundings against the system

System gives out more energy to the surroundings

qp is more negative than qv (more exothermic)

13

• H is more easily measured than H as

most reactions happen in open vessels.

i.e. at constant pressure.

• The absolute values of H and H cannot

be measured.

14

Exothermic and endothermic Exothermic and endothermic reactionsreactions

An exothermic reaction is a reaction that releases heat energy to the surroundings. (H = -ve)

An exothermic reaction is a reaction that releases heat energy to the surroundings. (H = -ve)


15

An endothermic reaction is a reaction that absorbs heat energy from the surroundings. (H = +ve)

An endothermic reaction is a reaction that absorbs heat energy from the surroundings. (H = +ve)


Exothermic and endothermic Exothermic and endothermic reactionsreactions

Check Point 6-1Check Point 6-1

16

Law of conservation of energyLaw of conservation of energy


The law of conservation of energy states that energy can neither be created nor destroyed,

but can be exchanged between a system and its surroundings

The law of conservation of energy states that energy can neither be created nor destroyed,

but can be exchanged between a system and its surroundings

17

Exothermic : -

P.E. of the system K.E. of the surroundings

Endothermic : -

K.E. of the surroundings P.E. of the system

18

Enthalpy changes related to breaking Enthalpy changes related to breaking and forming of bondsand forming of bonds

CH4 + 2O2 CO2 + 2H2O

6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)

19


In an exothermic reaction,

E absorbed to break bonds < E released as bonds are formed.

In an exothermic reaction,

E absorbed to break bonds < E released as bonds are formed.

20

Enthalpy changes related to breaking Enthalpy changes related to breaking and forming of bondsand forming of bonds


N2(g) + 2O2(g) 2NO2(g)

21


In an endothermic reaction,

E absorbed to break bonds > E released as bonds are formed.

In an endothermic reaction,

E absorbed to break bonds > E released as bonds are formed.


22

For non-gaseous reactions,

PV 0

H = U + PV U

For gaseous reactions,

H = U + PV = U + (n)RT (PV = nRT)

23

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

U = H – (n)RT

29810008.314

3)(1890

= 885 kJ mol1

Given : R = 8.314 J K1 mol1, T = 298 K

Q.2

24

Reaction coordinate

En

thalp

y

bond breaking

C(g) + 4H(g) + 4O(g)

bond forming

H = 890 kJ mol1

CH4(g) + 2O2(g)H1

CO2(g) + 2H2O(l)H2

25

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) 890

H / kJ mol1

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) 802

At 298K

At 373K

2H2O(l) 2H2O(g) +88 kJ

373K 373K

26

Reaction coordinate

En

thalp

y

CH4(g) + 2O2(g)

C(g) + 4H(g) + 4O(g)

CO2(g) + 2H2O(l)

298 K

H = 890 kJ mol1

H1

H2

27

Reaction coordinate

En

thalp

y

CH4(g) + 2O2(g)

C(g) + 4H(g) + 4O(g)

CO2(g) + 2H2O(l)

373 K

H = 890 kJ mol1

Assume constant H

H1’

H2’

28

Reaction coordinate

En

thalp

y

CH4(g) + 2O2(g)

C(g) + 4H(g) + 4O(g)

CO2(g) + 2H2O(l)

373 K

H = 890 kJ mol1

In fact, H depends on T

29

Reaction coordinate

En

thalp

y

CH4(g) + 2O2(g)

C(g) + 4H(g) + 4O(g)

CO2(g) + 2H2O(l)

CO2(g) + 2H2O(g)

H = 802 kJ mol1

H = +88 kJ mol1

373 K

30

6.6.33 Standard Standard

Enthalpy Enthalpy ChangesChanges

31

Standard enthalpy changesStandard enthalpy changes

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

H = -802 kJ mol-1 at 373 K

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

H = -890 kJ mol-1 at 298 K

6.3 Standard enthalpy changes (SB p.141)

32

Standard enthalpy changesStandard enthalpy changes

As enthalpy changes depend on temperature and pressure, it is necessary to define standard conditions:

1. elements or compounds in their normal physical states;2. a pressure of 1 atm (101325 Nm-2); and3. a temperature of 25oC (298 K)

1. elements or compounds in their normal physical states;2. a pressure of 1 atm (101325 Nm-2); and3. a temperature of 25oC (298 K)

Enthalpy change under standard conditions denoted by symbol: H

ø


33

Standard enthalpy change of reaction

The enthalpy change when the molar quantities of reactants as stated in the equation react under standard conditions.

2H2(g) + O2(g) 2H2O(l)

H = 572 kJ mol1

per mole of O2

34

Standard enthalpy change of reaction

2H2(g) + O2(g) 2H2O(l)

H = 572 kJ mol1

per mole of O2

H = 286 kJ mol1 H2(g) + O2(g) H2O(l)

2

1

per mole of H2 or H2O

H

depends on the equation

4H2(g) + 2O2(g) 4H2O(l)

H = 1144 kJ

35

Standard enthalpy change of formation

Hf

The enthalpy change when one mole of the substance is formed from its elements under standard conditions.

H2(g) + O2(g) H2O(l)2

1Hf [H2O] = 286 kJ mol1

Q.3

O2(g) O2(g) Hf [O2] = 0 kJ mol1

Hf [element] = 0 kJ mol1

36

C(graphite) C(diamond) Hf [diamond] = +1.9 kJ mol1

Most stable allotrope

37

Q.4

(i) C(graphite) + O2(g) CO2(g)

(ii) C(graphite) + 2H2(g) CH4(g)

(iii) Mg(s) + O2(g) MgO(s)2

1

(v) 2C(graphite) + 2H2(g) + O2(g) CH3COOH(l)

(iv) Na(s) + H2(g) + C(graphite) + O2(g) 2

1

2

3 NaHCO3(s)

38

mol 0.4961.0082

1.00 used (g)H of moles no.of 2

mol 0.248 mol 2

0.496 used (g)O of moles no.of 2

H2(g) + O2(g) H2O(l)2

1 qv = U = 140.3 kJ per g of H2

Q.5

n = 0 – 0.496 – 0.248 = -0.744 mol

39

H = U + nRT

K) (2981000

K mol kJ 8.31mol) (-0.744 kJ 140.3

-1-1

= -142.1 kJHeat released for the formation of 0.496 mol of

water

Molar Hf[H2O] = 1-mol kJ -286.5

mol 0.496kJ 142.1

Q.5

40

Standard enthalpy change of combustion

Hc

The enthalpy change when one mole of the substance undergoes complete combustion under standard conditions.

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)

Hc [C2H5OH(l)] = -1368 kJ mol1

41

-395.4

-393.5

C (diamond)

C (graphite)

Hc (kJ mol-1)Substance

ø


Reaction coordinate

En

thalp

y C(diamond) + O2(g)

CO2(g)

395.4

C(graphite) + O2(g)

393.5

1.9

Hf [diamond]

= +1.9 kJ mol1

42

Q.6

H

= Hf [CO(g)]

Hc [graphite]

H

= 2 Hc [H2(g)]

(b) 2H2(g) + O2(g) 2H2O(l)

(a) C(graphite) + O2(g) CO(g)2

1 Incomplete combustion

= 2 Hf [H2O(l)]

43

Q.6

(c) C(graphite) + O2(g) CO2(g)

H

= Hc [graphite]

= Hf [CO2(g)]

(d) CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

H

= Hc [CH4(g)]

Hf [CO2(g)]

2 Hf [H2O(l)]

Not formed from elements


44

Standard enthalpy changes of Standard enthalpy changes of neutralizationneutralization

Standard enthalpy change of neutralization (Hne

ut) is the enthalpy change when one mole of water is formed from the neutralization of an acid by an alkali under standard conditions.

Standard enthalpy change of neutralization (Hne

ut) is the enthalpy change when one mole of water is formed from the neutralization of an acid by an alkali under standard conditions.

ø

e.g. H+(aq) + OH-(aq) H2O(l)

Hneut = -57.3 kJ mol-1

ø


45


Standard enthalpy changes of Standard enthalpy changes of neutralizationneutralization

Enthalpy level diagram for the neutralization of a strong acid and a strong alkali

46

-57.1

-57.2

-52.2

-68.6

NaOH

KOH

NH3

NaOH

HCl

HCl

HCl

HF

Hneu AlkaliAcid ø


NH3(aq) + HCl(aq) NH4Cl (aq)

H+(aq) + OH-(aq) + Cl(aq) H2O(l) + Cl(aq) H2 = -57.3 NH3(aq) + H2O(l) NH4

+(aq) + OH(aq) H1 > 0

ø

Hneu = H1 + H2 = 52.2 kJ mol1

47

-57.1

-57.2

-52.2

-68.6

NaOH

KOH

NH3

NaOH

HCl

HCl

HCl

HF

Hneu AlkaliAcid ø


HF(aq) + NaOH(aq) NaF(aq) + H2O(l)

H+(aq) + OH-(aq) + Na+(aq) H2O(l) + Na+(aq) H2 = -57.3 HF(aq) H+(aq) + F(aq) H1 < 0

ø

Hneu = H1 + H2 = 68.6 kJ mol1

48

Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is dissolved in a specified number of moles of solvent (e.g. water) under standard conditions.

Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is dissolved in a specified number of moles of solvent (e.g. water) under standard conditions.

ø


Standard enthalpy change of Standard enthalpy change of solutionsolution

NaCl(s) + 10H2O(l) Na+(aq) + Cl-(aq)

Hsoln[NaCl(s)]= +2.008 kJ mol-1

ø

NaCl(aq) NaCl(aq)dilution H > 0

49

Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is dissolved to form an infinitely dilute solution under standard conditions.

Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is dissolved to form an infinitely dilute solution under standard conditions.

ø



NaCl(s) + water Na+(aq) + Cl-(aq)

øHsoln= +4.98 kJ mol-1

concentration 0

50

e.g. NaCl(s) + water Na+(aq) + Cl-(aq)

Hsoln= +4.98 kJ mol-1ø



Enthalpy level diagram for the dissolution of Na

Cl

+ 4.98 kJ mol1

51


Standard enthalpy change of Standard enthalpy change of solutionsolutione.g. LiCl(s) + water Li+(aq) + Cl-(aq)

Hsoln= -37.2 kJ mol-1ø

Enthalpy level diagram for the dissolution of Li

Cl in water

52

35.543.173.074.037.2+4.98+21.0+22.6

NH3

NaOHHCl

H2SO4

LiClNaCl

NaNO3

NH4Cl

Hsoln(kJ mol-1)Salt

ø



53

6.6.44Experimental DeteExperimental Determination of Enthrmination of Enthalpy Changes by Calpy Changes by C

alorimetryalorimetry

54

Experimental determination of enthalpy chaExperimental determination of enthalpy changes by calorimetrynges by calorimetry

Calorimeter is any set-up used for the determination of H.

By temperature measurement.

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.148)

H = qp = (m1c1 + m2c2)T

55

H = qp = (m1c1 + m2c2)Twhere

m1 is the mass of the reaction mixture,

m2 is the mass of the calorimeter,

c1 is the specific heat capacity of the reaction mixture,

c2 is the specific heat capacity of the calorimeter,

T is the temperature change of the reaction mixture.


56

Determination of enthalpy change of Determination of enthalpy change of neutralizationneutralization


57

If the reaction is fast enough, T1 T2

T0

T1

t1

T2

t2

H = (m1c1 + m2c2)(T1 – T0)

H (m1c1 + m2c2)(T2 – T0)

Check Point 6-4(a)Check Point 6-4(a)

58


Determination of enthalpy change Determination of enthalpy change of combustionof combustion

The Philip Harris calorimeter used for

determining the enthalpy change of

combustion of a liquid fuel

59

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151) Determination of enthalpy change Determination of enthalpy change

of combustionof combustion

A simple apparatus used to determine the enthalpy change of combustion of ethanol

60

Heat evolved = (m1c1 + m2c2) ΔT

Where

m1 is the mass of water in the calorimeter,

m2 is the mass of the calorimeter,

c1 is the specific heat capacity of the water,

c2 is the specific heat capacity of the calorimeter,

ΔT is the temperature change of the reaction


61

kJ 43.5 K) )(13Kg J g)(4.18 (800 q 1-1p

mol 0.326g 46.0

g 1.5 ethanol of moles of no.

1-c mol kJ 1330-

mol 0.0326kJ 43.5-

[ethanol]ΔH

Q.7(Example)

Check Point 6-4(c)Check Point 6-4(c)

heat given out

62

6.6.55 Hess’s LawHess’s Law

63

Hess’s Hess’s LawLawHess’s law of constant heat summation states that the total enthalpy change accompanying a chemical reaction

is independent of the route by which the chemical reaction takes place and

depends only on the difference between the total enthalpy of the reactants and that of the products.

Hess’s law of constant heat summation states that the total enthalpy change accompanying a chemical reaction

is independent of the route by which the chemical reaction takes place and

depends only on the difference between the total enthalpy of the reactants and that of the products.

6.5 Hess’s law (SB p.153)

64

Hess’s Hess’s LawLaw

A(HA) B(HB)Route

1H1

D

H4 H5

Route 3

H1 = HB – HA H1 = HB – HA


= H2 + H3

= H2 + H3

= H4 + H5= H4 + H5

CH2 H3

Route 2

65

Importance of Hess’s lawImportance of Hess’s lawThe enthalpy change of some chemical reactions cannot be determined directly because:

But the enthalpy change of such reactions can be determined indirectly by applying Hess’s Law.


• the reactions cannot be performed/controlled in the laboratory• the reaction rates are too slow• the reactions may involve the formation of

side products

66

Enthalpy change of formation of CO(g)Enthalpy change of formation of CO(g)

C(graphite) + ½O2(g)

CO(g)Hf [CO(g)]ø


due to further oxidation of CO to CO2

The reaction cannot be controlled.

Hf [CO(g)]ø cannot be determined directly

67

Enthalpy change of formation of CO(g)Enthalpy change of formation of CO(g)

= -393 - (-283 )= -110 kJ mol-1

Hc [graphite] = -393 kJ mol-1 ø

H2

+ ½O2(g)

CO2(g)H1

+ ½O2(g)


CO(g)Hf [CO(g)]ø

Hf [CO(g)] + H2 = H1Hf [CO(g)] + H2 = H1

ø

Hf [CO(g)] = H1 - H2

ø


Hc [CO(g)] = -283.0 kJ mol-1ø

68


Enthalpy cycle (Born-Haber cycle)Enthalpy cycle (Born-Haber cycle)

• Relate the various equations involved in a reaction

H2

+ ½O2(g)

CO2(g)H1

+ ½O2(g)


CO(g)Hf [CO(g)]ø

69

Steps for drawing Born-Haber cycleSteps for drawing Born-Haber cycle


CO(g)Hf [CO(g)]ø


1. Give the equation for the change being considered.

70


H2

+ ½O2(g)

CO2(g)H1

+ ½O2(g)


2. Complete the cycle by giving the equations for the

combustion reactions of reactants and products.


CO(g)Hf [CO(g)]ø

Hf [CO(g)]= Hc[graphite] - Hc[CO(g)]Hf [CO(g)]= Hc[graphite] - Hc[CO(g)]ø øø

71


H2

+ ½O2(g)

CO2(g)H1

+ ½O2(g)


2. Complete the cycle by giving the equations for the

combustion reactions of reactants and products.


CO(g)Hf [CO(g)]ø

Hf [CO(g)]= Hc[reactant] - Hc[product]Hf [CO(g)]= Hc[reactant] - Hc[product]ø øø

72

Calculation of standard enthalpy Calculation of standard enthalpy change of formation from standard change of formation from standard enthalpy changes of combustionenthalpy changes of combustion

Hf = Hc [reactants] - Hc [product]Hf = Hc [reactants] - Hc [product]ø øø

6B

73

Q.8

Hf [C4H10(g)]

= 4Hc[C(graphite)] + 5Hc[H2(g)] - Hc[C4H10(g)]

ø

ø øø

4C(graphite) + 5H2(g)

C4H10(g)

Hf [C4H10(g)]ø

4CO2(g)

4Hc

[graphite]

ø

+ 4O2(g)

5Hc [H2(g)]ø

+ 5H2O(l)

+ 2.5 O2(g)

Hc [C4H10(g)]ø

+ 6.5 O2(g)

= [4(-393)+ 5(-286) – (2877)] kJ mol1

= 125 kJ mol1

74

Q.8 Method B :

By addition and/or subtraction of equations with known

Hc

(1) C(graphite) + O2(g) CO2(g) 393Hc /kJ mol1

(2) H2(g) + O2(g) H2O(l) 28621

(3) C4H10(g) + O2(g) 4CO2(g) + 5H2O(l) 287721

6

Overall reaction : 4(1) + 5(2) – (3)

4C(graphite) + 5H2(g)

C4H10(g)

Hf [C4H10(g)]ø

Hf [C4H10(g)] = [4(-393) + 5(-286) (2877)] kJ mol1

= 125 kJ mol1

75


Enthalpy level Enthalpy level diagramdiagram• Relate substances together in terms of

enthalpy changes of reactions

Enthalpy level diagram for the

oxidation of C(graphite) to

CO2(g)

76

En

thalp

y /

kJ m

ol1

Steps for drawing enthalpy level diagram

1. Draw the enthalpy level of elements.

C(graphite) + O2(g)

77

En

thalp

y /

kJ m

ol1


2. Enthalpies of elements are arbitrarily taken as zero.

C(graphite) + O2(g)

78

En

thalp

y /

kJ m

ol1


3. Higher enthalpy levels are drawn above that of elements

C(graphite) + O2(g)

79

En

thalp

y /

kJ m

ol1


C(graphite) + O2(g)

4. Lower enthalpy levels are drawn below that of elements

Route 1

CO2(g)

Hc[graphite] = 393 kJ

80

En

thalp

y /

kJ m

ol1


C(graphite) + O2(g)

4. Lower enthalpy levels are drawn below that of elements

Route 1

CO2(g)


CO(g) + O2(g)21

Hc[CO(g)] = 283 kJ

Hf[CO(g)] = 110 kJ

Route 2

81


(c)The formation of ethyne (C2H2(g) can be represented by the following equation:

2C(graphite) + H2(g) C2H2(g)

(i) Draw an enthalpy cycle relating the above equation to carbon dioxide and water.

(ii) Calculate the standard enthalpy change of formation of ethyne.

(Given: Hc [C(graphite)] = -393.5 kJ mol-1;

Hc [H2(g)] = -285.8 kJ mol-1;

Hc [C2H2(g)] = -1299 kJ mol-1)ø

ø

ø

82



+ H2O(l)

Hf

2CO2(g)

2Hc[graphite]

+ 2O2(g)

Hc[H2(g)]

+ 0.5O2(g)

Hc[C2H2(g)]

+ 2.5O2(g)

By Hess’s law,Hf

+ Hc[C2H2(g)] =

2Hc[graphite]

+ Hc[H2(g)]

Hf

- Hc[C2H2(g)]

= 2Hc[graphite]

+ Hc[H2(g)]

= [2(393.5) + (285.8) –(1299)] kJ mol1

= +226.2 kJ mol1

83

En

thalp

y /

kJ m

ol1

2C(graphite) + H2(g) + 2.5O2(g)

(iii). Draw an enthalpy level diagram for the reaction using the enthalpy changes in (ii)

Route 1

2CO2(g) + H2O(l)

Route 2

Hc[H2(g)]

Hf[C2H2(g)]

C2H2(g) + 2.5O2(g)

2CO2(g) + H2(g) + 0.5O2(g)

2Hc[graphite] Hc[C2H2(g)]

84

6.6.66Calculations Calculations

involving involving Standard Enthalpy Standard Enthalpy

Changes of Changes of ReactionsReactions

85

Calculation of standard enthalpy Calculation of standard enthalpy change of reaction from standard change of reaction from standard enthalpy changes of formationenthalpy changes of formation

6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)

H from Hf

86

Q.9

NH3(g) + HCl(g) NH4Cl(s)

+ 0.5H2(g) + 0.5Cl2(g)

0.5N2(g) + 1.5H2(g)

Hf[NH4Cl(s)]

Hf[NH3(g)]

Hf[HCl(g)]

H

By Hess’s law,Hf[NH3(g)]

Hf[HCl(g)]

H

+ + = Hf[NH4Cl(s)]

Hf[NH4Cl(s)]

Hf[NH3(g)]

H

--= Hf[HCl(g)]

= [314 –(46) –(92)] kJ mol1 = 176 kJ mol1

87

Q.9

NH3(g) + HCl(g) NH4Cl(s)

+ 0.5H2(g) + 0.5Cl2(g)

0.5N2(g) + 1.5H2(g)

Hf[NH4Cl(s)]

Hf[NH3(g)]

Hf[HCl(g)]

H

By Hess’s law,Hf[NH3(g)]

Hf[HCl(g)]

H

+ + = Hf[NH4Cl(s)]

Hf[NH4Cl(s)]

Hf[NH3(g)]

H

--= Hf[HCl(g)]

Hreaction = Hf [products] - Hf [reactants]Hreaction = Hf [products] - Hf [reactants]ø ø

88

Q.10

4CH3NHNH2(l) + 5N2O4(l) 4CO2(g) + 9N2(g) + 12H2O(l)

4Hf[CO2(g)]

4C(graphite) + 12H2(g) + 4N2(g)

4Hf[CH3NHNH2(l)]

+ 5N2(g) + 10O2(g)

5Hf[N2O4(l)]

H

By Hess’s law,

= [4(393) + 12(286) - 4(+53) – 5(20)] kJ

= 5116 kJ

12Hf[H2O(l)]

H

Highly exothermic/ignites spontaneously

Used as rockel fuel in Apollo 11

89

Q.11

C3H6(g) + H2(g) C3H8(g)

H

(1) C3H6(g) + 4.5O2(g) 3CO2(g) + 3H2O(l) 2058

H /kJ mol1

(2) H2(g) + 0.5O2(g) H2O(l) 286

(3) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) 2220Overall reaction : (1) + (2) – (3)

= [(2058) + (286) – (2220)] kJ mol1

= 124 kJ mol1H

90

Q.11

C3H6(g) + H2(g) C3H8(g)

H

= [(2058) + (286) – (2220)] kJ mol1

= 124 kJ mol1H

Hc[C3H8(g)]

3H2O(l) + 3CO2 (g)

Hc[C3H6(g)]

Hf[H2O(l)]

+ H2O(l)

H

+ (2220) = (2058) + (286)

By Hess’s law,

91

Q.11

C3H6(g) + H2(g) C3H8(g)

H

Hc[C3H8(g)]

3H2O(l) + 3CO2 (g)

Hc[C3H6(g)]

Hc[H2(g)]

+ H2O(l)

Hf = Hc [reactants] - Hc [product]Hf = Hc [reactants] - Hc [product]ø øø

Hrx = Hc [reactants] - Hc [products]Hrx = Hc [reactants] - Hc [products]ø øø

92

Relative stability of compounds and Hf

Hf indicates the energetic stability of the compound with respect to its elements

Hf

< 0 energetically more stable than its elements

Hf

> 0 energetically less stable than its elements

Hf

Elements Compound

93

H2(g) + O2(g) H2O2(l) Hf [H2O2(l)] = 188 kJ mol1

H2O2(l) H2O(l) + 0.5O2(g)

Hrx = 98 kJ mol1

H2O(l) + 0.5O2(g)

En

thalp

y /

kJ m

ol1

H2(g) + O2(g)

H2O2(l)

Hrx= 98 kJ

Hf[H2O2(l)] = 188 kJ

94

En

thalp

y /

kJ m

ol1

H2(g) + O2(g)

H2O(l) + 0.5O2(g)

H2O2(l)

Hrx= 98 kJ

H2O2 is energetically stable w.r.t. H2 and O2

H2O2 is energetically unstable w.r.t. H2O and 0.5O2

Hf[H2O2(l)] = 188 kJ

95

En

thalp

y /

kJ m

ol1

H2(g) + O2(g)

H2O(l) + 0.5O2(g)

Hc[H2(g) ] = 286 kJ

H2O2(l)

Hrx= 98 kJ

H2 and O2 are energetically unstable w.r.t. H2O2 and H2O

Hf[H2O2(l)] = 188 kJ

96

C(graphite) + O2(g)

En

thalp

y /

kJ m

ol1

C(diamond) + O2(g)

CO2(g)

Hc[diamond] = 395 kJ

H = 2 kJ

C(diamond) C(graphite) H = 2 kJ mol1


Energetically unstable

97

C(graphite) + O2(g)

En

thalp

y /

kJ m

ol1

C(diamond) + O2(g)

CO2(g)


H = 2 kJ



Kinetically stable The rate of conversion is extremely low

98

C(graphite) + O2(g)

En

thalp

y /

kJ m

ol1

C(diamond) + O2(g)

CO2(g)

Hc[diamond] = 395 kJ Hc[graphite]

= 393 kJ

C(g) + O(g) + O(g)

bond breaking

bond forming

99

C(graphite) + O2(g)

En

thalp

y /

kJ m

ol1

C(diamond) + O2(g)

CO2(g)


H = 2 kJ


Rate of reaction depends on the ease of bond breaking in reactants (e.g. diamond)

The minimum energy required for a reaction to start is known as the activation energy, Ea

100

C(graphite) + O2(g)

En

thalp

y /

kJ m

ol1

C(diamond) + O2(g)

CO2(g)


H = 2 kJ



The extremely low rate is due to high Ea

101

C(graphite) + O2(g)

En

thalp

y /

kJ m

ol1

C(diamond) + O2(g)

CO2(g)


H = 2 kJ



Ea is always > 0 as bond breaking is endothermic

102

Ea tells how fast a reaction can proceed.

Hf

tells how far a reaction can proceed

Rate of reaction / kinetics / kinetic stability

Equilibrium / energetics / energetic stability

103

Energetic stabilityHf

Ea kinetic stability (rate of reaction)

Higher Ea higher kinetic stability of reactants w.r.t. products

Hf

> 0 lower energetic stability w.r.t. its elements

< 0 higher energetic stability w.r.t. its elementsHf

Lower Ea lower kinetic stability of reactants w.r.t. products

lower rate to give products

higher rate to give products

104

diamond

graphite

Diamond energetically unstable w.r.t. graphite

kinetically stable w.r.t. graphite

Graphite stable w.r.t. diamond energetically and kinetically

105

diamond graphiteextremely slowextremely slow

diamond

graphite

106

6.6.77Spontaneity of Spontaneity of

ChangesChanges

6.7 Entropy change (SB p.164)

107

Spontaneity : The state of being spontaneous

Spontaneous :

- self-generated

- natural

- happening without external influence

internal

external

external

externalexternal

boundary

108


A process is said to be spontaneous• If no external “forces” are required to keep t

he process going,

although external “forces” may be required to get the process started (Ea).

• The process may be a physical change or a chemical change

109


• Spontaneous physical change:

E.g. condensation of steam at 25°C• Spontaneous chemcial change:

E.g. burning of wood once the fire has started

Exothermic spontaneous ?

Endothermic not spontaneous ?

Q.12

110

NoDissolution of

NH4Cl in water at 25°C

NoMelting of ice at

25°C

YesCondensation of steam at 25°C

Yes

Yes

Yes

YesYesBurning of CO at

25°C

Spontaneous ?

(Yes / No)Exothermic ?

(Yes / No)Change

111


• Exothermicity is NOT the only reason for the spontaneity of a process

• Some spontaneous changes are endothermic

• E.g. Melting of ice

Dissolution of NH4Cl in water

112


Melting of ice

Dissolution of ammonium nitrate in

water

113


EntropyEntropy• Entropy is a measure of the randomness or

the degree of disorder (freedom) of a system

Solid Liquid Gas

Entropy increases

114


Entropy change (Entropy change (S)S)

• Entropy change means the change in the degree of disorder of a system

S = Sfinal - Sinitial

115

6.7 Entropy change (SB p.166)Positive entropy change (Positive entropy change (S S

> 0)> 0)

• Increase in entropy

• Sfinal > Sinitial

• Example:

Ice (low entropy) Water (high entropy)

S = Swater – Sice = +ve

116

6.7 Entropy change (SB p.166)Negative entropy change (Negative entropy change (S < S <

0)0)

• Decrease in entropy

• Sinitial > Sfianl

• Example:

Water (high entropy) Ice (low entropy)

S = Sice – Swater = -ve

Q.13

117

Changes S

CO(g) + O2(g) CO2(g)

H2O(g) H2O(l)

H2O(s) H2O(l)

NH4Cl(s) NH4Cl(aq)

118

Changes S

2CO(g) + O2(g) 2CO2(g) -ve

H2O(g) H2O(l) -ve

H2O(s) H2O(l) +ve

NH4Cl(s) NH4Cl(aq) +ve

119

Changes S

C(s) + O2(g) CO2(g)

SO2(g) + O2(g) SO3(g)

Diffusion of a drop of ink in water

diamond graphite

120

Changes S

C(s) + O2(g) CO2(g) +ve

2SO2(g) + O2(g) 2SO3(g) -ve

Diffusion of a drop of ink in water

+ve

diamond graphite +ve

2 2

121

Consider an isolated system which has no exchange of energy and matter with its surroundings

122

S = S2 – S1 > 0

S1 S2

S2 S1

S = S1 – S2 < 0

Which one is spontaneous ?

123

Spontaneous, S = S2 – S1 > 0

S1 S2

S2 S1

Not spontaneous, S = S1 – S2 < 0

124

Free adiabatic expansion of an ideal gas into a vacuum.

A molecular statistical interpretation

Slit is open

vacuum

Probability

16

1

2

1

2

1

2

1

2

1

125

Free adiabatic expansion of an ideal gas into a vacuum.

A molecular statistical interpretation

Slit is open

vacuum

Probability

23

23

108.1

106

10

1

2

1

1 mole

126

A spontaneous process taking place in an isolated system is always associated with an increase in entropy (I.e. S > 0)In closed system,

The spontaneity of a process depends on both H and S.

The driving force of a process is a balance of H and S.

6B

127


Ice (less entropy) Water (more entropy)

H is +ve not favourable

S is +ve favourable

Considering both S & H ,

the process is spontaneous

Q.14

128

Changes H SSpontaneous

(Yes / No)

H2O(g) H2O(l) at 25°C

H2O(g) H2O(l) at 110°C

H2O(s) H2O(l) at 25°C

H2O(s) H2O(l) at -10°C

129


(Yes / No)

H2O(g) H2O(l) at 25°C -ve -ve Yes

H2O(g) H2O(l) at 110°C -ve -ve No

H2O(s) H2O(l) at 25°C +ve +ve Yes

H2O(s) H2O(l) at -10°C +ve +ve No

130


(Yes / No)





Favourable

131


(Yes / No)





Not favourable

132


(Yes / No)





Spontaneity depends on temperature

133


(Yes / No)





Favourable

134


(Yes / No)





Not favourable

135


(Yes / No)





Spontaneity depends on temperature

136

Spontaneity of a process depends on

H, S & T

G = H –TS

G is the (Gibbs’) free energy

J. Willard Gibbs (1839 – 1903)

137

G = H –TS

Spontaneity depends on G

‘Free’ means the energy free for work

138

A spontaneous process is always associated with a decrease in the free energy of the system.

G < 0 spontaneous process

G > 0 not spontaneous process

Q.15

139

H S T G Results

+ve +ve high

+ve +ve low

-ve -ve high

-ve -ve low

G = H –TS

140

Spontaneous-velow-ve-ve

Not spontaneus+vehigh-ve-ve

Not spontaneous+velow+ve+ve

Spontaneous-vehigh+ve+ve

ResultsGTSH

G = H –TS

141

H S T G Results

-ve +ve high

-ve +ve low

+ve -ve high

+ve -ve low

G = H –TS

142

low-ve+veNot spontaneous+ve

high-ve+ve

low+ve-veSpontaneous-ve

high+ve-ve

ResultsGTSH

G = H –TS

143

G = H –TS

Q.16

144

Diamond Graphite

G = H –TS < 0

H < 0

S > 0

The process is spontaneous, although activation energy is required to start the conversion.

145

Spontaneous S = S2 – S1 > 0

S1 S2

The entropy of a system can be considered as a measure of the availability of the system to do work.

Before expansion, the system is available to do work.

After expansion, the system is not available to do work.

146

Spontaneous S = S2 – S1 > 0

S1 S2

The lower the entropy of a system(before expansion), the more available is the system to do work.

Thus, entropy is considered as a measure of the useless energy of a system.

147

G = H –TS

G = H – TS

H = G + TS

Useless energyTotal energy

Useful energy

148

H = G + TS

Useless energyTotal energy

Useful energy

If the universe is an isolated system

H is a constant and H is always zero

149

H = G + TSH = G + TS = 0

cosmic background radiation = 4K

Useful energy

Useless energy

= G + TS = 0S always >

0, G always < 0,

S always , useless energy always

G always , useful energy always

150

In an isolated system, entropy will only increase with time, it will not decrease with time.

The second law of thermodynamics

151

If the universe is an isolated system,

Suniverse = Ssystem + Ssurroundings > 0

the total entropy (randomness) of the universe will tend to increase to a maximum;

the total free energy of the universe will tend to decrease to a minimum.

152

As time increases, the universe will always become more disordered.

Entropy is considered as a measure of time.

Entropy can be used to distinguish between future and past.

153

Time can only proceed in one direction that results in an increase in the total entropy of the universe.

This is known as the arrow of time.

154

The history of the universeminimum entropy,

maximum free energy (singularity)

expanding

maximum entropy, minimum free energy

Big bang

Big chill

H = G +TS

T 0 K

T 1.41032 K

155

Planck’s units

Planck’s length

Planck’s time

1.416785(71) × 1032 K Planck’s

temperature 2

5

2 Gk

hcTP

h = Planck’s constant G = gravitational constant c = speed of light in vacuum k = Boltzmann constant

156

Planck’s units

1.416785(71) × 1032 K Planck’s

temperature 2

5

2 Gk

hcTP

Absolute hot beyond which all physical laws break down

Planck’s length

Planck’s time

157

Planck’s units

1.416785(71) × 1032 K Planck’s

temperature 2

5

2 Gk

hcTP

Planck’s length

Planck’s time

1.616252(81) × 10−35 m32 c

hGlP

Physical significance not yet

known

158

Planck’s units

1.416785(71) × 1032 K Planck’s

temperature 2

5

2 Gk

hcTP

Planck’s length

Planck’s time

1.616252(81) × 10−35 m32 c

hGlP

the diameter of proton20101

159

Planck’s units

1.416785(71) × 1032 K Planck’s

temperature 2

5

2 Gk

hcTP

Planck’s length

Planck’s time

1.616252(81) × 10−35 m32 c

hGlP

= 7.310-37 m22 times of Mr Chio’s

wavelength

160

1.616252(81) × 10−35 m32 c

hGlP

Planck’s units

1.416785(71) × 1032 K Planck’s

temperature 2

5

2 Gk

hcTP

It is the time required for light to travel, in a vacuum, a distance of 1 Planck length.

52 c

hGt p

5.39124(27) × 10−44 s

Planck’s length

Planck’s time

161

1.616252(81) × 10−35 m32 c

hGlP

Planck’s units

1.416785(71) × 1032 K Planck’s

temperature 2

5

2 Gk

hcTP

10-15 s femtosecond(飛秒 )10-18 s attosecond(阿托秒 )

52 c

hGt p

5.39124(27) × 10−44 s

Planck’s length

Planck’s time

Time taken for light to travel the length of 3 H atoms

162

Q.17(a)

The drop in temperature of the system is accompanied by the rise in temperature of its surroundings.

Ssystem < 0

Ssurroundings > 0


163

Q.17(b)

Ssystem < 0

Ssurroundings > 0


The drop in entropy of the system is at the cost of the rise in entropy of its surroundings.

164

6.8 Free energy change (SB p.170)


165

The END

166

State whether the following processes are exothermic or endothermic.

(a) Melting of ice.

(b) Dissolution of table salt.

(c) Condensation of steam.

Back

Answer


(a) Endothermic

(b) Endothermic

(c) Exothermic

167

(a)State the difference between exothermic and endothermic reactions with respect to

(i) the sign of H;

(ii) the heat change with the surroundings;

(iii) the total enthalpy of reactants and products.

Answer

(a) (i) Exothermic reactions: H = -ve; endothermic reactions: H = +ve

(ii) Heat is given out to the surroundings in exothermic reactions whereas heat is taken in from the surroundings in endothermic reactions.

(iii) In exothermic reactions, the total enthalpy of products is less than that of the reactants. In endothermic reactions, the total enthalpy is greater than that of the reactants.


168

(b)Draw an enthalpy level diagram for a reaction which is

(i) endothermic, having a large activation energy.

(ii) exothermic, having a small activation energy.

Answer


169


(b) (i)

170


(ii)

Back

171

(a)Why must the condition “burnt completely in oxygen” be emphasized in the definition of standard enthalpy change of combustion?

Answer


(a) If the substance is not completely burnt in excess oxygen, other products such as C(s) and CO(g) may be formed. The enthalpy change of combustion measured will not be accurate.

172

(b) The enthalpy change of the following reaction under standard conditions is –566.0 kJ.

2CO(g) + O2(g) 2CO2(g)

What is the standard enthalpy change of combustion of carbon monoxide?

Answer


(b) Standard enthalpy change of combustion of CO

= (-566.0) kJ

= -283.0 kJ21

173

(c)What terms may be given for the enthalpy change of the following reaction?

N2(g) + O2(g) NO2(g)21

Answer


(c) ½ Enthalpy change of combustion of nitrogen or enthalpy change of formation of nitrogen dioxide.

Back

174


Determine the enthalpy change of neutralization of 25 cm3 of 1.25 M hydrochloric acid and 25 cm3 of 1.25 M sodium hydroxide solution using the following data:

Mass of calorimeter = 100 g

Initial temperature of acid = 15.5 oC (288.5 K)

Initial temperature of alkali = 15.5 oC (288.5 K)

Final temperature of the reaction mixture = 21.6 oC (294.6 K)

The specific heat capacities of water and calorimeter are 4200 J kg-1 K-1 and 800 J kg-1 K-1 respectively.

Answer

175


Assume that the density of the reaction mixture is the same as that of water, i.e. 1 g cm-3.

Mass of the reaction mixture = (25 + 25) cm3 1 g cm-3 = 50 g = 0.05 kg

Heat given out = (m1c1 + m2c2) T

= (0.05 kg 4200 J kg-1 K-1 + 0.1 kg 800 J kg-1 K-1) (294.6 – 288.5) K

= 1769 J

H+(aq) + OH-(aq) H2O(l)

Number of moles of HCl = 1.25 mol dm-3 25 10-3 dm3 = 0.03125 mol

Number of moles of NaOH = 1.25 mol dm-3 25 10-3 dm3 = 0.03125 mol

Number of moles of H2O formed = 0.03125 mol

176

Back


Heat given out per mole of H2O formed

=

= 56608 J mol-1

The enthalpy change of neutralization is –56.6 kJ mol-1.

mol 0.03125J 1769

177


Determine the enthalpy change of combustion of ethanol using the following data:

Mass of spirit lamp before experiment = 45.24 g

Mass of spirit lamp after experiment = 44.46 g

Mass of water in copper calorimeter = 50 g

Mass of copper calorimeter without water = 380 g

Initial temperature of water = 18.5 oC (291.5 K)

Final temperature of water = 39.4 oC (312.4 K)

The specific heat capacities of water and copper calorimeter are 4200 J kg-1 K-1 and 2100 J kg-1 K-1 respectively. Answer

178


Heat evolved by the combustion of ethanol

= Heat absorbed by the copper calorimeter

= (m1c1 + m2c2) T

= (0.05 kg 4200 J kg-1 K-1 + 0.38 kg 2100 J kg-1 K-1) (312.4 – 291.5)K

= 21067 J

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)

Mass of ethanol burnt = (45.24 – 44.46) g = 0.78 g

Number of moles of ethanol burnt = = 0.017 mol 1mol g 46.0g 0.78

179


Heat given out per mole of ethanol

=

= 1239235 J mol-1

= 1239 kJ mol-1

The enthalpy change of combustion of ethanol is –1239 kJ mol-1.

There was heat loss by the system to the surroundings, and incomplete combustion of ethanol might occur. Also, the experiment was not carried out under standard conditions. Therefore, the experimentally determined value (-1239 kJ mol-1) is less than the theoretical value of the standard enthalpy change of combustion of ethanol (-1371 kJ mol-1).

mol 0.017J 21067

Back

180


0.02 mol of anhydrous ammonium chloride was added to

45 g of water in a polystyrene cup to determine the enthalpy change of solution of anhydrous ammonium chloride. It is found that there was a temperature drop from 24.5 oC to 23.0 oC in the solution.

Given that the specific heat capacity of water is 4200 J kg-1 K-1 and

NH4Cl(s) + aq NH4Cl(aq)

Calculate the enthalpy change of solution of anhydrous ammonium chloride.

(Neglect the specific heat capacity of the polystyrene cup.)

Answer

181


Heat absorbed = m1c1T ( c2 0)

= 0.045 kg 4200 J kg-1 K-1 (297.5 – 296) K

= 283.5 J (0.284 kJ)

Heat absorbed per mole of ammonium chloride =

= 14.2 kJ mol-1

The enthalpy change of solution of anhydrous ammonium chloride is +14.2 kJ mol-1.

mol 0.02kJ 0.284

Back

182

(a) A student tried to determine the enthalpy change of neutralization by putting 25.0 cm3 of 1.0 M HNO3 in a polystyrene cup and adding 25.0 cm3 of 1.0 M NH3 into it. The temperature rise recorded was 3.11 oC. Given that the mass of the polystyrene cup is 250 g, the specific heat capacities of water and the polystyrene cup are

4200 J kg-1 K-1 and 800 J kg-1 K-1 respectively. Determine the enthalpy change of neutralization of nitric acid and aqueous ammonia. (Density of water = 1 g cm-3)Answer


183


(a) Heat evolved = m1c1T + m2c2 T

= 0.050 kg 4200 J kg-1 K-1 3.11 K + 0.25 kg 800 J kg-1 K-1 3.11 K

= (653.1 + 622) J

= 1275.1 J

No. of moles of HNO3 used = 1.0 M 25 10-3 dm3

= 0.025 mol

184


(a) No. of moles of NH3 used = 1.0 M 25 10-3 dm3

= 0.025 mol

No. of moles of H2O formed = 0.025 mol

Back

Heat evolved per mole of H2O formed

= mol 0.025J 1275.1

= 51.004 kJ mol-1

The enthalpy change of neutralization of nitric acid and aqueous ammonia is –51.004 kJ mol-1.

185

(b)When 0.05 mol of silver nitrate was added to 50 g of water in a polystyrene cup, a temperature drop of 5.2 o

C was recorded. Assuming that there was no heat absorption by the polystyrene cup, calculate the enthalpy change of solution of silver nitrate.

(Specific heat capacity of water = 4200 J kg-1 K-1)

Answer


186


(b) Energy absorbed = mcT

= 0.05 kg 4200 J kg-1 K-1 5.2 K

= 1092 J

No. of moles of AgNO3 used = 0.05 mol

Energy absorbed per mole of AgNO3 used =

= 21.84 kJ mol-1

The enthalpy change of solution of silver nitrate is +21.84 kJ mol-1.

mol 0.05J 1092

Back

187

(c) A student used a calorimeter as shown in Fig. 6-15 to determine the enthalpy change of combustion of methanol. In the experiment, 1.60 g of methanol was used and 50 g of water was heated up, raising the temperature by 33.2 oC. Given that the specific heat capacities of water and copper calorimeter are 4200 J kg-1 K-1 and 2100 J kg-1 K-1 respectively and the mass of the calorimeter is 400 g, calculate the enthalpy change of combustion of methanol.

Answer


188


(c) Heat evolved = m1c1T + m2c2 T

= 50 g 4.18 J g-1 K-1 33.2 K + 400g 2.10 J g-1 K-1 33.2 K

= (6939 + 27888) J

= 34827 J

No. of moles of methanol used =

= 0.05 mol

Heat evolved per mole of methanol used =

= 696.5 kJ mol-1

The enthalpy change of combustion of methanol is –696.5 kJ mol-1.

1-mol g 32.0g 1.60

mol 0.05

J 34827

189


(a)Given the following thermochemical equation:

2H2(g) + O2(g) 2H2O(l)

(i) Is the reaction endothermic or exothermic?

(ii) What is the enthalpy change for the following reactions?

(1) 2H2O(l) 2H2(g) + O2(g)

(2) H2(g) + O2(g) H2O(l)

(iii) If the enthalpy change for the reaction H2O(l) H2O(g) is +41.1 kJ mol-1, calculate the H for 2H2

(g) + O2(g) 2H2O(g).

21

Answer

190


(a) (i) Exothermic

(ii) (1) +571.6 kJ mol-1

(2) –285.8 kJ mol-1

(iii)

H = [-571.6 + 2 (+41.1)] kJ mol-1 = -489.4 kJ mol-1

191


(b)Given the following information about the enthalpy change of combustion of allotropes of carbon:

Hc [C(graphite)] = -393.5 kJ mol-1

Hc [C(diamond)] = -395.4 kJ mol-1

(i) Which allotrope of carbon is more stable?

(ii) What is the enthalpy change for the following process?

C(graphite) C(diamond)

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Answer

192


(b) (i) Graphite

(ii)

H = [-393.5 – (-395.4)] kJ mol-1 = +1.9 kJ mol-1

193


(c)The formation of ethyne (C2H2(g) can be represented by the following equation:


(i) Draw an enthalpy level diagram relating the above equation to carbon dioxide and water.

(ii) Calculate the standard enthalpy change of formation of ethyne.

(Given: Hc [C(graphite)] = -393.5 kJ mol-1;

Hc [H2(g)] = -285.8 kJ mol-1;

Hc [C2H2(g)] = -1299 kJ mol-1)ø

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194


Given the following information, find the standard enthalpy change of the reaction:

C2H4(g) + H2(g) C2H6(g)

Hf [C2H4(g)] = +52.3 kJ mol-1

Hf [C2H6(g)] = -84.6 kJ mol-1

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Answer

195


Note: H1 = [Hf (reactants)] = Hf [C2H4(g)] + Hf [H2(g)]

H2 = [Hf (products)] = Hf [C2H6(g)]

Applying Hess’s law,

H1 + H = H2

H = H2 - H1

= Hf [C2H6(g)] – (Hf [C2H4(g)] + Hf [H2(g)])

= [-84.6 – (+52.3 + 0)] kJ mol-1 =-136.9 kJ mol-1

The standard enthalpy change of the reaction is –136.9 kJ mol-1.

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196



6PbO(s) + O2(g) 2Pb3O4(s)

Hf [PbO(g)] = -220.0 kJ mol-1

Hf [Pb3O4(g)] = -737.5 kJ mol-1

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Answer

197


Note: H1 = [Hf (reactants)] = 6 Hf [PbO(s)] + Hf [O2(g)]

H2 = [Hf (products)] = 2 Hf [Pb3O4(s)]


H1 + H = H2

H = H2 - H1

= 2 Hf [Pb3O4(s)] – (6 Hf [PbO(s)] + Hf [O2(g)])

= [2 (-737.5) – 6 (-222.0) – 0] kJ mol-1 =-155.0 kJ mol-1


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198



Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

Hf [Fe2O3(s)] = -822.0 kJ mol-1

Hf [CO(g)] = -110.5 kJ mol-1

Hf [CO2(g)] = -393.5 kJ mol-1

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Answer

199


Note: H1 = [Hf (reactants)] = Hf [Fe2O3(s)] + 2 Hf [CO(g)]

H2 = [Hf (products)] = 2 Hf [Fe(s)] + 3 Hf [CO2(g)]


H1 + H = H2

H = H2 - H1

= 2 Hf [Fe(s)] + 3 Hf [CO2(g)] - Hf [Fe2O3(s)] - 3 Hf [CO(g)]

= [2 (0) + 3 (-393.5) –(-822.0) – 3 (-110.5)] kJ mol-1

=-27.0 kJ mol-1


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200



4CH3 · NH · NH2(l) + 5N2O4(l)

4CO2(g) + 12H2O(l) + 9N2(g)

Hf [CH3 · NH · NH2(l)] = +53 kJ mol-1

Hf [N2O4(l)] = -20 kJ mol-1

Hf [CO2(g)] = -393.5 kJ mol-1

Hf [H2O(l)] = -285.8 kJ mol-1

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Answer

201


Note:H1 = [Hf (reactants)] = 4 Hf [CH3·NH ·NH2(l)] + 5 Hf [N2O4(l)]

H2 = [Hf (products)] = 4 Hf [CO2(g)] + 12 Hf [H2O(l)] + 9 Hf [N2(g)]


H1 + H = H2

H = H2 - H1

= (4 Hf [CO2(g)] + 12 Hf [H2O(l)] + 9 Hf [N2(g)] – (3 Hf [CH3·NH ·NH2(l)] + 5 Hf [N2O4(l)]

= [4 (-393.5) + 12 (-285.8) + 9 (0) – 4 (+53) – 5 (-20)] kJ mol-1

=- 5115.6 kJ mol-1


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202


Given the following information, find the standard enthalpy change of formation of methane gas.

C(graphite) + O2(g) CO2(g)


H2(g) + O2(g) H2O(l) Hc [H2(g)] = -285.8 kJ mol-1

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Hc [CH4(g)] = -20 kJ mol-1

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Answer

203


Direct measurement of ΔHf [CH4(g)] is impossible because carbon(graphite) and hydrogen do not combine directly, and methane does not decompose directly to form carbon(graphite) and hydrogen. Since methane contain carbon and hydrogen only, they can be related to carbon dioxide and water by the combustion of methane and its constituent elements as shown in the diagram below.

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Note: H1 = Hc [C(graphite)]

H2 = 2 Hc [H2(g)]

H3 = Hc [CH4(g)]


Hf [CH4(g)] + H3 = H1 + H2

Hf [CH4(g)] = H1 + H2 - H3

= Hc [C(graphite)] + 2 Hc [H2(g)] - Hc [CH4(g)]

= [-393.5 + 2 (-285.8) –(-890.4)] kJ mol-1

= -74.7 kJ mol-1

The standard enthalpy change of formation of methane gas is –74.7 kJ mol-1.

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205


Given the following information, find the standard enthalpy change of formation of methanol.

C(graphite) + O2(g) CO2(g)


H2(g) + O2(g) H2O(l) Hc [H2(g)] = -285.8 kJ mol-1

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)

Hc [C2H5OH(l)] = -1371 kJ mol-1

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Answer

206


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Note: H1 = 2 Hc [C(graphite)]

H2 = 3 Hc [H2(g)]

H3 = Hc [C2H5OH(l)]


Hf [C2H5OH(l)] + H3 = H1 + H2

Hf [C2H5OH(l)] = H1 + H2 - H3

= 2 Hc [C(graphite)] + 3 Hc [H2(g)] - Hc [C2H5OH(l)]

= [2 (-393.5) + 3 (-285.8) –(-1371)] kJ mol-1

= -273.4 kJ mol-1

The standard enthalpy change of formation of ethanol is –273.4 kJ mol-1.

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208


(a) Find the standard enthalpy change of formation of butane gas (C4H10(g)).

Given: Hc [C(graphite)] = -393.5 kJ mol-1

Hc [H2(g)] = -285.8 kJ mol-1

Hc [C4H10(g)] = -2877 kJ mol-1

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Answer

209


Hf [C4H10(g)]

= Hc [C(graphite)] 4 + Hc [H2(g)] 5 - Hc [C4H10(g)]

= [(-393.5) 4 + (-285.8) 5 – (-2877)] kJ mol-1

= -126 kJ mol-1

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(b) Find the standard enthalpy change of the reaction:

Br2(l) + C2H4(g) C2H4Br2(l)

Given: Hf [C2H4(g)] = +52.3 kJ mol-1

Hf [C2H4Br2(l)] = -80.7 kJ mol-1

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Answer

211


H

= [Hf (products)] - [Hf (reactants)]

= [-80.7 – (+52.3) – 0)] kJ mol-1

= -133 kJ mol-1

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212

Predict whether the following changes or reactions involve an increase or a decrease in entropy.• Dissolving salt in water to form salt solution• Condensation of steam on a cold mirror• Complete combustion of carbon• Complete combustion of carbon monoxide• Oxidation of sulphur dioxide to sulphur trioxide

Answer


(a) Increase

(b) Decrease

(c) Increase

(d) Decrease

(e) Decrease

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213


In the process of changing of ice to water, at what temperature do you think G equals 0?

Back

G equals 0 means that neither the forward nor the reverse process is spontaneous. The system is therefore in equilibrium. Melting point of ice is 0 oC (273 K) at which the process of changing ice to water and the process of water turning to ice are at equilibrium. At 0 oC, G of the processes equals 0.

Answer

214

(a)At what temperatures is the following process spontaneous at 1 atmosphere?

Water Steam

(b)What are the two driving forces that determine the spontaneity of a process?

Answer


(a) 100 oC

(b) Enthalpy and entropy

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(c)State whether each of the following cases is spontaneous at all temperatures, not spontaneous at any temperature, spontaneous at high temperatures or spontaneous at low temperatures.

(i) positive S and positive H

(ii) positive S and negative H

(iii) negative S and positive H

(iv) negative S and negative H

Answer

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(i) Spontaneous at high temperatures

(ii) Spontaneous at all temperatures

(iii) Not spontaneous at any temperature

(iv) Spontaneous at low temperatures


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6.5 Hess’s law (SB p.153)Enthalpy change of formation of Enthalpy change of formation of CaCOCaCO33(s)(s)

Ca(s) + C(graphite) + O223 CaCO3(s)

CaO(s) + CO2

(g)

H1H2

Hf [CaCO3(s)]

øHf [CaCO3(s)] = H1 + H2

= -1028.5 kJ mol-1 + (-178.0) kJ mol-1

= -1206.5 kJ mol-1

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Enthalpy change of hydration of Enthalpy change of hydration of MgSOMgSO44(s)(s)

aq

MgSO4(s) + 7H2O(l) MgSO4·7H2O(s)

Mg2+(aq) + SO42-(aq) + 7H2O(l)

ΔH

ΔH2 aq

ΔH1

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ΔH = enthalpy change of hydration of MgSO4(s)

ΔH1 = molar enthalpy change of solution of anhydrous magnesium sulphate(VI)

ΔH2 = molar enthalpy change of solution of magnesium

sulphate(VI)-7-water

ΔH = ΔH1 - ΔH2

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Documents

1 Energetics 6.1What is Energetics? 6.2Enthalpy Changes Related to Breaking and Forming of Bonds 6.3Standard Enthalpy Changes 6.4Experimental Determination