3 energetics rev

EnergeticsEnergetics

Standard Enthalpy Changespy g

As enthalpy changes depend on temperature and pressure. In order to standardize the data recorded it is necessary toIn order to standardize the data recorded, it is necessary to define the standard conditions:

1. elements or compounds in their normal physical states;2. a pressure of 1 atm (101325 Nm-2); and

03. a temperature of 250C (298 K)4. concentration of solution = 1 mol dm-3

Enthalpy change under standard conditions denoted by symbol: ΔH

Standard Enthalpy Change of NeutralizationStandard Enthalpy Change of Neutralization

The standard enthalpy change of neutralization (ΔHneut) is the h l h h l f i f d f henthalpy change when one mole of water is formed from the

neutralization of an acid by an alkali under standard conditions.

H+(aq) + OH-(aq) → H2O(l) ΔHθneut = -57.3 kJ mol-1

Standard Enthalpy Changes of neutralizationpy g

ΔHneuAlkaliAcid-57.1-57.3

NaOHKOH

HClHNO3

-52.2NH3HCl

Th l i ll if k id / lk li d bThe value is smaller if weak acids/alkalis are used because some energy is used for the complete dissociation/ionization of the weak acids/alkalis.

NH3(aq) + H2O(l) === NH4+(aq) + OH-(aq)

St d d E th l Ch f F tiStandard Enthalpy Change of Formation

The standard enthalpy change of formation (ΔHf) is the enthalpy change of the reaction when one mole of the compound in its standard state is formed from its constituent elements under standard conditionsstandard conditions.

Na(s) + ½Cl2(g) → NaCl(s) ΔHf = -411 kJ mol-1

1 mole1 mole

Standard Enthalpy Change of CombustionStandard Enthalpy Change of Combustion

The standard enthalpy change of combustion (ΔH ) of a substance isThe standard enthalpy change of combustion (ΔHc) of a substance is the enthalpy change when one mole of the substance is burntcompletely in oxygen under standard conditions.

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

ΔHc = -2220 kJ mol-1

Experimental Determination of Enthalpy ChangesExperimental Determination of Enthalpy Changes

Calorimeter = a container used for measuring the temperature change of solution

Hess’s LawHess s Law

A + B C + DRoute 1

A + B C + DΔH1

ΔH2 ΔH3

E2 3

Route 2ΔH1 = ΔH2 + ΔH3Route 2

Hess’s Law states that the total enthalpy change py gaccompanying a chemical reaction is independent ofthe route by which the chemical reaction takes place.

Why? Conservation of energy

Importance of Hess’s LawImportance of Hess s LawThe enthalpy change of some chemical reactions cannotbe determined directly from experiment because:

• the reactions cannot be performed in the

be determined directly from experiment because:

laboratory• the reaction rates are too slow• the reactions may involve the formation of side

products

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

Energetics of Formation of Ionic Compoundg p

N ( ) + ½Cl ( ) N Cl( )Δ Hf

ΔH θ 411 kJ l 1Na(s) + ½Cl2(g) → NaCl(s) ΔHfθ = -411 kJ mol-1

Na(g) Cl(g)

Na+(g) Cl-(g)

Standard Enthalpy Change of Atomization (ΔH t )

The enthalpy change when one mole of gaseousi f d f i l i h d fi d

Standard Enthalpy Change of Atomization (ΔH atom)

atoms is formed from its elements in the defined physical state under standard conditions.

Na(s) Na(g) ΔH atom [Na(s)] = +109 kJ mol-1

1/2 Cl2(g) Cl(g) ΔH atom [1/2Cl2(g)] = +121 kJ mol-1

Questions: Why are the changes endothermic?

What type of bond is broken in each case?yp

Ionization Enthalpy (ΔH )The amount of energy required to remove one

l f l f l f

Ionization Enthalpy (ΔH I.E.)

mole of electrons from one mole of atoms or ions in the gaseous state.

Na(g) Na+(g) + e- ΔH I.E [Na(g)] = +494 kJ mol-1

Mg(g) Mg+(g) + e- ΔH I.E [Mg(g)] = +736 kJ mol-1

Mg+(g) Mg2+(g) + e- ΔH I.E [Mg +(g)] = +1 450 kJ mol-1

Questions:Why are the changes endothermic?

Energy is needed to overcome the attractive force gybetween the positive nucleus and the negatively charged electrons.

Electron Affinity (ΔH )The energy change when one mole of electrons is added to one mole of atoms or ions in the gaseous

Electron Affinity (ΔH E.A.)

added to one mole of atoms or ions in the gaseous state.

First electron affinity

O( ) + O ( ) ΔH [O( )] 142 kJ l 1O(g) + e- O-(g) ΔH E.A [O(g)] = - 142 kJ mol-1

Second electron affinity

Questions: Why does the 1st E.A. of O is negative

O-(g) + e- O2-(g) ΔH E.A [O-(g)] = + 844 kJ mol-1

Q y gwhile the second one is positive?

Lattice Enthalpy (ΔH )The energy change when one mole of an ionic crystal is formed from its constituent ions in the

Lattice Enthalpy (ΔH L.E.)

crystal is formed from its constituent ions in the gaseous state under standard conditions

Na+ (g) + Cl-(g) NaCl(s) ΔH lattice [Na+Cl-(s)]

It is a measure of the strength of– + It is a measure of the strength of ionic bond.

+ –

Na+ (g) + Cl-(g) NaCl(s) ΔH lattice [Na+Cl-(s)]

L.E. can be determined indirectly by either:(1) calculations basing on the knowledge of electrostatics in

Ph i ( i i i h ) Th i lPhysics (assuming ions are point charges) – Theoretical Value or

(2) calculations basing on Hess’s Law --- Experimental Value

Born-Haber Cycle for the formation of sodium chloride

ΔH∅t [Na(s)]ΔH atom[Na(s)]

ΔHI.E.

Strength of ionic latticeStrength of ionic lattice1. Ionic Size

Greater the size → lower charge density → weaker attractionG e e e s e → owe c ge de s y → we e c o→ lower (less negative) lattice enthalpy

NaCl: 771 kJ mol-1; KCl: 701 kJ mol-1NaCl: -771 kJ mol 1; KCl: -701 kJ mol 1

2. Ionic ChargeHigher charge → stronger attraction → higher lattice enthalpy

CaO: -3513 kJ mol-1; CaCl2: -2237 kJ mol-1

Discrepancy between calculated and experimental valueDiscrepancy between calculated and experimental valueAssumption in calculating lattice enthalpy:. Ions are hard sphere. Completely transfer of electrons. Charge density distributes evenly on the sphere

compound calculated experimental difference

NaCl -770 -780 10

KCl -702 -711 9

AgCl -833 -905 72

AgI -778 -889 111

Incomplete transfer of electrons covalent charactersIncomplete transfer of electrons covalent characters

Bond Enthalpies Covalent CompoundsBond Enthalpies – Covalent Compounds

Bond Enthalpy (for covalent bond) is the energy associated with a chemical bond.

When a chemical bond is broken, a certain amount of energy is absorbed.

When a chemical bond is formed, a certain amount of energy is released.

Bond Dissociation EnthalpiesBond Dissociation EnthalpiesB.D.E of a certain bond is the amount of energy required to break one mole of that bond in a particular compound under standard conditionsstandard conditions.

Bond EnthalpiesAverage bond enthalpy (or simply bond enthalpy) is theAverage bond enthalpy (or simply bond enthalpy) is the average of the bond dissociation enthalpies required to break a particular chemical bond.

Bond LengthsBond length (for covalent bond)

Bond Lengths

Bond enthalpies and bond lengthsAny conclusion for the relationship between

Bond Bond length (nm) Bond enthalpy(kJ mol-1)

p g

relationship between bond length & bond enthalpy?

(kJ mol )

H-H 0.074 436

Usually a longer bond length corresponds to a l l f b d

Cl-ClBr-Br

I I

0.1990.2280 266

242193151 lower value of bond

enthalpy (weaker bond)I-I

H-FH-Cl

0.2660.0920.127

151565431H Cl

H-BrH-I

0.1270.1410.161

431364299

Distance between shared electrons pair and nuclei increases attraction decreases bond strength decreases

Covalent Radius(often referred as ‘Atomic radius’ ???)

Covalent Radius

The space occupied by an atom in a covalently bonded molecule in thecovalently bonded molecule in the direction of the covalent bond (generally taken as half of the bond length)

Calculated and experimentally determined bond lengthBond Calculated bond length (nm) Experimentally determined

bond length (nm)

C-OC-FC-Cl

0.1500.1490.176

0.1430.1380.177

C-BrC-CH Cl

0.1910.1540 136

0.1930.1540 128H-Cl

C-HN-Cl

0.1360.1140.173

0.1280.1090.174

Difference in electronegativities polar bond ionic character

d d h l h f l i di l iStandard Enthalpy Change of Solution -- dissolving

The standard enthalpy change of solution (ΔH ) is the enthalpyThe standard enthalpy change of solution (ΔHsoln) is the enthalpy change when one mole of a solute is completely dissolved in a sufficient large volume of solvent to form an infinitely dilute solution under standard conditions.

NaCl(s) + water → Na+(aq)+Cl-(aq) ΔH l =+3 9 kJ mol-1NaCl(s) + water → Na (aq)+Cl (aq) ΔHsoln +3.9 kJ mol

LiCl(s) + water → Li+(aq) + Cl-(aq) ΔHsoln=-37.2 kJ mol-1

Note that enthalpy changes of solution associate with physical changes.

According to Hess’s law,

The enthalpy change of solution is :ΔHsoln = ΔHhyd – ΔHlatticesoln hyd lattice

This change involves 2 processes:

1st : NaCl(s) ⎯→ Na+(g) + Cl–(g) ΔH = +776 kJ mol–1

The enthalpy change involved in this process is the reverseThe enthalpy change involved in this process is the reverse of lattice enthalpy. The lattice enthalpy is –776 kJ mol–1.

2nd : hydration enthalpy2nd : hydration enthalpy

Na+(g) + Cl–(g) ⎯→ Na+(aq) + Cl–(aq)

ΔHhyd = –772 kJ mol–1

NaCl(s) ⎯→ Na+(aq) + Cl–(aq) ΔHsoln = +776 – 772

= +4 kJ mol–1

Effect of charge and size of ions on ΔHhyd and ΔHlattice

Standard Enthalpy Changes

g hyd lattice

−+

ΔZZH −+ +

∝ΔrrlatticeH

−

+

+

+∝ΔZZ

hydH −+ rrhyd

Fast Prediction of solubility

Generalizations (Not group trend):

When there is a mismatch in ionic size, the salt is expected , pto be fairly soluble.

Reasons:Reasons:

i. The magnitude of lattice enthalpy is not great because of the large ion, making (r++ r-) relatively large.

ii. The magnitude of the hydration enthalpy is still large due to the presence of the small counter ion.

Relative Solubility of the alkalis metal halides

Group trend of solubility

−+

−+

+∝Δ

rrZZ

latticeH−

−

+

+

+∝Δrz

rz

hydH

LiINaIKIRbIRbICsI

less negativeweaker bondmore soluble

less negativesmaller attraction to waterless solubleless soluble

Solubility of Group I metal iodides – group trend


• For Group I iodide, cations are much smaller than anions

• The ΔHlattice is determined by the reciprocal of the sum of

cationic and anionic radii (i.e. )−+ + rr

1

⇒ Large anionic radius makes the relatively small cationic

radius insignificant w.r.t the sum of r+ and r–g

r+ + r- ~ r-

⇒ Down the group, increase in cationic size does not make

a significant decrease in the magnitude of ΔHlattice


• The hydration enthalpy is determined by −+ +∝Δ

rr11Hhyd

>>11

+

−+

Δr

H

rr

hyd1α

• An increase in cationic size causes ΔHhyd to become less

and less negative significantlyand less negative significantly

⇒ ΔHsoln becomes less and less exothermic

⇒The solubility of Group I metal iodides decreases down

the groupthe group


How about the solubility trend of Group I metal Fluorides?

For small anion like F-, as the size of the cation increases, the magnitude of the lattice enthalpies decrease quickly (become less g p q y (negative).

On the other hand the increase in cationic size does not cause aOn the other hand, the increase in cationic size does not cause a great decrease in the sum of the hydration enthalpies because of the great magnitude of hydration enthalpy of the small anion.

As a result, the solubility increases with an increase in cationic radius.


Questions:

1. The solubility of CsF is relatively high. Explain.y y g p

2. The solubility of LiF is relatively low. Explain.

3. The solubility of CsI is relatively low. Explain.

EntropyEntropyScientists want to find out what governs a spontaneous reaction:

• an exothermic reaction is a spontaneous reaction while an endothermic reaction is not (exothermicity)

ti ?any exception?

• Some spontaneous change is endothermic, e.g. melting of ice at room temperature.⇒ besides enthalpy change, there is another factor that

determine a chemical reactiondetermine a chemical reaction.⇒ Entropy

E i f h d h d fEntropy is a measure of the randomness or the degree of disorder of a system.

EntropyEntropyIn any spontaneous process, there is always an increase in the

(di d ) f h d i di Thentropy (disorder) of the system and its surroundings --- The second law of thermodynamics.

Melting of ice at room temperature → increase in entropy of water molecules

Dissolving of sodium chloride in water → increase in entropy

E Ch ΔSEntropy Change ΔS

ΔS = S SΔS = Sfinal - Sinitial

Entropy change is temperature dependent:Entropy change is temperature dependent:High temperature → entropy increasesLow temperature → entropy decreases

Unit of entropy: kJ mol-1 K-1

Free EnergyFree EnergyWhen predicting whether a reaction is spontaneous, we need to consider both enthalpy change and entropy change (at constant co s de bot e t a py c a ge a d e t opy c a ge (at co sta ttemperature).

A new term is developed to include both enthalpy and entropy →A new term is developed to include both enthalpy and entropy →free energy:

ΔG = ΔH – TΔS where ΔG = change in free energy

A process is spontaneous when ΔG is negative (ΔH is –ve, ΔS is +ve)( , )A process is not spontaneous when ΔG is positive.

3 energetics rev

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