Chemical Energetics

Why Study Energy Changes?

Why Study Energy Changes?

Why Study Energy Changes?

Why Study Energy Changes?

What is Enthalpy?

Absolute H cannot be
measured; only change in H
can be measured

H1

H2

Enthalpy, H

(heat content)

P

Q

In a chemical rxn, energy either (net) given out or (net) taken in from surroundings, principally in the form of heat.

Q is energetically
more stable than P.

What is Enthalpy Change?

ΔH

Enthalpy change, ΔH = Hproducts – Hreactants

HA + HB

reactants ( A + B )

products ( C + D )

Enthalpy, H

HC + HD

Consider: A + B → C + D

ΔH = ( HC + HD ) – ( HA + HB )

–ve (exothermic)

Why is there Enthalpy Change?

In most chemical reactions,
• bonds in reactant particles are broken &

energy is absorbed (endothermic process)

• new bonds are formed in product particles &

energy is given out (exothermic process)

Enthalpy change of a rxn
= difference between the quantity of heat

absorbed to break bonds in reactants and quantity of heat evolved during formation of bonds in products.

Exothermic Reactions

ΔH –ve

reactants

products

Enthalpy

Thermite rxn

Heat is released to surroundings

�

surroundings gains the heat released from rxn

�

Multiple Choice

For a exothermic reaction (net heat released), what happens to the temperature of the surroundings?

• It increases

• It decreases

• It remains the same

Exothermic Reactions

ΔH –ve

reactants

products

Enthalpy

Thermite rxn

Heat is released to surroundings

�

surroundings gains the heat released from rxn

�

temp. of surroundings rises

Endothermic Reactions

ΔH
+ve

products

reactants

Enthalpy

Ba(OH)2.8H2O

+ NH4SCN

Heat is absorbed from surroundings

�

surroundings releases the heat absorbed by the rxn

�

Endothermic Reactions

ΔH
+ve

products

reactants

Enthalpy

Ba(OH)2.8H2O

+ NH4SCN

Heat is absorbed from surroundings

�

surroundings releases the heat absorbed by the rxn

�

temp. of surroundings drops

Draw

Circle the correct characteristics of endothermic reactions

When quoting ΔH values, signs must

always be included.
exothermic rxns: negative sign
endothermic rxns: positive sign

Exothermic & Endothermic Reactions

Enthalpy Change of Reaction, ΔH

ΔH is the enthalpy change when molar

quantities of reactants, as specified by the balanced chemical equation, react to form

products

e.g. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

ΔH

= –890 kJ mol^-1

Unit of ΔH

=kJ
mol-1

Enthalpy Change of Reaction, ΔH

Ө

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

ΔH = –890 kJ mol^-1

According to the thermochemical eqn above,
890 kJ of heat is evolved when
1 mol of CH4 gas reacts with 2 mol of O2 gas
to form 1 mol of CO2gas & 2 mol of liquid H2O
under standard conditions.

The use of “per mole” in the unit of ΔH does not

imply “per mole of any particular substance

formed or used up”, but “per mole of equation”.

Properties of Enthalpy Change

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

ΔH = –890 kJ mol-1

Enthalpy depends on the quantities of substances present.

What if 2 mol of CH4(g) reacted with excess oxygen?

ΔH = 2(–890) UNITS?

Properties of Enthalpy Change

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

ΔH = –890 kJ mol^-1

Why?

What if 2 mol of CH4(g) reacted with excess oxygen?

ΔH = 2(–890) kJ

Properties of Enthalpy Change

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

ΔH = –890 kJ mol^-1

ΔH for a rxn is equal in magnitude, but
opposite in sign, to ΔH for the reverse rxn.

CO2(g) + 2H2O(l)→CH4(g) + 2O2(g)

ΔH = +890 kJ mol^-1

Properties of Enthalpy Change

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

ΔH = –890 kJ mol^-1

ΔH for a rxn depends on the physical state
of the reactants & products.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

ΔH = –802 kJ mol^-1

Important to specify the physical states of all
reactants & products in a thermochemical eqn

Recap: Properties of Enthalpy Change

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

ΔH

= –890 kJ mol^-1

ΔH for a rxn depends on the physical state
of the reactants & products.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

ΔH

= –802 kJ mol^-1

Important to specify the physical states of all
reactants & products in a thermochemical eqn

Multiple Choice

Why is there a Difference in ΔH?

• Energy is absorbed to convert 2 moles of H2O(l) to H2O(g).

• Energy is released to convert 2 moles of H2O(l) to H2O(g).

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH

= –890 kJ mol-1

Why is there a Difference in ΔH?

–890

CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)

–802

Energy / kJ mol-1

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH

= –890 kJ mol-1

Why is there a Difference in ΔH?

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH

= –802 kJ mol-1

–890

CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)

–802

Energy / kJ mol-1

CO2(g) + 2H2O(g)

+88

Open Ended

Given H2(g) + Cl2(g)  →  2HCl(g)               H =  −185 kJ mol−1,

Find enthalpy change of 2 mol of hydrogen reacting with excess chlorine.

Open Ended

Given H2(g) + Cl2(g)  →  2HCl(g)               H =  −185 kJ mol−1,

Find enthalpy change of 0.5 mol of chlorine reacting with excess hydrogen.

Open Ended

Given H2(g) + Cl2(g)  →  2HCl(g)               H =  −185 kJ mol−1,

Find enthalpy change of 2HCl(g)  →  H2(g) + Cl2(g)

Hess’ Law of
Constant Heat

Summation

…. used to determine enthalpy changes

of reactions that cannot be found
directly by experiment in a calorimeter

Hess’ Law of Constant Heat Summation

Hess' Law states that the enthalpy

change of a chemical reaction depends only on the initial and final
states of the

system and is independent of the pathway

taken.

Reactants

A + B

Products

C

D

E + F

path 2

ΔH

path 1

path 3

Enthalpy change for all 3 paths is the

same.

Initial
state

Final
state

Hess’ Law of Constant Heat Summation

Reactants

A + B

Products

C

D

ΔH1

E + F

path 2

ΔH

path 1

Initial
state

Final
state

Hess’ Law of Constant Heat Summation

By Hess’ Law,

ΔH =

ΔH1 + ΔH2

ΔH2

Reactants

A + B

Products

C

D

E + F

ΔH5

ΔH

path 1

Initial
state

Final
state

Hess’ Law of Constant Heat Summation

By Hess’ Law,

ΔH =

ΔH3 + ΔH4

path 3

– ΔH5

+ (–ΔH5)

ΔH

ΔH3

4

Reactants

A + B

Products

C

D

E + F

ΔH5

ΔH

path 1

Initial
state

Final
state

Hess’ Law of Constant Heat Summation

By Hess’ Law,

ΔH =

ΔH3 + ΔH4

path 3

– ΔH5

+ (–ΔH5)

ΔH

ΔH3

4

ΔH1 + ΔH2

=

(a) Write the equation (including state symbols)

for which the ΔH is to be determined

(b) Complete the cycle by filling in eqns which

correspond to data given (ensure all eqns are balanced)

(c) Write corresponding ΔH next to arrows (note

sign & value)

(d) Apply Hess’ Law to calculate the required ΔH

In Summary… Calculating ΔH using

Hess’ Law

Bond dissociation energy is the energy absorbed energy absorbed when one mole of particular covalent bonds between atoms in a gaseous molecule is broken.

Bond Dissociation Energy

Multiple Choice

Bond dissociation energy is the energy absorbed energy absorbed when one mole of particular covalent bonds between atoms in a gaseous molecule is broken. Hence, bond dissociation energy is always

• exothermic

• endothermic

Only 1 value of BDE for gaseous diatomic

molecules, e.g.

H–H(g) → 2H(g)

E(H–H) = +436 kJ mol-1

O=O(g) → 2O(g)

E(O=O) = +436 kJ mol-1

H–Cl(g) → H(g) + Cl(g) E(H–Cl) = +431 kJ mol-1

BDE for Diatomic Molecules

Polyatomic molecules may have 1st, 2nd, 3rd,

etc BDE. These 1st, 2nd, etc BDE values are
different because strength of a covalent
bond is influenced by neighbouring atoms
present, e.g.

H–OH(g) → H(g) + OH(g) ΔH = +494 kJ mol-1
O–H(g) → H(g) + O(g) ΔH = +430 kJ mol-1

BDE for Polyatomic Molecules

∴Average BDE of the O-H bond in H2O
= ½ [(+494) + (+430)] = +462 kJ mol-1

Bond Energy, E(X–X) or ΔHBE

Bond energy, E(X–X) or ΔHBE, is the

average energy absorbed when one mole of covalent bonds between atoms in a gaseous

molecule is broken

(all species being in the gas phase).

O─H bond energy in H2O

= ½ [(+494) + (+430)] = +462 kJ mol-1

The O–H bond energy in H2O is taken to be the
average of the separate BDE of the 2 O–H bonds.

Bond Energy, E(X–X)

Back to the earlier example…
H–OH(g) → H(g) + OH(g) ΔH = +494 kJ mol-1
O–H(g) → H(g) + O(g) ΔH = +430 kJ mol-1

​

The bond energy of a particular bond quoted in the Data Booklet represents the average BDE of that particular bond in the full range of
molecules that contain the bond.

E.g. from Data Booklet, E(O–H) = 460 kJ mol-1

Bond Energy, E(X–X)

found by considering the

BDE of O–H bond in

various compounds, e.g.

H2O (H–O–H),

H2O2 (H–O–O–H),

CH3O–H, etc

Bond Energy, E(X–X)

Bond energy values quoted
in Data Booklet organised
by

(a) Diatomic molecules

(b) Polyatomic molecules

- unambiguous bond

energies

- average bond energies

Bond energy gives information about the

strength of covalent bonds

– higher bond energy = stronger bond

Bond Energy & Bond Strength

Bond

Energy / kJ mol-1

Cl─Cl

244

Br─Br

193

I─I
151

strongest

weakest

Cl-Cl: ​244 kJ mol^-1 ; Br-Br: 193 kJ mol^-1 ; I-I: 151 kJ mol^-1

Calculation of ΔHr using Bond Energies

Bond energies can be used to estimate ΔH for rxns involving gaseous reactants & products.

ΔH = ΔH(bond breaking) + ΔH(bond forming)

energy absorbed

⇒ ΔH is +ve
(same sign as

ΔHBE)

energy evolved

⇒ ΔH is –ve

(opp. sign as ΔHBE)

ΔHr

Ө = ΣΔHBE(bds broken) + {–ΣΔHBE(bds formed)}

ΔHr

Ө = ΣΔHBE(bds broken) –ΣΔHBE(bds formed)