Download H 2 (g)

THERMOCHEMISTRY
Concept of Enthalpy
Important Terms

Heat is energy transferred between two bodies of
different temperatures

System is any specific part of the universe

Surroundings is everything that lies outside the
system

Open system is a system that can exchange
mass and energy with its surroundings

Closed system is a system that allows the
exchange of energy with its surroundings

Isolated system is a system that does not allow
the exchange of either mass or energy with its
surroundings

Energy is the ability to do work



SI unit of energy is kg m2 s-2 or Joule (J)
Non SI unit of energy is calorie (Cal)
1 Cal = 4.184 J
Thermochemistry


A study of heat change in chemical reactions.
Two types of chemical reactions:


Exothermic
Endothermic
Exothermic reactions

Enthalpy of products < Enthalpy of reactants, ΔH is
negative.

Energy is released from the system to the
surroundings.

Consider the following reaction:
A (g) + B (g) → C (g)
(reactants)
(product)
ΔH = ve
reactants
enthalpy
H= -ve
products
reaction pathway
Energy profile diagram for exothermic reaction
Endothermic Reactions

Enthalpy of products > enthalpy of reactants, ΔH is
positive

Energy is absorbed by the system from the surrounding

Consider the following reaction
A (g) + B (g) → C(g)
ΔH = + ve
(reactants)
(product)
Energy profile of diagram endothermic reactions
Enthalpy, H

The heat content of a system or total energy in the
system

Enthalpy, H of a system cannot be measured when
there is a change in the system.

Example: system undergoes combustion or ionisation.
Enthalpy of Reaction, ∆H and
Standard Condition

Enthalpy of reaction:
 The enthalpy change associated with a chemical
reaction.

Standard enthalpy, ∆Hº
 The enthalpy change for a particular reaction that
occurs at 298K and 1 atm (standard state)
Thermochemical Equation

The thermochemical equation shows the enthalpy changes.
Example : H2O(s) →
H2O(l)
ΔH = +6.01 kJ

1 mole of H2O(l) is formed from 1 mole of H2O(s) at 0°C,
ΔH = +6.01 kJ

However, when 1 mole of H2O(s) is formed from 1 mole of
H2O(l), the magnitude of ΔH remains the same with the
opposite sign of it.
H2O(l) → H2O(s) ΔH = 6.01 kJ
Types of Enthalpies

There are many kind of enthalpies such as:
 Enthalpy of formation
 Enthalpy of combustion
 Enthalpy of atomisation
 Enthalpy of neutralisation
 Enthalpy of hydration
 Enthalpy of solution
 Enthalphy of sublimation
Enthalpy of Formation, ∆Hf

The change of heat when 1 mole of a compound is
formed from its elements at their standard states.
H2 (g) + ½ O2(g) → H2O (l)

∆Hf = 286 kJ mol1
The standard enthalpy of formation of any element in
its most stable state form is ZERO.
∆H (O2 ) = 0
∆H (Cl2) = 0
Enthalpy of Combustion, ∆Hc

The heat released when 1 mole of substance is
burned completely in excess oxygen.
C(s) + O2(g) → CO2(g)
∆Hc = 393 kJ mol1
Enthalpy of Atomisation, Ha

The heat change when 1 mole of gaseous atoms is formed
from its element

Ha is always positive because it involves only breaking of
bonds
e.g:

Na(s)  Na(g)
Ha = +109 kJ mol-1
½Cl2(g)  Cl(g)
Ha = +123 kJ mol-1
Enthalpy of Neutralization, ∆Hn

The heat change when 1 mole of water, H2O is formed from the
neutralization of acid and base .

HCl(aq)+ NaOH(aq) → NaCl(aq) +H2O(aq)
ΔHn = 58 kJ mol1
Enthalpy of Hydration, Hhyd


The heat change when 1 mole of gaseous ions is
hydrated in water.
e.g:
Na+(g)  Na+(aq) Hhyd = 406 kJ mol-1
Cl-(g)  Cl-(aq)
Hhyd = 363 kJ mol-1
Enthalpy of Solution, Hsoln
 The
heat change when 1 mole of a substance is
dissolves in water.
 e.g:
KCl(s)  K+(aq) + Cl(aq) Hsoln = +690 kJ mol-1
Enthalpy of Sublimation, Hsubl
The heat change when one mole of a substance
sublimes (solid into gas).
H subl
I2 (s) →
I2(g)
Hsubl =
+106 kJ mol1
Calorimetry



A method used in the laboratory to measure the
heat change of a reaction.
Apparatus used is known as the calorimeter
Examples of calorimeter


Simple calorimeter
Bomb calorimeter
Simple calorimeter


The outer Styrofoam cup
insulate the reaction
mixture from the
surroundings (it is
assumed that no heat is
lost to the surroundings)
Heat release by the
reaction is absorbed by
solution and the
calorimeter
A bomb calorimeter
Important Terms in Calorimeter


Specific heat capacity, c
 Specific heat capacity, c of a substance is the amount of
heat required to raise the temperature of one gram of the
substance by one degree Celsius (Jg 1C1).
Heat capacity, C
 Heat capacity,C is the amount of heat required to raise
the temperature of a given quantity of the substance by
one degree Celsius (JC1)
Heat released by
substance
=
Heat absorbed
by calorimeter
q = mc∆T
q = heat released by substance
m= mass of substance
C= specific heat capacity
∆T = temperature change
Basic Principle in Calorimeter
Heat released
by a reaction
=
Heat absorbed
by surroundings
• Surroundings may refer to
the:
i. Calorimeter itself or;
ii. The water and calorimeter
• qreaction= mcΔT or CΔT
Example 1
In an experiment, 0.100 g of H2 and excess of O2 were compressed
into a 1.00 L bomb and placed into a calorimeter with heat capacity
of 9.08 x 104 J0C1. The initial temperature of the calorimeter was
25.0000C and finally it increased to 25.155 0C. Calculate the
amount of heat released in the reaction to form H2O, expressed in kJ
per mole.
Solution
Heat released = Heat absorbed by the
calorimeter
q
= C∆T
= (9.08 X 104 J0C-1) X (0.1550C)
= 1.41 X 104 J
= 14.1 kJ
H2(g) + ½O2(g) → H2O(c)
mole of H2 = 0.100
2.016
= 0.0496 mol
moles of H2O
=
mole of H2
0.0496 mol of H2O released 14.1 kJ energy
1 mol H2O released
14.1
= 0.0496
∆Heat of reaction, ∆H
= 284 kJ
= - 284 kJ mol1
kJ
Example 2
1. Calculate the amount of heat released in a reaction in
an aluminum calorimeter with a mass of 3087.0 g and
contains 1700.0 mL of water. The initial temperature of
the calorimeter is 25.0°C and it increased to 27.8°C.
Given:
Specific heat capacity of aluminum = 0.553Jg-1 °C-1
Specific heat capacity of water
= 4.18 Jg-1 °C-1
Water density = 1.0 g mL-1
ΔT = (27.8 -25.0 )°C = 2.8°C
Solution
Heat released
=
Heat absorbed by
aluminium
calorimeter
q = mwcwΔT + mcccΔT
= (1700.0 g)(4.18 Jg-1 °C-1)(2.8 °C) +
(3087.0 g)(0.553 Jg-1 °C-1)(2.8°C)
= 24676.71 J
= 24.7 kJ
+ Heat absorbed by
water
HESS’S LAW
Hess Law


Hess’s Law states that when reactants are converted to
products, the change in enthalpy is the same whether the
reaction takes place in one step or in the series of steps.
The enthalpy change depends only on the nature of the
reactants and products and is independent of the route
taken.
H1
A
H 2
C
B

H 3
H1  H2  H3
Algebraic Method
Step 1
i. List all the thermochemical equations involved
i.C
O
 CO
(S)
2( g )
2( g )
1 O
H O
2
2( g )
2( g )
2 ( g)
iii.C H
7 O
 2CO
 3H O
2
2 6( g )
2( g )
2( g )
2 ( g)
ii.H
H  - 393kJmol- 1
H  -286kJmol- 1
H  -1560kJmol
-1
Algebraic Method
Step 1
i. List all the thermochemical equations involved
i.C
O
 CO
(S)
2( g )
2( g )
1 O
H O
2
2( g )
2( g )
2 ( g)
iii.C H
7 O
 2CO
 3H O
2
2 6( g )
2( g )
2( g )
2 ( g)
ii.H
H  - 393kJmol- 1
H  -286kJmol- 1
H  -1560kJmol
-1
ii. Write the enthalpy of formation reaction for C2H6
H  ?
f
C  3H
   C H
( s)
2( g )
2 6( g )
iii. Add the given reactions so that the result is the desired
reaction.
(i )  2
2C( S )  2O2( g )  2CO2( g )
H1  2  -393 kJ
(ii )  3
3 H 2( g )  3 O2( g )  3 H 2O( g )
2
H 2  3  -286kJ
reverse (iii) 2CO2(g)  3 H 2O( g )  C 2 H 6( g )  7 O2( g ) H3  1560kJ
2
_______________________________________________________________
H f  ?
2C( s )  3 H 2( g )    C 2 H 6( g )
H f  H1  H 2  H3
 -84kJ
- 84kJ
Energy Cycle Method
Draw the energy cycle and apply Hess’s Law to calculate the
unknown value.
HOf
2C (s )
+
3H2 (g)
C2H6 (g)
H O2 = 3(-286)
HO1
2O2 (g)
3/2 O2 (g)
= 2(-393)
2CO2 (g) + 3H2O (g)
7/2 O2 (g)
HO3 = - (-1560)
°
°
°
°
ΔH f = 2( H1 ) + 3(H 2 ) + ΔH3
 -786 - 858  1560
1
 -84 kJmol
Example 1
The thermochemical equation of combustion of carbon
monoxide is shown as below.
C(s) + ½ O2(g)  CO(g)
=?

H
given :


H
C(s) + O2(g)  CO2(g)
∆H= -394 kJ mol-1

CO(s) + ½ O2(g)  CO2(g)
∆H=H-283
kJ mol-1
Calculate the enthalpy change of the combustion of carbon to
carbon monoxide.
Example 2
Calculate the standard enthalpy of formation of methane if the
enthalpy of combustion of carbon, hydrogen and methane are
as follows:
= -393 kJ mol-1
H c ∆H [C(s)]
∆H [H2(s)]
= -293 kJ mol-1

H c
-1
∆H
[CH
]
=
-753
kJ
mol
4(s)
H c
Example 3
 Standard enthalpy of formation of ammonia, hydrogen
chloride and ammonium chloride is -46.1 kJ mol-1, -92.3 kJ
mol-1, 314.4 kJ mol-1 respectively. Write the thermochemical
equation for the formation of each substance and calculate
the enthalpy change for the following reaction.
NH3(g) + HCl (g)  NH4Cl(s)
Exercise
1.Calculate the enthalpy of formation of benzene
if :
∆H (CO2(g) ) = -393.3 kJ mol-1
∆H (H2O(l) ) = -285.5 kJ mol-1
∆H (C6H6(l) ) = -3265.3 kJ mol-1

H f
H f
H f
Born-Haber
Cycle
Lattice Energy, Hlattice

is the energy required to completely separate one mole of a
solid (ionic compound) into gaseous ions
e.g:
NaCl(s)  Na+(g) + Cl-(g)

Na+(g) + Cl-(g)  NaCl(s)

Hlattice = +771 kJ mol-1
(lattice dissociation)
Hlattice = -771 kJ mol-1
(lattice formation)
The magnitude of lattice energy increases as
 the ionic charges increase
 the ionic radii decrease

There is a strong attraction between small ions and
highly charged ions so the H is more negative.

H for MgO is more negative than H for Na2O
because Mg2+ is smaller in size and has bigger charge
than Na+, therefore
Hºlattice (MgO) > Hºlattice (Na2O)
Hydration Process of Ionic Solid

Na+ and Cl- ions in the solid crystal are separated from
each other and converted to the gaseous state (Hlattice)

The electrostatic forces between gaseous ions and polar
water molecules cause the ions to be surrounded by water
molecules (Hhydr)
Hsoln = Hlattice + Hhdyr
Na+ and Cl- ion in
the gaseous state
Heat of Solution
Na+ and Cl- ion in
the solid state
Hydrated Na+ and Cl- ion
Born-Haber Cycle




The process of ionic bond formation occurs in a few stages. At
each stage the enthalpy changes are considered.
The Born Haber cycle is often used to calculate the lattice
energy of an ionic compound.
In the Born-Haber cycle energy diagram, by convention,
positive values are denoted as going upwards, negative values
as going downwards.
Consider the enthalpy changes in the formation of sodium
chloride.
Example :
Na(s) 

i.
ii.
iii.
iv.
v.
vi.
1
2
Cl2(g)  NaCl(s)
Given;
Enthalpy of formation NaCl
Enthalpy of sublimation of Na
First ionization energy of Na
Enthalpy of atomization of Cl
Electron affinity of Cl
Lattice energy of NaCl
=
=
=
=
=
=
-411 kJmol-1
+108 kJmol-1
+500 kJmol-1
+122 kJmol-1
-364 kJmol-1
?
Example: A Born-Haber cycle for NaCl
energy
Na+(g) + e
+ Cl(g)
Ionisation
Energy of Na
Na(g)
Electron Affinity of Cl
Na+(g) + Cl- (g)
+ Cl(g)
HaCl
+ve
Na(g) + ½ Cl2(g)
Lattice energy
HaNa
E=0
-ve
Na(s) + ½ Cl2(g)
From Hess’s Law:
Hf
Hf NaCl
NaCl(s)
NaCl
= HaNa + HaCl +IENa +
EACl + Lattice Energy
Calculation:
H0f  HS  IE  Ha( Cl )  EA  Hlattice
Hlattice  H0f  HS  IE  Ha( Cl )  EA
Hlattice  411kJ   108 kJ  500 kJ  122kJ   364 kJ
Hlattice  777 kJ
Exercise:

i.
ii.
iii.
iv.
v.
vi.
Construct a Born-Haber cycle to explain why ionic compound
NaCl2 cannot form under standard conditions. Use the data
below:
Enthalpy of sublimation of sodium
=
+108 kJmol-1
First ionization energy of sodium
=
+500 kJmol-1
Second ionization energy of sodium
=
+4562 kJmol-1
Enthalpy of atomization of chlorine
=
+121kJmol-1
Electron affinity of chlorine
=
-364 kJmol-1
Lattice energy of NaCl2
=
-2489 kJmol-1