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al/physical/energetic_(1/11)
Energy Change & Chemical Reaction
(1) The Concept of Conservation of Energy
The total energy is unchanged during a chemical reaction. Energy may be exchanged between
the materials and the surroundings but the total energy of the materials and the surroundings
remain _____________.
H=U+PV
Where  H = enthalpy change
 U = internal energy change
P  V = work done by the system
(2) Energy Changes in Reactions
An exothermic reaction is one that releases heat
energy, the temperature of the reaction mixture is
observed to ___________.
Enthalpy, H = heat content (energy stored in a
substance)
Enthalpy
reactant
Hr
H
 H = Hp – Hr = - ve
Where  H
Hp
Hr
= Enthalpy change
= Enthalpy of products
= Enthalpy of reactants
An endothermic reaction is one that absorbs heat
energy, the temperature of the reaction mixture is
observed to ___________.
 H = Hp – Hr = + ve
Where  H
Hp
Hr
Hp
product
Enthalpy
product
Hp
= Enthalpy change
= Enthalpy of products
= Enthalpy of reactants
H
Hr
reactant
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(3) Thermochemical Equation & Heat of Reaction
Consider the reaction between hydrogen gas and chlorine gas to form hydrogen chloride gas.
H2 (g) + Cl2 (g)  2 HCl (g)
1
1
H2 (g) + Cl2 (g)  HCl (g)
2
2
 H = -183 kJ
 H = -91.5 kJ
 H = Enthalpy of reaction (or heat of reaction)
(4) Heat Change and Bond Energy
(a) Energy change in a reaction
Any chemical change involves a change in bonding between atoms or ions. Some bonds in
reactants are broken and new bonds are formed to give the product.
Breaking of bonds absorbs energy and is an ______________ change.
Formation of bonds releases energy and is an _____________ change.
Consider the reaction, H2 (g) + Cl2 (g)  2 HCl (g)
Enthalpy
2 H + 2 Cl
H2 + Cl2
HCl + HCl
(b) Bond Energy
Bond energy is the energy absorb (or __________) when one mole of a particular bond is
broken (or __________).
Bond
Bond Energy
(kJmol-1)
H-H
435
Bond
Bond Energy
(kJmol-1)
H-Cl
431
Cl-Cl
244
O-H
464
C-C
347
C=C
611
C-H
414
CC
835
C-Cl
338
O=O
498
C-O
360
NN
945
C=O
736
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Example (1)
Calculate the approximate heat of reaction of the following equation from bond energy values:
H2 (g) + Cl2 (g)  2 HCl (g)
Example (2)
The thermochemical equation for the reaction between hydrogen and bromine is given below:
H2 (g) + Br2 (g)  2 HBr (g)
 H = -101 kJ
Given that the H-H and Br-Br bond energies are 435 and 192 kJmole-1 respectively, calculate
the bond energy of H-Br bond.
Example (3)
Account for the fact that chlorine gas (Cl2) is much more reactive than nitrogen gas (N2) in
ordinary conditions, in terms of bond energy
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Standard Enthalpy Changes
(1) Standard Conditions
The standard conditions for
 temperature at ______ K (25 oC)
 pressure at ONE atmosphere.
 concentration at ONE mole dm-3 (M) for solution.
Any enthalpy change measured under these conditions is called standard enthalpy change and
given the symbol  H or ______.
Consider the reaction,
2 H2 (g) + O2 (g)  2 H2O (l)
 H298 = -575 kJ mol-1
Calculate the values of  H298 of the following
1
(i)
H2O (l)  H2 (g) + O2 (g)
2
1
(ii)
2 H2 (g) + O2 (g) 
H2O (g)
2
Given that heat of vaporization of water: H2O (l)  H2O (g)
 H298 = + 44 kJ mol-1
(2) Standard Heat of Combustion
The heat of combustion of a substance is the heat change, which occurs when _____ mole of it
is completely burned in oxygen.
C2H6 (g) + 3
1
O2 (g)  2 CO2 (g) + 3 H2O (l)
2
 Hc = -1542 kJ mol-1
To determine the heat of combustion,  Hc by burning a known mass of substance
Example (1)
When 1.20 g of ethanol, C2H5OH, was completely burned in excess oxygen, the heat produced
raised the temperature of 300 g of water by 28.3 oC.
(a)
Calculate the heat of combustion of ethanol.
(b)
Write a thermochemical equation for the reaction.
(c)
Draw an energy level diagram to represent the change.
(Assume there were no heat losses to the surroundings and given that the specific heat capacity
of water is 4.18 Jg-1 K-1)
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(3) Standard Heat of Neutralization
The Heat of neutralization is the heat change when _____ mole of water is produced form the
neutralization of an acid by an alkali.
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
 Hn = -57.3 kJ mol-1
Example (1)
To determine heat of neutralization of HCl (aq) by NaOH (aq)
Initial temperature of alkali = initial temperature of acid = 25.0 oC
Final temperature of mixture = 31.6 oC
Calculate the heat of neutralization of hydrochloric acid by sodium hydroxide solution.
(Assume all resulting solution to have a density of 1.0 g cm-3 and specific heat capacity of 4.2
Jg-1K-1. Neglect the heat absorbed by the polystyrene cup and that lost to the surroundings.)
(a)
Constant heat of neutralization for strong acid and strong alkali reactions
For all neutralization, the only reaction take place is:
(b)
Heat of neutralization for weak acids and alkalis
The heats of neutralization for reactions involving weak acids or weak alkalis are either
_________ or _________ than -57.3 kJ mol-1.
Solutions of weak acids and weak alkalis contain mainly __________________
molecules (e.g. for 2 M ethanoic acid, only 4 out of 10,000 molecules ionize).
CH3COOH
CH3COO-
+
+
H
The molecules must be converted to hydrated ions before neutralization takes place.
But ionization involves a heat change which may be either exothermic or endothermic
(usually, this is ____________).
Consequently, the heats of neutralization are different from the value of –57.3 kJ mol-1.
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(4) Standard Heat of Solution
The standard heat of solution ( Hsoln) is the enthalpy change when ONE mole of a
solute is dissolved to form an infinitely dilute solution under standard conditions.
For example, the dissolution of sodium chloride is endothermic. The standard heat of
solution of sodium chloride is +3.9 kJmol-1 under standard conditions. There is a
temperature drop when sodium chloride is dissolved in water.
NaCl (s) + water  Na+ (aq) + Cl- (aq)
 Hsoln = +3.9 kJmol-1
(5) Standard Heat of Formation
(a)
The standard heat of formation of a substance is the heat evolved or absorbed when
_____ mole of the substance is formed from its ELEMENTs under standard conditions.
Mg (s) + 1/2 O2 (g)  MgO (s)
 H = -602 kJ mol-1
Hence,  Hf [MgO (s)] = -602 kJ mol-1
The heat of formation of an element in its normal physical state under standard
conditions is defined as zero.
(b)
To measure standard heat of formation:
(1) By direct combustion of the substance.
(2) By indirect method,
Example (1)
Calculate the heat of formation of carbon monoxide.
C (graphite) + 1/2 O2 (g)  CO (g)
 Hf [CO (g)] = ???
Given that:
C (graphite) + O2 (g)  CO2 (g)
 Hc [C (graphite)] = -393 kJ mol-1
CO (g) + 1/2 O2 (g)  CO2 (g)
 Hc [CO (g)] = -283 kJ mol-1
(Hint: draw the energy level diagram of the above chemical changes)
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Hess's Law
(1)
Hess's Law: Constant Heat Summation
Energy change in converting reactants, A + B, to products, X + Y, is EQUAL, regardless of the
route by which the chemical change occurs provided the initial and final conditions are the
same.
Route 1:
Route 2:
A+BX+Y
A+BIX+Y
 H1
 H2
By Hess's Law:  H1 =  H2
Energy Level Diagram:
Use of Hess's Law: To find the standard heat of a reaction, which cannot be measured directly.
Example (1)
To determine the standard heat of formation of butane
4 C (graphite) + 5 H2 (g)  C4H10 (g)
Give that
 Hc [C (graphite)]
 Hc [H2 (g)]
 Hc [C4H10 (g)]
 Hf [C4H10 (g)] = ???
= -393 kJ mol-1
= -286 kJ mol
= -2877 kJ mol
(a) Draw the energy level diagram for all the reactions.
(b) Draw energy cycle (Born-Haber Cycle) diagram.
(c) Calculate the  Hf [C4H10 (g)] by Hess's Law.
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Example (2)
The Apollo 11 project landed the first man on the moon on 21 July 1969.
During this project, engines of the lunar module used methylhydrazine (CH3NHNH2) and
dinitrogen tetratoxide (N2O4). These liquids were carefully chosen since they ignite
spontaneously and very exothermically on contact.
What is the enthalpy change for their reaction,  Hr (CH3NHNH2)?
4 CH3NHNH2 (l) + 5 N2O4 (l)  4 CO2 (g) + 12 H2O (l) + 9 N2 (g)
Given that
 Hf (CH3NHNH2 (l))
 Hf (N2O4 (l))
 Hf (CO2 (g))
 Hf (H2O(l))
= + 53 kJ mol-1
= -20 kJ mol-1
= -393 kJ mol-1
= -286 kJ mol-1
(a) Draw the Born-Haber Cycle for the reaction.
(b) Calculate the standard heat of reaction By Hess's Law.
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(2)
Standard Heat of formation and Relative Stability of Compounds
(a)
Energetically stable compounds
Consider an exothermic reaction
C (graphite) + O2 (g)  CO2 (g)
 H = -393 kJ mol-1
 Hf [CO2 (g)] is negative.
The reaction is exothermic.
CO2 (g) is at a lower energy level than C (graphite) and O2 (g).
CO2 (g) is _________ stable than C (graphite) and O2 (g).
Usually a compound is formed exothermically from its elements, standard heat of
formation is __________, the compound is energetically more stable than the elements
from which they are formed.
Example (1)
The standard heat of formation of hydrogen peroxide, H2O2 (l), is -188 kJmol-1. From
this we would expect hydrogen peroxide, H2O2 (l) to be stable.
But, hydrogen peroxide, H2O2 (l), decompose fairly readily into water and oxygen!
Why?
Given that:
H2 (g) + O2 (g)  H2O2 (l)
 Hf [H2O2 (l)] = -188 kJmol-1
1
H2 (g) + O2 (g)  H2O (l)
 Hf [H2O (l)] = -286 kJmol-1
2
1
H2O2 (l)  H2O (l) + O2 (g)
 Hr = ???
2
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(b)
Energetically Unstable Compounds
Consider standard heat of formation of ethyne (C2H2) gas,
2 C (graphite) + H2 (g)  C2H2 (g)
 H = +227 kJ mol-1
Therefore, ethyne gas, C2H2 (g), is formed endothermically from its elements and is
energetically __________ with respect to its elements.
However, ethyne gas can be stored for long time at room temperature and pressure and it
does not decompose at ordinary conditions or in the absence of catalyst. Why?
(Hint: ethyne gas begins to decompose at high temperature and pressure or in the
presence of catalyst.)
Hence, ethyne gas, C2H2 (g), with positive heat of formation,  Hf = +ve, is said to be
kinetically stable or with high _________ stability. That is the decomposition rate of it
is ________ at ordinary conditions.
(3)
Predicting whether reactions will occur
Exothermic reactions are more likely to ________ than endothermic reactions.
Reaction that occur spontaneously are often ____________.
Limitations of heat of formation data
(a)
 Hf is no guide to the speed of a reaction.
C (diamond)  C (graphite)
 Hr = -2 kJmol-1
(b)
Predictions from  Hf values relate to standard conditions only.
This situation may be very ____________ under different conditions.
(c)
The process of dissolving glucose in water is endothermic, however, the solid
dissolves readily and easily.
Although the enthalpy change is positive, there is an enormous increase in
disorder (randomness or entropy) as the solid dissolves, causing the reaction to
occur spontaneously.
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Example (4)
Petrol contains mainly the hydrocarbon octane, C8H18 (l).
Standard heat of combustion of octane,  Hc [C8H18 (l)] = -5513 kJ mol-1
(i)
Would you expect petrol to react with oxygen?
(ii)
Why are petrol/ oxygen mixtures stable at room temperature before a spark is
applied?
(iii)
How does the spark initiate a reaction between petrol and oxygen?