Energetics
5.1 Endothermic and
Exothermic Reactions
Endothermic rxn  heat is taken in from the
surroundings (rxn vessel gets cooler)
Exothermic rxn rxn that result in the release of heat
(rxn vessel gets warmer)
Enthalpy change amount of heat energy taken in/
given out in a chemical rxn (ΔH)
Positive for endo. Rxn
Negative for exo. rxn
Exothermic Rxn
Endothermic Rxn
Stability
Exo. produces a more stable product
Ex. Cdiamond Cgraphite
mol-1
ΔH= -1.9 kJ
Kinetic vs. Thermodynamic stability
Graphite is more thermodynamically more stable
than diamond
Diamond is kinetically stable
What does this mean?
Activation Energy
Define Activation Energy
Activation energy
High AE= faster or slower rxn?
Can the conditions be altered?
If yes, in what ways?
Does how endo- or exothermic a rxn is tell us
how quickly the rxn will run?
Ex. Which rxn will run faster?
-52 kJ mol-1
-16 kJ mol-1
Do endo. or exo. rxns usually occur
spontaneously? (under normal conditions)
5.2 Calculations of enthalpy
changes from experimental data
Specific heat capacity (c) energy required
to raise the temp. of 1 g of substance by 1 K
(1°C) or, the energy to raise 1 kg of
substance by 1 K.
Units-> J g-1 K-1
J g-1 °C-1
kJ kg-1 K-1
J kg-1 K-1
Calculating c
q= mcΔT
q= heat energy
m= mass
ΔT= change in temperature
How does heat capacity affect how easily
a substance can be heated?
Can c be calculated for a substance
undergoing cooling?
Measuring enthalpy change of
combustion rxns
Worked ex. Page 185
Literature value for ΔH ethanol is -1371 kJ mol-1
What are some reasons the calculated value is
different?
A bomb calorimeter could have been used so that
the system was heavily insulated and provide a
plentiful supply of oxygen
Enthalpy changes in solution
General method for measuring
Measure known amounts of reagents
Record initial temps
Mix in a polystyrene cup
Record max/ min temperatures observed
Assume that c for the final solution is the
same as water
ΔH of Solutions:
Definitions
Enthalpy change of neutralisation (ΔHn) enthalpy
change when 1 mol of water molecules are formed
when acid reacts with alkali under standard conditions
H+(aq)
+ OH-(aq)  H2O(l)
Enthalpy change of solution (ΔHsol) the enthalpy
change when 1 mol of solute is dissolved in excess
solvent to form a solution of ‘infinite dilution’ under
standard conditions
NH4NO3(s)
NH4+(aq) + NO3-(aq)
5.3 Hess’s Law
The enthalpy change accompanying a chemical reaction is
independent of the pathway between the initial and final states
What does this mean?
Ex. Find ΔHr for the reaction of AB
Knowns:
ΔHr= ΔH1 +ΔH2
AC =ΔH1
BC= ΔH2
What is C?
What needs to change about the BC step?
Hess’s law:
Definitions
State function pathway does not matter
Standard conditions pressure= 1 atm (or,
1.01E5 Pa), 298K (or, 25°C)
Standard enthalpy change (ΔHrΘ) the enthalpy
change when molar amounts of reactants as
shown in the stoichiometric equation react
together under standard conditions to give
products (Θ= under standard conditions)
Working out enthalpy changes
Hess’s law can be used to determine
enthalpy changes of unknows from knowns
P. 194 Worked example
5.4 Bond enthalpies
The enthalpy change when 1 mole of covalent
bonds, in a gaseous molecule, are broken under
standard conditions
(aka bond energy)
Ex.
mol-1
Then enthalpy of H-H bond is 436 kJ
H2(g)  2H(g)
ΔHΘ= +436 kJ mol-1
How many H-H bonds were broken?
Bond Enthalpy
What state must a substance be in to calculate bond
enthalpy?
Consider this process:
Br2(l)  2Br(g)
Br-Br= 193 kJ mol-1
What is the ΔHΘ?
+224 kJ mol-1
Why is this higher than the bond enthalpy?
The reactants are not in a gaseous state
We must also account for the energy required for vaporisation of
the reactants
This process is called atomisation
Bond breaking
Bond breaking is…
endothermic or exothermic?
Endothermic! What does that mean about
ΔH?
Positive! What will bond making be?
Exothermic with a negative ΔH
Average bond enthalpy
The average amount of energy required
to break 1 mole of covalent bonds, in a
gaseous molecule under standard
conditions
These are the values used to calculate
bond enthalpies
Using bond enthalpies to work
out enthalpy changes in a rxn
1. Must draw out the structural formulas for rxn
2. Imagine the rxn happening and ALL bonds being broken
1. Add up the total energy for all broken bonds
3. Draw in all the bonds formed in products
1. Add up the total energy of all bonds made
4. Determine signs for the total enthalpy changes
1. Broken positive
2. Made negative
5. Add the changes to get the overall enthalpy change of the rxn
Example
Consider the rxn between ethene and bromine,
to produce 1,2-dibromoethane,
C2H4(g) + Br2(g) C2H4Br2(g)
What bonds are broken?
What bonds are made?
Follow your steps!
Using a cycle
Same concept as previous calculations, but
a process is drawn out to see all the steps
P. 203
HL2
5.5 Calculating enthalpy changes:
Definitions
Standard enthalpy change of combustion (ΔHcΘ) the
enthalpy change when 1 mole of a substance is
completely burnt in oxygen under standard conditions.
If ΔHcΘ is always negative, what does this mean?
Standard enthalpy change of formation (ΔHfΘ) the
enthalpy change when 1 mole of the substance is
formed from its elements in their standard states under
standard conditions
Endo and exo rxns are dependent on the type of substance
ΔHfΘ for any element in its standard state is zero
Using ΔHcΘ to calculate
enthalpy change
Method 1: Construct an enthalpy cycle
P. 208
Method 2: rearrange the equations to give the
overall equations related to the enthalpy change
P. 209
Method 3: use an enthalpy level diagram for
calculations
P. 210
Method 4: use the equation,
ΔHr = ΣΔHc (reactants)- ΣΔHc (products)
Using
Θ
ΔHf to
calculate other
enthalpy chages
Method 5: similar to method 1, but used for formation
rather than combustion
P. 214
Method 6: refer method 2 (be sure equations are
running in the correct direction)
Method 7: draw enthalpy level diagram for formation
(method 3)
Method 8: use the equation,
ΔHr = ΣΔHf (products)- ΣΔHf (reactants)
Choosing your method
Choose a method based on the data you are
given, NOT on what needs to be found
If needing the enthalpy of combustion and given
the enthalpy of formation, use one of the
methods 5-8
Once the basic principle of the methods are
understood, there is no need to have any
distinctions between them
5.6 Enthalpy changes for ionic
compounds
First ionisation energy
Second ionisation energy
First electron affinity enthalpy change when one
electron is added to each atom in 1 mol of gaseous
atoms under standard conditions (always
EXOTHERMIC)
X (g) +e-  X- (g)
Second electron affinity (always ENDOTHERMIC)
why?
Lattice enthalpy(ΔHΘlatt) the enthalpy change when
1 mol of an ionic compound is broken apart into iest
constituent gaseous ions under standard conditions
Born-Haber cycles
Enthalpy level diagram breaking down the formation of
an ionic compound into a series of simpler steps
1. put the equation for the enthalpy of formation
2. add lattice enthalpy
3. convert to gaseous form (why?)
Two steps
Must convert ALL reactants to gaseous form
Connect the cycle by adding the electrons removed from
one reactant to the more electronegative reactant
Draw a Born-Haber cycle
P. 219-221
Na and Cl example
Comparisons of lattice enthalpy
P. 224
What is lattice enthalpy the result of?
Electrostatic attractions of + and – ions
If the attractions of great, will more or less energy need to be
supplied to break the bonds?
More
Effect of charge and size
How does the charge of the ions effect lattice enthalpy?
The higher the ion charge, the greater the lattice enthalpy
•
Does NaCl or MgCl2 have great lattice enthalpy?
MgCl2
•
How does size effect lattice enthalpy?
The larger the ions the weaker the forces, the smaller the lattice
enthalpy
•
Which has the larger lattice enthalpy, CsCl or NaCl?
NaCl
Theoretical vs. experimental
Theoretical assumes a totally ionic model
What is this?
Bonding is solely due to attractive forces between
oppositely charged ions
Experimental use the Born-Haber cycle
to find
These are compared to determine how
ionic a particular compound is
How to use theoretical and
experimental values
If values are exactly the same, complete ionic bonding
is suggested
If values are significantly different, it is suggested that
the bonding has a significant degree of covalency
Ex. Silver iodide
Theoretical value/ kJ mol-1  736
Experimental value/ kJ mol-1 876
What do the values suggest?
Covalent character
What is covalent character the result of?
Polarisation of the negative ion by the positive
one
How does size of the anion effect this?
The polarisation effect is greater
t
5.7 EN ropY
A measure of randomness or disorder of a system
Especially significant in the case of endothermic
processes occurring at standard conditions (ice melting
at room temp, water evaporating, NaCl dissolving in
water etc)
Endo rxns can only occur if there is an increase in
entropy
Represented by S
Units: J K-1 mol-1
Standard entropy
Represented by SΘ
Positive ΔSΘ indicates increased entropy
Less order
Negative ΔSΘ indicates decreased entropy
More order
Predicting sign of entropy change
Which state of matter has the higher entropy?
Gases
Which state has the least entropy?
Solids
Period 2 entropies
Elem Li
Be B
C
N2 O2 F2 Ne
ent
State Solid Solid Solid Solid Gas Gas Gas Gas
SΘ / J 29
10
6
6
192 205 203 146
K-1
mol-1
Predicting sign
Must consider whether the system’s disorder
increases or decreases
Good to consider whether moles of gas have
increased or decreased
What would an increase in moles of gas mean?
If moles of gas remain constant, our prediction
of a change in entropy would be approximately
zero
Calculating entropy change for
a rxn
ΔSΘ= ΣSΘproducts -- ΣSΘreactants
5.8 Spontaneity
Spontaneous reaction one that occurs
without any outside influence
Predicting spontaneity
A reaction being spontaneous does not mean it will
run quickly!
Whether a rxn is spontaneous or not under a certain
set of conditions can be deduced by looking at the
change in the “entropy of the Universe”
ΔSUniverse = ΔSsuroundings + Δssystem
If ΔSUniverse is positive, the entropy of the universe
increases and the rxn occurs spontaneously
When heat is given out in a rxn, the entropy of the
surroundings get hotter
Gibbs free energy
Represented by ΔG
Also called free energy change
ΔG = ΔH – TΔS
ΔH and ΔS are referring to the system
Under standard conditions this symbol is used: ΔGΘ
For a reaction to be spontaneous, ΔG for the rxn must
be NEGATIVE
Units kJ mol-1
Calculating ΔG
Method 1 use ΔGΘ = ΔHΘ – TΔSΘ
Method 2 use ΔGΘ = ΣΔGfΘ(products) – ΣΔGfΘ(reactant)
Standard free energy of formation the free energy
change for the formation of 1 mol of substance from its
elements in their standard state under standard conditions
Temp must be in K
If no temp is given and it is standard conditions, assume
298K
Non-spontaneous rxns
If a rxn is non-spontaneous, does that mean
it will never happen?
NO
What will make it run?
Outside influence such as; temp, catalysts, etc.
Effects of temp on spontaneity
Refer to the equation:
ΔG = ΔH – TΔS
If ΔS is positive, temp must be high or low
to be spontaneous?
High
ΔG must be negative to be spontaneous,
thus TΔS must be higher than ΔH
Temp and spontaneity
If ΔS is negative --TΔS will be positive and the
rxn cannot be spontaneous
Endo rxns will only occur spontaneously in
entropy is increased and temp is significantly high
Exo rxns will always be spontaneous at some temp
If the rxn involves a decrease in entropy the rxn
will be spontaneous at a lower temp; becomes less
spontaneous as temp increases
Signs
ΔH
ΔS
--TΔS
ΔG
Spontaneous?
--
+
--
Negative
At all temps
+
+
--
Becomes more neg Becomes more
as temp increases spontaneous as
temp increases
--
--
+
Becomes less neg
as temp increases
Becomes less
spontaneous as
temp increases
+
--
+
Positive
Never