2. When substances participate in chemical
reactions chemical energy is released,
stored, or converted to other form of
energies.
3. Thermodynamic System
The substance under study
in which a change occurs is
called the thermodynamic
system (or just system).
Thermodynamic
Surroundings
Everything else in the
vicinity is called the
thermodynamic
surroundings (or just the
surroundings).
4. THERMODYNAMICS
First Law Energy can be neither created nor
destroyed but it can be converted from one form to
another.
Energy changes all chemical reactions are
accompanied by some form of energy change
changes can be very obvious (e.g. coal
burning) but can go unnoticed
Exothermic Energy is given out
Endothermic Energy is absorbed
5. Examples Exothermic combustion of fuels
respiration (oxidation of
carbohydrates)
reaction of the zinc with hydrochloric
acid
Endothermic photosynthesis
thermal decomposition of calcium
carbonate
reaction of the ammonium-nitrate by
the bariumoxide
6. Enthalpy a measure of the heat content of a
substance at constant pressure
•you cannot measure the actual enthalpy of a
substance
• you can measure an enthalpy CHANGE
written as the symbol ΔΗ , “delta H ”
Enthalpy change (ΔΗ) = Enthalpy of products -
Enthalpy of reactants
REACTION CO-ORDINATE
ENTHALPY
THERMODYNAMICS - ENTHALPY CHANGES
7. Why a
standard? enthalpy values vary according to the conditions
a substance under these conditions is said to be in its standard state ...
Pressure:- 100 kPa (1 atmosphere)
A stated temperature usually 298K (25°C)
• as a guide, just think of how a substance would be under normal lab conditions
• assign the correct subscript [(g), (l) or (s) ] to indicate which state it is in
• any solutions are of concentration 1 mol dm-3
• to tell if standard conditions are used we modify the symbol for ΔΗ .
Enthalpy Change Standard Enthalpy Change
(at 298K)
STANDARD ENTHALPY CHANGES
8. STANDARD ENTHALPY OF FORMATION
Definition The enthalpy change when ONE MOLE of a compound is formed
in its
standard state from its elements in their standard states.
Symbol ΔΗ°f
Values Usually, but not exclusively, exothermic
Example(s) C(graphite) + O2(g) ———> CO2(g)
H2(g) + ½O2(g) ———> H2O(l)
2C(graphite) + ½O2(g) + 3H2(g) ———> C2H5OH(l)
Notes Elements In their standard states have zero enthalpy of formation.
Carbon is usually taken as the graphite allotrope.
9. Definition The enthalpy change when ONE MOLE of a substance undergoes complete
combustion under standard conditions. All reactants and products are in
their standard states.
Symbol ΔΗ°c
Values Always exothermic
Example(s) C(graphite) + O2(g) ———> CO2(g)
H2(g) + ½O2(g) ———> H2O(l)
C2H5OH(l) + 3O2(g) ———> 2CO2(g) + 3H2O(l)
Notes Always only ONE MOLE of what you are burning
To aid balancing the equation, remember...
you get one carbon dioxide molecule for every carbon atom in the original
and one water molecule for every two hydrogen atoms
When you have done this, go back and balance the oxygen.
STANDARD ENTHALPY OF COMBUSTION
10. “The enthalpy change is independent of the path taken”
How The enthalpy change going from
A to B can be found by adding the
values of the enthalpy changes for
the reactions A to X, X to Y and Y to B.
ΔΗr = ΔΗ1 + ΔΗ2 + ΔΗ3
HESS’S LAW
11. “The enthalpy change is independent of the path taken”
How The enthalpy change going from
A to B can be found by adding the
values of the enthalpy changes for
the reactions A to X, X to Y and Y to B.
ΔΗr = ΔΗ1 + ΔΗ2 + ΔΗ3
If you go in the opposite direction of an arrow, you subtract the value of
the enthalpy change.
e.g. ΔΗ2 = - ΔΗ1 + ΔΗr - ΔΗ3
The values of ΔΗ1 and ΔΗ3 have been subtracted because the route
involves going in the opposite direction to their definition.
HESS’S LAW
12. Sample calculation
Calculate the standard enthalpy change for the following reaction, given that the
standard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286,
+33 and -173 kJ mol-1 respectively; the value for oxygen is ZERO as it is an element
2H2O(l) + 4NO2(g) + O2(g) ———> 4HNO3(l)
By applying Hess’s Law ... The Standard Enthalpy of Reaction ΔΗ°r will be...
PRODUCTS REACTANTS
[ 4 x ΔΗf of HNO3 ] minus [ (2 x ΔΗf of H2O) + (4 x ΔΗf of NO2) + (1 x ΔΗf of O2)
]
ΔΗ°r = 4 x (-173) - 2 x (-286) + 4 x (+33) + 0
ANSWER = - 252 kJ
Enthalpy of reaction from enthalpies of formation
ΔΗ = Σ ΔΗf of products – Σ ΔΗf of reactants
17. Heats of Formation, ΔH°f
The enthalpy change when one mole of a compound is formed
from the elements in their standard states
° = standard conditions
• Gases at 1 atm pressure
• All solutes at 1 M concentration
• Pure solids and pure liquids
f = a formation reaction
• 1 mole of product formed
• From the elements in their standard states (1 atm, 25°C)
For all elements in their standard states, ΔH°f = 0
→ C9H12NO3(s)
O2(g)
N2(g) +
H2(g) +
Cgr + 1/2
9 6 3/2
What’s the formation reaction for adrenaline,
C9H12NO3(s)?
