Thermochemical
Principles
Author: J R Reid
Data book Enthalpies
We can’t measure enthalpies of specific reactants
or products directly, but we can measure enthalpy
changes.
Rather than listing the enthalpy change for every
possible reaction, data books list the standard
enthalpy of combustion and the standard enthalpy
of formation of compounds, and we can use these
figures to work out the enthalpy change for other
reactions.
Enthalpy change of combustion Hc
Enthalpy change of combustion Hc
the enthalpy change for the complete
combustion of 1 mole of a compound in
excess pure oxygen under standard
conditions of 298k and 1 atmosphere
pressure.
C2H5OH(l) + 3O2(g)
2CO2(g) +3H2O(l) Hc (ethanol)
= - 1367 kJ mol-1
Hc are always negative - an exothermic reaction. A fuel
contains lots of energy which is released to the environment
when more stable products are formed.
C2H5OH(l) + 3O2(g)
2CO2(g) +3H2O(l) Hc (ethanol)
= - 1367 kJ mol-1
A fuel has high chemical
potential Energy
C2H5OH(l) + 3O2(g)
Energy
Hc (ethanol)
= - 1367 kJ mol-1
stable
products
2CO2(g) +3H2O(l)
Enthalpy of Formation of a Compound Hf
Enthalpy change of formation Hf :
the enthalpy change for the formation of 1 mole of a compound
from its elements under standard conditions of 298k and 100
kPa atmosphere pressure.
2C(s) +3H2(g) + 0.5O2  C2H5OH(l) Hf (ethanol)
= -1367 kJ mol-1
Mg(s) + Cl2(g)  MgCl2(s)
Hf (MgCl2(s) )
= - 641 kJ mol-1
Hf are almost always negative, since stable compounds have
less energy, are more stable, than their elements
Hf of elements are all of course 0, since to form themselves
from themselves takes no energy !!!
Mg(s) + Cl2(g)
MgCl2(s)
Hf (MgCl2(s) )
= - 641 kJ mol-1
Elements have high
chemical potential Energy
Mg(s) + Cl2(g)
Energy
Hf (MgCl2(s) )
stable
products
= - 641 kJ mol-1
MgCl2(s)
If Hf of the reactants and products of a reaction are
known, then Hr of the reaction can be calculated
For the reaction below a = moles of reactant A etc
aA + bB
cC + dD
Hr
Then
H = c.Hf (C) + d.Hf (D) - a.Hf (A) - b.Hf (B)
Hr = sum Hf of products - sum Hf reactants
(Use correct number of moles as in the equation,
because all Hf are per MOLE )
An example of the equation
Hr = sum Hf of products - sum Hf reactants
Hf for reactants and products in the equation below are
known. Calculate Hcombustion ethanol. Show that:
Hr = sum Hf of products - sum Hf reactants
We use Hess’s Law, and construct a cycle. If we know the
energy by 1 route we can can calculate the energy by another
route that ends at the same chemical.
Step 1
Write down the equation for the reaction which we are asked to
calculate the unknown H
Hc (ethanol)
C2H5OH(l) + 3O2(g)
2CO2(g) +3H2O(l)
STEP 2
We are given Hf for the reactants. We can connect
the combustion equation to the elements contained
in the reactants.
Make sure that the equation for Hf is
balanced for moles of each element
Hc (ethanol)
C2H5OH(l) + 3O2(g)
2CO2(g) +3H2O(l)
Hf(ethanol)
2C(s) +3H2(g) + 0.5O2
STEP 3
We are given Hf for the products too; we can
connect the products to the elements contained in
the compounds
Hc (ethanol)
C2H5OH(l) + 3O2(g)
Hf(ethanol)
2CO2(g) +3H2O(l)
2. Hf (CO2)
3.Hf (H2O)
2C(s) +3H2(g) + 0.5O2
STEP 4
We can now calculate Hc (ethanol); by applying
Hess’s law.
Route 1
Hc (ethanol)
C2H5OH(l) + 3O2(g)
2CO2(g) +3H2O(l)
2. Hf (CO2)
Hf(ethanol)
Route 2
3.Hf (H2O)
2C(s) +3H2(g) + 0.5O2
Enthalpy Change by Route 1 = Enthalpy Change by Route 2
Hc (ethanol) = - Hf(Ethanol) + 2. Hf (CO2) + 3.Hf (H2O)
Enthalpy change of neutralisation
Ho(neutralisation)
Enthalpy change of neutralisation:
the enthalpy change when 1 mole of water is
produced by the neutralisation of an acid by an alkali
under standard conditions of 100 kPa pressure and
at 298 k
NaOH(aq) +HCl(aq) = NaCl(aq) + H2O(l) Ho(neut.)= - 57.6 kJmol-1
KOH(aq)+ 0.5H2SO4(aq) =KHSO4(aq)+H2O(l) Ho(n)= -57.6 kJ mol-1
These enthalpies of neutralization are the same because the
reaction is the same. The acids and alkalis are strong, so they
are totally ionised; Na+, K+, Cl-, SO42- and HSO4- are spectator
ions.
H+(aq) + OH-(aq)
H2O(l) Ho(n) = - 57.6 kJ mol-1
Calculate Hro for the reaction below, given the
enthalpies of combustion / kJ mol-1; C2H4 -1411,
H2 - 286, C2H6 - 1560
Route 1
C 2H 4 + H 2
Hro
C 2H 6
Route 2
- 2 x 1411
- 286
-1560
2 CO2 + 2 H2O H2O
Enthalpy Change by Route 1 = Enthalpy Change by Route 2
Hro = - 2 x 1411 - 286 - ( - 1560)
= -1548 kJ
Enthalpy change of solution Hsolution
Enthalpy change of solution Hsolution:
the enthalpy change when 1 mole of ionic solid
dissolves to form a dilute 1 molar solution in water at
298 k and 100 kPa pressure.
Dissolving an ionic solid is a chemical process.
Ionic bonds between the positive and negative ions must be
broken, to free the ions from the giant lattice structure; this is a
highly endothermic process.
This breaking of the ionic lattice is only possible in water as
strong ion - dipole bonds are formed between water and the
ions. This is a highly exothermic process, and is called
hydration.
Hydration of ions
The breaking of the lattice as an ionic solid dissolves in water
is possible because strong ion - dipole bonds are formed
between water molecules and free ions, releasing energy.
This is called hydration, and is highly exothermic. Hydration
enthalpy offsets the endothermic breaking of the ionic bonds
in the lattice.
Water has a dipole, the O atom carries a slight negative
charge and the H atom a slight positive charge
+
-
Na+
O
H
H
one ion -dipole chemical bond
in Na+(H2O)6 i.e. Na+(aq)
+
+
-
O
H
H
+
Cl-
one ion -dipole chemical bond
in Cl-(H2O)6 i.e. Cl-(aq)
Enthalpy of solution of sodium chloride is slightly
endothermic, because the lattice enthalpy is
slightly bigger than the hydration enthalpy
Na+(g) + Cl-(g) + aq
Energy
Lattice enthalpy
(breaks up ions)
Hydration of ions, forms
ion-dipole strong bonds
Na+(aq) + Cl-(aq)
NaCl(s) + aq
Na+
Cl-
+ 6H2O
+ 6H2O
hydration
hydration
Hsolution(NaCl)
Na+(H2O)6 = Na+(aq)
Cl-(H2O)6 = Cl-(aq)
Solid Sodium Chloride in water
Na+ ion
Water
molecule
Cl- ion
One Cl- is being removed by water
molecules. Water molecules do not yet
completely surround the Cl- ion
Sodium chloride crystal
dissolving in water
Chloride ion
hydrated by water.
Note that the
positive H atom is
orientated to the
Cl- ion - it is now a
Cl-(aq) ion
A water
molecule
Experimental Determination of enthalpy changes
If in a reaction in solution, m g of water is warmed through
T oC and c is the specific heat capacity of water
Then the heat absorbed by the water is q, and
q = m. c. T
If n moles of product are produced then the
standard molar enthalpy change, i.e. the heat
change per mole is Hr ,
Hr = - q
n
and Hr = -
= - m.c. T
n
m.c. T
n .1000
J per mole
kJ per mole
If the temperature falls, the reaction is endothermic and H
has a positive sign.
Enthalpy of Neutralisation of NaOH and HCl
Initial
Temperature =
22.35 oC
Enthalpy of Neutralisation of NaOH and HCl
Final
temperature =
29.85 oC
Calculation of the Enthalpy of Neutralisation
H+(aq) + OH-(aq) = H2O(l)
Mass of liquid (water) being warmed = 50 + 20 = 70 g
Temperature rise of liquid & calorimeter = 29.85 - 22.35 = 7.5 oC
Heat capacity of calorimeter = C(cal) = 78.2 J oC-1
Heat to warm calorimeter = C(cal) .T = 78.2 x 7.5 = 586.5 J
Specific Heat capacity of water = 4.184 J g-1oC-1
Heat to warm up water = q = m.c. T
q = 70 x 4.184 x 7.5 = 2196.6 J
Total heat out in reaction = qtot = 586.5 + 2196.6 = 2783.1 J
Calculation of the Enthalpy of Neutralisation
H+(aq) + OH-(aq) = H2O(l)
20 mL of 3M HCl and 50 mL of 1 M NaOH were mixed
Moles of H+ = 0.02 x 3 = 0.06 ; Moles of OH- = 0.05 x 1 = 0.05
Thus H+ is in excess, and 0.05 moles of water will be produced
Hneutralisation = total heat out per mole of water produced
H is negative, since the temperature has risen
Hneutralisation = - qtot / n = - 2783.1 / 0.05 J mol-1
Hneutralisation = - 55662 J mol-1
Hneutralisation = - 55.6 kJ mol-1
An Energy level diagram for the Enthalpy of
Neutralisation of hydrochloric acid by sodium hydroxide
Ionic Equation
NaOH(aq) +HCl(aq)
Energy
H+(aq) + OH-(aq)
Hneutralisation = - 55.6 kJ mol-1
NaCl(aq) +H2O(l)
H2O(l)
The neutralisation equation can be written:
Na+ (aq) + OH- (aq) + H+ (aq) + Cl- (aq) = Na+ (aq) + Cl- (aq) +H2O(l)
But Na+ (aq) and Cl- (aq) are spectator ions and contribute
nothing to the enthalpy change; they are ignored in the ionic
equation.
Enthalpy of Solution of Ammonium Nitrate
The initial temperature =23.15
oC
ammonium nitrate
in glass vial
Crushing device
Enthalpy of Solution of Ammonium Nitrate
The final temperature is 19.05 oC
- the reaction is endothermic
Calculation of the Enthalpy of Solution
NH4NO3 (s) +aq = NH4+(aq) + NO3-(aq)
Mass of liquid (water) being cooled = 60 g
Temperature fall of liquid & calorimeter = 23.15 - 19.05 = 4.1 oC
Heat capacity of calorimeter = C(cal) = 153 J oC-1
Heat to cool calorimeter = C(cal) x T = 153 x 4.1 = 627.3 J
Specific Heat capacity of water = 4.184 J g-1oC-1
Heat to cool water = q = m.c. T
q = 60 x 4.184 x 4.1 = 1029.3 J
Total heat removed in reaction = qtot = 627.3 + 1029.3= 1656.6 J
Calculation of the Enthalpy of Solution
NH4NO3 (s) +aq = NH4+(aq) + NO3-(aq)
60 g of water was mixed with 5.00 g of pure
ammonium nitrate - Mr= 80.04
Moles of NH4NO3 = 5 / 80.04 = 0.0625
H is positive, since the temperature has fallen. Heat has been
taken from the water and calorimeter by the chemicals.
Hsolution = total heat absorbed per mole of NH4NO3
Hsolution = + qtot / n = + 1656.6 / 0.0625 J mol-1
Hsolution = + 26505.6 J mol-1
Hsolution = + 26.5 kJ mol-1
An Energy level diagram for the Enthalpy of
Solution of ammonium nitrate
NH4+(aq) + NO3-(aq)
Energy
Hsolution = + 26.5 kJ mol-1
NH4NO3 (s) + aq
The ionic lattice in the giant structure has to be
broken. With ammonium nitrate this endothermic
process is larger than the exothermic hydration
enthalpies of the ammonium and nitrate ions. So
dissolving the solid is overall endothermic.
The Bomb Calorimeter
How accurate enthalpy changes are measured
Stirrer
Insulated
container
Ignition
heater
Water
Ignition
wires
Thermometer
High pressure
oxygen
Steel
Bomb
Reactants
in sample
cup
The operation of a Bomb Calorimeter
• Benzoic acid is used to determine the heat capacity of the
apparatus. It has an accepted value for its enthalpy of
combustion.
• A known mass of benzoic acid is burnt and the temperature
rise measured. The heat capacity of the ‘Bomb’ can be
calculated.
• To determine the enthalpy of combustion of a new substance
a known mass is burnt. The conditions are kept as similar as
possible to the standardization experiment; i.e. same
temperature rise.
• The Enthalpy of Combustion is then calculated
Bond Energies (enthalpies)
The bond energy is the enthalpy change to break 1 mole
of specified covalent bonds in a gaseous molecule under
standard conditions of 298k and 100 kPa pressure.
H2(g)
2H(g)
H-H
2H.
H = + 435.9 kJ mole-1
H= + 435.9 kJ mole-1
The H-H Bond Energy is + 435.9 kJ mole-1
Bond energies are always positive, since they are the
energies to break covalent bonds. I.e to break the
electrostatic attraction between a shared pair of
electrons and 2 adjacent positive nuclei.
Average bond energies
These are averages of bond energies for a particular
bond, taken from a wide range of compounds
containing that bond.
The Average Bond Energy is the average energy to
break 1 mole of specified bonds from a wide number
of compounds at 298 k and 1 atmosphere pressure.
•They are averages, and the actual bond energy in a
molecule may not equal the average bond energy.
•Factors that change actual bond energies from the
average are the electro-negativities of adjacent atoms.
•When one breaks up methane into C(g) + 4 H(g) atoms 4
C-H bonds are broken; the average bond energy is the
total enthalpy for breaking up methane /4
•The average bond energy for a C-H bond is quoted as
413 kJ mole-1, which is the average for methane, ethane
and a host of other hydrocarbons etc.
If all the Bond energies (B.E.) are known for the
reactants and products of a reaction then Hr
can be calculated. If B.E.(A-B) is the bond energy
in the molecule A-B etc.
Hr
A-B + C-D
A-C + B-D
B.E.(C-D)
B.E.(A-B)
A B C D
B.E.(A-C)
B.E.(B-D)
Gaseous
elements
By Hess’s law; energy change route 1 = energy change route 2
Hr = B.E.(A-B) + B.E.(C-D) - B.E.(A-C) - B.E.(B-D)
Calculation of Hr from the bond energies
of reactants and products
Estimate the H for the reaction, given the Average Bond
Enthalpies / kJ mol-1 C=C 699, C-C 368, C-H 435, C-F 484, F-F 155
H
H
C=C
H
F
F
H
Hr
F
H
H
C
C
H
Break
ALL
4 C-H = + 4 x 435
1 C=C = + 699
F
H
bonds
1 F-F = +155
4 C-H = + 4 x 435
1 C-C = + 369
2 C-F = + 2 x 484
2C
+
4H
+
2F
Enthalpy change by route 1 = change by route 2
- (4 x 435) - 369 - (2 x 484)
Hr = + 4 x 435 + 699 + 155
= - 478 kJ
H
An estimate, as average bond energies used
Route 1
H
C=C
H
H
Hr
F
F
F
H
ALL
4 C-H = + 4 x 435
C
C
H
Route 2
Break
H
F
H
bonds
1 F-F = +155
1 C=C = + 699
4 C-H = + 4 x 435
1 C-C = + 369
2 C-F = + 2 x 484
2C
+
4H
+
2F
The - sign arise as bonds are formed in 1,2-difluoroethane
Using Bond Enthalpies (B.E.) to calculate the
enthalpy of combustion of ethanol
2 C + 6H + 7O
BREAK 1 C-C, 5 C-H, 1
C-O and 1 O-H BONDS
HH
Endothermic(+)
H-C-C-O-H 3 O=O
HH
Hc(ethanol)
MAKE 4 C=O, and
6O-H BONDS
Exothermic (-)
2 O=C=O + 3 H-O-H
B.E./ kJ: C-H +413, C-C 348, O=O 498, C-O 360, C=O 743,O-H 463
Hc(ethanol) = + 348 + 5 x 413 +360 + 463 - 4 x 743 - 6 x 463
= -1368 kJ mol-1
Enthalpy and Changes of State
SOLID
H FUSION (+ve)
LIQUID
- H FUSION (+ve)
- H VAP
H SUBLIMATION
H VAPORISATION (+ve)
GAS
Liquids have higher kinetic energies than solids. Heat
energy must be supplied to change a solid into a liquid,
H FUSION is thus +ve. Similarly gases have more KE than
liquids, and so H VAPORISATION is also +ve.
Enthalpy and Changes of State
H2(g) + 1/2 O2(g)
H
=
H
-- 286 kJ mol-1
=
-- 242 kJ mol-1
H2O(g)
H Vaporisation (Water)
H2O(l)
By Hess’s law
- 242 =
- 286 + H Vaporisation (Water)
H Vaporisation (Water) = - 242 +286
= + 44 kJ mol-1
4th April 2012
Enthalpy of combustion
• AIM – to calculate enthalpy of
combustion from experimental data
Complete investigation 4.2
from p.168