Download File - Garbally Chemistry

THERMOCHEMISTRY
KNOCKHARDY PUBLISHING
Definitions
An exothermic reaction is one where heat is given
out. E.g. burning of a fuel ΔH = negative
An endothermic reaction is one where heat is
taken in.
ΔH = positive
Definitions
Heat of reaction
The heat of reaction is the heat change which
takes place when substances react in molar
quantities according to a balanced equation.
Definitions
Heat of Combustion.
The heat of combustion of a substance is the
heat change when one mole of a substance
is completely burned in excess oxygen
Bomb Calorimeter
• A closed metal calorimeter, called a bomb calorimeter, is used
to measure heats of combustion, to compare efficiencies of fuels
and to measure the energy content of foodstuffs. The bomb is a
thick-walled steel vessel into which oxygen at high pressure can
be pumped. This ensures that the substance burns in an excess of
oxygen and the normal oxide is formed. The sample is held in a
platinum crucible and ignited electrically. The platinum will not
burn and will not react with the sample even under the
conditions of high temperature and pressure which exist inside
the bomb during reaction. The bomb is immersed in a vessel of
water in a container designed to eliminate Energy transfer to or
from the surroundings. The bomb and surrounding water may
then be considered to be the universe, the combustion reaction
taking place in the bomb is the system and the water and
insulated container with lid act as the surroundings.
Bomb calorimeter
Definitions
Calorific value of food. This is the heat released when 100g
of food is burned.
Kilogram Calorific value of a fuel. This is the heat
released when 1 kg of a fuel is burned.
Heat of Neutralisation.
This is the heat change when one mole of H+ ions from an acid
react with one mole of OH- ions from a base to form one mole of
water.
Bond Energy
Bond Energy.
The bond energy is the energy required to
break one mole of covalent bonds and to
completely separate the atoms.
In a reaction, energy is absorbed to break
bonds and energy is released when new
bonds are formed.
Enthalpy of reaction from bond enthalpies
Calculate the enthalpy change for the hydrogenation of ethene
DH2
1 x C=C bond
4 x C-H bonds
1 x H-H bond
@ 611
@ 413
@ 436
= 611 kJ
= 1652 kJ
= 436 kJ
Total energy to break bonds of reactants
= 2699 kJ
DH3
= 346 kJ
= 2478 kJ
1 x C-C bond
6 x C-H bonds
@ 346
@ 413
Total energy to break bonds of products
Applying Hess’s Law
DH1 = DH2 – DH3
= 2824 kJ
= (2699 – 2824) = – 125 kJ
Heat of NEUTRALISATION
This is the heat change when one mole of H+
ions from an acid react with one mole of OHions from a base to form one mole of water.
HEAT OF NEUTRALISATION
Definition
The HEAT change when ONE MOLE of water is formed from
its ions in dilute solution.
Values
Exothermic
Equation
H+(aq)
Notes
A value of -57kJ mol-1 is obtained when
strong acids react with strong alkalis.
+ OH¯(aq)
———>
H2O(l)
See later slides for practical details of measurement
MEASURING Heat of Neutralisation
Example 2
25cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial
temperature of both solutions was 20°C. The highest temperature reached by the
solution was 33°C. Calculate the Heat of Neutralisation. [The specific heat capacity (c)
of water is 4.18 kJ K -1 kg -1]
NaOH
+
Temperature rise (DT)
Volume of resulting solution=
Equivalent mass of water
(density is 1g per cm3)
HCl
——>
NaCl
+
= 306K – 293K
25 + 25 = 50cm3
= 50g
H2O
=
= 13K
0.05 dm3
= 0.05 kg
Heat absorbed by the water (q)
=
m x c x DT =
Moles of HCl reacting
Moles of NaOH reacting
Moles of water produced
=
=
2 x
2 x
Enthalpy change per mol (DH)
= heat energy / moles of water
0.05 x 4.18 x 13 = 2.717 kJ
25/1000 =
25/1000 =
=
= 54.34 kJ mol -1
0.05 mol
0.05 mol
0.05 mol
= 2.717 / 0.05
Definitions
Heat Of Formation
The Heat of Formation of a compound is the heat
change that takes place when one mole of a substance
is formed from its elements in their standard states.
STANDARD HEAT OF FORMATION
Definition
The hEAT change when ONE MOLE of a compound is formed in its
standard state from its elements in their standard states.
Symbol
DH°f
Values
Usually, but not exclusively, exothermic
Example(s)
C(graphite) + O2(g) ———> CO2(g)
H2(g) +
½O2(g)
———>
2C(graphite) + ½O2(g) +
Notes
H2O(l)
3H2(g) ———> C2H5OH(l)
Only ONE MOLE of product on the RHS of the equation
Elements In their standard states have zero enthalpy of formation.
Carbon is usually taken as the graphite allotrope.
Hess’s Law
The heat change for a reaction is equal to the sum of
heat changes for intermediate reactions.
HESS’S LAW
“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.
DHr = DH1 + DH2 +
DH3
If you go in the opposite direction of an arrow, you subtract the value of
the enthalpy change.
e.g.
DH2 = - DH1
+ DHr
- DH3
The values of DH1 and DH3 have been subtracted because the route
involves going in the opposite direction to their definition.
Enthalpy of reaction from enthalpies of combustion
Sample calculation
Calculate the standard heat of formation of methane; the standard heatsof combustion
of carbon, hydrogen and methane are -394, -286 and -890 kJ mol-1 .
C(graphite)
+
2H2(g)
———>
CH4(g)
DH =  DHc of reactants –  DHc of products
By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be...
REACTANTS
PRODUCTS
[ (1 x DHc of C) + (2 x DHc of H2) ]
DH°r =
1 x (-394)
+
2 x (-286)
minus
[ 1 x DHc of CH4O ]
-
1 x (-890)
ANSWER = - 74 kJ mol-1
THERMODYNAMICS
First Law
Energy
changes
Examples
Energy can be neither created nor destroyed but
It can be converted from one form to another
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
Exothermic
combustion of fuels
respiration (oxidation of carbohydrates)
Endothermic
photosynthesis
thermal decomposition of calcium carbonate
STANDARD HEAT OF FORMATION
Definition
The hEAT change when ONE MOLE of a compound is formed in its
standard state from its elements in their standard states.
Symbol
DH°f
Values
Usually, but not exclusively, exothermic
Example(s)
C(graphite) + O2(g) ———> CO2(g)
H2(g) +
½O2(g)
———>
2C(graphite) + ½O2(g) +
Notes
H2O(l)
3H2(g) ———> C2H5OH(l)
Only ONE MOLE of product on the RHS of the equation
Elements In their standard states have zero enthalpy of formation.
Carbon is usually taken as the graphite allotrope.
THERMODYNAMICS - ENTHALPY CHANGES
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 DH , “delta H ”
Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants
THERMODYNAMICS - ENTHALPY CHANGES
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 DH , “delta H ”
Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants
ENTHALPY
Enthalpy
REACTION CO-ORDINATE
Enthalpy of reactants > products
DH = - ive
EXOTHERMIC
Heat given out
THERMODYNAMICS - ENTHALPY CHANGES
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 DH , “delta H ”
ENTHALPY
Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants
ENTHALPY
Enthalpy
REACTION CO-ORDINATE
Enthalpy of reactants > products
DH = - ive
EXOTHERMIC
Heat given out
REACTION CO-ORDINATE
Enthalpy of reactants < products
DH = + ive
ENDOTHERMIC
Heat absorbed
STANDARD ENTHALPY CHANGES
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 DH .
Enthalpy Change
Standard Enthalpy Change
(at 298K)
STANDARD HEAT OF FORMATION
Definition
The hEAT change when ONE MOLE of a compound is formed in its
standard state from its elements in their standard states.
Symbol
DH°f
Values
Usually, but not exclusively, exothermic
Example(s)
C(graphite) + O2(g) ———> CO2(g)
H2(g) +
½O2(g)
———>
2C(graphite) + ½O2(g) +
Notes
H2O(l)
3H2(g) ———> C2H5OH(l)
Only ONE MOLE of product on the RHS of the equation
Elements In their standard states have zero enthalpy of formation.
Carbon is usually taken as the graphite allotrope.
STANDARD HEAT OF COMBUSTION
Definition
The heat change when ONE MOLE of a substance undergoes complete
combustion in excess Oxygen under standard conditions. All reactants
and products are in their standard states.
Symbol
DH°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 on the LHS of the equation
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.
HEAT OF NEUTRALISATION
Definition
The HEAT change when ONE MOLE of water is formed from
its ions in dilute solution.
Values
Exothermic
Equation
H+(aq)
Notes
A value of -57kJ mol-1 is obtained when
strong acids react with strong alkalis.
+ OH¯(aq)
———>
H2O(l)
See later slides for practical details of measurement
BOND DISSOCIATION ENTHALPY
Definition
Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms.
Values
Endothermic - Energy must be put in to break any chemical bond
Examples
Cl2(g)
Notes
•
•
•
•
———>
2Cl(g)
O-H(g) ———>
O(g) + H(g)
strength of bonds also depends on environment; MEAN values quoted
making bonds is exothermic as it is the opposite of breaking a bond
for diatomic gases, bond enthalpy = 2 x enthalpy of atomisation
smaller bond enthalpy = weaker bond = easier to break
BOND DISSOCIATION ENTHALPY
Definition
Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms.
Values
Endothermic - Energy must be put in to break any chemical bond
Examples
Cl2(g)
Notes
•
•
•
•
———>
O-H(g) ———>
2Cl(g)
O(g) + H(g)
strength of bonds also depends on environment; MEAN values quoted
making bonds is exothermic as it is the opposite of breaking a bond
for diatomic gases, bond enthalpy = 2 x enthalpy of atomisation
smaller bond enthalpy = weaker bond = easier to break
Mean Values
Average (mean)
values are quoted
because the actual
value depends on
the environment of
the bond i.e. where
it is in the molecule
H-H
C-C
C=C
CC
C-O
C=O
C-H
C-N
C-F
C-Cl
436
346
611
837
360
743
413
305
484
338
H-F
H-Cl
H-Br
H-I
H-N
H-O
H-S
H-Si
P-H
O-O
562
431
366
299
388
463
338
318
322
146
UNITS = kJ mol-1
N-N
N=N
NN
P-P
F-F
Cl-Cl
Br-Br
I-I
S-S
Si-Si
163
409
944
172
158
242
193
151
264
176
HESS’S LAW
“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.
DHr = DH1 + DH2 +
DH3
HESS’S LAW
“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.
DHr = DH1 + DH2 +
DH3
If you go in the opposite direction of an arrow, you subtract the value of
the enthalpy change.
e.g.
DH2 = - DH1
+ DHr
- DH3
The values of DH1 and DH3 have been subtracted because the route
involves going in the opposite direction to their definition.
HESS’S LAW
“The enthalpy change is independent of the path taken”
Use
applying Hess’s Law enables one to calculate enthalpy changes from
other data, such as...
changes which cannot be measured directly e.g. Lattice Enthalpy
enthalpy change of reaction from bond enthalpy
enthalpy change of reaction from DH°c
enthalpy change of formation from DH°f
Enthalpy of reaction from bond enthalpies
Theory
Imagine that, during a reaction, all the bonds of reacting species are broken
and the individual atoms join up again but in the form of products. The
overall energy change will depend on the difference between the energy
required to break the bonds and that released as bonds are made.
energy released making bonds > energy used to break bonds ... EXOTHERMIC
energy used to break bonds > energy released making bonds ... ENDOTHERMIC
Step 1
Energy is put in to break bonds to form separate, gaseous atoms
Step 2
The gaseous atoms then combine to form bonds and energy is released
its value will be equal and opposite to that of breaking the bonds
Applying Hess’s Law
DHr =
Step 1 + Step 2
Enthalpy of reaction from bond enthalpies
Alternative view
Step 1
Energy is put in to break bonds
to form separate, gaseous atoms.
Step 2
Gaseous atoms then combine
to form bonds and energy is
released; its value will be equal
and opposite to that of breaking
the bonds
DHr = Step 1
-
ATOMS
SUM OFTHE BOND
ENTHALPIES OF
THE REACTANTS
REACTANTS
DH
Step 2
Because, in Step 2 the route involves
going in the OPPOSITE DIRECTION
to the defined change of bond enthalpy,
it’s value is subtracted.
SUM OFTHE
BOND
ENTHALPIES OF
THE PRODUCTS
PRODUCTS
DH =  bond enthalpies –  bond enthalpies
of reactants
of products
Enthalpy of reaction from bond enthalpies
Calculate the enthalpy change for the hydrogenation of ethene
Enthalpy of reaction from bond enthalpies
Calculate the enthalpy change for the hydrogenation of ethene
DH2
1 x C=C bond
4 x C-H bonds
1 x H-H bond
@ 611
@ 413
@ 436
= 611 kJ
= 1652 kJ
= 436 kJ
Total energy to break bonds of reactants
= 2699 kJ
DH3
= 346 kJ
= 2478 kJ
1 x C-C bond
6 x C-H bonds
@ 346
@ 413
Total energy to break bonds of products
Applying Hess’s Law
DH1 = DH2 – DH3
= 2824 kJ
= (2699 – 2824) = – 125 kJ
Enthalpy of reaction from enthalpies of formation
If you formed the products from their elements you should need the same
amounts of every substance as if you formed the reactants from their elements.
Enthalpy of formation tends to be an exothermic process
Enthalpy of reaction from enthalpies of formation
Step 1
Energy is released as reactants
are formed from their elements.
ELEMENTS
Step 2
Energy is released as products
are formed from their elements.
SUM OFTHE
ENTHALPIES OF
FORMATION OF
THE REACTANTS
DHr =
or
- Step 1
+
Step 2
Step 2
-
Step 1
In Step 1 the route involves going in the
OPPOSITE DIRECTION to the defined
enthalpy change, it’s value is subtracted.
REACTANTS
DH
PRODUCTS
DH =  DHf of products –  DHf of reactants
SUM OFTHE
ENTHALPIES
OF
FORMATION
OF THE
PRODUCTS
Enthalpy of reaction from enthalpies of formation
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)
DH =  DHf of products –  DHf of reactants
By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be...
PRODUCTS
REACTANTS
[ 4 x DHf of HNO3 ] minus
DH°r =
4 x (-173)
-
[ (2 x DHf of H2O) + (4 x DHf of NO2) + (1 x DHf of O2) ]
2 x (-286)
ANSWER = - 252 kJ
+
4 x (+33)
+
0
Enthalpy of reaction from enthalpies of combustion
If you burned all the products you should get the same amounts of oxidation products
such a CO2 and H2O as if you burned the reactants.
Enthalpy of combustion is an exothermic process
Enthalpy of reaction from enthalpies of combustion
Step 1
Energy is released as reactants
undergo combustion.
Step 2
Energy is released as products
undergo combustion.
DHr
REACTANTS
DH
PRODUCTS
=
Step 1 - Step 2
Because, in Step 2 the route involves
going in the OPPOSITE DIRECTION to the
defined change of Enthalpy of
Combustion, it’s value is subtracted.
SUM OFTHE
ENTHALPIES OF
COMBUSTION OF
THE PRODUCTS
OXIDATION
PRODUCTS
DH =  DHc of reactants –  DHc of products
SUM OFTHE
ENTHALPIES OF
COMBUSTION
OF THE
REACTANTS
Enthalpy of reaction from enthalpies of combustion
Sample calculation
Calculate the standard enthalpy of formation of methane; the standard enthalpies of
combustion of carbon, hydrogen and methane are -394, -286 and -890 kJ mol-1 .
C(graphite)
+
2H2(g)
———>
CH4(g)
DH =  DHc of reactants –  DHc of products
By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be...
REACTANTS
PRODUCTS
[ (1 x DHc of C) + (2 x DHc of H2) ]
DH°r =
1 x (-394)
+
2 x (-286)
minus
[ 1 x DHc of CH4O ]
-
1 x (-890)
ANSWER = - 74 kJ mol-1
MEASURING ENTHALPY CHANGES
Calorimetry
involves the practical determination of enthalpy changes
usually involves heating (or cooling) known amounts of water
water is heated up reaction is EXOTHERMIC
water cools down reaction is ENDOTHERMIC
Calculation
The energy required to change the temperature
of a substance can be calculated using...
q = m x c x DT
where
q
m
c
DT
=
=
=
=
heat energy
mass
Specific Heat Capacity
change in temperature
kJ
kg
kJ K -1 kg -1 [ water is 4.18 ]
K
MEASURING ENTHALPY CHANGES
Calorimetry
involves the practical determination of enthalpy changes
usually involves heating (or cooling) known amounts of water
water is heated up reaction is EXOTHERMIC
water cools down reaction is ENDOTHERMIC
Calculation
The energy required to change the temperature
of a substance can be calculated using...
q = m x c x DT
where
Example
q
m
c
DT
=
=
=
=
heat energy
mass
Specific Heat Capacity
change in temperature
kJ
kg
kJ K -1 kg -1 [ water is 4.18 ]
K
On complete combustion, 0.18g of hexane raised the temperature of
100g water from 22°C to 47°C. Calculate its HEAT of combustion.
Heat absorbed by the water (q) = 0.1 x 4.18 x 25 = 10.45 kJ
Moles of hexane burned
= mass / Mr
= 0.18 / 86
= 0.00209
Enthalpy change = heat energy / moles
= 10.45 / 0.00209
ANS
= 5000 kJ mol -1
MEASURING ENTHALPY CHANGES
Example 1 - graphical
The temperature is taken every half
minute before mixing the reactants.
Reactants are mixed after three minutes.
Further readings are taken every half
minute as the reaction mixture cools.
Extrapolate the lines as shown and
calculate the value of DT.
MEASURING ENTHALPY CHANGES
Example 1 - graphical
The temperature is taken every half
minute before mixing the reactants.
Reactants are mixed after three minutes.
Further readings are taken every half
minute as the reaction mixture cools.
Extrapolate the lines as shown and
calculate the value of DT.
Example calculation
When 0.18g of hexane underwent complete combustion, it raised the temperature of
100g (0.1kg) water from 22°C to 47°C. Calculate its HEAT of combustion.
Heat absorbed by the water (q) = m C DT
= 0.1 x 4.18 x 25 = 10.45 kJ
Moles of hexane burned
= mass / Mr
Enthalpy change = heat energy / moles
= 0.18 / 86
= 0.00209
= 10.45 / 0.00209 = 5000 kJ mol -1
MEASURING ENTHALPY CHANGES
Example 2
25cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial
temperature of both solutions was 20°C. The highest temperature reached by the
solution was 33°C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat
capacity (c) of water is 4.18 kJ K -1 kg -1]
NaOH
+
HCl
——>
NaCl
+
H2O
MEASURING ENTHALPY CHANGES
Example 2
25cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial
temperature of both solutions was 20°C. The highest temperature reached by the
solution was 33°C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat
capacity (c) of water is 4.18 kJ K -1 kg -1]
NaOH
+
Temperature rise (DT)
Volume of resulting solution=
Equivalent mass of water
(density is 1g per cm3)
Heat absorbed by the water (q)
HCl
——>
NaCl
= 306K – 293K
25 + 25 = 50cm3
= 50g
=
m x c x DT =
+
H2O
=
= 13K
0.05 dm3
= 0.05 kg
0.05 x 4.18 x 13 = 2.717 kJ
MEASURING ENTHALPY CHANGES
Example 2
25cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial
temperature of both solutions was 20°C. The highest temperature reached by the
solution was 33°C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat
capacity (c) of water is 4.18 kJ K -1 kg -1]
NaOH
+
Temperature rise (DT)
Volume of resulting solution=
Equivalent mass of water
(density is 1g per cm3)
HCl
——>
NaCl
+
= 306K – 293K
25 + 25 = 50cm3
= 50g
H2O
=
= 13K
0.05 dm3
= 0.05 kg
Heat absorbed by the water (q)
=
m x c x DT =
Moles of HCl reacting
Moles of NaOH reacting
Moles of water produced
=
=
2 x
2 x
Enthalpy change per mol (DH)
= heat energy / moles of water
0.05 x 4.18 x 13 = 2.717 kJ
25/1000 =
25/1000 =
=
= 54.34 kJ mol -1
0.05 mol
0.05 mol
0.05 mol
= 2.717 / 0.05
REVISION CHECK
What should you be able to do?
Explain the difference between an endothermic and exothermic reaction
Understand the reasons for using standard enthalpy changes
Recall the definitions of enthalpy of formation and combustion
Write equations representing enthalpy of combustion and formation
Recall and apply Hess’s law
Recall the definition of bond dissociation enthalpy
Calculate standard enthalpy changes using bond enthalpy values
Calculate standard enthalpy changes using enthalpies of formation and combustion
Know simple calorimetry methods for measuring enthalpy changes
Calculate enthalpy changes from calorimetry measurements
CAN YOU DO ALL OF THESE?
YES
NO
You need to go over the
relevant topic(s) again
Click on the button to
return to the menu
WELL DONE!
Try some past paper questions
ENTHALPY
CHANGES
THE END
© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING