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About This Presentation
Chgcvyv
Slide Content
CHEMISTRY for Engineering
CHEM1071
CHAPTER 6
Thermochemistry
CHEM1071
CHAPTER 6
Thermochemistry
Energy is the capacity to do work or to produce heat.
•Radiant energy comes from the sun and is earth’s primary
energy source
•Thermal energy is the energy associated with the random
motion of atoms and molecules
•Chemical energy is the energy stored within the bonds of
chemical substances
•Nuclear energy is the energy stored within the collection of
neutrons and protons in the atom
•Potential energy is the energy available by virtue of an
object’s position
•Kinetic energy: Due to the motion of an object.
•Work (w): force acting over a distance.
(6.1) the nature of energy
•Radiant energy comes from the sun and is earth’s primary
energy source
•Thermal energy is the energy associated with the random
motion of atoms and molecules
•Chemical energy is the energy stored within the bonds of
chemical substances
•Nuclear energy is the energy stored within the collection of
neutrons and protons in the atom
•Potential energy is the energy available by virtue of an
object’s position
•Kinetic energy: Due to the motion of an object.
•Work (w): force acting over a distance.
(6.1) the nature of energy
Heat (q) : the transfer of thermal energy between two
bodies that are at different temperatures (units: J or kJ).
Calorie (Cal), 1 cal = 4.184 J
Temperature (T) is a measure of the average KE of
particles in solids, liquids and gases. It is the degree
of coldness or hotness of a body. (units: C or K)
Temperature is a measure of the thermal energy.
�����������= ������� �����??????
➢Note: Temperature is not energy, but a
measure of it. Thermal energy is
energy (energy associated with the
random motion of atoms and molecules
that is responsible for its temperature)
(6.1) the nature of energy
bodies that are at different temperatures (units: J or kJ).
Calorie (Cal), 1 cal = 4.184 J
Temperature (T) is a measure of the average KE of
particles in solids, liquids and gases. It is the degree
of coldness or hotness of a body. (units: C or K)
Temperature is a measure of the thermal energy.
�����������= ������� �����??????
➢Note: Temperature is not energy, but a
measure of it. Thermal energy is
energy (energy associated with the
random motion of atoms and molecules
that is responsible for its temperature)
(6.1) the nature of energy
(6.1) the nature of energy
•Heat flows from a hot body to a cold body until the two
bodies have equal temperature.
Pathway: specific conditions that define the path by which
energy is transferred.
▪Energy change is independent of the pathway.
▪Work and heat are dependent on the pathway.
Energy is a state function (state property)
It doesn’t depend on how the system arrived at present
state, but only on the characteristics of the present state
•Heat flows from a hot body to a cold body until the two
bodies have equal temperature.
Pathway: specific conditions that define the path by which
energy is transferred.
▪Energy change is independent of the pathway.
▪Work and heat are dependent on the pathway.
Energy is a state function (state property)
It doesn’t depend on how the system arrived at present
state, but only on the characteristics of the present state
Thermochemistry:is the study of heat changes
accompanying various physical and chemical
transformations.
Heat is lost or gainedin a chemical or physical change.
Heat releasedby one object is absorbedby another object.
System: A part of the universe which is
under observation, OR The part of the
universe in which we are interested. e.g.
reactants and products of a reaction.
Surroundings: The remaining portion of
the universe which is not a part of the
system e.g. rooms, buildings around etc.
System + surrounding = UNIVERSE
(6.2) Energy changes in Chemical Reactions
accompanying various physical and chemical
transformations.
Heat is lost or gainedin a chemical or physical change.
Heat releasedby one object is absorbedby another object.
System: A part of the universe which is
under observation, OR The part of the
universe in which we are interested. e.g.
reactants and products of a reaction.
Surroundings: The remaining portion of
the universe which is not a part of the
system e.g. rooms, buildings around etc.
System + surrounding = UNIVERSE
(6.2) Energy changes in Chemical Reactions
Open system: A type of system in which both
mass and energy can move (enter or exit) freely.
Example: a reaction vessel, or a beaker with
reagents.
Closed system: A type of system in which mass
can not pass through the boundary but energy
can move (enter or exit) freely. Example; Bomb
calorimeter in which energy can move freely but no
matter can enter or exit.
Isolated system: Nothing can enter or leave. The
system that has neither mass nor energy
exchange with its surroundings. A substance
contained in the ideal thermos flask is an example of
the isolated system.
(6.2) Energy changes in Chemical Reactions
mass and energy can move (enter or exit) freely.
Example: a reaction vessel, or a beaker with
reagents.
Closed system: A type of system in which mass
can not pass through the boundary but energy
can move (enter or exit) freely. Example; Bomb
calorimeter in which energy can move freely but no
matter can enter or exit.
Isolated system: Nothing can enter or leave. The
system that has neither mass nor energy
exchange with its surroundings. A substance
contained in the ideal thermos flask is an example of
the isolated system.
(6.2) Energy changes in Chemical Reactions
Exothermic process Endothermic process
any process that gives off heat –
transfers thermal energy from the
system to the surroundings.
any process in which heat has to
be supplied to the system from
the surroundings.
H
2O(g) → H
2O(l) + energy energy + H
2O (s) → H
2O (l)
energy+ 2HgO(s) → 2Hg(l) + O
2(g)2H
2(g) + O
2(g) → 2H
2O(l) + energy
(6.2) Energy changes in Chemical Reactions
any process that gives off heat –
transfers thermal energy from the
system to the surroundings.
any process in which heat has to
be supplied to the system from
the surroundings.
H
2O(g) → H
2O(l) + energy energy + H
2O (s) → H
2O (l)
energy+ 2HgO(s) → 2Hg(l) + O
2(g)2H
2(g) + O
2(g) → 2H
2O(l) + energy
(6.2) Energy changes in Chemical Reactions
Examples of chemical processes (reactions):
CH
4(g) + 2O
2(g) → CO
2(g) + 2H
2O(g) q= -890 kJ (exothermic )
Ba(OH)
2.8H
2O (s) + 2NH
4NO
3 (s)→ 2NH
3(g) + 10H
2O(l) +Ba(NO
3)
2(aq)
q= +170.4 kJ (endothermic)
Examples of physical processes:
H
2O
(s) → H
2O(l) (fusion) q= +6.02 kJ/mol (endothermic )
H
2O
(l) → H
2O(s) (freezing) q= -6.02 kJ/mol (exothermic )
I
2 (s) → I
2(g) (sublimation) q= +62 kJ/mol (endothermic )
Br
2 (l) → Br
2(g) (vaporization) q= +31 kJ/mol (endothermic )
CH
3OH
(g) → CH
3OH(l) (condensation) q= -38 kJ/mol (exothermic )
(6.2) Energy changes in Chemical Reactions
CH
4(g) + 2O
2(g) → CO
2(g) + 2H
2O(g) q= -890 kJ (exothermic )
Ba(OH)
2.8H
2O (s) + 2NH
4NO
3 (s)→ 2NH
3(g) + 10H
2O(l) +Ba(NO
3)
2(aq)
q= +170.4 kJ (endothermic)
Examples of physical processes:
H
2O
(s) → H
2O(l) (fusion) q= +6.02 kJ/mol (endothermic )
H
2O
(l) → H
2O(s) (freezing) q= -6.02 kJ/mol (exothermic )
I
2 (s) → I
2(g) (sublimation) q= +62 kJ/mol (endothermic )
Br
2 (l) → Br
2(g) (vaporization) q= +31 kJ/mol (endothermic )
CH
3OH
(g) → CH
3OH(l) (condensation) q= -38 kJ/mol (exothermic )
(6.2) Energy changes in Chemical Reactions
The energy released or absorbed as heat comes from the
difference in potential energies between products and reactants
Schematic of Exothermic and Endothermic Processes
(6.2) Energy changes in Chemical Reactions
difference in potential energies between products and reactants
Schematic of Exothermic and Endothermic Processes
(6.2) Energy changes in Chemical Reactions
Thermodynamics: is the scientific study of the
interconversions of heat and other kinds of energy
State functions are properties that are determined by
the state of the system, regardless of how that
condition was achieved.
e.g. energy, pressure, volume, temperature
(6.3) Introduction to Thermodynamics
Potential energy of hiker 1 and hiker 2 is the
same even though they took different paths.
??????� = �
����� – �
�������
??????� = �
����� – �
�������
??????� = �
����� – �
�������
??????� = �
����� – �
�������
interconversions of heat and other kinds of energy
State functions are properties that are determined by
the state of the system, regardless of how that
condition was achieved.
e.g. energy, pressure, volume, temperature
(6.3) Introduction to Thermodynamics
Potential energy of hiker 1 and hiker 2 is the
same even though they took different paths.
??????� = �
����� – �
�������
??????� = �
����� – �
�������
??????� = �
����� – �
�������
??????� = �
����� – �
�������
A thermodynamic “State function”
State functions are only depend on the current
thermodynamic state of the system.
How the system attained that state does not matter
Example: compression of a gas inside a piston
The state function
only depends on the
initial and final state
of the system.
And is independent
of the path followed
in getting from one
to the other
(6.3) Introduction to Thermodynamics
State functions are only depend on the current
thermodynamic state of the system.
How the system attained that state does not matter
Example: compression of a gas inside a piston
The state function
only depends on the
initial and final state
of the system.
And is independent
of the path followed
in getting from one
to the other
(6.3) Introduction to Thermodynamics
First law of thermodynamics – energy can be
converted from one form to another, but cannot be
created or destroyed.
??????�
�??????���� + ??????�
������������ = �
��
??????�
�??????���� = – ??????�
������������
??????
�??????
� + ��
� → �??????�
� + �??????
��Exothermic chemical reaction!
Chemical energy lost by combustion
system =
Energy gained by the
surroundings
(6.3) Introduction to Thermodynamics
converted from one form to another, but cannot be
created or destroyed.
??????�
�??????���� + ??????�
������������ = �
��
??????�
�??????���� = – ??????�
������������
??????
�??????
� + ��
� → �??????�
� + �??????
��Exothermic chemical reaction!
Chemical energy lost by combustion
system =
Energy gained by the
surroundings
(6.3) Introduction to Thermodynamics
The internal energy of a system changes when the system
exchanges heat with the surroundings
Or work is done by the system on the surrounding or by the
surrounding on the system
Or exchange of both heat and work
ΔU
system = q + w
Change in internal
energy of the system
Heat absorbed or
released by the system
(heat exchange between
system and surrounding)
Work done by or
on the system
Internal energy U of a system is the sum of kinetic and
potential energies of all particles in the system
U
system = KE + PE
(6.3) Introduction to Thermodynamics
exchanges heat with the surroundings
Or work is done by the system on the surrounding or by the
surrounding on the system
Or exchange of both heat and work
ΔU
system = q + w
Change in internal
energy of the system
Heat absorbed or
released by the system
(heat exchange between
system and surrounding)
Work done by or
on the system
Internal energy U of a system is the sum of kinetic and
potential energies of all particles in the system
U
system = KE + PE
(6.3) Introduction to Thermodynamics
Thermodynamic quantity consists of two parts:
▪a number giving the magnitude of the change.
▪a sign indicating the direction of the flow
+ve sign indicates that the system’s energy is increasing
–ve sign indicates that the system’s energy is decreasing.
Table 6.1 Sign Conventions for Work and Heat
Process Sign
(W) Work done by the system on the surroundings−
(W) Work done on the system by the surroundings+
(q) Heat absorbed by the system from the
surroundings (endothermic process)
+
(q) Heat absorbed by the surroundings from the
system (exothermic process)
−
(6.3) Introduction to Thermodynamics
▪a number giving the magnitude of the change.
▪a sign indicating the direction of the flow
+ve sign indicates that the system’s energy is increasing
–ve sign indicates that the system’s energy is decreasing.
Table 6.1 Sign Conventions for Work and Heat
Process Sign
(W) Work done by the system on the surroundings−
(W) Work done on the system by the surroundings+
(q) Heat absorbed by the system from the
surroundings (endothermic process)
+
(q) Heat absorbed by the surroundings from the
system (exothermic process)
−
(6.3) Introduction to Thermodynamics
Work is defined as the force acting over the distance.
W= F d
But , P = F/A
W = P A d
W = P A h = PV
P is pressure , A is area, d is h is the
piston moving a distance in picture,
V is the change in volume.
ΔU
system = q + w
A gaseous system does work on the
surrounding when expanding,
or work is done on the system by
surrounding when compressing the
gaseous system.
(6.3) Introduction to Thermodynamics
W= F d
But , P = F/A
W = P A d
W = P A h = PV
P is pressure , A is area, d is h is the
piston moving a distance in picture,
V is the change in volume.
ΔU
system = q + w
A gaseous system does work on the
surrounding when expanding,
or work is done on the system by
surrounding when compressing the
gaseous system.
(6.3) Introduction to Thermodynamics
*PV* work =� = −????????????�
V
sys = V
f – V
i
in expansion: V
f is larger and V
i is smaller ,,,
thus V
sys is positive and w is negative value.
in compression: V
f is smaller and V
i is larger ,,,
thus V
sys is negative and w is positive value.
Unit of energy is J
1J= 1kg.m
2
/s
2
Work Done on the System
W = � × �
W = – � ??????�
� × � =
�
�
�
× �
�
= � × �=??????
??????� > �
– �??????� < �
??????
�??????� < �
Work is not a state function.Δ =w –
final initial
ww
(6.3) Introduction to Thermodynamics
V
sys = V
f – V
i
in expansion: V
f is larger and V
i is smaller ,,,
thus V
sys is positive and w is negative value.
in compression: V
f is smaller and V
i is larger ,,,
thus V
sys is negative and w is positive value.
Unit of energy is J
1J= 1kg.m
2
/s
2
Work Done on the System
W = � × �
W = – � ??????�
� × � =
�
�
�
× �
�
= � × �=??????
??????� > �
– �??????� < �
??????
�??????� < �
Work is not a state function.Δ =w –
final initial
ww
(6.3) Introduction to Thermodynamics
A gas expanding against pressure p, W is a negative
quantity as required, since work flows out of the system.
when a gas is compressed, ∆V is a negative quantity, which
makes W a positive quantity (work flows into the system).
W= −�??????�
Volume (V) : units is L or m
3
Pressure (P) : has many common units
SI unit is N/m
2
= 1 Pa
The unit atm is also widely used
1 atm = 101325 10
5
Pa = 101.325 kPa
1 atm = 760 torr = 760 mmHg
For a gaseous system: W= -PV atm.L (or L.atm)
1L.atm = 101.3 J
(6.3) Introduction to Thermodynamics
quantity as required, since work flows out of the system.
when a gas is compressed, ∆V is a negative quantity, which
makes W a positive quantity (work flows into the system).
W= −�??????�
Volume (V) : units is L or m
3
Pressure (P) : has many common units
SI unit is N/m
2
= 1 Pa
The unit atm is also widely used
1 atm = 101325 10
5
Pa = 101.325 kPa
1 atm = 760 torr = 760 mmHg
For a gaseous system: W= -PV atm.L (or L.atm)
1L.atm = 101.3 J
(6.3) Introduction to Thermodynamics
Example 6.1
A certain gas expands in volume from 2.0 L to 6.0 L at
constant temperature.
Calculate the work done by the gas if it expands
(a)against a vacuum
(b)against a constant pressure of 1.2 atm
Strategy
A simple sketch of the
situation is helpful here:
The work done in gas expansion is equal to the product of the
external, opposing pressure and the change in volume.
What is the conversion factor between L · atm and J?
(6.3) Introduction to Thermodynamics
A certain gas expands in volume from 2.0 L to 6.0 L at
constant temperature.
Calculate the work done by the gas if it expands
(a)against a vacuum
(b)against a constant pressure of 1.2 atm
Strategy
A simple sketch of the
situation is helpful here:
The work done in gas expansion is equal to the product of the
external, opposing pressure and the change in volume.
What is the conversion factor between L · atm and J?
(6.3) Introduction to Thermodynamics
Solution
a)Because the external pressure is zero, no work is done in
the expansion.
?????? = – �??????�
= –��.� – �.�= �
b)The external, opposing is 1.2 atm, so
?????? = –�??????�
= –�.� ����.� – �.� ??????
= – �.�??????⋅ ���
To convert the answer to joules, we write
??????=–�.�??????⋅���×
���.�??????
�??????⋅���
=–�.�×��
�
??????
Check
Because this is gas expansion (work is done by the system on the
surroundings), the work done has a negative sign.
(6.3) Introduction to Thermodynamics
a)Because the external pressure is zero, no work is done in
the expansion.
?????? = – �??????�
= –��.� – �.�= �
b)The external, opposing is 1.2 atm, so
?????? = –�??????�
= –�.� ����.� – �.� ??????
= – �.�??????⋅ ���
To convert the answer to joules, we write
??????=–�.�??????⋅���×
���.�??????
�??????⋅���
=–�.�×��
�
??????
Check
Because this is gas expansion (work is done by the system on the
surroundings), the work done has a negative sign.
(6.3) Introduction to Thermodynamics
The work done when a gas is compressed in a cylinder like
that shown in Figure 6.5 is 462 J.
During this process, there is a heat transfer of 128 J from the
gas to the surroundings.
Calculate the energy change for this process.
Example 6.2
Strategy
Compression is work done on the gas, so what is the sign for
w?
Heat is released by the gas to the surroundings.
- Is this an endothermic or exothermic process?
- What is the sign for q?
(6.3) Introduction to Thermodynamics
that shown in Figure 6.5 is 462 J.
During this process, there is a heat transfer of 128 J from the
gas to the surroundings.
Calculate the energy change for this process.
Example 6.2
Strategy
Compression is work done on the gas, so what is the sign for
w?
Heat is released by the gas to the surroundings.
- Is this an endothermic or exothermic process?
- What is the sign for q?
(6.3) Introduction to Thermodynamics
Solution
To calculate the energy change of the gas, we need Equation
(6.1). Work of compression is positive and because heat is
released by the gas, q is negative. Therefore, we haveΔ= +
= – 128 J + 462 J
= 334 J
U q w
As a result, the energy of the gas increases by 334 J.
(6.3) Introduction to Thermodynamics
To calculate the energy change of the gas, we need Equation
(6.1). Work of compression is positive and because heat is
released by the gas, q is negative. Therefore, we haveΔ= +
= – 128 J + 462 J
= 334 J
U q w
As a result, the energy of the gas increases by 334 J.
(6.3) Introduction to Thermodynamics
Chemistry in Action: Making Snow
?????? � = � + ??????
� = �
?????? < �,??????� < �
??????� = ?????? ??????�
??????� < �,����!
(6.3) Introduction to Thermodynamics
?????? � = � + ??????
� = �
?????? < �,??????� < �
??????� = ?????? ??????�
??????� < �,����!
(6.3) Introduction to Thermodynamics
Example
calculate the work associated with the expansion of a gas
from 46 L to 64 L at a constant external pressure of 15 atm.
solution: ?????? = − �∆� = -15 atm (64L- 46L)
= -270 L.atm
note that since the gas expands, it does work on its
surrounding.
Example: calculate ∆E for a system undergoing
endothermic process in which 15.6 kJ of heat flows and
where 1.4 kJ work done on the system.
Using equation;
∆?????? = � + ??????= 15.6 + 1.4 =17.0 kJ
(6.3) Introduction to Thermodynamics
calculate the work associated with the expansion of a gas
from 46 L to 64 L at a constant external pressure of 15 atm.
solution: ?????? = − �∆� = -15 atm (64L- 46L)
= -270 L.atm
note that since the gas expands, it does work on its
surrounding.
Example: calculate ∆E for a system undergoing
endothermic process in which 15.6 kJ of heat flows and
where 1.4 kJ work done on the system.
Using equation;
∆?????? = � + ??????= 15.6 + 1.4 =17.0 kJ
(6.3) Introduction to Thermodynamics
Example: A system absorbs 850 J of heat while doing 3.35 kJ of
work on the surrounding. Calculate the internal energy change
for the system
ΔE
system = q + w
q= +850 J= +850 J
�??????
��
�
??????
= +0.85 kJ
w = -3.35 kJ
ΔE
system = q + w = + 0.85 kJ + (-3.35 kJ)= - 2.50 kJ
Example: Surrounding do 0.4440 kJ of work on a system at an
external pressure of 2.19 atm. If the initial volume of the system
was 2.85 L. Calculate the final volume
??????=+�.�����??????×
��
�
??????
�??????
×
??????���
���.�??????
=+�.���??????���
w = -PV = -P(V
f - V
i)
+4.383 L atm = - (2.19 atm) (V
f – 2.85 L)
V
f = 0.85 L ➔ V
f V
i compression
(6.3) Introduction to Thermodynamics
work on the surrounding. Calculate the internal energy change
for the system
ΔE
system = q + w
q= +850 J= +850 J
�??????
��
�
??????
= +0.85 kJ
w = -3.35 kJ
ΔE
system = q + w = + 0.85 kJ + (-3.35 kJ)= - 2.50 kJ
Example: Surrounding do 0.4440 kJ of work on a system at an
external pressure of 2.19 atm. If the initial volume of the system
was 2.85 L. Calculate the final volume
??????=+�.�����??????×
��
�
??????
�??????
×
??????���
���.�??????
=+�.���??????���
w = -PV = -P(V
f - V
i)
+4.383 L atm = - (2.19 atm) (V
f – 2.85 L)
V
f = 0.85 L ➔ V
f V
i compression
(6.3) Introduction to Thermodynamics
What is enthalpy? It is a property of a substance
that allows us to quantify the heat flow into or out of a
system in a process that occurs at a constant
pressure.
U= q + w
At constant pressure ΔH = q
p
.. And w = -PV
U= ΔH -PV
H = U+ PV
ΔHis the Enthalpy change of a reaction
ΔHis the heatgiven off or absorbed during a
reaction at constant pressure.
(6.4) Enthalpy of Chemical Reactions
that allows us to quantify the heat flow into or out of a
system in a process that occurs at a constant
pressure.
U= q + w
At constant pressure ΔH = q
p
.. And w = -PV
U= ΔH -PV
H = U+ PV
ΔHis the Enthalpy change of a reaction
ΔHis the heatgiven off or absorbed during a
reaction at constant pressure.
(6.4) Enthalpy of Chemical Reactions
In the case in which the products of the reaction have a greater
enthalpy, ΔH will be positive (endothermic reaction)
If the enthalpy of the products is less than that of the reactants,
ΔH will be negative. (exothermic reaction)
For a chemical reaction or a physical change :
ΔH= H
products – H
reactantsHH
H
products reactants
>
Δ> 0 HH
H
products reactants
<
Δ< 0
(6.4) Enthalpy of Chemical Reactions
enthalpy, ΔH will be positive (endothermic reaction)
If the enthalpy of the products is less than that of the reactants,
ΔH will be negative. (exothermic reaction)
For a chemical reaction or a physical change :
ΔH= H
products – H
reactantsHH
H
products reactants
>
Δ> 0 HH
H
products reactants
<
Δ< 0
(6.4) Enthalpy of Chemical Reactions
Thermodynamical Equations
Heat is absorbed by the system from the
surroundings, and the change in enthalpy
(from reactants to products) is a positive
value.
Is ΔH negative or positive?
System absorbs heat
Endothermic0H Δ>
6.01 kJ are absorbed for every 1 mole of ice that melts
at 0 C and 1 atm() ()→ Δ =
22
HO HO 6.01 kJmolsl H
(6.4) Enthalpy of Chemical Reactions
Heat is absorbed by the system from the
surroundings, and the change in enthalpy
(from reactants to products) is a positive
value.
Is ΔH negative or positive?
System absorbs heat
Endothermic0H Δ>
6.01 kJ are absorbed for every 1 mole of ice that melts
at 0 C and 1 atm() ()→ Δ =
22
HO HO 6.01 kJmolsl H
(6.4) Enthalpy of Chemical Reactions
Heat is absorbed by the system from
the surroundings, and the change in
enthalpy (from reactants to products)
is a positive value.
Is ΔH negative or positive?
System gives off heat
ExothermicH Δ< 0
890.4 kJ are released for every 1 mole of methane that is
combusted at 0 C and 1 atm() () () ()
4 2 2 2
CH + 2O CO 2HO →+g g g l Δ = – 890.4kJmolH
(6.4) Enthalpy of Chemical Reactions
the surroundings, and the change in
enthalpy (from reactants to products)
is a positive value.
Is ΔH negative or positive?
System gives off heat
ExothermicH Δ< 0
890.4 kJ are released for every 1 mole of methane that is
combusted at 0 C and 1 atm() () () ()
4 2 2 2
CH + 2O CO 2HO →+g g g l Δ = – 890.4kJmolH
(6.4) Enthalpy of Chemical Reactions
H
2O (s) H
2O (l)H = + 6.01 kJ
The stoichiometric coefficients always refer to the number
of moles of a substance
•If you reverse a reaction, the sign of H changes.
H
2O (l) H
2O (s)H = - 6.01 kJ
2H
2O (s) 2H
2O (l)H = 2 x 6.01 = 12.0 kJ
Twice as much energy is required to melt 2 moles of
iced water
Because the direction of heat flow reversed
Characteristics of H of a reaction
•If you multiply both sides of the equation by a factor n,
then ΔH must change by the same factor n.
(6.4) Enthalpy of Chemical Reactions
2O (s) H
2O (l)H = + 6.01 kJ
The stoichiometric coefficients always refer to the number
of moles of a substance
•If you reverse a reaction, the sign of H changes.
H
2O (l) H
2O (s)H = - 6.01 kJ
2H
2O (s) 2H
2O (l)H = 2 x 6.01 = 12.0 kJ
Twice as much energy is required to melt 2 moles of
iced water
Because the direction of heat flow reversed
Characteristics of H of a reaction
•If you multiply both sides of the equation by a factor n,
then ΔH must change by the same factor n.
(6.4) Enthalpy of Chemical Reactions
Exercise:
B
2H
6(g) + 3O
2(g) → B
2O
3(s) + 3H
2O
(g) H= -2035 kJ
What is H when 10.0 g of B
2H
6 is mixed with 10.0 g O
2?
H
2O (s) H
2O (l)H = 6.01 kJ/mol
The physical states of all reactants and products
must be specified in thermochemical equations.
H
2O (l) H
2O (g)H = 44.0 kJ/mol
Because the of difference in physical states
Characteristics of H of a reaction
(6.4) Enthalpy of Chemical Reactions
B
2H
6(g) + 3O
2(g) → B
2O
3(s) + 3H
2O
(g) H= -2035 kJ
What is H when 10.0 g of B
2H
6 is mixed with 10.0 g O
2?
H
2O (s) H
2O (l)H = 6.01 kJ/mol
The physical states of all reactants and products
must be specified in thermochemical equations.
H
2O (l) H
2O (g)H = 44.0 kJ/mol
Because the of difference in physical states
Characteristics of H of a reaction
(6.4) Enthalpy of Chemical Reactions
Example 6.3
Given the thermochemical equation
2SO
2(g) + O
2(g) →2SO
3(g) H= -198.2 kJ/mol
Calculate the heat evolved when 87.9 g of SO
2 (molar
mass=64.07 g/mol) is converted to SO
3.
Solution
We need to first calculate the number of moles of SO
2
in 87.9 g of the compound and then find the number of
kilojoules produced from the exothermic reaction. The
sequence of conversions is as follows:
(6.4) Enthalpy of Chemical Reactions
Given the thermochemical equation
2SO
2(g) + O
2(g) →2SO
3(g) H= -198.2 kJ/mol
Calculate the heat evolved when 87.9 g of SO
2 (molar
mass=64.07 g/mol) is converted to SO
3.
Solution
We need to first calculate the number of moles of SO
2
in 87.9 g of the compound and then find the number of
kilojoules produced from the exothermic reaction. The
sequence of conversions is as follows:
(6.4) Enthalpy of Chemical Reactions
��??????� �� ??????�
2 → ����� �� ??????�
2 → ���������� �� ℎ�??????� �����??????���
Therefore, the enthalpy change for this reaction is given by:
????????????=��.����
�×
������
�
��.�����
�
×
–���.��??????
������
�
= – ��� �??????
and the heat released to the surroundings is 136 kJ.
Check
Because 87.9 g is less than twice the molar mass of SO
2
(264.07 g)
as shown in the preceding thermochemical equation, we
expect the heat released to be smaller than 198.2 kJ.
(6.4) Enthalpy of Chemical Reactions
2 → ����� �� ??????�
2 → ���������� �� ℎ�??????� �����??????���
Therefore, the enthalpy change for this reaction is given by:
????????????=��.����
�×
������
�
��.�����
�
×
–���.��??????
������
�
= – ��� �??????
and the heat released to the surroundings is 136 kJ.
Check
Because 87.9 g is less than twice the molar mass of SO
2
(264.07 g)
as shown in the preceding thermochemical equation, we
expect the heat released to be smaller than 198.2 kJ.
(6.4) Enthalpy of Chemical Reactions
A Comparison of ΔHand ΔU() () () ()
22
2
2Na + 2HO 2NaOH + H Δ = – 367.5 kJmol
Δ= Δ– Δ At 25 C, 1 mole H = 24.5 L at 1 atm
= 1 atm 24.5 L = 2.5 kJ
= – 367.5kJmol – 2.5 kJ mol = – 370.0 kJmo
→
o
s l aqgH
U H PV
PΔV
ΔU l
Remember convert 24.5 L. atm. into KJ
(6.4) Enthalpy of Chemical Reactions
22
2
2Na + 2HO 2NaOH + H Δ = – 367.5 kJmol
Δ= Δ– Δ At 25 C, 1 mole H = 24.5 L at 1 atm
= 1 atm 24.5 L = 2.5 kJ
= – 367.5kJmol – 2.5 kJ mol = – 370.0 kJmo
→
o
s l aqgH
U H PV
PΔV
ΔU l
Remember convert 24.5 L. atm. into KJ
(6.4) Enthalpy of Chemical Reactions
Example 6.4
Calculate the change in internal energy
when 2 moles of CO are converted to 2
moles of CO
2 at 1 atm and 25°C:
2CO(g) + O
2(g) → 2CO
2(g)
H = -198.2 kJ/mol
P(V) = RT(n)
Strategy
We are given the enthalpy change, ∆??????, for the
reaction and are asked to calculate the change
in internal energy, ∆�.
Therefore, we need Equation (6.10).
What is the change in the number of moles of gases?
ΔH is given in kilojoules, so what units should we use for R?
(6.4) Enthalpy of Chemical Reactions
Calculate the change in internal energy
when 2 moles of CO are converted to 2
moles of CO
2 at 1 atm and 25°C:
2CO(g) + O
2(g) → 2CO
2(g)
H = -198.2 kJ/mol
P(V) = RT(n)
Strategy
We are given the enthalpy change, ∆??????, for the
reaction and are asked to calculate the change
in internal energy, ∆�.
Therefore, we need Equation (6.10).
What is the change in the number of moles of gases?
ΔH is given in kilojoules, so what units should we use for R?
(6.4) Enthalpy of Chemical Reactions
Solution
From the chemical equation we see that 3 moles of gases
are converted to 2 moles of gases so that
n = number of moles of product gas – number of moles
reactant gases
= 2 − 3
= − 1
Using 8.314 J/K.mol for R and T = 298K in Equation (6.10),
we write( ) ( )()
U HRTnΔ=Δ– Δ
1 kJ
=– 566.0 kJ mol – 8.314JK mol 298 K –1
1000 J
=– 563.5 kJ mol
(6.4) Enthalpy of Chemical Reactions
From the chemical equation we see that 3 moles of gases
are converted to 2 moles of gases so that
n = number of moles of product gas – number of moles
reactant gases
= 2 − 3
= − 1
Using 8.314 J/K.mol for R and T = 298K in Equation (6.10),
we write( ) ( )()
U HRTnΔ=Δ– Δ
1 kJ
=– 566.0 kJ mol – 8.314JK mol 298 K –1
1000 J
=– 563.5 kJ mol
(6.4) Enthalpy of Chemical Reactions
6.5 Calorimetry
The specific heat (s) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the substance
by one degree Celsius.
Table 6.2 The Specific Heats of
Some Common Substances( )
( )
( )
( )
o
Specific Heat
Substances
Jg.C
2
22
AI 0.900
Au 0.129
C graphics 0.720
C diamond 0.502
Cu 0.385
Fe 0.444
Hg 0.139
Pb 0.158
HO 0.184
CHOH ethanol2.46
The heat capacity (C) of a substance is the amount of heat (q)
required to raise the temperature of a given quantity (m) of
the substance by one degree Celsius.= C m s
Heat (q) absorbed or released:final initial
= Δ
= Δ
Δ= –
q m s t
q C t
t t t
The specific heat (s) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the substance
by one degree Celsius.
Table 6.2 The Specific Heats of
Some Common Substances( )
( )
( )
( )
o
Specific Heat
Substances
Jg.C
2
22
AI 0.900
Au 0.129
C graphics 0.720
C diamond 0.502
Cu 0.385
Fe 0.444
Hg 0.139
Pb 0.158
HO 0.184
CHOH ethanol2.46
The heat capacity (C) of a substance is the amount of heat (q)
required to raise the temperature of a given quantity (m) of
the substance by one degree Celsius.= C m s
Heat (q) absorbed or released:final initial
= Δ
= Δ
Δ= –
q m s t
q C t
t t t
Example 6.5
A 466-g sample of water is heated from 8.50
C to 74.60
C.
Calculate the amount of heat absorbed (in kilojoules) by
the water.
Strategy
We know the quantity of water and the specific heat of
water. With this information and the temperature rise, we
can calculate the amount of heat absorbed (q)
6.5 Calorimetry
A 466-g sample of water is heated from 8.50
C to 74.60
C.
Calculate the amount of heat absorbed (in kilojoules) by
the water.
Strategy
We know the quantity of water and the specific heat of
water. With this information and the temperature rise, we
can calculate the amount of heat absorbed (q)
6.5 Calorimetry
Solution
Using Equation (6.12), we write( )( )( )
5
= Δ
= 466 g 4.184 JgC 74.60C – 8.50C
1 kJ
= 1.29 × 10J×
1000 J
= 129 kJ
o o o
q mst
Check
The units g and º C cancel, and we are left with the
desired unit kJ. Because heat is absorbed by the water
from the surroundings, it has a positive sign.
6.5 Calorimetry
Using Equation (6.12), we write( )( )( )
5
= Δ
= 466 g 4.184 JgC 74.60C – 8.50C
1 kJ
= 1.29 × 10J×
1000 J
= 129 kJ
o o o
q mst
Check
The units g and º C cancel, and we are left with the
desired unit kJ. Because heat is absorbed by the water
from the surroundings, it has a positive sign.
6.5 Calorimetry
6.5 Calorimetry
Constant pressure calorimetry
coffee cup calorimeter
Used to find ΔH for reaction in solutions
ΔH =q
p
Heat released by reaction = heat
absorbed by calorimeter + solution
Constant volume calorimetry
bomb calorimeter
Used to study reactions with gases (combustion)
P ΔV = 0
ΔE = q
v
heat released by reaction = heat absorbed by
calorimeter + water
Constant pressure calorimetry
coffee cup calorimeter
Used to find ΔH for reaction in solutions
ΔH =q
p
Heat released by reaction = heat
absorbed by calorimeter + solution
Constant volume calorimetry
bomb calorimeter
Used to study reactions with gases (combustion)
P ΔV = 0
ΔE = q
v
heat released by reaction = heat absorbed by
calorimeter + water
Constant-Volume Calorimetry( )
sys water bomb rxn
rxn water bomb
water
bomb
= + +
= 0
= – +
= × s× Δ
= × Δ
sys
bomb
qqqq
q
qqq
qm t
qCt
Reaction at Constant V H q
H q
rxn
rxn
Δ
Δ~
ΔU = q
rxn or ΔE = q
rxn
6.5 Calorimetry
sys water bomb rxn
rxn water bomb
water
bomb
= + +
= 0
= – +
= × s× Δ
= × Δ
sys
bomb
qqqq
q
qqq
qm t
qCt
Reaction at Constant V H q
H q
rxn
rxn
Δ
Δ~
ΔU = q
rxn or ΔE = q
rxn
6.5 Calorimetry
Example 6.6:
A quantity of 1.435 g of naphthalene
C
10H
8 pungent-smelling substance used
in moth repellents, was burned in a
constant-volume bomb calorimeter.
Consequently, the temperature of the water rose from 20.28
C to 25.95C.
If the heat capacity of the bomb plus water was 10.17 kJ/C
Calculate the heat of combustion of naphthalene on a
molar basis; that is, find the molar heat of combustion
6.5 Calorimetry
A quantity of 1.435 g of naphthalene
C
10H
8 pungent-smelling substance used
in moth repellents, was burned in a
constant-volume bomb calorimeter.
Consequently, the temperature of the water rose from 20.28
C to 25.95C.
If the heat capacity of the bomb plus water was 10.17 kJ/C
Calculate the heat of combustion of naphthalene on a
molar basis; that is, find the molar heat of combustion
6.5 Calorimetry
Strategy
Knowing the heat capacity and the temperature rise,
How do we calculate the heat absorbed by the calorimeter?
What is the heat generated by the combustion of 1.435 g of
naphthalene?
What is the conversion factor between grams and moles of
naphthalene?
Solution
The heat absorbed by the bomb and water is equal to the
product of the heat capacity and the temperature change.
From Equation (6.16), assuming no heat is lost to the
surroundings, we write( )( )
o o o
qt
C
cal cal
= CΔ
= 10.17KJ25.95C–20.28C
= 57.66 kJ
6.5 Calorimetry
Knowing the heat capacity and the temperature rise,
How do we calculate the heat absorbed by the calorimeter?
What is the heat generated by the combustion of 1.435 g of
naphthalene?
What is the conversion factor between grams and moles of
naphthalene?
Solution
The heat absorbed by the bomb and water is equal to the
product of the heat capacity and the temperature change.
From Equation (6.16), assuming no heat is lost to the
surroundings, we write( )( )
o o o
qt
C
cal cal
= CΔ
= 10.17KJ25.95C–20.28C
= 57.66 kJ
6.5 Calorimetry
.
sys cal rxn cal rxn
Because = + = 0, = – qqqqq The heat change of the reaction is −57.66 kJ. This is
the heat released by the combustion of 1.435 g of
C
10H
8; therefore, we can write the conversion factor as108
– 57.66 kJ
1.435 g CH
The molar mass of naphthalene is 128.2 g, so the heat
of combustion of 1 mole of naphthalene is108
– 57.66 kJ
molar heat of combustion =
1.435g CH
108
128.2 g CH
×
108
3
1 mol CH
= – 5.151 × 10kJmol
6.5 Calorimetry
sys cal rxn cal rxn
Because = + = 0, = – qqqqq The heat change of the reaction is −57.66 kJ. This is
the heat released by the combustion of 1.435 g of
C
10H
8; therefore, we can write the conversion factor as108
– 57.66 kJ
1.435 g CH
The molar mass of naphthalene is 128.2 g, so the heat
of combustion of 1 mole of naphthalene is108
– 57.66 kJ
molar heat of combustion =
1.435g CH
108
128.2 g CH
×
108
3
1 mol CH
= – 5.151 × 10kJmol
6.5 Calorimetry
Check
Knowing that the combustion reaction is exothermic and
that the molar mass of naphthalene is much greater than
1.4 g, is the answer reasonable? Under the reaction
conditions, can the heat change (-57.66 kJ) be equated
to the enthalpy change of the reaction?
6.5 Calorimetry
Knowing that the combustion reaction is exothermic and
that the molar mass of naphthalene is much greater than
1.4 g, is the answer reasonable? Under the reaction
conditions, can the heat change (-57.66 kJ) be equated
to the enthalpy change of the reaction?
6.5 Calorimetry
Constant-Pressure Calorimetry
No heat enters or leaves!( )
sys water cal rxn
sys
rxn water cal
water
cal cal
= + +
= 0
= – +
= × × Δ
= × Δ
qqqq
q
qqq
qm s t
qCt
Reaction at Constant Prxn
Δ= H q
6.5 Calorimetry
No heat enters or leaves!( )
sys water cal rxn
sys
rxn water cal
water
cal cal
= + +
= 0
= – +
= × × Δ
= × Δ
qqqq
q
qqq
qm s t
qCt
Reaction at Constant Prxn
Δ= H q
6.5 Calorimetry
Example 6.7:
A lead (Pb) pellet having a mass of 26.47 g
at 89.98
C was placed in a constant-
pressure calorimeter of negligible heat
capacity containing 100.0 mL of water.
The water temperature rose from 22.50
C
to 23.17
C.
What is the specific heat of the lead pellet?
Strategy
A sketch of the initial and
final situation is as follows:
6.5 Calorimetry
A lead (Pb) pellet having a mass of 26.47 g
at 89.98
C was placed in a constant-
pressure calorimeter of negligible heat
capacity containing 100.0 mL of water.
The water temperature rose from 22.50
C
to 23.17
C.
What is the specific heat of the lead pellet?
Strategy
A sketch of the initial and
final situation is as follows:
6.5 Calorimetry
We know the masses of water and the lead pellet as well as the
initial and final temperatures. Assuming no heat is lost to the
surroundings, we can equate the heat lost by the lead pellet to
the heat gained by the water. Knowing the specific heat of
water, we can then calculate the specific heat of lead.
Solution
Treating the calorimeter as an isolated system (no heat lost
to the surroundings), we write
�
�� + �
??????��
= �
or �
�� = – �
??????��
The heat gained by the water is given by
�
??????�� = �� ?????? �
Or q
H2O = - q
Pb
where m and s are the mass and specific heat andfinal initial
Δ= – t tt
6.5 Calorimetry
initial and final temperatures. Assuming no heat is lost to the
surroundings, we can equate the heat lost by the lead pellet to
the heat gained by the water. Knowing the specific heat of
water, we can then calculate the specific heat of lead.
Solution
Treating the calorimeter as an isolated system (no heat lost
to the surroundings), we write
�
�� + �
??????��
= �
or �
�� = – �
??????��
The heat gained by the water is given by
�
??????�� = �� ?????? �
Or q
H2O = - q
Pb
where m and s are the mass and specific heat andfinal initial
Δ= – t tt
6.5 Calorimetry
Therefore,( )( )( )
o o o
2
HO
= 100.0 g 4.184J gC 23.17C – 22.50C
= 280.3 J
q
Because the heat lost by the lead pellet is equal to the
heat gained by the water, q
Pb = −280.3 J. Solving for the
specific heat of Pb, we write( )()( )
Pb
= Δ
– 280.3 J = 26.47g 23.17C – 89.98C
= 0.158J gC
oo
o
qmst
s
s
6.5 Calorimetry
o o o
2
HO
= 100.0 g 4.184J gC 23.17C – 22.50C
= 280.3 J
q
Because the heat lost by the lead pellet is equal to the
heat gained by the water, q
Pb = −280.3 J. Solving for the
specific heat of Pb, we write( )()( )
Pb
= Δ
– 280.3 J = 26.47g 23.17C – 89.98C
= 0.158J gC
oo
o
qmst
s
s
6.5 Calorimetry
Table 6.3 Heats of Some Typical Reactions Measured
at Constant Pressure( )
→
→
→
2
+–
2
22
Heat of neutralizationHCl() + NaOH () Na Cl ( ) + HO ()–56.2
Heat of ionizationHO() H() + OH() 56. 2
Heat of fusion HO() HO) 6.01
Heat of v
Type of
Example
kJ/ molReaction
aq aq aq l
laqaq
s(l
H
→
→
22
2
aporizationHO() HO() 44.0*
Heat of reactionMgCl() + 2Na() 2Nacl() + Mg()–180.2
lg
sl ss
Some Heats of Reaction
6.5 Calorimetry
at Constant Pressure( )
→
→
→
2
+–
2
22
Heat of neutralizationHCl() + NaOH () Na Cl ( ) + HO ()–56.2
Heat of ionizationHO() H() + OH() 56. 2
Heat of fusion HO() HO) 6.01
Heat of v
Type of
Example
kJ/ molReaction
aq aq aq l
laqaq
s(l
H
→
→
22
2
aporizationHO() HO() 44.0*
Heat of reactionMgCl() + 2Na() 2Nacl() + Mg()–180.2
lg
sl ss
Some Heats of Reaction
6.5 Calorimetry
Example 6.8:
A quantity of 1.0010
2
mL of 0.500 M HCl was mixed with
1.0010
2
mL of 0.500M NaOH in a constant-pressure
calorimeter of negligible heat capacity. The initial
temperature of the HCl and NaOH solutions was the
same, 22.50 C, and the final temperature of the mixed
solution was 25.86 C. Calculate the heat change for the
neutralization reaction on a molar basis() () () ()→
2
NaOH + HCl NaCl + HO aqaq aq l
Assume that the densities and specific heats of the
solutions are the same as for water( )
o
1.00 gmL and 4.184Jg C, respectively
6.5 Calorimetry
A quantity of 1.0010
2
mL of 0.500 M HCl was mixed with
1.0010
2
mL of 0.500M NaOH in a constant-pressure
calorimeter of negligible heat capacity. The initial
temperature of the HCl and NaOH solutions was the
same, 22.50 C, and the final temperature of the mixed
solution was 25.86 C. Calculate the heat change for the
neutralization reaction on a molar basis() () () ()→
2
NaOH + HCl NaCl + HO aqaq aq l
Assume that the densities and specific heats of the
solutions are the same as for water( )
o
1.00 gmL and 4.184Jg C, respectively
6.5 Calorimetry
Strategy
Because the temperature rose, the neutralization reaction
is exothermic.
How do we calculate the heat absorbed by the combined
solution?
What is the heat of the reaction?
What is the conversion factor for expressing the heat of
reaction on a molar basis?
Solution
Assuming no heat is lost to the surroundings,sys soln rxn rxn soln soln
= + = 0,so = – , where qqq qq q
is the heat absorbed by the combined solution. Because
the density of the solution is 1.00 g/ mL, the mass of a
100-mL solution is 100 g.
6.5 Calorimetry
Because the temperature rose, the neutralization reaction
is exothermic.
How do we calculate the heat absorbed by the combined
solution?
What is the heat of the reaction?
What is the conversion factor for expressing the heat of
reaction on a molar basis?
Solution
Assuming no heat is lost to the surroundings,sys soln rxn rxn soln soln
= + = 0,so = – , where qqq qq q
is the heat absorbed by the combined solution. Because
the density of the solution is 1.00 g/ mL, the mass of a
100-mL solution is 100 g.
6.5 Calorimetry
( )( )( )
soln
22
3
= Δ
= 1.00 × 10g + 1.00 × 10 g 4.184 Jg. °C 25.86° C – 22.50°C
= 2.81 × 10 J
= 2.81 kJ
qmst Because = – , = – 2.81 kJ.
rxn soln rxn
qqq From the molarities given, the number of moles of both
HCl and NaOH in 1.0010
2
mL solution is: =
0.500mol
0.100L 0.0500mol
1L
Therefore, the heat of neutralization when 1.00 mole of
HCl reacts with 1.00 mole of NaOH is–2.81 kJ
heat of neutralization = = –56.2kJmol
0.0500 mol
6.5 Calorimetry
soln
22
3
= Δ
= 1.00 × 10g + 1.00 × 10 g 4.184 Jg. °C 25.86° C – 22.50°C
= 2.81 × 10 J
= 2.81 kJ
qmst Because = – , = – 2.81 kJ.
rxn soln rxn
qqq From the molarities given, the number of moles of both
HCl and NaOH in 1.0010
2
mL solution is: =
0.500mol
0.100L 0.0500mol
1L
Therefore, the heat of neutralization when 1.00 mole of
HCl reacts with 1.00 mole of NaOH is–2.81 kJ
heat of neutralization = = –56.2kJmol
0.0500 mol
6.5 Calorimetry
6.6 Standard Enthalpy of Formation and Reaction
Because there is no way to measure the absolute value of the
enthalpy of a substance, must I measure the enthalpy change
for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of
formation (
f)
as a reference point for all enthalpy expressions.
Standard enthalpy of formation (
f)
is the heat change that results when one mole of a compound
is formed from its elements at a pressure of 1 atm.
▪The standard enthalpy of formation of any element in its most
stable form is zero.() ( )
() ( )
00
f 2 f
00
f 3 f
ΔO = 0 ΔC,graphite = 0
ΔO = 142 KJmol ΔC,diamo nd = 1.90 KJmol
HH
HH
Because there is no way to measure the absolute value of the
enthalpy of a substance, must I measure the enthalpy change
for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of
formation (
f)
as a reference point for all enthalpy expressions.
Standard enthalpy of formation (
f)
is the heat change that results when one mole of a compound
is formed from its elements at a pressure of 1 atm.
▪The standard enthalpy of formation of any element in its most
stable form is zero.() ( )
() ( )
00
f 2 f
00
f 3 f
ΔO = 0 ΔC,graphite = 0
ΔO = 142 KJmol ΔC,diamo nd = 1.90 KJmol
HH
HH
The standard Enthalpy of Formation ( H
f) of a
substance is the enthalpy change (∆H) that
accompanies the formation of one mole of
compound from its elements in their reference form
and their standard state.
H
f unit is kJ/mol
Definition of standard states:
For a compound
-For Gaseous substance is a pressure of 1 atm
-For Pure solid or liquid is pure solid or liquid
-For a solute in a solution is a concentration of 1 M
For elements
- The form of the element exists under 1 atm and 25 C
6.6 Standard Enthalpy of Formation and Reaction
f) of a
substance is the enthalpy change (∆H) that
accompanies the formation of one mole of
compound from its elements in their reference form
and their standard state.
H
f unit is kJ/mol
Definition of standard states:
For a compound
-For Gaseous substance is a pressure of 1 atm
-For Pure solid or liquid is pure solid or liquid
-For a solute in a solution is a concentration of 1 M
For elements
- The form of the element exists under 1 atm and 25 C
6.6 Standard Enthalpy of Formation and Reaction
Reference form of an element:
chosen (usually the most stable form) allotrope at 25C
e.g. O
2(g) not O
3(g)
C(s, graphite) not C (s, diamond)
S(s,rhombic)
P(s, white)
See table 6.4
element H
f(kJ/mol)element H
felement H
f
Al(s) 0 Cd (s) 0 Ba (s) 0
Ca (s) 0 Cu (s) 0 Fe (s) 0
Br
2(l) 0 Br
2(g) -3 Br
2(aq) -121
Cl
2(g) 0 O
2(g) 0 O
3(g) 143
F
2(g) 0 F
2(aq) -333Hg (l) 0
P (s) white0 P (s) red -18 P (s) black-39
P
4(g) 59 C (s) graphite0 C (s) diamond2
6.6 Standard Enthalpy of Formation and Reaction
chosen (usually the most stable form) allotrope at 25C
e.g. O
2(g) not O
3(g)
C(s, graphite) not C (s, diamond)
S(s,rhombic)
P(s, white)
See table 6.4
element H
f(kJ/mol)element H
felement H
f
Al(s) 0 Cd (s) 0 Ba (s) 0
Ca (s) 0 Cu (s) 0 Fe (s) 0
Br
2(l) 0 Br
2(g) -3 Br
2(aq) -121
Cl
2(g) 0 O
2(g) 0 O
3(g) 143
F
2(g) 0 F
2(aq) -333Hg (l) 0
P (s) white0 P (s) red -18 P (s) black-39
P
4(g) 59 C (s) graphite0 C (s) diamond2
6.6 Standard Enthalpy of Formation and Reaction
Chemical equations for ∆H
o
f of substances
∆H
o
f for NH
4BrO
3(s)
�
�
N
2 (g) + 2H
2(g) +
�
�
????????????
2 (l) +
�
�
O
2 (g) → NH
4BrO
3(s)
∆H
o
f for Hg
4I
2(s)
2Hg(l) + I
2(g) → Hg
2I
2(s)
∆H
o
f for CH
3CN(l)
2C(s,graphite) +
�
�
N
2 (g) +
�
�
H
2(g) → CH
3CN(l)
∆H
o
f for O
3(g)
�
�
O
2 (g) → O
3(g)
∆H
o
f for O(g)
�
�
O
2 (g) → O(g)
∆H
o
f for N
2(l)
N
2(g) → N
2(l)
6.6 Standard Enthalpy of Formation and Reaction
o
f of substances
∆H
o
f for NH
4BrO
3(s)
�
�
N
2 (g) + 2H
2(g) +
�
�
????????????
2 (l) +
�
�
O
2 (g) → NH
4BrO
3(s)
∆H
o
f for Hg
4I
2(s)
2Hg(l) + I
2(g) → Hg
2I
2(s)
∆H
o
f for CH
3CN(l)
2C(s,graphite) +
�
�
N
2 (g) +
�
�
H
2(g) → CH
3CN(l)
∆H
o
f for O
3(g)
�
�
O
2 (g) → O
3(g)
∆H
o
f for O(g)
�
�
O
2 (g) → O(g)
∆H
o
f for N
2(l)
N
2(g) → N
2(l)
6.6 Standard Enthalpy of Formation and Reaction
Enthalpy calculations:
▪When the reaction is reversed the magnitude of H remains
the same but its sign changes
▪When the balanced equation is multiplied by an integer , H
should be multiplied by the same integer.
▪H
f of elements in their standard states is zero.
▪H of a reaction can be calculated by:
H (reaction) = nH
f (products) - nH
f (reactants)
Examples:
1
2
N
2 (g) +
3
2
H
2(g) → NH
3(g) H
f= -46 kJ mol
-1
N
2 (g) + 3H
2(g) → 2NH
3(g) H= 2 mol -46 kJ mol
-1
=-92 kJ
2Fe(s) +
3
2
O
2(g) → Fe
2O
3(g) H
f= -826 kJ mol
-1
6.6 Standard Enthalpy of Formation and Reaction
▪When the reaction is reversed the magnitude of H remains
the same but its sign changes
▪When the balanced equation is multiplied by an integer , H
should be multiplied by the same integer.
▪H
f of elements in their standard states is zero.
▪H of a reaction can be calculated by:
H (reaction) = nH
f (products) - nH
f (reactants)
Examples:
1
2
N
2 (g) +
3
2
H
2(g) → NH
3(g) H
f= -46 kJ mol
-1
N
2 (g) + 3H
2(g) → 2NH
3(g) H= 2 mol -46 kJ mol
-1
=-92 kJ
2Fe(s) +
3
2
O
2(g) → Fe
2O
3(g) H
f= -826 kJ mol
-1
6.6 Standard Enthalpy of Formation and Reaction
Example 1:
Find the enthalpy of the following reaction by knowing the
enthalpy of formation
CH
4(g) + 2O
2(g) → CO
2(g) + 2H
2O(l)
H
f (CH
4)= -75 kJ mol
-1
, H
f (CO
2)= -393.5 kJ mol
-1
, H
f
(H
2O)= -286 kJ mol
-1
Answer :
H (reaction) = nH
f (products) - nH
f (reactants)
H= [H
f (CO
2) + 2 H
f (H
2O) ] - [H
f (CH
4) + 2 H
f (O
2)]
= [-393.5 kJ mol
-1
+ 2× -286 kJ mol
-1
] – [-75 kJ mol
-1
+ 2 × 0 kJ]
= [-965.5 kJ] + [75 kJ] = -890.5 kJ
Calculating H
rxnfrom H
f
H (reaction) = nH
f (products) - n H
f (reactants)
Or,,, H
rxn = n
p (H
f )
p - n
r (H
f )
r
6.6 Standard Enthalpy of Formation and Reaction
Find the enthalpy of the following reaction by knowing the
enthalpy of formation
CH
4(g) + 2O
2(g) → CO
2(g) + 2H
2O(l)
H
f (CH
4)= -75 kJ mol
-1
, H
f (CO
2)= -393.5 kJ mol
-1
, H
f
(H
2O)= -286 kJ mol
-1
Answer :
H (reaction) = nH
f (products) - nH
f (reactants)
H= [H
f (CO
2) + 2 H
f (H
2O) ] - [H
f (CH
4) + 2 H
f (O
2)]
= [-393.5 kJ mol
-1
+ 2× -286 kJ mol
-1
] – [-75 kJ mol
-1
+ 2 × 0 kJ]
= [-965.5 kJ] + [75 kJ] = -890.5 kJ
Calculating H
rxnfrom H
f
H (reaction) = nH
f (products) - n H
f (reactants)
Or,,, H
rxn = n
p (H
f )
p - n
r (H
f )
r
6.6 Standard Enthalpy of Formation and Reaction
Example 2:
Find the enthalpy of the following reaction by knowing the
enthalpy of formation
C
6H
12O
6 (s) + 6O
2(g) → 6CO
2(g) + 6H
2O(l)
H
f (C
6H
12O
6 )= -1275 kJ mol
-1
, H
f (CO
2)= -393.5 kJ mol
-1
, H
f (H
2O)= -286 kJ mol
-1
Answer :
H
comb = n
p (H
f )
p - n
r (H
f )
r
H
comb = [(6 mol-393.5 kJ mol
-1
) + (6 mol -286 kJ mol
-1
)]
- [(1 mol -1275 kJ mol
-1
) + (6 mol 0 kJ mol
-1
)]
H
comb = [-4077] – [-1275 ]
= -2802 kJ
6.6 Standard Enthalpy of Formation and Reaction
Find the enthalpy of the following reaction by knowing the
enthalpy of formation
C
6H
12O
6 (s) + 6O
2(g) → 6CO
2(g) + 6H
2O(l)
H
f (C
6H
12O
6 )= -1275 kJ mol
-1
, H
f (CO
2)= -393.5 kJ mol
-1
, H
f (H
2O)= -286 kJ mol
-1
Answer :
H
comb = n
p (H
f )
p - n
r (H
f )
r
H
comb = [(6 mol-393.5 kJ mol
-1
) + (6 mol -286 kJ mol
-1
)]
- [(1 mol -1275 kJ mol
-1
) + (6 mol 0 kJ mol
-1
)]
H
comb = [-4077] – [-1275 ]
= -2802 kJ
6.6 Standard Enthalpy of Formation and Reaction
Example 3:
Find the enthalpy of formation H
f of C
2H
4 giving that:
H
comb= -1411.1 kJ mol
-1
, H
f (CO
2)= -393.5 kJ mol
-1
,
H
f (H
2O)= -286 kJ mol
-1
C
2H
4 (g) + 3O
2(g) → 2CO
2(g) + 2H
2O(l)
Answer:
H
comb = n
p (H
f )
p - n
r (H
f )
r
-1411.1 kJ mol
-1
= [(2 mol-393.5 kJ mol
-1
)+(2 mol -286 kJ mol
-1
)]
- [(H
f of C
2H
4 ) + (3 mol 0 kJ mol
-1
)]
-1411.1 = -1358.8 - H
f of C
2H
4
H
f of C
2H
4 = +52.3 kJ/mol
6.6 Standard Enthalpy of Formation and Reaction
Find the enthalpy of formation H
f of C
2H
4 giving that:
H
comb= -1411.1 kJ mol
-1
, H
f (CO
2)= -393.5 kJ mol
-1
,
H
f (H
2O)= -286 kJ mol
-1
C
2H
4 (g) + 3O
2(g) → 2CO
2(g) + 2H
2O(l)
Answer:
H
comb = n
p (H
f )
p - n
r (H
f )
r
-1411.1 kJ mol
-1
= [(2 mol-393.5 kJ mol
-1
)+(2 mol -286 kJ mol
-1
)]
- [(H
f of C
2H
4 ) + (3 mol 0 kJ mol
-1
)]
-1411.1 = -1358.8 - H
f of C
2H
4
H
f of C
2H
4 = +52.3 kJ/mol
6.6 Standard Enthalpy of Formation and Reaction
Example 4:The Ostwald process for the commercial production of
nitric acid from ammonia and oxygen involves the following steps
4NH
3(g) + 5O
2(g) → 4NO(g) + 6H
2O(g)
2NO(g) + O
2(g) → 2NO
2(g)
3NO
2(g) + H
2O(l)→ 2HNO
3(aq) + NO(g)
a)Use the values of H
f to calculate the value of H for each of
the preceding reactions.
H
fNO (g)= 90 kJ/mol H
f NH
3 (g)= -46 kJ/mol
H
f(H
2O(g) =-242 kJ/mol H
f H
2O(l) = -286 kJ/mol
H
f NO
2(g)= +34 kJ/mol H
fHNO
3(aq)= -207 kJ/mol
6.6 Standard Enthalpy of Formation and Reaction
nitric acid from ammonia and oxygen involves the following steps
4NH
3(g) + 5O
2(g) → 4NO(g) + 6H
2O(g)
2NO(g) + O
2(g) → 2NO
2(g)
3NO
2(g) + H
2O(l)→ 2HNO
3(aq) + NO(g)
a)Use the values of H
f to calculate the value of H for each of
the preceding reactions.
H
fNO (g)= 90 kJ/mol H
f NH
3 (g)= -46 kJ/mol
H
f(H
2O(g) =-242 kJ/mol H
f H
2O(l) = -286 kJ/mol
H
f NO
2(g)= +34 kJ/mol H
fHNO
3(aq)= -207 kJ/mol
6.6 Standard Enthalpy of Formation and Reaction
Exercise: Using enthalpies of formation . Calculate the
standard change in enthalpy for the thermite reaction
2Al(s) + Fe
2O
3(s) → Al
2O
3(s) + 2Fe(s)
6.6 Standard Enthalpy of Formation and Reaction
standard change in enthalpy for the thermite reaction
2Al(s) + Fe
2O
3(s) → Al
2O
3(s) + 2Fe(s)
6.6 Standard Enthalpy of Formation and Reaction
Exercise:
Consider the reaction
2ClF
3(g) + 2NH
3(g) → N
2(g) + 6HF(g) + Cl
2(g) H= -1196 kJ
Calculate H
f for ClF
3(g)
H
f(HF)= -271 kJ/mol H
f (NH
3)= -46 kJ/mol
6.6 Standard Enthalpy of Formation and Reaction
Consider the reaction
2ClF
3(g) + 2NH
3(g) → N
2(g) + 6HF(g) + Cl
2(g) H= -1196 kJ
Calculate H
f for ClF
3(g)
H
f(HF)= -271 kJ/mol H
f (NH
3)= -46 kJ/mol
6.6 Standard Enthalpy of Formation and Reaction
•Hess’slawstatesthatwhenreactantsareconvertedto
products,thechangeinenthalpyisthesamewhether
thereactiontakesplaceinonesteporaseriesofsteps
•EnthalpyisastatefunctionorPathIndependent,it
doesn’tmatterhowyougetthere,onlywhereyou
startandend.
•For a chemical equation that can be written as the sum of
two or more steps, the enthalpy change for the overall
reaction is the sum of the enthalpy changes for the
individual steps.
H
rxn= H
1 + H
2 + H
3 + ….
6.6 Standard Enthalpy of Formation and Reaction
products,thechangeinenthalpyisthesamewhether
thereactiontakesplaceinonesteporaseriesofsteps
•EnthalpyisastatefunctionorPathIndependent,it
doesn’tmatterhowyougetthere,onlywhereyou
startandend.
•For a chemical equation that can be written as the sum of
two or more steps, the enthalpy change for the overall
reaction is the sum of the enthalpy changes for the
individual steps.
H
rxn= H
1 + H
2 + H
3 + ….
6.6 Standard Enthalpy of Formation and Reaction
Consider the formation of carbon dioxide (CO
2).
Direct Method
C
(s) + O
2
(g) → CO
2
(g)
C
(s) + ½ O
2 → CO
(g)
CO + ½ O
2 → CO
2(g)
∆H = - 108.98 KJ
∆H = - 284.52 KJ
C
(s) + O
2
(g) → CO
2
(g) ∆H = - 393.50 KJ
Adding:
Indirect Method
∆H = - 393.5 KJ
6.6 Standard Enthalpy of Formation and Reaction
Hess’s Law
2).
Direct Method
C
(s) + O
2
(g) → CO
2
(g)
C
(s) + ½ O
2 → CO
(g)
CO + ½ O
2 → CO
2(g)
∆H = - 108.98 KJ
∆H = - 284.52 KJ
C
(s) + O
2
(g) → CO
2
(g) ∆H = - 393.50 KJ
Adding:
Indirect Method
∆H = - 393.5 KJ
6.6 Standard Enthalpy of Formation and Reaction
Hess’s Law
Application of Hess’s law
RULES:
1.Balance individual equations.
2.If a equation is reversed the sign of H must also be
reversed.
3.If an entire equation is multiplied or divided by n , H
must also be multiplied or divided by n
4.Cross out substances that appear on both sides of
equation
5.Write overall reaction and add up H values for
individual steps.
6.6 Standard Enthalpy of Formation and Reaction
RULES:
1.Balance individual equations.
2.If a equation is reversed the sign of H must also be
reversed.
3.If an entire equation is multiplied or divided by n , H
must also be multiplied or divided by n
4.Cross out substances that appear on both sides of
equation
5.Write overall reaction and add up H values for
individual steps.
6.6 Standard Enthalpy of Formation and Reaction
Chemistry in Action: Bombardier Beetle Defense()() () () ()
()() () ()
() ()
()
()
() ()
→
→
→
→
0
64 22 642 22
0
64 642 2 2
0
22 2 2
22 2
CHOH + HO CHO+2HO Δ = ?
CHOH CHO + H Δ = 177 kJ mol
1 HO HO + O Δ = –94.6 kJ /mol
2
1H + O HO Δ = – 286 kJ /mol
2
g
g
aq aq aq lH
aq aqgH
aq l H
g IH o
Δ = 177 – 94.6 – 286 = – 204 kJmolH
Exothermic!
6.6 Standard Enthalpy of Formation and Reaction
()() () ()
() ()
()
()
() ()
→
→
→
→
0
64 22 642 22
0
64 642 2 2
0
22 2 2
22 2
CHOH + HO CHO+2HO Δ = ?
CHOH CHO + H Δ = 177 kJ mol
1 HO HO + O Δ = –94.6 kJ /mol
2
1H + O HO Δ = – 286 kJ /mol
2
g
g
aq aq aq lH
aq aqgH
aq l H
g IH o
Δ = 177 – 94.6 – 286 = – 204 kJmolH
Exothermic!
6.6 Standard Enthalpy of Formation and Reaction
Example 6.9
Calculate the standard enthalpy of formation of
acetylene (C
2H
2) from its elements:( ) () ()→
2 22
2C graphite + H CH gg
The equations for each step and the corresponding
enthalpy changes are()( ) () ()
() () () ()
() () () () ()
→
→
→
l
2 2 rxn
2 2 2 rxn
22 2 2 2 rxn
a Cgraphite + O CO Δ = – 393.5kJ/mol
1
b H + O HO Δ = – 285.8 kJ/mo
2
c 2CH + 5O 4CO + 2HO Δ = – 2598.8kJ/mol
g g H
g g l H
g g g lH
6.6 Standard Enthalpy of Formation and Reaction
Calculate the standard enthalpy of formation of
acetylene (C
2H
2) from its elements:( ) () ()→
2 22
2C graphite + H CH gg
The equations for each step and the corresponding
enthalpy changes are()( ) () ()
() () () ()
() () () () ()
→
→
→
l
2 2 rxn
2 2 2 rxn
22 2 2 2 rxn
a Cgraphite + O CO Δ = – 393.5kJ/mol
1
b H + O HO Δ = – 285.8 kJ/mo
2
c 2CH + 5O 4CO + 2HO Δ = – 2598.8kJ/mol
g g H
g g l H
g g g lH
6.6 Standard Enthalpy of Formation and Reaction
Strategy
Our goal here is to calculate the enthalpy change for the
formation of C
2H
2 from its elements C and H
2. The reaction
does not occur directly, however, so we must use an indirect
route using the information given by Equations (a), (b), and (c).
Solution
Looking at the synthesis of C
2H
2, we need 2 moles of
graphite as reactant. So we multiply Equation (a) by 2 to
get()( ) () () ( )
→
2 2 rxn
d 2C graphite + 2O 2CO Δ = 2– 393.5 kJ/ mol
= – 787.0 kJ/ mol
g gH
Next, we need 1 mole of H
2 as a reactant and this is
provided by Equation (b). Last, we need 1 mole C
2H
2 of as
a product.
6.6 Standard Enthalpy of Formation and Reaction
Our goal here is to calculate the enthalpy change for the
formation of C
2H
2 from its elements C and H
2. The reaction
does not occur directly, however, so we must use an indirect
route using the information given by Equations (a), (b), and (c).
Solution
Looking at the synthesis of C
2H
2, we need 2 moles of
graphite as reactant. So we multiply Equation (a) by 2 to
get()( ) () () ( )
→
2 2 rxn
d 2C graphite + 2O 2CO Δ = 2– 393.5 kJ/ mol
= – 787.0 kJ/ mol
g gH
Next, we need 1 mole of H
2 as a reactant and this is
provided by Equation (b). Last, we need 1 mole C
2H
2 of as
a product.
6.6 Standard Enthalpy of Formation and Reaction
Equation (c) has 2 moles of C
2H
2 as a reactant so we
need to reverse the equation and divide it by 2:() () () () () ( )
→
2 2 22 2 rxn
51
e 2CO + HOCH + O Δ = 2598.8 kJ / mol
22
= 1299.4 kJ/ mol
g l g gH
Adding Equations (d), (b), and (e) together, we get( ) () ()
() () ()
() () () ()
→
→
→
2 2 rxn
2 2 2 rxn
2 2 22 2 rxn
2C graphite + 2O 2CO Δ = – 787.0 kJ / mol
1
H + O HO Δ = – 285.8 kJ / mol
2
5
2CO + HO CH + O Δ = 1299.4 kJ / mol
2
g g H
g g l H
g l g gH
( ) () ()
→
2 22 rxn
2C graphite + H CH Δ = 226.6 kJ/ molg g H
6.6 Standard Enthalpy of Formation and Reaction
2H
2 as a reactant so we
need to reverse the equation and divide it by 2:() () () () () ( )
→
2 2 22 2 rxn
51
e 2CO + HOCH + O Δ = 2598.8 kJ / mol
22
= 1299.4 kJ/ mol
g l g gH
Adding Equations (d), (b), and (e) together, we get( ) () ()
() () ()
() () () ()
→
→
→
2 2 rxn
2 2 2 rxn
2 2 22 2 rxn
2C graphite + 2O 2CO Δ = – 787.0 kJ / mol
1
H + O HO Δ = – 285.8 kJ / mol
2
5
2CO + HO CH + O Δ = 1299.4 kJ / mol
2
g g H
g g l H
g l g gH
( ) () ()
→
2 22 rxn
2C graphite + H CH Δ = 226.6 kJ/ molg g H
6.6 Standard Enthalpy of Formation and Reaction
Therefore,
f
Δ = 226.6 kJ/molH
This value means that when 1mole of C
2H
2 is synthesized
from 2 moles of C(graphite) and 1mole of H
2, 226.6 kJ of
heat are absorbed by the reacting system from the
surroundings. Thus, this is an endothermic process.
6.6 Standard Enthalpy of Formation and Reaction
f
Δ = 226.6 kJ/molH
This value means that when 1mole of C
2H
2 is synthesized
from 2 moles of C(graphite) and 1mole of H
2, 226.6 kJ of
heat are absorbed by the reacting system from the
surroundings. Thus, this is an endothermic process.
6.6 Standard Enthalpy of Formation and Reaction
Example 6.10
The thermite reaction involves
aluminum and iron(III) oxide() () () ()→
23 23
2Al + FeO AlO + 2Fe s s s l
This reaction is highly exothermic
and the liquid iron formed is used to
weld metals.
Calculate the heat released in
per gram of Al reacted with
23 f
FeO. The Δ for Fe () is 12.40
kJ / mol.
Hl
The molten iron formed
in a thermite reaction is
run down into a mold
between the ends of
two railroad rails. On
cooling, the rails are
welded together.
6.6 Standard Enthalpy of Formation and Reaction
The thermite reaction involves
aluminum and iron(III) oxide() () () ()→
23 23
2Al + FeO AlO + 2Fe s s s l
This reaction is highly exothermic
and the liquid iron formed is used to
weld metals.
Calculate the heat released in
per gram of Al reacted with
23 f
FeO. The Δ for Fe () is 12.40
kJ / mol.
Hl
The molten iron formed
in a thermite reaction is
run down into a mold
between the ends of
two railroad rails. On
cooling, the rails are
welded together.
6.6 Standard Enthalpy of Formation and Reaction
Strategy
The enthalpy of a reaction is the difference between the sum of
the enthalpies of the products and the sum of the enthalpies of
the reactants.
The enthalpy of each species (reactant or product) is given by
its stoichiometric coefficient times the standard enthalpy of
formation of the species.
Solution
Using the given ∆??????
??????
°
value for Fe(l) and other values in
Appendix 3 and ∆??????
??????
°
Equation (6.18), we write( ) () () ( )
( )( ) ()( )
rxn f 23 f f f 23
Δ = ΔAlO+ 2Δ – 2ΔAl + ΔFeO
= – 1669.8 kJ / mol + 212.40kJ / mol – 20 + – 822.2kJ / mol
= – 822.8 kJ/mol
HH HFe H H
6.6 Standard Enthalpy of Formation and Reaction
The enthalpy of a reaction is the difference between the sum of
the enthalpies of the products and the sum of the enthalpies of
the reactants.
The enthalpy of each species (reactant or product) is given by
its stoichiometric coefficient times the standard enthalpy of
formation of the species.
Solution
Using the given ∆??????
??????
°
value for Fe(l) and other values in
Appendix 3 and ∆??????
??????
°
Equation (6.18), we write( ) () () ( )
( )( ) ()( )
rxn f 23 f f f 23
Δ = ΔAlO+ 2Δ – 2ΔAl + ΔFeO
= – 1669.8 kJ / mol + 212.40kJ / mol – 20 + – 822.2kJ / mol
= – 822.8 kJ/mol
HH HFe H H
6.6 Standard Enthalpy of Formation and Reaction
This is the amount of heat released for two moles of Al
reacted. We use the following ratio– 822.8 kJ
2 mol Al
to convert to kJ/g Al , The molar mass of Al is 26.98 g, so– 822.8 kJ1 mol Al
heat released per gram of Al = ×
2 mol Al26.98 g Al
= 15.25 kJ/g
Check
Is the negative sign consistent with the exothermic nature of the
reaction? As a quick check, we see that 2 moles of Al weigh
about 54 g and give off about 823 kJ of heat when reacted with
Fe
2O
3. Therefore, the heat given off per gram of Al reacted is
approximately
= - 15.25KJ/g
6.6 Standard Enthalpy of Formation and Reaction
reacted. We use the following ratio– 822.8 kJ
2 mol Al
to convert to kJ/g Al , The molar mass of Al is 26.98 g, so– 822.8 kJ1 mol Al
heat released per gram of Al = ×
2 mol Al26.98 g Al
= 15.25 kJ/g
Check
Is the negative sign consistent with the exothermic nature of the
reaction? As a quick check, we see that 2 moles of Al weigh
about 54 g and give off about 823 kJ of heat when reacted with
Fe
2O
3. Therefore, the heat given off per gram of Al reacted is
approximately
= - 15.25KJ/g
6.6 Standard Enthalpy of Formation and Reaction
Example : Our reaction of interest is:
N
2(g) + 2O
2(g) → 2NO
2(g) H
1= 68 kJ
This reaction can also be carried out in two steps:
N
2(g) + O
2(g) → 2NO(g) H
2= 180 kJ
2NO (g) + O
2(g) → 2NO
2(g) H
3= -112 kJ
N
2(g) + 2O
2(g) → 2NO
2(g) H
1= H
2 + H
3
6.6 Standard Enthalpy of Formation and Reaction
N
2(g) + 2O
2(g) → 2NO
2(g) H
1= 68 kJ
This reaction can also be carried out in two steps:
N
2(g) + O
2(g) → 2NO(g) H
2= 180 kJ
2NO (g) + O
2(g) → 2NO
2(g) H
3= -112 kJ
N
2(g) + 2O
2(g) → 2NO
2(g) H
1= H
2 + H
3
6.6 Standard Enthalpy of Formation and Reaction
H
2O
(g)+ C
(s)→ CO
(g)+ H
2(g) ∆H
1= ??
Use these equations to calculate the molar enthalpy
change which produces hydrogen gas.
C(s) + ½ O
2(g) → CO(g) ∆H
2= -110.5kJ
H
2(g) + ½ O
2(g) → H
2O(g) ∆H
3= -241.8kJ
Example:
C(s) + ½ O
2(g) → CO(g) ∆H
2 = -110.5kJ
H
2O(g) → H
2(g) + ½ O
2(g) -∆H
3 = +241.8kJ
_____________________________________
C(s) + H
2O(g) → H
2(g) + CO(g) ∆H
1 =+131.3kJ
6.6 Standard Enthalpy of Formation and Reaction
2O
(g)+ C
(s)→ CO
(g)+ H
2(g) ∆H
1= ??
Use these equations to calculate the molar enthalpy
change which produces hydrogen gas.
C(s) + ½ O
2(g) → CO(g) ∆H
2= -110.5kJ
H
2(g) + ½ O
2(g) → H
2O(g) ∆H
3= -241.8kJ
Example:
C(s) + ½ O
2(g) → CO(g) ∆H
2 = -110.5kJ
H
2O(g) → H
2(g) + ½ O
2(g) -∆H
3 = +241.8kJ
_____________________________________
C(s) + H
2O(g) → H
2(g) + CO(g) ∆H
1 =+131.3kJ
6.6 Standard Enthalpy of Formation and Reaction
CH
4 (g) + 4Cl
2(g) → CCl
4(l) + 4HCl (g)
C(graphite) + 2H
2(g) → CH
4(g) ∆H = -74.9 kJ
C(graphite) + 2Cl
2(g) → CCl
4(l) ∆H = -139 kJ
½ H
2(g) + ½Cl
2(g) → HCl(g) ∆H = -92.3 kJ
Example : Calculate ∆H
for the reaction
CH
4 (g) → C(graphite) + 2H
2(g) ∆H = -(-74.9 kJ)
C(graphite) + 2Cl
2(g) → CCl
4 (l) ∆H = -139 kJ
2 H
2(g) + 2Cl
2(g) → 4HCl(g) ∆H = 4(-92.3 kJ)
___________________________________________
CH
4(g) + 4Cl
2(g) → CCl
4(l) + 4HCl(g) ∆H
rxn =-433 kJ
6.6 Standard Enthalpy of Formation and Reaction
4 (g) + 4Cl
2(g) → CCl
4(l) + 4HCl (g)
C(graphite) + 2H
2(g) → CH
4(g) ∆H = -74.9 kJ
C(graphite) + 2Cl
2(g) → CCl
4(l) ∆H = -139 kJ
½ H
2(g) + ½Cl
2(g) → HCl(g) ∆H = -92.3 kJ
Example : Calculate ∆H
for the reaction
CH
4 (g) → C(graphite) + 2H
2(g) ∆H = -(-74.9 kJ)
C(graphite) + 2Cl
2(g) → CCl
4 (l) ∆H = -139 kJ
2 H
2(g) + 2Cl
2(g) → 4HCl(g) ∆H = 4(-92.3 kJ)
___________________________________________
CH
4(g) + 4Cl
2(g) → CCl
4(l) + 4HCl(g) ∆H
rxn =-433 kJ
6.6 Standard Enthalpy of Formation and Reaction
Example
Diborane B
2H
6 is a highly reactive compound .
Calculate H for the synthesis of borane from its
elements according to the following equation:
2B (s) + 3H
2 (g) → B
2H
6 (g)
∆H
1
Using the following data
2B(s) +
�
�
O
2(g) → B
2O
3(s) ∆H
2 = -1273 kJ
B
2H
6 (g) + 3O
2 (g) → B
2O
3(s) + 3H
2O(g) ∆H
3 = -2035 kJ
H
2(g) + ½ O
2(g) → H
2O(l) ∆H
4 = -286 kJ
H
2O(l) → H
2O (g) ∆H
5 = -44 kJ
6.6 Standard Enthalpy of Formation and Reaction
Diborane B
2H
6 is a highly reactive compound .
Calculate H for the synthesis of borane from its
elements according to the following equation:
2B (s) + 3H
2 (g) → B
2H
6 (g)
∆H
1
Using the following data
2B(s) +
�
�
O
2(g) → B
2O
3(s) ∆H
2 = -1273 kJ
B
2H
6 (g) + 3O
2 (g) → B
2O
3(s) + 3H
2O(g) ∆H
3 = -2035 kJ
H
2(g) + ½ O
2(g) → H
2O(l) ∆H
4 = -286 kJ
H
2O(l) → H
2O (g) ∆H
5 = -44 kJ
6.6 Standard Enthalpy of Formation and Reaction
Solution:
2B(s) +
3
2
O
2(g) → B
2O
3(s) ∆H
2 = -1273kJ
B
2O
3(s) + 3H
2O(g) → B
2H
6(g) + 3O
2(g) -∆H
3= -(-2035 kJ)
3 H
2(g) + 3× ½ O
2(g) → 3 H
2O(l) 3 ∆H
4 =3× -286 kJ
3 H
2O(l) → 3 H
2O (g) 3×∆H
5 = 3× -44 kJ
2B(s) + 3H
2(g) → B
2H
6 (g)
∆H
1= ∆H
2 + (-∆H
4) +3∆H
4 + 3×∆H
5
H
1 = -1273 + 2035 – 858 – 132 = +36 kJ
2B(s) +
�
�
O
2(g) → B
2O
3(s) ∆H
2 = -1273 kJ
B
2H
6 (g) + 3O
2 (g) → B
2O
3(s) + 3H
2O(g) ∆H
3 = -2035 kJ
H
2(g) + ½ O
2(g) → H
2O(l) ∆H
4 = -286 kJ
H
2O(l) → H
2O (g) ∆H
5 = -44 kJ
6.6 Standard Enthalpy of Formation and Reaction
2B(s) +
3
2
O
2(g) → B
2O
3(s) ∆H
2 = -1273kJ
B
2O
3(s) + 3H
2O(g) → B
2H
6(g) + 3O
2(g) -∆H
3= -(-2035 kJ)
3 H
2(g) + 3× ½ O
2(g) → 3 H
2O(l) 3 ∆H
4 =3× -286 kJ
3 H
2O(l) → 3 H
2O (g) 3×∆H
5 = 3× -44 kJ
2B(s) + 3H
2(g) → B
2H
6 (g)
∆H
1= ∆H
2 + (-∆H
4) +3∆H
4 + 3×∆H
5
H
1 = -1273 + 2035 – 858 – 132 = +36 kJ
2B(s) +
�
�
O
2(g) → B
2O
3(s) ∆H
2 = -1273 kJ
B
2H
6 (g) + 3O
2 (g) → B
2O
3(s) + 3H
2O(g) ∆H
3 = -2035 kJ
H
2(g) + ½ O
2(g) → H
2O(l) ∆H
4 = -286 kJ
H
2O(l) → H
2O (g) ∆H
5 = -44 kJ
6.6 Standard Enthalpy of Formation and Reaction
Example : Given the following data
P
4(s) + 6Cl
2(g) → 4PCl
3(g) H= -1225.6 kJ
P
4(s) + 5O
2(g) → P
4O
10(s) H= -2967.3 kJ
PCl
3(g) +
1
2
O
2(g) → Cl
3PO(g) H= -285.7 kJ
PCl
3(g) + Cl
2(g) → PCl
5(g) H= -84.2 kJ
Calculate H for the reaction:P
4O
10(s) + 6PCl
5(g) → 10Cl
3PO(g)
P
4O
10(s) → P
4(s) + 5O
2(g) - H= -(-2967.3) kJ
10 PCl
3(g) +10(
1
2
O
2) (g) → 10 Cl
3PO(g) 10H= 10-285.7 kJ
6PCl
5(g) → 6 PCl
3(g) + 6Cl
2(g) -6H=-6 -84.2 kJ
P
4(s) + 6Cl
2(g) → 4PCl
3(g) H= -1225.6 kJ
P
4O
10(s) + 6PCl
5(g) → 10Cl
3PO(g) H= 5104 kJ
6.6 Standard Enthalpy of Formation and Reaction
P
4(s) + 6Cl
2(g) → 4PCl
3(g) H= -1225.6 kJ
P
4(s) + 5O
2(g) → P
4O
10(s) H= -2967.3 kJ
PCl
3(g) +
1
2
O
2(g) → Cl
3PO(g) H= -285.7 kJ
PCl
3(g) + Cl
2(g) → PCl
5(g) H= -84.2 kJ
Calculate H for the reaction:P
4O
10(s) + 6PCl
5(g) → 10Cl
3PO(g)
P
4O
10(s) → P
4(s) + 5O
2(g) - H= -(-2967.3) kJ
10 PCl
3(g) +10(
1
2
O
2) (g) → 10 Cl
3PO(g) 10H= 10-285.7 kJ
6PCl
5(g) → 6 PCl
3(g) + 6Cl
2(g) -6H=-6 -84.2 kJ
P
4(s) + 6Cl
2(g) → 4PCl
3(g) H= -1225.6 kJ
P
4O
10(s) + 6PCl
5(g) → 10Cl
3PO(g) H= 5104 kJ
6.6 Standard Enthalpy of Formation and Reaction
Example : Calculate ∆H for the reaction:
NO(g) + O(g) → NO
2(g) given the following information:
NO(g) + O
3(g) → NO
2(g) + O
2(g) ∆H = -198.9 kJ
O
3(g) →
�
�
O
2(g) ∆H = -142.3 kJ
O
2(g) → 2O(g) ∆H = 495.0 kJ
NO(g) + O
3(g) → NO
2(g) + O
2(g) ∆H = -198.9 kJ
�
�
O
2(g) → O
3(g) - ∆H = -(-142.3) kJ
�
�
2O(g) →
�
�
O
2(g) −
�
�
∆H = −
�
�
495.0 kJ
NO(g) + O(g) → NO
2(g) ∆H
rxn= -304.2 kJ
6.6 Standard Enthalpy of Formation and Reaction
NO(g) + O(g) → NO
2(g) given the following information:
NO(g) + O
3(g) → NO
2(g) + O
2(g) ∆H = -198.9 kJ
O
3(g) →
�
�
O
2(g) ∆H = -142.3 kJ
O
2(g) → 2O(g) ∆H = 495.0 kJ
NO(g) + O
3(g) → NO
2(g) + O
2(g) ∆H = -198.9 kJ
�
�
O
2(g) → O
3(g) - ∆H = -(-142.3) kJ
�
�
2O(g) →
�
�
O
2(g) −
�
�
∆H = −
�
�
495.0 kJ
NO(g) + O(g) → NO
2(g) ∆H
rxn= -304.2 kJ
6.6 Standard Enthalpy of Formation and Reaction
Exercise:
Calculate H for the following reaction:
4NH
3(g) + 5O
2(g) → 4NO(g) + 6H
2O(g)
Data:
N
2(g) + O
2(g) → 2NO(g) H = 180.6 kJ
N
2(g) + 3H
2(g) → 2NH
3 (g) H = -91.8 kJ
2H
2(g) + O
2(g) → 2H
2O(g) H = -483.7 kJ
6.6 Standard Enthalpy of Formation and Reaction
Calculate H for the following reaction:
4NH
3(g) + 5O
2(g) → 4NO(g) + 6H
2O(g)
Data:
N
2(g) + O
2(g) → 2NO(g) H = 180.6 kJ
N
2(g) + 3H
2(g) → 2NH
3 (g) H = -91.8 kJ
2H
2(g) + O
2(g) → 2H
2O(g) H = -483.7 kJ
6.6 Standard Enthalpy of Formation and Reaction
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