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THEMODYNAMICS AND THERMOCHEMISTRY
(SILBERBERG CHAPTER 6)
1
...
With ever increasing energy costs
it becoming more and more important to minimize energy use and thus save energy
...
e
...
g
...
g
...

Firstly we are going to concentrate on certain aspects of thermodynamics (which is regarded
as the science of energy it all its forms) and then we will focus on thermo-chemistry, which
is the subsection of thermodynamics that describes reaction heat effects
...

To be able to understand the typical questions that need to be answered, it is required to be
aware of the various forms of energy associated with a specific quantity of matter, namely:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)

Internal energy
Displacement work
Kinetic energy
Potential energy
Flow work
Technical work
Heat
Chemical energy

The creation of energy to nuclear reactions or due to the effects of external electrical or
magnetic fields will not be considered in this course
...
Open and Closed systems
A system is the quantity of matter of a constant identity or of a constant area that is selected
...
e
...
All the real or imaginary surfaces that separate the system from the surroundings
are known as the system boundaries
...
e
...

Surrounding
Closed
System

System
Sisteem

System boundary

Figure 2
...
Energy

...
2: Closed System

A closed system consists of a fixed mass of matter and no mass is able to move across the
system boundary
...
If energy is also cannot pass the system boundary, them
the system is said to be isolated
...


...
3: Isolated System

Bunsen burner

Figure 2
...


An open system (or a control volume) is a constant volume in space where both mass and
energy is able to pass though the system boundary
...
An open system is also known as a flow system
...
5: Open System
The format of the thermodynamic equations differs for an open
system and a closed system
...

3
...
[See figure 3
...

There exists no instrument that is able to calculate the
internal energy on macroscopic level
...

Let E = total internal energy of a body [J]
m = total mass of a body [kg]

Figure 3
...
1)

To calculate E we use the fact that dE is an exact differential due to the fact that
internal energy is a point function
...
The
change in the property is thus only dependent on the position of the starting and
finishing conditions
...
2 the internal energy change between state 1 and state 2 is the same
although two different routes were taken: 1ABC2 or 1D2
...
Internal energy is thus path independent
...
2: Internal Energy as a point function

Figure 3
...
e
...
2)

 m3 
with v = specific volume 
 kg  and T = absolute temperature (K)



For an ideal gas:
dei = cv dT

(3
...
4)

T1

with cv being the specific heat capacity at constant volume
...
Sometimes the specific heat is expressed on a molar basis
kg °C
kg K

3|Page

J
J
or
, which then leads to ei having the
molK
kmolK
units J/mol or J/kmol
...

In such cases equation (3
...
5)

Remember ei is the specific internal energy in J/kg
...
6)

Only changes in internal energy can be calculated with the aid of equations (3
...
6)
...
6) needs to be replaced
with the total number of mole to obtain the total internal energy in J
...
4)
...

dx

F

x2
x1

Figure 3
...
The force F is not constant and changes as the
compression increases
...
7)

x2

W1-2 = - ∫ F dx [J]

(Displacement work)

(3
...

A
Thus F = PA [N]

(3
...
8) thus becomes:
x2

W1-2 = -

∫ PA dx

x1

2

= − ∫ P dV [J]

(3
...
11)

J 
 kg 



W1− 2
m

(3
...
10) dV is the volume in m3 that the piston has moved
...
The displacement work is thus the
area under the P vs
...
5)
...


P

P2

b

2
a

P2

1

2

PdV
∫ Pdυ
1

V2

V1 V

Figure 3
...
10) and (3
...
13)

dw = - Pdv

J 
 kg



(3
...
e
...

Kinetic Energy

The kinetic energy of a particle can be calculated as follows:
KE =
ke =

(

1
δ mU2
2

)

KE
U2
=
mass
2

[J]

(3
...
16)

U

δm

Figure 3
...


5|Page

2
δ mg

z

1
δ mg

Figure 3
...
17)

PE
= Pe = gz
mass

J
 kg 
 

(3
...
19)

Flow energy = PV

[J]

(3
...
Mechnanical energy is added to the
control volume through the rotating axes
...

Note:

Work that is removed from the system,
i
...
work conducted by the system is negative
...
e
...


+

Heat
From experience it is known that heat is naturally transferred from a body at a higher
temperature to a body of lower temperature
...
This process continues until the temperature of the two bodies are the same
and equilibrium has been established
...
It is known that:
Q α ∆T

Q α m

and

where Q = heat
∴Q = c m ∆T

[J]

(3
...
Equation (3
...
22)

Q = m ∫ c dT
Or

[J]
[J]

(3
...
24)

J 
 kg



(3
...
23) and (3
...
26)

J 
 kg



(3
...
28)

7|Page

sR =

SR
mass

J 
 kg



(3
...
28):

∆h R
∆m

J
 J 
= specific reaction heat of a chemical reaction 
 or  kg 
 kmol 
 

= the number of kmol (or kg) matter participating in the reaction with
the base upon which ∆h R is calculated
...
(Compare to sign convention for heat)
...
∆m is always negative as a specific species of molecules is always
used up in chemical reactions
...
The First Law of Thermodynamics: The Energy Equation:
Formulation:

The first law of thermodynamics can not be proven: It is based on an assumption yet
no process is allowed to contradict it
...

1) Energy can not be created or destroyed
...

2) Heat is a form of energy and can be created from mechanical, electrical or
other forms of energy
...

3) There exists no method to build a machine that is able to continuously
deliver work without using an equivalent amount of the same of other type
of energy, i
...
the working of the “Perpetual Motion” of the first type is
impossible
...

The first law can thus be regarded as a general law regarding the conservation of
energy
...
1)

In order to apply equation (4
...
Within the time
limits of this module, only a couple of special cases of the application of the first law
will be investigated
...
e
...


(2)

Kinetic and potential energy effects are negligible
...
e
...


In this case the first law can be written as follows:
E2 – E1 = Q 1-2 + W1-2

[J]

(4
...
(E2 – E1) is the change in internal energy of the mass in the system and
Q1-2 is the heat transferred to the system during the change from state 1 to state 2
...
2)
can be written as:
ei 2 – ei 1 = q1-2 + w1-2

J
 kg 
 

(4
...
3) in differential form, which can then be
integrated:
dei = dq + dw

J
 kg 
 

(4
...


The Heat Capacity:
The Heat Capacity at Constant Volume, cv
As mentioned previously, heat capacity is not constant for a particular body
but also depends on the state change the body is undergoing
...
24):

9|Page

J
 kg 
 

dq = cv dT

(4
...
6)

Combining equations (4
...
6):
J
 kg 
 

dei = cv dT

(4
...


 J 
 J 
∴ cv = heat capacity at constant volume units 
 of  kg k 
 kg °C 


This relationship between dei and cv can also be derived from the well-known
Gay-Lussac/ Joule experiment
...
e
...

In the first law of thermodynamics (equation 4
...
5)

The ideal gas law:
Pv = RT

(4
...
7)

Substitute into equation (4
...
8)

∴ cp – cv = R

(4
...
If the heat capacity is given per mol or kmol, it is known as the
kmol °C
molar heat capacity
...
10)

Where Μr is the molecular weight in [

kg
]
kmol

From equation (4
...
11)
(4
...
13)

Simple Examples of the First Law of Thermodynamics
Example 4
...
During the process the liquid looses 500 kJ heat to the environment while the
stirrer applied 100 kJ of work on the liquid
...
1
...


11 | P a g e

1
...
If the tank contains 20
...
1

1
...
2 dei = cv dT
dE = m cv dT
E2 – E1 = m cv [T2 – T1]
∴ [T2 – T1] =
=

E 2 − E1
m cv

− 400 k J kg K 1000 J
x
20 kg ⋅ 4100 J
1k J

= -4,88 K
The temperature thus decreases with 4,88 °C
...
2:
A mass of oxygen is compressed in a frictionless piston-cylinder set-up from an
initial state of V = 0,14 m3, P = 1,5 bar abs and 150 °C to a final volume of 0,13 m3,
still at 1,5 bar abs
...
Assume the ideal gas law holds true and
J
the cv of oxygen is 649

...
2
The mass is constant and can be calculated from the initial conditions:
m=

PV1M r 1
...
14m3 × 32 g / mol
1
=
= 191
...
1911kg
RoT
8
...
K × 423

Now the temperature can be calculated with the aid of the ideal gas law
...
13m3
= ⇒ T2 = T1 2 = 423K ×
= 392
...
14m3
The change in internal energy can now be calculated:
E2 − E1 = m × cv × (T2 − T1 )
J
(392
...
K
= −3745 J = −3
...
1911kg × 649

The pressure is constant and thus the work done can be calculated:
W1-2 = − ∫ PdV = − P(V2 − V1 )
N

= - 1,5 x 105 [0,13 - 0,14]  2 ⋅ m 3 
m


= 1500 J = 1
...
75kJ
Q1-2

= Q1-2 + 1
...
25 J

(energy from the system to the environment)

Simple State Changes for Ideal Gasses in a Closed System:
1st law: dei = dq + dw
If the only work done by or on the system is displacement work (piston – cylinder
system):
dei = dq - Pdv

(4
...
15)

13 | P a g e

With R the specific gas constant:

Ro  J 
M  kg K 



Isochoric or Constant Volume Changes, V = constant
...
3: Isochoric state change
The displacement work:
w = − ∫ Pdv = 0

(4
...

For the change of process from (1) to (2)
ei2- ei1 = q1-2 = cv (T2- T1)

(4
...
However, the
accompanying pressure increase has no influence on the internal energy (for
an ideal gas)
...
18)

If the Isochoric process proceeds in the opposite direction, i
...
from (2) to (1),
both the pressure and the temperature will decrease and heat needs to be
removed from the gas
...


If an isobaric change from state (1) to state (2) takes places, as shown in
Figure 4
...
e
...


14 | P a g e

P

gas
dei = positief
q1-2 2

1

-w
v

Figure 4
...
19)

= R (T1 - T2)

(4
...

The work is done at the cost of energy added:
q1-2 = cp (T2 – T1)

J
 kg 
 

(4
...
22)

If the opposite state change takes place i
...
from (2) to (1) the following
occurs:
-q
P

1

dei negatief
+w
gas
q2-1

2

w
v

Figure 4
...
e
...
23)

(4
...
25)

(negative

because T2 > T1)
Isothermal State Changes, T = constant
Due to the fact that the temperature remains constant, there is no change in
internal energy, thus
ei1 = ei2 of E1 = E2

(4
...
6: Isothermal process
The displacement work from (1) to (2) is:
2

w1-2 = − ∫ Pdv
1

According to the first law: 0 = q1-2 + w1-2
2
J
∴ w1− 2 = − ∫ Pdv = −q1− 2  
 kg 
1

(4
...
28)

1

but P1v1 = Pv (because T remains constant)
∴P =

P1v1
v

(4
...
29) into (4
...
30)

P 
= P1 v1 ℓn  2 
P 
 1

(4
...
32)

J
 kg 
 

(4
...

Adiabatic State Change, q = 0
If a gas is expanded in a completely isolated cylinder (q = 0), the work done
by the system done can only be changed into internal energy, which must
thus decrease
...
7: Adiabatic state change
1st law for a closed system:
dei = dq + dw
∴dei = dw

J
 kg 
 

(4
...


17 | P a g e

Examples of Simple State Changes
Example 4
...
The combustion process then takes place at constant
pressure until the volume has doubled
...
Calculate the change in internal energy
...
2)

∴ E2 - E1 = 24000 – 8500 = 15 500 J
The internal energy has thus increased with 15500 J
...
4: Constant Volume Process
kg
) at 17 bar abs and 150 oC is cooled to 15 oC at
kmol
constant volume
...

the system) and the final pressure
...
5: Isothermal Process
Air at 1
...
Calculate the
work required, as well as the heat transferred if the process is regarded to be
reversible
...

kg K
kg K

Solution:
P 14

(2)

(1)

1,06

V

cp – cv = R

(4
...
32)
14  J ⋅ K 
1,06  kg K 



J
kg

 bar 
 bar 



[work conducted on the air]

P 
q1− 2 = RT ℓn  1 
P 
 2

(4
...
06   J ⋅ K 
= 287 ⋅ 288 ℓn 
 

 14   kg K 
= - 213,3

J
 bar 
 bar  = - 213 300 kg



kJ
[heat transferred to the surroundings]
kg

19 | P a g e

Example 4
...
You can assume no friction losses
and that the air acts like an ideal gas
...

Solution:
P 14

(2)

Voorbeeld 5

1,06

(1)

(3)

V

w2-3 = cv (T3 – T2)
T2 = 288 K, T3 is unknown
...
The Energy Equation for Open or Flow systems
The Enthalpy

Just as internal energy is an important quantity for closed systems, so enthalpy is
important in flow systems
...
1)
(5
...
3)

Pv = RT for ideal gases

∴

h2 – h1 = cv (T2 – T1) + R (T2 – T1)

(5
...
5)

Or in differential form
dh = Cp dT
or

(5
...
7)

J
 kg 
 
[J]

Equations (5
...
7) are exact for ideal gases as well as real gases as long as
pressure remains constant
...
1: Steady state conditions


 ɺ ɺ ɺ

 J 
U2
U2
ɺ
ɺ
m h1 + 1 + gz1  + Q + Wt + S R = m h2 + 2 + gz 2    of [W ]

 s
2
2





(5
...
8) can also be written in specific energy format:

21 | P a g e

2
U 12
U2
h1 +
+ gz1 + q + wt + s R = h2 +
+ gz 2
2
2

J
kg

(5
...
1
Referring to figure 5
...
Assume the
temperature of the water remains constant
...

T2 = 25 °
C
V2 = 6 m/s
P2 = 1 atm
(2)
20 m

(1)
ɺ
m = 600 kg/min
V1 = 0,3 m/s
P1 = 1 atm
T1 = 25 °C

Figure 5
...
1
Solution:
From equation (5
...
2
An airstream needs to be heated from 20 °C to 300 °C
...
The
normal cubic meter (Nm3) is calculated at 1 atm and 0oC
...

Solution:
2,0 N/m3
(1)

T2 = 300 °
C
(2)

T1 = 20 °
C
ɺ
Q

Convert the volumetric flow rate to mass flow rate:
ɺ
m=

ɺ
PV
1,01325 ⋅10 5 x 2,0x 29
=
Ro
8314 ⋅ 293
T
Mr

= 2,412

N m 3 kmol K kg
m 2 min J kmol K

kg
kg
= 0,0402
min
s

Assume ideal gas conditions without losses
...
No technical work or chemical energy is applicable
...
9)
(5
...
10)

Assumptions:
Constant temperature ⇒ ei1 = ei2
No heat, work or chemical energy changes, thus q = 0, wt = 0, sR = 0
Frictionless flow

23 | P a g e

Uncompressible flow: v1 = v2 = v
Equation (5
...
11)

In the case of uncompressible flow, we usually work with the density ρ =
2
U 12
P2 U 2
Therefore:
+
+ gz1 = +
+ gz 2
ρ
2
ρ
2

P1

1
v

(5
...
3
Water flows through the system shown in figure 5
...

Estimate the pressure at point (1) if friction losses are ignored
...

20 mm binnedeursnee pyp

(2)
P2 = 1 atm abs 50 m
(1)

10 mm binnedeursnee pyp

Figure 5
...
3
Solution:
Water can be regarded as incompressible
...
Consider steady state flow
...
12)

Assume z1 = 0 and thus z2 = 50 m
Investigate the system between (1) and (2)
...
P1 can now be
solve:

24 | P a g e

 m2 
1,013 ⋅ 105 N m 3 1,062 m 2
m2
+ 9,81 ⋅ 0 =
+
+ 9,81 ⋅ 50 2
 s2 
999 m 2 kg
2 s2
s
 
1,06 2
4,24 2 
5
∴ P1 = 1,013 ⋅10 + 999
+ 9,81x50 −

2 
 2
P1 4,242
+
999
2

=5,83 x 105

N
abs = 5,83 bar abs
...
Heat of Reaction and Chemical Changes
Enthalpy of Reaction
Enthalpy is a state function (just as internal energy) which means that a change in
enthalpy only depends on the difference between the enthalpy of the start and the end
conditions
...
1)

= H products − H reactants

(7
...

In an exothermic reaction heat is liberated and thus the enthalpy of the products is
lower than the enthalpy of the reactants
...
Examples of
exothermic reactions include:

1
H2 (g ) + O 2 (g ) → H2 O(l)
2

Δh = 285,8

kJ
mol

kJ
5
1
3
C 3H5 (NO 3 )3 (l) → 3CO 2 (g ) + H2 O(g ) + O 2 (g ) + N2 (g )
Δh = 5,72x10 3
So
2
4
2
mol
metimes the exothermic reaction heat of reaction is written as follows:
1
H 2 (g ) + O 2 (g ) → H 2 O(l ) + 285,8 k J
2
In an endothermic reaction, heat is absorbed and thus the enthalpy of the products is
higher than the enthalpy of the reactants
...
Examples of
endothermic reactions include:

N 2 (g ) + O 2 (g ) → 2 NO(g )

∆h = 180,6

kJ
mol N 2

which can also be written as:

180,6 kJ + N 2 (g ) + O 2 (g ) → 2 NO(g )
[In both the above mentioned cases we have accepted that the products and the
reactants are at the same temperature]
...
For example:
C 4 H10 (l ) +

13
O 2 (g ) → 4CO 2 (g ) + 5H 2 O(l )
2

Δhrxn = Δh c
where ∆h c is the heat of combustion
...

c

Heat of Formation
If 1 mole of a compound is formed from its elements, the heat of reaction is
known as the heat of formation, for example:

1
K(s) + Br2 (l) → KBr(s)
2

Δh = Δh f

with ∆h f being the heat of formation
...

f

Latent Heat of Fusion
If 1 mole of a compound melts, the enthalpy change is known as the latent
heat of fusion for example:
Na Cl(s) → NaCl(ℓ)

∆h = ∆h fus

The opposite of the melting process is the solidification process
...

Note that the latent heat of fusion and the latent heat of evaporation are often
kJ
given in
and in all cases you need to check that the units you are using
kg
are consistent
...
g
...


Latent Heat of Evaporation
When 1 mole of a compound is evaporated, the enthalpy change is known as
the latent heat of evaporation, for example:
H 2 O(ℓ) → H 2 O (g )

∆h = ∆h vap

The same principle apply here as for the latent heat of evaporation
...


26 | P a g e

•

Method of writing: Reactants on the left, products on the right for example:

CH 4 (g ) + H 2 O(g) ⇔ CO(g ) + 3H 2 (g )
•

Indicate the phases present:
g = vapor / gas
ℓ or l = liquid

s = solid
•

Conditions of phase, temperature and pressure must also be given
...


•

If the reaction proceeds in the opposite direction, the sing of the reaction heat
changes
...
Types if calorimeters
include constant pressure calorimeter, constant volume calorimeter and a flow
calorimeter
...


Example 6
...
64 grams is heated in a test tube to 100
...
Without energy losses, the solid is placed in a constant
pressure calorimeter containing 50 grams of water
...
1 to 28
...
What is the heat capacity of the solid? Assume all
heat liberated from the solid is absorbed by the water and that the heat capacity at
kJ
constant pressure of the water is 4,184

...
We
know the heat capacity at constant pressure for water and can thus calculate the heat
absorbed by the water
...

The heat absorbed by the water:

Q water = m H 2O c p H 2 O ∆Twater
27 | P a g e



kJ
= 0,050⋅4,184⋅(28,49-25,1) kg ⋅
⋅ K  = 0,7092 kJ =709,2 J
kg K 

Heat liberated by the solid:



J
Q = m C ∆T = -709,2 [J] = 0,02564⋅c⋅[28,49-100,0] kg ⋅
⋅ K
kg K 

c=

J
J
− 709,2
= 386,8
kg K
0,02564 ⋅ (− 71,51) kg K

Calculating the Heat of Reaction with the Use of the Hess’s Law
1
Consider the reaction: C(s ) + O 2 → CO(g )
2

(a)

It is very difficult to convert graphite completely to CO without the formation of
some CO2
...
However, it is possible to study the complete combustion of graphite
and CO in a bomb calorimeter
...

Example 6
...
An
engineer is investigating the possibility to convert the CO and NO to less harmfull
gasses using the following reaction:
1
CO(g ) + NO(g ) → CO 2 (g ) + N 2 (g )
2

∆h rxn = ?

Given the following information, calculate ∆hrxn for the proposed reaction:
A:

1
CO(g ) + O 2 (g ) → CO 2 (g )
2

B:

N 2 (g ) + O 2 (g ) → 2NO(g )

∆h rxn A = −283,0

∆h rxn B = 180,6

kJ
mol

kJ
mol N 2

Solution:
We need to write reaction A and B so that the sum or difference gives us the required
reaction:
1
[2NO → N 2 (g ) + O 2 (g )]
2

X:

=−

180,6
2

1
∆h rxn = − ∆h rxn B
X
2
kJ
= −90,3
mol NO

1
CO(g ) + O 2 (g ) → CO 2 (g )
2

Y:

X + Y: CO(g ) + NO(g ) →

∆h rxn Y = −283,0

kJ
mol NO
kJ
mol

1
N (g ) + CO 2 (g )
2 2

∆h rxn = ∆h rxn (X ) + ∆h rxn (Y) = - 90,3 + (- 283,0) = - 373,3

kJ
mol

Standard Heat of Reaction:
The standard conditions for calculating the heat of reaction are:
•

For an ideal gas the pressure is 1 atm and it is assumed that the gas obeys the
ideal gas law
...


29 | P a g e

Standard Heat of Formation ∆h o
f
The heat of formation is a special type of reaction heat for the formation of a
compound from its elements
...
By convention, the heat of formation means that 1
mole of a compound is formed from its elements
...
In this case ∆h o = −74,9

...
It is
possible to determine the heat of formation of all compounds at 25 oC and 1 atm abs
...
Table 6
...


Table 6
...

Compound
C (granite)
C (diamond)
CO (g)
CO2 (g)
CH4 (g)
CH3OH ( ℓ )
C2H4 (g)
C2H6 (g)
C3H8 (g)
C4H10 (g)
C2H5OH ( ℓ )
HCN (g)
CS2 ( ℓ )
N2 (g)
NH3 (g)
NO (g)
NO2 (g)
N2O (g)
N2O4 (g)

 kJ 
∆h o 

f
 mol 
0
1,9
-110,5
-393,5
-74,9
-238,6
52,26
-84,68
-103,8
-125,6
-277,7
135
87,9
0
-46,11
90,25
33,18
82,05
9,16

Compound
H2 (g)
HF (g)
HCl (g)
HBr (g)
HI (g)
H2O (g)
H2O ( ℓ )
H2S (g)
Cl2 (g)
Cl (g)
S (rhombic)
SO2 (g)
SO3 (g)
Ca (s)
CaO (s)
CaCO3 (s)
Na (s)
Na (g)
NaCl (s)

 kJ 
∆h o 

f
 mol 
0
-271,1
-92,31
-36,40
26,48
-241,8
-285,8
-20,63
0
121,0
0
-296,8
-395,7
0
-635,1
-1206,9
0
107,8
-411,1

Table 6
...
For most compounds ∆h o is negative, which indicates exothermic heat of
f
formation
...
Compounds with a positive heat of formation indicate an
endothermic heat of formation
...


30 | P a g e

With the use of the tabulated standard heats of formation, the standard heats of
reaction can be calculated with the use of the Law of Hess
...
3)

[J]

(7
...
4) can be used because ∆h o is a point function, i
...
it is path
f
independent
...
3: Use of the Standard Heat of Formation
Nitric acid is an important compound in the chemical industry because it is used in
the production of fertilizers, pigments, explosives etc
...

Calculate the heat of reaction
...
Obtain the required heats of formation from table 6
...
4: Use of the Standard Heat of Formation
The standard heat of formation of H2O (l) is -285,8

kJ
and the latent heat of
mol

kJ
by 25 °C and 1 atm abs
...


vaporization is 2442

Solution:
Firstly we need to convert the information to a common basis as the latent heat of
kJ
vaporization is given per kg
...
g
...
) petrol etc
...

Due to the economic importance of combustion, there is an alternative series of
information available, namely the standard heat of combustion
...

The standard heats of combustion do not have the same standard conditions as the
standard heats of formation and the following convention needs to be taken into
account when using the standard heats of combustion
...
(HCl (aq) is an initively dilute solution of HCl in
water)
...

The reference temperature and pressure remain 25 °C and 1 atm abs
...
[As stated
above, in this module, we will only consider oxygen as oxidation reagent]
...

For fuels such as coal and oil, the standard heat of combustion is often called the
heating value
...
In reality, the water usually does not condense and it
remains in the combustion gasses
...

The nett or lower heating value [LHV] with water as a gas/vapor
...
)
Just as in the case of the standard heat of formation, there are tables of data available
in the literature with standard heats of combustion
...
5: Heat of Formation: Use of Hess’s Law
Calculate the standard heat of formation of C2H2 (g) with the following given
information:

∆h o = −1300,5
c

Acetylene

C2H2 (g):

Graphite

C (s): ∆h o = −393,8
c
H2 (g) ∆h o = −286,0
c

kJ
mol

(H 2 O

as ℓ )

kJ
mol

Hydrogen

kJ
mol

(H 2 O

as ℓ )

Solution:
Formation reaction for C2H2 (g): 2 C (s) + H2 (g) → C2H2 (g) ∆h o = ?
f
We can use the law of Hess and write the reactions as follows:
A:

1
kJ
C 2 H 2 (g ) + 2 O 2 (g ) → 2CO 2 (g ) + H 2 O(ℓ ) ∆h o = −1300,5
2
mol

B:

C (s ) + O 2 (g ) → CO 2 (g ) ∆h o = −393,8

C:

1
kJ
H 2 (g ) + O 2 (g ) → H 2 O(ℓ ) ∆h o = −286,0
2
mol

kJ
mol

- A + 2B + C ≅ 2C (s) + H2 (g) → C2H2 (g)
According to the law of Hess:

 kJ 
∆hrxn = ∆ho = − [ −1300,5] + 2 ( −393,8 ) + ( −286,0 ) = 226,9 
f
 mol 


Example 6
...


35 | P a g e



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