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UNIT II
THERMODYNAMICS
Questions:
1. The subject of chemical thermodynamics. The main definitions: the system the parameters of the
condition, the functions of the condition.
2. The 1st Law of Thermodynamics. Internal energy. Enthalpy.
3. The Hess's Law and its effects.
4. The IInd Law of Thermodynamics. Entropy Gibbs free energy.
5. What is the spontaneous process? The criteria and directions of spontaneous processes.
6. The constant of the chemical equilibrium.
7. The application of the main thermodynamic laws for (he living organisms and bio systems.
During chemical changes energy may be evolved or absorbed in various forms as heat, light, electricity,
etc. Quantitative investigation of the thermal changes accompanying by chemical reactions constitutes the
subject of a special branch of chemistry known as thermochemistry.
It is important for both practical and theoretical reasons. Practical applications include measuring the
energy value of fuels, and determining energy requirement of industrial processes. Theoretical consideration
includes calculation of energy changes for hypothetical reactions and prediction whether a particular reaction
can take place or not: its feasibility.
The subjects of thermodynamics are a system, the condition, the parameters and functions of the
condition.
The system is a body or the group of bodies mentally separated from the environment.
SYSTEM
OPEN
CLOSED
ISOLATED
The open system is a system which exchanges energy and substance with environment (ex.: a man).
The dosed system is a system which exchanges energy only (ex.: the healed stone).
The isolated system is a system, which exchanges nothing (the thermoflask).
SYSTEM
HOMOGENEOUS
HETEROGENEOUS
The homogeneous system consists of one phase. The heterogeneous system consists of two or more
phases.
The condition of the system is any physical or chemical property of the system, such as thermal
capacity/eternal energy, entropy, etc.
The parameter of the system is a thermal property which is easily expressed by the practical way (p,
n, T).
The parameters of the system are interconnected to each other by the equation of state of a gas:
PV = nRT
n is the number of moles, R is the ideal gas constant, R = 8.3 1J/mol∙K.
The parameters describing the conditions of the system are functions of the condition. Their main
property is independence on ways of their changes.
PARAMETERS OF THE SYSTEM
EXTENSIVE
INTENSIVE
The extensive parameters depend on mass of the system (thermal capacity, internal energy, volume,
entropy). The intensive parameters do not depend on mass of the system (temperature, pressure).
The transition of the system from one state to another is called the thermodynamic process.
PROCESS
SPONTANEOUS
NONSPONTANEOUS
Any system consists of moving particles, such as atoms, ions and molecules. The quantitative
characteristic of the movement is the particle energy. The total combination of all types of particle energy in
the system is called the internal energy of the system (U). Internal energy is a function of condition. The
change of U does not depend on intermediate stages of the process.
It follows from the Law of Conservation of Energy: if a certain quantity of heal is evolved (or
absorbed) during the formation of a chemical compound, the same quantity of heat will be absorbed (or
evolved) upon the decomposition of that compound into its initial substances.
The Ist Law of Thermodynamics is a special case of the law of conservation of energy for macroscopic
systems.
§1. The lst Law of Thermodynamics
The heat which was delivered to the system is used for the change of internal energy and carrying out
work against external forces
Or
The energy does not appear from anywhere and does not disappear, but it turns from one type into
another.
Q = ΔU + A
Q = (U2 – U1) + p ∙ (V2 – V1)
A is a work V1 (U1) and V2 (U2) are the volume (internal energy) of the system in the initial and final
conditions.
In case of V = constant (isochoric process), ΔV = 0.0L
QV = (U2 – U1) = ΔU
In case of p = constant (isobaric process) Qp = (U2 – U1) + p ∙ (V2 – V1) = (U2 + pV2) – (U1 + pV1) =
H2 – H1 = ΔH
The 1st Law is used for calculation of the thermal effect of the chemical reactons.
Since energy is absorbed or evolved during chemical reactions most frequently in the from of heat, all
reactions during which energy is liberated are called exothermic (ΔH < 0 or Q > 0). Reactions in which
energy is absorbed are termed endothermic (ΔH > 0 or Q < 0).
Chemical equations in which energy is absorbed during the reaction is indicated are called
thermochical reactions.
A + B → AB ± Q
For instance, a) CS2 (s) + 3O2 = CO2 (g) + 2SO2 (g) + 1101.8kJ; It is the exothermal reaction. b)
CaCO3 (s) = CaO (s) + CO2 (g) – 57.17kJ; It is the endothermic reaction.
If p, T = constant, Qp = - ΔH.
§2. The Hess's Law
The main principle underlying all thermochemical calculations was established by the Russian
Academician G. Hess in 1840. This principle, known as the Hess's Law, may be formulated as follow:
The thermal effect of a reaction depend only on the initial and final conditions of the reacting
substances, but not on the intermediate stages of the process.
This can be explained by an example. Carbon dioxide can be obtained from carbon, carbon monoxide
and oxygen in various routes, for instance:
First route:
C (s) + O2 (g) = CO2 (g) + 941kcal.
Second route:
C (s) + l/2O2 (g) = CO (g) + 26.4kcal
CO (g) + 1/2O2 (g) = CO2 (g) + 67.6kcal
Thus, the total amount of heat evolved during the formation of one gram-molecule of carbon dioxide is
the same in both cases, although the reaction took place in the two stages in the second case.
The Hess's Law makes it possible to calculate the thermal effect of the reaction when they cannot be
measured directly for some reason of other.
1. The thermal effect of a chemical reaction equals the sum of the formation heats of the reactants.
ΔH0298 = ΣΔH0f,products – ΣΔH0f,reactants
ΔH0f is in the reference book or tables.
The amount of heat evolved or absorbed during the formation of one mole of a chemical compound
from its simple substances is called the heat of formation of the compound.
The formation heat of the simple substances (H; Cl2; O2; Na and so on) is equal to 0.0kJ.
2. The thermal effect of a chemical reaction equals the sum of the combustion heat of the readouts
minus the sum of the combustion heat of the products.
ΔH0298 = ΣΔH0comb,products – ΣΔH0comb,reactants
The amount of heat evolved during the full combustion of one mole of chemical compound in pure
oxygen to burning product is called the heat of combustion of a compound.
The combustion heat of the burring products (CO2; H2O; HCl and N2) is equal to 0.0kJ.
The heat of combustion of food products in living organisms is a source of energy which gives us a
life.
Table 1
The combustion heat of food products
PRODUCT
PROTEINS
CONTENT, %
FATS
CARBOHYDRATES
CALORICITY
kJ/g
kcal/g
proteins
100
17.0
4.0
fats
100
38.0
9.0
carbohydrates
100
17.0
4.0
bread
6.5
1.0
40.1
7.95
1.90
sausages
12.3
25.3
11.59
2.77
cheese
23.4
30.0
15.52
3.71
milk
2.8
3.2
4.7
2.43
0.58
egg
12.7
11.5
0.7
6.57
1,57
beef
18.9
12.4
7.89
1.87
pork
14.6
33,0
14.85
3.55
apple
0.4
11.3
1.92
0.46
orange
0.8
8.6
1.59
0.38
grapes
0.4
17.5
2.89
0.69
nuts
16.1
66.9
9.9
29.46
7.04
3. The heat of formation of a complex substance from simple substances squats its heat of
decomposition with the opposite sign.
§3. The IInd Law of Thermodynamics
The 1st Law of Thermodynamics does not allow to judge about an ability of processes to the
spontaneous flow. But the IInd Law does.
What is the motive force of the spontaneous processes? Get to know entropy. Entropy is a measure of
the disorder of the system.
R
Equation of Boltsman S = ———— ∙ lnW
NA
NA is Avogadro's number (NA = 6.02 ∙ 1023 mol'), W is the number of microconditions which
determine the given macrocondition.
At the standard temperature and pressure (STP) (T = 298K and p =; 1 arm =-• 760 fTun Hg =
101.3kPa) entropy is marked S°298 and is called standard entropy.
The Second Law of Thermodynamics:
In the isolated systems all spontaneous processes take place in the direction of increasing entropy. The
change of entropy value (at constant temperature) is equal the ratio of the heat to the absolute temperature.
Q
ΔS ≥ ————
T
The changes of entropy and enthalpy are the motive force of a chemical reaction.
If ΔS>0 and ΔH<0, it means that the process is spontaneous.
§4. United the 1st and the IInd Law of Thermodynamics
The function which includes entropy and enthalpy is called Gibbs free energy (p, t= const).
ΔG = ΔH – T ∙ ΔS
Any reaction is spontaneous, if it takes place with decreasing energy. Hence, a spontaneous direct
process is characterized by ΔG<0, a process which can go to the opposite direction spontaneously is in case
of ΔG>0 and if ΔG=0, the process is in the equilibrium.
The dependence of maximum work of a reaction upon equilibrium constant was established by Van't
Hoff in 1885:
ΔG0 = - RT ∙ lnK = - 2.303 RT ∙ lgK
§5. Bioenergetics
Many of the reactions that take place in living organisms require a source of free energy to drive them.
The immediate source of this energy in heterotrophic organisms, which include animals, fungi, and most
bacteria, is the sugar glucose.
Oxidation of glucose to carbon dioxide and water is accompanied by a large negative free energy
change
C6H12O6 + O2 → 6CO2 + 6H2O; ΔG0 = – 2880 kJ mol-1 (40)
Of course, it would not do to simply "burn" the glucose in the normal way; the energy change would be
wasted as heat, and rather too quickly for the well-being of the organism. Effective utilization of this free
energy requires a means of capturing it from the glucose and then releasing it in small amounts when and
where it is needed. This is accomplished by breaking down the glucose in a series of a dozen or more steps in
which the energy liberated in each stage is captured by an “energy carrier” molecule, of which the most
important is adenosine diphosphate, known as ADP. At each step in the breakdown of glucose, an ADP
molecule reacts with inorganic phosphate (denoted by Pi) and changes into ATP:
ADP + Pi → ATP
ΔGº = +30kJ∙mol-1
The 30kJ∙mol-1 of free energy stored in each ATP molecule is released when the molecule travels to a
site where it is needed and loses one of its phosphate groups, yielding inorganic phosphate and ADP, which
eventually finds its way back the site of glucose metabolism for recycling back into ATP. The complete
breakdown of one molecule of glucose is coupled with the production of 38 molecules of ATP according to
the overall reaction:
C6H12O6 + 6O2 + 38Pi + 38ADP → 38ATP + 6CO2 + 44 H20
For each mole of glucose metabolized, 38 x (30 kJ) = 1140 kJ of free is captured as ATP, representing
an energy efficiency of 1140/2880 = 0.4. That is, 40% of the free energy obtainable from the oxidation of
glucose is made available to drive other metabolic processes. The rest is liberated as heat.
Where does the glucose come from? Animals obtain their glucose from their food, especially cellulose
and starches that, like glucose, have the empirical formula {CH2O}. Animals obtain this food by eating
plants or other animals. Ultimately, all food comes from plants, most of which are able to make their own
glucose from CO2 and H2O through the process of photosynthesis. This is just the reverse of Eq 40 in which
the free energy is supplied by the quanta of light absorbed b y chlorophyll and other photosynthetic pigments.
The photosynthesis-respiration free energy cycle
The zigzags represent the individual steps of energy capture by ADP and its delivery to metabolic
processes by ATP. “Thank a green plant every day”
This describes aerobic respiration, which evolved after the development of photosynthetic life on Earth
began to raise the concentration of atmospheric oxygen. Oxygen is a poison to most life processes at the
cellular level, and it is believed that aerobic respiration developed as a means to protect organisms from this
peril. Those that did not adapt have literally “gone underground” and constitute the more primitive anaerobic
bacteria.
The function of oxygen in respiration is to serve as an acceptor of the electrons that glucose loses when
it undergoes oxidation. Other electron acceptors can fulfill the same function when oxygen is not available,
but none yields nearly as much free energy. For example, if oxygen cannot be supplied to mammalian muscle
cells as rapidly as it is needed, they switch over to an anaerobic process yielding lactic acid instead of CO2:
C6H12O6 + 2ADP → 2CH3CH(OH)COOH
ΔGº = –218 kJ∙mol–1
In this process, only (2 x 30 kJ) = 60 kJ of free energy is captured so the efficiency is only 28% on the
basis of this reaction, and it is even lower in relation to glucose. In “aerobic” exercising, one tries to maintain
sufficient lung capacity and cardiac output to supply oxygen to muscle cells at a rate that promotes the
aerobic pathway.
§6. Examples of thermodynamic problems
1.
Find the enthalpy of hydrogen peroxide formation (ΔHf, 298), if the thermochemical equations
are given:
H2O2 = H2O + ½O2 + 98.23kJ (1)
H2 + ½O2 = H2O + 284.24kJ (2)
Solution:
We should find the thermal effect of the reaction which cannot be carried out:
H2 + O2 = H2O2 + Q
Subtract (1) from (2): H2 + ½O2 – H2O = H2O – H2O – ½O2 + 284.24kJ – 98.23kJ
We will have: H2 + O2 = H2O2 + 186.01kJ
Answer: ΔH°f298 = – Q = – 186.01kJ
2.
Calculate the thermal effect of the reaction: Fe2O3(s) + 3CO(g) = 2Fe + 3CO2(g).
Solution:
Using the reference book:
ΔH°f298 (Fe2O3(s)) = – 822.2kJ/mol,
ΔH°f298 (CO(g)) = – 110.5kJ/mol,
ΔH°f298 (CO2(g)) = – 393.5kJ/mol.
In harmony to the Hess’s Law effect: ΔH = Σ ΔHf,products – Σ ΔHf,reactants
ΔH298 = [2∙ΔH°f298(Fe(s)) + 3∙ΔH°f298(CO2(g))] – [ΔH°f298(Fe2O3(s)) + 3∙ΔH°f298 (CO(g))]
So
ΔH298 = [2 ∙ 0 + 3 ∙ (–393.5)] – [(–822.2) + 3 ∙ (–110.5)] = –26.8kJ
Answer: ΔH°298 = –26.8kJ. It is exothermic reaction.
3.
Find the equilibrium constant of the reactinon: NH3(g) + HCl(g) = NH4Cl(s) at 298K.
Solution:
Calculate Gibbs free energy of the reaction:
ΔG298 = ΣΔG298, products – ΣΔG298, reactants
In accordance to the reference book: ΔG298(NH3(g)) = –16 7 kJ/mol
ΔG298(HCl(g)) = –95.4kJ/mol and ΔG298(NH4Cl(s)) = –203.7kJ/mol.
So,
ΔG298 = ΔG298(NH4Cl(s)) – [ΔG298(NH3(g)) + ΔG298(HCl(g))] = –91.6kJ
Then
ΔG298 = –2.303RT∙lgK
–91600J = –2,303 ∙ 8.31 ∙ 298K ∙ lgK
lgK = 16 or K = 1016
Answer: K = 1016. Ir means that the equilibrium is shifted to the products direction very much.
4.
Define if the reaction takes place at 25ºC:
2NO2 = N2O4 + 57.99kJ
Solution:
S298(NO2) = 240.2J/mol ∙ K and S298(N2O4) = 303.8J/mol ∙ K
ΔG = ΔH – T ∙ ΔS
So, in harmony with thermochemical equation ΔH = – Q = –57.99kJ
S298 = S298(N2O4) – 2 ∙ S298(NO2) = 303.8 – 2 ∙ 240.2 = –176.6J/K
Substitute these values in the United Ist and IInd Law of Thermodynamics:
ΔG298 = – 57990J – 298K ∙ (– 176.6J/K) = –5363.2J = –5.3632kJ
Answer: ΔG < 0, but it has a small value, it means that the reaction is not practicable at 298K.
§7. The practical work
The experimental determination of the thermal effects
The determination of the thermal effect of the chemical reactions is a problem of Thermochemistry.
Thermochemical methods have a great importance not only for chemistry but in medicine and biology.
Energy is necessary for living organisms to accomplish a work, to support the invariable temperature of
the body and etc. It is produced by means of exothermic redox reactions which take place in the cells.
The calorimeter is used for determination of thermal effect. The simplest calorimeter which is used in
the chemical laboratory consists of a vessel immersed into an external shell. A calorimeter has a
thermometer, a stirrer (glass rod) and a funnel for leading of solutions into the vessel.
The quantity of heat, which evolved (absorbed) in a calorimeter, is calculated by the total specific heat
capacity and the change of temperature.
Experimental work №1
The determination of the neutralization heat in the reaction between strong acid and alkali
The aim.
To determine the neutralization heat of the reaction between hydrochloric acid and alkali.
The neutralization heat is a quantity of heat which is evolved in interaction of one mole-equivalent of
an acid with one mole-equivalent of an alkali.
The neutralization of the one mole-equivalent of (he strong acid by the strong base is accompanied by
the same exothermic effect corresponding to the formation of one mole liquid water from H+(aq) – ions
(H3O+) and OH–(aq), according to the equation:
H+(aq) + OH–(aq) = H2O(liquid) + 55,9kJ/mole
The same value of the neutralization heat is not observed in the reactions between weak acid and weak
base, because in those reactions the thermal effect of dissociation reaction adds to the thermal effect of
neutralization reaction.
Course of work
Pour 50ml 0.1N alkali solution into weighted calorimetric vessel and immerse thermometer into it and
two minutes later read and record the temperature of the solution.
Then measure off 60ml 0.1N acid solution into another vessel and record its temperature. After that,
mix both solutions in the calorimetric vessel (an acid solution is poured with the funnel). Read and record the
highest temperature after mixing (t2).
Table 2
Record the results:
№
symbol
value
units
1. The mass ofcalorimetric vessel
m1
g
2. The concentrations of the solutions
CN
mole/L
or N
3. The volume of the alkali solution
V1
ml
4. The volume of the acid solution
V2
ml
5. The temperature of alkali solution
talkali
°C
6. The temperature of acid solution
tacid
°C
7. The average value of temperature, t1 = 0.5 (talkali + tacid)
t1
ºC
8. The temperature after mixing
t2
ºC
9. The total mass of the solution after mixing; m2 = ρ(V1+V2)
m2
g
Calculations
1.
The heat evolving in the calorimetric vessel:
g = (m1 ∙ c1 + m2 ∙ c2) ∙ Δt
here Δt = t2 – t1
c1 = 0.753J/ºC∙g is the specific heat capacity of glass,
c2 = 4.184J/°C∙g is the specific heat capacity of the solution.
ρ = l.0g/ml is the density.
2.
The neutralization heat is calculated for one mole-equivalent of alkali because an acid is in an
excess.
g ∙ 1000
Q = –––––––––, J/mole
CN ∙ V1
3.
The relative mistake of the experiment.
|Qtheory – Qexperement|
|55.9 ∙ 103 J/mole – Qexperement|
% = –––––––––––––––––– ∙ 100% = –––––––––––––––––––––––––––– ∙ 100%
Qtheory
55.9 ∙ 103J/mole
4.
Record all experimental and calculation results. Make a conclusion.
Experimental work №2
Determination ofthc heat of dissolution process
The aim.
To determine the KCl dissolving heat.
The dissolving heat (Qdis) is the amount of heat evolved or absorbed during the solving of the one
mole of a compound in a big amount of a solvent.
Course of work.
Pour 100ml of water into the weighted calorimetric vessel.
Weight off 2g KCl to an approximation of 0.0lg. Record the temperature of water which is in the
calorimeter and then put the salt into the vessel. Stir the mixture by the glass rod.
The temperature of the solution will be depressed during the dissolving and then will be increased.
Record the minimum temperature after full dissolving.
Table 3
The results:
№
symbol
value
units
1.
The mass of calorimetric vessel
m1
g
2.
The mass of water
m2
g
3.
The mass of KCl
m3
g
4.
The temperature of water
t1
°C
5.
The temperature of the solution
t2
°C
Calculations
1. The dissolving heat.
Δt ∙ c ∙ Mr
Qdis = ––————— , Mr is the molecular weight of salt.
m3
Here Δt = t2 – t1 and c = m1 ∙ c1 + m2 ∙ c2
c1 and c2 are the specific heat capacity of glass and solution (look at the previous work).
2. The relative mistake.
|Qtheory – Qpractical|
|–17577 J/mole – Qpractica|
% = ————–––––––––––––– ∙ 100% = ——————————––––– ∙ 100%
Qtheory
3. Make a conclusion.
|–17577J/mole|
§8. Problems
1. Calculate the thermal effect of the reaction:
CaO + 3C = CaC2 + CO2.
2. The heat of the reaction between 2.1g Fe and sulfur is equal to 3.77kJ. Calculate the formation heat
of FeS. (-100kJ/mole).
3. The thermal effect of the reaction
3CaO + P2O5(solid) = Ca3(PO4)2(solid) is (–739kJ).
Define ΔH°f,298 of calcium phosphate. (–4137.5kJ/mole)
4. Calculate the value of the thermal effect for two reactions which take place in our organisms:
a) C6H12O6(solid)= 2C2H5OH(liquid) + 2CO2(g)
b) C6H12O6(solid) + 6O2 = 6CO2(g) + 6H2O(liquid)
Which brings the greater amount of energy? (b)
5. Calculate ΔG° of the following reaction:
CaCO3 = CaO + CO2(g) at 25, 500 and 1500ºC
( 129.1 U, 50.7kJ.-114.0kJ)
6. Using the reference books calculate the equilibrium constants of the following reactions:
a) H2O(g) + CO(g) ↔ CO2(g) + H2(g)
b) CO2(g) + C(g) ↔ 2CO(g)
c)N2(g) + 3H2(g) ↔ 2NH3(g)
at 298K and 1000K.
a) 1.1 ∙ 105; 0.91;
b)7.5 ∙ 10–22; 1.4;
c)7.4 ∙ l05; 3.0 ∙ 10–6.
Formula
ΔH°f,298,
kJ/mole
Sº298,
J/mole∙K
ΔG°298,
kJ/mole
O2(g.)
P2O5(s.)
H2O(l.)
H2O(g.)
CaCO3(s.)
CaO(s.)
CO2(s.)
FeO(s.)
C(s.)
Fe(s.)
HCl(g.)
Cl2(g.)
NO(g.)
NO2(g.)
Fe2O3(s.)
H2(g.)
NH4NO3(s.)
0
-1492
-285,8
-241,8
-1207
-635,5
-393,5
-264,8
0
0
-92,3
0
90,3
33,5
-822,2
0
-365,4
205,0
114,5
70,1
188,7
88,7
39,7
213,7
60,8
5,7
27
186,8
222,9
210,6
240,2
87,4
130,5
151
0
-1348,8
-237,3
-228,6
-1127,7
-604,2
-394,4
-244,3
0
0
-95,2
0
86,6
51,5
-740,3
0
-183,8
N2O(g.)
CO(g.)
Ca3(PO4)3(s.)
H2S(g.)
SO2(g.)
Al(s.)
Al2O3(s.)
SO3(g.)
Al2(SO4)3(s.)
N2O4(g.)
82,0
-110,5
-4125
-21,0
-296,9
0
-1676
-395,8
-3441,8
9,6
219,9
197,5
236
205,7
248,1
28
50,9
256,7
239,2
303,8
104,1
-137,1
-4195,3
-33,8
-300,2
0
-1582
-371,2
-3100,87
98,4
PbS(s.)
PbO(s.)
Hg2Cl2(s.)
HgCl2(s.)
CaC2(s.)
N2(g.)
NH3(g.)
MgO(s.)
MgCO3(s.)
NiO(s.)
CH4(g.)
C2H6(g.)
C2H4(g.)
CH3OH(l.)
C2H2(g.)
C2H5OH(l.)
C6H12O6(s.) (glucose)
C6H6(l.)
CH3COOH(l.)
C3H8(g.)
-101
-219,3
-266
-225
-59,83
0
-46,2
-601,8
-1096
-239,7
-74,9
-89,7
52,3
-238,57
226,8
-277,6
-1273,0
82,9
-484,09
-103,85
91
66,1
192
146
69,96
191.5
192,6
26,9
66
38,0
186,2
229,5
219,4
126,78
200,8
160,7
––
269,2
159,83
269,91
-128,1
-189,1
-323,2
-268,5
-64,85
0
-16,7
-569,6
-1115,7
-211,6
-50,8
-32,9
68,1
-166,77
209.2
-174,8
-919,5
129,7
-389,36
-23,53