Samacheer Class 11 Chemistry - Thermodynamics

The first law of thermodynamics states that “the total energy of an isolated system remains constant though it may change from one form to another”
(or)
Energy can neither be created nor destroyed but may be converted from one form to another.

The heat changes in chemical reactions are equal to the difference in internal energy (∆U) or heat content (∆H) of the products and reactants, depending upon whether the reaction is studied at constant volume or constant pressure. Since ∆U and ∆H are functions of the state of the system, the heat evolved or absorbed in a given reaction depends only on the initial state and final state of the system and not on the path or the steps by which the change takes place.

Entropy is a measure of the molecular disorder (randomness) of a system.
The thermodynamic definition of entropy is concerned with the change in entropy that occurs as a result of a process.
It is defined as, dS = dq rev /T

• When both ∆H and ∆S are positive, the reaction is not feasible.
• When both ∆H and ∆S are negative, the reaction is not feasible.
• When ∆H decreases but ∆S increases, the reaction is feasible.

Gibbs free energy is defined as the part of the total energy of a system that can be converted (or) available for conversion into work.
G = H -TS,
where G = Gibb’s free energy
H = enthalpy
T = temperature
S = entropy

The heat of combustion of a substance is defined as “The change in enthalpy of a system when one mole of the substance is completely burnt in excess of air or oxygen”. It is denoted by ∆H C.

The heat capacity for 1 mole of substance, is called molar heat capacity (cm). It is defined as “The amount of heat absorbed by one mole of the substance to raise its temperature by 1 kelvin”.
The SI unit of molar heat capacity is JK -1 mol -1

The calorific value is defined as the amount of heat produced in calories (or joules) when one gram of the substance is completely burnt. The SI unit of calorific value is J kg -1. However, it is usually expressed in cal g -1.

The enthalpy of neutralization is defined as the change in enthalpy of the system when one gram equivalent of an acid is neutralized by one gram equivalent of a base or vice versa in dilute solution.
H + (aq) + OH – (aq) → H 2 O (l) = 57.32 kJ.

Lattice energy is defined as the amount of energy required to completely remove the constituent ions from its crystal lattice to an infinite distance. It is also referred as lattice enthalpy.

• The variables like P. V, T, and ‘n’ that are used to describe the state of the system are called state functions. e.g. pressure, volume, temperature, internal energy, enthalpy, and free energy.
• A path function is a thermodynamic property of the system whose value depends on the path or manner by which the system goes from its initial to the final state. e.g., work (w) and heat (q).

∆G = -2.303 RT logK eq
Substituting the known values of R, T and K eq
R = 8.314(JK -1 mol -1 )
T = 300 K
K eq = 10
∆G = –2.303 8.314(JK -1 ) 300(K) log 10
= -5744.14 J/mol
= -5.744 KJ/mol
∆G < 0, then it is spontaneous.

* Enthalpy of neutralization of a strong acid by a strong base is always a constant and it is equal to -57.32 kJ irrespective of which acid or base is used.
* Because strong acid or strong base means it is completely ionized in solution state. For e.g., NaOH (strong base) is neutralized by HCl (strong acid), due to their complete ionization, the net reaction takes place is only water formation.
So the enthalpy of neutralization is always constant for strong acid by a strong base.
H + Cl – + Na + OH – → Na + Cl – + H 2 O
H + NO 3 + + K + OH – → K + NO 3 + + H 2 O
(Net reaction) H + + OH – → H 2 O ∆H = -57.32 kJ

The third law of thermodynamics states that the entropy of pure crystalline substance at absolute zero is zero. Otherwise, it can be stated as it is impossible to lower the temperature of an object to absolute zero in a finite number of steps. Mathematically,
lim T→0 S = 0 for a perfectly ordered crystalline state.

Step 1:
Solid calcium is converted to gaseous state (Enthalpy of atomization)
Ca (S) → Ca (g) Ca(g) ∆H° a = 178 KJ/mol
Step 2:
The calcium is converted to ionic form
(divalent cation): (Ionisation enthalpy)
Ca → Ca + + e – ∆H° IE = 590 kJ
Ca → Ca 2+ + e – ∆H° IE = 590 KJ
Step 3:
Atomisation of chlorine molecule to chlorine atom: (Atomisation enthalpy)
\(\frac{1}{2}\)Cl 2 → Cl ∆H° Cl-Cl = 121 KJ
For two chlorine atoms required, atomistion energy = 2 × 121 = 242 KJ/mol
Step 4:
Convrsion of chlorine atom into ion:(Electron affinity)
Cl + e – → Cl –; ∆H° ea = – 364KJ
Step 5:
Finally the two ions join together by lattice energy as: (here two Cl – ions are involved)
Ca 2+ + 2Cl – → CaCl 2
Hence Lattice energy = Heat of formation – heat of atomization – dissociation energy – (sum of ionization energies) – (sum of electron affinities)
Since, heat of formation (i.e standard enthalpy of formation ) of CaCl 2 = – 796KJ/mol
Lattice energy = -796 -178 -242 -(590 + 1145) – (2 × – 364) = -2223KJ/mol

1. Entropy statement:
Whenever a spontaneous process takes place, it is accompanied by an increase in the total entropy of the universe”.
2. Kelvin-Planck statement:
it is impossible to take heat from a hotter reservoir and convert it completely into work by a cyclic process without transferring a part of heat to a cooler reservoir.
3. Efficiency statement:
Even an ideal, frictionless engine cannot convert 100% of its input heat into work.
Efficiency = \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}\)
% Efficiency = \(\left[\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}\right]\) x 100
% Efficiency = \(\left[\frac{\text { Output }}{\text { Input }}\right]\) x 100
% Efficiency < 100%
4. Clausius statement:
Heat flows spontaneously from hot objects to cold objects and to get it to flow in the opposite direction, we have to spend some work.

* A reaction that occurs under the given set of conditions without any external driving force is called a spontaneous reaction.
* The spontaneity of any process depends on three different factors.
* If the enthalpy change of a process is negative, then the process is exothermic and may be spontaneous. (∆H is negative)
* If the entropy change of a process is positive, then the process may occur spontaneously. (∆S is positive)
* The gibbs free energy which is the combination of the above two (∆H – T∆S) should be negative for a reaction to occur spontaneously, i.e. the necessary condition for a reaction to be spontaneous is ∆H – T∆S < 0
* For spontaneous process,
∆S total > 0, ∆G < 0, ∆S < 0

* The internal energy of a system is an extensive property. It depends on the number of substances present in the system.
* The internal energy of a system is a state function. It depends only upon the state variables (T, P, V. n) of the system.
* The change in internal energy is as ∆U = U 2 – U 1
* In a cyclic process, there is no energy change. ∆U (cyclic) = 0.
* If the internal energy of the system at the final state (U f ) is less than the internal energy of the
system at its initial state (U i ), then ∆U would be negative.
* if U f < U i ∆U = U f – U i = -ve
* if U f > U i ∆U = U f – U i = +ve

The calorimeter is used for measuring the amount of heat change in a chemical or physical change. In calorimetry, the temperature change in the process is measured which is directly proportional to the heat capacity. By using the expression C = \(\frac{q}{m \Delta T}\), we can calculate the amount of heat change in the process. Calorimetric measurements are made under two different conditions
(i) At constant volume (qV)
(ii) At constant pressure (qp)
(A) ∆U Measurements:
For chemical reactions, heat evolved at constant volume, is measured in a bomb calorimeter.
The inner vessel (the bomb) and its cover are made of strong steel. The cover is fitted tightly to the vessel by means of metal lid and screws.
A weighed amount of the substance is taken in a platinum cup connected with electrical wires for striking an arc instantly to kindle combustion. The bomb is then tightly closed and pressurized with excess oxygen. The bomb is immersed in water, in the inner volume of the calorimeter. A stirrer is placed in the space between the wall of the calorimeter and the bomb, so that water can be stirred, uniformly. The reaction is started by striking the substance through electrical heating.
A known amount of combustible substance is burnt in oxygen in the bomb. Heat evolved during the reaction is absorbed by the calorimeter as well as the water in which the bomb is immersed. The change in temperature is measured using a Beckman thermometer. Since the bomb is sealed its volume does not change and hence the heat measurements is equal to the heat of combustion at a constant volume (∆U) c.
The amount of heat produced in the reaction (∆U) c is, equal to the sum of the heat abosrbed by the calorimeter and water.
Heat absorbed by the calorimeter
q 1 = k.∆T
where k is a calorimeter constant equal to mc Cc ( me is mass of the calorimeter and Cc is heat capacity of calorimeter)
Heat absorbed by the water q 2 = m w C w ∆T
where mw is molar mass of water
C w is molar heat capacity of water (4,184 kJ K -1 mol -1 )
Therefore ∆U c = q 1 + q 2
= k.∆T + m w C w ∆T
= (k + m w C w )∆T
Calorimeter constant can be determined by burning a known mass of standard sample (benzoic acid) for which the heat of combustion is known (-3227 kJ mol -1 ). The enthalpy of combustion at constant pressure of the substance is calculated from the equation
∆H° c (pressure) = ∆U° c (Vol) + ∆n g RT

Work involved in expansion and compression processes:
In most thermodynamic calculations we are dealing with the evaluation of work involved in the expansion or compression of gases. The essential condition for expansion or compression of a system; is that there should be difference between external pressure (P ext ) and internal pressure (P int ).
For understanding pressure-volume work, let us consider a cylinder which contains V moles of an ideal gas fitted with a frictionless piston of cross sectional area A. The total volume of the gas inside is V and pressure of the gas inside is P int. If the external pressure P ext is greater than P int, the piston moves inward till the pressure inside becomes equal to P ext. Let this change be achieved in a single step
and the final volume be V f.In this case, the work is done on the system (+w). It can be calculated as follows
w = -F.∆x ……….(1)
where dx is the distance moved by the piston during the compression and F is the force acting on the gas.
F = P ext A ……….(2)
Substituting (2) in (1)
w = – P ext. A. ∆x
A.∆x = change in volume = V f – V i
w = -P ext.{V f – V i ) ……….(3)
w = -P ext. (-∆V) …………(4)
= P ext.∆V
Since work is done on the system, it is a positive quantity.
If the pressure is not constant, but changes during the process such that it is always infinitesimally greater than the pressure of the gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, dV. In such a case we can calculate the work done on the gas by the relation
W rev = – \(\int_{V_{i}}^{v_{f}}\) p ext dV.
In a compression process, Pext the external pressure is always greater than the pressure of the system.
i.e., P ext = (P int + dP)
In an expansion process, the external pressure is always less than the pressure of the system
i.e., P ext = (P int – dP)
When pressure is not constant and changes in infinitesimally small steps (reversible conditions) during compression from V i to V f. the P-V plot looks like in figure. Work done on the gas is represented by the shaded area.
In general case we can write,
P ext = (P int ± dP). Such processes are called reversible processes. For a compression process work can be related to internal pressure of the system under reversible conditions by writing equation
W rev = – \(\int_{V_{i}}^{v_{f}}\) P int dV
For a given system with an ideal gas
P int V = nRT
p int = \(\frac{n R T}{V}\)
W rev = – \(\int_{V_{i}}^{v_{f}}\) \(\frac{n R T}{V}\) dV
W rev = – nRTln(\(\frac{V_{f}}{V_{i}}\))
W rev = -2.303nRTlog\(\frac{V_{f}}{V_{i}}\) ………….(5)
If V f > V i (expansion), the sign of work done by the process is negative.
If V f < V i (compression) the sign of work done on the process is positive.

1. When the system at constant pressure undergoes changes from an initial state with H 1, U 1, V 1 and P parameters to a final state with H 2, U 2, V 2 and P parameters, the change in enthalpy ∆H, is given by
AH = U + PV
2. At initial state H 1 = U 1 + PV 1 ………(1)
At final state H 1 = U 1 + PV 1 ……..(2)
(2) – (1) ⇒ (H 2 – H 1 ) = (U 2 – U 1 ) + P(V 2 – V 1 )
∆H = ∆U + P∆V …………(3)
Considering ∆U = q + w; w = -P∆V
∆H = q + w + P∆V
∆H = q p – P∆V+ P∆V
∆H = q p …………(4)
q p is the heat absorbed at constant pressure and is considered as heat content.
3. Consider a closed system of gases which are chemically reacting to produce product gases at constant temperature and pressure with V. and as the total volumes of the reactant and product gases respectively, and n1 and n f are the number of moles of gaseous reactants
and products. Then,
For reactants: P V i = n i RT
For products: P V f = n f RT
Then considering reactants as initial state and products as final state,
P (V i – V i ) = (n i – n i ) RT
P∆V = ∆n g RT
∆H = ∆U + P∆V
∆H = ∆U + ∆n g RT ……….(5)

The formation of NaCl can be considered in five steps. The sum of the enthalpy changes of these steps is equal to the enthalpy change for the overall reaction from which the lattice enthalpy of NaCl is calculated.
Let us calculate the lattice energy of sodium chloride using Bom-Haber cycle
∆H = heat of formation of sodium chloride = -411.3 kJmol -1
∆H 1 = heat of sublimation of Na (s) = 108.7 kJmol -1
∆H 2 = ionisation energy of Na (s) = 495. kJ mol -1
∆H 3 = dissociation energy of Cl 2(g) = 244 kJ mol -1
∆H 4 = Electron affinity of Cl (g) = -349 kJ mol -1
U = lattice energy of NaCl
∆H f = ∆H 1 + ∆H 2 + ∆H 3 + ∆H 4 + U
∴ U = (∆H f ) – (∆H 1 + ∆H 2 + \(\frac{1}{2}\)∆H 3 + ∆H 4 )
=» U = (-411.3) – (108.7 + 495.0 + 122 – 349)
U = (-411.3) – (376.7)
∴ U = -788kJ mol -1

Characteristics of Gibbs free energy:
1. Gibbs free energy is defined as the part of the total energy of a system that can be converted (or) available for conversion into work.
G = H – TS ………..(1)
Where
H = enthalpy, T = temperature and S = entropy
2. G is a state function and is a single value function.
3. G is an extensive property, whereas ∆G becomes intensive property for a closed system. Both G and ∆G values correspond to the system only.
4. ∆G gives criteria for spontaneity at constant pressure and temperature.
* If ∆G is negative (∆G < O), the process is spontaneous.
* If ∆G is positive (∆G > O), the process is non-spontaneous.
* If ∆G is zero (AG = O), the process is equilibrium.
5. For any system at constant pressure and temperature,
∆G = ∆H – T∆S ……….. (2)
We know AH = ∆U + P∆V
∆G = ∆U + P∆V – T∆S ………(3)
6. For the first law of thermodynamics, ∆U = q + w
∆G= q+ w+P∆V – T∆S …………(4)
For second law of thermodynamics, ∆S = \(\frac {q}{T}\)
∆G = q + w + P∆V – T\(\frac {q}{T}\)
∆G = w + P∆V …………(5)
∆G = – w – P∆V ……….(6)
7. – P∆V represent the work done due to expansion against constant external pressure. Therefore, it is clear that the decrease in free energy (-∆G) accompanying a process taking place at constant temperature and pressure is equal to the maximum work obtainable from the system other than the work of expansion.
8. Unit of Gibb’s free energy is J mol -1

Given
n = 2 moles
V i = 500 ml = 0.5 lit
V f = 2 lit
T = 25°C = 298 K
w = -2303 nRTIog (\(\frac{V_{f}}{V_{i}}\))
w = -2.303 × 2 × 8.314 × 298 × log (4)
w = -2.303 × 2 × 8.314 × 298 × 0.6021
w = -6871 J
w = -6.871 kJ.

Given,
T i = 298 K
T f = 298.45 K
k = 2.5 kJ K
m = 3.5g
M m = 28
heat evolved = k∆T
∆H C = k (T f – T i )
∆H C = 2.5 kJ K’ (298.45 – 298) K -1
∆H C = 1.125 kJ
∆H C = \(\frac {1.125}{3.5}\) x 28 kJ mol -1
∆H C = 9 kJ mol -1

Given
T sys = 77° C = (77 + 273) = 350 K
T surr = 33°C = (33 + 273) = 360 K
q = 245J
∆S sys = \(\frac{q}{T_{s y}}=\frac{-245}{350}\) = -0.7 JK -1
∆S surr = \(\frac{q}{T_{s \mathrm{sr}}}=\frac{+245}{306}\) = +0.8 JK -1
∆S univ = ∆S sys + ∆S surr
∆S univ = -0.7 JK -1 + 0.8 JK -1
∆S univ = 0.1 JK -1

Given
n = 1 mole
P = 4.1 atm
V = 2 Lit
T = ?
q = 3710 J
∆S = \(\frac{q}{T}\)
∆S = \(\frac{q}{\frac{p v}{n R}}\)
∆S = \(\frac{n R q}{P V}\)
∆S = \(\frac{1 \times 0.082 \text { lit atm } K^{-1} \times 3710 J}{4.1 \text { atm } \times 2 \text { lit }}\)
∆S = 37.10 JK -1

Given,
∆H f (NaCl) = 30.4 kJ = 30400 J mol -1
∆S f (NaCl) = 28.4 JK -1 mol -1
∆S -1 = \(\frac{\Delta H_{f}}{T_{f}}\)
T f = \(\frac{\Delta H_{f}}{\Delta S_{f}}\)
T f = \(\frac{30400 J m o l^{-1}}{28.4 J K^{-1} m o l^{-1}}\)
T f = 1070.4 K

Given:
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
∆H°C = 2220.2kJ mol -1 ……………..(1)
C + O 2 → CO 2
∆H° f = -393.5 kf mol -1 ……….(2)
H 2 + \(\frac{1}{2}\)O 2 → H 2 O
∆H° f = -285.8 kJ mol -1 ……………(3)
3C + 4H 2 → C 3 H 8
∆H° c =?
(2) × (3)
⇒ 3C + 3O 2 → 3CO 2
∆H° f = -1180.5 kJ ………..(4)
(3)× (4)
⇒ 4H 2 + 2O 2 → 4H 2 O
∆H° f = 1143.2 kJ ……..(5)
(4) + (5) – (1)
⇒ 3C + 3O 2 + 4H 2 + 2O 2 + 3CO 2 + 4H 2 O → 3CO 2 + 4H 2 O + C 3 H 8 + 5O 2
∆H° f = -1180.5 – 1143.2 – (-2220.2) kJ
3C + 4H 2 → C 3 H 8
∆H° f = -103.5 kJ
Standard heat of formation of propane is ∆H° f(C 3 H 8 ) = -103.5 kJ

Given,
∆H = 30.56 kJ mol -1
∆S = 6.66 x 10 -3 kJK -1 mol -1
T = ? at which ∆G =0
∆G = ∆H – T∆S
0 = ∆H – T∆S
T = \(\frac{\Delta H}{\Delta S}\)
T = \(\frac{30.56 \mathrm{~kJ} \mathrm{~mol}^{-1}}{66.6 \times 10^{-3} \mathrm{~kJ} K^{-1} \mathrm{~mol}^{-1}}\)
T = 4589 K
(i) At 4589K;
∆G = 0 the reaction is in equilibrium.
(ii) at temperature below 4598 K
∆H > T∆S
∆G = ∆H – T∆S > 0, the reaction in the forward direction, is non spontaneous. In other words the reaction occurs in the backward direction.

Given,
T = 298K; ∆U = -742.4 kJ mol -1
∆H = ?
∆H = ∆U + ∆n (g) RT
∆H = ∆U + (n p – n r ) RT
∆H = – 742.4 +[2 – \(\frac {3}{2}\)] x 8.314 x 10 -3 x 298
∆H = -742.4 + (0.5 x 8.314 x 10 -3 x 298)
∆H = -742.4 + 1.24.
∆H = -741.16 kJ mol -1

Given
E C-H = 414 kJ mol -1
E C-C = 347 kJ mol -1
E C-C = 6l8 kJ mol -1
E H-H = 435 kJ mol -1
∆H r = Σ(Bond energy) r – Σ(Bond energy) p
∆H r = (E C=C + 4E C-H + E H-H ) – (E C-C + 6E C-H )
∆H r = (618 + (4 × 414) + 315) – (347 + (6 ×414))
∆H r = 2709 – 2831
∆H r = -122 kJ mol -1

∆H f = ∆H 1 + ∆H 2 + ∆H 3 +∆H 4 + u
-795 = 121 + 2422 + 242.8 + (2 × -355) + u
-795 = 2785.8 – 710 + u
-795 = 2075.8 + u
u = -795 – 2075.8
u = -2870.8 kJ mol -1

Given
T= 33 K
N 2 O 4 ⇌ 2NO 2
Initial concentration: 100% 0
Concentration dissociated 50% – Concentration remaining at equilibrium 50% 100%
K eq = \(\frac{100}{50}\) = 2
∆G° = -2.303 RT log K eq
∆G°= -2.303 × 8.314 × 33 × log 2
∆G° = -190.18 Jmol -1

Given,
∆G f °(SO 2 ) = – 297 KJ mol -1
∆G f °(SO 2 ) = – 297 KJ mol -1
SO 2 + \(\frac{1}{2}\)O 2 → SO 3 ∆H r ° = ?
∆H r ° = (∆H f °) compound – ∑(∆H f °) elements
∆H r ° = ∆H f ° (SO 3 ) – [∆H f ° (SO 2 ) + \(\frac{1}{2}\) ∆H f ° (O 2 )]
∆H r ° = – 396 kJ mol -1 – (- 297 kJ mol -1 + 0)
∆H r ° = – 396 kJ mol -1 +297
∆H r ° = – 99 kJmol -1

Given,
T = 298 ∆T
∆H = 400 J mol -1 = 400 J mol -1
∆S = 0.2 JK -1 mol -1
∆G = ∆H – T∆S
if T = 2000 K
∆G = 400 – (0.2 × 2000) = 0
if T > 2000 K
∆G will be negative. The reaction would be spontaneous only beyond 2000 K

Given:
T = 298 K
∆G° r = -13.6 kJ mol -1
= – 13600 J mol -1
∆G° = – 2.303 RT log K eq
log K eq = \(\frac{-\Delta G^{0}}{2.303 R T}\)
log K eq = \(\frac{13.6 \mathrm{~kJ} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \times 10^{-3} \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K}}\)
log K eq = 2.38
K eq = antilog (2.38)
K eq = 239.88

Given
∆H C(CH 4 ) = – 890 kJ mol -1
∆H C(C 2 H 4 ) = -1423 kJ mol -1
Let the mixture contain x lit of CH 4 and (3.67 – x) lit of ethylene.
CH 4 + 2O 2 → CO + 2H 2 O
x lit x lit
C 2 H 4 + 3O 2 → 2 CO 2 + 2H 2 O
(3.67 -x) lit 2 (3.67 -x) lit
Volume of Carbondioxide formed = x + 2(3.67 – x) = 6.11 lit
x + 7.34 – 2x = 6.11
7.34 – x = 6.11
x = 1.23 lit
Given mixture contains 1.23 lit of methane and 2.44 lit of ethylene, hence