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

Gibbs free energy is defined as G = H - TS

The first law of thermodynamics, also known as the law of conservation of energy, states that "The total energy of an isolated system remains constant though it may change from one form to another."
The mathematical statement of the First Law is: ΔU=q+w
Where q - the amount of heat supplied to the system; w - work done on the system

(i) Free energy is defined as G = H - TS. 'G' is a state function.
(ii) G- Extensive property; ΔG - intensive property. When mass remains constant between initial and final states of system.
(iii) 'G' has a single value for the thermodynamic state of the system.
(iv) G and ΔG values correspond to the system only.

(v) Gibbs free energy and the net work done by the system:
For any system at constant pressure and temperature
ΔG = ΔH - TΔS .....(1)
We know that,
ΔH = ΔU + PΔV
ΔG =ΔU + PΔV-TΔS
from first law of thermodynamics
ΔU = q +w
from second law of thermodynamics
Δ S=\(\frac{q}{T}\) Δ G=q+w+PΔ V-T\((\frac{q}{T}) \)
Δ G = w+PΔV
-ΔG = -w - PΔV ......(2)
But -PΔV represents the work done due to expansion against a constant external pressure.

(i) Heat evolved at constant volume, is measured in a bomb calorimeter.
(ii) Apparatus setup: 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.
(iii) Experiment: 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.
(iv) Calculation: 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 (ΔU)_c
The amount of heat produced in the reaction (ΔU)_c is equal to the sum of the heat absorbed by the calorimeter and water.
Heat absorbed by the calorimeter q₁ = k.ΔT
where k is a calorimeter constant equal to m_c C_c (m_c is mass of the calorimeter and C_c is heat capacity of calorimeter)
Heat absorbed by the water q₂ = m_w C_w ΔT
where m_w is molar mass of water
C_w is molar heat capacity of water (4,184 kJ K^-1mol^-1)
Therefore ΔU_c = q₁ + q₂
=k.ΔT + m _w C_w ΔT
=(k+m _w Cw) ΔT
Calorimeter constant can be determined by burning a know.n mass of standard sample (benzoic acid) for which the heat of combustion is known (-3227 kJmol^-1)
The enthalpy of combustion at constant pressure of the substance is calculated from the equation (7.17)
\(\Delta { H }_{ C(pressure) }^{ 0 }=\Delta { U }_{ C(vol) }^{ o }+\Delta { n }_{ g }RT\)

T =298 K
ΔH 400 J mol^-1 = 400 J mol^-1
ΔS= 0.2 J K^-1 mol^-1
ΔG =ΔH-TΔS
if T =2000 K
ΔG = 400 - (0.2 \(\times\) 2000) = 0
if T > 2000 K
ΔG will be negative.
The reaction would be spontaneous only beyond 2000K.

\(\Delta { H }_{ f }^{ 0 }\)(SO₂) = - 297 kJ mol^-1
\(\Delta { H }_{ f }^{ 0 }\)(SO₃) = - 396 kJ mol^-1
SO₂ + \(\frac{1}{2}\)O₂⟶SO₃ \(\Delta { H }_{ r}^{ 0 }\)=?
\(\Delta { H }_{ r}^{ 0 }\)=\((\Delta { H }_{ f }^{ 0 })\)_compound-Σ(ΔH_f)_elements
\(\Delta { H }_{ r}^{ 0 }\)=​​​​\(\Delta { H }_{ f }^{ 0 }\)(SO₃)-(\(\Delta { H }_{ f }^{ 0 }\)(SO₂) +\(\frac{1}{2}\)\((\Delta { H }_{ f }^{ 0 })\)(O₂))
\(\Delta { H }_{ r}^{ 0 }\)=- 396 kJ mol^-1 - (-297 kJ mol^-1 + 0)
\(\Delta { H }_{ r}^{ 0 }\)=-396 kJ mol^-1 + 297
\(\Delta { H }_{ r}^{ 0 }\)=-99 kJ mol^-1

n = 1mole
P = 4.1 atm
V= 2 L
T=?
q=3710 J
ΔS=\(\frac{q}{T}\)
ΔS=\(\frac { q }{ \left( \frac { PV }{ nR } \right) } \)
ΔS=\(\frac{nRq}{PV}\)
ΔS=\(\frac { 1\times 0.082\quad lit\quad atm{ K }^{ -1 }\times 3710J }{ 4.1\quad atm\times 2\quad lit } \)
ΔS​​​​​​​=37.10JK^-1