T_sys=77°C = (77 + 273) = 350 K
T_surr=33°C = (33 + 273) = 306 K
q=245 J
ΔS_sys=\(\frac{q}{T_{sys}}=\frac{-245}{350}=-0.7JK^{-1}\)
ΔS_surr=\(\frac{q}{T_{sys}}=\frac{+245}{350}=0.8JK^{-1}\)
ΔS_univ=ΔS_sys+ΔS_surr
ΔS_univ=-0.7 JK^-1+ 0.8 JK^-1
ΔS_univ=0.1 JK^-1.

(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.

When the system at constant pressure undergoes changes from an initial state with H₁, U₁ and, V₁ to a final state with H₂, U₂ and V₂ the change in enthalpy ΔH, can be calculated as follows:
H = U+PV
In the initial state
H₁ = U₁ +PV₁ ........(1)
In the final state
H₂ = U₂ +PV₂ ........(2)
change in enthalpy is (2) - (1)
(H₂ - H₁) = (U₂ - U₁) + P(V₂ - V₁)
ΔH = ΔU+PΔV
As per first law of thermodynamics,
ΔV = q+w
Equation (3) becomes
ΔH = q+w+PΔV
w = -PΔV
ΔH = qp-PΔV+PΔV
ΔH = qp....(4)
q_p - is the heat absorbed at constant pressure and is p considered as heat content. Consider a closed system of gases which are chemically reacting to form gaseous products at constant temperature and pressure with V_i and V_f as the total volumes of the reactant and product gases respectively, and n _i and n_f as the number of moles of gaseous reactants and products, then,
For reactants (initial state) :
PVi = ni RT ...(5)
For products (final state) :
PVf = nfRT .........(6)
(6) - (5)
P(Vf - Vi) = (nf - ni) RT
PΔV = Δn(g) RT ...........(7)
Substituting in (7) in (3)
ΔH = ΔV +Δn(g) RT ........(8)

(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)
\(\Delta { H }_{ C(pressure) }^{ 0 }=\Delta { U }_{ C(vol) }^{ o }+\Delta { n }_{ g }RT\)

Born - Haber cycle for the formation of CaCl₂
Born - Haber cycle is used to calculate the lattice enthalpy of CaCl₂ 

Δ^oH₁ - Enthalpy change for the sublimation of Ca_(s) to Ca_(g)
Δ^oH₂ - Enthalpy change for dissociation of Cl_2(g) to 2Cl_(g)
Δ^oH₃ - Ionisation energy for Ca_(g) to Ca²⁺_(g)
Δ^oH₄ - Electron affinity for the conversion of 2Cl_(g) to 2Cl^-_(g)
Δ^oH₅ - Lattice enthalpy for the formation of solid CaCl₂·
Δ^oH_f =Δ^oH₁+Δ^oH₂+Δ^oH₃+Δ^oH₄+Δ^oH₅