T = 298K ; ΔU= - 742.4 kJ mol^-1
ΔH=?
ΔH=ΔU+Δn_gRT
ΔH=ΔU+(n_p-n_r)RT
ΔH=-742.4 +\((2-\frac{3}{2})\)\(\times\)8.314 \(\times\) 10^-3 \(\times\) 298
=-742.4 + (0.5 \(\times\) 8.314 \(\times\)10^-3 \(\times\)298)
=-742.4 + 1.24
=-741.16 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

(i) Kelvin-Planck statement: It is impossible to construct a machine that absorbs heat from a hot source and converts it completely into work by a cyclic process without transferring a part of heat to a cold sink.
(ii) Clausius statement: It is impossible to transfer heat from a cold reservoir to a hot reservoir without doing some work.
(iii) Entropy statement: The entropy of an isolated system increases during a spontaneous process

Strong acids and strong bases exist in the fully ionised form in aqueous solutions as below:
H₃O⁺ + CI^- + Na⁺ + OH^- ⟶ Na⁺ + CI + 2H₂O
(or)
H₃O⁺_(aq)+OH^-_(aq)⟶2H₂O(1)⁰
ΔH^o =-57.32 KJ.
The H⁺ions produced in water by the acid molecules exist as H₃O⁺. Thus, enthalpy change of neutralisation is essentially due to enthalpy change per mole of water formed from H₃O⁺ and OH- ions. Therefore, irrespective of the chemical nature, the enthalpy of neutralisation of strong acid by strong base is a constant value which is equal to -57.32 KJ

(i) State function: A state function is a thermodynamic property of a system, which has a specific value for a given state and does not depend on the path (or manner) by which the particular state is reached.
Example: Pressure (P), Volume (V), Temperature(T)
(ii) Path functions: A path function is a thermodynamic property of the system whose value depends on the path by which the system changes from its initial to final states.
Example: Work (w), Heat (q).