(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

ΔH_f (NaCl)=30.4 kJ = 30400 J mol^-1
ΔS_f (NaCl)=28.4 JK^-1mol^-1
T_f=?
ΔS_f =\(\frac { \Delta { H }_{ f } }{ { T }_{ f } } \)
T_f=\(\frac { \Delta { H }_{ f } }{ \Delta { S }_{ f } } \)
T_f=\(\frac { 30400J{ mol }^{ -1 } }{ 28.4J{ K }^{ -1 }mol^{ -1 } } \)
T_f=1070.4K

E_C-H=414 kJ mol^-1
E_C-C=347 kJ mol^-1
E_C=C=618 kJ mol^-1
E_H-H=435 kJ mol^-1

ΔH_r= Σ(Bond energy)_r -Σ(Bond energy)p
ΔH_r= (E_C=C + 4E_C-H + E_H-H) - (E_C-C + 6E_C-H)
ΔH_r= (618 + (4 \(\times\) 414) + 435) - (347 +(6 \(\times\) 414))
ΔH_r= 2709 - 2831
ΔH_r​​​​​​​= -122 kJ mol^-1

\(\Delta { H }_{ f }^{ 0 }\)(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

T=298K
\(\Delta { G }_{ r }^{ 0 }\)=- 13.6 kJ mol^-1
=-13600 J mol^-1
ΔG⁰=- 2.303 RT log K_eq
log K_eq=\(\frac { 13.6kJmol^{ -1 } }{ 2.303\times 8.314\times { 10 }^{ -3 }JK^{ -1 }mol^{ -1 }\times 298K } \)
log K_eq = 2.38
K_eq = antilog (2.38)
K_eq = 239.88.