From third law of thermodynamics, it can be explained that entropy of a perfectly crystalline substance is zero at zero kelvin.
\(\triangle G=0\) (because the reaction is in equilibrium)
\(\triangle G^{ \ominus }\)= - 2.303 RT log K
= - 2.303 x 8.314 JK^-l mol^-1x 300 K log 10²
= - 11.488 kJ mol^-1

(i) First law of thermodynamics: It states that energy can neither be created nor be destroyed. The energy of an isolated sytem is constant.
\(\triangle U=q+w\)
(ii) It is defined as the amount of heat evolved or absorbed when one mole of the compound is formed from its constituent elements in their standard states

The reaction involved is
Al₂O₃(s) + 6Na(s)\(\longrightarrow \)2Al(S) + 3Na₂O(s)
\(\triangle G^{ \ominus }=\sum { \triangle _{ f } } G^{ \ominus }(p)-\sum { \triangle _{ f } } G^{ \ominus }(r)\)
\(=[2\triangle _{ f }G^{ \ominus } \ AI(s)+3\triangle _{ f }G^{ \ominus }Na_{ 2 }O(s)]-[\triangle _{ f }G^{ \ominus }Al_{ 2 }O(s)+6\triangle _{ f }G^{ \ominus }Na(s)]\)
[2 x 0 + 3x (-377)] - [1582 + 6 x 0]
= 451 KJ mol^-1
This means that sodium can not reduce aluminium oxide(Al₂O₃ ) to aluminium metal because \(\triangle G^{ \ominus }\) comes out to be positive.

\(\sum { { \triangle } } _{ f }G^{ \ominus }_{ (products) }-\sum { { \triangle } } _{ f }G^{ \ominus }_{ (reactants) }\)
\(\{ 4 \ mol\times { 4 }_{ f }G^{ \ominus }NO(g)+6 \ mol\times { \triangle }_{ f }G^{ \ominus }{ H }_{ 2 }O(l)\} \)
\(-[4mol\times 86.7 \ KJ \ mol^{ -1 })+6\times (-237.2 \ KJ \ mol^{ -1 })]\)
\(-[4 \ mol\times (-16.8 \ KJ \ mol^{ -1 })+5\times zero]\)
\(=(364.8 \ KJ \ -1423.2 \ KJ)+67.2 \ KJ=-1009.2 \ KJ\)
since \({ \triangle }G^{ \ominus }\) is negative, the reaction is feasible in the forward direction.

(i) Total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (p _i ,V _i) to ( p _f ,V _f). Reversible work is represented by the combined areas ABC and BCV_iV_f.

(ii) Work against constant pressure, p_f is represented by the area BCV_iV_f .Work(i) > Work (ii) 

2 NaHCO₃(s) \(\longrightarrow \) Na₂CO₃ (s) + CO₂(g) + H₂O(g)
\({ \triangle }_{ r }{ H }^{ o }={ \triangle }_{ f }{ H }^{ o }({ Na }_{ 2 }{ CO }_{ 3 })+{ \triangle }_{ f }{ H }^{ o }({ CO }_{ 2 })+{ \triangle }_{ r }{ H }^{ o }({ H }_{ 2 }O)-2{ \triangle }_{ f }{ H }^{ o }({ S }_{ m }^{ o }(NaHCO_{ 3 })\)
\(=-1130+(-393.51)+(-241.82)-2X(-947.7)\)
\( =-1766.23+1895.4=129.17KJ{ mol }^{ -1 }\)
\( { \triangle }_{ r }{ S }^{ o }={ \triangle }_{ r }{ S }_{ m }^{ o }({ Na }_{ 2 }{ CO }_{ 3 })+{ S }_{ m }^{ o }({ CO }_{ 2 })+{ S }_{ m }^{ o }({ H }_{ 2 }O)-2{ S }_{ m }^{ o }(NaHCO_{ 3 })\)
\(=136.0+188.83+231.74-2\times102.1\)
\( =538.57-204.2=334.37J{ K }^{ -1 }{ mol }^{ -1 }\)
From second law of thermodynamics \({ \triangle }_{ r }{ S }^{ o }=\frac { { \triangle }_{ r }{ H }^{ o } }{ T } \)
 \(\therefore T=\frac { { \triangle }_{ r }{ H }^{ o } }{ { \triangle }_{ r }{ S }^{ o } } =\frac { 129.17 }{ 334.37\times{ 10 }^{ -3 } } =386.3 \ K\)
Reaction will be spontaneous above 386.3 K.

A heat capacity of water = heat capacity of calorimeter, the heat gained by water = heat lost by calorimeter
\(=125\times (296.5-286.4)\times 4.184 \ J=5282J=5.282kJ\)

\(\text{ we know } \triangle _{ r }G^{ \ominus }=-2.303 \ RT \ \log { { K }_{ P } }\) \(and R=8.314 \ JK^{ -1 }mol^{ -1 }\)
\(\text{ therefore,}\triangle _{ r }G^{ \ominus }=-2.303(831JK^{ -1 }mol^{ -1 })\times (298K)(\log { 2.47\times { 10 }^{ -29 }) }\)
= 163000 J mol^–1
= 163 kJ mol^–1.