\(\triangle _{ r }{ G }^{ \ominus }\) = -RT InK, thus if \(\triangle { G }^{ \ominus }\) is less than zero i.e., it is negative, then InK will be positive and hence K will be greater than one.

For equilibrium reaction,
\(H_{ 2 }O(s)\leftrightharpoons H_{ 2 }O(l)\)
\(\Delta G=0\Rightarrow \Delta G=\Delta H-T\Delta S\)
\(0=\Delta H-T\Delta S\)
\( or \ \Delta S=\frac { \Delta H }{ T } =\frac { 6.06 \ kJmol^{ -1 } }{ 273 }\)
\(=0.02219 \ kJK^{ -1 }mol^{ -1 }=22.2J \ K^{ -1 }mol^{ -1 }\)

We are given
(i) \(C(s)+\frac { 1 }{ 2 } O_{ 2 }(g)\longrightarrow CO_{ 2 }(g); \ \Delta H=-110.5 \ kJ\)
(ii) \(H_{ 2 }(g)+\frac { 1 }{ 2 } O_{ 2 }(g)\longrightarrow H_{ 2 }O(g); \ \Delta H=-243.0 \ kJ\)
Subtracting Eq. (ii) from Eq. (i), we get
\(C(s)+H_{ 2 }O(g)\longrightarrow CO(g)+H(g);\)
\(\Delta H=+132.5 \ kJ\)

Haber's process of synthesis for ammonia is
\({ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightarrow 2N{ H }_{ 3 }(g)\)
\(\triangle { n }_{ g }=2-(1+3)=-2\)
\( But\quad \triangle H=\triangle U+\triangle { n }_{ g }RT\)
\( \therefore \ \triangle H=\triangle U-2RT\)

\(0.532 g C_6 H_6=0.532 / 78 \mathrm{~mole}=0.00682 \mathrm{~mole} \)
\( \therefore \Delta U=-\frac{22.3}{0.00682} \mathrm{kJmol}^{-1}=-3269.8 \mathrm{kJmol}^{-1}\)
Calculate \(\Delta \)H as in Solved Problem above.
\(\Delta \)H = -3269 KJ

CH₄ > C₂H₄ > C₄H₁₀

\(\triangle H=\triangle U+\triangle (pV)\)
For an ideal gas, \(pV=RT\)
\(\therefore \ \triangle H=\triangle U+\triangle (RT)=\triangle U+R\triangle T\)
Since T is constant, \(\triangle T=0\)
\(\therefore \ \triangle H=0\)

Given, \(\frac { 1 }{ 2 } { N }_{ 2 }(g)+{ \frac { 3 }{ 2 } }{ H }_{ 2 }(g)\longrightarrow { NH }_{ 3 }(g);\)
\({ \Delta }_{ f }{ H }^{ \Theta }=-91.8 \ kJ \ { mol }^{ -1 }\)
(\({ \Delta }_{ f }{ H }^{ \Theta }\) means enthalpy of formation of 1 mole of NH₃)
\(\therefore\) Enthalpy change for the formation of 2 moles of NH₃
\({ N }_{ 2 }(g)+{ 3H }_{ 2 }(g)\rightarrow { 2NH }_{ 3 }(g);\)
\({ \Delta }_{ f }{ H }^{ \Theta }=2\times -91.8=-183.6 \ kJ \ { mol }^{ -1 }\)
And for the reverse reaction. \(2{ NH }_{ 3 }(g)\longrightarrow { N }_{ 2 }(g)+{ 3H }_{ 2 }(g);{ \Delta }_{ f }{ H }^{ \Theta }=+183.6 \ kJ \ { mol }^{ -1 }\)  Hence, the value of \({ \Delta }_{ f }{ H }^{ \Theta }\) for NH₃ is +183.6 kJ mol^-1

At constant pressure \(q_{ p }=\triangle U+p\triangle V.\)
But \(\triangle U+p\triangle V=\triangle H\).
\(\therefore \ q_{ p }=\triangle H\). As \(\triangle H\) is a state function, therefore, \(q_{ p }\) is a state function.

18.0 g H₂O = 1 mol H₂O
Enthalpy change for vaporising 1 mole of H₂O = 40.79 KJ
\(\therefore \) Enthalpy change for vaporising 2 moles of 
H₂O = 2 x 40.79 KJ = 81.58 KJ
Standard enthalpy of vaporisation at 100^oC and 1 bar pressure,
\({ \triangle }_{ vap }{ H }^{ \circ }=+40.79 \ KJ \ { mol }^{ -1 }\)

Given, mass of Al = 60.0g
Molar mass of Al = 27g mol\(^{ -1 }\)
Molar heat capacity, C = 24Jmol\(^{ -1 }\) K\(^{ -1 }\)
\(\triangle T=55^{ \circ }C-35^{ \circ }C=20^{ \circ }C \ or \ 20K\)
\(Heat, q=n.C.\triangle T\)
\( q=\frac { 60 }{ 27 } \times 24Jmol​​​​^{ -1 }K^{ -1 }\times 20K\left( n=\frac { 60 }{ 27 } mol \right)\)
\( =1066.66J=1.067kJ\)

1 mole of CCI₄ = 154 g
Heat required for vapourising 154 g CCI₄ = 30.5 kJ
\(\therefore\) Heat required for vapourising
\(284 \ g \ CC{ I }_{ 4 }=\frac { 30.5\times 284 }{ 154 } kJ=56.25kJ\)

\(N_{ 2 }{ O }_{ 4 }(g)\leftrightharpoons 2NO_{ 2 }(g)\)
If N₂O₄ is 50% dissociated,d, the mole fraction of both the substances is given by
\(x_{ N_{ 2 }O_{ 4 } }=\frac { 1-0.5 }{ 1+0.5 } \Rightarrow x_{ NO_{ 2 } }=\frac { 2\times 0.5 }{ 1+0.5 }\)
\( p_{ N_{ 2 }O_{ 4 } }=\frac { 0.5 }{ 1.5 } \times 1atm,\ p_{ NO_{ 2 } }=\frac { 1 }{ 1.5 } \times 1atm\)
The equilibrium constant K_p is given by
\(K_{ p }=\frac { (p_{ NO_{ 2 } })^{ 2 } }{ p_{ { N }_{ 2 }{ O }_{ 4 } } } =\frac { 1.5 }{ (1.5)^{ 2 }(0.5) } =1.33atm\)
Since,
\(\Delta _{ r }G^{ \circ }=-RT \ ln \ K_{ p }\)
\( \Delta _{ r }G^{ \circ }=(-8.314 \ JK^{ - } \ mol^{ - })\times (333K)\times (2.303)\times (0.1239)\)
\( =-763.8 \ kJmol^{ -1 }\)