\(when \ the \ reaction \ is \ carried \ \ out \ at \ 0^{ \circ }K\)
\( or \ \triangle S=0\)
\((b)\triangle G=\triangle H\)
\( { H }_{ 2 }O(s)\rightleftharpoons { H }_{ 2 }O(l)\)
\({ \triangle }_{ fus }H=6.0 \ KJ \ mol^{ -1 }\)
\(=6000 \ J \ mol^{ -1 }\)
\( { T }_{ f }=0^{ \circ }C=(0+273)=273K\)
\( { \triangle }_{ fus }S=\frac { { \triangle }_{ fus }H }{ { T }_{ f } } =\frac { 6000J{ mol }^{ -1 } }{ 273K } 21.98J{ K }^{ -1 }{ mol }^{ -1 }\)

(a)\(q+w=\triangle U\)
As \(\triangle U\) is a state function hence, q + w is a state function.
(b) A real crystal has some disorder due to the presence of defects in its structural arrangement whereas ideal crystal does not have any disorder. Hence, a real crystal has more entropy than an ideal crystal.

\(C(s)+2H_{ 2 }(g)\rightarrow { CH }_{ 4 }(g) \ { \triangle }_{ r }H=-74.8KJ \ ......(i)\)
\(C(s)\rightarrow C(g) \ { \triangle }_{ r }H^{ \ominus }=+719.6 \ KJ \ ......(ii)\)
\({ H }_{ 2 }(g)\rightarrow 2H(g) \ { \triangle }_{ r }H^{ \ominus }=+435.4 \ KJ .....(iii)\)
\( C(s)+2H_{ 2 }(g)\rightarrow C(s)+4H(g) \ .......(iv)\)
\(C(s)+2H_{ 2 }(g)\rightarrow { CH }_{ 4 }(g) \ ...........(v)\)
\( subtract \ (v)from \ (iv)0=C(s)+4H(g)-{ CH }_{ 4 }(g)\)
\( { \triangle }_{ r }H^{ \ominus }=719.6+2(435.4)-(-74.8)\)
\( { CH }_{ 4 }(g)=C(s)+4H(g)\)
\(\triangle H=+1665.2 \ KJ\)
This gives the enthapy of dissociation of four moles of C-H bonds(called enthalpy of atomisation)
Hence bond energy for C - H bond = \(\frac { 1665.2 }{ 4 } =416.3 \ KJ \ mol^{ -1 }\)

The chemical equation for the reaction is:
H2(g) + Cl(g) \(\longrightarrow \) 2HCL(g)
The enthalpy of reaction is:
\(\triangle _{ r }H=\sum { B.E.Of\quad reactants } -\sum { B.E \ Of \ products } \)
\(=\left[ B.E.of \ H-H \ bond \ +B.E \ of \ Cl-Cl \ bond \right] -\left[ 2\times B.E \ of \ H-Cl \ bond \right] \)
= (433 - 244) - 2 x 431) = 433 + 244 - 862
= -185 KJ

The equation representing the enthalpy of formation of methane is:
\(C(s)_{ 2 }(g)\longrightarrow CH_{ 4 }(g); \ \triangle HH=-78.84 \ KJ\)
\(\triangle H=78.84\quad KJ; \ \triangle ^{ ng }=1-2=-1 \ mol\)
\(R=8.314\times { 10 }^{ -3 }KJ \ { K }^{ -1 }mol^{ -1 },T=300K\)
According to the relation, \(\triangle H=\triangle U+\triangle ^{ ng }RT\)
\(\triangle U=\triangle H+\triangle ^{ ng }RT\)
= (-78.84 KJ) - (1 mol) x 8.314 x 10^-3 kJ K^-1 mol^-1) x 300K
= - 78.84 - 2.49 = - 81.35 kJ

The chemical equation representing the oxidation of liquid benzene is :
C₆H₆(l) + \(\frac { 15 }{ 2 } { O }_{ 2 }(g)\longrightarrow 6CO_{ 2 }(g)+3H_{ 2 }O(l)\)
\(\triangle H=-3728KJ; \ { \triangle }^{ ng }=6-\frac { 15 }{ 2 } =-\frac { 3 }{ 2 } mol\)
\(R=8.314\times { 10 }^{ -3 }KJ \ { K }^{ -1 }mol^{ -1 },T=300K\)
According to the relation, \(\triangle H=\triangle U+\triangle ^{ ng }RT\)
\((-3728 \ KJ) \ =\triangle U+\left( -\frac { 3 }{ 2 } mol \right) \times (8.314\times { 10 }^{ -3 } \ KJ \ { K }^{ -1 }mol^{ -1 })\times (300 \ K)\)
\((-3278 \ KJ) \ =\triangle U-3.7413 \ KJ\)
\(\triangle U=-3278+3.7413=-3724.2587 \ KJ\)

N₂O₄(g) + 3co(g) \(\longrightarrow \) N₂O(g) + 3CO₂(g)
Given that ; \(\triangle _{ f }\) HCO (g) = -110 KJ mol^-1; \(\triangle _{ f }\)HCO₂(g) = -393 KJ  mol^-1
\(\triangle _{ f }\)HN₂O(g) = 81 KJ mol^-1 \(\triangle _{ f }\) HN₂O₄(g) = 9.4 KJ  mol^-1
Enthalpy of reaction (\(\triangle _{ f }\)H) = [81 + 3(-393)] - [9.7 + 3(-110)] = [81 - 1179] - [9.7-330] = 778  -KJ  mol^-1

The change may be represented as:
H₂O (l) (10^{\(\circ \)} C) \(\underrightarrow { \triangle H } \) H2O(s) (-10^{\(\circ \)} C)
\(\downarrow \) \(\triangle { H }_{ 1 }\)                         \(\downarrow \)\(\triangle { H }_{ 3 }\)
H₂O (l) (0^{\(\circ \)} C) \(\underrightarrow { \triangle { H }_{ 2 } } \) H₂O(s) (0^{\(\circ \)} C)
According to Hess's Law;
\(\triangle \)H = \(\triangle \)H₁ +\(\triangle \)H₂ +\(\triangle \)H₃ 
\(\triangle \)H₁ = 75.3J mol^-1 
\(\triangle \)H₂ (solidification) = -6.03 KJ mol^-1 K^-1 (10K) = 753 J mol^-1 
(sign changed)
\(\triangle \)H₃ = 36.8 J  mol^-1 K^-1 (-10K) = -368 J  mol^-1 
\(\triangle \)H = (753 - 6030-368) J  mol^-1 = -5645 J mol^-1
\(\therefore \) = - 5.645 KJ mol^-1

Required reaction for the formation of methanol is as follows.
\(C(s)+2H_{ 2 }(g)+\frac { 1 }{ 2 } O_{ 2 }(g)\rightarrow CH_{ 3 }OH(l);\Delta _{ f }H^{ 0 }=?\)
Multiplying Eq. (iii) by 2 we have
\( 2H_{ 2 }(g)+O_{ 2 }(g)\rightarrow 2H_{ 2 }O(l);\Delta _{ f }H^{ 0 }=-572 \ kJ \ mol^{ -1 }\)
Summing up the Eqs. (ii)and (iv), we get
\( C(s)+2H_{ 2 }(g)+20_{ 2 }(g)\rightarrow CO_{ 2 }(g)+2H_{ 2 }O(l);\Delta _{ f }H^{ 0 }=-965 \ kJ \ mol^{ -1 }\)
Reversing Eq. (i), we get
\( CO_{ 2 }(g)+2H_{ 2 }O(g)\rightarrow CH_{ 3 }OH(l)+\frac { 3 }{ 2 } O_{ 2 }(g);\Delta _{ r }H^{ 0 }=+726 \ kJ \ mol^{ -1 }\)
Adding Eqs. (v) and(vi) we get the required equation
\(C(s)+2H_{ 2 }(g)+\frac { 1 }{ 2 } O_{ 2 }(g)\rightarrow CH_{ 3 }OH(l);\Delta _{ f }H^{ 0 }=-965+726 \ kJ \ mol^{ -1 }\)

(i) \(\triangle \)G = -2.303 RT log K. Thus, when \(\triangle \)G^o < 0, K > 1.
(ii) Under ordinary conditions, the average energy of the reactants may be less than threshold energy. They require some activation energy to initiate the reaction.
(iii) △G = △H - T△S. At low temperature, T△S is small. Hence, △H dominates. At high temperature, T△S is large, i.e. △S dominates the value of △G.

\(\mathrm{NH}_4 \mathrm{NO}_3(s) \longrightarrow \mathrm{N}_2 \mathrm{O}(g)+2 \mathrm{H}_2 \mathrm{O}(l) \)
\( \Delta H_{\text {Reaction }}=\Delta H_{\text {Products }}-\Delta H_{\text {Reactants }} \)
\( =\Delta H_{\mathrm{N}_2 \mathrm{O}}+2 \times \Delta H_{\mathrm{H}_2 \mathrm{O}}-\Delta \mathrm{H}_{\mathrm{NH}_4 \mathrm{NO}_3} \)
\( =81.46+2(-285.8)-(-367.57) \)
\( \Delta H=-122.560 \mathrm{~kJ} \)
\( \Delta H=\Delta E+\Delta n R T \)
\( -122560=\Delta E+1 \times 8.314 \times 298 \)
\( \Delta E=-125037 \mathrm{~J} \)
\(=-125.037 \mathrm{~kJ}\)

\(3 C (graphite) + 4H2(g) -> C3H8 (g)\)
\(\triangle _{ r }S=\sum { { s }_{ m }^{ o } } (products)-{ S }_{ m }^{ o }(reactants){ 3 }{ H }_{ 8 }(g)-{ 3S }_{ m }^{ o }[C(graphite)]+{ 4S }_{ m }^{ o }[{ H }_{ 2 }]\)
\(\triangle _{ r }S={ S }_{ m }^{ o }[{ C }]\)
= (270.2 - 3 x 5.70 - 4 x 130.7)JK^-1mol^-1
= (270.2 - 17.10 - 522.80) JK^-1 mol^-1
= (270.2 - 539.90)
= -269.7 JK^-1mol-1
\(\triangle _{ r }{ H }^{ o }={ \triangle }_{ f }{ H }^{ o }product-{ \triangle }_{ f }{ H }^{ o }reactants\\ \triangle _{ r }{ H }^{ o }={ \triangle }_{ f }{ H }^{ o }({ C }_{ 3 }{ H }_{ 8 })-4{ \triangle }_{ f }{ H }^{ o }[{ H }_{ 2 }(g)]\)
\( -3{ \triangle }_{ f }{ H }^{ o }[C(graphite)]\)
\(\triangle _{ r }{ H }^{ o }=-103.8KJ{ mol }^{ -1 }\)
\(\triangle _{ r }{ G }^{ o }=\triangle _{ r }{ H }^{ o }-T\triangle _{ r }{ S }^{ o }\)
\( =\left( -103.8-\frac { 298X(-269.7) }{ 1000 } \right) KJ \ { mol }^{ -1 }\)
\(=(-103.80+80.370)KJ{ mol }^{ -1 }=-23.43KJ{ mol }^{ -1 } \)

For vaporization of diethyl ether
 \(\therefore \ \triangle _{ vap }{ S }^{ o }=\frac { \triangle _{ vap }{ H }^{ o } }{ T }\)
\( \triangle _{ vap }{ H }^{ o }=26.0kJ \ mo{ l }^{ -1 }, \ T=273+35=308K\)
\(\triangle _{ vap }{ S }^{ o }=\frac { 26.0\times{ 10 }^{ 3 }J{ mol }^{ -1 } }{ 308K } =84.4J{ K }^{ -1 }{ mol }^{ -1 }\)

Energy taken by a man = 10000 kJ
Change in internal energy per day = 12500 - 10000 = 2500 kJ
The energy is lost by the man as he expends more energy than he takes.Now 100g of sugar corresponds to energy = 1632 kJ loss in energy.
2000g of sugar corresponds to energy = \(\frac { 1632\times 2000 }{ 100 } \)
= 32640 kJ
Number of days required to lose 2000g of weight or 32640 kJ of energy = \(\frac { 32640 }{ 2500 } \) = 13days