The formation reaction of benezene is given by :
\(6 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(1) ;\) \(\Delta_{f} H^{\ominus}=?\) ......(i)
The enthalpy of combustion of 1 mol of benzene is :
\(\mathrm{C}_{6} \mathrm{H}_{6}(1)+\frac{15}{2} \mathrm{O}_{2} \rightarrow 6 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(1)\);
\(\Delta_{C} H^{\ominus}=-3267 \mathrm{~kJ} \mathrm{~mol}^{-1}\) ......(ii)
The enthalpy of formation of 1 mol of CO₂(g) :
\(\mathrm{C}(\text { graphite })+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\) \(\Delta_{f} H^{\ominus}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) ........(iii)
The enthalpy of formation of 1 mol of H₂O(1) is :
\(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(1)\); \(\Delta_{C} H^{\ominus}=-285.83 \mathrm{~kJ} \mathrm{~mol}^{-1}\) ........(iv)
multiplying eqn. (iii) by 6 and eqn. (iv) by 3 we get:
\( 6 \mathrm{C}(\text { graphite })+6 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_{2}(\mathrm{~g}) ; \Delta_{f} H^{\ominus} =-2361 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
\( 3 \mathrm{H}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{H}_{2} \mathrm{O}(1) \Delta_{f} H^{\ominus}=-857.49 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
Summing up the above two equations :
\( 6 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(\mathrm{~g})+\frac{15}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_{2}(\mathrm{~g}) +3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \)
\(\Delta_{f} H^{\ominus}=-3218.49 \mathrm{~kJ} \mathrm{~mol}^{-1}\) .........(v)
Reversing equation (ii);
\(6 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(1)+\frac{15}{2} \mathrm{O}_{2}\)
\(\Delta_{f} H^{\ominus}=-3267.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) .........(vi)
Adding equations (v) and (vi), we get
\(6 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(1) \Delta_{c} H^{\ominus}=-48.51 \mathrm{~kJ} \mathrm{~mol}^{-1} \) ......(iv)

The change take place as follows:
Step - 1  : 1 mol H₂O (1, 100°C) ⟶ 1 mol (1, 0°C) Enthalpy change ΔH₁
Step - 2  : 1 mol H₂O (1, 0°C) ⟶1 mol H₂O( S, 0°C) Enthalpy change ΔH₂
Total enthalpy change will be - \(\Delta \mathrm{H}=\Delta \mathrm{H}_{1}+\Delta \mathrm{H}_{2}\)
\(\Delta \mathrm{H}_{1}=-(18 \times 4.2 \times 100) \mathrm{J} \mathrm{mol}^{-1}\)
= - 7560 J mol^-1 = - 7.56 k J mol^-1
\(\Delta \mathrm{H}_{2}\) =  - 6.00 kJ mol^-1
Therefore,
Δ H = - 7.56 kJ mol^-1 + (-6.00 kJ mol^-1)
= -13.56 kJ mol^-1
There is negligible change in the volume during the change form liquid to solid state.
Therefore, pΔv = Δng RT = 0
ΔH = ΔU = - 13.56kJ mol^-1

\((i) \ CCI_{ 4 }(l)\rightarrow CCI_{ 4 }(g);\Delta _{ vap }{ H }^{ o }=+30.5 \ kJ \ mol^{ -1 }\)
\( (ii) \ C(s)+2CI_{ 2 }(g)\rightarrow CCI_{ 4 }(l);\Delta _{ f }{ H }^{ o }=-135.5 \ kJ \ mol^{ -1 }\)
\((iii) \ C(s)\rightarrow C(g);\Delta _{ a }{ H }^{ o }=715.0 \ kJ \ mol^{ -1 }\)
\( (iv) \ CI_{ 2 }(g)\rightarrow 2Ci(g);\Delta _{ a }{ H }^{ o }=242 \ kJ \ mol^{ -1 }\)
Multiplying Eq. (iv) by 2, we get
\( (v) \ 2CI_{ 2 }(g)\rightarrow 4CI(g);\Delta _{ a }{ H }^{ o }= \ 484.0 \ kJ \ mol^{ -1 }\)
Adding Eqs. (iii) and (v), we get
\((vi) \ C(s)+2CI_{ 2 }(g)\rightarrow C(g)+4CI(g);\Delta { H }= \ 1199 \ kJ \ mol^{ -1 }\)
Reversing Eqs. (i) and (ii), we get
\((vii) \ CCI_{ 4 }(g)\rightarrow CCI_{ 4 }(l);\Delta { H }=-30.5 \ kJ \ mol^{ -1 }\)
\( (viii) \ CCI_{ 4 }(l)\rightarrow C(s)+2CI_{ 2 }(g);\Delta { H }=+135.5 \ kJ \ mol^{ -1 }\)
Adding Eqs. (vi),(vii) and (viii), we get
\( CCI_{ 4 }(g)\rightarrow C(g)+4CI(g);\Delta { H }=1304 \ kJ \ mol^{ -1 }\)
Bond enthalpy of C−Ci bond in
\( CCI_{ 4 }=\frac { 1304 }{ 4 } =326 \ kJ \ mol^{ -1 }\)
(\(\because\) There are four C-CI bonds in CCI₄, )
\(\quad \quad CI\\ \quad \quad \quad |\\ CI-C-CL\\ \quad \quad \quad |\\ \quad \quad CI\)

\(N_{ 2 }{ O }_{ 4 }(g)\leftrightharpoons 2NO_{ 2 }(g)\)
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,
\(\Delta _{ r }G^{ \circ }=-RT \ ln \ K_{ p }\)
\( \Delta _{ r }G^{ \circ }=(-8.314 \ JK^{ - } \ mol^{ - })\times (333K)\times (2.303)\times (0.1239)\)
\( =-763.8 \ kJmol^{ -1 }\)

One decides the spontaneity of a reaction by considering
\(\triangle { S }_{ total }(\triangle { S }_{ sys }+\triangle { S }_{ surr })\). For calculating ∆Ssurr, we have to consider the heat absorbed by the surroundings which is equal to – \({ \triangle }_{ r }{ H }^{ \circleddash }\). At temperature T, entropy change of the surroundings is
\(\triangle { S }_{ surr }=\frac { { \triangle }_{ r }{ H }^{ \circleddash } }{ T } \)  (At constant pressure)
\( =\frac { (-1648\times { 10 }^{ 3 }J{ mol }^{ -1 } }{ 298K } \)
= 5530.20 JK^-1mol^-1
Thus, total entropy change for this reaction
\(\triangle { S }_{ total }=5530 \ J{ K }^{ -1 }{ mol }^{ -1 }+(-549.4 \ J{ K }^{ -1 }{ mol }^{ -1 })\)
\(=4980.6\ J{ K }^{ -1 }{ mol }^{ -1 }\)
This shows that the above reaction is spontaneous.

(i) The change H₂O (l) → H₂O (g)
∆H = ∆U + ∆ngRT or ∆U = ∆H – ∆ng RT, substituting the values,
we get ∆U = 41.00 kJ mol^–1 – 1 × 8.3 J mol^–1 K^–1 × 373 K
= 41.00 kJ mol^-1 – 3.096 kJ mol^-1 = 37.904 kJ mol^–1

We can represent the process of evaporation as
\({ H }_{ 2 }O(l)\overset { Vaporisation }{ \longrightarrow } { H }_{ 2 }O(g)\)
No. of moles in 18 g H₂O(l) is
\(=\frac { 18 \ g }{ 18 \ g \ { mol }^{ -1 } } =1 \ mol\)
Heat supplied to evaporate18g water at 298 K = n × \({ \Delta }_{ vap }{ H }^{ \Theta }\)
= (1 mol) × (44.01 kJ mol^–1) = 44.01 kJ
(assuming steam behaving as an ideal gas).
∆_vapU = \({ \Delta }_{ vap }{ H }^{ \Theta }\) – p∆V = \({ \Delta }_{ vap }{ H }^{ \Theta }\) – ∆ng RT
∆_vapH^V – ∆ng RT = 44.01 kJ –(1)(8.314 JK^–1mol^–1)(298K)(10–3kJ J^–1)
∆vapUV = 44.01 kJ – 2.48kJ = 41.53 kJ

Use the following steps to solve out such problems.
Step I   Write the balanced equation
\({ NH }_{ 2 }CN(s)+\frac { 3 }{ 2 } { O }_{ 2 }(g)\rightarrow { N }_{ 2 }(g)+{ CO }_{ 2 }(g)+{ H }_{ 2 }O(l)\)
Step II  Calculate \({ \Delta n }_{ g }\)
Difference of moles of gaseous products and
reactants, \({ \Delta n }_{ g }={ n }_{ p }-{ n }_{ r }=2-\frac { 3 }{ 2 } =\frac { 1 }{ 2 } =0.5mol\)
Step III Calculate \(\Delta H\) by using the formula
\(\Delta H=\Delta U+{ \Delta n }_{ g }RT\)
\(\Delta H=-742.7 \ kj \ { mol }^{ -1 }\)
\( +(0.5mol\times 8.314\times { 10 }^{ -3 }kj \ { mol }^{ -1 }\times 298K)\)
During calculation always remember the units of different quantities, i.e. 
\(\Delta H,\Delta U\) and R must be the same.

Given, q = +701 J (heat is absorbed, hence q is positive).
W = -394 J (work is done by the system, hence W is negative). By first law of thermodynamics;
Internal energy change,
\(\Delta U=q+W\)
\( =+701J+(-394J)=+307J\)
Hence, internal energy of the system increases by 307J.

Suppose q is the quantity of heat from the reaction mixture and C_V is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter.
q = C_V × ∆T
Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
q = –C_V × ∆T = – 20.7 kJ/K × (299 – 298) K = – 20.7 kJ
(Here, negative sign indicates the exothermic nature of the reaction) Thus, ∆U for the combustion of the 1g of graphite = – 20.7 kJK^–1 For combustion of 1 mol of graphite,
\(=\frac { 12.0gmol^{ -1 }\times (-20.7kJ) }{ 1g }\)
= – 2.48 ×102 kJ mol^–1 , Since ∆ n_g = 0, ∆ H = ∆ U = – 2.48 ×102 kJ mol^–1

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

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
\(C(s)+2H_{ 2 }(g)+\frac { 1 }{ 2 } O_{ 2 }(g)\rightarrow CH_{ 3 }OH(l);\Delta _{ f }H^{ 0 }=-965+726 \ kJ \ mol^{ -1 }\)

For the given reaction,
2A(g)+B(g)→2D(g)
Δng=2−(3)=–1 mole
Substituting the value of ΔUθ the expression of Δ H:
Δ H^θ=Δ U^θ+ ΔngRT
=(−10.5kJ)+(−1)(8.314×10⁻³kJ K⁻¹mol⁻¹)(298 K)
=−10.5 kJ⁻².48 kJ
ΔH^θ=−12.98 kJ
Substituting the value of ΔH^θ and ΔS^θ in the expression of ΔG^θ:
ΔG^θ=ΔH^θ=TΔS^θ
=−12.98 kJ−(298K)(−44.1JK⁻¹)
=−12.98 kJ+13.14 kJΔ G^θ= +0.16kJ
Since Δ Gθ for the reaction is positive, the reaction will not occur spontaneously.

We have q = – w = p_ex (10 – 2) = 0(8) = 0 No work is done; no heat is absorbed.

(i) ∆ U = w_ad, wall is adiabatic
(ii) ∆ U = – q, thermally conducting walls
(iii) ∆ U = q – w, closed system

We have q = – w = 2.303 nRT log V_f/V_s
= 2.303 x 1 x 0.8206 x 298 x log 10/2
= 2.303 x 0.8206 x 298 x log 5
= 2.303 x 0.8206 x 298 x 0.6990
= 393.66 L atm

We have q = – w = p_ex (8) = 8 litre-atm

\(\Delta G^{ θ }=-2.303 \ RTlogK_{ c }\)
Given, R = 8.314 JK^-1 mol^-1 T = 300K, K_c = 10
\(\Delta G^{ θ }=-2.303\times 8.314 \ JK^{ -1 }mol^{ -1 }\times \ 300K \ \times log \ 10\)
\( =-5744.14 \ J \ mol^{ -1 } \ \ [\therefore log10=1]\)

Gibbs free energy,
\(\Delta G=\Delta H-T\Delta S\)
0 = 400 kJmol^-1-T x 0.2kJK^-1mol^-1
Temperature, \(T=\frac { 400kJmol^{ -1 } }{ 0.2kJK^{ -1 }mol^{ -1 } } =2000 \ K\)
Therefore, above 2000 K, the reaction will become spontaneous.

We know,
\(\log \ K=\frac { -\Delta _{ r }G^{ \circ } }{ 2.303RT } =\frac { (-13.6 \times 10^{3}J \ mol^{-1}) }{ 2.303( 8.314 JK^{-1} mol^{-1}) (298 K) } =2.38\)
Hence, K = antilog 2.38 = 2.4 x 10²

(i) After freezing, the molecules attain an ordered state and therefore, entropy decreases.
(ii) At 0 K, the contituent particles are static and entropy is minimum. If temperature is raised to 115 K, these begin to move and oscillate about their equilibrium positions in the lattice and system becomes more disordered, therefore entropy increases.
(iii) Reactant, NaHCO₃ is a solid and it has low entropy. Among products there are one solid and two gases. Therefore, the products represent a condition of higher entropy.
(iv) Here one molecule gives two atoms i.e., number of particles increases leading to more disordered state. Two moles of H atoms have higher entropy than one mole of dihydrogen molecule.

Heat of reaction,
\({ \Delta }_{ r }{ H }^{ o }=\Sigma { \Delta }_{ f }{ H }_{ products }^{ o }-\Sigma { \Delta }_{ f }{ H }_{ reactants }^{ o }\)
\(= \left[ { \Delta }_{ f }{ H }^{ o }({ N }_{ 2 }O)+3{ \Delta }_{ f }{ H }^{ o }({ CO }_{ 2 }) \right] -\left[ { \Delta }_{ f }{ H }^{ o }({ N }_{ 2 }O_{ 4 })+3{ \Delta }_{ f }{ H }^{ o }({ CO }) \right]\)
\( = \left[ 81+(3\times -393) \right] -\left[ 9.7+(3\times -110) \right] kJ\)
\( = -777.7 \ kJ \ \approx \ -778 \ kJ\)

Enthalpy change for the formation of 1 mole of H₂O(l),
\(H_{ 2 }(g)+\frac { 1 }{ 2 } O_{ 2 }(g)\rightarrow H_{ 2 }O(l);\Delta _{ f }H^{ \circ }=-286kJ \ mol^{ -1 }\)
Energy released in the above reaction, is absorbed by the surroundings.
It means
 \(q_{ surr }=+286kJ \ mol^{ -1 }\)
\(\Delta S=\frac { q_{ surr } }{ T } =\frac { +286kJ \ mol^{ -1 } }{ 298 \ K } \)
\( =0.9597kJ \ K^{ -1 }mol^{ -1 }\)
\(=959.7 \ JK^{ -1 }mol^{ -1 }\)

For an isolated system, \(\Delta U=0\) and for a spontaneous process, total entropy change must be positive. For example, consider the diffusion of two gases A and B into each other in a closed container which is isolated from the surroundings. The two gases A and B are separated by a movable partition. When partition is removed, the gases begin to diffuse into each other and the system becomes more disordered. It shows that \(\Delta S>0\) and \(\Delta U=0\) for this process.
Moreover, 
\(\Delta S=\frac { q_{ rev } }{ T } =\frac { \Delta H }{ T }\)
\( \Delta S=\frac { \Delta U+p\Delta V }{ T } =\frac { p\Delta V }{ T } (\because \Delta U=0)\)

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

Given, \({ N }_{ 2 }(g)+{ 3H }_{ 2 }(g)\longrightarrow 2{ NH }_{ 3 }(g); \ { \Delta }_{ r }{ H }^{ o }\)
= -92.4 kJ mol^-1 
Chemical reaction for the enthalpy of formation of NH₃ (g) is as follows.
\(\frac { 1 }{ 2 } { N }_{ 2 }(g)+\frac { 3 }{ 2 } { H }_{ 2 }(g)\longrightarrow { NH }_{ 3 }(g)\)
Therefore, \({ \Delta }_{ f }{ H }^{ o }=\frac { -92.4 }{ 2 } =-46.2 \ kJ\ { mol }^{ -1 }\)

In the given reaction, a molecule of Cl₂ is formed from its two gaseous atoms and the energy is released with the formation of bond. Hence, ΔH is -ve. In this reaction, randomness (entropy) also decreases because 2 mole atoms of Cl have more randomness than one mole molecule of chlorine. Hence, ΔS is -ve.

The positive value of Δ_rH indicates that heat is absorbed during the formation of NO_(g). This means that NO_(g) has higher energy than the reactants (N₂ and O₂). Hence, NO_(g) is unstable.
The negative value of Δ_rH indicates that heat is evolved during the formation of NO_2(g) from NO_(g) and O_2(g). The product, NO_2(g) is stabilized with minimum energy.
Hence, unstable NO_(g) changes to stable NO_2(g).

The reaction for the combustion of carbon into CO₂ is
C(s) + O₂(g) ⟶ CO₂(g); △H = -393.5 KJ mol^-1
Heat released in the formation of 44 g CO₂ = 393.5 KJ
∴∴ Heat released in the formation of 35.2 g CO₂
\(=\frac{393 KJ × 35.2 g}{44 g} \)= 314.8 KJ