Sol. No, there is no enthalpy change in a cyclic process because the system returns to the initial state.

Sol. The coffee held in a cup is an open system because it can exchange matter (water vapour) and energy (heat) with the surroundings.

Sol. The change in internal energy is a state function and it depends upon the temperature only. In the isothermal expansion the temperature remains constants. Therefore $\Delta E=0$ under isothermal conditions.

Sol. $\Delta H=\Delta E$ during a process which is carried out in a closed vessel $(\Delta v=0)$ or number of moles of gaseous products $=$ number of moles of gaseous reactants or the reaction does not involve any gaseous reactant or product.

Sol. $E=\frac{3}{2} R T$ Mono-atomic gas.

Sol. (i) If work is done on the system, internal energy will increase.

(ii) If work is done by the system, internal energy will decrease.

Sol. The heat released in the above two reactions will be different. It will be greater in reaction, (i) because when water (g) condense to form water (l), heat is released.

Sol. 18 times.

Sol. since $\Delta_{r} H^{o}$ is +ve i.e., enthalpy of formation of $N O$ is positive,

so $\mathrm{NO}(g)$ is unstable. But $\mathrm{NO}_{2}(g)$ is formed.

Sol. $\Delta G^{\circ}=-2.303 R T \log K .$ Hence, $\log k=0$ or $K=1$

Sol. – For a spontaneous process, $\Delta G<0 .$ Also entropy change (\DeltaS) during polymerization is negative. According to Gibbs Helmholtz equation,

$\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$

$\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative.

Sol. The reaction is spontaneous in the backward direction, therefore, $\Delta G$ is positive in the forward direction. Now, forward reaction is exothermic, therefore the entropy change for the forward direction should be negative and large $(T \Delta S>\Delta H)$

Sol. Order of increasing randomness

1 Mole of solid $x<1$ mole of liquid $x<1$ mole of gas

Sol.

Sol. $\Delta G=\Delta H-T \Delta S$ where $\Delta G, H$ and $\Delta S$ are free energy change, enthalpy change and entropy change respectively.

Sol. – The reaction will occur spontaneously only when $T \Delta S>\Delta H$

$\Delta G=\Delta H-T \Delta S=(+)-T(+)$

For $\Delta G$ to be negative, $T \Delta S$ must be $>\Delta H$

Sol. $-(1) \quad \Delta S=-v e$

(ii) $\quad \Delta S=+v e$

(iii) $\quad \Delta \mathrm{S}=-\mathrm{ve}$

Sol. (i) $\quad \Delta S<O$

(ii) $\quad \Delta S>O$

(iii) $\quad \Delta S>O$

(iv) $\quad \Delta S>O$

Sol. Negative

Sol. Open system : (i) Human being (ii) The earth (v) A satellite in orbit,

Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon.

Isolated system : (vi) Coffee in thermos flask

Sol. Internal energy : The energy of a thermodynamic system under given conditions is called internal energy. It is made up of kinetic and potential energy of constituent particles.

Internal energy change is measure at constant volume.

$\Delta U$ is measured in bomb calorimeter. Enthalpy is defined as heat content of the system $H=U+P V$

Enthalpy change is measured at constant pressure

$\Delta H=\Delta U+P \Delta V$

Sol. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g) \Delta H=-92.38 k_{\circlearrowright}$

$\Delta n=2-4=-2$

$\Delta H=\Delta U+\Delta n R T$

$-92.38 k J=\Delta U-2 \times 8.314 \times 10^{-3} k J \times 298 k$

$\Delta U=-92.38 k J+4.955=-87.425 k J m o l^{-1}$

Sol. (i) $\quad \Delta H=\Delta U+\Delta n R T$

(ii) $\quad \Delta n=1-3=-2$

Sol.

Sol. (i) Here, $q=0$

\[ \therefore \Delta U=q+w=0+w_{a d}=w_{a d} \]

As no heat is absorbed by the system, the wall is adiabatic.

(ii) Here, $w=0, q=-q$

As heat is taken out, the system must be having thermally conducting walls.

(iii) As work is done by the system on absorbing heat, it must be a closed system.

Sol. For an ideal gas, from kinetic theory of gases, the average kinetic energy per mole $\left(E_{k}\right)$ of the gas at any temperature $T K,$ is given by $E_{k}=3 / 2 R T$

At $(T+1) K,$ the kinetic energy per mole $\left(E_{k}\right)=3 / 2 R(T+1)$ Therefore, increase in the average kinetic energy of the gas for $1^{\circ} \mathrm{C}(\text { or } 1 \mathrm{K})$ rise in temperature $\Delta E_{k}=3 / 2 R(T+1)-3 / 2 R T=3 / 2 R$

When the gas is heated to raise its temperature by $1^{\circ} \mathrm{C},$ the increase in its internal energy is equal to the increase in kinetic energy, i.e., $\Delta U=\Delta E_{K}$

Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$

$\Rightarrow C_{v}=\Delta E_{K}$

Therefore, $C_{v}=3 / 2 R$

Sol. $q=c \times \Delta T, \quad c=n \times C_{m}$

Here, $C_{m}=24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; n=\frac{60}{27}=2.22 \mathrm{mol}$

\[ c=2.22 \mathrm{mol} \times 24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}=53.28 \mathrm{JK}^{-1} \]

Now, $q=53.28 \mathrm{JK}^{-1} \times \Delta T$

$=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ or $1.065 \mathrm{kJ}$

Sol. $q=m \times s \times\left(t_{2}-t_{1}\right)$

$q=125 g \times 4.18 J / g \times(286.4-296.5)$

$q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$

Sol. – Molar heat capacity $=$ specific heat capacity $\times$ molar mass

So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass.

Molar heat capacity of $L i(s)=3.57 \times 7=25.01 J \mathrm{mol}^{-1} K^{-1}$

Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$

Molar heat capacity of $K(s)=0.756 \times 39=29.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$

Molar heat capacity of $R b(s)=0.363 \times 85=30.88 J \mathrm{mol}^{-1} \mathrm{K}^{-1}$

Molar heat capacity of $C s(s)=0.242 \times 133=32.2 \mathrm{J} \mathrm{mol}^{-1}$

The trend is that there is continuous increase of molar heat capacity with increase in atomic mass. By plotting graph between molar heat capacity and atomic mass, the molar heat capacity of $F r$

(atomic mass $=223$ ) would be $33.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$

Sol. Average bond enthalpy is average heat required to break 1 mole of particular bond in various molecules (polyatomic). In diatomic molecule bond enthalpy has fixed value, $e . g . .$ in $H_{2}$ molecule $436 k J$ is bond enthalpy.

$416 \mathrm{kJ} \mathrm{mol}^{-1}$

$H-O-H \rightarrow H(g)+O H(g) ; \Delta H=498 k_{\mathrm{d}}$

$O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$

Average bond enthalpy $\Delta H(O-H)=\frac{498+430}{2}=464 k J$

Sol. $\Delta_{v a p} H^{\ominus}$ of $C O=+6.04 \mathrm{kJmol}^{-1}$

Molar mass of $C O=28 g \mathrm{mol}^{-1}$

Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$

Energy required for vapourising $2.38 g$ of $\mathrm{CO}$

$=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$

Sol. $C(s)+O_{2}(g) \rightarrow C O_{2}(g) \Delta H=-393.5 k_{0} J m o l^{-1}$

Heat released for the formation of $44 g(1 \mathrm{mol})$ of

$C O_{2}=-393.5 k_{U}$

Heat released for the formation of $35.2 g$ of $C O_{2}$

$\frac{-393.5 \times 35.2}{44}=-314.8 k J$

Sol. $\Delta H$ and $\Delta U$ are related as

$\Delta H=\Delta U+\Delta n_{g} R T$

For the reaction

$N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$

$\Delta n_{g}=2-(1+3)=-2 m o l, T=298 K$

$\Delta H-92.38 k J=-92380 J, R=8.314 J K^{-1} \mathrm{mol}^{-1}$

$-92380=\Delta U-2 \times 8.314 \times 298$

$-92380=\Delta U-4955$

$\Delta U=-92380+4955=-87425 J=-87.425 k J$

Sol. We know

$\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ or $\log K=\frac{-\Delta_{r} G}{2.303 R T}$

$\Delta_{r} G^{o}=-8.1 k J m o l^{-1}, T=1000 K$

$R=8.314 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}$

$\log K=-\frac{-8.1}{2.303 \times 8.314 \times 10^{-3} \times 1000}$ or $K=2.64$

Sol. $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{o}=-243 k J m o l^{-1}$

Molar mass of phosphorus $=30 \mathrm{gmol}^{-1}$

Mass of $P=10.32 \mathrm{g}$

Moles of $P=\frac{10.32}{31}=0.333 \mathrm{mol}$

Enthalpy change for 2 mole of $P=-243 \mathrm{kJ}$

Enthalpy change for 0.333 mole of $P=-\frac{243}{2} \times 0.333$

$=-40.46 \mathrm{kJ}$

Sol. Molar mass of benzene

$\left(C_{6} H_{6}\right)=(6 \times 12)+(6 \times 1)=78$

$\therefore$ Energy required to vapourise $100 g$ benzene

$=\frac{30.8}{78} \times 100=39.487 k_{U}=39487 J$

Given that power $=\frac{\text { Energy }}{\text { time }} \Rightarrow$ time $=\frac{\text { energy }}{\text { power }}$

Time $=\frac{39487 \mathrm{J}}{100 \mathrm{Js}^{-1}}=394.87 \mathrm{s}$

Sol. Molar mass of $C O=12+16=28$

$\Rightarrow \Delta H$ during vapourisation of $28 g=6.04 \mathrm{kJ}$

$\Delta H$ during vapourisation of $2.38 g \mathrm{CO}=\frac{6.04}{28} \times 2.38$

$=0.5134 \mathrm{kJ}$

Thus, enthalpy change $=513.4 \mathrm{J}$

Sol. $-\Delta H_{\text {reaction }}=-2.05 \times 10^{3} \mathrm{kJmol}$

$\Delta H=8 B_{C-H}+2 B_{C-C}+5 B_{O=O}-6 B_{C=O}-8 B_{O-H}$

$-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$

$-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$

$-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$

$-2050 \mathrm{kJ}=-4152 \mathrm{kJ}+5 \mathrm{B}_{\mathrm{O}=0}$

$5 B_{O=O}=-2050+4152=+2102$

$B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$

Sol.

Sol. (i) $\quad \frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g)$

\[ \Delta_{f} H^{o}=-92.8 k J m o l^{-1} \]

(ii) $\quad H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l(a q)$

\[ =\Delta_{d i s s} H^{o}=-75.2 k J m o l^{-1} \]

Adding eq. ( i ) and (ii)

$\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$

$\Delta H^{\circ}=-168.0 \mathrm{kJ} \mathrm{mol}^{-1}$

$\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$

$-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$

The enthalpy of formation of $H_{3} O^{+}(a q)$ in dilute solution may be

taken as zero i.e., $\Delta_{f} H\left[H_{3} O^{+}(a q)\right]=0$

$-168.0=0+\Delta_{f} H^{\circ}\left(\mathrm{C} l^{-}\right)-\left[\frac{1}{2} \times 0+\frac{1}{2} \times 0\right]$

$\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$

Sol. The enthalpy change for the reaction:

$\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$

$\Delta H^{\circ}=\Delta_{a} H^{\circ}(C)+4 \Delta_{a} H^{\circ}(C l)-\Delta_{f} H^{\circ}\left(C C l_{4}\right)(g)$

$\Delta_{a} H^{o}(C l)=715.0 \mathrm{kJ} \mathrm{mol}^{-1}$

$\Delta_{a} H^{o}(C l)=\frac{1}{2} \times 242=121 k J m o l^{-1}$

Let us first calculate $\Delta_{f} H^{\circ} \mathrm{CCl}_{4}(g)$

$C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{\circ}=-135.5 k J m o l^{-1}$

$\mathrm{CCl}_{4}(l) \rightarrow \mathrm{CCl}_{4}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=30.5 \mathrm{kJ} \mathrm{mol}^{-1}$

Adding $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{o}=-105.0 \mathrm{kJ} \mathrm{mol}^{-1}$

$\therefore \Delta H^{\circ}=715.0+4 \times 121-(-105.0)=1304.0 \mathrm{kJ} \mathrm{mol}^{-1}$

This enthalpy change corresponds to breaking four $C-C l$ bonds

Bond enthalpy of $\mathrm{C}-\mathrm{Cl}$ bond $=\frac{1304.0}{4}$

Sol. We know

$\Delta_{r} G^{\circ}=\Delta_{r} H^{o}-T \Delta_{r} S^{\circ}$

$\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$

$\Delta_{r} S^{o}=197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}$

$\Delta_{r} G^{\circ}=491.18 k J \mathrm{mol}^{-1}-298 \mathrm{Kx}$

$\left(197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$

$=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$

Since Gibbs energy change is positive, therefore, at the reaction is not possible.

Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero

$\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$

$\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$

$=\frac{491.18 \mathrm{kJ} \mathrm{mol}^{-1}}{197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=2484.8 \mathrm{K}$

Therefore, the reaction will be spontaneous above $2484.8 \mathrm{K}$

$\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$

Sol. Heat capacity of water $=1 \mathrm{cal} g^{-1}=4.184 \mathrm{Jg}^{-1}$

$\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$

$\Delta T=286.4-296.5=-10.1 K$

$\therefore$ Heat change $(q)$ for calorimeter $=$ Heat capacity $\times \Delta T$

\[ \Rightarrow q=523 \times(-10.1)=-5282.3 \mathrm{J}=-5.282 \mathrm{kJ} \]

Thus, calorimeter loses $5.282 \mathrm{kJ}$ of heat during dissolution

of $20 g$ of $N H_{4} N O_{3}$ in $125 g$ water Molecular mass of

$N H_{4} N O_{3}$

$=(2 \times 14)+(4 \times 1)+(3 \times 16)=80$

Enthalpy of solution of $N H_{4} N O_{3}=\frac{5.282}{20} \times 80$

$=21.129 \mathrm{kJ} \mathrm{mol}^{-1}$

Sol. $\Delta T=300.78-294.05=6.73 K$

Heat transferred $=$ Heat capacity $\times \Delta T$

$=\left(8.93 \mathrm{kJ} K^{-1}\right) \times(6.73 \mathrm{K})=60.0989 \mathrm{kJ}$

Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$

$\therefore$ Internal energy change $(\Delta E)$ during combustion of one mole of

octane $=\frac{60.0989}{1.250} \times 114=5481.02 k J \mathrm{mol}^{-1}$

Reaction of combustion of octane:

$C_{8} H_{18}(l)+\frac{25}{2} O_{2}(g) \rightarrow 8 C O_{2}(g)+9 H_{2} O(l)$

$\Delta n_{g}=8-\frac{25}{2}=-\frac{9}{2}$

$\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$

$\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$

$=\left(5481.02 \times 10^{3}\right)+(-4.5 \times 8.314 \times 300.78)$

$=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}$

$=5492273.082 \mathrm{Jmol}^{-1}=5492.27 \mathrm{kJ} \mathrm{mol}^{-1}$

Enthalpy of combustion of octane

Sol. $\Delta G^{\circ}=-2.303 R T \log K_{p}$ or $\log K_{p}=\frac{-\Delta G^{\circ}}{2.303 \times R T}$

$=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$

$\Rightarrow K_{p}=$ antilog $0.4230=2.649$

Sol. $-$ (i) $\quad S=+v e$ because liquid changes to more disordered gaseous state.

(ii) $\quad \Delta S=+v e$ because aqueous solution has more disorder than solid.

(iii) $\quad \Delta S=-v e$ because gas is changing to less disorder solid.

(iv) because graphite has more disorder than diamond.

Sol. As there is little order in gases are compared to liquids, therefore, entropy of gas decreases enormously on

-condensation into a liquid. Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$

Therefore, the decrease in entropy when a gas condenses into a liquid is much more as compared to decrease in entropy when a liquid solidifies.

Sol. (i) Entropy increases due to more freedom of movement of

$I_{2}$ molecules upon dissolution.

(ii) Due to settling of solid $A g C l$ from solution, entropy decreases.

(iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. Thus, entropy increases.

Sol. $-\Delta H_{v a p}=26.0 \mathrm{kJ} \mathrm{mol}^{-1}=26000 \mathrm{J} \mathrm{mol}^{-1}$

$T_{b}=35+273=308 K$

$\Delta S_{v a p .}=\frac{\Delta H_{v a p .}}{T_{b}}=\frac{\left(26000 \mathrm{J} \mathrm{mol}^{-1}\right)}{(308 \mathrm{K})}=84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \mathrm{since}$

the condensation of diethyl ether is the reverse process, therefore,

$\Delta S_{\text {condensation }}=-84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$

Sol. $\Delta_{v a p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}, T_{b}=373 \mathrm{K}$

$\Delta_{v a p} S=\frac{\Delta_{v a p} H}{T_{b}}=\frac{40.63 \times 1000 \mathrm{J} \mathrm{mol}^{-1}}{373 \mathrm{K}}=109 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$

Entropy change for evaporation of $36 g$ of wate

$\frac{109}{18} \times 36=218 \mathrm{JK}^{-1}$

Sol. $\mathrm{SiH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SiO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$

$\Delta_{r} G^{\circ}=\Delta_{f} G^{\circ}\left(S i O_{2}\right)+2 \Delta_{f} G^{\circ}\left(H_{2} O\right)-\left[\Delta_{f} G^{\circ}\left(S i H_{4}\right)\right]+$

$2 \Delta_{f} G^{\circ}\left(O_{2}\right)$

$=-805.0+2(-228.6)-[+52.3+2(0)]$

$=-805.0-457.2-52.3=-1314.5 k J$

since $\Delta_{r} G^{\circ}$ is negative, the reaction will be spontaneous.

Sol. We know,

$\Delta_{r} G^{\circ}=-2.303 R T \log K$

$R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$

$\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$

$\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$

Sol. $\Delta G_{r}^{o}=\left[2 \Delta G_{f}^{o}\left(N O_{2}(g)\right)\right]-\left[2 \Delta G_{f}^{O}(N O(g))+\Delta G_{f}^{O}\left(O_{2}\right)\right]$

$=[2 \times 51.3]-[2 \times 86.55+0]$

$|=(102.6)-(173.10)=-70.50 k_{0} \mathrm{J} \mathrm{mol}^{-1}$

since, the value of $\Delta G_{r}^{\circ}$ is negative, therefore, the reaction is

spontaneous.

Sol. $\Delta G^{\circ}=-2.303 R T \log K$

$=-2.303 \times 8.314 \times 298 \times \log \left(1.8 \times 10^{-7}\right)=38484.4$

Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$

$\therefore \Delta S^{\circ}=\frac{\Delta H^{\circ}-\Delta G^{\circ}}{T}=\frac{28400-38484.4}{298}$

$=-33.84 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$

Sol. $\Delta S_{\text {Reaction }}=\Sigma S_{m(\text { products })}^{\circ}-\Sigma S_{m(\text { reactants })}^{\circ}$

$S_{m C_{3} H_{8}(g)}-\left[3 \times S_{m}^{\circ} C_{(g r a p h i t e)}+4 \times S_{m H_{2}(g)}^{\circ}\right]$

$=270.2-[(3 \times 5.74)-(4 \times 130.68)] J K^{-1} \mathrm{mol}^{-1}$

$=270.2-17.22-522.72=-269.72 J K^{-1} \mathrm{mol}^{-1}$

Sol. (i) The process, $2 \mathrm{Al}_{2} \mathrm{O}_{3} \longrightarrow 4 \mathrm{Al}+3 \mathrm{O}_{2}$ is non-spontaneous

because $\Delta G_{r}$ is $+22500 \mathrm{kJ} .$ When this process is coupled with

combustion of $C$ to $C O_{2} .$ The net free energy change is calculated

as

Hence it is non-spontaneous. Thus $A l_{2} O_{3}$ cannot be reduced by $C$

(ii) Let us now calculate the $\Delta G$ value for reduction of $P b O$

$2 P b O \longrightarrow 2 P b+O_{2} ; \Delta+120 k J$

$C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$

On adding, $2 P b O+C \longrightarrow 2 P b+C O_{2}$

$\Delta G=120-380=-260 k J$

Hence $P b O$ can bereduced by carbon because $\Delta \mathrm{G}$ of couple reaction is $-v e$

Sol. According to Gibbs Helmholtz equation,

$\Delta G=\Delta H-T \Delta S$

For, $\Delta G=0$

$0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or $T=\frac{\Delta H}{\Delta S}$

Here, $\Delta H=30.56 \mathrm{kJ} \mathrm{mol}^{-1}=30560 \mathrm{J} \mathrm{mol}^{-1}$

$\Delta S=66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$

$\therefore \quad T=\frac{30560}{66.0}=463 \mathrm{K}$

(i) At $463 K,$ the reaction will be at equilibrium because $\Delta G$ is

zero.

(ii) Below this temperature, $\Delta G$ will be $+$ve because both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are positive and $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})$

Therefore, the reaction will not be spontaneous below this temperature.

Sol. (i) $\quad \Delta G_{f}^{\circ}=\left[\Delta G_{f}^{\circ} C O_{2}(g)+2 \times \Delta G_{f}^{\circ} H_{2} O(g)\right]$

$-\left[\Delta G_{f}^{\circ} C H_{4}(g)+2 \Delta G_{f}^{\circ} O_{2}(g)\right]$

$=[-394.36+\{2 \times(-228.57)\}-[-50.72+0]$

$=-800.78 \mathrm{kJ} \mathrm{mol}^{-1}$

(ii) $\Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} O(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$

$\quad-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(\mathrm{s})+2 \Delta G_{f}^{\circ} \mathrm{H}^{+}(a q)\right]$

$=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$

$=+21.83 \mathrm{kJ} \mathrm{mol}^{-1}$

(ii) $\quad \Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} \mathrm{O}(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$

$-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(s)+2 \Delta G_{f}^{\circ} H^{+}(a q)\right]$

$=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$

$=+21.83 \mathrm{kJ} \mathrm{mol}^{-1}$

Sol. (i) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { product })-\Sigma S^{\circ}(\text { reactants })$

$=\left(2 S_{F e_{2} O_{3}}^{\circ}\right)-\left(4 S_{F e(s)}^{\circ}+3 S_{O_{2}}^{\circ}(g)\right)$

$=(2 \times 87.4)-(4 \times 27.28+3 \times 205.14)$

$=(174.8)-(109.12+615.42)$

$=(174.8)-(724.54)=-549.74 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$

(ii) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { products })-\Sigma S^{\circ}(\text { reactans })$

$=\left(S_{C a(O H)_{2}}^{\circ}(a q)+S_{H_{2}(g)}^{\circ}-\left(S_{C a(s))}^{\circ}+2 S_{H_{2} O(l)}^{\circ}\right)\right.$

$=\{-74.50+130.68\}-\{41.42+(2 \times 69.91)\}$

$\Rightarrow 56.18-(41.42+139.82) \Rightarrow-125.06 J K^{-1} \mathrm{mol}^{-1}$

Sol. $\Delta G=\Delta H-T \Delta S$

At equilibrium $\Delta G=0$ so that

$T=\frac{\Delta H}{\Delta S}=\frac{-10000 J m o l^{-1}}{-33.3 J K^{-1} m o l^{-1}}=300.3 \mathrm{K}$

(i) For spontaneity from left to right, $\Delta G$ should be $-v e$ for the given reaction. This will be so if $T<300.3 \mathrm{K}$

(ii) For reverse reaction to occur, should be tve for forward reaction. This will be so if

Sol. – The required equation for the formation of $C H_{3} O H(l)$ is :

$C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$

The given equations are:

(i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$

(i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$

$\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$

(ii) $\quad C(g)+O_{2}(g) \longrightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o r^{1}$

(iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$

Multiply eqn. (iii) by 2 and add to eqn. (ii)

(ii) $C_{( \text {graphite) } }+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 \mathrm{kJ} \mathrm{mol}^{-1}$

(iii) $\times 2: 2 H_{2}(g)+O_{2}(g) \rightarrow 2 H_{2} O(l) ; \Delta_{r} H^{o}=-572 k J m o r^{1}$

(iv) $\quad C(g)+2 H_{2}(g)+2 O_{2}(g) \longrightarrow C O_{2}(g)+2 H_{2}^{\circ} O$

$\Delta_{r} H^{\circ}=-965 \mathrm{kJmol}^{-1}$

Subtract eq. ( $i \text { ) from eq. ( } i v)$

(i) $\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(l)$

$\Delta_{r} H^{\circ}=-726 \mathrm{kJmol}^{-1}$

$C_{(G r a p h i t e)}+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$

$\Delta_{r} H^{\circ}=-239 k J m o l^{-1}$

Sol. (i) $\quad H_{2} O(l) \rightleftharpoons H_{2} O(g)$

since process occurs at constant $T$ and constant $P$ Thus heat change refers to $\Delta H$

$\therefore \quad \Delta H=+22.2 k_{0} J$

(ii) Calculation of $w$

Mol. of water vaporised $=\frac{10}{18}=0.56$

$\therefore W=-p_{e x} \Delta V \quad\left(\Delta V=\frac{n R T}{P_{e x t}}\right)$

$=-p_{e x} \frac{n R T}{P_{e x t}}=-n R T$

$=-0.56 \times 8.314 \times 10^{-3} \times 373.15=-1.737 k J$

(iii) Calculation of $\Delta U$

$\Delta H=\Delta U+P \Delta V$

$\Delta U=\Delta H-P \Delta V=\Delta H-n R T$

$=22.2-1.737=20.463 \mathrm{kJ}$

Sol. First method: by using the relation

$\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$

Here, we are given

Sol. $C_{p}=\left(1.0 \mathrm{cal} K^{-1} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=18.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$

$\Delta H_{v}=\left(5397 \operatorname{cal} g^{-1}\right)\left(180 g m o l^{1}\right)=97146 \mathrm{calmol}^{1}$

$\Delta H_{f}=\left(79.7 \mathrm{cal} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=1435 \mathrm{cal} \mathrm{mol}^{-1}$

The process consists of the following reversible steps :

(i) $\quad H_{2} \mathrm{O}_{( \text {steam) } }$ at $100^{\circ} \mathrm{C} \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$ at $100^{\circ} \mathrm{C}$

$\Delta S_{1}=-\frac{\Delta H_{v}}{T_{b}}=-\frac{9714.6 \mathrm{cal} \mathrm{mol}^{-1}}{373 \mathrm{K}}$

$-26.0445 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$

(ii) $\quad H_{2} O(l)$ at $100^{\circ} C \longrightarrow H_{2} O(l)$ at $0^{\circ} C$

$\Delta S_{2} \int_{T_{1}}^{T_{2}} C_{p(\text {liquid})} \frac{d T}{T}=C_{p(\text {liquid})} \ln \frac{T_{2}}{T_{1}}$

$=2.303\left(18 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) \log \frac{273}{373}$

$=-5.62 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$

(iii) $\quad H_{2} O(l)$ at $0^{\circ} C \rightarrow H_{2} O(s)$ at $0^{\circ} C$

$\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$