Thermodynamics

N₂(g, 1 atom) $\buildrel \, \over \longrightarrow $ N₂(g, 5 atom)

Here pressure increases. When pressure increases then the molecules of will come closer and intermoleculer distance decreases, so entropy will also decreases and $\Delta S\, < \,0$.

A₂ (g) $\rightleftharpoons$ 2 A (g)

Assume initially concentration of A₂ = [A₂] = 1 m

at equilibrium [A₂] = 1 $ \times $ ${{80} \over {100}}$ = 0.8 M

and 20% of [A₂] = 1 $ \times $ ${{20} \over {100}}$ = 0.2 M

$\therefore\,\,\,\,$ [A] = 2 $ \times $ 0.2 = 0.4 M

Equilibrium constant

K = ${{{{[A]}^2}} \over {[{A_2}]}}$ = ${{{{\left[ {0.4} \right]}^2}} \over {\left[ {0.8} \right]}}$ = 0.2

$\Delta $G^o = $-$ RT $\ell $nK

= $-$ 8.314 $ \times $ 320 $ \times $ $\ell $n(0.2)

= 4281 J/mol.

We know from thermodynamics,

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

$ \Rightarrow \,\,\,\, - RT\ln K = \Delta {H^o} - T\Delta {S^o}$

$ \Rightarrow \,\,\,\,\ln K = - {{\Delta {H^o}} \over R} \times {1 \over T} + {{\Delta {S^o}} \over R}.......\left( 1 \right)$

We know, equation of straight line is

$y = mx + c\,\,\,.........\,\,\,\left( 2 \right)$

by comparing (1) and (2) we get,

slope (m) = $-$ ${{\Delta {H^o}} \over R}$ and intercept (c) = ${{\Delta {S^o}} \over R}$

For exothermic reaction $\Delta {H^o} = - Ve$

$ \Rightarrow \,\,\,\, - \Delta {H^o} = + ve$

$ \Rightarrow \,\,\,\,{{ - \Delta {H^o}} \over R} = + ve$

$\therefore\,\,\,$ Slope is positive, so, graph A and B is possible.

Now if entropy $\left( {\Delta {S^o}} \right)$ is positive then intercept $\left( {{{\Delta {S^o}} \over R}} \right)$ is positive. It is represented by graph A.

If entropy $\left( {\Delta {S^o}} \right)$ is negative then intercept $\left( {{{\Delta {S^o}} \over R}} \right)$ is negative. It is represented by graph B.

Formula of Heat of combination is

$\Delta H = \Delta u\,\, + \,\,\Delta ng\,\,RT$

Where, $\Delta H$ $=$ Heat of combination at constant pressure

$\Delta u\, = $ Heat at constant volume

$\Delta {n_g}$ change in number of moles for gaseous molecule.

R = gas constant

T = Temperature.

The Required reaction

C₆H₆(l) + ${{15} \over 2}$ O₂(g) $ \to $ 6CO₂(g) + 3H₂O(l)

Here O₂ and CO₂ are gaseous molecules so to calculate $\Delta {n_g}$ we only consider those.

$\therefore\,\,\,$ $\Delta {n_g}$ $=$ 6 $-$ ${{15} \over 2}$ = $-$ ${3 \over 2}$

Given $\Delta $u = $-$ 3263.9 kJ mol^$-$1 R = 8.314 JK$-$ mol^$-$1

R = 8.314 JK^$-$1 mol^$-$1

= 8.314 $ \times $ 10^$-$3 kJ K^$-$1 mol^$-$1

T = 25^o C

= 25 + 273 K

= 298 K

So, $\Delta H$ = $-$ 3263.9 + $\left( { - {3 \over 2}} \right)$ $ \times $ 8.314 $ \times $ 10^$-$3 $ \times $ 298

= $-$ 3267.6 kJ mol^$-$1

Given

(i)   2Fe₂O₃(s) $ \to $ 4Fe(s) + 3O₂(g);

$\Delta $_rG^o = + 1487.0 kJ mol^-1

(ii)   2CO(g) + O₂(g) $ \to $ 2CO₂(g);

$\Delta $_rG^o = $-$ 514.4 kJ mol^-1

Multiply reaction (ii) with 3 and we get,

(iii)   6CO(g) + 3O₂ (g)    $\buildrel \, \over \longrightarrow $    6CO₂(g) ;

$\Delta $_rG^o = 3 $ \times $ $-$ 514.4 = $-$ 1543.2 kJ mol^$-$1

Now add reaction (i)   with  (iii), we get

2Fe₂O₃(s) + 6CO(g)    $\buildrel \, \over \longrightarrow $    4Fe(s) + 6CO₂(g)

$\therefore\,\,\,$ Free energy change of the reaction,

$\Delta $_rG^o  =  1487.0 $-$ 1543.2  =  $-$ 56.2 kJ/mol

S($l$)   $\buildrel \, \over \longrightarrow $   S(g)

$\Delta $G^o = $\Delta $_fG^o (Vapour) $-$ $\Delta $_f G^o (liquid)

= 103 $-$ 100.7

= 2.3 kcal / mol

= 2.3 $ \times $ 10³ cal/mol.

$\Delta $G^o = $-$ RT$l$nk

$ \Rightarrow $$\,\,\,$ 2.3 $ \times $ 10³ = $-$ 2.303 $ \times $ 2 $ \times $ 500 log K

$ \Rightarrow $$\,\,\,$ log K = $-$ 1

$ \Rightarrow $$\,\,\,$ K = 0.1 atm

Given,

$\Delta $U_BC = $-$ 5 kJ/mol

$\Delta $U_AB = q_AB + W_AB

= 2 + ($-$ 5)

= $-$ 3 kJ/mol.

For cyclic process,

$\Delta $U = 0

$ \Rightarrow $$\,\,\,\,$ $\Delta $U_AB + $\Delta $U_BC + $\Delta $U_CA = 0

$ \Rightarrow $$\,\,\,\,$ $\Delta $U_CA = $-$($-$3) $-$ ($-$ 5) = 8 kJ/mol

As, $\,\,\,\,\,\,$ $\Delta $U_CA = q_CA + W_CA

$ \Rightarrow $$\,\,\,\,$ 8 = q_CA + 3

$ \Rightarrow $$\,\,\,\,$ q_CA = + 5 kJ/mol.

$\Delta $H = $\Delta $$\mu $ + $\Delta $ng RT

$\therefore\,\,\,$ $\Delta $H = $\Delta $$\mu $

When, $\Delta $ng RT = 0

$ \Rightarrow $ $\Delta $ng = 0

For this reaction,

2HI(g) $ \to $ H₂ (g) + I₂ (g)

$\Delta $ng = (1 + 1) $-$ 2 = 0

$\therefore$ $\Delta $H = $\Delta $$\mu $

For the reaction at equilibrium

$ \mathrm{A} \rightleftharpoons \mathrm{P} $

At any temperature:

Concentration of $A$ at equilibrium $[A]_{e q}<5$ Concentration of $\mathrm{B}$ at equilibrium $[\mathrm{P}]_{e q}>5$

Equilibrium constant $\left(\mathrm{K}_{e q}\right)=\frac{[\mathrm{P}]_{e q}}{[\mathrm{~A}]_{e q}}$

$ \mathrm{K}_{e q}>1 $

$ \text{Also,}~~ \frac{\ln \mathrm{K}_{\mathrm{T}_1}}{\ln \mathrm{K}_{\mathrm{T}_2}}>\frac{\mathrm{T}_2}{\mathrm{~T}_1} $

$ \begin{aligned} & \text{If}~\mathrm{T}_2 >\mathrm{T}_1 \\\\ & \text{Hence}~\frac{\ln \mathrm{K}_{\mathrm{T}_1}}{\ln \mathrm{K}_{\mathrm{T}_2}} >\frac{\mathrm{T}_2}{\mathrm{~T}_1}>1 \\\\ & \frac{\mathrm{K}_{\mathrm{T}_1}}{\mathrm{~K}_{\mathrm{T}_2}} >1 \\\\ & \mathrm{~K}_{\mathrm{T}_1} >\mathrm{K}_{\mathrm{T}_2} \end{aligned} $

It can be clearly seen from diagram that concentration of reactant is lesser and that of product is higher at $T_1$ compared to $T_2$. Hence,

equilibrium constant is higher at $\mathrm{T}_1$ than $\mathrm{T}_2$. The reaction is exothermic.

i.e., $\Delta \mathrm{H}^{\circ}<0$ for forward reaction

The expression for the spontaneity and Gibbs free energy of reaction:

$ \begin{aligned} \Delta \mathrm{G}^{\circ} & =\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \\\\ \Delta \mathrm{S}^{\circ} & =\frac{\Delta \mathrm{H}^{\circ}-\Delta \mathrm{G}^{\circ}}{\mathrm{T}} \end{aligned} $

Since, both $\Delta \mathrm{G}^{\circ}$ and $\Delta \mathrm{H}^{\circ}$ are less than zero $\left\{\Delta \mathrm{H}^{\circ}<0\right.$ and $\left.\Delta \mathrm{G}^{\circ}<0\right\}$

If $\Delta \mathrm{H}^{\circ}>\Delta \mathrm{G}^{\circ}$, then $\Delta \mathrm{S}^{\circ}>0$

Otherwise,

If $\Delta \mathrm{H}^{\circ}<\Delta \mathrm{G}^{\circ}$, then $\Delta \mathrm{S}^{\circ}<0$

we know,

$ \begin{aligned} \Delta \mathrm{G}^{\circ} & =-\mathrm{RT} \ln \mathrm{K} . \\\\ \Delta \mathrm{G}_1^{\circ} & =-\mathrm{RT}_1 \ln \mathrm{K}_{\mathrm{T}_1} \\\\ \text { and } \Delta \mathrm{G}_2^{\circ} & =-\mathrm{RT} \ln \mathrm{K}_{\mathrm{T}_2} \end{aligned} $

Its given, $\ln \mathrm{K}_{\mathrm{T}_1}>\ln \mathrm{K}_{\mathrm{T}_2}$

$ \begin{gathered} \Delta \mathrm{G}_1^{\circ}<\Delta \mathrm{G}_2^{\circ} \\\\ \Delta \mathrm{H}_{\mathrm{T}_1}^{\circ}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{T}_1}^{\mathrm{o}}<\Delta \mathrm{H}_{\mathrm{T}_2}^{\mathrm{o}}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{T}_2}^{\mathrm{o}} \end{gathered} $

This is possible only if $\Delta S^{\circ}<0$

Hence,

$ \begin{array}{r} \Delta \mathrm{G}^{\circ}<0, \Delta \mathrm{S}^{\circ}<0 \\\\ \Delta \mathrm{H}^{\circ}<0, \Delta \mathrm{S}^{\circ}<0 \end{array} $

In the given curve AC represents isochoric process as volume at both the points is same i.e., V₁

Similarly, AB represents isothermal process (as both the points are at T₁ temperature) and BC represents isobaric process as both the points are at p₂ pressure.

Now, (i) for option (a)

q_AC = $\Delta$U_BC = nC_v(T₂ $-$ T₁)

where, n = number of moles

C_v = specific heat capacity at constant volume

However, W_AB $\ne$ p₂(V₂ $-$ V₁) instead

${W_{AB}} = nR{T_1}\ln \left( {{{{V_2}} \over {{V_1}}}} \right)$

So, this option is incorrect.

(ii) For option (b)

q_BC = $\Delta$H_AC = nC_p(T₂ $-$ T₁)

where, C_p = specific heat capacity at constant pressure

Likewise,

W_BC = $-$p₂(V₁ $-$ V₂)

Hence, this option is correct.

(iii) For option (c)

as nC_p(T₂ $-$ T₁) < nC_v(T₂ $-$ T₁)

so, $\Delta$H_CA < $\Delta$U_CA

and q_AC = $\Delta$U_BC

Hence, this option is also correct.

(iv) For option (d)

Although q_BC = $\Delta$H_AC

but $\Delta$H_CA $\le$ $\Delta$U_CA

Hence, this option is incorrect.

From first law of thermodynamics, we have

$\Delta$U = q + w

$\bullet$ For state A to B : q = +5 J, w = $-$8 J and

$\Delta$U_AB = 5 + ($-$8) = $-$3 J

$\bullet$ For state B to A : q = $-$3 J

Since the internal energy is a state function, we have $\Delta$U_BA = $-$ $\Delta$U_AB. Therefore,

$\Delta$U_BA = q + w

3 = $-$3 + w $\Rightarrow$ w = 6 J

Positive value of work indicates that 6 J of work is done by the surrounding on gas.

In an isothermal expansion process, the temperature of the system remains constant throughout the process. Since, for an ideal gas, U depends only on temperature, we have $\Delta$U = 0. The enthalpy change of the system in isothermal expansion is also zero as

$\Delta$H = $\Delta$U + nR$\Delta$T = 0 + 0 = 0

The entropy change foe an isothermal process is given by

$\Delta S = nR\ln \left( {{{{V_f}} \over {{V_i}}}} \right) \ge 0$

(1) From 5$^\circ$C to 0$^\circ$C :

${H_1} = {C_P} \times (\Delta T)$

$ = 75.3 \times (278 - 273)$

= 377 J/mol

= 0.377 KJ/mol

(2) Phase change (0$^\circ$C to 0$^\circ$C) :

${H_2} = - \Delta {H_{fusion}}$

= $-$ 6 KJ/mol

(3) 0$^\circ$C to $-$5$^\circ$C (Ice phase) :

${H_3} = - {C_{P(s)}} \times \Delta T$

$ = - 36.8 \times (273 - 268)$

= $-$0.184 KJ/mol

$\therefore$ ${H_{Total}} = {H_1} + {H_2} + {H_3}$

$ = 0.377 - 6 - 0.184$

= $-$5.837 KJ/mol

Given reaction,

A(g)  $ \to $  A($l$) ;

We know,

$\Delta $H = $\Delta $V + $\Delta $n_g RT

Given, $\Delta $H = $-$3RT

From reaction,

$\Delta $n_g = 0$-$ 1 = $-1$

$\therefore\,\,\,$ $\Delta $H = $\Delta $U $-$ RT

$ \Rightarrow $ $\,\,\,$ $-$ 3RT = $\Delta $U $-$ RT

$ \Rightarrow $$\,\,\,$ $\Delta $U = $-$ 2RT

$\therefore\,\,\,$ $\left| {\Delta U} \right|$ = 2RT

and $\left| {\Delta H} \right|$ = 3RT

$\therefore\,\,\,$ $\left| {\Delta H} \right|$ > $\left| {\Delta U} \right|$

From 1st law of thermodynamics

ΔU = q + w

For adiabatic process :

q = 0

$ \therefore $ ΔU = w

So change in internal energy (ΔU) is equal to adiabatic work.

C(graphite) + O₂(g) $ \to $ CO₂ (g);
${\Delta _r}{H^o}$ = - 393.5 kJ mol^-1 .............(1)

${{\rm H}_2}(g)$ + ${1 \over 2}{O_2}(g)$$\to {{\rm H}_2}{\rm O}(l)$
${\Delta _r}{H^o}$ = - 285.8 kJ mol^-1 .........(2)

$C{O_2}(g)$ + $2{{\rm H}_2}{\rm O}(l) \to$ $C{H_4}(g)$ + $2{O_2}(g)$
${\Delta _r}{H^o}$ = + 890.3 kJ mol^-1 .........(3)

C(graphite) + 2H₂(g) $ \to $ CH₄ (g);    $\Delta $H = ? ........ (4)

You can see,

[Eq. (1) + Eq. (3)] + [2 × Eq. (2)] = Eq. (4)

$ \therefore $ [ $\Delta $_rH₁ + $\Delta $_rH₃ ] + [2 $ \times $ $\Delta $_rH₂] = $\Delta $H

$ \Rightarrow $ $\Delta $H = [(-393.5) + (890.3)] + [2(-285.8)] = -74.8 kJ / mol

$\Delta {S_{surr}} = - {{\Delta H} \over {{T_{surr}}}}$

For endothermic, if T_surr increases, unfavourable change in entropy of the surroundings decreases.

For exothermic, if T_surr increases, favourable change in entropy of the surroundings decreases.

Option (A): Correct. For compression of gas, $V_2 < V_1$, thus, from $w=-p \Delta V$, work done on the gas will be positive. Below graph shows the pressure-volume curve for work done during reversible and irreversible compression process. Area under curve during irreversible process is greater than the area under curve during reversible process. Therefore, work done on the gas in irreversible process is greater than in reversible process, that is, $w_{\text {irr }}>w_{\text {rev }}$.


Option (B): Correct. For work done in free expansion, $p_{\text {ext }}=0$, therefore, $w=0$. For an adiabatic process, $q=0$ and for isothermal process, $\Delta T=0$. Therefore, from first law of thermodynamics, $\Delta U=q+w=0$.

Option (C): Correct. Below graph represents the expansion under adiabatic (AC) and isothermal condition (AB). Therefore, $\left|w_{\mathrm{AB}}\right|>\left|w_{\mathrm{AC}}\right|$.

We have

C(graphite) $\to$ C(diamond)

For solid at constant temperature, Gibbs energy can be calculated as

$dG = \Delta Vdp$

$\int\limits_{\Delta {G_1}}^{\Delta {G_2}} {d(\Delta {G_T}) = \int\limits_{{p_1}}^{{p_2}} {\Delta Vdp} } $

$\Delta {G_2} - \Delta {G_1} = \Delta V({p_2} - {p_1})$

$(2.9 \times {10^3} - 0) = 2 \times {10^{ - 6}}({p_2} - 1)$

$({p_2} - 1) = {{2.9 \times {{10}^3}} \over {2 \times {{10}^{ - 6}}}}$ Pa

$({p_2} - 1) = {{2.9 \times {{10}^3}} \over {2 \times {{10}^{ - 6}} \times {{10}^5}}}$ bar

p₂ = 14501 bar

For reaction : 2H₂O₂(1) $\rightleftharpoons$ 2H₂O(1) + O₂(g)

Given, p = 1 bar, T = 300 K, n = 100 mol, R = 8.3 J K^$-$ mol^$-$1

We know that

w = p$\Delta$V = nRT

= 100 $\times$ 8.3 $\times$ 300

= 249000 J or 249 kJ

Carbon is a good reducing agent for oxides. The reason why carbon reduces metal oxide spontaneously at T < 1200 K is that $ - {{\ln {K_p}} \over T}$ line for CO has a negative slope.

We know that $\Delta$G = $\Delta$H $-$ T$\Delta$S

At low temperature : $\Delta$G is positive (non-spontaneous process), so $\Delta$H is positive and $\Delta$S is positive (T$\Delta$S < $\Delta$H as T is low).

At high temperature : $\Delta$G is negative (spontaneous process), so $\Delta$H is positive and $\Delta$S is positive (T$\Delta$S > $\Delta$H as T is high).

We have

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

$\Delta$G$^\circ$ = $-$ 29.8 $-$ 298 ($-$ 0.1) = 0

$\Delta$G$^\circ$ = $-$ 2.302 RT log K_eq

log K_eq = 0 $\Rightarrow$ K_eq = 1

Given

$C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right);$

$\Delta H = - 393.5\,\,kJ\,mo{l^{ - 1}}......\left( i \right)$

$CO\left( g \right) + {1 \over 2}{O_2}\left( g \right) \to C{O_2}\left( g \right);$

$\Delta H = - 283.5\,kJ\,mo{l^{ - 1}}\,\,\,\,\,....\left( {ii} \right)$

$\therefore$ Heat of formation of

$Co = eqn\left( i \right) - eqn\left( {ii} \right)$

$ = - 393.5 - \left( { - 283.5} \right)$

$ = - 110\,kJ$

Work done in an irreversible isothermal process is given by :

$ \begin{aligned} \mathrm{W}_{\mathrm{irr}} & =-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\\\ & =-3.0 \mathrm{~atm}(2.0 \mathrm{~L}-1.0 \mathrm{~L}) \\\\ & =-3.0 \times 1.0 \mathrm{~atm} \mathrm{~L} \\\\ \mathrm{~W}_{\mathrm{irr}} & =-3.0 \mathrm{~L} \mathrm{~atm} \\\\ \mathrm{~W}_{\mathrm{irr}} & =-3.0 \times 101.3 \mathrm{~J} \,\,\,\,\,\,\,\,\,\ [1 \mathrm{~L} \mathrm{~atm}=101.3] \\\\ & =-303.9 \mathrm{~J} \end{aligned} $

At a constant pressure,

$ \begin{aligned} & \Delta \mathrm{U}=\mathrm{q}_{\mathrm{p}}+\mathrm{W}_{\text {irr }}=\mathrm{q}_{\mathrm{p}}-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\\\ & \Delta \mathrm{U}=\Delta \mathrm{H}-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\\\ & \Delta \mathrm{H}=\Delta \mathrm{U}+\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right) \end{aligned} $

Since, reaction takes place at constant temperature; hence, $\Delta \mathrm{U}=0$.

$ \begin{aligned} \Delta \mathrm{H} & =\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right)=-\mathrm{W}_{\mathrm{irr}}=303.9 \mathrm{~J} \\\\ \Delta \mathrm{H}_{\mathrm{sys}} & =303.9 \mathrm{~J} \\\\ \Delta \mathrm{S}_{\text {sys }} & =\frac{\Delta \mathrm{H}_{\text {sys }}}{\mathrm{T}} \text { and } \\\\ \end{aligned} $

Finally, the change in entropy of the surroundings (ΔS_surr) is equal to the negative of the change in entropy of the system, because the system and its surroundings form an isolated system :

ΔS_surr = -ΔS_sys

$ \begin{aligned} \Delta \mathrm{S}_{\text {surr }} & =-\frac{\Delta \mathrm{H}_{\text {sys }}}{\mathrm{T}}=-\frac{303.9 \mathrm{~J}}{300 \mathrm{~K}} \\\\ & =-1.013 \mathrm{~J} \mathrm{~K}^{-1} \end{aligned} $

$\Delta {G^ \circ }_{NO\left( g \right)}$

$\,\,\,\,\,\,\,\,\,\,\,\,\, = 86.6kJ/mol = 86600J/mol;$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$${G^ \circ }_{N{O_2}\left( g \right)} = x\,J/mol$

$T = 298,\,{K_p} = 1.6 \times {10^{12}}$

$\Delta {G^ \circ } = - RT\,\ln \,{K_p}$

Given equation,

$2NO\left( g \right) + {O_2}\left( g \right)\,\rightleftharpoons\,2N{O_2}\left( g \right)$

$2\Delta {G^ \circ }_{N{O_2}} - 2\Delta {G^ \circ }_{NO}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)$

$2\Delta {G^ \circ }_{N{O_2}} - 2 \times 86600$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)$

$2\Delta {G^ \circ }_{N{O_2}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = 2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)$

$\Delta {G^ \circ }_{N{O_2}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = {1 \over 2}\left[ {2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)} \right]$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = 0.5\left[ {2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)} \right]$

(A) $\to$ (R and T)

Freezing of water,

H₂O_(l) $\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{273\,K}^{1\,atm}} $ H₂O_(s)

The system is cooled i.e.; heat is released during the process

so, q < 0

$\mathop {Water}\limits_{(Less\,volume)} $ $\rightleftharpoons$ $\mathop {Ice}\limits_{(More\,volume)} $ + heat

Volume is increased i.e.; $\Delta$V = +ve.

w = $-$P$\Delta$V = $-$ve

i.e.; w < 0 (expansion)

Entropy of system is decreased, $\Delta$S_sys < 0.

$\Delta$U = q + w

As q < 0, w < 0 so, $\Delta$U < 0.

At equilibrium, $\Delta$G = 0.

(B) $\to$ (P, Q and S)

Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions,

w = 0, q = 0 so, $\Delta$U = 0

For expansion, $\Delta$S_sys > 0 as entropy increases.

$\Delta G = - nRT\ln {{{V_2}} \over {{V_1}}}$

For expansion, V₂ > V₁

$\Delta$G = $-$ve i.e.; $\Delta$G < 0.

(C) $\to$ (P, Q and S)

Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container.

q = 0 (isolated)

w = $-$P$\Delta$V

w = 0 ($\because$ $\Delta$V = 0)

$\Delta$S_sys > 0 (mixing of gases)

$\Delta$U = q + w = 0

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

$\Delta$G = q_p $-$ T$\Delta$S (at constant P, T)

$\Delta$G = 0 $-$ T$\Delta$S = $-$T$\Delta$S

$\Delta$G < 0 ($\because$ $\Delta$S_sys > 0)

(D) $\to$ (P, Q, S and T)

$\mathop {{H_{2(g)}}}\limits_{(300\,K)} \mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{Cool}^{Heat,\,1\,atm}} \mathop {{H_{2(g)}}}\limits_{(600\,K)} $

Internal energy (U), entropy (S) and free energy (G) are state functions which depend only upon the state of the system and do not depend upon the path by which the state is attained.

Thus, $\Delta$U = 0, $\Delta$S = 0 and $\Delta$G = 0

Work and heat are path functions but the same path is retraced so, q = 0 and w = 0.

${C_2}{H_5}OH\left( \ell \right) + 3{O_2}\left( g \right)$

$ \to 2C{O_2}\left( g \right) + 3{H_2}O\left( \ell \right)$

Bomb calorimeter gives $\Delta U$ of the reaction

Given, $\Delta U = - 1364.47\,kJ\,mo{l^{ - 1}}$

$\Delta {n_g} = - 1$

$\Delta H = \Delta U + \Delta {n_g}RT$

$ = - 1364.47 - {{1 \times 8.314 \times 298} \over {1000}}$

$ = - 1366.93\,kJ\,mo{l^{ - 1}}$

To analyze the options provided, let's discuss the changes in entropy for the system and the surroundings separately during the phase change from liquid water (H₂O(l)) to water vapor (H₂O(g)) at 100°C and 1 atmosphere pressure. This is a typical boiling process.

First, let's consider the change in entropy, $\Delta S_{\text{system}}$, of the water itself (the system). When water boils and changes from a liquid to a gas, the molecules spread out significantly, occupying a larger volume and showing greater randomness in their motion. This transition from a more ordered phase (liquid) to a less ordered phase (gas) results in an increase in entropy. Thus, $\Delta S_{\text{system}}>0$.

Next, let's consider the change in entropy of the surroundings, $\Delta S_{\text{surroundings}}$. When the water boils at 100°C, it must absorb heat to do so. This heat is typically taken from the surroundings. As the surroundings lose heat, according to the formula for entropy change $\Delta S = \frac{q}{T}$, where $q$ is the heat transferred and $T$ is the temperature, the surroundings lose entropy if the process is happening at a constant temperature as typically assumed in such cases. Therefore, here $\Delta S_{\text{surroundings}}<0$ as heat loss by surroundings (negative $q$) results in a decrease in entropy.

Combining these insights, we find:

• The entropy of the system increases ($\Delta S_{\text{system}}>0$).
• The entropy of the surroundings decreases ($\Delta S_{\text{surroundings}}<0$).

Therefore, the correct answer is: Option B: $\Delta S_{\text{system}}>0$ and $\Delta S_{\text{surroundings}}<0$.

By I^st law of thermodynamics, q = $\Delta $E – W

At const T, ∆E = 0

q = – W

Heat absorbed = 208 J

$ \therefore $ q = +208 J

W = – 208 J

When dealing with solutions and thermodynamics, particularly with ideal solutions, several key aspects must be considered to understand the behavior of the system and its interactions with the surroundings. Let's analyze each option in the context of forming an ideal solution from benzene and naphthalene at room temperature:

Option A: $\Delta G$ is positive

For an ideal solution, the Gibbs free energy change ($\Delta G$) upon mixing can be described as follows:

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

An ideal solution has the characteristic that there is no enthalpy change upon mixing ($\Delta H = 0$) because the intermolecular forces between unlike molecules are assumed to be equal to those between like molecules. Additionally, when two or more components form an ideal solution, the entropy of the system ($\Delta S$) increases due to the mixing of different molecules, which is a manifestation of increased randomness or disorder.

Since $\Delta H = 0$ and $\Delta S > 0$ for an ideal solution, applying these to the Gibbs free energy equation yields:

$\Delta G = 0 - T\Delta S = -T\Delta S$

Given that $T > 0$ and $\Delta S > 0$, $\Delta G$ will ultimately be negative, indicating a spontaneous process. Therefore, Option A is false.

Option B: $\Delta S_{system}$ is positive

As mentioned earlier, the entropy of the system increases in the formation of an ideal solution. This is because the different molecules of benzene and naphthalene mix, increasing the randomness or disorder of the system. Therefore, Option B is true.

Option C: $\Delta S_{surroundings}$ = 0

In the formation of an ideal solution, there is no heat exchange with the surroundings since $\Delta H = 0$. Without a heat exchange, there is no change in the entropy of the surroundings, as entropy change in the surroundings is generally associated with heat exchange according to the relation $\Delta S = \frac{q}{T}$ for reversible processes. Therefore, Option C is true.

Option D: $\Delta H$ = 0

In the context of an ideal solution, it is assumed that the enthalpy change ($\Delta H$) is zero because the energy required to break interactions among similar molecules is exactly balanced by the energy released from the formation of new interactions between different types of molecules. This implies no net heat effect. Therefore, Option D is true.

In summary, Options B, C, and D are true, while Option A is false for the ideal solution of benzene and naphthalene at room temperature.

From the graph, as the term pV increases, the temperature increases; and when pV decreases the temperature decreases.

The term PV increases K$\to$L, so this involves heating and is an isobaric process.

It increases from N$\to$K, so this involves heating and is an isochoric process.

It decreases from M$\to$N, so this involves cooling and is an isobaric process.

It decreases from L$\to$M, so this involves cooling and is an isochoric process.

From the above solution, we have

From K$\to$L : heating (isobaric)

From L$\to$M : cooling (isochoric)

From M$\to$N : cooling (isobaric)

From N$\to$K : heating (isochoric)

To find the standard enthalpy of combustion of glucose, we will first write the balanced chemical equation for the combustion of glucose. The chemical formula for glucose is $ C_6H_{12}O_6 $. Combustion involves the reaction of a substance with oxygen to produce carbon dioxide (CO₂) and water (H₂O) typically. The reaction for the combustion of glucose can be represented as:

$ C_6H_{12}O_6 (s) + 6O_2 (g) \rightarrow 6CO_2 (g) + 6H_2O (l) $

Each molecule of glucose (C6H12O6) produces six molecules of CO₂ and six molecules of H₂O. Now, we use the given standard enthalpies of formation (ΔH_f°) for these substances to find the enthalpy of the reaction. The enthalpy of a reaction can be calculated using the formula:

$ \Delta H_{\text{reaction}} = \sum \Delta H_{f,\text{products}} - \sum \Delta H_{f,\text{reactants}} $

For the given reaction:

• ΔH_f° of CO₂(g) = –400 kJ/mol
• ΔH_f° of H₂O(l) = –300 kJ/mol
• ΔH_f° of glucose(s) = –1300 kJ/mol

The sum of the standard enthalpies of formation of the products:

$ [6 \times (-400) \text{ kJ/mol}] + [6 \times (-300) \text{ kJ/mol}] = (-2400 + -1800) \text{ kJ} = -4200 \text{ kJ} $

The sum of the standard enthalpies of formation of the reactants (glucose as only part contributing with enthalpy value other than zero):

$ [-1300 \text{ kJ/mol}] + [6 \times 0 \text{ kJ/mol for } O_2] $

Therefore, the standard enthalpy of combustion of glucose is:

$ \Delta H_{\text{reaction}} = (-4200 \text{ kJ}) - (-1300 \text{ kJ}) = -2900 \text{ kJ} $

So, -2900 kJ of energy is released per mole of glucose combusted at standard conditions. The standard enthalpy of combustion per gram of glucose can be calculated using the molar mass of glucose. Glucose has a molar mass of approximately 180 g/mol.

$ \Delta H_{\text{combustion per gram}} = \frac{-2900 \text{ kJ/mol}}{180 \text{ g/mol}} = -16.11 \text{ kJ/g} $

Thus, the correct answer is -16.11 kJ per gram of glucose, which corresponds to option C –16.11 kJ.

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

$\,\,\,\,\,\,\,\,\, - RT\,\ell nK = \Delta {H^ \circ } - T\Delta {S^ \circ }$

$\ell nK = - {{\Delta {H^ \circ } - T\Delta {S^ \circ }} \over {RT}}$

For isothermal process T₁ = T₂. Work done in isothermal process is less than adiabatic process. In case of isothermal process, the temperature remains constant so there is no change in the internal energy, whereas in case of adiabatic process expansion occurs through internal energy.

As entropy is a state function and is additive

$\Delta {S_{X \to Z}} = \Delta {S_{X \to Y}} + \Delta {S_{Y \to Z}}$

On moving from Y to Z, the work done is zero as the volume is kept constant (isochoric process), so

${W_{X \to Y \to Z}} = {W_{X \to Y}}$

We know that

Binding energy = Total energy of reactants $-$ Total energy of products

So, we have

225 = 2(C $-$ C) + 1(H $-$ H) $-$ 2(C $-$ H) + 1(C $\equiv$ C)

225 = 1410 + 330 $-$ 2 $\times$ 350 + (C $\equiv$ C)

Solving, we get the value of C $\equiv$ C as 815 kJ mol^$-$1.

Entropy change for an isothermal reversible process is given by

$\Delta S = nR{\mkern 1mu} ln{{{V_2}} \over {{V_1}}}$

$ = 2 \times 8.314 \times 2.303log\,{{100} \over {10}}$

$ = 38.3\,J\,mo{l^{ - 1}}\,\,{K^{ - 1}}$

Analysis of each transformation:

(A) CO₂(s) $\to$ CO₂(g)

This transformation represents the sublimation of solid carbon dioxide (dry ice) into gaseous carbon dioxide. This process is:

Phase transition (p): The substance changes from solid to gas phase.

$\Delta H$ is positive (r): Sublimation is an endothermic process, meaning it absorbs heat from the surroundings.

$\Delta S$ is positive (s): The entropy increases as the substance moves from a more ordered solid state to a less ordered gaseous state.

(B) CaCO₃(s) $\to$ CaO(s) + CO₂(g)

This transformation is the thermal decomposition of calcium carbonate to form calcium oxide and carbon dioxide. This process involves:

$\Delta H$ is positive (r): Decomposition reactions are typically endothermic, requiring energy to break chemical bonds.

$\Delta S$ is positive (s): The process results in an increase in entropy, primarily because a gas (CO₂) is produced from a solid compound.

(C) 2H $\to$ H₂(g)

This is the formation of hydrogen gas (H₂) from atomic hydrogen (H). This transformation:

$\Delta S$ is negative (t): Two individual hydrogen atoms combine into a diatomic molecule, decreasing the system's entropy since fewer particles are free to move independently.

(D) P_{(white, solid)} $\to$ P_{(red, solid)}

Crystalline solid white phosphorous is converted to amorphous red phosphorous. This process involves a phase change. Since, white and red phosphorous are allotropic form of phosphorous, i.e., both are composed of phosphorous atoms but the connectivity between atom differ, it is therefore regarded as allotropic change.

The white form of phosphorous has lot of angular strain as compared to stable red phosphorous. Hence, randomness of system due to angular strain decreases as red phosphorus is formed. The entropy change is negative $(\Delta S<0)$

Option Evaluation:

Upon reviewing each transformation and comparing the mappings:

Option A agrees completely with the above analysis with proper mapping for each statement.

Thus, Option A is correct:

A $\to$ p,r,s; B $\to$ r,s; C $\to$ t; D $\to$ p,q,t

At equilibrium $\Delta G = 0$

Hence, $\Delta G = \Delta H - {T_e}\Delta S = 0$

$\therefore$ $\,\,\,\,\,\,\Delta H = {T_e}\Delta S\,\,\,$

or $\,\,\,\,{T_e} = {{\Delta H} \over {\Delta S}}\,\,$

For a spontaneous reaction

$\Delta G$ must be negative

which is possible only if $\Delta H < T\Delta S$

or $\,\,\,T > {{\Delta H} \over {\Delta S}};{T_e} < T$

${N_2} + 3{H_2} \to 2N{H_3}\,$

$\,\Delta H = 2 \times - 46.0\,\,kJ\,mo{l^{ - 1}}$

Let $x$ be the bond enthalpy of $N-H$ bond then

[Note : Enthalpy of formation or bond formation enthalpy is given which is negative but the given reaction involves bond breaking hence values should be taken as positive. ]

$\Delta H = \sum \, $ Bond energies of products

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \sum \, $ Bond energies of reactants

$2 \times - 46 = 712 + 3 \times \left( {436} \right) - 6x;$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\, - 92 = 2020 - 6x$

$6x = 2020 + 92$

$ \Rightarrow 6x = 2112$

$ \Rightarrow x = + 352\,kJ/mol$

An intensive property is a property that is independent of the amount of mass or size of the system. In contrast, an extensive property is dependent on the size or amount of mass in the system. Let's analyze each option based on this definition:

• Molar Conductivity: Molar conductivity, denoted as $\Lambda_m$, is a measure of the ionic conductivity of an electrolyte solution divided by the concentration of the electrolyte. Since it is normalized by the amount of substance, it does not depend on the total amount of the substance present in the solution. Thus, molar conductivity is an intensive property.
• Electromotive Force (EMF): EMF is the voltage generated by a battery or by the magnetic force according to Faraday’s Law of electromagnetic induction. It refers to a potential difference and does not depend on the quantity of material or size of the battery. Therefore, EMF is considered an intensive property.
• Resistance: Resistance, denoted by $R$, measures how much a material opposes the flow of electric current. It depends on the material’s length, cross-sectional area, and resistivity. For instance, a longer wire has greater resistance. Therefore, resistance is an extensive property.
• Heat Capacity: Heat capacity is the amount of heat required to change the temperature of a substance by a certain temperature interval. It is dependent on the amount of substance present, and therefore, it is an extensive property. The specific heat capacity, however, is intensive since it is the heat capacity per unit mass.

From the options provided:

• Option A (Molar Conductivity) is intensive.
• Option B (Electromotive Force) is intensive.
• Option C (Resistance) is extensive.
• Option D (Heat Capacity) is extensive.

Therefore, the intensive properties among the given options are Option A (Molar Conductivity) and Option B (Electromotive Force).

The bond energy of a C-C (carbon-carbon) single bond is a measure of the strength of that bond, or how much energy is required to break one mole of such bonds in a gaseous substance. This value is a fundamental concept in chemistry, particularly in the study of organic molecules and reactions.

For a C-C single bond, the bond energy is typically around 83 kcal/mol (kcal per mole). This places the value closest to option C amongst the given choices.

Therefore, the correct answer is:

Option C: 100

The standard molar enthalpy of formation, denoted as $\Delta H_f^\circ$, is defined as the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states under standard conditions (298 K and 1 atm pressure). By definition, the standard molar enthalpy of formation of a pure element in its most stable form at 298 K is zero.

Looking at the options:

• Br₂ (g) - Bromine's standard state at room temperature is liquid (Br₂(l)), not gas.
• Cl₂ (g) - Chlorine's standard state at room temperature is a gas, which means the standard molar enthalpy of formation for Cl₂(g) is indeed zero.
• H₂O (g) - Water in gaseous form is not a basic element but a compound of hydrogen and oxygen, so its enthalpy of formation is not zero.
• CH₄ (g) - Methane, represented here, is also a compound composed of carbon and hydrogen, thus its enthalpy of formation is also not zero.

Therefore, the correct option is Option B, Cl₂ (g), as chlorine gas is a diatomic molecule and an element in its standard state at 298 K, and as such, its standard molar enthalpy of formation is zero.

Given, for reaction

$(i)$ $\,\,\,\,\,\,{H_2}O\left( \ell \right) \to {H^ + }\left( {aq.} \right) + O{H^ - }\left( {aq.} \right);$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H_r} = 57.32\,kJ$

$(ii)$ $\,\,\,\,\,\,{H_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right) \to {H_2}O\left( \ell \right);$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H_r} = - 286.20\,kJ$

For reaction $(i)$

$\Delta {H_r} = \Delta {H^ \circ }_f\left( {{H^ + }.aq} \right) + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)$

$57.32 = 0 + \Delta {H^ \circ }_f\left( {O{H^{ - 1}},aq} \right) - $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)\,\,\,\,\,\,...\left( {iii} \right)$

For reaction $(ii)$

$\Delta {H_r} = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right) - \Delta {H^ \circ }_f\left( {{H_2},g} \right) - $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{1 \over 2}\Delta {H^ \circ }_f\left( {{O_2},g} \right)$

$ - 286.20 = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)$

On replacing this value in equ. $(iii)$ we have

$57.32 = \Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \left( { - 286.20} \right)$

$\Delta {H^ \circ }_f = - 286.20 + 57.32$

$ = - 228.88\,kJ$

State functions are properties of a system that depend only on the current state of the system, not on the path used to get to that state. They are intrinsic properties of the system and include properties like pressure, temperature, volume, internal energy, enthalpy, entropy, and Gibbs free energy.

Option A: Internal Energy

Internal energy, denoted as $U$, is a state function. It is the total energy contained within the system, including kinetic and potential energy at the molecular level. Internal energy changes in a system only depend on the initial and final states of the system, regardless of the process or path taken to achieve these states. Therefore, it fits the criteria of a state function.

Option B: Irreversible Expansion Work

Irreversible expansion work is not a state function. Work, in general, is a path function because it depends on the path taken during a process. Whether the path involves irreversible or reversible processes, work includes energy transfer that depends significantly on how that transfer is carried out. Therefore, irreversible expansion work depends on the specific details of the process and is not solely determined by the initial and final states of the system.

Option C: Reversible Expansion Work

Similarly to irreversible expansion work, reversible expansion work is also a path function and not a state function. Even though reversible processes are ideal and involve quasi-static changes that maintain the system in near equilibrium throughout, the work done (or reversible expansion work) during such processes still depends on the specific path taken. This includes how slowly the process is carried out and the intermediate steps, distinguishing it from a state function.

Option D: Molar Enthalpy

Molar enthalpy, denoted as $H$, is a state function. It is defined as the sum of the internal energy $U$ of a system plus the product of the pressure $P$ and volume $V$ of the system, multiplied by the number of moles $n$, expressed as:

$H = U + PV$

Since $U$, $P$, and $V$ are all state functions, their combination into enthalpy continues to be dependent solely on the state of the system, not on the path taken to reach that state.

Conclusion:

Among the given options, Internal energy (Option A) and Molar enthalpy (Option D) are state functions, while both Irreversible expansion work (Option B) and Reversible expansion work (Option C) are not state functions. They are path functions.