Thermodynamics

Given,

$ \begin{aligned} \Delta H & =+30.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta S & =0.06 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\ p & =1 \mathrm{~atm} \\ \Delta G & =0 \end{aligned} $

We know that,

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

if, we put temperature $227^{\circ} \mathrm{C}$ or

$ 227+273=500 \mathrm{~K} $

In equation then the value of free energy is obtained will be zero.

$ \begin{aligned}Hence,\,\, & \Delta G=30-500 \times 0.06 \\ & \Delta G=30-30=0 \end{aligned} $

and nature of reaction will be non-spontaneous.

This is an irreverseable process as gas is expanding against a constant external process.

Work done in irreverseable process

W = - P_ext$\Delta $V

= - 1 bar $ \times $ 9 L

= - 10⁵ Pa $ \times $ 9 $ \times $ 10^-3 m³

= - 9 $ \times $ 10² N-m

= - 900 J

= - 0.9 kJ

Kirchoff's Equation :

${{\Delta {H_{{T_2}}} - \Delta {H_{{T_1}}}} \over {{T_2} - {T_1}}} = \Delta \left( {{C_p}} \right)$

Sublimation of iodine :

I₂(s) $ \to $ I₂(g)

${{\Delta {H_{250^\circ C}} - \Delta {H_{200^\circ C}}} \over {250 - 200}} = {C_p}\left[ {{I_2}(g)} \right] - {C_p}\left[ {{I_2}(s)} \right]$

${{\Delta {H_{250^\circ C}} - 24} \over {50}} = \left( {0.031 - 0.055} \right)$

$ \Rightarrow $ ${\Delta {H_{250^\circ C}}}$ = 22.8 cal g^–1

We know,

$\Delta $H - $\Delta $U = $\Delta $n_gRT

C₇H₁₆($l$) + 11O₂(g) $ \to $ 7CO₂(g) + 8H2O($l$)

Here $\Delta $n_g = 7 - 11 = - 4

$ \therefore $ $\Delta $H - $\Delta $U = - 4RT

A reaction is spontaneous if $\Delta $G is negative.

and we know that

$\Delta $G = $\Delta $H – T$\Delta $S

If $\Delta $H = –ve and $\Delta $S = + ve then at all the temperature the process will be spontaneous.

Here heat is released so q is negative.

$ \therefore $ q = - 2 kJ

Work done on the system, w = 10 kJ

From first law of thermodynamics,

$\Delta $U = q + w = -2 + 10 = 8 kJ

(A) q + w = $\Delta $E, state function

(B) q, Path function

(C) w, Path function

(D) H – TS = G, State function

For ideal gas,

$\Delta $U = nC_v$\Delta $T

= 5 $ \times $ 28 $ \times $ (200 - 100)

= 14 kJ

We know,

PV = nRT

$ \therefore $ $\Delta $(PV) = nR$\Delta $T = 5 $ \times $ 8 $ \times $ 100 = 4 kJ

From 1^st law of thermodynamics we know,

$\Delta $U = q + W

Option A :

In adiabatic process exchage of heat = 0

$ \therefore $ q = 0

$ \therefore $ From 1st law of thermodynaics, $\Delta $U = W

So option A is wrong.

Option B :

U is a state function. In cyclic process, initial state and final state both are same. So change in all the state function in cyclic process will be zero.

$ \therefore $ $\Delta $U = 0

$ \therefore $ From 1st law of thermodynaics,

q + W = 0 $ \Rightarrow $ q = -W

So option B is correct..

Option C :

In isochoric process volume (V) is constant. So dV = 0.

We know, W = $ - \int {{P_{ex}}} dV$

$ \therefore $ W = 0

$ \therefore $ From 1st law of thermodynaics,

$\Delta $U = q

So option C is correct.

Option D :

In isothermal process temerature (T) is constant. So dT = 0.

We know, $\Delta $U = nC_vdT

$ \therefore $ $\Delta $U = 0

$ \therefore $ From 1st law of thermodynaics,

q + W = 0 $ \Rightarrow $ q = -W

So option D is correct.

Give that,

n = 3

T₁ = 300

T₂ = 1000

C_p = 23 + 0.01T

We know,

$\Delta $H = $\int\limits_{{T_1}}^{{T_2}} {n{C_p}dT} $

= $\int\limits_{300}^{1000} {3\left( {23 + {T \over {100}}} \right)dT} $

= $3\left[ {23T + {{{T^2}} \over {200}}} \right]_{300}^{1000}$

= 3[ 23(1000 - 300 + ${{1 \over {200}}}$((1000)² - (300)²)]

= 3[ 23 $ \times $ 700 + ${{700 \times 1300} \over {200}}$ ]

= 61950 J

= 61.95 kJ

$ \simeq $ 62 kJ

In reaction (i), the product is CO₂

If we add reaction (ii) and (iii) we get,
      C + O₂  $ \to $  CO₂
here also product is CO₂ from same reactant C and O₂.

So according to hess law, we can say
      x = y + z

For isothermal process of ideal gas,

PV = constants = K

$ \therefore $  P = ${k \over v}$

So, the graph between P and ${1 \over v}$ is straight line passing through the origin. So, graph (A) is correct.

As PV = K so the P and V curve is hyperbola. So, graph (B) is wrong.

In isothermal PV = constant.

So, graph (C) is right.

We know internal energy (u) = ${f \over 2}$ nRT. Internal energy is function of temperature (T) only. So, U does not change when volume(V) changes. So, graph (D) is wrong.

C_P does not changes with change in pressure.

We know,

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

For a reaction to be spontaneous $\Delta $G^o must be negative i.e.,
T$\Delta $S^o > $\Delta $H^o

$ \Rightarrow $ T > ${{\Delta H^\circ } \over {\Delta S^\circ }}$

$ \Rightarrow $ T > ${{491.1 \times 1000} \over {198}}$

$ \Rightarrow $ T > 2480.3 K

2H₂O $\rightleftharpoons$ H₃O⁺ + OH^$-$

Here K_eq = K_w(H₂O)

At 298 K, K_w(H₂O) = 10^-14

$ \therefore $ K_eq = 10^-14

$\Delta $G^o = -RTln(K_eq)

= -RTln(K_w)

= -2.303RT log (10^-14)

= -2.303 $ \times $ 8.314 $ \times $ 298 $ \times $ $ \times $ (-14)

= 80 kJ/mol

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

Given that A and B are non-zero constants, i.e., A = $\Delta $H^o, B = $\Delta $S^o

If $\Delta $H^o is positive means reaction is endothermic.

X $\rightleftharpoons$ Y

K_eq = ${{\left[ Y \right]} \over {\left[ X \right]}}$

If K_eq > 1 $ \Rightarrow $ [Y] > [X]

and K_eq < 1 $ \Rightarrow $ [Y] < [X]

We know, $\Delta $G^o = -RT ln(K_eq)

So when K_eq > 1 then $\Delta $G^o < 0 and Y is major.

And when K_eq < 1 then $\Delta $G^o > 0 and X is major.

Temperature at which G^o = 0 is

120 $ - {3 \over 8}$ T = 0

$ \Rightarrow $ T = 320 K

For T > 320 K

$\Delta $G^o < 0 and Y is major.

For T < 320 K

$\Delta $G^o > 0 and X is major.

When two blocks comes in contact with each other and attain thermal equilibrium then

final temperature of the blocks,

T_f = ${{{T_1} + {T_2}} \over 2}$

$\Delta $S = $\Delta $S₁ + $\Delta $S₂

= C_p $\ln \left( {{{{T_f}} \over {{T_1}}}} \right)$ + C_p $\ln \left( {{{{T_f}} \over {{T_2}}}} \right)$

= C_p $\ln \left( {{{{T_1} + {T_2}} \over {2{T_1}}}} \right)$ + C_p $\ln \left( {{{{T_1} + {T_2}} \over {2{T_2}}}} \right)$

= C_p $\ln \left( {{{{{\left( {{T_1} + {T_2}} \right)}^2}} \over {4{T_1}{T_2}}}} \right)$

N₂(g) + 3H₂(g) $\rightleftharpoons$ 2NH₃(g); $\Delta $n_g < 0

Work done on isothermal irreversible for ideal gas

= $-$P_ext (V₂ $-$ V₁)

= $-$4 N/m² (1 m³ $-$ 5m³)

= 16 Nm

Isothermal process for ideal gas

$\Delta $U = 0

q = $-$ w

= $-$ 16 Nm

= $-$ 16 J

Heat used to increase temperature of Al

q = n C_m $\Delta $T

16 J = 1 $ \times $ 24 ${J \over {mol.K}}$ $ \times $ $\Delta $T

$\Delta $T = ${2 \over 3}$K

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

T = ${{\Delta H} \over {\Delta S}} = {{200} \over {40}} = 5K$

$\Delta {S_1} = {{\Delta {H_{fusion}}} \over {273}} = {{334} \over {273}} = 1.22$

$\Delta {S_2} = 4.2\ell N\left( {{{363} \over {273}}} \right) = 1.31$

$\Delta {S_3} = {{\Delta {H_{vap}}} \over {373}} = {{2491} \over {373}} = 6.67$

$\Delta {S_4} = 2.0\ell n\left( {{{383} \over {373}}} \right) = 0.05$

$\Delta {S_{total}} = 9.26\,$ kJ Kg^$-$1 K^$-$1

Work done in isothermal process,

W = $-$ nRT ln${{{V_2}} \over {{V_1}}}$

$ \therefore $   $\left| W \right|$ = nRT ln${{{V_2}} \over {{V_1}}}$

$ \Rightarrow $   $\left| W \right|$ = nRT (lnV₂ $-$ lnV₁)

$ \Rightarrow $   $\left| W \right|$ = nRT lnV₂ $-$ nRT lnV₁

given that $V$₂ $=$ $V$

$ \therefore $   $\left| W \right|$ = nRT lnV $-$ nRT lnV₁

By comparing with the straight line equation,

y = mx + C

We get slope is nRT and intercept $-$ nRT lnV₁ in $\left| W \right|$ and lnV graph.

As T₂ > T₁   So,

Slope nRT₂ > nRT₁

and intercept

$-$ nRT₂ lnV₁ < $-$ nRT₁ lnV₁

So, we can say

(1)   slope of T₂ line is more then T₁

(2)  intercept of T₁ line is less negative than T₂ line, and intercept of T₁ can't be positive, it can be 0 or less than 0 as $-$ nRT₁lnV₁ always $ \le $ 0

The standard enthalpy of formation is defined as standard enthalpy change for formation of 1 mole of a substance from its elements, present in their most stable state of aggregation.

${3 \over 2}$O₂(g) $ \to $ O₃(g);

${1 \over 8}$S₈(s) + O₂(g) $ \to $ SO₂(g)

In the above two reactions standard enthalpy of reaction is equal to standard enthalpy of formation.

The explanation of given statements are as follows :

(a) U_rms is inversely proportional to the square root of its molecular mass.

${U_{rms}} = \sqrt {{{3RT} \over M}} $

Hence, option (a) is correct.

(b) When temperature is increased four times then U_rms become doubled.

${U_{rms}} = \sqrt {{{3R} \over M} \times 4T} $

${U_{rms}} = 2 \times \sqrt {{{3RT} \over M}} $

Hence, option (b) is correct.

(c) and (d) E_av is directly proportional to temperature but does not depends on its molecular mass at a given temperature as ${E_{av}} = {3 \over 2}KT$. If temperature raised four times than E_av becomes four time multiple.

Thus, option (c) is incorrect and option (d) is correct.

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$