Thermodynamics

${C_2}{H_5}OH(I) + 3{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(I)$

$\Delta U = - 1364.47$ kJ/mol,

$ \Rightarrow \Delta {n_g} = - 1$

T = 25$^\circ$C = 298 K

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

$\Delta H = - 1364.47 + {{ - 1 \times 8314 \times 298} \over {1000}}$

[Here, value of R in unit of J must be converted into kJ]

$ = - 1364.47 - 2.4776 = - 1366.95$ kJ/mol

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

For a system, total entropy change is $\Delta$S_total,

$\Delta$H_total = 0

$\therefore$ $\Delta$G_system = $-$T$\Delta$S_total

$\therefore$ ${{\Delta {G_{system}}} \over {\Delta {S_{total}}}} = - T$

Thus, (a) is true,

(b) For isothermal reversible process, $\Delta$E = 0.

By first law of thermodynamics,

$\Delta$E = q $\times$ W

$\therefore$ W_reversible = $-$q = $-$$\int_{{V_t}}^{{V_f}} {p\,dV} $

W_reversible = $ - nRT\ln {{{V_f}} \over {{V_t}}}$

Thus, (b) is true.

(c) $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ $

$\Delta G^\circ = - nRT\ln \,K = - RT\ln K$ (for n = 1)

$\therefore$ $ - RT\ln K = \Delta H^\circ - T\Delta S^\circ $

$\therefore$ $\ln K = - \left( {{{\Delta H^\circ - T\Delta S^\circ } \over {RT}}} \right)$

Thus, (c) is incorrect.

(d) $\Delta G^\circ = - RT\ln K$

$\therefore$ $\ln K = - {{\Delta G^\circ } \over {RT}}$ $\therefore$ $K = {e^{ - \Delta G^\circ /RT}}$

Thus, (d) is also true.

∵ According to first law of thermodynamics, the total energy of an isolated system remains constant, through it may change from one form to another. Thus, statement (c) is correct. Hence, option (c) is the correct answer.

(i) $2 \mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{Cl}_2(\mathrm{~g}), \Delta n_g=1-2=-1$

So, $\Delta S<0$

(ii) $\mathrm{CO}_2(g) \longrightarrow \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) ; \Delta n_g=\left(1+\frac{1}{2}\right)-1=\frac{1}{2}$

So, $\Delta S>0$

(a) $\Delta G=\Delta G^{\circ}$ for a standard system of STP condition $\left(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)$.

The corrected statements of options-(a), (c) and (d) will be like,

(b) At equilibrium state, $\Delta G=0$.

(c) $\Delta G$ measures spontaneity of a reaction.

(d) When $\Delta G>0$, the reaction will proceed in backward direction.

∵ Given, Molar heat capacity for oxygen

$ \begin{aligned} & =20.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\ Q & =n C \Delta T \end{aligned} $

where, $Q=$ value of $\Delta U$ in J

$ \begin{aligned} n & =\text { number of moles }(=1) \\ C & =\text { heat capacity } \\ \Delta T & =\text { temperature } \\ Q & =1 \times 20.8 \times 60 \end{aligned} $

Thus, $\Delta T=(-) 20$ to $(+) 40=60^{\circ} \mathrm{C}$

Hence, for one mole of oxygen, value of $\Delta U=60 \times 20.8=1248 \mathrm{~J}$ Hence, option (d) is the correct answer.

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.

Statement (a) is correct. Rest of the all statements are incorrect. These are explained as follows:

(a) Cl₂(g) $\to$ 2Cl(g)

As randomness is increasing. Therefore, $\Delta$S is positive for this reaction.

$\therefore$ Statement is correct.

(b) In closed container, $\Delta$V = 0, hence work done is zero. There is no heat exchange. Hence, $\Delta$E = q + W = 0.

$\therefore$ Statement is incorrect.

(c) $\Delta$G will be zero only when equilibrium is reached.

$\therefore$ Statement is incorrect.

(d) $\Delta$G$^\circ$ = $-$RT ln K_eq, not a function of pressure. Thus, this statement is also incorrect.

At top of a mountain,

T₁ = 0 + 273 K, ${p_1} = {{710} \over {760}}$ atm and density = $\rho$₁

At bottom of mountain,

T₂ = 303 K, p₂ = 1 atm and density = $\rho$₂

$\because$ $pV = {w \over M}RT$ or $p = \rho {{RT} \over M}$

$\therefore$ ${{{\rho _1}} \over {{\rho _2}}} = {{{p_1}} \over {{T_1}}} \times {{{T_2}} \over {{p_2}}}$

or, ${{{\rho _1}} \over {{\rho _2}}} = {{710/760 \times 303} \over {273 \times 1}} = {{1.04} \over 1}$

Heat change at constant pressure means enthalpy change ($\Delta$_rH = q_p).

Heat of formation of CO is written as:

$C(s) + {1 \over 2}{O_2}(g) \to CO(g)$; ${\Delta _f}{H^\Theta }$ = ?

Now, we have,

I. $C(g) + {O_2}(g) \to C{O_2}(g)$; ${\Delta _c}H_1^\Theta $ = $-$393.5 kJ/mol

II. $CO(g) + {1 \over 2}{O_2}(g) \to C{O_2}(g)$; ${\Delta _c}H_2^\Theta $ = $-$ 283 kJ/mol

Substracting appropriately (i.e. I-II), we get,

$C(s) + {1 \over 2}{O_2}(g) \to CO(g)$; ${\Delta _f}H_{}^\Theta $

From Hess's law, ${\Delta _f}H_{}^\Theta = {\Delta _c}H_1^\Theta - {\Delta _c}H_2^\Theta $

$ \Rightarrow {\Delta _f}H_{}^\Theta = - 393.5 - ( - 283) = 110.5$ kJ/mol

Now, calculation of the heat of formation at constant volume means that we have to calculate change in internal energy (i.e. $\Delta$_rU)

Using, ${\Delta _r}H = {\Delta _r}U + p\Delta V$ [For a chemical reaction]

$ = {\Delta _r}U + \Delta {n_g}RT$ [As pV = nRT]

[$\Delta$n_g = gaseous moles of products $-$ gaseous moles of reactant]

$ \Rightarrow {\Delta _r}U = {\Delta _r}H - \Delta {n_g}RT$

Now, putting the values : $\Delta {n_g} = 1 - {1 \over 2} = {1 \over 2}$, T = 298 K,

R = 8.314 J/K/mol and ${\Delta _f}{H^\Theta }$ = $-$110.5 kJ/mol

$ \Rightarrow {\Delta _f}{H^\Theta } = - 110.5 - {1 \over 2} \times 8.314 \times 2.98 \times {10^{ - 3}}$

= $-$111.7 kJ/mol