Thermodynamics

Heat at constant pressure,

$ \begin{aligned} & q_p=n C_p \Delta T \\ & \Rightarrow \quad 1 \times 10.314 \times 1=10.314 \mathrm{~J} \end{aligned} $

Enthalpy change $\Delta H$

$ \Delta H=q_p=10.314 \mathrm{~J} / \mathrm{mol} $

Given, $\Delta S=-42.4 \mathrm{JK}^{-1}$

$ \Delta H=-41.2 \mathrm{~kJ}=-41200 \mathrm{~J} $

Using the equation, $\Delta G=\Delta H-T \Delta S$

$ T_{\mathrm{eq}}=\frac{-41200 \mathrm{~J}}{-42.4 \mathrm{JK}^{-1}} \approx 971.8 \mathrm{~K} $

A reversible process is one that can be reversed by an infinitesimal change in a condition, allowing the system and surroundings to be restored to their original states. Such process is occur at equilibrium.

I. Vaporisation of a liquid at its boiling point This is a reversible process because the liquid and vapour are in equilibrium at the boiling point.

II. Expansion of gas into vacuum This is an irreversible, spontaneous process as the system is not in equilibrium and cannot be reversed without external work.

III. Transformation of a solid into liquid at its melting point This is a reversible processes as the solid and liquid phases are in equilibrium at the melting point.

IV. Neutralisation of an acid by a base This is an irreversible, spontaneous reaction that proceeds to completion.

Therefore, only processes I and III are reversible.

$ \text { Gibbs' free energy is given by } $

$ \begin{array}{rlrl} \Delta G^{\circ} & =-R T \ln (K) \\ \Rightarrow \quad-115000 \mathrm{~J} & =-8.314 \times 298 \times \ln K_p \\ \Rightarrow \quad \log _{10}\left(K_p\right) & =\frac{46.42}{2.303} \\ & \log _{10}\left(K_p\right) =+20.15 \end{array} $

$ \begin{aligned} &\text { Work done in isothermal reversible expansion, }\\ &\begin{aligned} W & =-n R T \ln \left(\frac{p_2}{p_1}\right) \\ W & =-(1)(8.3)(300) \ln \left(\frac{2}{20}\right)=5730 \mathrm{~J} / \mathrm{mole} \text { or } 5.73 \mathrm{~kJ} / \mathrm{mol} \\ x & =5.73 \end{aligned} \end{aligned} $

$ \text { } \begin{array} GGiven,\,\,{r} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^{\circ}=-286 \mathrm{~kJ}\,\,\,\,\,\,\,\,\,\,\, \ldots(\mathrm{i}) \end{array} $

$ \begin{aligned} \mathrm{C}(s)+\mathrm{O}_2(g) & \longrightarrow \mathrm{CO}_2(g);\Delta H^{\circ} =-394 \mathrm{~kJ} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$ \mathrm{CH}_4(g)+2 \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^{\circ}=-890 \mathrm{~kJ}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

$ \begin{aligned} &\text { Reverse reaction given in Eq. (ii), }\\ &\begin{aligned} & \mathrm{CO}_2(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{~s})+\mathrm{O}_2(g) ; \Delta H^{\circ}=+394 \mathrm{~kJ} \,\,\,\,\,\ldots(\mathrm{iv}) \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Now, add Eq. (iii) and (iv), }\\ &\begin{gathered} \mathrm{CH}_4(g)+2 \mathrm{O}_2(g)+\mathrm{CO}_2(g) \longrightarrow \mathrm{CO}_2(s)+2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{C}(s)+\mathrm{O}_2(g) \\ \mathrm{CH}_4(g)+\mathrm{O}_2(g) \longrightarrow \mathrm{C}(s)+2 \mathrm{H}_2 \mathrm{O}(l) \\ \text { So, } \Delta H^{\circ}=-890+394=-496 \mathrm{~kJ} \end{gathered} \end{aligned} $

Given, temperature $ =300 \mathrm{~K} $, number of moles $ (n)=3 $

Change in volume $ (\Delta V)=-\frac{1}{2}(V) $ where

$ V $ is initial volume and external pressure, $ p_{\text {ext }}=6 \mathrm{~atm} $.

Pressure at 300 K of gas, $ p=3 \mathrm{~atm} $

Work done in isothermal compression is given by

$ \begin{array}{l} W=-p_{\mathrm{ext}} \Delta V \\\\ W=-p_{\mathrm{ext}}\left(-\frac{V}{2}\right) \end{array} $

Using ideal gas equation $ p V=n R T $,

$ V=\frac{n R T}{p} $

Substitute the value of $ V $ in Eq. (ii)

$ W=p_{\mathrm{ext}} \frac{(n R T)}{2 p} $

Substitute all values in Eq. (iv)

$ \begin{array}{l} W=\frac{6 \times 3 \times 0.082 \times 300}{2 \times 3} \times 101.3 \mathrm{~J} \\\\ W=7475.94 \mathrm{~J} \text { or } W=7.476 \mathrm{~kJ} \end{array} $

Given values:

$\Delta_{f} H^{\ominus}$ of $AO(s) = -635 \, \text{kJ/mol}$

$\Delta_{f} H^{\ominus}$ of $BO_{2}(g) = x \, \text{kJ/mol}$

$\Delta_{f} H^{\ominus}$ of $ABO_{3}(s) = -1210 \, \text{kJ/mol}$

$\Delta_{r} H^{\ominus} = 175 \, \text{kJ/mol}$ for the reaction:

$ ABO_{3}(s) \longrightarrow AO(s) + BO_{2}(g) $

The enthalpy change for the reaction, $\Delta_{r} H^{\ominus}$, is calculated using the formula:

$ \Delta_{r} H^{\ominus} = \left[ \Delta_{f} H^{\ominus}(AO) + \Delta_{f} H^{\ominus}(BO_{2}) \right] - \Delta_{f} H^{\ominus}(ABO_{3}) $

Substitute the known values:

$ 175 = [-635 + x] - [-1210] $

Simplifying:

$ 175 = x + 1210 - 635 $

$ 175 = x + 575 $

$ x = 175 - 575 $

$ x = -400 \, \text{kJ/mol} $

Therefore, the value of $x$ is $-400 \, \text{kJ/mol}$.

According to given data

$ \begin{array}{l} \mathrm{C}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \\\\ \Delta H_{C}=-393 \mathrm{~kJ} / \mathrm{mol} \\\\ \mathrm{H}_{2}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \\\\ \Delta H_{C}=-286 \mathrm{~kJ} / \mathrm{mol} \end{array} $

Multiply Eq. (ii) by (2),

$ \begin{array}{r} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \\\\ \Delta H_{C}=-572 \mathrm{~kJ} / \mathrm{mol} . \end{array} $

For combustion of $ \mathrm{CH}_{3} \mathrm{OH}(l) $

$ \begin{array}{r} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\\\ \Delta H_{C}=-726 \mathrm{~kJ} / \mathrm{mol} \\\\ \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}+\frac{3}{2} \mathrm{O}_{2} \end{array} $

$ \Delta H_{C}=+726 \mathrm{~kJ} / \mathrm{mol} $

Add Eqs. (i), (iii) and (iv), we get

$ \begin{aligned} \mathrm{C}(\mathrm{~s})+2 \mathrm{H}_{2}(g) & +\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH} \\\\ \Delta H_{C} & =-393+(-572)-(-726) \\\\ & =-239 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

Given,

Initial volume $\left(V_i\right)=1.0 \mathrm{~L}$

Final volume $\left(V_f\right)=10.0 \mathrm{~L}$

$n$ (number of moles) $=1$

Temperature $=61 \mathrm{~K}$

Now, workdone for isothermal process is given by

$ W=-n R T \times 2.303 \log _{10}\left(\frac{V_f}{V_i}\right) $

Substitute all the values,

$ \begin{aligned} W & =-2.303 \times 1 \times 0.0821 \times 61 \times \log _{10}\left(\frac{10}{1}\right) \\\\ & =-11.52 \mathrm{~L} \mathrm{~atm} \end{aligned} $

Given reaction is

$ \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \longrightarrow 2 \mathrm{NH}_3 $

Two moles of ammonia are formed during the reaction. Thus, the reaction for formation of one mole of ammonia is given

$ \Delta_f H^{\ominus}=-46.2 \mathrm{~kJ} / \mathrm{mol} $

The standard enthalpy of reaction on is

$ \begin{aligned} \Delta_r H^{\ominus} & =2 \times \Delta_f H^{\ominus} \\ & =2 \times-46.2 \\ & =-92.4 \mathrm{~kJ} \end{aligned} $

Work done for isothermal and reversible process is given by

$ W=-n R T \ln \frac{p_1}{p_2} $

$ \begin{aligned} & =-2.303 n R T \log \frac{p_1}{p_2} \\ & =-2.303 \times 2 \times 8.3 \times T \log \frac{10}{1} \\ & {\left[\begin{array}{l} \text { where, } n=\text { no. of moles } \\ R=\text { gas constant } \\ T=\text { temperature } \\ p_1=\text { initial pressure } \\ p_2=\text { final pressure } \end{array}\right] } \\ & =-38.2 \times T \mathrm{~J} \text { or }-3.82 \times 10^{-2} \times T \mathrm{~kJ} \end{aligned} $

$ \begin{aligned} & \mathrm{X}_2 \mathrm{O}_4(l) \longrightarrow 2 \mathrm{XO}_2(g) \\ & \Delta U=2.1 \mathrm{k} \mathrm{cal} \\ & \Delta S=20 \mathrm{cal} / \mathrm{K} \\ & R=2 \mathrm{cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \end{aligned} $

Using formula,

$ \begin{aligned} \Delta H & =\Delta U+\Delta n_g R T \\ & =2.1+(2-0) \times 2 \times 298 \\ & =2.1 \times 10^3+(2 \times 2 \times 298)=3292 \mathrm{cal} \\ & =3.292 \mathrm{kcal} \end{aligned} $

Gibbs energy change can be calculated as

$ \begin{aligned} \Delta \mathrm{G} & =\Delta H-T \Delta S \\ & =3292-(298 \times 20)=-2668 \mathrm{cal} \\ or\,\,\,\,\,& -2.67 \mathrm{kcal} \end{aligned} $

The properties which do not depend upon the quantity or size of matter are called intensive properties. Among the given list molar volume (I), temperature (VI) and density (VII) are intensive properties.

Given,

$ \begin{aligned} & \mathrm{C}+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{CO} ; \Delta H_f=-110 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{C}+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2 ; \Delta H_f^1=-393 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

To find enthalpy of combustion of CO .

i.e. $\mathrm{CO}+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{CO}_2 ; \Delta H_{\mathrm{CO}}=$ ?

$ \begin{aligned} \Delta H_{\mathrm{CO}} & =H_f^1-H_f \\ & =-393-(-110) \\ & =-283 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

Given the following information about enthalpies of formation at 298 K:

For the formation of $\mathrm{N}_2 \mathrm{O}$:

$ \mathrm{N}_2 + \frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{N}_2 \mathrm{O}; \quad \Delta H_{f(\mathrm{N}_2 \mathrm{O})} = 82 \, \mathrm{kJ/mol} $

For the formation of $\mathrm{NO}$:

$ \frac{1}{2} \mathrm{N}_2 + \frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{NO}; \quad \Delta H_{f(\mathrm{NO})} = 90 \, \mathrm{kJ/mol} $

To find the enthalpy change for the reaction:

$ \mathrm{N}_2 \mathrm{O}(g) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) $

We use the formula for calculating the enthalpy change of the reaction based on the enthalpies of formation:

$ \Delta H = \sum \Delta H_f (\text{products}) - \sum \Delta H_f (\text{reactants}) $

Applying this to the given reaction:

$ \Delta H = \left( 2 \times \Delta H_{f(\mathrm{NO})} \right) - \left( \Delta H_{f(\mathrm{N}_2 \mathrm{O})} + \frac{1}{2} \Delta H_{f(\mathrm{O}_2)} \right) $

Since the formation enthalpy of a pure element in its standard state, like $\mathrm{O}_2$, is zero:

$ \Delta H = 2 \times 90 \, \mathrm{kJ/mol} - 82 \, \mathrm{kJ/mol} $

Calculating gives:

$ \Delta H = 180 \, \mathrm{kJ/mol} - 82 \, \mathrm{kJ/mol} = 98 \, \mathrm{kJ/mol} $

Therefore, the enthalpy change of the reaction is $+98 \, \mathrm{kJ/mol}$.

We can solve this using the formula for the work done by an ideal gas during an isothermal, reversible expansion:

$ W = -nRT \ln\left(\frac{V_2}{V_1}\right) $

where:

• $ W $ is the work done by the gas (given as $-16.5\,\text{kJ}=-16500\,\text{J}$),

• $ n = 2.5\,\text{mol} $,

• $ R = 8.314\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1} $,

• $ V_1 = 2\,\text{dm}^3 $ and $ V_2 = 20\,\text{dm}^3 $.

Follow these steps:

Write down the work formula:

$ W = -nRT\ln\left(\frac{V_2}{V_1}\right) $

Solve for the temperature $ T $:

$ T = -\frac{W}{nR\ln\left(\frac{V_2}{V_1}\right)} $

Substitute the known values:

$ T = -\frac{-16500}{2.5 \times 8.314 \times \ln\left(\frac{20}{2}\right)} $

Calculate the volume ratio and its natural logarithm:

$ \frac{V_2}{V_1} = \frac{20}{2} = 10, \quad \text{so} \quad \ln(10) \approx 2.3026. $

Now compute the temperature:

$ T = \frac{16500}{2.5 \times 8.314 \times 2.3026}. $

First, compute the product in the denominator:

$ 2.5 \times 8.314 = 20.785, $

and then

$ 20.785 \times 2.3026 \approx 47.88. $

Finally, calculate $ T $:

$ T \approx \frac{16500}{47.88} \approx 344.8 \, \text{K}. $

Rounding off to the nearest value gives approximately $ 345 \, \text{K} $.

Thus, the correct option is Option C: 345.

Enthalpy of formation of $\mathrm{D}_2 \mathrm{O}$ will be

$ \begin{aligned} & \mathrm{D}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{D}_2 \mathrm{O} \\ & \Delta H=\Delta H_p \text { (Products) } \\ & =400+\frac{1}{2} \times 498-490 \times 2 \\ & \Delta H_r \text { (Reactants) } \\ & =400+249-980=-331 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

$ \begin{aligned} & \Delta H=\Delta U+\Delta(p V) \\ & \Delta H=\Delta U+\Delta n(R T) \\ & \quad p=1 \mathrm{~atm}, T=0^{\circ} \mathrm{C}=273 \mathrm{~K} \end{aligned} $

$ \begin{aligned}As,& \Delta H=\Delta U \\ & \Delta U=-41 \mathrm{~kJ} / \mathrm{mol}\,\,\,\,\,\,\,\,\,\,\,(given) \end{aligned} $

Therefore, $\Delta H=-41 \mathrm{~kJ} / \mathrm{mol}$

Graphite is the most stable state of carbon and its $\Delta H_f^{\circ}$ is considered as zero.

( ∵ has more van der Waals' force) also $\Delta H_f^{\circ}$ for $\mathrm{C}_{60}>\Delta H_f^{\circ}$ for diamond. $\Delta H_f^{\circ}$ values of graphite, diamond and $\mathrm{C}_{60}$ are $0,1.9$ and $38.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta H_f^{\circ}$ values are in order.

Fullerene $>$ Diamond $>$ Graphite.

In an adiabatic expansion,

$ p V^{\top}=\text { constant } $

where, $p=$ pressure

$V=$ volume

$\gamma=$ ratio of specific heat at constant pressure and constant volume $\left(\frac{C_p}{C_V}\right)$

Differentiating above equation,

$ \begin{aligned} & d p \cdot V^\gamma+\gamma V^{\gamma-1} \cdot p d V=0 \\ & \frac{d p}{p}=\frac{-\gamma d V}{V} \\ & \Rightarrow \quad \frac{d p}{p}=-(1.4) \times 5 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-7 \% \end{aligned} $

∴ Percentage change in pressure is $7 \%$.

Given,

$ \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; $

Also, $\quad 2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow 2 \mathrm{H}_2(g)+\mathrm{O}_2(g)$;

$ \Delta H_1^{\prime}=285 \times 2=+570 \mathrm{~kJ} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \mathrm{N}_2 \mathrm{O}_5(g)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_3(l) \text {; } $

$ \Delta H_2=-76.6 \mathrm{~kJ} $

Also,

$ \begin{gathered} 4 \mathrm{HNO}_3(l) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_5(g)+2 \mathrm{H}_2 \mathrm{O}(l) ; \\ \Delta H_2^{\prime}=76.6 \times 2=+153.2 \mathrm{~kJ} \quad \mathrm{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{gathered} $

$ \begin{aligned} & \mathrm{N}_2(g)+3 \mathrm{O}_2(g)+\mathrm{H}_2(g) \longrightarrow 2 \mathrm{HNO}_3(l) ; \\ & \Delta H_3=-348.2 \mathrm{~kJ} \end{aligned} $

Also,

$ \begin{aligned} 2 \mathrm{~N}_2(g)+6 \mathrm{O}_2(g)+2 \mathrm{H}_2(g) & \longrightarrow 4 \mathrm{HNO}_3(l) \\ \Delta H_3^{\prime} & =-348.2 \times 2 \\ & =-696.4 \mathrm{~kJ} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

On adding Eqs. (i), (ii) and (iii), we get

$ \begin{aligned} & 2 \mathrm{~N}_2(g)+5 \mathrm{O}_2(g) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_5(g) ; \\ & \Delta H=\Delta H_1^{\prime}+\Delta H_2^{\prime}+\Delta H_3^{\prime} \\ & \Delta H=570+153.2-696.4 \\ & \Delta H=723.2-696.4 \\ & \Delta H=26.8 \mathrm{~kJ} . \end{aligned} $

$ \begin{aligned} & \text { } \mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+}+2 e^{-} ; \\ & \begin{array}{l} \Delta H_1=\mathrm{IE}_1+\mathrm{IE}_2 \\ 2 \mathrm{~F}+2 e^{-} \longrightarrow 2 \mathrm{~F}^{-} ; \Delta H_2=2 \mathrm{EA}_1 \\ \mathrm{Mg}+2 \mathrm{~F}^{-} \longrightarrow \mathrm{Mg}^{2+}+2 \mathrm{~F}^{-} ; \Delta H=\text { ? } \\ \Delta H=\Delta H_1+\Delta H_2=737+1451+2(-328) \\ =1532 \mathrm{~kJ} / \mathrm{mol} \end{array} \end{aligned} $

$\Delta U=q+w$

$\Delta U$ will be same for path 1,2 and 3 because it depends on initial and final positions and they are same for all of them.

Work done is area under $p \Delta V$ curve. Greater the area more will be the work done. Work is a path function.

Since the area under $p \Delta V$ curve is largest for path 3 and smallest for path 1, therefore, $w_1

The standard enthalpy of formation of all elements in their standard states is zero. Since the natural standard state of carbon is graphite, the enthalpy of formation of graphite is zero, as there is no change involved in its formation.

$2 \mathrm{Na}(s)+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow2 \mathrm{NaOH}(a q)+\mathrm{H}_2(g) $

1 mole of $\mathrm{Na}=23 \mathrm{~g}$ of Na

4 moles of $\mathrm{Na}=92 \mathrm{~g}$ of Na

2 moles of Na reacts with 2 moles of $\mathrm{H}_2 \mathrm{O}$ to form

1 mole of $\mathrm{H}_2$.

4 moles of Na will react with 4 moles of $\mathrm{H}_2 \mathrm{O}$ to form 2 moles of $\mathrm{H}_2$.

Work done by expansion of gas is given by $-p \Delta V$

$ p \Delta V=\Delta n_g R T $

At constant external pressure,

$ \begin{aligned} \Delta V=V_{\mathrm{H}_2}-0 & =V_{\mathrm{H}_2} \\ p \Delta V & =\Delta n_g R T \\ p \Delta V=\Delta n_g R T & =2 \times 8.314 \times 300=4.98 \mathrm{~kJ} \end{aligned} $

Therefore, the work done is 4988.4 J .

Since it is work of expansion so will be with negative sign (-4988.4 J).

∵ 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.