Thermodynamics

$\Delta G=\Delta G^{\circ}+R T \text { in } Q$

where,

$\Delta G=$ free energy, $\Delta G^{\circ}=$ standard free energy

$R=$ gas constant, $T=$ temperature in Kelvin

$Q=$ reaction quotient at equilibrium.

$\Delta G^{\circ}=-R T \operatorname{ln} K$

where, $K=$ equilibrium constant

$\begin{array}{ll} \therefore & \Delta G=-R T \ln +R T \ln Q \\ \text { or } & \Delta G=-R T \ln \frac{K}{Q} \end{array}$

Hence, option (a) is the correct.

Using gas equation, $p V=\Delta n_g R T$

where, $p V=$ work done $(W)$

Thus, work done $(W)=\Delta n_g R T$

where, $\Delta n_g=$ number of moles of gaseous products number of moles of gaseous reactants.

Temperature, $T=25+273=298 \mathrm{~K}$

For the reaction,

$\begin{aligned} & \mathrm{C}_4 \mathrm{H}_{10}(g)+\frac{13}{2} \mathrm{O}_2(g) \longrightarrow 4 \mathrm{CO}_2(g)+5 \mathrm{H}_2 \mathrm{O}(l) \\ & \Delta n_g=4-\left(1+\frac{13}{2}\right)=-\frac{7}{2} \end{aligned}$

Thus,

$W =-\left[-\frac{7}{2} \mathrm{~mol} \times 0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K}\right]$

$=85.6 \mathrm{~L} \mathrm{~atm}$

Hence, option (b) is the correct.

External energy (U) is a state function as it depends only upon the states of the system (i.e. initial and final states) and does not depend on path temperature (T) is an intensive property as it does not depends on the quantity.

Thus, both assertion and reason are correct but given reason is not the correct answer for the given assertion.

Hence, option (b) is the correct answer.

For free expansion, $W=0$ and for adiabatic process, $q=0$.

$\because \Delta U=q+W=0$, this means that internal energy remains constant. Therefore, $\Delta T=0$ in ideal gas as there is no intermolecular attraction. Hence, when such a gas expands under adiabatic conditions into a vacuum, no heat is absorbed or evolved. Since, no external work is done to separate the molecules.

For any chemical reaction,

$\Delta H=\Delta E+\Delta n_g R T$

Where,

$\Delta n_g=$ total number of moles of gaseous products $-$ total number of moles of gaseous reactants

For the given reaction,

$\begin{aligned} & \mathrm{Fe}_2 \mathrm{O}_3(s)+3 \mathrm{H}_2(g) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{H}_2 \mathrm{O}(l) \\ & \Delta n_g=0-3=-3 \\ & \therefore \quad \Delta H=\Delta E+(-3) R T \\ & \Delta H=\Delta E-3 R T \\ \end{aligned}$