Answer

(c) Thermodynamics is not concerned with the rate at which a reaction proceeds. Thermodynamics deals with the energy change, feasibility and extent of a reaction, but not with the rate and mechanism of a process.

• (a) Thermodynamics is concerned with energy changes involved in a chemical reaction because it studies how energy is transferred and transformed during chemical processes.
• (b) Thermodynamics is concerned with the extent to which a chemical reaction proceeds because it involves the study of equilibrium states and how far a reaction will go before reaching equilibrium.
• (d) Thermodynamics is concerned with the feasibility of a chemical reaction because it determines whether a reaction can occur spontaneously based on the changes in enthalpy, entropy, and free energy.

Answer

(c) For a closed vessel made of copper, no matter can exchange between the system and the surroundings but energy exchange can occur through its walls.

Presence of reaction species in a covered beaker-closed system and exchange of matter as well as energy-open-system. Presence of reactant in a closed vessel closed system and presence of reactant in thermos flask-isolated system.

• (a) The presence of reacting species in a covered beaker is not an example of an open system because a covered beaker restricts the exchange of matter with the surroundings, making it a closed system.

• (b) There is no exchange of matter between the system and the surroundings in a closed system; only energy exchange can occur. Therefore, the statement is incorrect.

• (d) The presence of reactants in a thermos flask or any other closed insulated vessel is an example of an isolated system, not a closed system, because it prevents the exchange of both matter and energy with the surroundings.

Answer

(d) The state of a gas can be described by quoting the relationship between pressure, volume, temperature and amount. The ideal gas equation is

$$ p V=n R T $$

Thus, $p, V, T$ and $n$ describe the state of the system.

• (a) pressure, volume, temperature: This option is incorrect because it does not include the amount of gas (n), which is a crucial variable in describing the state of a gas. The ideal gas law ( pV = nRT ) requires the amount of gas to fully describe the state.

• (b) temperature, amount, pressure: This option is incorrect because it does not include the volume (V) of the gas. The volume is necessary to describe the state of the gas according to the ideal gas law ( pV = nRT ).

• (c) amount, volume, temperature: This option is incorrect because it does not include the pressure (p) of the gas. Pressure is a key variable in the ideal gas law ( pV = nRT ) and is needed to fully describe the state of the gas.

Answer

(c) The volume of gas is reduced to half from its original volume. The specific heat will be remain constant.

Specific heat is an intensive property depending only on the nature of the gas.

• (a) The specific heat does not reduce to half because specific heat is an intensive property, meaning it does not depend on the amount of substance or its volume.

• (b) The specific heat does not double because it is an intrinsic property of the material and remains constant regardless of changes in volume or mass.

• (d) The specific heat does not increase four times because it is independent of the volume of the gas and is determined by the nature of the gas itself.

Answer

(c) Given that, the complete combustion of one mole of butane is represented by thermochemical reaction as

$$ C_{4} H_{10}(g)+\frac{13}{2} O_{2}(g) \rightarrow 4 CO_{2}(g)+5 H_{2} O(l) $$

We have to take the combustion of one mole of $C_{4} H_{10}$ and $\Delta_{C} H$ should be negative and have a value of $2658 ~kJ ~mol^{-1}$.

• Option (a) is incorrect because it represents the combustion of 2 moles of butane, not 1 mole. Additionally, the enthalpy change should be for 1 mole of butane, not 2 moles.

• Option (b) is incorrect because the enthalpy change value is halved to $-1329.0 ~kJ ~mol^{-1}$, which does not match the given enthalpy change of $-2658.0 ~kJ ~mol^{-1}$ for the combustion of one mole of butane.

• Option (d) is incorrect because the enthalpy change is given as positive ($+2658.0 ~kJ ~mol^{-1}$), whereas it should be negative ($-2658.0 ~kJ ~mol^{-1}$) for an exothermic reaction like combustion.

Answer

(b) The reaction is $CH_{4}(g)+2 O_{2}(g) \rightarrow CO_{2}(g)+2 H_{2} O(l)$

$$ \Delta n_{g} =\left(n_{p}-n_{r}\right)_{g}=1-3=-2 $$

$$ \Delta_{f} H^{-s} =\Delta_{f} U^{s}+\Delta n_{g} R T $$

$$\text { As } \quad \Delta_{g} =-2 $$

$$\therefore \quad\Delta_{f} H^{s} <\Delta_{f} U^{s} $$

• (a) Zero: The enthalpy change of formation, $(\Delta_{f} H^{s})$, is not zero because the formation of $(CH_{4}(g))$ from its elements involves a significant energy change, as indicated by the given $(\Delta_{f} U^{s})$ value of $(-393 , kJ , mol^{-1})$.

• (c) $(>\Delta_{f} U^{s})$: For $(\Delta_{f} H^{s})$ to be greater than $(\Delta_{f} U^{s})$, the term $(\Delta n_{g} RT)$ would need to be positive. However, in this case, $(\Delta n_{g})$ is negative $(\Delta n_{g} = -2)$, making $(\Delta n_{g} RT)$ a negative value, which decreases $(\Delta_{f} H^{s})$ relative to $(\Delta_{f} U^{s})$.

• (d) Equal to $(\Delta_{f} U^{s})$: $(\Delta_{f} H^{s})$ is not equal to $(\Delta_{f} U^{s})$ because the term $(\Delta n_{g} RT)$ is non-zero. The difference in the number of moles of gas between reactants and products $(\Delta n_{g} = -2)$ results in a non-zero correction term, making $(\Delta_{f} H^{s})$ different from $(\Delta_{f} U^{s})$.

Answer

(c) Free expansion, $W=0$

Adiabatic process, $q=0$

$\Delta U=q+W=0$, this means that internal energy remains constant. Therefore, $\Delta T=0 \mathrm{ln}$ ideal gas 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.

• Option (a): This option states that $( q=0 )$, $( \Delta T \neq 0 )$, and $( W=0 )$. In a free expansion of an ideal gas under adiabatic conditions, no work is done $( W=0 )$ and no heat is transferred $( q=0 )$. Since the internal energy of an ideal gas depends only on temperature, and there is no change in internal energy $( \Delta U = 0 )$, the temperature must remain constant $( \Delta T = 0 )$. Therefore, $( \Delta T \neq 0 )$ is incorrect.

• Option (b): This option states that $( q \neq 0 )$, $( \Delta T = 0 )$, and $( W=0 )$. In an adiabatic process, by definition, there is no heat transfer $( q=0 )$. Therefore, $( q \neq 0 )$ is incorrect.

• Option (d): This option states that $( q=0 )$, $( \Delta T < 0 )$, and $( W \neq 0 )$. In a free expansion, no work is done $( W=0 )$. Therefore, $( W \neq 0 )$ is incorrect. Additionally, since $( \Delta U = 0 )$ in an adiabatic free expansion, the temperature change $( \Delta T )$ must be zero, not less than zero. Therefore, $( \Delta T < 0 )$ is also incorrect.

Answer

(b) The correct option is W (reversible) $<W$ (irreversible). This is because area under the curve is always more in irreversible compression as can be seen from given figure.

$p V$-plot when pressure is not constant and changes in finite steps during compression from initial volume, $V_{i}$ to final volume, $V_{f}$. Work done on the gas is represented by the shaded area.

$P V$-plot when pressure is not constant and changes in infinite steps (reversible conditions) during compression from initial volume, $V_{i}$ to final volume, $V_{f}$. Work done on the gas is represented by the shaded area.

• Option (a) $W$ (reversible) $=W$ (irreversible): This option is incorrect because, in a reversible process, the work done is typically less than in an irreversible process due to the system being in equilibrium at each infinitesimal step. In an irreversible process, there are finite steps and often higher external pressures, leading to more work done on the gas.

• Option (c) $W$ (reversible) $=W$ (irreversible): This option is incorrect for the same reason as option (a). The work done in a reversible process is less than in an irreversible process because the reversible process involves infinitesimally small changes, maintaining equilibrium, and thus requires less work.

• Option (d) $W$ (reversible) $=W$ (irreversible) $+p_{\mathrm{ex}} \cdot \Delta V$: This option is incorrect because it suggests that the work done in a reversible process is equal to the work done in an irreversible process plus an additional term involving external pressure and volume change. However, in reality, the work done in a reversible process is inherently less than in an irreversible process due to the nature of the infinitesimal steps and equilibrium conditions in the reversible process. The additional term does not accurately represent the relationship between the two types of work.

Answer

(c) The entropy change can be calculated by using the expression

$$ \Delta S=\frac{q_{\mathrm{rev}}}{T} $$

When water freezes in a glass beaker, $\Delta S$ (system) decreases because molecules in solid ice are less random than in liquid water. However, when water freezes to ice, heat is released which is absorbed by the surroundings.

Hence, entropy of the surrounding increases.

• (a) $\Delta S$ (system) decreases but $\Delta S$ (surroundings) remains the same: This is incorrect because when water freezes, heat is released to the surroundings, which increases the entropy of the surroundings. Therefore, $\Delta S$ (surroundings) does not remain the same; it increases.

• (b) $\Delta S$ (system) increases but $\Delta S$ (surroundings) decreases: This is incorrect because the entropy of the system (water) decreases when it freezes, as the molecules become more ordered in the solid state. Additionally, the entropy of the surroundings increases due to the release of heat.

• (d) $\Delta S$ (system) decreases but $\Delta S$ (surroundings) also decreases: This is incorrect because while the entropy of the system decreases as water freezes, the entropy of the surroundings increases due to the release of heat. Both cannot decrease simultaneously in this process.

Answer

(c) The algebraic relationships of the given reaction is equation (a) -equation (b) = equation (c)

(a) $C$ (graphite) $+O_{2}(g) \rightarrow CO_{2}(g) ; \Delta_{r} H=x ~kJ ~mol^{-1}$

(b) $C$ (graphite) $+\frac{1}{2} O_{2}(g) \rightarrow CO(g) ; \Delta_{r} H=y ~kJ ~mol^{-1}$

$CO(g)+\frac{1}{2} O_{2}(g) \rightarrow CO_{2}(g) ; \Delta_{r} H=z ~kJ ~mol^{-1}$

Hence, $x-y=z$ or $x=y+z$

• Option (a) $z=x+y$: This is incorrect because, according to Hess’s Law, the enthalpy change for the overall reaction should be the sum of the enthalpy changes of the individual steps. Here, the correct relationship derived from the given equations is $x = y + z$, not $z = x + y$.

• Option (b) $x=y-z$: This is incorrect because, based on the given thermochemical equations, the correct relationship is $x = y + z$. Rearranging this equation would give $y = x - z$, not $x = y - z$.

• Option (d) $y=2z-x$: This is incorrect because, from the correct relationship $x = y + z$, rearranging terms would give $y = x - z$. The expression $y = 2z - x$ does not align with the derived relationship from the given equations.

Answer

(c) Same bonds are formed in reaction (1) and (2) but no bonds are broken in reaction (1) whereas bonds in the reactant molecules are broken in reaction (2). As energy is absorbed when bonds are broken, energy released in reaction (1) is greater than that in reaction (2) hence, $x>y$

• Option (a) $x=y$: This is incorrect because the two reactions involve different initial states of carbon. In reaction (1), carbon is in the gaseous state, while in reaction (2), carbon is in the graphite form. The energy required to convert graphite to gaseous carbon is not accounted for in reaction (1), leading to different enthalpy changes.

• Option (b) $x=2y$: This is incorrect because the relationship between the enthalpy changes of the two reactions is not a simple multiple. The enthalpy change in reaction (2) includes the energy required to break the bonds in $H_2$ and convert graphite to gaseous carbon, which is not directly proportional to the enthalpy change in reaction (1).

• Option (d) $x<y$: This is incorrect because, as explained, reaction (1) does not involve breaking any bonds, whereas reaction (2) does. Breaking bonds requires energy, so the enthalpy change for reaction (2) (y) is less negative (or more positive) than for reaction (1) (x), making $x$ greater than $y$.

Answer

(c) Combustion of elements to form a compound can be exothermic or endothermic. e.g., $C+O_{2} \rightarrow CO_{2}$ is exothermic.

whereas, $\mathrm{C}+2 \mathrm{~S} \rightarrow \mathrm{CS}_{2}$ is endothermic.

Hence, enthalpy of formation can be positive or negative.

• (a) is always negative: This is incorrect because the enthalpy of formation can be either exothermic (negative) or endothermic (positive). Not all formation reactions release energy; some absorb energy.

• (b) is always positive: This is incorrect because the enthalpy of formation can be either positive or negative. While some formation reactions absorb energy, others release energy.

• (d) is never negative: This is incorrect because the enthalpy of formation can indeed be negative. Many formation reactions are exothermic, meaning they release energy and have a negative enthalpy change.

Answer

(a) Enthalpy of sublimation of a substance is equal to enthalpy of fusion + enthalpy of vaporisation.

Sublimation is, direct conversion of solid to vapour. solid $\rightarrow$ vapour

Writing in two steps, we have solid $\rightarrow$ liquid $\rightarrow$ vapour,

solid $\rightarrow$ liquid requires enthalpy of fusion

liquid $\rightarrow$ vapour requires enthalpy of vaporisation

• (b) Enthalpy of fusion: This option is incorrect because the enthalpy of sublimation involves both the phase change from solid to liquid (enthalpy of fusion) and the phase change from liquid to vapor (enthalpy of vaporization). Enthalpy of fusion alone only accounts for the energy required to convert a solid to a liquid, not to a vapor.

• (c) Enthalpy of vaporisation: This option is incorrect because the enthalpy of sublimation includes the energy required for both the solid to liquid transition (enthalpy of fusion) and the liquid to vapor transition (enthalpy of vaporization). Enthalpy of vaporization alone only accounts for the energy required to convert a liquid to a vapor, not from a solid to a vapor.

• (d) Twice the enthalpy of vaporisation: This option is incorrect because the enthalpy of sublimation is not simply twice the enthalpy of vaporization. It is the sum of the enthalpy of fusion and the enthalpy of vaporization, which are distinct and separate energy changes.

Answer

(b) $\Delta G$ gives a criteria for spontaneity at constant pressure and temperature.

(i) If $\Delta G$ is negative $(<0)$, the process is spontaneous.

(ii) If $\Delta G$ is positive ( $>0$ ), the process is non-spontaneous.

(iii) If $\Delta G$ is zero then reaction is equilibrium.

• Option (a) is correct because $\Delta \mathrm{G}$ is indeed zero for a reversible reaction, indicating that the system is at equilibrium.

• Option (c) is correct because $\Delta \mathrm{G}$ being negative indicates that the reaction is spontaneous.

• Option (d) is correct because $\Delta \mathrm{G}$ being positive indicates that the reaction is non-spontaneous.

Answer

$(a, d)$

Thermodynamics deals with interrelation of various forms of energy and their transformation into each other. It also deals with thermal or mechanical equilibrium. However, it does not tell anything about the rate of reaction.

• Option (b) is incorrect because thermodynamics does not focus on processes involving only a few molecules; it generally deals with macroscopic systems containing a large number of molecules.
• Option (c) is incorrect because thermodynamics does not concern itself with the rate at which energy transformations occur; that is the domain of kinetics.

Answer

$(a, b)$

Exothermic reactions are those reactions which are accompanied by the evolution of heat.

e.g.,

$$ \begin{aligned} & C(s)+O_{2}(g) \rightarrow CO_{2}(g)+393.5 ~kJ \\ & H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)+285.8 ~kJ \end{aligned} $$

$q_{p}$ and $\Delta_{r} H$ are negative for exothermic reaction.

• (c) $q_{p}$ will be positive: This option is incorrect because in an exothermic reaction, the system loses heat to the surroundings. Therefore, the heat exchanged at constant pressure ($q_{p}$) is negative, not positive.

• (d) $\Delta_{r} H$ will be positive: This option is incorrect because the enthalpy change ($\Delta_{r} H$) for an exothermic reaction is negative. This is because the system releases heat, resulting in a decrease in enthalpy.

Answer

(c, $d)$

Options (a) and (b) can neither occur by themselves nor by initiation, (c) can occur by itself,

(d) occur on initiation. Flowing of heat from warmer to colder body, expanding of gas and burning of carbon to give carbon dioxide, all are spontaneous process.

• Option (a) is incorrect because heat naturally flows from a warmer body to a colder body, not the other way around. The flow of heat from a colder to a warmer body would require external work or intervention, making it non-spontaneous.

• Option (b) is incorrect because a gas naturally tends to expand and fill the available volume due to the increase in entropy. A gas contracting into one corner of a container would require external work or intervention, making it non-spontaneous.

Answer

$(c, d)$

Given that, the work of reversible expansion under isothernal condition can be calculated by using the expression

$$ \begin{aligned} W & =-n R T \ln \frac{V_{f}}{V_{i}} \\ V_{f} & =10 V_{i} \\ T_{2} & =600 \mathrm{~K} \\ T_{1} & =300 \mathrm{~K} \end{aligned} $$

Putting these values in above expression

$$ \begin{aligned} & W_{600 \mathrm{~K}}=1 \times R \times 600 \mathrm{~K} \ln \frac{10}{1} \\ & W_{300 \mathrm{~K}}=1 \times R \times 300 \mathrm{~K} \ln \frac{10}{1} \\ & \text { Ratio }=\frac{W_{600 \mathrm{~K}}}{W_{300 \mathrm{~K}}}=\frac{1 \times R \times 600 \mathrm{~K} \ln \frac{10}{1}}{1 \times R \times 300 \mathrm{~K} \ln \frac{10}{1}}=\frac{600}{300}=2 \end{aligned} $$

For isothermal expansion of ideal gases, $\Delta U=0$. Since, temperature is constant this means there is no change in internal energy Therefore $\Lambda =0$

• Option (a): Work done at $600 \mathrm{~K}$ is 20 times the work done at $300 \mathrm{~K}$ is incorrect because the work done is directly proportional to the temperature. Since $600 \mathrm{~K}$ is twice $300 \mathrm{~K}$, the work done at $600 \mathrm{~K}$ is only twice the work done at $300 \mathrm{~K}$, not 20 times.

• Option (b): Work done at $300 \mathrm{~K}$ is twice the work done at $600 \mathrm{~K}$ is incorrect because, as mentioned, the work done is directly proportional to the temperature. Therefore, the work done at $600 \mathrm{~K}$ should be twice the work done at $300 \mathrm{~K}$, not the other way around.

Answer

$(a, c)$

For the reaction,

$$ 2 \mathrm{Zn}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{ZnO}(\mathrm{s}) ; \Delta H=-693.8 \mathrm{~kJ} \mathrm{~mol}^{-1} $$

As we know that,

$$ \Delta H=H_{p}-H_{R} $$

A negative value of $\Delta H$ shows that $H_{R}>H_{P}$ or $H_{P}<H_{R}$, i.e., enthalpy of two moles of $\mathrm{ZnO}$ is less than the enthalpy of two moles of zinc and one mole of oxygen by 693.8kJ. As $H_{R}>H_{P}$, $693.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ of energy is evolved in the reaction.

• Option (b) is incorrect because the enthalpy of two moles of ZnO is not more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ. The negative value of ΔH indicates that the enthalpy of the products (ZnO) is less than the enthalpy of the reactants (Zn and O₂).

• Option (d) is incorrect because 693.8 kJ/mol of energy is not absorbed in the reaction. The negative ΔH value indicates that energy is released (evolved) during the reaction, not absorbed.

Answer

Given that, quantity of water $=18.0 \mathrm{~g}$, pressure $=1$ bar

As we know that, $18.0 gH_{2} O=1$ mole $H_{2} O$

Enthalpy change for vaporising 1 mole of $\mathrm{H}_{2} \mathrm{O}=40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\therefore$ Enthalpy change for vaporising 2 moles of $\mathrm{H}_{2} \mathrm{O}=2 \times 40.79 \mathrm{~kJ}=81.358 \mathrm{~kJ}$

Standard enthalpy of vaporisation at $100^{\circ} \mathrm{C}$ and 1 bar pressure, $\Delta_{\text {vap }} H^{\circ}=+40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Answer

One mole of acetone requires less heat to vaporise than 1 mole of water. Hence, acetone has less enthalpy of vaporisation and water has higher enthalpy of vaporisation. It can be represented as $\left(\Delta H_{V}\right)$ water $>\left(\Delta H_{V}\right)$ acetone.

Answer

No, the $\Delta_{r} H^{\mathrm{s}}$ for the given reaction is not same as $\Delta_{r} H^{\mathrm{s}}$. The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states (reference states) is called standard molar enthalpy of formation, $\Delta_{f} H^{\mathrm{s}}$.

$$ Ca(s)+C(s)+\frac{3}{2} O_{2}(g) \rightarrow CaCO_{3}(~s) ; \Delta_{f} H^{s} $$

This reaction is different from the given reaction.

Hence,

$\Delta_{r} H^{\circ} \neq \Delta_{f} H^{\circ}$

Answer

Given, $\frac{1}{2} ~N_{2}(g)+\frac{3}{2} H_{2}(g) \rightarrow NH_{3}(g) ; \Delta_{f} H^{s}=-91.8 ~kJ ~mol^{-1}$

$\left(\Delta_{f} H^{s}\right.$ means enthalpy of formation of 1 mole of $\left.\mathrm{NH}_{3}\right)$

$\therefore \quad$ Enthalpy change for the formation of 2 moles of $\mathrm{NH}_{3}$

$$ N_{2}(g)+3 H_{2}(g) \rightarrow 2 NH_{3}(g) ; \Delta_{f} H^{s}=2 \times-91.8=-183.6 ~kJ ~mol^{-1} $$

And for the reverse reaction,

$$ 2 NH_{3}(g) \rightarrow N_{2}(g)+3 H_{2}(g) ; \Delta_{f} H^{s}=+183.6 ~kJ ~mol^{-1} $$

Hence, the value of $\Delta_{f} H^{\mathrm{s}}$ for $\mathrm{NH}_{3}$ is $+183.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Answer

In general, if enthalpy of an overall reaction $A \rightarrow B$ along one route is $\Delta_{r} H$ and $\Delta_{r} H_{1}, \Delta_{r} H_{2}, \Delta_{r} H_{3} \ldots$ representing enthalpies of reaction leading to same product $B$ along another route, then we have

$$ \Delta_{r} H=\Delta_{r} H_{1}+\Delta_{r} H_{2}+\Delta_{r} H_{3}+\ldots $$

Note For a general reaction Hess’s law of constant heat summation can be represented as

Thinking Process

To solve this problem, keep in mind that in methane all the four $\mathrm{C}-\mathrm{H}$ bonds are identical in bond length and energy. However, the energies required to break the individual $\mathrm{C}-\mathrm{H}$ bonds in each successive step differ. In such cases, we use mean bond enthalpy of $\mathrm{C}-\mathrm{H}$ bond i.e., $\Delta_{\mathrm{C}-H} H^{\mathrm{s}}=\frac{1}{4}\left(\Delta_{\mathrm{a}} H^{\mathrm{s}}\right)$

Answer

In $CH_{4}$, there are four $C-H$ bonds. The enthalpy of atomisation of 1 mole of $CH_{4}$ means dissociation of four moles of $C-H$ bond.

$$ \begin{aligned} \therefore \quad \mathrm{C}-\mathrm{H} \text { bond energy per } \mathrm{mol} & =\frac{1665 \mathrm{~kJ}}{4 \mathrm{~mol}} \\ & =416.25 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$

Thinking Process

This question is based upon the concept of Born-Haber cycle as well as Hess’s law. Following steps are used to solve this problem.

(i) $\mathrm{Na}(\mathrm{s}) \rightarrow \mathrm{Na}(g) ; \Delta_{\text {sub }} H^{\mathrm{s}}$

(ii) $\mathrm{Na}(g) \rightarrow \mathrm{Na}^{+}(g)+\mathrm{e}^{-}(g)$; $I E$

(iii) $\frac{1}{2} Br_{2}(g) \rightarrow Br(g) ; \Delta_{\text {diss }} H^{s}$

(iv) $\mathrm{Br}(g)+\mathrm{e}^{-}(g) \rightarrow \mathrm{Br}^{-}(g) ; \Delta_{\mathrm{eg}} H^{\mathrm{s}}$

(v) Applying Hess’s law $\Delta_{f} H^{\mathrm{s}}=\Delta_{\text {sub }} H^{\mathrm{s}}+I \mathrm{E}+\Delta_{\text {diss }} H^{\mathrm{s}}+\Delta_{\text {eg }} H^{\mathrm{s}}+U$

Answer

Given that, $\Delta_{\text {sub }} H^{\mathrm{s}}$ for $\mathrm{Na}$ metal $=108.4 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$

IE of $\mathrm{Na}=496 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}, \Delta_{\text {eg }} H^{\mathrm{s}}$ of $\mathrm{Br}=-325 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}, \Delta_{\text {diss }} H^{\mathrm{s}}$ of $\mathrm{Br}=192 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}, \Delta_{f} H^{\mathrm{s}}$ for $\mathrm{NaBr}=-360.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Born-Haber cycle for the formation of $\mathrm{NaBr}$ is as

By applying Hess’s law,

$$ \begin{aligned} \Delta_{f} H^{\mathrm{s}} & =\Delta_{\text {sub }} H^{\mathrm{s}}+I E+\Delta_{\text {diss }} H^{\mathrm{s}}+\Delta_{\mathrm{eg}} H^{\mathrm{s}}+U \\ -360.1 & =108.4+496+96+(-325)-U \\ U & =+735.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$

Answer

The mixing of two gases have $\Delta H$ equal to zero. Therefore, it is spontaneous process because energy factor has no role to play but randomness increases i.e., randomness factor favours the process.

Answer

Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. The mathematical relation which relates these three parameters is

$$ \Delta S=\frac{q_{\text {rev }}}{T} $$

Here, $\Delta S=$ change in entropy

$q_{\text {rev }}=$ heat of reversible reaction

$T=$ temperature

Answer

Yes, the temperature of system and surroundings be the same when they are in thermal equilibrium.

Note Thermal equilibrium is defined as when two physical systems are brought into a connection that does not allow transfer of matter between them, and does not allow transfer of energy between them, such a connection is said to permit transfer of energy as heat, and is called diathermal.

If a diathermal connection is made between two physical systems and the making of the connection is followed by no change of state of either, then the two systems are said to be in relation of thermal equilibrium. It obeys zeroth law of thermodynamics.

Answer

For the reaction, $N_{2} O_{4}(g) \rightleftharpoons 2 NO_{2}(g), K_{p}=0.98$

As we know that $\quad \Delta_{r} \mathrm{G}^{\mathrm{s}}=-2.303 \mathrm{RT} \log K_{p}$

Here, $K_{p}=0.98$ i.e., $K_{p}<1$ therefore, $\Delta_{r} G^{\circ}$ is positive, hence the reaction is non-spontaneous.

Answer

The net enthalpy change, $\Delta H$ for a cyclic process is zero as enthalpy change is a state function, i.e., $\Delta H($ cycle $)=0$

Answer

The standard molar entropy of $H_{2} O(l)$ is $70 JK^{-1} ~mol^{-1}$. The solid form of $H_{2} O$ is ice. In ice, molecules of $H_{2} O$ are less random than in liquid water.

Thus, molar entropy of $H_{2} O(s)<$ molar entropy of $H_{2} O(l)$. The standard molar entropy of $H_{2} O(s)$ is less than $70 ~J ~K^{-1} ~mol^{-1}$.

Answer

State functions are those values which depend only on the state of the system and not on how it is reached e.g., enthalpy, entropy, temperature and free energy. Path functions are those values which depend on the path of the system. e. $g$, heat and work.

Answer

Amount of heat required to vaporise one mole of a liquid at constant temperature and under standard pressure ( 1 bar ) is called its molar enthalpy of vaporisation $\Delta_{\text {vap }} H^{s}$. Molar enthalpy of vaporisation of water is more than that of acetone because there is strong hydrogen bonding in $\mathrm{H}_{2} \mathrm{O}$ molecule.

Answer

Gibbs energy for a reaction in which all reactants and products are in standard state. $\Delta_{r} G^{\circ}$ is related to the equilibrium constant of the reaction as follows

$$ \Delta_{r} G=\Delta_{r} G^{\mathrm{\circ}}+R T \ln K $$

$$ \begin{aligned} \text{At equilibrium} \quad 0 & =\Delta_{r} G^{\circ}+R T \ln K \quad (\therefore \Delta_r G= 0)\\ \ \text{or} \quad \Delta_{r} G^{\mathrm{\circ}} & =-R T \ln K \\ \Delta_{r} G^{\mathrm{\circ}} & =0 \text { when } K=1 \end{aligned} $$

For all other values of $K, \Delta_{r} G^{\circ}$ will be non-zero.

Answer

For isolated system there is no transfer of energy as heat, i.e., $q=0$ and there is no transfer of energy as work. i.e., $W=0$. According to the first law of thermodynamics

$$ \begin{aligned} & \Delta U=q+W \\ & \Delta U=0+0=0 \end{aligned} $$

Answer

The two conditions under which heat becomes independent of path are

(i) when volume remains constant

(ii) when pressure remains constant

Explanation

(i) At constant volume By first law of thermodynamics, $\Delta U=q+W$ or $q=\Delta U-W$. But $W=-p \Delta V$ Hence, $q=\Delta U+p \Delta V$. But as volume remains constant $\Delta V=0$

$\therefore q_{V}=\Delta U$ but $\Delta U$ is a state function.

Hence, $q_{v}$ is a state function.

(ii) At constant pressure As we know, $q_{p}=\Delta U+p \Delta V$. But $\Delta U+p \Delta V=\Delta H$.

$\therefore q_{p}=\Delta H$. As $\Delta H$ is a state function therefore, $q_{p}$ is a state function.

Answer

Work done of a gas in vacuum, $W=-p_{\text {ext }}\left(V_{2}-V_{1}\right)$. As $p_{\text {ext }}=0$ so $W=-0(5-1)=0$ As internal energy of an ideal gas depends only on temperature, therefore, for isothermal expansion of an ideal gas, internal energy remains constant,

i.e.,

$$ \Delta U=0 . $$

It is to be remember that as $H=U+p V, \Delta H=\Delta(U+p V)=\Delta U+p \Delta V=\Delta U+n R(\Delta T)$. For isothermal process, $\Delta T=0$ and also $\Delta U=0$, as stated above, therefore, $\Delta H=0$.

Answer

For water, molar heat capacity, $C_{p}=18 \times$ Specific heat, $c$

$$ \begin{aligned} C_{p} & =18 \times c \text { Specific heat } \\ c & =4.18 \mathrm{Jg}^{-1} \mathrm{~K}^{-1} \quad \text{(for work)}\\ \text { Heat capacity, } \quad C_{p} & =18 \times 4.18 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ & =75.24 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \end{aligned} $$

Answer

Given that, $C_{v}=$ heat capacity at constant volume, $C_{p}=$ heat capacity at constant pressure Difference between $C_{p}$ and $C_{v}$ is equal to gas constant $(R)$.

$$ \begin{aligned} \therefore \quad C_{p}-C_{v} & =n R \quad \text { (where, } n=\text { no. of moles) } \\ & =10 \times 4.184 \mathrm{~J} \\ & =41.84 \mathrm{~J} \end{aligned} $$

Answer

Given that, enthalpy of combustion of $1 \mathrm{~g}$ graphite $=20.7 \mathrm{~kJ}$

Molar enthalpy change for the combustion of graphite, $\Delta H=$ enthalpy of combustion of $1 \mathrm{~g}$

graphite $\times$ molar mass

$$ \begin{aligned} & \Delta H=-20.7 \mathrm{kJg}^{-1} \times 12 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \Delta H=-2.48 \times 10^{2} \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$

Negative sign in the value of $\Delta H$ indicates that the reaction is exothermic.

Thinking Process

To, calculate the enthalpy change, use the following formula when the reactants, and products are in gas phase as

$$ \Delta_{r} H^{\mathrm{-}}=\Sigma \text { bond energy (reactants) }-\Sigma \text { bond energy (products) } $$

Answer

Given that, bond energy of $\mathrm{H}_{2}=435 \mathrm{~kJ} \mathrm{~mol}^{-1}$

bond energy of $\mathrm{Br}_{2}=192 \mathrm{~kJ} \mathrm{~mol}^{-1}$

bond energy of $\mathrm{HBr}=368 \mathrm{~kJ} \mathrm{~mol}^{-1}$

For the reaction

$$ \begin{aligned} & H_{2}(g)+Br_{2}(g) \rightarrow 2 HBr(g) \\ & \Delta_{r} H^{-}=\Sigma B.E(\operatorname{Reactants})-\Sigma B.E(\text { Products }) \\ &=B.E\left(H_{2}\right)+B.E\left(Br_{2}\right)-2 B.E(HBr) \\ &=435+192-(2 \times 368) kJ mol^{-1} \\ &=-109 ~kJ ~mol^{-1} \end{aligned} $$

Answer

Given that, $1 \text{mol of} \quad CCl_{4}=154 ~g$

$$ \begin{aligned} & \Delta_{\text {vap }} H \text { for } 154 ~g CCl_{4}=30.5 ~kJ \\ \therefore \quad & \Delta_{\text {vap }} H \text { for } 284 ~g CCl_{4}=\frac{30.5 \times 284}{154} ~kJ=56.25 ~kJ \end{aligned} $$

Answer

Given that,

$$ 2 H_{2}(g)+O_{2}(g) \rightarrow 2 H_{2} O(l), \Delta_{r} H^{\circ}=-572 ~kJ ~mol^{-1} $$

Enthalpy of formation is the enthalpy change of the reaction when 1 mole of the compound is formed from its elements then

$$ H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l), \Delta_{r} H^{\circ}=? $$

This can be obtained by dividing the given equation by 2 .

Therefore,

$$ \Delta_{f} H^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right)=-\frac{572 \mathrm{kJmol}^{-1}}{2}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1} $$

Answer

Suppose total volume of the gas is $V_{i}$ and pressure of the gas inside cylinder is $p$. After compression by constant external pressure, $\left(p_{\text {ext }}\right)$ in a single step, final volume of the gas becomes $V_{f}$.

Then volume change, $\Delta v=\left(V_{f}-V_{i}\right)$

If $W$ is the work done on the system by movement of the piston, then

$$ \begin{aligned} & W=p_{\text {ext }}(-\Delta V) \\ & W=-p_{\text {ext }}\left(V_{f}-V_{i}\right) \end{aligned} $$

This can be calculated from $p-V$ graph as shown in the figure. Work done is equal to the shaded area $A B V_{f} V_{i}$

The negative sign in this expression is required to obtain conventional sign for $W$ which will be positive. Because in case of compression work is done on the system, so $\Delta V$ will be negative.

Answer

When compression is carried out in infinite steps with change in pressure, it is a reversible process. The work done can be calculated from $p-V$ plot as shown in the given figure. Shaded area under the curve represents the work done on the gas.

Answer

Representation of potential energy/enthalpy change in the following processes

(a) Throwing a stone from the ground to roof.

(b) $\frac{1}{2} H_{2}(g)+\frac{1}{2} Cl_{2}(g) \rightleftharpoons HCl(g) ; \Delta_{r} H^{\circ}=-92.32 ~kJ ~mol^{-1}$

Energy increases in (a) and it decreases in (b) process. Hence, in process (b), enthalpy change is the contributing factor to the spontaneity.

Answer

No, enthalpy is one of the contributing factors in deciding spontaneity but it is not the only factor. Another contributory factor, entropy factor has also to be taken into consideration.

Answer

The given diagram represent that the process is carried out in infinite steps, hence it is isothermal reversible expansion of the ideal gas from pressure $2.0 \mathrm{~atm}$ to $1.0 \mathrm{~atm} 298 \mathrm{~K}$.

$$ \begin{aligned} & W=-2.303 n R T \log \frac{p_{1}}{p_{2}} \\ & W=-2.303 \times 1 \mathrm{~mol} \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{Klog} 2 \quad\left(\because \frac{p_{1}}{p_{2}}=\frac{2}{1}\right) \\ & W=-2.303 \times 1 \times 8.314 \times 298 \times 0.3010 \mathrm{~J} \\ & W=-1717.46 \mathrm{~J} \end{aligned} $$

Answer

In the first case, as the expansion is against constant external pressure

$$ \begin{aligned} W & =-{\text {ext }}\left(V{2}-V_{1}\right)=-2 \operatorname{bar} \times(50-10) \mathrm{L} \\ & =-80 \mathrm{~L} \text { bar } \quad \quad (1L bar = 100 J)\\ & =-80 \times 100 \mathrm{~J} \\ & =-8 \mathrm{~kJ} \end{aligned} $$

If the given expansion was carried out reversibly, the internal pressure of the gas should be greater than the external pressure at every stage. Hence, the work done will be more.

Answer

A. $\rightarrow(5)$

B. $\rightarrow(4)$

C. $\rightarrow(6)$

D. $\rightarrow(1)$

E. $\rightarrow(7,11,12)$

F. $\rightarrow(2)$

G. $\rightarrow(3)$

H. $\rightarrow(10)$

I. $\rightarrow(8)$

J. $\rightarrow$ (9)

K. $\rightarrow(1,12,13)$

L. $\rightarrow(7,11)$

Correct Matching can be done as

Answer

A. $\rightarrow(2) \quad$

B. $\rightarrow(3) \quad$

C. $\rightarrow(1)$

A. When a liquid vaporises, i.e., liquid $\rightarrow$ vapour, entropy increase i.e., $\Delta S=$ positive

B. When $\Delta H=$ positive, i.e., energy factor opposes. The process is non-spontaneous at all temperatures if entropy factor also opposes the process, i.e., $\Delta S=$ negative

C. In the reversible expansion of an ideal gas, the system remains in equilibrium at every stage. Hence, $\Delta S=0$

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(2)$

A. When $\Delta_{r} G^{\circ}$ is positive, reaction is non-spontaneous at all temperatures

B. When $\Delta_{r} G^{\circ}$ is positive at high temperature means the reaction is non-spontaneous at high temperature.

C. When $\Delta_{r} H^{\mathrm{\circ}}=$ negative means it favours, $\Delta_{r} S^{\mathrm{s}}=$ positive means it also favours. $\Delta_{r} G^{\circ}=$ negative means reaction is spontaneous at all temperatures.

Answer

A. $\rightarrow(2,4)$

B. $\rightarrow(2)$

C. $\rightarrow(3)$

D. $\rightarrow(1)$

A. Entropy of vaporisation is always positive. It is equal to $\Delta H_{\text {vap }} / T_{b}$

B. $\Delta_{r} G^{\circ}=-R T \ln K$

If $K$ is positive, $\Delta_{r} G^{\circ}=$ negative and reaction is spontaneous.

C. Crystalline solid state has lowest entropy.

D. During adiabatic expansion of an ideal gas, $q=0$. Hence, $\Delta U=q+W$ gives $\Delta U=W$, i.e., work done is at the cost of internal energy which decreases.

Answer

(b) Both assertion and reason are true but reason is not the correct explanation of assertion.

Correct explanation In a combustion reaction, sum of enthalpies of reactants is greater than the sum of the enthalpies of products.

Answer

(b) Both assertion and reason are true but reason is not the correct explanation of assertion.

Spontaneous processes are accompanied by decrease in energy and increase in randomness.

Answer

(a) Both assertion and reason are true and reason is the correct explanation of assertion. When a liquid crystallises, entropy decreases because in crystalline form the molecules are more ordered as compared to the liquid.

Answer

From the first law of thermodynamics, $\quad q=\Delta U+p \Delta V$

If the process carried out at constant volume, $\Delta V=0$

Hence,

$ \quad\quad $[Here, $q_v=$ Heat absorbed at constant volume, $\Delta U=$ change in internal energy] $$q_v=\Delta U$$

Similarly,

Here, $q_p=$ heat absorbed at constant pressure

$\Delta H=$ enthalpy change of the system.

Enthalpy change of a system is equal to the heat absorbed or evolved by the system at constant pressure.

As we know that at constant pressure, $\Delta H=\Delta U+p \Delta V$

where, $\Delta V$ is the change in volume.

This equation can be rewritten as $\Delta H=\Delta U+p\left(V_f-V_i\right)=\Delta U+\left(p V_f-p V_i\right)$

where, $V_i=$ initial volume of the system $V_f=$ final volume of the system

But for the ideal gases, $$ \begin{aligned} & p V=n R T \end{aligned} $$

So that $$ V_1=n_1 R T$$

and $$p V_2=n_2 R T$$

where, $n_1=$ number of moles of the gaseous reactants $n_2=$ number of moles of the gaseous products.

Substituting these values in Eq. (i), we get

$$ \begin{aligned} & \Delta H=\Delta U+\left(n_{2} R T-n_{1} R T\right) \\ & \Delta H=\Delta U+\left(n_{2}-n_{1}\right) R T \\ or & \quad \Delta H=\Delta U+\Delta n_{g} R T \end{aligned} $$

where, $\Delta n_{g}=n_{2}-n_{1}$ is the difference between the number of moles of the gaseous products and gaseous reactants.

Putting the values of $\Delta H$ and $\Delta U$ we get

$$ q_{p}=q_{v}+\Delta n_{g} R T $$

Note Conditions under which $q_{p}=q_{v}$ or $\Delta H=\Delta U$

(i) When reaction is carried out in a closed vessel so that volume remains constant i.e., $\Delta V=0$

(ii) When reaction involves only solids or liquids or solutions but no gaseous reactant or product. This is because the volume changes of the solids and liquids during a chemical reaction are negligible.

(iii) When reaction involves gaseous reactants and products but their number of moles are equal (i.e., $n_{p}=n_{r}$ ) e.g.,

$$ \begin{array}{r} H_{2}(~g)+Cl_{2}(~g) \longrightarrow 2 HCl(g) \\ C(s)+O_{2}(~g) \longrightarrow CO_{2}(~g) \end{array} $$

Since, $a_{p}$ is different from $q_{v}$ only in those reactions which involves gaseous reactants and products and $\left(n_{p}\right)$ gaseous $\neq\left(n_{r}\right)$ gaseous.

Answer

Extensive properties Those properties whose value depends on the quantity or size of matter present in the system is known as extensive properties.

e.g., mass, internal energy, heat capacity.

Intensive properties Those properties which do not depend on the quantity or size of matter present are known as intensive properties. e.g., pressure, molar heat capacity, density, mole fraction, specific heat, temperature and molarity.

Mole fraction or molarity of a solution is same whether we take a small amount of the solution or large amount of the solution.

Ratio of two extensive properties is always intensive.

$$ \frac{\text { Extensive }}{\text { Extensive }}=\text { Intensive } $$

So, mole fraction and molarity are intensive properties.

$$ \begin{gathered} \text { e.g, mole fraction }=\frac{\text { Moles of the component }}{\text { Total no. of moles }}=\frac{\text { Extensive }}{\text { Extensive }} \\ \text {and molarity }=\frac{\text { Mole }}{\text { Volume }}=\frac{\text { Extensive }}{\text { Extensive }} \end{gathered} $$

Answer

The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state. For the reaction

$$ \mathrm{Na}^{+} \mathrm{Cl}^{-}(s) \rightarrow \mathrm{Na}^{+}(g)+\mathrm{Cl}^{-}(g) ; \Delta_{\text {lattice }} H^{\mathrm{s}}=+788 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1} $$

Since, it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber cycle.

Let us now calculate the lattice enthalpy of $\mathrm{Na}^{+} \mathrm{Cl}^{-}$(s) by following steps given below

(i) $\mathrm{Na}^{+}(\mathrm{s}) \rightarrow \mathrm{Na}(g)$; Sublimation of sodium metal, $\Delta_{\text {sub }} H^{\mathrm{s}}=108.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(ii) $\mathrm{Na}(g) \rightarrow \mathrm{Na}^{+}(g)+e^{-}(g)$; The ionisation of sodium atoms, ionisation enthalpy $\Delta_{i} H^{\mathrm{s}}=496 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iii) $\frac{1}{2} Cl_{2}(g) \rightarrow Cl(g)$; The dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy $\frac{1}{2} \Delta_{\text {bond }} H^{s}=121 ~kJ ~mol^{-1} $

(iv) $\mathrm{Cl}(g)+e^{-}(g) \rightarrow \mathrm{Cl}^{-}(g)$; electron gained by chlorine atoms. The electron gain enthalpy, $\Delta_{\text {eg }} H^{\mathrm{S}}=-348.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Enthalply diagram for lattice enthalpy of $\mathrm{NaCl}$

(v) $\mathrm{Na}^{+}(g)+\mathrm{Cl}^{-}(g) \longrightarrow \mathrm{Na}^{+} \mathrm{Cl}^{-}(s)$

The sequence of steps is shown in given figure and is known as Born-Haber cycle. The importance of the cycle is that, the sum of the enthalpy changes round a cycle is zero.

Applying Hess’s law, we get

$$ \Delta_{\text {lattice }} H^{\mathrm{S}}=411.2+108.4+121+496-348.6 $$

$\Delta_{\text {latitice }} H^{\mathrm{s}}=+788 \mathrm{~kJ}$

Answer

Gibbs free energy is that thermodynamic quantity of a system, the decrease in whose value during a process is equal to the maximum possible useful work that can be obtained from the system.

Mathematically, this results may be derived as follows

The relationship between heat absorbed by a system $q$, the change in its internal energy, $\Delta U$ and the work done by the system is given by the equation of the first law of thermodynamics, therefore,

$$ q=\Delta U+W_{\text {expansion }}+W_{\text {non-expansion }} \quad …(i) $$

Under constant pressure condition, the expansion work is given by $p \Delta V$.

$$ \begin{array}{rlr} \therefore \quad & q =\Delta U+p \Delta V+W_{\text {nonexpansion }} \quad (\because \Delta U+p \Delta V=\Delta H) \\ & =\Delta H+W_{\text {non-expansion }} \quad …(ii)& \end{array} $$

For a reversible change taking place at constant temperature,

$$ \Delta S=\frac{q_{\text {rev }}}{T} \text { or } q_{\text {rev }}=T \Delta S \quad …(iii) $$

Substituting the value of $q$ from Eq. (iii) in Eq. (ii), we get

$$ \begin{aligned} T \Delta S & =\Delta H+W_{\text {non-expansion }} \\ or \quad \Delta H-T \Delta S & =-W_{\text {non expansion }} \quad …(iv) \end{aligned} $$

For a change taking place under conditions of constant temperature and pressure,

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

Substituting this value in equation (iv), we get

$$ \Delta G=-W_{\text {non-expansion }} \quad …(v) $$

Thus, free energy change can be taken as a measure of work other than the work of expansion. For most changes, the work of expansion can not be converted to other useful work, whereas the non-expansion work is convertible to useful work.

Rearranging equation $(\mathrm{v})$, it may write as

$$ -\Delta G=W_{\text {non-expansion }}=W_{\text {useful }} $$

As $-\Delta G=W_{\text {useful }}$ therefore, $\Delta G$ has the same units as those of work i.e., joule

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

If $\Delta H=$ positive and $\Delta S=$ positive, then

$\Delta G$ will be negative i.e., process will be spontaneous only when $T \Delta S>\Delta H$ in magnitude, which will be so when temperature is high.

Answer

(i) Total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from $\left(p_{i}, V_{i}\right)$ to $\left(p_{f}, V_{f}\right)$. Reversible work is represented by the combined areas $A B C$ and $B C V_{i} V_{f}$.

(ii) Work against constant pressure, $p_{f}$ is represented by the area $B C V_{i} V_{f}$. Work (i) $>$ work (ii)