2 nd Law of thermodynamics and Entropy Questions in English

(C) For an isothermal reversible expansion,the entropy change is given by the formula: $\Delta S = n R \ln \frac{V_2}{V_1} = 2.303 n R \log \frac{V_2}{V_1}$.
Given values: $n = 2 \, \text{mol}$,$V_1 = 10 \, L$,$V_2 = 100 \, L$,and $R = 8.314 \, J \, K^{-1} \, \text{mol}^{-1}$.
Substituting the values: $\Delta S = 2.303 \times 2 \times 8.314 \times \log \frac{100}{10}$.
$\Delta S = 2.303 \times 2 \times 8.314 \times \log(10)$.
Since $\log(10) = 1$,we get $\Delta S = 2.303 \times 2 \times 8.314 \times 1 = 38.297 \approx 38.3 \, J \, K^{-1}$.

(D)
According to the second law of thermodynamics,whenever a spontaneous process takes place,it is always accompanied by an increase in the total entropy of the universe (system and surroundings).
Thus,the correct option is $(d)$.

(D)
For a spontaneous process,the total entropy change of the universe must be positive.
This is defined as $\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} > 0$.
Since the system and its surroundings together constitute the universe,the sum of their entropy changes must be greater than zero for the process to be spontaneous.

(C) For an isothermal process,the heat exchanged with the surroundings is $q = -w = P_{ext} \Delta V$.
Given $P_{ext} = 4 \, atm$ and $\Delta V = (3.0 \, L - 2.0 \, L) = 1.0 \, L$.
$q = 4 \, atm \times 1.0 \, L = 4 \, L \, atm$.
Using the conversion $1 \, L \, atm = 101.3 \, J$,we get $q = 4 \times 101.3 = 405.2 \, J$.
The entropy change of the surroundings is given by $\Delta S_{surr} = -\frac{q_{sys}}{T}$.
Since the process is isothermal and the system absorbs heat from the surroundings,$q_{surr} = -q_{sys} = -405.2 \, J$.
$\Delta S_{surr} = \frac{q_{surr}}{T} = \frac{-405.2 \, J}{350 \, K} \approx -1.158 \, J \, K^{-1}$.
The nearest integer is $-1$.

(A) The relationship between enthalpy change of fusion $(\Delta H_{fus})$,entropy change of fusion $(\Delta S_{fus})$,and melting point $(T_{mp})$ is given by the formula: $\Delta S_{fus} = \frac{\Delta H_{fus}}{T_{mp}}$.
Given: $\Delta H_{fus} = 30.4 \, kJ \, mol^{-1} = 30400 \, J \, mol^{-1}$ and $\Delta S_{fus} = 28.4 \, J \, K^{-1} \, mol^{-1}$.
Substituting the values into the equation: $28.4 = \frac{30400}{T_{mp}}$.
Solving for $T_{mp}$: $T_{mp} = \frac{30400}{28.4} \approx 1070.42 \, K$.
Rounding to the nearest integer,the melting point is $1070 \, K$.

(B) Entropy $(S)$ is a measure of the randomness or disorder of a system.
$A$. Evaporation of a liquid to vapour increases disorder,so entropy increases.
$B$. Lowering the temperature of a crystalline solid towards $0 \ K$ decreases the molecular motion and disorder,so entropy decreases.
$C$. $2 NaHCO_{3(s)} \rightarrow Na_2CO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$. Here,solid reactants produce gaseous products,leading to an increase in the number of moles of gas and thus an increase in entropy.
$D$. $Cl_{2(g)} \rightarrow 2 Cl_{(g)}$. One mole of gaseous $Cl_2$ produces two moles of gaseous $Cl$ atoms,increasing the number of particles and disorder,so entropy increases.
Therefore,processes $A, C,$ and $D$ result in an increase in entropy.

(B) Entropy is a state function, so the change in entropy depends only on the initial and final states, not on the path taken.
For the process $A \rightarrow B$, we can write the path as $A \rightarrow C \rightarrow D \rightarrow B$.
Therefore, $\Delta S_{(A \rightarrow B)} = \Delta S_{(A \rightarrow C)} + \Delta S_{(C \rightarrow D)} + \Delta S_{(D \rightarrow B)}$.
Since $\Delta S_{(D \rightarrow B)} = -\Delta S_{(B \rightarrow D)}$, we have:
$\Delta S_{(A \rightarrow B)} = \Delta S_{(A \rightarrow C)} + \Delta S_{(C \rightarrow D)} - \Delta S_{(B \rightarrow D)}$
Substituting the given values:
$\Delta S_{(A \rightarrow B)} = 50 + 30 - 20 = 60 \text{ e.u.}$

(A) $STATEMENT-1$ is true because work can be completely converted into heat (e.g.,through friction),but heat cannot be completely converted into work without leaving some effect elsewhere.
$STATEMENT-2$ is the Kelvin-Planck statement of the Second Law of Thermodynamics,which explains this inherent asymmetry.
Therefore,$STATEMENT-2$ is the correct explanation for $STATEMENT-1$.

(B) At $T = 100^{\circ} C$ and $1 \ atm$,the process $H_2O(\ell) \rightarrow H_2O_{(g)}$ is at equilibrium.
For a phase change,the entropy of the system increases as liquid converts to gas,so $\Delta S_{\text{system}} > 0$.
Since the process is at equilibrium,$\Delta G = 0$,which implies $\Delta H = T \Delta S_{\text{system}}$.
Because evaporation is an endothermic process,$\Delta H > 0$,meaning the system absorbs heat from the surroundings.
Therefore,the surroundings lose heat,making $q_{\text{surr}} < 0$,which leads to $\Delta S_{\text{surroundings}} = \frac{q_{\text{surr}}}{T} < 0$.

(B) The total entropy change $\Delta S_{\text{total}}$ is the sum of entropy changes for each step of the process:
$1$. Heating ice from $-5^{\circ} C$ $(268 \ K)$ to $0^{\circ} C$ $(273 \ K)$: $\Delta S_1 = \int_{268 \ K}^{273 \ K} \frac{C_{p,m}(\text{ice})}{T} \ dT$
$2$. Melting ice at $0^{\circ} C$ $(273 \ K)$: $\Delta S_2 = \frac{\Delta H_{m, \text{fusion}}}{273 \ K}$
$3$. Heating water from $0^{\circ} C$ $(273 \ K)$ to $100^{\circ} C$ $(373 \ K)$: $\Delta S_3 = \int_{273 \ K}^{373 \ K} \frac{C_{p,m}(\text{water})}{T} \ dT$
$4$. Vaporizing water at $100^{\circ} C$ $(373 \ K)$: $\Delta S_4 = \frac{\Delta H_{m, \text{vaporisation}}}{373 \ K}$
$5$. Heating steam from $100^{\circ} C$ $(373 \ K)$ to $110^{\circ} C$ $(383 \ K)$: $\Delta S_5 = \int_{373 \ K}^{383 \ K} \frac{C_{p,m}(\text{steam})}{T} \ dT$
Summing these gives the expression in option $B$.

(B) For the reaction: $\frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightleftharpoons XY_5$
$\Delta S_{rxn}^0 = S^0(XY_5) - [\frac{1}{2} S^0(X_2) + \frac{5}{2} S^0(Y_2)]$
$\Delta S_{rxn}^0 = 110 - [(\frac{1}{2} \times 70) + (\frac{5}{2} \times 50)] = 110 - [35 + 125] = 110 - 160 = -50 \ J \ K^{-1} \ mol^{-1}$
At equilibrium,$\Delta G^0 = 0$,so $\Delta H^0 = T \Delta S^0$
Given $\Delta H^0 = -35 \ kJ \ mol^{-1} = -35000 \ J \ mol^{-1}$
$T = \frac{\Delta H^0}{\Delta S^0} = \frac{-35000}{-50} = 700 \ K$

(C) For a reaction to be spontaneous according to the $2^{nd}$ law of thermodynamics: $\Delta G < 0$.
Since $\Delta G = \Delta H - T \Delta S$,we have $\Delta H - T \Delta S < 0$.
At equilibrium,$\Delta G = 0$,so $\Delta H = T_e \Delta S$,which implies $T_e = \frac{\Delta H}{\Delta S}$.
For the reaction to be spontaneous,$\Delta H - T \Delta S < 0$,which means $T \Delta S > \Delta H$.
Since $\Delta S$ is positive,$T > \frac{\Delta H}{\Delta S}$.
Therefore,$T > T_e$.

(A) Statement-$1$: At equilibrium,the system reaches a state of maximum entropy,which implies that the change in total entropy (system + surroundings) is zero,i.e.,$\Delta S_{total} = 0$. Thus,Statement-$1$ is correct.
Statement-$2$: The change in entropy for a reversible process is given by $\Delta S = \frac{q_{rev}}{T}$. If the same amount of heat $(q)$ is supplied,then $\Delta S \propto \frac{1}{T}$. This means that at a lower temperature $(T)$,the value of $\Delta S$ will be higher,not less. Thus,Statement-$2$ is incorrect.

(C) For an isothermal reversible expansion of an ideal gas,the entropy change is given by the formula: $\Delta S = nR \ln(\frac{V_2}{V_1})$.
Given values are: $n = 2 \ mol$,$R = 8.314 \ J \ mol^{-1} K^{-1}$,$V_1 = 10 \ L$,$V_2 = 100 \ L$.
Substituting these values into the formula:
$\Delta S = 2 \times 8.314 \times \ln(\frac{100}{10})$
$\Delta S = 2 \times 8.314 \times \ln(10)$
$\Delta S = 2 \times 8.314 \times 2.303$
$\Delta S = 38.29 \ J K^{-1} \approx 38.3 \ J K^{-1}$.

(B) Entropy $(S)$ is a measure of the randomness or disorder of a system. $\Delta S$ is positive when the disorder of the system increases.
$1$. $H_2O(\ell) \rightarrow H_2O_{(s)}$: Liquid to solid represents a decrease in disorder,so $\Delta S < 0$.
$2$. $H_2O(\ell) \rightarrow H_2O_{(g)}$: Liquid to gas represents a significant increase in disorder,so $\Delta S > 0$.
$3$. $N_2(g, 1 \ atm) \rightarrow N_2(g, 10 \ atm)$: Increasing pressure at constant temperature decreases the volume available to gas molecules,decreasing disorder,so $\Delta S < 0$.
$4$. $Fe_{(s)}, 400 \ K \rightarrow Fe_{(s)}, 300 \ K$: Decreasing temperature decreases the kinetic energy and vibrational motion of atoms in the solid,decreasing disorder,so $\Delta S < 0$.
Therefore,the correct option is $B$.

(A) The total entropy change is given by the formula: $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$.
Given $\Delta S_{sys} = \Delta S^{\circ} = -36 \ J \ K^{-1}$.
The entropy change of the surroundings is calculated as: $\Delta S_{surr} = -\frac{\Delta H^{\circ}}{T}$.
Convert $\Delta H^{\circ}$ to $J$: $\Delta H^{\circ} = -208.6 \ kJ = -208600 \ J$.
$\Delta S_{surr} = -\frac{-208600 \ J}{298 \ K} = 700 \ J \ K^{-1}$.
Therefore,$\Delta S_{total} = -36 \ J \ K^{-1} + 700 \ J \ K^{-1} = 664 \ J \ K^{-1}$.

(D) Entropy $(S)$ is a measure of the randomness or disorder of a system. A decrease in entropy $(\Delta S < 0)$ occurs when the number of moles of gaseous products is less than the number of moles of gaseous reactants, or when a gas is converted into a liquid or solid.
$A$: $2H_2O_{2(l)} \longrightarrow 2H_2O_{(l)} + O_{2(g)}$. Here, $0$ moles of gas become $1$ mole of gas. $\Delta S > 0$.
$B$: $H_{2(g)} \longrightarrow 2H_{(g)}$. Here, $1$ mole of gas becomes $2$ moles of gas. $\Delta S > 0$.
$C$: $CaCO_{3(s)} \longrightarrow CaO_{(s)} + CO_{2(g)}$. Here, $0$ moles of gas become $1$ mole of gas. $\Delta S > 0$.
$D$: $2H_{2(g)} + O_{2(g)} \longrightarrow 2H_2O_{(l)}$. Here, $3$ moles of gas become $0$ moles of gas (liquid). $\Delta S < 0$.
Therefore, reaction $D$ exhibits a decrease in entropy.

(B) The entropy change of the surrounding $(\Delta S_{surr})$ is given by the formula: $\Delta S_{surr} = \frac{-q_{sys}}{T}$.
Given that the reaction releases heat,$q_{sys} = -525 \ kJ = -525000 \ J$.
Therefore,the heat absorbed by the surrounding is $q_{surr} = +525000 \ J$.
The temperature $T = 300 \ K$.
Substituting the values: $\Delta S_{surr} = \frac{525000 \ J}{300 \ K} = 1750 \ J \ K^{-1}$.

(D) Entropy $(S)$ is a measure of the randomness or disorder of a system.
For a given substance,the entropy follows the order: $S_{(gas)} > S_{(liquid)} > S_{(solid)}$.
$A$ decrease in entropy occurs when the system moves from a state of higher disorder to a state of lower disorder.
In the process $H_2O_{(g)} \longrightarrow H_2O_{(\ell)}$,the substance changes from a gaseous state (high entropy) to a liquid state (lower entropy).
Therefore,this transformation exhibits a decrease in entropy.

(B) The total entropy change is given by the formula: $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}$.
Given $\Delta S_{\text{sys}} = 32 \ JK^{-1}$.
The entropy change of the surroundings is given by $\Delta S_{\text{surr}} = -\frac{\Delta H}{T}$.
Given $\Delta H = -150 \ kJ = -150000 \ J$ and $T = 300 \ K$.
$\Delta S_{\text{surr}} = -\frac{-150000 \ J}{300 \ K} = 500 \ JK^{-1}$.
Therefore,$\Delta S_{\text{total}} = 32 \ JK^{-1} + 500 \ JK^{-1} = 532 \ JK^{-1}$.

(C) For any reversible process,the system is always in equilibrium with its surroundings.
By definition,the total entropy change of the universe $(\Delta S_{total})$ for any reversible process is zero.
$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} = 0$.

(B) The process of melting ice is a phase transition occurring at constant temperature and pressure,which is an equilibrium process.
For the process $H_2O_{(s)} \rightleftharpoons H_2O_{(l)}$,the entropy change is given by the formula $\Delta S = \frac{\Delta H_{\text{fusion}}}{T}$.
Given,heat of fusion $\Delta H_{\text{fusion}} = 80 \ J \ g^{-1}$.
Temperature $T = 0^{\circ} C = 273 \ K$.
Substituting the values,$\Delta S = \frac{80 \ J \ g^{-1}}{273 \ K} \approx 0.293 \ J \ g^{-1} K^{-1}$.

(C) Entropy $(S)$ is a measure of the randomness or disorder of a system.
In the reaction $NaNO_{3_{(s)}} \longrightarrow Na^{+}_{(aq)} + NO_{3^{-(aq)}}$,a solid substance dissolves to form aqueous ions.
The aqueous phase possesses higher disorder and randomness compared to the solid phase.
Therefore,the entropy of the system increases in this process.
In other options,either the number of gas moles decreases or the system transitions from a more disordered state to a more ordered state (like precipitation),leading to a decrease in entropy.

(B) At the boiling point,the process of vaporisation is in equilibrium,so $\Delta G = 0$.
Using the relation $\Delta G = \Delta H - T\Delta S$,we get $\Delta H = T\Delta S$.
Therefore,$T = \frac{\Delta H}{\Delta S}$.
Given $\Delta H = 30 \ kJ \ mol^{-1} = 30000 \ J \ mol^{-1}$ and $\Delta S = 75 \ J \ K^{-1} \ mol^{-1}$.
$T = \frac{30000 \ J \ mol^{-1}}{75 \ J \ K^{-1} \ mol^{-1}} = 400 \ K$.

(B) The entropy change of the surrounding is given by the formula: $\Delta S_{surr} = -\frac{\Delta H_{sys}}{T}$.
Given,$\Delta H_{sys} = +7 \ kJ \ mol^{-1} = +7000 \ J \ mol^{-1}$ and $T = 300 \ K$.
Substituting the values: $\Delta S_{surr} = -\frac{7000 \ J \ mol^{-1}}{300 \ K} = -23.33 \ J \ K^{-1} \ mol^{-1}$.
Thus,the correct option is $B$.

(B) According to the second law of thermodynamics,the change in entropy $(\Delta S)$ for a reversible process is defined as the ratio of the heat exchanged reversibly $(q_{rev})$ to the absolute temperature $(T)$ at which the process occurs.
$\Delta S = \frac{q_{rev}}{T}$

(A) The enthalpy of vaporization $(\Delta H_{vap})$ is $+35.3 \ kJ/mol = +35300 \ J/mol$.
The boiling point $(T)$ is $80^{\circ} C = 80 + 273 = 353 \ K$.
For the transition of liquid to vapour,the entropy change is $\Delta S_{vap} = \frac{\Delta H_{vap}}{T} = \frac{35300 \ J/mol}{353 \ K} = +100 \ J/mol \cdot K$.
The question asks for the entropy change in the transition of vapour to liquid (condensation),which is the reverse process.
Therefore,$\Delta S_{cond} = -\Delta S_{vap} = -100 \ J/mol \cdot K$.

(A) We know that the change in entropy is given by $\Delta S = \frac{q_{rev}}{T}$.
Given that the quantity of heat $q_{rev}$ is the same in both cases,we have $\Delta S_{1} = \frac{q_{rev}}{T_{1}}$ and $\Delta S_{2} = \frac{q_{rev}}{T_{2}}$.
Since $T_{1} > T_{2}$,the denominator in the expression for $\Delta S_{1}$ is larger than the denominator in the expression for $\Delta S_{2}$.
Therefore,$\Delta S_{1} < \Delta S_{2}$.

(B) The process is the condensation of benzene vapor to liquid,which is the reverse of vaporization.
The enthalpy of condensation is $\Delta H_{\text{cond}} = -\Delta H_{\text{vap}} = -35.3 \ kJ \ mol^{-1} = -35.3 \times 10^{3} \ J \ mol^{-1}$.
The boiling point is $T_b = 80 + 273 = 353 \ K$.
The entropy change for the phase transition is given by $\Delta S = \frac{\Delta H_{\text{cond}}}{T_b}$.
$\Delta S = \frac{-35.3 \times 10^{3} \ J \ mol^{-1}}{353 \ K} = -100 \ J \ K^{-1} \ mol^{-1}$.

(B) Entropy is the measure of the degree of disorder (or randomness) of a system.
In general,the entropy of a substance follows the order: $Gas > Liquid > Solid$.
Among the given options,$Hydrogen$ $gas$ is in the gaseous state,while $Liquid$ $nitrogen$ is a liquid,$Mercury$ is a liquid,and $Diamond$ is a solid.
Therefore,$Hydrogen$ $gas$ has the highest entropy.

(D) According to the second law of thermodynamics,the entropy of the universe always increases in the course of every spontaneous (natural) change.

(B) The order of entropy for states of matter is $Gas > Liquid > Solid$.
Sublimation is the process where a solid directly converts into a gas.
Since the transition from solid to gas involves the greatest increase in disorder,the sublimation of naphthalene shows the maximum increase in entropy.

(A) According to the second law of thermodynamics,all spontaneous processes are thermodynamically irreversible and are accompanied by a net increase in entropy.
Therefore,for all spontaneous processes,the total entropy change (sum of the entropy changes of the system and the surroundings) is positive.
This implies that the entropy of the universe is continuously increasing.

(D) The relationship between Gibbs free energy,enthalpy,and entropy is given by the equation: $\Delta G = \Delta H - T\Delta S$.
Rearranging for entropy change: $\Delta S = \frac{\Delta H - \Delta G}{T}$.
Given: $\Delta H = -241.60 \ kJ \ mol^{-1} = -241600 \ J \ mol^{-1}$,$\Delta G = -228.4 \ kJ \ mol^{-1} = -228400 \ J \ mol^{-1}$,and $T = 300 \ K$.
Substituting the values: $\Delta S = \frac{-241600 - (-228400)}{300} \ J \ K^{-1}$.
$\Delta S = \frac{-241600 + 228400}{300} \ J \ K^{-1} = \frac{-13200}{300} \ J \ K^{-1} = -44 \ J \ K^{-1}$.

(B) Entropy change $(\Delta S)$ is negative when the randomness or disorder of the system decreases.
$I)$ Sublimation of dry ice $(CO_2(s) \rightarrow CO_2(g))$: Disorder increases,so $\Delta S > 0$.
$II)$ Freezing of water $(H_2O(l) \rightarrow H_2O(s))$: The system becomes more ordered,so $\Delta S < 0$.
$III)$ Crystallisation of the dissolved substance: The solute particles move from a disordered solution state to an ordered crystal lattice,so $\Delta S < 0$.
$IV)$ Burning of rocket fuel: This process involves combustion,which produces gaseous products and releases heat,leading to an increase in disorder,so $\Delta S > 0$.
Therefore,processes $II$ and $III$ have a negative entropy change.

(A) $(I)$ The total entropy change $(\Delta S_{\text{total}})$ is given by $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}$. Therefore,$\Delta S_{\text{sys}} = \Delta S_{\text{total}} - \Delta S_{\text{surr}}$. The statement $I$ is incorrect.
$(II)$ For the process $A_{(l)} \rightarrow A_{(s)}$,the system transitions from a liquid to a solid state. Since the disorder decreases,the entropy change is negative (entropy decreases). The statement $II$ is correct.
$(III)$ The $SI$ unit of entropy $(S)$ is $J K^{-1} mol^{-1}$. The statement $III$ is correct.

(B) Entropy $(S)$ is a measure of the randomness or disorder of a system.
In reaction $I$,a solid decomposes into a solid and a gas. Since the number of gaseous moles increases,entropy increases.
In reaction $II$,one mole of gaseous $Cl_2$ molecule dissociates into two moles of gaseous $Cl$ atoms. The increase in the number of particles leads to an increase in entropy.
In reaction $III$,liquid water freezes into solid ice. This process involves a decrease in randomness,so entropy decreases.
Therefore,entropy increases in reactions $I$ and $II$.

(C) Entropy is the measure of disorder or randomness in a system.
For a given substance,the entropy follows the order: $S_{\text{gas}} > S_{\text{liquid}} > S_{\text{solid}}$.
In option $A$,two gas molecules combine to form one gas molecule,leading to a decrease in the number of particles and thus a decrease in entropy.
In option $B$,gas converts to solid,which significantly decreases the randomness,so entropy decreases.
In option $C$,liquid converts to gas,which increases the degree of randomness,so entropy increases.
In option $D$,gas and solid reactants form a solid product,resulting in a decrease in the number of gaseous moles,leading to a decrease in entropy.

(A) Entropy is a measure of the randomness or disorder of a system.
In the process $H_2O(l) \longrightarrow H_2O(s)$,the liquid state (more disordered) changes to the solid state (more ordered).
Since the degree of disorder decreases during freezing,the entropy of the system decreases.
In other options,such as evaporation or heating,the disorder increases,leading to an increase in entropy.

(B) The given reaction is the vaporization of water at its boiling point: $H_2O_{(l)} \longrightarrow H_2O_{(g)}$.
This process occurs at equilibrium at $T=100^{\circ} C$ and $P=1 \ atm$.
For any process at equilibrium,the change in entropy of the universe is zero: $\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} = 0$.
Therefore,$\Delta S_{\text{system}} = -\Delta S_{\text{surroundings}}$.
Since vaporization involves a transition from a liquid to a gas,the disorder increases,so $\Delta S_{\text{system}} > 0$.
Consequently,$\Delta S_{\text{surroundings}}$ must be negative,i.e.,$\Delta S_{\text{surroundings}} < 0$.

(B) Entropy is a measure of the disorder or randomness of a system.
When a substance changes from a liquid state to a solid state,the particles become more ordered,leading to a decrease in entropy $(\Delta S < 0)$.
In the freezing of water,liquid water molecules (which are more disordered) transition into a rigid crystalline structure of ice (which is more ordered).
Therefore,the change in entropy is negative.
Hence,option $(B)$ is correct.

(D) Entropy $(S)$ is a measure of the randomness or disorder of a system. $A$ decrease in entropy occurs when the system becomes more ordered,such as when gas molecules are consumed to form a solid.
In option $D$,$CaO_{(s)} + CO_{2(g)} \longrightarrow CaCO_{3(s)}$,one mole of gas $(CO_2)$ is consumed to form a solid product $(CaCO_3)$.
Since the number of moles of gaseous species decreases from $1$ to $0$,the randomness of the system decreases,leading to a decrease in entropy $(\Delta S < 0)$.

(D) $Entropy$ $(S)$ is a measure of the randomness or disorder of a system. $A$ positive change in entropy $(\Delta S > 0)$ occurs when the system becomes more disordered,such as during a phase transition from liquid to gas.
In the reaction $H_2O_{(l)} \longrightarrow H_2O_{(g)}$,water changes from a liquid state to a gaseous state. Since gas molecules have much higher freedom of movement and randomness compared to liquid molecules,the entropy of the system increases.
In the other options,the reactions involve a decrease in the number of moles of gas or a transition to a more ordered state (like liquid to solid or gas to solid),which results in a negative entropy change.

(A) The entropy change of the system is given by the formula: $\Delta S_{sys} = \frac{q_{rev}}{T}$.
Assuming the process is reversible,the heat absorbed $q_{sys}$ is calculated as:
$q_{sys} = \Delta S_{sys} \times T$
$q_{sys} = 5 \ J \ K^{-1} \ mol^{-1} \times 300 \ K = 1500 \ J \ mol^{-1}$.
To convert the value into $kJ \ mol^{-1}$,divide by $1000$:
$q_{sys} = \frac{1500}{1000} \ kJ \ mol^{-1} = 1.5 \ kJ \ mol^{-1}$.

(B) Standard molar entropy is a measure of the disorder or randomness of a substance at $1 \ bar$ pressure and a specified temperature.
For substances in the same physical state (gas),entropy generally increases with increasing molecular complexity and molecular mass.
Comparing the molar masses:
$CO_{(g)} = 28 \ g/mol$
$CO_{2(g)} = 44 \ g/mol$
$SO_{2(g)} = 64 \ g/mol$
$SO_{3(g)} = 80 \ g/mol$
Since $SO_{3(g)}$ has the highest molecular mass and the greatest number of atoms among the given options,it possesses the highest degree of vibrational and rotational complexity,leading to the highest standard molar entropy.
Therefore,the correct option is $B$.

(B) $(a) H_{2(g)} \longrightarrow 2H_{(g)}: \Delta n_g > 0$,so $\Delta S > 0$.
$(b) H_2O_{(l)} \longrightarrow H_2O_{(s)}:$ The degree of randomness of $H_2O$ molecules decreases during the freezing of water into ice,so $\Delta S < 0$.
$(c)$ With an increase in temperature,the kinetic energy $(KE)$ of $H_2O$ molecules in ice increases,so $\Delta S > 0$.
$(d) 2NaHCO_{3(s)} \longrightarrow Na_2CO_{3(s)} + CO_{2(g)} + H_2O_{(g)}:$ Here,$\Delta n_g > 0$,so $\Delta S > 0$.
Therefore,entropy decreases when liquid water crystallizes into ice.

(C) For reaction $(I)$:
$2 Cl_{(g)} \longrightarrow Cl_{2(g)}$
$\Delta n_g = 1 - 2 = -1$
Since the number of moles of gas decreases,the entropy decreases,so $\Delta S < 0$.
For reaction $(II)$:
$CO_{2(g)} \longrightarrow CO_{(g)} + \frac{1}{2} O_{2(g)}$
$\Delta n_g = (1 + 0.5) - 1 = 0.5$
Since the number of moles of gas increases,the entropy increases,so $\Delta S > 0$.
Therefore,the signs are negative and positive.

(B) Given,
$Heat \ (q) = 1 \ kJ = 1000 \ J$
$Temperature \ (T) = 3 \ K$
The change in entropy is given by the formula:
$\Delta S = \frac{q}{T}$
Substituting the values:
$\Delta S = \frac{1000 \ J}{3 \ K} = 333.3 \ J K^{-1}$

(C) Entropy $(S)$ increases when the disorder of the system increases,such as phase changes from liquid to gas or an increase in the number of gaseous moles.
$(i)$ $H_2O_{(l)} \rightarrow H_2O_{(g)}$: Phase change from liquid to gas increases entropy.
$(ii)$ $C_{(s)} + CO_{2(g)} \rightarrow 2CO_{(g)}$: The number of gaseous moles increases from $1$ to $2$,so entropy increases.
$(iii)$ $2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$: Gaseous reactants form a liquid product,so entropy decreases.
$(iv)$ $N_{2(g)} + O_{2(g)} \rightarrow \text{Mixture of } N_2 \text{ and } O_2$: Mixing of gases increases the randomness of the system,so entropy increases.
Therefore,reactions $(i)$,$(ii)$,and $(iv)$ show an increase in entropy.