2 nd Law of thermodynamics and Entropy Questions in English

(C) For a reversible process,the change in entropy is defined by the relation $dS = \frac{dq_{rev}}{T}$.
Integrating this expression for a finite change,we get $\Delta S = \int \frac{dq_{rev}}{T}$.
Since $T$ is constant for an isothermal reversible process,$\Delta S = \frac{1}{T} \int dq_{rev}$.
As $\int dq_{rev} = q$,the expression becomes $\Delta S = \frac{q}{T}$.

(D) According to the second law of thermodynamics,for a spontaneous process,the entropy of the universe must increase. Since disorder is a measure of entropy,an increase in disorder implies $\Delta S > 0$. Therefore,the change is spontaneous.

(D) The correct answer is $(D)$. According to the second law of thermodynamics,heat spontaneously flows from a body at a higher temperature to a body at a lower temperature.

(B) Mixing of non-reacting gases increases the randomness of the system,which leads to an increase in entropy $(\Delta S > 0)$.

(B) For an isolated system,the total energy remains constant,but for any irreversible process,the entropy of the system must increase.
According to the second law of thermodynamics,$\Delta S_{total} > 0$ for all spontaneous or irreversible processes.
Therefore,the correct option is $(b)$.

(D) The entropy change for the reaction is calculated using the formula: $\Delta S^o = \sum \nu_p S^o_p - \sum \nu_r S^o_r$.
For the reaction $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$,the expression is: $\Delta S^o = 2S^o(HCl) - [S^o(H_2) + S^o(Cl_2)]$.
Substituting the given values: $\Delta S^o = 2 \times 186.7 - (130.6 + 223.0)$.
$\Delta S^o = 373.4 - 353.6 = 19.8 \, J \, K^{-1} \, mol^{-1}$.

(A) Entropy is a measure of the degree of randomness or disorder in a system.
In the solid state,molecules are held in a fixed,ordered lattice structure,resulting in the lowest entropy.
In the liquid state,molecules have more freedom of movement than in the solid state.
In the gaseous state (steam),molecules move randomly and have the highest entropy.
Therefore,$Ice$ is the least $random$ state of water.

(B) In an adiabatic process,there is no exchange of heat between the system and the surroundings,so $q = 0$.
For a reversible adiabatic process,the change in entropy is defined as $\Delta S = \frac{q_{rev}}{T}$.
Since $q = 0$,it follows that $\Delta S = 0$.
Therefore,the correct option is $(B)$.

(C) Entropy is a measure of the degree of randomness or disorder in a system.
In the solid state,particles are held in a fixed lattice structure,resulting in the lowest entropy.
In the liquid state,particles have more freedom of movement than in solids,leading to higher entropy.
In the gaseous state,particles are in constant,random motion with the greatest freedom,resulting in the highest entropy.
Therefore,the correct answer is $(C)$.

(C) The process of vaporisation is the reverse of the process of condensation.
For a reversible process,the entropy change of the reverse reaction is equal in magnitude but opposite in sign to that of the forward reaction.
Given: $\Delta S_{\text{vap}} = 88.3 \ J/mol \ K$.
Therefore,$\Delta S_{\text{cond}} = -\Delta S_{\text{vap}} = -88.3 \ J/mol \ K$.

(B) The change in entropy,$\Delta S$,is related to the change in the number of moles of gaseous species $(\Delta n_g)$.
$A$. $\Delta n_g = 0 - 0.5 = -0.5$
$B$. $\Delta n_g = 1 - 0 = +1$
$C$. $\Delta n_g = 1 - 1 = 0$
$D$. $\Delta n_g = 2 - 2 = 0$
Since reaction $B$ involves the production of a gas from a solid,it results in the largest increase in disorder,making $\Delta S$ maximum.

(A) The entropy of vaporization $(\Delta S_{vap})$ is calculated using the formula: $\Delta S_{vap} = \frac{\Delta H_{vap}}{T_b}$.
Given,$\Delta H_{vap} = 186.5 \ kJ \ mol^{-1} = 186500 \ J \ mol^{-1}$ and $T_b = 373 \ K$.
Therefore,$\Delta S_{vap} = \frac{186500 \ J \ mol^{-1}}{373 \ K} = 500 \ J \ K^{-1} \ mol^{-1}$.

(A) According to the $2^{nd}$ law of thermodynamics,for any spontaneous process,the total entropy change of the universe is positive.
$\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surrounding}} > 0$
Therefore,the entropy of the universe is continuously increasing and tends towards a maximum value.

(B) For an irreversible process,the total entropy of the system and surroundings increases,i.e.,$\Delta S_{total} > 0$.

(C) The change in entropy $(\Delta S)$ for chemical reactions is typically calculated at constant temperature and pressure.
These are the standard conditions under which most laboratory reactions are performed.
Furthermore,thermodynamic data,such as standard molar entropy,are tabulated at these specific conditions.

(B) The ability to perform work is related to the availability of energy. Entropy is a measure of the disorder or randomness of a system. According to the second law of thermodynamics,as entropy increases,the energy becomes less available to do useful work. Therefore,when the value of entropy is greater,the ability to perform work is minimum.

(C) The entropy change for a phase transition at equilibrium is given by the formula: $\Delta S_{vap} = \frac{\Delta H_{vap}}{T}$.
Given,$\Delta H_{vap} = 37.3 \ kJ \ mol^{-1} = 37300 \ J \ mol^{-1}$ and $T = 373 \ K$.
Substituting the values: $\Delta S_{vap} = \frac{37300 \ J \ mol^{-1}}{373 \ K} = 100 \ J \ mol^{-1} K^{-1}$.

(D) The process of melting involves the transition from a highly ordered solid state to a more disordered liquid state.
Since entropy is a measure of the randomness or disorder of a system,the transition from solid to liquid results in an increase in disorder.
Therefore,$\Delta S > 0$,which means entropy increases.

(C) For a spontaneous process,the total entropy change of the system and its surroundings must be positive,i.e.,$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$. Therefore,the entropy of the universe increases.

(B) Entropy $(S)$ is a measure of the degree of randomness or disorder in a system.
For a process,the change in entropy is given by $\Delta S$.
If $\Delta S > 0$ (positive value),it indicates an increase in the randomness or disorder of the system.
Therefore,the correct option is $(B)$.

(B) The entropy of vaporization $(\Delta S_{vap})$ is calculated using the formula: $\Delta S_{vap} = \frac{\Delta H_{vap}}{T}$.
Given: $\Delta H_{vap} = 386 \ kJ \ mol^{-1}$ and $T = 298 \ K$.
$\Delta S_{vap} = \frac{386}{298} \approx 1.295 \ kJ \ K^{-1} \ mol^{-1}$.
Therefore,the correct option is $(B)$.

(B) Entropy $(S)$ is a measure of the disorder or randomness of a system.
$(a)$ $H_{2(g)} \to 2H_{(g)}$: The number of moles of gas increases $(1 \to 2)$,leading to an increase in disorder,so $\Delta S > 0$.
$(b)$ $N_{2(g)} (1 \ atm) \to N_{2(g)} (8 \ atm)$: Increasing the pressure of a gas restricts the volume available to the molecules,decreasing the disorder and thus $\Delta S < 0$.
$(c)$ $2SO_{3(g)} \to 2SO_{2(g)} + O_{2(g)}$: The number of moles of gas increases $(2 \to 3)$,leading to an increase in disorder,so $\Delta S > 0$.
$(d)$ $C_{(diamond)} \to C_{(graphite)}$: Graphite is more disordered than diamond,so $\Delta S > 0$.

(A) The boiling point of water is $T = 373 \ K$.
The molar mass of water $(H_2O)$ is $M = 18 \ g/mol$.
The heat exchanged per mole $(\Delta H_{vap})$ is $900 \ J/g \times 18 \ g/mol = 16200 \ J/mol$.
The change in entropy is given by $\Delta S_{vap} = \frac{\Delta H_{vap}}{T}$.
$\Delta S_{vap} = \frac{16200 \ J/mol}{373 \ K} \approx 43.4 \ J \ K^{-1} \ mol^{-1}$.

(C) When two different gases are mixed at constant temperature and pressure,the process is spontaneous and irreversible.
Mixing leads to an increase in the randomness or disorder of the system.
According to the second law of thermodynamics,for any spontaneous process,the entropy of the system increases.
Therefore,the entropy of the mixture increases.

(A) The temperatures in Kelvin are:
$T_H = 150 + 273 = 423\,K$
$T_L = 25 + 273 = 298\,K$
Heat absorbed from the reservoir,$Q_H = 500\,J$.
For a Carnot engine (no frictional losses),the efficiency $\eta$ is given by:
$\eta = \frac{W}{Q_H} = \frac{T_H - T_L}{T_H}$
$W = Q_H \times \left( \frac{T_H - T_L}{T_H} \right)$
$W = 500 \times \left( \frac{423 - 298}{423} \right)$
$W = 500 \times \left( \frac{125}{423} \right) \approx 147.75\,J$.
Rounding to the nearest provided option,the work done is $147.7\,J$.

(C) The formation reaction of $CO_{2(g)}$ is:
$C_{(s)} + O_{2(g)} \to CO_{2(g)}$
The standard entropy of formation $\Delta_f S^o$ is calculated as:
$\Delta_f S^o = S^o(CO_{2(g)}) - [S^o(C_{(s)}) + S^o(O_{2(g)})]$
Substituting the given values:
$\Delta_f S^o = 213.5 - (5.690 + 205) \ J \ K^{-1} \ mol^{-1}$
$\Delta_f S^o = 213.5 - 210.690 \ J \ K^{-1} \ mol^{-1}$
$\Delta_f S^o = 2.81 \ J \ K^{-1} \ mol^{-1}$
Therefore,the correct option is $C$.

(C) The entropy change for a phase transition is given by $\Delta S = \frac{\Delta H_{vap}}{T_b}$.
Given that $1 \, \text{mole}$ of $H_2O$ has a mass of $18 \, g$,the total enthalpy of vaporization for $1 \, \text{mole}$ is $\Delta H = 2.257 \, kJ/g \times 18 \, g = 40.626 \, kJ = 40626 \, J$.
The temperature of boiling $T_b = 373 \, K$.
Therefore,$\Delta S = \frac{40626 \, J}{373 \, K} \approx 108.9 \, J \, K^{-1}$.

(C) The change in entropy is defined by the formula $\Delta S = \frac{q_{rev}}{T}$.
Here,$q_{rev}$ is the heat exchanged in Joules $(J)$ and $T$ is the temperature in Kelvin $(K)$.
For one mole of a substance,the unit of molar entropy is $J\,K^{-1}\,mol^{-1}$.
Therefore,the correct option is $C$.

(B) The process is $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$.
Entropy change is given by $\Delta S = \frac{\Delta H_{vap}}{T}$.
Given $\Delta H_{vap} = 2.257 \, kJ/g$.
Since the molar mass of water is $18 \, g/mol$,the enthalpy of vaporization per mole is $\Delta H_{vap} = 2.257 \times 18 \, kJ/mol = 40.626 \, kJ/mol$.
At $T = 373 \, K$,$\Delta S = \frac{40.626 \, kJ/mol}{373 \, K} \approx 0.109 \, kJ/mol \cdot K$.

(A) During the boiling process,the substance transitions from the liquid state to the gaseous state: $Liquid \xrightarrow{} Vapour$.
Since gas molecules have much higher randomness and disorder compared to liquid molecules,the entropy of the system increases $( \Delta S > 0 )$.

(C) Entropy $(\Delta S^o)$ is a measure of the disorder or randomness of a system.
For a reaction,$\Delta S^o > 0$ when the disorder increases,which typically occurs when the number of moles of gas increases or when a solid dissolves into aqueous ions.
In option $(C)$,$NaNO_{3(s)}$ (a solid) dissolves to form $Na^{+}_{(aq)}$ and $NO_{3(aq)}^{-}$ (aqueous ions),which increases the disorder of the system.
In options $(A)$,$(B)$,and $(D)$,the number of gas moles decreases or the system moves from a more disordered state to a more ordered state,resulting in $\Delta S^o < 0$.

(A) Entropy is a measure of the degree of randomness or disorder in a system.
In general,the entropy of a substance increases in the order: $Solid < Liquid < Gas$.
Among the given options:
$1$. $Hydrogen$ $(H_2)$ is a gas.
$2$. $Water$ $(H_2O)$ is a liquid.
$3$. $Graphite$ $(C)$ is a solid.
$4$. $Mercury$ $(Hg)$ is a liquid.
Since gases have the highest degree of molecular disorder,$Hydrogen$ has the highest entropy.

(B) The standard entropy change for a reaction is calculated as: $\Delta S^{o} = \sum S^{o}_{products} - \sum S^{o}_{reactants}$.
Given: $S^{o}(H_{2}O) = 70 \ J \ K^{-1} \ mol^{-1}$,$S^{o}(H^{+}) = 0 \ J \ K^{-1} \ mol^{-1}$,and $S^{o}(OH^{-}) = -10.7 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values into the formula:
$\Delta S^{o} = 70 - [0 + (-10.7)]$
$\Delta S^{o} = 70 + 10.7 = 80.7 \ J \ K^{-1} \ mol^{-1}$.

(D) The change in entropy for a phase transition at constant temperature is given by $\Delta S = \frac{\Delta H_{vap}}{T}$.
Given: Heat of vaporisation of water $= 540\,cal/g$.
Since the molar mass of water $(H_2O)$ is $18\,g/mol$,the molar heat of vaporisation is $\Delta H_{vap} = 540\,cal/g \times 18\,g/mol = 9720\,cal/mol$.
The temperature $T = 100 + 273 = 373\,K$.
Therefore,$\Delta S = \frac{9720\,cal/mol}{373\,K} \approx 26.06\,cal\,mol^{-1}\,K^{-1}$.

(B) Given that,$T_1 = 500 \ K$ (source temperature) and $T_2 = 300 \ K$ (sink temperature).
By using the formula for efficiency of a $Carnot$ engine: $\eta = \frac{T_1 - T_2}{T_1}$.
Substituting the values: $\eta = \frac{500 - 300}{500} = \frac{200}{500} = 0.4$.
Therefore,the efficiency is $0.4$.

(D) Entropy is the measure of randomness or disorder in a system.
In the states of matter,the randomness of molecules increases in the order: $Solid < Liquid < Gas$.
Since water vapours represent the gaseous state of water,they possess the highest degree of randomness.
Therefore,the entropy is maximum for $Water \ vapours$.

(A) The formation reaction of $CO_{2(g)}$ is: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$.
The standard entropy of formation $\Delta_f S^{\circ}$ is calculated as: $\Delta_f S^{\circ} = S^{\circ}(CO_{2(g)}) - [S^{\circ}(C_{(s)}) + S^{\circ}(O_{2(g)})]$.
Substituting the given values: $\Delta_f S^{\circ} = 213.5 - [5.740 + 205]$.
$\Delta_f S^{\circ} = 213.5 - 210.740 = 2.76 \ J \ K^{-1} \ mol^{-1}$.

(A) Entropy is a measure of the randomness or disorder of a system.
In the gaseous state,particles have the highest freedom of movement and the least amount of intermolecular forces compared to liquid or solid states.
Since steam is water in the gaseous state,it possesses the highest degree of randomness.
Therefore,entropy is maximum in the case of steam.

(C) The criterion for the spontaneity of any process in terms of entropy is based on the Second Law of Thermodynamics.
For a process to be spontaneous,the total entropy change of the universe must be positive.
$\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} > 0$.
This indicates that the entropy of the system and its surroundings considered together must increase for an irreversible (spontaneous) process.

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

(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: $\Delta H = 6 \ kJ \ mol^{-1} = 6000 \ J \ mol^{-1}$ and $\Delta S = 16 \ J \ mol^{-1} \ K^{-1}$.
$T = \frac{6000 \ J \ mol^{-1}}{16 \ J \ mol^{-1} \ K^{-1}} = 375 \ K$.

(C) The conversion of graphite into diamond involves a transition from a less dense,more disordered structure (graphite) to a more dense,highly ordered structure (diamond).
Since the entropy $(S)$ is a measure of the disorder or randomness of a system,the transition to a more ordered state (diamond) results in a decrease in entropy.
Therefore,the change in entropy,$\Delta S = S_{\text{diamond}} - S_{\text{graphite}}$,is negative.

(A) According to the $II$ law of thermodynamics,the entropy of the universe is continuously increasing. Entropy is a measure of the disorder or randomness of a system. Therefore,while the total energy remains constant (First Law),the disorder of the universe is increasing toward a maximum.

(B) Entropy is a measure of the disorder or randomness of a system.
For the same substance,the entropy follows the order: $S_{(gas)} > S_{(liquid)} > S_{(solid)}$.
Comparing the given options:
$SO_2Cl_{2(s)}$,$SO_2Cl_{2(l)}$,and $SO_2Cl_{2(g)}$ are different states of the same compound. Among these,the gaseous state has the highest entropy.
Comparing $SO_2Cl_{2(g)}$ and $SO_{2(g)}$,entropy also depends on the complexity of the molecule. $SO_2Cl_{2}$ is a larger,more complex molecule than $SO_2$,meaning it has more vibrational and rotational degrees of freedom.
Therefore,$SO_2Cl_{2(g)}$ has the largest entropy per mole.

(A) The entropy change for the reaction is given by $\Delta S^{o} = \sum S^{o}_{products} - \sum S^{o}_{reactants}$.
For the reaction $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$,the equation is $\Delta S^{o} = [2 \times S^{o}(HBr)] - [S^{o}(H_2) + S^{o}(Br_2)]$.
Substituting the given values: $20.1 = [2 \times 198.5] - [130.6 + S^{o}(Br_2)]$.
$20.1 = 397 - 130.6 - S^{o}(Br_2)$.
$20.1 = 266.4 - S^{o}(Br_2)$.
$S^{o}(Br_2) = 266.4 - 20.1 = 246.3 \, JK^{-1}$.

(A) The process of phase change (vaporization) occurs at equilibrium,so $\Delta G = 0$.
Using the relation $\Delta G = \Delta H - T\Delta S$,we get $\Delta S = \frac{\Delta H}{T}$.
Given: $\Delta H = 30 \ kJ/mol = 30000 \ J/mol$ and $T = 27 + 273 = 300 \ K$.
Substituting the values: $\Delta S = \frac{30000 \ J/mol}{300 \ K} = 100 \ J/mol \ K$.

(A) The entropy change $\Delta S$ for a phase transition is given by $\Delta S = \frac{q_{rev}}{T} = \frac{m \times L_f}{T}$.
Given: Mass $m = 1 \, kg = 1000 \, g$,Latent heat $L_f = 80 \, cal \, g^{-1} = 80 \times 4.2 \, J \, g^{-1} = 336 \, J \, g^{-1}$,Temperature $T = 0 \, ^\circ C = 273 \, K$.
Substituting the values: $\Delta S = \frac{1000 \, g \times 336 \, J \, g^{-1}}{273 \, K} = \frac{336000}{273} \, J \, K^{-1} = 1230.67 \, J \, K^{-1}$.

(C) The entropy change for the vaporization of water is given by the formula: $\Delta S = \frac{\Delta H_{vap}}{T_{vap}}$.
Given: $\Delta H_{vap} = 40.8 \, kJ \, mol^{-1} = 40.8 \times 1000 \, J \, mol^{-1}$ and $T = 373 \, K$.
Substituting the values: $\Delta S = \frac{40.8 \times 1000 \, J \, mol^{-1}}{373 \, K}$.
Calculation: $\Delta S = 109.38 \, J \, K^{-1} \, mol^{-1}$.

(A) The reaction is: $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$
The formula for the change in standard entropy is: $\Delta S^o = \sum S^o(\text{products}) - \sum S^o(\text{reactants})$
$\Delta S^o = [2 \times S^o(HCl)] - [S^o(H_2) + S^o(Cl_2)]$
Substitute the given values: $\Delta S^o = [2 \times 0.19] - [0.13 + 0.22]$
$\Delta S^o = 0.38 - 0.35 = 0.03 \, kJ \, K^{-1} \, mol^{-1}$
Convert to $J \, K^{-1} \, mol^{-1}$ by multiplying by $1000$: $0.03 \times 1000 = 30 \, J \, K^{-1} \, mol^{-1}$.

(D) The change in entropy $(\Delta S)$ for a phase transition is given by the formula: $\Delta S = \frac{\Delta H_{fus}}{T_f}$.
Given: $\Delta H_{fus} = 6 \, kJ \ mol^{-1} = 6000 \, J \ mol^{-1}$ and $T_f = 0 \, ^\circ C = 273 \, K$.
Substituting the values: $\Delta S = \frac{6000 \, J \ mol^{-1}}{273 \, K} \approx 21.98 \, J \ K^{-1} \ mol^{-1}$.