2 nd Law of thermodynamics and Entropy Questions in English

(B) Entropy $(S)$ is a measure of the randomness or disorder of a system.
$(I)$ Melting of a solid: The transition from solid to liquid increases the freedom of movement of particles,thus increasing entropy.
$(II)$ Mixing of gases: Mixing increases the number of possible arrangements of gas molecules,leading to an increase in entropy.
$(III)$ Compression of a gas: Compressing a gas reduces the volume available to the molecules,decreasing their randomness and thus decreasing entropy.
$(IV)$ Expansion of a gas: Expanding a gas increases the volume available to the molecules,increasing their randomness and thus increasing entropy.
Therefore,processes $(I)$,$(II)$,and $(IV)$ are associated with an increase in entropy.

(D) process is non-spontaneous if it does not occur on its own under the given conditions.
$1$. Mixing of gases is a spontaneous process due to an increase in entropy.
$2$. The reaction of sodium with water is highly exothermic and spontaneous.
$3$. The reaction of sulfuric acid with lime (calcium hydroxide) is a neutralization reaction,which is spontaneous.
$4$. Boiling of water requires the temperature to be at its boiling point ($100\,^oC$ at $1$ atm). At $60\,^oC$ and $1$ atm,water does not boil spontaneously. Thus,this process is non-spontaneous.

(C) The entropy change for vaporization is given by the formula: $\Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T}$.
Given: $\Delta H_{\text{vap}} = 40.8\, kJ\, mol^{-1} = 40800\, J\, mol^{-1}$ and $T = 100 + 273 = 373\, K$.
Substituting the values: $\Delta S_{\text{vap}} = \frac{40800}{373} \approx 109.4\, J\, K^{-1}\, mol^{-1}$.

(D) For a process to be spontaneous,the total entropy change of the universe,which is the sum of the entropy change of the system and the surroundings,must be positive. Mathematically,$\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$.

(A) The entropy change of the surroundings is given by the formula $\Delta S_{surr} = -\frac{\Delta H_{sys}}{T}$.
For an exothermic reaction,the enthalpy change of the system $\Delta H_{sys}$ is negative.
Therefore,$\Delta S_{surr} = -\frac{(\text{negative value})}{T}$,which results in a positive value.
Thus,$\Delta S_{surr}$ is always positive for an exothermic reaction.

(D) When a solid melts,it undergoes a phase transition from a highly ordered state to a less ordered state.
Since entropy $(S)$ is a measure of the randomness or disorder of a system,the transition from solid to liquid results in an increase in the randomness of the particles.
Therefore,$S$ increases during the melting process.

(C) Entropy is a measure of the randomness or disorder of a system. $\Delta S < 0$ implies a decrease in entropy,which occurs when a system becomes more ordered.
In option $A$,solid to gas transition increases disorder.
In option $B$,dissolution increases entropy due to mixing.
In option $D$,mixing of gases increases entropy.
In option $C$,two moles of gaseous reactants form one mole of solid product. Since the solid state is much more ordered than the gaseous state,the entropy of the system decreases.

(A) Entropy is a measure of the degree of randomness or disorder in a system.
In the solid state (ice),molecules are held in a fixed,ordered lattice,whereas in the liquid state (water),molecules have more freedom of movement.
Therefore,the entropy of ice is less than that of water.
The reason provided,that ice has an open cage-like structure due to hydrogen bonding,is also a correct statement.
This cage-like structure restricts the movement of water molecules,which is the underlying cause for the lower entropy of ice compared to water.
Thus,the Reason correctly explains the Assertion.

(A) At room temperature,water is more stable than ice because ice spontaneously melts into liquid water. Thus,the Assertion is true.
Entropy is a measure of the randomness or disorder of a system. Liquid water has a more disordered structure compared to the highly ordered crystalline structure of ice,meaning liquid water has higher entropy. Thus,the Reason is true.
According to the second law of thermodynamics,for a spontaneous process at constant temperature and pressure,the Gibbs free energy change $\Delta G = \Delta H - T\Delta S$ must be negative. Since the melting of ice is spontaneous at room temperature,the increase in entropy $(\Delta S > 0)$ contributes to making $\Delta G$ negative,thereby explaining the stability of the liquid state. Therefore,the Reason correctly explains the Assertion.

(D) For the reaction $2H_{(g)} \rightarrow H_{2(g)}$,two moles of gaseous atoms combine to form one mole of a gaseous molecule.
Since the number of moles of gas decreases,the randomness or disorder of the system decreases.
Therefore,the change in entropy $(\Delta S)$ is negative.

(A) Entropy $(S)$ is a state function and its value depends on the state of the system,which is defined by variables like temperature $(T)$,pressure $(P)$,and volume $(V)$.
Since $S = f(T, P, V)$,entropy is a function of temperature.
Similarly,the change in entropy $(\Delta S = S_2 - S_1)$ also depends on the initial and final states of the system,which are defined by temperature.
Therefore,both $S$ and $\Delta S$ are functions of temperature.

(A) $(i)$ After freezing,the molecules attain an ordered state,therefore,entropy decreases.
$(ii)$ At $0 \, K$,the constituent particles are static and entropy is minimum. If the temperature is raised to $115 \, K$,these particles begin to move and oscillate about their equilibrium positions in the lattice,and the system becomes more disordered. Therefore,entropy increases.
$(iii)$ The reactant,$NaHCO_3$,is a solid and has low entropy. Among the products,there is one solid and two gases. Therefore,the products represent a condition of higher entropy.
$(iv)$ Here,one molecule gives two atoms,i.e.,the number of particles increases,leading to a more disordered state. Two moles of $H$ atoms have higher entropy than one mole of $H_2$ molecule.

(N/A) The spontaneity of a reaction is determined by the total entropy change,$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$.
To calculate $\Delta S_{surr}$,we consider the heat released to the surroundings,which is equal to $-\Delta_r H^\Theta$. At constant pressure and temperature $T$,the entropy change of the surroundings is given by:
$\Delta S_{surr} = -\frac{\Delta_r H^\Theta}{T}$
Substituting the given values:
$\Delta S_{surr} = -\frac{(-1648 \times 10^3 \ J \ mol^{-1})}{298 \ K} = 5530 \ J \ K^{-1} \ mol^{-1}$
Now,the total entropy change for the reaction is:
$\Delta_r S_{total} = \Delta S_{sys} + \Delta S_{surr} = -549.4 \ J \ K^{-1} \ mol^{-1} + 5530 \ J \ K^{-1} \ mol^{-1} = 4980.6 \ J \ K^{-1} \ mol^{-1}$
Since $\Delta_r S_{total} > 0$,the reaction is spontaneous.

(C) For an isolated system,the total energy remains constant,so $\Delta U = 0.$
According to the second law of thermodynamics,for any spontaneous process in an isolated system,the entropy of the system must increase.
Therefore,$\Delta S > 0$ for a spontaneous process in an isolated system.

(A) The formation of $1 \ mol$ of $H_2O_{(l)}$ releases $286 \ kJ \ mol^{-1}$ of heat,meaning $q_{sys} = -286 \ kJ \ mol^{-1}$.
Since the process is exothermic,the surroundings absorb this heat: $q_{surr} = +286 \ kJ \ mol^{-1} = +286000 \ J \ mol^{-1}$.
Standard conditions imply a temperature of $T = 298 \ K$.
The entropy change for the surroundings is given by $\Delta S_{surr} = \frac{q_{surr}}{T}$.
$\Delta S_{surr} = \frac{286000 \ J \ mol^{-1}}{298 \ K} = 959.73 \ J \ K^{-1} \ mol^{-1}$.

(N/A) Entropy $(S)$ is a measure of the degree of randomness or disorder in a system. For a spontaneous process,the total entropy change of the system and surroundings must be positive,i.e.,$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$.
Spontaneity is determined by the Gibbs free energy change $(\Delta G)$. $A$ reaction is spontaneous if $\Delta G < 0$. The relationship is given by $\Delta G = \Delta H - T\Delta S$.
For an exothermic reaction $(\Delta H < 0)$,if $\Delta S > 0$,the reaction is spontaneous at all temperatures. If $\Delta S < 0$,it is spontaneous only at low temperatures.
For an endothermic reaction $(\Delta H > 0)$,if $\Delta S > 0$,it is spontaneous at high temperatures. If $\Delta S < 0$,the reaction is non-spontaneous at all temperatures.

(N/A) Entropy $(S)$ is a measure of the degree of randomness or disorder in a system. For a spontaneous process,the total entropy of the system and surroundings must increase,i.e.,$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$.
Spontaneity refers to the tendency of a process to occur without continuous external influence. While exothermic reactions (where $\Delta H < 0$) are often spontaneous,enthalpy change alone is not sufficient to predict spontaneity. The second law of thermodynamics states that for a spontaneous process,the entropy of the universe increases.
The Gibbs free energy change $(\Delta G = \Delta H - T\Delta S)$ is the definitive criterion for spontaneity at constant temperature and pressure:
$1$. If $\Delta G < 0$,the process is spontaneous.
$2$. If $\Delta G > 0$,the process is non-spontaneous.
$3$. If $\Delta G = 0$,the system is at equilibrium.

(N/A) Entropy is a measure of the degree of randomness or disorder in a system.
Entropy is lowest in the solid state and highest in the gaseous state. When heat is added to a system,it increases molecular motion,leading to increased randomness.
The change in entropy $(\Delta S)$ is defined as the heat added to the system reversibly $(q_{rev})$ divided by the absolute temperature $(T)$: $\Delta S = \frac{q_{rev}}{T}$.
For a process to be spontaneous,the total entropy change of the universe must be positive: $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$.
$1$. If $\Delta S_{total} > 0$,the process is spontaneous.
$2$. If $\Delta S_{total} < 0$,the process is non-spontaneous.
$3$. If $\Delta S_{total} = 0$,the system is in equilibrium.
Entropy is a state function,meaning its value depends only on the initial and final states of the system.

(A) The change in entropy for a phase transition is given by the formula: $\Delta S_{vap} = \frac{\Delta H_{vap}}{T_b}$.
Given,$\Delta H_{vap} = 30.779 \ kJ/mol$ and $T_b = 353 \ K$.
Substituting the values: $\Delta S_{vap} = \frac{30.779 \ kJ/mol}{353 \ K} = 0.0872 \ kJ \ K^{-1} \ mol^{-1}$.

(A) The change in entropy for a phase transition is given by the formula: $\Delta S_{vap} = \frac{n \times \Delta H_{vap}}{T}$.
Given:
$n = 3 \ mol$
$\Delta H_{vap} = 40.668 \ kJ/mol$
$T = 373 \ K$
Substituting the values:
$\Delta S_{vap} = \frac{3 \ mol \times 40.668 \ kJ/mol}{373 \ K} = \frac{122.004 \ kJ}{373 \ K} = 0.327 \ kJ/K$.

(A) The process of melting ice at $273 \ K$ to water at $273 \ K$ is spontaneous because the surrounding temperature $(298 \ K)$ is higher than the melting point of ice $(273 \ K)$.
For the process $H_2O(s) \rightarrow H_2O(l)$ at $273 \ K$:
$\Delta H = 6.025 \ kJ \ mol^{-1} = 6025 \ J \ mol^{-1}$.
$\Delta S_{sys} = \frac{\Delta H}{T} = \frac{6025 \ J \ mol^{-1}}{273 \ K} = 22.07 \ J \ K^{-1} \ mol^{-1}$.
For the surrounding,$\Delta S_{surr} = \frac{-\Delta H}{T_{surr}} = \frac{-6025 \ J \ mol^{-1}}{298 \ K} = -20.22 \ J \ K^{-1} \ mol^{-1}$.
Total entropy change $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} = 22.07 - 20.22 = 1.85 \ J \ K^{-1} \ mol^{-1}$.
Since $\Delta S_{total} > 0$,the process is spontaneous.

(A) For the mixing of two gases,the enthalpy change is $\Delta H = 0$.
According to the Gibbs free energy equation,$\Delta G = \Delta H - T\Delta S$.
Since $\Delta H = 0$,the equation becomes $\Delta G = -T\Delta S$.
When two gases mix,the entropy of the system increases,meaning $\Delta S > 0$.
Therefore,$\Delta G$ becomes negative $(\Delta G < 0)$,which indicates that the process is spontaneous.

(N/A) Heat has a randomizing influence on a system,and temperature is the measure of the average chaotic motion of particles in the system.
The mathematical relation that relates these three parameters is $\Delta S = \frac{q_{rev}}{T}$.
Here,$\Delta S$ is the change in entropy.
$q_{rev}$ is the heat exchanged in a reversible process.
$T$ is the absolute temperature.

(N/A) The standard molar entropy of $H_{2}O_{(l)}$ is $70 \ J \ K^{-1} \ mol^{-1}$.
The solid form of $H_{2}O$ is ice. In ice,molecules of $H_{2}O$ are held in a rigid crystalline structure,making them less random and disordered than in liquid water.
Entropy is a measure of the randomness or disorder of a system. Since the solid state has a more ordered arrangement of molecules compared to the liquid state,the entropy of the solid is lower.
Therefore,the standard molar entropy of $H_{2}O_{(s)}$ will be less than $70 \ J \ K^{-1} \ mol^{-1}$.

(A) The melting point of ice is $273 \ K$.
At $275 \ K$,which is above the melting point,the process of ice melting into water is spontaneous.
For a spontaneous process,the change in entropy of the system is positive.
Therefore,$\Delta S > 0$ (Positive).

(B) When an ideal gas expands into a vacuum,the external pressure $P_{ext} = 0$.
Since work done $w = -P_{ext} \Delta V$,the work done is $0$.
For an adiabatic process in a vacuum,$q = 0$.
According to the first law of thermodynamics,$\Delta U = q + w = 0$.
For an ideal gas,$\Delta U$ depends only on temperature,so $\Delta T = 0$.
However,the expansion into a vacuum is an irreversible process.
The entropy change for an ideal gas is given by $\Delta S = nR \ln \frac{V_2}{V_1}$.
Since $V_2 > V_1$,$\Delta S > 0$.

(B) The $2^{nd}$ law of thermodynamics helps in predicting the spontaneity of a process based on entropy changes,but it does not provide any information regarding the rate or speed at which the chemical reaction proceeds.

(B) The second law of thermodynamics is primarily understood using the state functions $Entropy$ $(S)$ and $Gibbs$ $Free$ $Energy$ $(G)$.
$Entropy$ helps determine the spontaneity of a process in an isolated system,while $Gibbs$ $Free$ $Energy$ is used for systems at constant temperature and pressure.

(B) The entropy change $(\Delta S)$ for a reversible process is defined as the ratio of the heat absorbed reversibly $(q_{rev})$ to the absolute temperature $(T)$.
Thus,the mathematical equation is $\Delta S = \frac{q_{rev}}{T}$.

Entropy is a thermodynamic state function that measures the degree of randomness or disorder in a system.
It is denoted by the symbol $S$.
For a reversible process,the change in entropy $(dS)$ is defined as the ratio of heat absorbed reversibly $(dq_{rev})$ to the absolute temperature $(T)$:
$dS = \frac{dq_{rev}}{T}$.
The unit of entropy is $J \ K^{-1} \ mol^{-1}$.

(B) When an ideal gas expands into a vacuum,the process is known as free expansion.
Since the external pressure $P_{ext} = 0$,the work done $w = -P_{ext} \Delta V = 0$.
For an ideal gas,internal energy depends only on temperature,so $\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$,which implies $q = 0$.
Since the process is adiabatic $(q = 0)$ and involves no work $(w = 0)$,the change in entropy for an ideal gas during isothermal expansion is given by $\Delta S = nR \ln \frac{V_2}{V_1}$.

(B) The second law of thermodynamics helps in predicting the spontaneity of a process but it does not provide any information regarding the rate at which the reaction proceeds.

(B) The second law of thermodynamics is primarily concerned with the spontaneity of processes and the direction of energy flow.
It introduces the concept of $Entropy$ $(S)$ to measure the disorder of a system.
Furthermore,$Gibbs$ free energy $(G)$ is used to predict the spontaneity of a reaction at constant temperature and pressure,defined as $G = H - TS$.

(B) The entropy change $(\Delta S)$ for a reversible process is defined as the ratio of the heat absorbed reversibly $(q_{rev})$ to the absolute temperature $(T)$ at which the process occurs.
The mathematical expression is: $\Delta S = \frac{q_{rev}}{T}$.

(B) Entropy is a thermodynamic state function that serves as a measure of the degree of randomness or disorder in a system.

(N/A) $(i)$ Entropy is a measure of disorder or randomness in a system.
$(ii)$ It is a state function,meaning its value depends only on the initial and final states of the system.
$(iii)$ It is an extensive property,as its value depends on the amount of matter present in the system.

(N/A) The entropy of $1 \ mol$ of a substance at standard state is known as standard molar entropy or absolute entropy.

(N/A) As the temperature increases,the kinetic energy of the molecules increases,leading to an increase in their translational,rotational,and vibrational motions. This results in greater disorder in the system,which corresponds to an increase in entropy.

(N/A) $(i)$ Standard molar entropy
$(ii)$ $J \cdot K^{-1} \cdot mol^{-1}$
$(iii)$ Endothermic

(A) The correct matches are:
$a-2$: Conversion of liquid to vapor increases disorder,so $\Delta S = (+)$.
$b-3$: For a reaction to be non-spontaneous at all temperatures where $\Delta H = (+)$,the entropy change $\Delta S$ must be negative $((-))$ because $\Delta G = \Delta H - T\Delta S$. If $\Delta S = (-)$,then $\Delta G$ will always be positive.
$c-1$: For a reversible process,the entropy change of the system and surroundings is zero,but specifically for the reversible expansion of an ideal gas in an isolated system,$\Delta S = 0$ is often considered in specific contexts,though generally,it refers to the equilibrium state.
Correct sequence: $a-2, b-3, c-1$.

(C) $(i)$ Entropy is a measure of the degree of disorder or randomness in a system.
$(ii)$ It is a state function,meaning its value depends only on the initial and final states of the system.
$(iii)$ It is an extensive property,as its value depends on the amount of matter present in the system.

(N/A) The entropy of $1 \ mol$ of a substance at standard state is known as standard molar entropy or absolute entropy.

(N/A) As the temperature increases,the kinetic energy of the molecules increases,which leads to an increase in the translational,rotational,and vibrational motions of the particles. This results in greater disorder in the system,and therefore,the entropy increases.

(A) The entropy change for a system can be calculated as:
$\Delta S = n C_{v} \ln \frac{T_{2}}{T_{1}} + n R \ln \frac{V_{2}}{V_{1}} \dots (I)$
For an isochoric process,$V_{2} = V_{1}$,so $\ln \frac{V_{2}}{V_{1}} = \ln(1) = 0$.
For a diatomic gas,$C_{v} = \frac{5R}{2}$.
Substituting the values into expression $(I)$:
$\Delta S = 1 \times \frac{5 \times 8.314}{2} \times \ln \frac{500}{300}$
$\Delta S = 2.5 \times 8.314 \times 0.5108$
$\Delta S = 10.61\,J/K$

(C) An isolated system is defined as a system that cannot exchange energy or matter with its surroundings.
For an isolated system,the total internal energy remains constant,so $ \Delta U = 0 $.
According to the second law of thermodynamics,for a spontaneous process in an isolated system,the total entropy of the system must increase,so $ \Delta S_{total} > 0 $.
Note that $ \Delta G $ is typically defined for systems at constant temperature and pressure,which is not applicable to an isolated system in the same way; however,the primary criterion for spontaneity in an isolated system is $ \Delta S > 0 $.

(A) . Freezing of water to ice at $0^{\circ} C$ involves a phase change from liquid to solid,which decreases disorder,so $\Delta S < 0$.
$B$. Freezing of water to ice at $-10^{\circ} C$ is a spontaneous process where liquid water turns to solid ice,decreasing entropy,so $\Delta S < 0$.
$C$. In the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the number of moles of gas decreases from $4$ to $2$,leading to a decrease in entropy,so $\Delta S < 0$.
$D$. Adsorption of $CO_{(g)}$ on a lead surface restricts the movement of gas molecules,decreasing entropy,so $\Delta S < 0$.
$E$. Dissolution of $NaCl$ in water increases the disorder of the system as the crystal lattice breaks down into ions,so $\Delta S > 0$.
Therefore,entropy decreases in processes $A, B, C,$ and $D$.

(B) Entropy is a measure of the randomness or disorder of a system. For gaseous molecules,entropy generally increases with an increase in the complexity and size of the molecule,as there are more degrees of freedom for rotation and vibration.
Comparing the given molecules:
$H_2$ (diatomic,small)
$CH_4$ (polyatomic,$5$ atoms)
$C_2H_2$ (polyatomic,$4$ atoms)
$C_2H_6$ (polyatomic,$8$ atoms)
Since $C_2H_6$ has the largest number of atoms and the most complex structure among the choices,it possesses the highest degree of randomness and thus the largest entropy value.
The standard molar entropy values $(J \ K^{-1} \ mol^{-1})$ at $298 \ K$ are:

$C_2H_2$	$200.8$
$H_2$	$130.58$
$C_2H_6$	$229.5$
$CH_4$	$186.2$

(B) When the partition is removed,each gas expands to occupy the total volume $V_{total} = V_{1} + V_{2}$.
For an ideal gas undergoing isothermal expansion,the change in entropy is given by $\Delta S = n R \ln \left( \frac{V_{final}}{V_{initial}} \right)$.
Since entropy is an extensive property,the total change in entropy is the sum of the entropy changes for each gas:
$\Delta S_{total} = \Delta S_{He} + \Delta S_{Ne}$
$\Delta S_{total} = n_{1} R \ln \left( \frac{V_{1}+V_{2}}{V_{1}} \right) + n_{2} R \ln \left( \frac{V_{1}+V_{2}}{V_{2}} \right)$
Thus,the correct option is $B$.

(A) The entropy change $\Delta S$ for a phase transition is given by the formula $\Delta S = \frac{\Delta H}{T}$.
Given,latent heat of fusion $\Delta H = 6 \, kJ \, mol^{-1} = 6000 \, J \, mol^{-1}$.
The temperature of melting of ice at $1 \, atm$ is $T = 0^{\circ} C = 273 \, K$.
Substituting the values: $\Delta S = \frac{6000 \, J \, mol^{-1}}{273 \, K} \approx 21.97 \, J \, K^{-1} \, mol^{-1}$.
Rounding to the nearest integer,we get $22 \, J \, K^{-1} \, mol^{-1}$.
Thus,the correct option is $A$.