2 nd Law of thermodynamics and Entropy Questions in English

(B) For an ideal gas,the entropy change $\Delta S$ is given by the general expression:
$\Delta S = n C_p \ln \left( \frac{T_f}{T_i} \right) + n R \ln \left( \frac{p_i}{p_f} \right)$
For an isothermal process,the temperature remains constant,so $T_i = T_f$.
This implies $\ln \left( \frac{T_f}{T_i} \right) = \ln(1) = 0$.
Therefore,the expression simplifies to:
$\Delta S = n R \ln \left( \frac{p_i}{p_f} \right)$

(A) For a reaction,$\Delta G^o = \Delta H^o - T \Delta S^o$. If $\Delta S^o > 0$,then as temperature $T$ increases,the term $-T \Delta S^o$ becomes more negative,causing $\Delta G^o$ to decrease sharply.
$\Delta S^o$ is positive when the number of moles of gaseous products is greater than the number of moles of gaseous reactants (i.e.,$\Delta n_g > 0$).
In option $A$: $C_{(graphite)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}$,$\Delta n_g = 1 - 0.5 = 0.5 > 0$.
In option $B$: $CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}$,$\Delta n_g = 1 - 1.5 = -0.5 < 0$.
In option $C$: $Mg_{(s)} + \frac{1}{2} O_{2(g)} \to MgO_{(s)}$,$\Delta n_g = 0 - 0.5 = -0.5 < 0$.
In option $D$: $\frac{1}{2} C_{(graphite)} + \frac{1}{2} O_{2(g)} \to \frac{1}{2} CO_{2(g)}$,$\Delta n_g = 0.5 - 0.5 = 0$.
Thus,only reaction $A$ has a positive $\Delta S^o$.

(C) The process of melting ice at $0 \ ^\circ C$ $(273 \ K)$ is a reversible phase transition.
For a reversible process,the entropy change is given by $\Delta S = \frac{\Delta H_{fus}}{T}$.
Given $\Delta H_{fus} = 1.435 \ kcal/mol = 1435 \ cal/mol$.
Temperature $T = 0 \ ^\circ C = 273 \ K$.
Therefore,$\Delta S = \frac{1435 \ cal/mol}{273 \ K} = 5.256 \ cal/(mol \ K) \approx 5.26 \ cal/(mol \ K)$.

(D) The process is the phase transition of liquid water to steam at a constant temperature.
Given: $\Delta H_{vap} = 30 \ kJ \ mol^{-1} = 30000 \ J \ mol^{-1}$.
Temperature $T = 27 \ ^oC = 27 + 273 = 300 \ K$.
The entropy change is given by the formula: $\Delta S_{vap} = \frac{\Delta H_{vap}}{T}$.
Substituting the values: $\Delta S_{vap} = \frac{30000 \ J \ mol^{-1}}{300 \ K} = 100 \ J \ mol^{-1} \ K^{-1}$.

(B) The spontaneity of a process depends on the tendency to acquire a minimum energy state and maximum randomness (entropy).
For a spontaneous process in an isolated system,the total change in entropy $(\Delta S_{total})$ must be greater than zero,i.e.,$\Delta S > 0$.

(C) For a reaction to be at equilibrium,$\Delta G = 0$.
Since $\Delta G = \Delta H - T \Delta S$,at equilibrium,$\Delta H = T \Delta S$.
For the reaction $\frac{1}{2} X_2 + \frac{3}{2} Y_2 \to XY_3$,$\Delta H = -30 \ kJ = -30000 \ J$.
Calculating $\Delta S$ for the reaction:
$\Delta S = S^{\circ}(XY_3) - [\frac{1}{2} S^{\circ}(X_2) + \frac{3}{2} S^{\circ}(Y_2)]$
$\Delta S = 50 - [\frac{1}{2} \times 60 + \frac{3}{2} \times 40] \ J \ K^{-1} \ mol^{-1}$
$\Delta S = 50 - [30 + 60] = -40 \ J \ K^{-1} \ mol^{-1}$.
At equilibrium,$T = \frac{\Delta H}{\Delta S} = \frac{-30000 \ J}{-40 \ J \ K^{-1}} = 750 \ K$.

(A) The entropy change for an isothermal reversible process is given by the formula: $\Delta S = n R \ln \frac{V_2}{V_1}$
Substituting the given values: $n = 2 \, \text{mol}$,$R = 8.314 \, J \, K^{-1} \, \text{mol}^{-1}$,$V_1 = 10 \, dm^3$,$V_2 = 100 \, dm^3$.
$\Delta S = 2 \times 8.314 \times 2.303 \log \left( \frac{100}{10} \right)$
$\Delta S = 2 \times 8.314 \times 2.303 \times \log(10)$
$\Delta S = 2 \times 8.314 \times 2.303 \times 1 = 38.29 \, J \, K^{-1} \approx 38.3 \, J \, K^{-1}$

(D) The entropy change for the reaction is given by the formula: $\Delta S^{\circ} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants}$
For the reaction $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$,the expression is: $\Delta S^{\circ} = 2 S^{\circ}_{HCl} - (S^{\circ}_{H_2} + S^{\circ}_{Cl_2})$
Substituting the given values: $\Delta S^{\circ} = 2 \times 186.7 - (130.6 + 223.0)$
$\Delta S^{\circ} = 373.4 - 353.6 = 19.8 \ J K^{-1} mol^{-1}$

(D) The change in entropy for a phase transition is given by the formula: $\Delta S = \frac{\Delta H_{vap}}{T_b}$
Given:
Mass of $H_2O = 9.0 \, g$
Latent heat of vaporization $= x \, J / g$
Total enthalpy of vaporization $\Delta H_{vap} = 9.0 \, g \times x \, J / g = 9x \, J$
Boiling point $T_b = 100 \, ^\circ C = 100 + 273 = 373 \, K$
Substituting these values into the formula:
$\Delta S = \frac{9x}{373} \, J / K$
Since $\frac{9}{373} = \frac{1}{2} \times \frac{18}{373}$,we can write:
$\Delta S = \frac{1}{2} \times \frac{18x}{373} \, J / K$
Thus,the correct option is $D$.

(A) The entropy change of fusion is calculated using the formula: $\Delta S_{fusion} = \frac{\Delta H_{fusion}}{T_{mp}}$.
Given,$\Delta H_{fusion} = 6.01 \, kJ \, mol^{-1} = 6010 \, J \, mol^{-1}$.
The melting point of ice $(T_{mp})$ is $273 \, K$.
Therefore,$\Delta S_{fusion} = \frac{6010 \, J \, mol^{-1}}{273 \, K} \approx 22 \, J \, K^{-1} \, mol^{-1}$.

(B) The general formula for entropy change is $\Delta S = nC_V \ln \left(\frac{T_2}{T_1}\right) + nR \ln \left(\frac{V_2}{V_1}\right)$.
Given:
$n = 1 \ mol$
$T_1 = 25 \ ^{\circ}C = 298 \ K$
$T_2 = 100 \ ^{\circ}C = 373 \ K$
$C_V = 5 \ cal / mol \cdot K$
$R = 2 \ cal / mol \cdot K$
$V_1 = 1 \ L$
$V_2 = 10 \ L$
Substituting the values:
$\Delta S = 1 \times 5 \ln \left(\frac{373}{298}\right) + 1 \times 2 \ln \left(\frac{10}{1}\right)$
$\Delta S = 5 \ln \left(\frac{373}{298}\right) + 2 \ln 10$.

(C) For any process,the entropy change of the universe is given by $\Delta S_{univ} = \Delta S_{syst} + \Delta S_{surr}$.
For a reversible process,$\Delta S_{univ} = 0$.
For an irreversible (spontaneous) process,$\Delta S_{univ} > 0$.
In an adiabatic process,$\Delta S_{syst}$ is zero only if the process is reversible. If the adiabatic process is irreversible,$\Delta S_{syst} > 0$. Thus,the statement that $\Delta S_{syst} = 0$ always in an adiabatic process is incorrect.
For an isochoric process $(V = \text{constant})$,the entropy change is given by $\Delta S_{syst} = nC_v \ln(\frac{T_2}{T_1})$,which is correct.

(D) For a heat engine operating between temperatures $T_H = 500 \ K$ and $T_L = 300 \ K$,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_L}{T_H} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$.
The efficiency is also defined as $\eta = 1 - \frac{Q_L}{Q_H}$,where $Q_H = 100 \ J$ is the heat absorbed and $Q_L$ is the heat rejected.
According to the Second Law of Thermodynamics,the efficiency of any heat engine cannot exceed the Carnot efficiency: $1 - \frac{Q_L}{Q_H} \leq 1 - \frac{T_L}{T_H}$.
This simplifies to $\frac{Q_L}{Q_H} \geq \frac{T_L}{T_H}$.
Substituting the values: $\frac{Q_L}{100} \geq \frac{300}{500} = 0.6$.
Therefore,$Q_L \geq 60 \ J$.
This means the heat rejected must be at least $60 \ J$. Any value less than $60 \ J$ is impossible. Among the given options,$20 \ J$ is less than $60 \ J$,making it an impossible amount of heat rejected.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1 = 500 \ K$ and $T_2 = 400 \ K$.
$\eta = 1 - \frac{400}{500} = 1 - 0.8 = 0.2$.
The efficiency is also defined as $\eta = 1 - \frac{Q_2}{Q_1}$,where $Q_1 = 100 \ J$ is the heat absorbed and $Q_2$ is the heat rejected.
For a reversible engine,$\frac{Q_2}{Q_1} = \frac{T_2}{T_1} = 0.8$,so $Q_2 = 0.8 \times 100 \ J = 80 \ J$.
According to the Second Law of Thermodynamics,for any real engine,the efficiency $\eta_{real} \le \eta_{Carnot}$.
$1 - \frac{Q_2}{Q_1} \le 0.2 \implies \frac{Q_2}{Q_1} \ge 0.8$.
$Q_2 \ge 0.8 \times 100 \ J = 80 \ J$.
Therefore,any value of $Q_2$ less than $80 \ J$ is impossible. Thus,$10 \ J$ is impossible.

(B) The change in entropy of the system is given by $\Delta S_{system} = m \times (s_2 - s_1) = 100 \ kg \times (0.4 - 0.3) \ kJ \ kg^{-1} \ K^{-1} = 10 \ kJ \ K^{-1}$.
The change in entropy of the surrounding is given by $\Delta S_{surrounding} = S_{2,surr} - S_{1,surr} = 75 \ kJ \ K^{-1} - 80 \ kJ \ K^{-1} = -5 \ kJ \ K^{-1}$.
The entropy change of the universe is $\Delta S_{universe} = \Delta S_{system} + \Delta S_{surrounding} = 10 \ kJ \ K^{-1} + (-5 \ kJ \ K^{-1}) = 5 \ kJ \ K^{-1}$.

(B) The process is the phase transition of ice to water at its melting point: $H_2O_{(s)} \longrightarrow H_2O_{(\ell)}$.
At equilibrium,the change in entropy is given by $\Delta S = \frac{\Delta H}{T}$.
Given $\Delta S = S_{H_2O(\ell)} - S_{H_2O(s)} = 60.01 - 38.2 = 21.81 \ J \ mol^{-1} \ K^{-1}$.
Given temperature $T = 273 \ K$.
Therefore,$\Delta H = \Delta S \times T = 21.81 \times 273 = 5954.13 \ J \ mol^{-1}$.

(B) Entropy change $(\Delta S)$ is positive when the disorder of the system increases.
In the reaction $H_2O_{(s)} \to H_2O_{(l)}$,the substance changes from a solid state (ordered) to a liquid state (disordered).
Since $S_{(liquid)} > S_{(solid)}$,the change in entropy $\Delta S = S_{(liquid)} - S_{(solid)}$ is positive $(+ve)$.
In other options,the number of moles of gas decreases or the system becomes more ordered,leading to a negative entropy change.

(C) Entropy is a measure of the randomness or disorder of a system.
For a given substance,the entropy follows the order: $S_{gas} > S_{liquid} > S_{solid}$.
Additionally,entropy increases with an increase in temperature.
Comparing the options:
$A$ and $B$ are solid and liquid at $0\ ^\circ C$,which have lower entropy.
$D$ is liquid at $100\ ^\circ C$,which has higher entropy than solid or liquid at $0\ ^\circ C$.
$C$ is water vapor (gas) at $100\ ^\circ C$. Since gases have significantly higher entropy than liquids due to the high degree of molecular freedom,$H_2O\, (g, 100\ ^\circ C, 1\ atm)$ has the maximum molar entropy.

(A) The change in entropy for an ideal gas is given by the formula: $\Delta S = n C_p \ln(\frac{T_2}{T_1}) - nR \ln(\frac{P_2}{P_1})$.
Given: $n = 3 \ mol$,$C_p = 7 \ cal/mol \cdot K$,$T_1 = 300 \ K$,$T_2 = 1000 \ K$,$P_1 = 0.1 \ atm$,$P_2 = 1 \ atm$,$R \approx 2 \ cal/mol \cdot K$.
Substituting the values:
$\Delta S = 3 \times 7 \times 2.3 \log(\frac{1000}{300}) - 3 \times 2 \times 2.3 \log(\frac{1}{0.1})$
$\Delta S = 21 \times 2.3 \log(3.33) - 6 \times 2.3 \log(10)$
Using $\log(3.33) \approx 0.52$ and $\log(10) = 1$:
$\Delta S = 48.3 \times 0.52 - 13.8 \times 1$
$\Delta S = 25.116 - 13.8 = 11.316 \ cal/K$.
The approximate value is $11 \ cal/degree$.

(A) The relationship between total entropy change and system entropy change is given by: $\Delta S_{Total} = \Delta S_{System} + \Delta S_{Surroundings}$.
We know that $\Delta S_{Surroundings} = -\frac{\Delta H_{System}}{T}$.
Substituting the values: $\Delta S_{Surroundings} = -\frac{2000 \ kJ/mol}{400 \ K} = -5 \ kJ/mol \cdot K$.
Now,$\Delta S_{Total} = \Delta S_{System} + \Delta S_{Surroundings}$.
$-40 \ kJ/mol \cdot K = \Delta S_{System} + (-5 \ kJ/mol \cdot K)$.
$\Delta S_{System} = -40 + 5 = -35 \ kJ/mol \cdot K$.

(B) For a reversible process,the change in entropy is defined as $dS = \frac{dq_{rev}}{T}$.
In an adiabatic process,there is no exchange of heat with the surroundings,so $dq_{rev} = 0$.
Therefore,$\Delta S = 0$ for a reversible adiabatic process,which is also known as an isentropic process.

(B) Entropy $(S)$ is a measure of the randomness or disorder of a system.
$1$. Process $(a)$: $Br_{2(l)} \to Br_{2(g)}$. Liquid to gas transition increases disorder. $\Delta S > 0$.
$2$. Process $(b)$: $H_2O_{(s)} \to H_2O_{(g)}$. Solid to gas transition increases disorder. $\Delta S > 0$.
$3$. Process $(c)$: $N_2 \left[ 1 \ atm, 100 ^oC \right] \to N_2 \left[ 1 \ atm, 150 ^oC \right]$. Heating a gas increases the kinetic energy and randomness of molecules. $\Delta S > 0$.
$4$. Process $(d)$: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$. Here,$4 \ moles$ of gas are converted to $2 \ moles$ of gas. The number of moles decreases,so disorder decreases. $\Delta S < 0$.
$5$. Process $(e)$: $CaCO_{3(s)} \to CaO_{(s)} + CO_{2(g)}$. $A$ solid produces a gas,which significantly increases disorder. $\Delta S > 0$.
Therefore,entropy increases in processes $(a)$,$(b)$,$(c)$,and $(e)$.

(B) The change in entropy for the reaction is calculated using the formula:
$\Delta S^o = \sum \nu_p S^o_{products} - \sum \nu_r S^o_{reactants}$
Substituting the given values:
$\Delta S^o = (4 \times S_C^o + 5 \times S_D^o) - (2 \times S_A^o + 3 \times S_B^o)$
$\Delta S^o = (4 \times 200 + 5 \times 150) - (2 \times 100 + 3 \times 120)$
$\Delta S^o = (800 + 750) - (200 + 360)$
$\Delta S^o = 1550 - 560 = 990 \ J \ K^{-1}$

(B) Entropy $(S)$ is a measure of the randomness or disorder of a system.
Processes that lead to a more ordered state result in a decrease in entropy $(\Delta S < 0)$.
$(A)$ Raising the temperature of a solid increases the vibrational motion of particles,leading to an increase in entropy.
$(B)$ Crystallisation of a liquid into a solid involves the transition from a disordered liquid state to a highly ordered crystalline solid state,which results in a decrease in entropy.
$(C)$ The decomposition of $2NaHCO_{3(s)}$ produces gaseous products ($CO_2$ and $H_2O$),which significantly increases the disorder of the system.
$(D)$ The dissociation of $H_{2(g)}$ into $2H_{(g)}$ increases the number of particles and the randomness of the system,leading to an increase in entropy.
Therefore,the correct option is $B$.

(A) For a phase transition at equilibrium,the entropy change is given by the formula:
$\Delta S = \frac{\Delta_{vap} H}{T_{BP}}$
Given $\Delta_{vap} H = 39.2 \, kJ \, mol^{-1} = 39.2 \times 10^3 \, J \, mol^{-1}$ and $T = 373 \, K$.
Substituting the values:
$\Delta S = \frac{39.2 \times 10^3}{373} \approx 105.09 \, J \, K^{-1} \, mol^{-1}$.
Rounding to the nearest integer,we get $105 \, J \, K^{-1} \, mol^{-1}$.

(B) The entropy change $\Delta S$ is related to the change in the number of moles of gaseous species,$\Delta n_g$.
If $\Delta n_g < 0$,the entropy of the system decreases,resulting in a negative $\Delta S$.
For the reaction $2SO_{2_{(g)}} + O_{2_{(g)}} \to 2SO_{3_{(g)}}$,the change in gaseous moles is $\Delta n_g = 2 - (2 + 1) = -1$.
Since $\Delta n_g$ is negative,the reaction involves a decrease in disorder,leading to a negative value for $\Delta S$.

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

(B) The entropy of vaporisation $(\Delta S_{vap})$ is calculated using the formula: $\Delta S_{vap} = \frac{\Delta H_{vap}}{T}$.
Given,molar heat of vaporisation $\Delta H_{vap} = 9710\,cal/mol$.
The temperature $T = 100\,^{\circ}C = 100 + 273 = 373\,K$.
Substituting the values: $\Delta S_{vap} = \frac{9710\,cal/mol}{373\,K} \approx 26.03\,cal/K/mol$.
Rounding to the nearest integer,we get $26\,cal/K/mol$.

(D) The process is the freezing of water at its melting point,which is a reversible process at equilibrium.
$\Delta H_{\text{freezing}} = -\Delta H_{\text{fusion}} = -80 \ cal/g \times 10 \ g = -800 \ cal$.
The temperature $T = 0 \ ^\circ C = 273 \ K$.
$\Delta S_{\text{freezing}} = \frac{\Delta H_{\text{freezing}}}{T} = \frac{-800 \ cal}{273 \ K} \approx -2.93 \ cal/K$.
To convert to $J/K$,use the conversion factor $1 \ cal = 4.184 \ J$.
$\Delta S = -2.93 \ cal/K \times 4.184 \ J/cal \approx -12.25 \ J/K$.

(C) The formula for entropy change is $\Delta S = nC_V \ln \left( \frac{T_2}{T_1} \right) + nR \ln \left( \frac{V_2}{V_1} \right)$.
Since the process is isochoric,the volume remains constant,so $\Delta S = nC_V \ln \left( \frac{T_2}{T_1} \right)$.
Given:
$n = 2 \, \text{moles}$
$C_V = \frac{3}{2} R$ (for a monoatomic gas)
$T_1 = 200 + 273 = 473 \, \text{K}$
$T_2 = 300 + 273 = 573 \, \text{K}$
Substituting the values:
$\Delta S = 2 \times \left( \frac{3}{2} R \right) \times \ln \left( \frac{573}{473} \right)$
$\Delta S = 3 R \ln \left( \frac{573}{473} \right)$.

(B) At the melting point,the process of fusion is in equilibrium,so $\Delta G = 0$.
Using the relation $\Delta G = \Delta H - T\Delta S$,we get $T = \frac{\Delta H_{fusion}}{\Delta S_{fusion}}$.
Given $\Delta H_{fusion} = 9.95 \ kJ/mol = 9950 \ J/mol$ and $\Delta S_{fusion} = 35.7 \ J/K \cdot mol$.
$T = \frac{9950 \ J/mol}{35.7 \ J/K \cdot mol} \approx 278.7 \ K$.

(B) . The conversion of diamond to graphite involves a structural change where entropy generally increases,so $\Delta S > 0$.
$B$. When the pressure of a gas increases at constant temperature,the gas molecules are compressed into a smaller volume,leading to a decrease in the number of available microstates and a decrease in randomness. Thus,$\Delta S < 0$.
$C$. Increasing the temperature of a gas increases the kinetic energy and the randomness of the molecules,leading to an increase in entropy,so $\Delta S > 0$.
$D$. The dissociation of $H_2$ gas into $2H$ atoms increases the number of moles of gas particles,which increases the disorder of the system,so $\Delta S > 0$.

(B) The entropy change for the reaction is calculated using the formula:
$\Delta S^o = \sum S^o(\text{products}) - \sum S^o(\text{reactants})$
For the reaction $CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)}$:
$\Delta S^o = [S^o(CO_2) + 2 \times S^o(H_2O)] - [S^o(CH_4) + 2 \times S^o(O_2)]$
Substituting the given values:
$\Delta S^o = [213.6 + 2 \times 69.9] - [186.2 + 2 \times 205.2]$
$\Delta S^o = [213.6 + 139.8] - [186.2 + 410.4]$
$\Delta S^o = 353.4 - 596.6$
$\Delta S^o = -242.8 \ J \ K^{-1} \ mol^{-1}$

(D) The change in entropy is given by the formula $\Delta S = \frac{q_{rev}}{T}$.
Given,$\Delta S = 0.836 \ J \ K^{-1}$ and $q_{rev} = 0.3344 \ J$.
Rearranging the formula to solve for temperature $T$: $T = \frac{q_{rev}}{\Delta S}$.
Substituting the values: $T = \frac{0.3344}{0.836} = 0.4 \ K$.

(B) For an isothermal process,the change in entropy $\Delta S$ is given by the formula: $\Delta S = nR \ ln \ (P_1 / P_2)$.
Given that the gas is expanded to half of its initial pressure,$P_2 = P_1 / 2$,so $P_1 / P_2 = 2$.
For $n = 1 \ mol$,the expression becomes: $\Delta S = 1 \times R \times ln \ (2)$.
Substituting the values: $\Delta S = 8.314 \times 0.693 = 5.76 \ J \ K^{-1} \ mol^{-1}$.

(D) Entropy change $(\Delta S)$ is related to the change in the number of moles of gaseous species. $A$ decrease in the number of gaseous moles leads to a negative entropy change.
$A$: $CaSO_{4(s)} \to CaO_{(s)} + SO_{3(g)}$ (Number of gaseous moles increases from $0$ to $1$,$\Delta S > 0$)
$B$: $CO_{2(s)} \to CO_{2(g)}$ (Solid to gas transition,$\Delta S > 0$)
$C$: $I_{2(s)} \to I_{2(aq)}$ (Solid to aqueous transition,$\Delta S > 0$)
$D$: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ (Number of gaseous moles decreases from $4$ to $2$,$\Delta S < 0$)
Therefore,the synthesis of ammonia results in a decrease in entropy.

(A) When two blocks of the same mass and same heat capacity $C_p$ at temperatures $T_1$ and $T_2$ are brought into contact,the final equilibrium temperature $T_f$ is given by $T_f = \frac{T_1 + T_2}{2}$.
The change in entropy for the first block is $\Delta S_1 = C_p \ln \left( \frac{T_f}{T_1} \right) = C_p \ln \left( \frac{T_1 + T_2}{2T_1} \right)$.
The change in entropy for the second block is $\Delta S_2 = C_p \ln \left( \frac{T_f}{T_2} \right) = C_p \ln \left( \frac{T_1 + T_2}{2T_2} \right)$.
The total entropy change is $\Delta S_{total} = \Delta S_1 + \Delta S_2 = C_p \ln \left( \frac{(T_1 + T_2)^2}{4T_1T_2} \right) = 2C_p \ln \left( \frac{T_1 + T_2}{2\sqrt{T_1T_2}} \right)$.

(C) For the fusion process at the melting point,the relationship between enthalpy of fusion,entropy of fusion,and melting temperature is given by: $\Delta_{fus}H = T_{mp} \times \Delta_{fus}S$.
Given: $\Delta_{fus}H = 600 \ cal \ mol^{-1}$,$\Delta_{fus}S = 2 \ cal \ mol^{-1} \ K^{-1}$,and $T_{mp} = x \ K$.
Substituting the values: $600 = x \times 2$.
Solving for $x$: $x = 600 / 2 = 300 \ K$.

(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 disorder of the system increases.
$A$. $H_2O_{(g)} \longrightarrow H_2O_{(l)}$: Gas to liquid transition decreases disorder,so $\Delta S < 0$.
$B$. $H_2O_{(l)} \longrightarrow H_2O_{(s)}$: Liquid to solid transition decreases disorder,so $\Delta S < 0$.
$C$. $N_2(g) + 3H_2(g) \longrightarrow 2NH_3(g)$: The number of moles of gas decreases from $4$ to $2$,which decreases disorder,so $\Delta S < 0$.
$D$. $2O_3(g) \longrightarrow 3O_2(g)$: The number of moles of gas increases from $2$ to $3$. An increase in the number of gaseous molecules leads to an increase in disorder,so $\Delta S > 0$.

(A) According to the $2^{nd}$ law of thermodynamics,for any spontaneous or irreversible process,the entropy change of the system $(dS)$ is always greater than the heat exchange divided by temperature $(\frac{dq}{T})$.
This is expressed by the Clausius inequality: $dS > \frac{dq_{irr}}{T}$.
For a reversible process,$dS = \frac{dq_{rev}}{T}$.

(A) When an egg is boiled,the proteins inside the egg undergo denaturation.
This process involves the unfolding of the highly ordered,compact protein structure into a more disordered,random coil structure.
Since entropy is a measure of the randomness or disorder of a system,the transition from a highly ordered state to a more disordered state results in an increase in entropy $( \Delta S > 0 )$.

(B) According to the $2^{nd}$ Law of Thermodynamics,for any spontaneous process in an isolated system,the total entropy change $(\Delta S_{total})$ must be greater than zero,i.e.,$\Delta S_{total} > 0$.
For an irreversible spontaneous process,the entropy of the universe (system surroundings) increases.
Specifically,for an isolated system,$\Delta S_{system} > 0$ for an irreversible spontaneous process.
Option $B$ correctly states that for an irreversible spontaneous process,the entropy of an isolated system increases.

(C) During the melting of ice $(H_2O(s))$ to water $(H_2O(l))$,the solid structure of ice breaks down into a more disordered liquid state.
Entropy is a measure of the randomness or disorder of a system.
Since the liquid state has more disorder than the solid state,the entropy of the system increases during the phase transition from solid to liquid.
Therefore,$\Delta S > 0$.

(C) The change in entropy $\left( \Delta S \right)$ for an ideal gas at constant pressure is given by the formula: $\Delta S = n C_{p,m} \ln \left( \frac{T_2}{T_1} \right)$.
Given values are: $n = 2 \ mol$,$C_{p,m} = \frac{5}{2} R$,$T_1 = 300 \ K$,and $T_2 = 600 \ K$.
Substituting these values into the formula:
$\Delta S = 2 \times \left( \frac{5}{2} R \right) \ln \left( \frac{600 \ K}{300 \ K} \right)$.
$\Delta S = 5 R \ln (2)$.
Therefore,the correct option is $C$.

(A) The process is the fusion of ice: $H_2O_{(s)} \rightarrow H_2O_{(l)}$ at $T = 0 \ ^\circ C = 273 \ K$.
The molar entropy change is $\Delta S_{fus} = S_{product} - S_{reactant} = 60 - 38.2 = 21.8 \ J/mol \ K$.
For a reversible phase transition at equilibrium,$\Delta S = \frac{\Delta H}{T}$.
Therefore,$\Delta H = \Delta S \times T = 21.8 \ J/mol \ K \times 273 \ K = 5951.4 \ J/mol$.

(C) The reaction is: $4 Fe_3O_{4(s)} + O_{2(g)} \to 6 Fe_2O_{3(s)}$
Given $\Delta_rS^o = -266 \ J \ K^{-1}$
$S_m^o(O_2) = 205 \ J \ K^{-1} \ mol^{-1}$
$S_m^o(Fe_2O_3) = 87 \ J \ K^{-1} \ mol^{-1}$
Using the formula: $\Delta_rS^o = \sum \nu_p S_m^o(products) - \sum \nu_r S_m^o(reactants)$
$-266 = [6 \times S_m^o(Fe_2O_3)] - [4 \times S_m^o(Fe_3O_4) + 1 \times S_m^o(O_2)]$
$-266 = (6 \times 87) - (4 \times S_m^o(Fe_3O_4) + 205)$
$-266 = 522 - 4 \times S_m^o(Fe_3O_4) - 205$
$4 \times S_m^o(Fe_3O_4) = 522 - 205 + 266$
$4 \times S_m^o(Fe_3O_4) = 583$
$S_m^o(Fe_3O_4) = \frac{583}{4} = 145.75 \ J \ K^{-1} \ mol^{-1}$

(A) For any reversible adiabatic process,the change in entropy $(\Delta S)$ is defined by the relation $\Delta S = \int \frac{dq_{rev}}{T}$.
Since the process is adiabatic,there is no heat exchange with the surroundings,meaning $dq_{rev} = 0$.
Therefore,$\Delta S = \frac{0}{T} = 0\, cal/K$.

(A) spontaneous process is one that occurs naturally without external intervention. According to the second law of thermodynamics,for any spontaneous process in an isolated system,the total entropy change must be positive $(\Delta S_{total} > 0)$. Since real-world spontaneous processes involve an increase in entropy,they are thermodynamically irreversible. Therefore,all spontaneous processes are thermodynamically irreversible.

(B) For a reaction at equilibrium or standard conditions where $\Delta G = 0$ is assumed for the phase/chemical change context provided in typical textbook problems of this type,we use $\Delta S = \frac{\Delta H}{T}$.
Given $\Delta H = -31.3\,kcal = -31300\,cal$ and $T = 25 + 273 = 298\,K$.
Therefore,$\Delta S = \frac{-31300}{298}\,cal\,K^{-1}$.

(C) The thermodynamic definition of entropy change at constant temperature is given by $\Delta S = \frac{q_{rev}}{T}$.
In terms of energy,the total energy of a system can be divided into available energy (Gibbs free energy,$G$) and unavailable energy $(T \cdot \Delta S)$.
Therefore,the unavailable energy is equal to $T \cdot \Delta S$,which implies that entropy is the measure of the unavailable energy per unit temperature,i.e.,$\Delta S = \frac{\text{Unavailable Energy}}{T}$.
Thus,option $(c)$ is the correct statement.