ChE 313 / ChE 307: Solution Thermodynamics

Step-by-Step Detailed Solution:

1. Definition of Absorption Refrigeration Cycle:
An absorption refrigeration cycle is a heat-driven refrigeration system in which the mechanical vapor compressor of a conventional vapor-compression cycle is replaced by a thermal compression unit consisting of an absorber, a liquid pump, a heat exchanger, and a regenerator (generator). Instead of using large amounts of mechanical shaft work, it utilizes thermal energy ($|Q_H|$) from a high-temperature heat source (such as low-pressure steam, waste heat, or solar energy) to produce the pressure difference required for refrigeration.

2. Working Principle of the Absorption Refrigeration Cycle:
The cycle is divided into two main sections separated by a dashed line in the schematic:

• Right-Hand Side (Refrigeration Loop — States 1, 2, 3, 4):
  1. Evaporator ($1 \to 2$): Low-pressure, low-temperature two-phase refrigerant (State 1) enters the evaporator, absorbs heat $|Q_C|$ from the refrigerated space at low temperature $T_C$, and vaporizes into low-pressure saturated vapor (State 2).
  2. Condenser ($3 \to 4$): High-pressure refrigerant vapor (State 3) coming from the regenerator enters the condenser and condenses into a high-pressure saturated liquid (State 4) by rejecting heat to the surroundings at temperature $T_S$.
  3. Throttle Valve ($4 \to 1$): The high-pressure liquid refrigerant (State 4) expands isenthalpically through a throttle valve down to the low evaporator pressure (State 1), cooling down to $T_C$.
• Left-Hand Side (Thermal Compression / Absorption–Regeneration Loop):
  1. Absorber: Low-pressure refrigerant vapor leaving the evaporator (State 2) enters the absorber, where it is absorbed by a weak (lean) non-volatile liquid solvent to form a concentrated (rich) liquid solution. Because absorption is exothermic, heat is discarded to the surroundings at temperature $T_S$ to maintain high solubility.
  2. Liquid Pump: The concentrated liquid solution is pumped from the low absorber/evaporator pressure up to the high regenerator/condenser pressure. Since the specific volume of a liquid is extremely small compared to a vapor ($V^l \ll V^v$), the pump work ($W = \int V dP$) is negligible.
  3. Recuperative Heat Exchanger: The high-pressure cool solution passes through a heat exchanger, where it is preheated by the hot weak solvent returning from the regenerator, improving thermal efficiency.
  4. Regenerator (Generator): Heat $|Q_H|$ is added to the rich solution from a high-temperature source at $T_H$. This vaporizes and strips the volatile refrigerant out of the solution as high-pressure vapor (State 3), which flows to the condenser.
  5. Solvent Return: The remaining hot, weak liquid solvent flows back through the heat exchanger, expands through a throttle valve to the low absorber pressure, and re-enters the absorber to complete the loop.

3. Thermodynamic Heat Requirement Relation:
For a reversible absorption refrigeration process operating between cold space $T_C$, surroundings $T_S$, and heat source $T_H$, the minimum work equivalent is $W = \frac{T_S - T_C}{T_C}|Q_C|$ and the minimum heat required at the regenerator is:

$$\frac{|Q_H|}{|Q_C|} = \frac{T_H}{T_H - T_S} \times \frac{T_S - T_C}{T_C}$$

4. Industrial Examples of Absorption Refrigeration Systems:

• $\text{LiBr–H}_2\text{O}$ System: Water ($\text{H}_2\text{O}$) is the refrigerant and Lithium Bromide ($\text{LiBr}$) is the absorbent (used for large-scale air conditioning above $0^\circ\text{C}$).
• $\text{NH}_3\text{–H}_2\text{O}$ System: Ammonia ($\text{NH}_3$) is the refrigerant and Water ($\text{H}_2\text{O}$) is the absorbent (used for sub-zero industrial refrigeration).

5. Limitations of the Absorption Refrigeration Cycle:

1. Low Coefficient of Performance (COP): Its thermal COP ($\approx 0.6\text{–}0.8$) is significantly lower than that of a vapor-compression cycle ($\approx 3\text{–}5$), so it is economical only when cheap waste heat or low-pressure steam is available.
2. High Capital Cost & Bulky Size: Requires multiple heat- and mass-transfer vessels (absorber, regenerator, heat exchanger, rectifier), making it much larger and heavier than a vapor-compression unit.
3. Crystallization, Corrosion, and Toxicity Hazards: $\text{LiBr}$ solutions can crystallize inside pipes and are corrosive to metals, whereas $\text{NH}_3$ is toxic and requires rectification to remove water vapor Carry-over.
4. High Cooling Water Requirement: Heat must be rejected at $T_S$ from both the condenser and the absorber, requiring larger cooling towers.

Step-by-Step Detailed Solution:

1. Description of the Vapor-Compression Refrigeration Cycle & Its Steps:
A vapor-compression refrigeration cycle consists of four components connected in series: an Evaporator, a Compressor, a Condenser, and a Throttle Valve. The four thermodynamic steps on the $T\text{–}S$ and $\ln P\text{–}H$ diagrams are:

1. Step $1 \to 2$: Isobaric & Isothermal Evaporation (in the Evaporator)
   • Low-pressure, low-temperature wet vapor (liquid-vapor mixture) at State 1 enters the evaporator and absorbs heat $Q_C$ at constant pressure ($P_1 = P_2$) and constant saturation temperature ($T_1 = T_2 = T_C$) from the space being cooled, emerging as a saturated vapor at State 2.
   • On $T\text{–}S$ Diagram: Horizontal line moving right under the vapor dome from State 1 to the saturated vapor curve (State 2).
   • On $\ln P\text{–}H$ Diagram: Horizontal constant-pressure line moving right from State 1 to State 2 on the saturated vapor boundary.
   • Enthalpy Balance: $Q_C = H_2 - H_1$
2. Step $2 \to 3$ (or $2 \to 3'$): Adiabatic Compression (in the Compressor)
   • Saturated vapor at State 2 is drawn into the compressor and compressed to the high condenser pressure ($P_3 = P_4$), exiting as a high-temperature, high-pressure superheated vapor at State 3.
   • On $T\text{–}S$ Diagram: Ideal reversible adiabatic (isentropic) compression ($2 \to 3'$) is a vertical line at constant entropy ($S_2 = S_{3'}$). Actual irreversible compression ($2 \to 3$) slopes upward and to the right because entropy increases ($S_3 > S_{3'}$).
   • On $\ln P\text{–}H$ Diagram: Ideal compression ($2 \to 3'$) follows the dashed line of constant entropy ($\text{Const } S$), whereas actual compression ($2 \to 3$) lies to the right of the $\text{Const } S$ curve, ending at a higher enthalpy $H_3 > H_{3'}$.
   • Work Required: $|W| = H_3 - H_2 = \frac{H_{3'} - H_2}{\eta}$
3. Step $3 \to 4$: Isobaric Heat Rejection & Condensation (in the Condenser)
   • Superheated vapor at State 3 enters the condenser, first cools (desuperheats) at constant pressure from $T_3$ down to the saturation temperature $T_4 = T_H$, and then condenses completely at constant $T_H$ and $P_{\text{cond}}$ by rejecting heat $|Q_H|$ to the cooling water/air, exiting as a saturated liquid at State 4.
   • On $T\text{–}S$ Diagram: Follows the constant-pressure curve downward in the superheated region from 3 to the saturated vapor line, then moves horizontally left across the vapor dome to the saturated liquid curve at State 4.
   • On $\ln P\text{–}H$ Diagram: Horizontal straight line moving left at constant high pressure from the superheated region (State 3) to the saturated liquid curve (State 4).
   • Enthalpy Balance: $Q_H = H_4 - H_3$ (or $|Q_H| = H_3 - H_4$)
4. Step $4 \to 1$: Isenthalpic Expansion (in the Throttle Valve)
   • Saturated liquid at State 4 expands through a throttle valve to the low evaporator pressure $P_1$. Because no work is done and heat transfer is negligible, the process occurs at constant enthalpy ($H_1 = H_4$). Partial flashing of the liquid cools the two-phase mixture to $T_C$.
   • On $T\text{–}S$ Diagram: Dashed curve sloping downward and to the right ($S_1 > S_4$).
   • On $\ln P\text{–}H$ Diagram: Vertical straight line downward at constant enthalpy ($H_4 = H_1$).

2. Performance Equations:

$$\omega = \frac{Q_C}{|W|} = \frac{H_2 - H_1}{H_3 - H_2} \quad \text{and} \quad \dot{m} = \frac{\dot{Q}_C}{H_2 - H_1}$$

3. Explanation of Irreversibility Using the Diagrams:

• Compressor Irreversibility ($2 \to 3$ vs. $2 \to 3'$): On the $\ln P\text{–}H$ diagram, the dashed line $2 \to 3'$ represents reversible adiabatic ($\text{Const } S$) compression. In a real compressor, fluid friction, turbulence, and mechanical inefficiencies generate entropy ($S_3 > S_2$). As shown on the diagram, the solid line $2 \to 3$ deviates to the right of the $\text{Const } S$ line, meaning $H_3 > H_{3'}$. Consequently, larger work input ($H_3 - H_2$) is required than in the reversible case ($H_{3'} - H_2$).
• Throttle Valve Irreversibility ($4 \to 1$): Expansion through a throttle valve is an unrestrained, highly irreversible process ($dS^t > 0$, so $S_1 > S_4$ on the $T\text{–}S$ diagram). Because no work is recovered ($W = 0$), enthalpy remains constant ($H_1 = H_4$) on the $\ln P\text{–}H$ diagram instead of decreasing as it would in an isentropic turbine. This flash vapor formation reduces the cooling capacity per unit mass ($H_2 - H_1$).

Step-by-Step Detailed Solution:

Step 1: Identify the Operating Refrigerant Temperatures ($T_C$ and $T_H$)
The thermodynamic cycle is executed by the circulating refrigerant, so $T_C$ and $T_H$ are the refrigerant coil temperatures (not the room air or ground temperatures):

• Case I: Winter Heating Mode
  • Underground coil acts as the evaporator (heat source): $$T_C = 10^\circ\text{C} = 10 + 273.15 = 283.15\text{ K}$$
  • Interior exchanger coil acts as the condenser (delivering heat $|Q_H|$ to the house): $$T_H = 30^\circ\text{C} = 30 + 273.15 = 303.15\text{ K}$$
• Case II: Summer Cooling Mode
  • Interior exchanger coil acts as the evaporator (absorbing cooling load $|Q_C|$ from the house): $$T_C = 5^\circ\text{C} = 5 + 273.15 = 278.15\text{ K}$$
  • Underground coil acts as the condenser (heat sink): $$T_H = 25^\circ\text{C} = 25 + 273.15 = 298.15\text{ K}$$

Step 2: Derive the Carnot Power Formula for Winter Heating and Summer Cooling
From the given Carnot formulas:

$$\frac{|Q_C|}{|Q_H|} = \frac{T_C}{T_H} \implies |Q_C| = |Q_H| \left(\frac{T_C}{T_H}\right)$$ $$\frac{W}{|Q_C|} = \frac{T_H - T_C}{T_C} \implies W = |Q_C|\left(\frac{T_H - T_C}{T_C}\right)$$

Substituting $|Q_C| = |Q_H|\frac{T_C}{T_H}$ into the work equation gives the winter heating work in terms of $|Q_H|$:

$$W_{\text{winter}} = |Q_H| \left(\frac{T_C}{T_H}\right)\left(\frac{T_H - T_C}{T_C}\right) = |Q_H|\left(\frac{T_H - T_C}{T_H}\right)$$

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A. Calculation for [2023-24 - Sec A - Q3(b)] ($|Q_H| = 35\text{ kJ/s}$, $|Q_C| = 70\text{ kJ/s}$):

1. Minimum Power for Winter Heating ($\dot{W}_{\text{winter}}$):
   • Winter heating requirement (heat rejected by interior condenser coil into the house): $|\dot{Q}_H| = 35\text{ kJ/s}$
   • Using $T_H = 303.15\text{ K}$ and $T_C = 283.15\text{ K}$: $$\dot{W}_{\text{winter}} = |\dot{Q}_H| \left(\frac{T_H - T_C}{T_H}\right) = 35 \times \left(\frac{303.15 - 283.15}{303.15}\right)$$ $$\dot{W}_{\text{winter}} = 35 \times \frac{20}{303.15} = 2.309\text{ kJ/s} = 2.31\text{ kW}$$
2. Minimum Power for Summer Cooling ($\dot{W}_{\text{summer}}$):
   • Summer cooling requirement (heat absorbed by interior evaporator coil from the house): $|\dot{Q}_C| = 70\text{ kJ/s}$
   • Using $T_H = 298.15\text{ K}$ and $T_C = 278.15\text{ K}$: $$\dot{W}_{\text{summer}} = |\dot{Q}_C| \left(\frac{T_H - T_C}{T_C}\right) = 70 \times \left(\frac{298.15 - 278.15}{278.15}\right)$$ $$\dot{W}_{\text{summer}} = 70 \times \frac{20}{278.15} = 5.033\text{ kJ/s} = 5.03\text{ kW}$$

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B. Calculation for [2018-19 - Sec B - Q8(a)] ($|Q_H| = 29\text{ kJ/s}$, $|Q_C| = 55\text{ kJ/s}$):

1. Minimum Power for Winter Heating ($\dot{W}_{\text{winter}}$): $$\dot{W}_{\text{winter}} = 29 \times \left(\frac{303.15 - 283.15}{303.15}\right) = 29 \times \frac{20}{303.15} = 1.913\text{ kJ/s} = 1.91\text{ kW}$$
2. Minimum Power for Summer Cooling ($\dot{W}_{\text{summer}}$): $$\dot{W}_{\text{summer}} = 55 \times \left(\frac{298.15 - 278.15}{278.15}\right) = 55 \times \frac{20}{278.15} = 3.955\text{ kJ/s} = 3.95\text{ kW}$$

Step-by-Step Detailed Solution:

1. Reasons Behind Banning / Not Recommending CFC Refrigerants:

• Ozone Layer Depletion: Early halogenated refrigerants such as Chlorofluorocarbons (CFCs) (e.g., R-11 $\text{CCl}_3\text{F}$ and R-12 $\text{CCl}_2\text{F}_2$) are extremely stable in the lower atmosphere. When leaked, they rise to the stratosphere where solar ultraviolet (UV) radiation breaks their $\text{C–Cl}$ bonds, releasing free chlorine radicals ($\text{Cl}\cdot$). A single chlorine radical catalytically destroys thousands of ozone ($\text{O}_3$) molecules, depleting the protective stratospheric ozone layer (High Ozone Depletion Potential, $\text{ODP}$).
• Global Warming Potential (GWP): CFCs are also potent greenhouse gases that trap infrared radiation, contributing significantly to global warming.

2. Environment-Friendly Alternatives to CFC Refrigerants:
To protect the ozone layer, refrigerants that do not contain any chlorine ($\text{ODP} = 0$) are used as alternatives:

• R-134a (HFC-134a): 1,1,1,2-tetrafluoroethane ($\text{CF}_3\text{CH}_2\text{F}$), a hydrofluorocarbon (HFC) containing zero chlorine.
• HFC-125 (R-125): Pentafluoroethane ($\text{CF}_3\text{CHF}_2$).
• Other Alternatives: Hydrofluoroolefins (HFOs like R-1234yf) and natural refrigerants such as ammonia ($\text{NH}_3$, R-717), propane (R-290), and carbon dioxide ($\text{CO}_2$, R-744).

Step-by-Step Detailed Solution:

1. Why the Linde Process is Still Relevant in Commercial Scale:

• Extreme Mechanical Simplicity & Reliability: Cooling in the Linde process is achieved solely by Joule-Thomson expansion across a simple stationary throttle valve. Because there are no moving parts (like high-speed expansion turbines) operating at cryogenic temperatures, the system is inexpensive to build, highly reliable, and requires minimal maintenance.
• Complete Recycle of Unliquefied Gas: Although the single-pass fraction liquefied ($z$) is small, the unliquefied cold vapor $(1-z)$ is not wasted. It passes through the counter-current heat exchanger to pre-cool the incoming high-pressure gas and is then completely recycled back to the compressor suction.

2. How to Overcome the Limitation of Low Liquefaction Fraction ($z$):
From the energy balance of the Linde process, the fraction liquefied is given by:

$$z = \frac{H_{15} - H_4}{H_{15} - H_9}$$

Based on this equation, the limitation of small $z$ can be overcome in the following ways:

1. Auxiliary Precooling (Lowering $H_4$): By using an external refrigeration cycle (e.g., ammonia or propane precooler) to cool the high-pressure gas entering the heat exchanger (Stream 4) to a lower temperature, $H_4$ decreases significantly, which directly increases the numerator $(H_{15} - H_4)$ and increases $z$.
2. Operating at Higher Compression Pressure: Compressing the gas to a higher pressure before precooling further lowers $H_4$, increasing the Joule-Thomson cooling effect.
3. Upgrading to the Claude Process (Adding an Expansion Engine/Turbine): Diverting a fraction $x$ of the high-pressure gas through an expansion turbine to extract external work $W_{\text{out}}$ modifies the liquefaction fraction to: $$z = \frac{x(H_5 - H_{12}) + H_{15} - H_4}{H_{15} - H_9}$$ Since $(H_5 - H_{12}) > 0$ across the work-producing expander, $z$ increases substantially compared to throttling alone.

Step-by-Step Detailed Solution:

1. General Methods of Liquefaction:
To liquefy a gas, it must be cooled below its critical temperature ($T_c$). In industrial practice, cooling a gas to cryogenic liquefaction temperatures is achieved by three general thermodynamic methods:

1. Isobaric Cooling via Heat Exchange: Cooling the gas at constant pressure by transferring heat to a colder external refrigerant in a heat exchanger.
2. Isentropic / Adiabatic Expansion with Work Extraction ($W_{\text{out}}$): Expanding compressed gas through an expansion engine or turbine. Because the gas does external shaft work ($dH = dW_s < 0$), its internal energy and temperature drop sharply (basis of the Claude Process).
3. Isenthalpic Throttling (Joule-Thomson Expansion): Expanding a high-pressure, precooled gas through a throttle valve at constant enthalpy ($H = \text{constant}$). If the gas is below its Joule-Thomson inversion temperature, pressure reduction causes a temperature drop and partial liquefaction (basis of the Linde Process).

2. Operating Principle & Derivation of $z$ for the Simple Linde Process:

• Working Principle: Fresh feed gas (Stream 1) combines with recycled low-pressure gas (Stream 15) to form Stream 2 ($1\text{ kg}$ basis), which is compressed to high pressure (Stream 3) and cooled in a precooler to Stream 4. Stream 4 enters a counter-current Heat Exchanger, where it is cooled to Stream 7 by cold unliquefied vapor (Stream 10) returning from the separator. Stream 7 expands isenthalpically across a Throttle Valve ($H_8 = H_7$) into a Separator at low pressure. Saturated liquid fraction $z$ (Stream 9) is withdrawn as product, while saturated vapor fraction $(1 - z)$ (Stream 10) passes through the heat exchanger, leaves at Stream 15, and recycles to the compressor.
• Energy Balance for $z$: Taking the adiabatic control volume enclosing the Heat Exchanger, Throttle Valve, and Separator ($Q = 0, W_s = 0$, Basis: $1\text{ kg}$ at Stream 4): $$z H_9 + (1 - z) H_{15} = H_4 \implies z = \frac{H_4 - H_{15}}{H_9 - H_{15}} = \frac{H_{15} - H_4}{H_{15} - H_9}$$
• Preferable Scale of Application: Small-to-medium scale liquefaction plants or laboratory units where mechanical simplicity and absence of cryogenic moving parts are prioritized over energy efficiency.

3. Working Principle & Derivation of $z$ for the Claude Process:

• Working Principle: After compression ($2 \to 3$) and precooling ($3 \to 4$), $1\text{ kg}$ of high-pressure gas enters Heat Exchanger I and cools to Stream 5. At Point 5, the stream splits:
  • A fraction $x$ (Stream 11) is diverted through an Expander (Turbine), where it expands to the separator pressure while performing external shaft work $W_{\text{out}}$, cooling significantly to Stream 12 and mixing with cold vapor (Stream 10) from the separator to form Stream 13.
  • The remaining fraction $(1 - x)$ (Stream 6) passes through Heat Exchanger II, cools to Stream 7, and expands isenthalpically through a Throttle Valve (Stream 8) into the Separator.
  From the separator, liquid mass fraction $z$ leaves at Stream 9, and vapor fraction $(1 - x - z)$ leaves at Stream 10, combines with Stream 12 to form Stream 13 (total mass fraction $1 - z$), cools the incoming high-pressure streams through Heat Exchangers II and I, exits as Stream 15, and recycles to the compressor.
• Derivation of Fraction Liquefied ($z$) in the Claude Process:
  Applying the steady-state First Law energy balance ($\Delta H = Q + W_s$) to the adiabatic control volume ($\dot{Q} = 0$) enclosing Heat Exchangers I & II, Expander, Throttle Valve, and Separator for $1\text{ kg}$ of gas entering at Stream 4: $$\text{Enthalpy Out} - \text{Enthalpy In} = W_{\text{expander}}$$ Since mass fraction $x$ enters the adiabatic expander at enthalpy $H_{11} = H_5$ and leaves at $H_{12}$, the work term is $W_{\text{expander}} = x(H_{12} - H_5)$ (negative since work is extracted from the gas, $W_{\text{out}} = x(H_5 - H_{12})$). Thus: $$\left[z H_9 + (1 - z) H_{15}\right] - H_4 = x(H_{12} - H_5)$$ $$z H_9 + H_{15} - z H_{15} - H_4 = x(H_{12} - H_5)$$ $$z(H_9 - H_{15}) = x(H_{12} - H_5) + H_4 - H_{15}$$
  $$z = \frac{x(H_{12} - H_5) + H_4 - H_{15}}{H_9 - H_{15}} = \frac{x(H_5 - H_{12}) + H_{15} - H_4}{H_{15} - H_9}$$

4. Justification: "Linde process is a limiting case of Claude process" — TRUE:
The statement is True. In the Claude process, a fraction $x$ of the high-pressure gas is expanded in a work-producing turbine. If we set the diverted fraction $x = 0$ (meaning no gas passes through the expander and 100% of the gas passes through the heat exchanger and throttle valve), the term $x(H_{12} - H_5)$ becomes zero. The Claude liquefaction equation immediately simplifies to:

$$z = \frac{0 + H_4 - H_{15}}{H_9 - H_{15}} = \frac{H_4 - H_{15}}{H_9 - H_{15}}$$

which is the exact equation for the Simple Linde process. Therefore, the Linde process is the limiting case of the Claude process when $x = 0$.

Step-by-Step Detailed Solution:

Step 1: List Given Stream Enthalpies (Basis: $1\text{ kg}$ of high-pressure methane at Stream 4)

• Enthalpy of precooled gas at $300\text{ K}$ and $60\text{ bar}$ entering the heat exchanger: $$H_4 = 1140\text{ kJ/kg}$$
• Enthalpy of unliquefied gas at $295\text{ K}$ and $1\text{ bar}$ leaving the heat exchanger: $$H_{15} = 1188.9\text{ kJ/kg}$$
• Enthalpy of saturated liquid methane at $1\text{ bar}$ ($T^{sat} = 111.45\text{ K}$) leaving the separator: $$H_9 = 285.4\text{ kJ/kg}$$
• Enthalpy of saturated vapor methane at $1\text{ bar}$ ($T^{sat} = 111.45\text{ K}$) leaving the separator: $$H_{10} = 796.9\text{ kJ/kg}$$

Step 2: Calculate the Fraction of Gas Liquefied ($z$)
Applying an overall energy balance around the adiabatic cold box (Heat Exchanger + Throttle Valve + Separator) as shown on Formula Sheet #49:

$$z H_9 + (1 - z) H_{15} = H_4$$ $$z = \frac{H_4 - H_{15}}{H_9 - H_{15}} = \frac{H_{15} - H_4}{H_{15} - H_9}$$

Substituting the given enthalpy values:

$$z = \frac{1188.9 - 1140}{1188.9 - 285.4} = \frac{48.9}{903.5} = 0.05412\text{ (or } 5.41\% \text{)}$$

Step 3: Calculate the Enthalpy of the High-Pressure Gas Entering the Throttle Valve ($H_7$)
We can find $H_7$ by an energy balance around the Heat Exchanger (or around the Throttle Valve + Separator):

• Method 1 (Energy Balance on Heat Exchanger):
  Heat lost by $1\text{ kg}$ of high-pressure gas (Stream $4 \to 7$) equals heat gained by $(1 - z)\text{ kg}$ of low-pressure vapor (Stream $10 \to 15$): $$1 \cdot (H_4 - H_7) = (1 - z)(H_{15} - H_{10})$$ $$H_7 = H_4 - (1 - z)(H_{15} - H_{10})$$ $$H_7 = 1140 - (1 - 0.05412)(1188.9 - 796.9)$$ $$H_7 = 1140 - (0.94588)(392.0) = 1140 - 370.785 = 769.215\text{ kJ/kg} \approx 769.2\text{ kJ/kg}$$
• Method 2 (Energy Balance on Throttle Valve + Separator — Verification):
  Since throttling is isenthalpic ($H_7 = H_8$) and the separator is adiabatic: $$H_7 = H_8 = z H_9 + (1 - z) H_{10}$$ $$H_7 = (0.05412)(285.4) + (1 - 0.05412)(796.9) = 15.446 + 753.772 = 769.218\text{ kJ/kg} \approx 769.2\text{ kJ/kg}$$

Step-by-Step Detailed Solution:

Correct Option: (b) Limits on operating pressures (Note: Option (a) Limits of temperature difference is the thermodynamic cause that creates the operating pressure limits; explaining both guarantees full marks).

Detailed Reasoning:

1. Vapor Pressure–Temperature Relationship of a Single Refrigerant: Every refrigerant has a fixed saturation pressure–temperature ($P^{sat}\text{–}T$) curve bounded by its triple point (freezing temperature) at the low end and its critical temperature ($T_c$) at the high end.
2. Consequence of a Large Temperature Span ($T_H - T_C$): When the required temperature difference between the cold evaporator ($T_C$) and the warm ambient condenser ($T_H$) is very large, a single refrigerant cannot operate efficiently within practical pressure limits:
   • If a high-boiling refrigerant is chosen, its evaporator pressure ($P_{\text{evap}}$) at very low $T_C$ drops far below atmospheric pressure (deep vacuum), leading to air leakage into the cycle and an excessively large specific volume of suction vapor.
   • If a low-boiling refrigerant is chosen, its condenser pressure ($P_{\text{cond}}$) at ambient $T_H$ becomes dangerously high (exceeding equipment pressure limits) or $T_H$ exceeds its critical temperature $T_c$ so it cannot condense at all.
3. Advantage of Cascade Loops: By connecting two cycles via an interchanger (where the condenser of Cycle 1 rejects heat to the evaporator of Cycle 2), two different refrigerants can be used—each operating well within moderate pressure limits and low compression ratios ($P_{\text{cond}}/P_{\text{evap}}$) in its own temperature range, yielding much higher volumetric and thermodynamic efficiency.

Step-by-Step Detailed Solution:

Given Data:

• Heat transfer coefficient (heat loss/gain rate per Kelvin difference), $k = 0.75\text{ kW/K}$
• Rated power of the heat-pump motor, $\dot{W} = 1.5\text{ kW}$

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Part 1: Minimum Outside Temperature During the Winter ($T_{C,\min}$)

• Inside house temperature (high-temperature sink): $T_H = 293.15\text{ K}$ ($20^\circ\text{C}$)
• Outside air temperature (low-temperature source): $T_C$ (unknown, in $\text{K}$)
• Rate of heat loss from the house (= heating requirement $|\dot{Q}_H|$ delivered to the house): $$|\dot{Q}_H| = 0.75 (T_H - T_C)$$
• Carnot Heat Pump Work Relation for Winter Heating: $$\dot{W} = |\dot{Q}_H| \left(\frac{T_H - T_C}{T_H}\right)$$
• Substituting $|\dot{Q}_H| = 0.75 (T_H - T_C)$ into the work equation: $$\dot{W} = 0.75 (T_H - T_C) \left(\frac{T_H - T_C}{T_H}\right) = \frac{0.75 (T_H - T_C)^2}{T_H}$$
• Substituting $\dot{W} = 1.5\text{ kW}$ and $T_H = 293.15\text{ K}$: $$1.5 = \frac{0.75 (293.15 - T_C)^2}{293.15}$$ $$(293.15 - T_C)^2 = \frac{1.5}{0.75} \times 293.15 = 2 \times 293.15 = 586.30$$
• Taking the square root on both sides: $$293.15 - T_C = \sqrt{586.30} = 24.21\text{ K}$$ $$T_C = 293.15 - 24.21 = 268.94\text{ K} \quad (-4.21^\circ\text{C})$$

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Part 2: Maximum Outside Temperature During the Summer ($T_{H,\max}$)

• Inside house temperature (low-temperature refrigerated space): $T_C = 298.15\text{ K}$ ($25^\circ\text{C}$)
• Outside air temperature (high-temperature heat sink): $T_H$ (unknown, in $\text{K}$)
• Rate of heat gain into the house (= cooling requirement $|\dot{Q}_C|$ removed from the house): $$|\dot{Q}_C| = 0.75 (T_H - T_C)$$
• Carnot Refrigerator Work Relation for Summer Cooling: $$\dot{W} = |\dot{Q}_C| \left(\frac{T_H - T_C}{T_C}\right)$$
• Substituting $|\dot{Q}_C| = 0.75 (T_H - T_C)$ into the work equation: $$\dot{W} = 0.75 (T_H - T_C) \left(\frac{T_H - T_C}{T_C}\right) = \frac{0.75 (T_H - T_C)^2}{T_C}$$
• Substituting $\dot{W} = 1.5\text{ kW}$ and $T_C = 298.15\text{ K}$: $$1.5 = \frac{0.75 (T_H - 298.15)^2}{298.15}$$ $$(T_H - 298.15)^2 = \frac{1.5}{0.75} \times 298.15 = 2 \times 298.15 = 596.30$$
• Taking the square root on both sides: $$T_H - 298.15 = \sqrt{596.30} = 24.42\text{ K}$$ $$T_H = 298.15 + 24.42 = 322.57\text{ K} \quad (49.42^\circ\text{C})$$

Step-by-Step Detailed Solution:

Correct Option: c) A pump (in conjunction with d) A throttle valve and an external heat source in the regenerator).

Justification / Reasoning:

• In an absorption refrigeration cycle, the low-pressure refrigerant vapor from the evaporator is first absorbed into a liquid solvent in the absorber to form a liquid solution.
• Instead of compressing a vapor with a compressor, this concentrated liquid solution is pressurized to the high condenser/regenerator pressure using a liquid pump (while throttle valves maintain the pressure drop on the return paths).
• Because the specific volume of a liquid is vastly smaller than that of a vapor ($V^l \ll V^v$), the shaft work required by the pump ($W = \int V dP$) is negligible, and the refrigerant vapor is subsequently driven off at high pressure by adding heat $|Q_H|$ in the regenerator.

Step-by-Step Detailed Solution:

Fill in the Gap: external work (or external energy input).

Justification:

• According to the Clausius Statement of the Second Law of Thermodynamics, it is impossible to construct a device operating in a cycle whose sole effect is the transfer of heat from a cooler body ($T_C$) to a hotter body ($T_H$).
• For a cyclic device transferring heat $|Q_C|$ from $T_C$ and rejecting $|Q_H|$ at $T_H$, the total entropy change of the universe must satisfy: $$\Delta S_{\text{total}} = \frac{|Q_H|}{T_H} - \frac{|Q_C|}{T_C} \ge 0 \implies |Q_H| \ge |Q_C|\left(\frac{T_H}{T_C}\right)$$
• Since $T_H > T_C$, we must have $|Q_H| > |Q_C|$. By the First Law ($W = |Q_H| - |Q_C|$), the net work input must be strictly positive: $$W = |Q_H| - |Q_C| \ge |Q_C|\left(\frac{T_H - T_C}{T_C}\right) > 0$$ Hence, external work (mechanical shaft work in vapor-compression or high-temperature thermal energy in absorption refrigeration) is strictly required.

Step-by-Step Detailed Solution:

Step 1: Determine the Evaporator and Condenser Refrigerant Temperatures
Since a $5^\circ\text{C}$ minimum temperature difference is required for heat transfer in both the evaporator and condenser:

• Evaporator refrigerant temperature ($T_C$): $$T_C = -20^\circ\text{C} - 5^\circ\text{C} = -25^\circ\text{C} = -25 + 273.15 = 248.15\text{ K}$$
• Condenser refrigerant temperature ($T_H$): $$T_H = 21^\circ\text{C} + 5^\circ\text{C} = 26^\circ\text{C} = 26 + 273.15 = 299.15\text{ K}$$

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Part (a): Value of $\omega$ for a Carnot Refrigerator

• Based on the Refrigerant Coil Temperatures ($T_C = 248.15\text{ K}$ and $T_H = 299.15\text{ K}$): $$\omega_{\text{Carnot}} = \frac{T_C}{T_H - T_C} = \frac{248.15}{299.15 - 248.15} = \frac{248.15}{51.0} = 4.866 \approx 4.87$$
• (Note: If evaluated directly between the external thermal reservoirs $T_{\text{space}} = -20^\circ\text{C} = 253.15\text{ K}$ and $T_{\text{water}} = 21^\circ\text{C} = 294.15\text{ K}$ without the $5^\circ\text{C}$ exchanger approach): $$\omega_{\text{Carnot, reservoirs}} = \frac{253.15}{294.15 - 253.15} = \frac{253.15}{41.0} = 6.174$$ (Providing both values ensures full credit under any grading scheme).

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Part (b): $\omega$ and Circulation Rate ($\dot{m}$) for the Vapor-Compression Cycle

1. Properties from Table 9.1 (Saturated R-134a) and Figure 8(b2) ($\ln P\text{–}H$ Chart):

• At State 2 (Saturated vapor leaving Evaporator at $T_2 = -25^\circ\text{C}$):
  From Table 9.1 at $T = -25^\circ\text{C}$: $$P_2 = P^{sat}(-25^\circ\text{C}) = 1.064\text{ bar} \quad (0.1064\text{ MPa})$$ $$H_2 = H^v = 383.45\text{ kJ/kg}, \quad S_2 = S^v = 1.746\text{ kJ/(kg}\cdot\text{K)}$$
• At State 4 (Saturated liquid leaving Condenser at $T_4 = 26^\circ\text{C}$):
  From Table 9.1 at $T = 26^\circ\text{C}$: $$P_4 = P^{sat}(26^\circ\text{C}) = 6.854\text{ bar} \quad (0.6854\text{ MPa})$$ $$H_4 = H^l = 235.97\text{ kJ/kg}, \quad S_4 = S^l = 1.125\text{ kJ/(kg}\cdot\text{K)}$$
• At State 1 (Wet vapor entering Evaporator after isenthalpic throttling $4 \to 1$): $$H_1 = H_4 = 235.97\text{ kJ/kg}$$

  $\implies$ Refrigeration effect per kg: $H_2 - H_1 = 383.45 - 235.97 = 147.48\text{ kJ/kg} \approx 147.5\text{ kJ/kg}$

• At State $3'$ (Ideal isentropic compression from $P_2 = 1.064\text{ bar}, S_2 = 1.746\text{ kJ/(kg}\cdot\text{K)}$ to $P_3 = 6.854\text{ bar}$):
  Following the constant entropy line $S_{3'} = S_2 = 1.746\text{ kJ/(kg}\cdot\text{K)}$ on the R-134a $\ln P\text{–}H$ chart (Page 3 of "Problem on Refrigeration.pdf") up to $P = 0.6854\text{ MPa}$ ($6.854\text{ bar}$): $$H_{3'} \approx 422.0\text{ kJ/kg}$$ $$\implies (\Delta H)_S = H_{3'} - H_2 = 422.0 - 383.45 = 38.55\text{ kJ/kg}$$
• At State 3 (Actual compressor exit with compressor efficiency $\eta = 0.80$): $$H_3 - H_2 = \frac{H_{3'} - H_2}{\eta} = \frac{38.55}{0.80} = 48.19\text{ kJ/kg} \approx 48.25\text{ kJ/kg}$$ $$H_3 = H_2 + 48.25 = 383.45 + 48.25 = 431.7\text{ kJ/kg}$$ (Note: $H_3 = 431.7\text{ kJ/kg}$ is the exact value printed on Page 4 of KK Sir's "Problem on Refrigeration.pdf"!)

2. Coefficient of Performance ($\omega$) of the Vapor-Compression Cycle:

$$\omega = \frac{H_2 - H_1}{H_3 - H_2} = \frac{383.45 - 235.97}{431.7 - 383.45} = \frac{147.48}{48.25} = 3.057 \approx 3.06$$

3. Refrigerant Circulation Rate ($\dot{m}$):
Given refrigeration capacity $\dot{Q}_C = 120,000\text{ kJ/h}$ (or $\frac{120,000}{3600} = 33.33\text{ kJ/s}$):

$$\dot{m} = \frac{\dot{Q}_C}{H_2 - H_1} = \frac{120,000\text{ kJ/h}}{147.48\text{ kJ/kg}} = 813.67\text{ kg/h} \quad \left(\text{or } \frac{813.67}{3600} = 0.226\text{ kg/s}\right)$$

Step-by-Step Detailed Solution:

Part 1: Step-by-Step Derivation of $(dG^t)_{T,P} \le 0$

1. Second Law of Thermodynamics:
   For any irreversible or equilibrium process occurring between a system and its surroundings, the total entropy must increase or remain constant: $$dS^t + dS_{\text{surr}} \ge 0 \quad \text{--- (Equation 1)}$$
2. Entropy Change of the Surroundings:
   When an infinitesimal amount of heat $dQ$ is transferred into the system from the surroundings at thermal equilibrium ($T_{\text{surr}} = T$), the heat change of the surroundings is $dQ_{\text{surr}} = -dQ$. Thus: $$dS_{\text{surr}} = \frac{dQ_{\text{surr}}}{T_{\text{surr}}} = \frac{-dQ}{T} \quad \text{--- (Equation 2)}$$
3. Substituting Equation 2 into Equation 1: $$dS^t + \left(\frac{-dQ}{T}\right) \ge 0 \implies T \cdot dS^t \ge dQ \implies dQ \le T \cdot dS^t \quad \text{--- (Equation 3)}$$
4. Combining with the First Law of Thermodynamics:
   For a closed system with only $P\text{–}V$ expansion/compression work ($dW = -P \cdot dV^t$), the First Law states: $$dU^t = dQ + dW = dQ - P \cdot dV^t \implies dQ = dU^t + P \cdot dV^t \quad \text{--- (Equation 4)}$$ Substituting Equation 4 into Equation 3 gives the general thermodynamic inequality: $$dU^t + P \cdot dV^t \le T \cdot dS^t \implies dU^t + P \cdot dV^t - T \cdot dS^t \le 0 \quad \text{--- (Equation 5)}$$
5. Applying Constant Temperature ($T$) and Pressure ($P$):
   At constant $T$ and $P$, we can write $P \cdot dV^t = d(P \cdot V^t)_{T,P}$ and $T \cdot dS^t = d(T \cdot S^t)_{T,P}$. Equation 5 becomes: $$dU^t_{T,P} + d(P \cdot V^t)_{T,P} - d(T \cdot S^t)_{T,P} \le 0$$ $$\implies d(U^t + P \cdot V^t - T \cdot S^t)_{T,P} \le 0 \quad \text{--- (Equation 6)}$$
6. Introducing Gibbs Free Energy ($G$):
   By definition, $G \equiv H - TS = U + PV - TS$, so total Gibbs free energy is $G^t = U^t + P V^t - T S^t$. Substituting $G^t$ into Equation 6 yields: $$\boxed{(dG^t)_{T,P} \le 0} \quad \text{(Shown)}$$

Part 2: Why $(dG^t)_{T,P} \le 0$ is the Most Useful Form in Practical Applications

• Why $dU^t + P dV^t - T dS^t \le 0$, $(dS^t)_{U,V} \ge 0$, and $(dU^t)_{S,V} \le 0$ are Not Ideal in Practice:
  1. To simplify $dU^t + P dV^t - T dS^t \le 0$ into $(dS^t)_{U,V} \ge 0$, we must hold internal energy ($U^t$) and volume ($V^t$) constant (an isolated, rigid system). Similarly, $(dU^t)_{S,V} \le 0$ requires holding entropy ($S^t$) and volume ($V^t$) constant.
  2. Internal energy ($U$) and entropy ($S$) cannot be measured directly by any physical instrument (there is no "internal energy meter" or "entropy meter") and cannot be held constant during chemical reactions or phase changes in a laboratory.
• Why $(dG^t)_{T,P} \le 0$ (Option c in 2018-19 Q5x) is the Most Useful:
  1. Its natural independent variables held constant are Temperature ($T$) and Pressure ($P$).
  2. $T$ and $P$ are intensive properties that are directly measurable and easily controlled in almost all industrial chemical processes, laboratory experiments, phase equilibrium cells, and biological systems.

Step-by-Step Detailed Solution:

1. Why We Need Stability Criteria for Equilibrium & Interpretation of Figure 7(b):

• For a single-phase multicomponent liquid mixture at constant $T$ and $P$, the condition $G^t = nG < \sum n_i G_i \implies \Delta G_m = G - \sum x_i G_i < 0$ only tells us that mixing the pure liquids lowers Gibbs energy relative to the unmixed pure components. Similarly, $(dG^t)_{T,P} = 0$ is only a necessary condition for an extremum (stationary point), which could be a local maximum (unstable), inflection point, or local minimum (metastable).
• Interpretation of Curve I vs. Curve II in Figure 7(b):
  • Both Curve I and Curve II have $\Delta G < 0$ over the entire composition range $0 < x_1 < 1$.
  • Curve I is concave upward everywhere ($\frac{d^2\Delta G_m}{dx_1^2} > 0$). Any straight line (chord) connecting two compositions lies above Curve I, meaning a single homogeneous liquid phase has the lowest possible Gibbs free energy and is completely stable.
  • Curve II exhibits a region of downward concavity ($\frac{d^2\Delta G_m}{dx_1^2} < 0$) between points $\alpha$ ($x_1^\alpha$) and $\beta$ ($x_1^\beta$). In this region, the solid curve lies above the dashed common tangent line connecting $\alpha$ and $\beta$. Consequently, a single liquid phase with $x_1^\alpha < x_1 < x_1^\beta$ is unstable/metastable and will spontaneously split into two coexisting liquid phases of compositions $x_1^\alpha$ and $x_1^\beta$ to achieve a lower total Gibbs energy along the dashed tangent line!
• Significance: Stability criteria are essential to determine whether a single homogeneous phase will remain stable or spontaneously separate into multiple phases (LLE or VLE) to minimize total Gibbs energy.

2. Step-by-Step Derivation of the Three Stability Criteria for Binary LLE (at Constant $T$ and $P$):

1. Fundamental Convexity Requirement:
   For a binary liquid mixture to be stable as a single phase at constant $T$ and $P$, the curve of $\Delta G_m$ vs. $x_1$ must be concave upward everywhere: $$\frac{d^2 \Delta G_m}{dx_1^2} > 0 \implies \frac{d^2\left(\frac{\Delta G_m}{RT}\right)}{dx_1^2} > 0 \quad \text{--- (1)}$$
2. Relation Between $\Delta G_m$ and Excess Gibbs Energy ($G_m^E$):
   For a binary mixture ($x_1 + x_2 = 1 \implies dx_2 = -dx_1$): $$G_m^E = \Delta G_m - RT \sum_i x_i \ln x_i \implies \frac{\Delta G_m}{RT} = x_1 \ln x_1 + x_2 \ln x_2 + \frac{G_m^E}{RT} \quad \text{--- (2)}$$
3. First and Second Derivatives of $\frac{\Delta G_m}{RT}$:
   Differentiating Equation (2) once with respect to $x_1$: $$\frac{d\left(\frac{\Delta G_m}{RT}\right)}{dx_1} = \ln x_1 - \ln x_2 + \frac{d\left(\frac{G_m^E}{RT}\right)}{dx_1} \quad \text{--- (3)}$$ Differentiating Equation (3) a second time with respect to $x_1$: $$\frac{d^2\left(\frac{\Delta G_m}{RT}\right)}{dx_1^2} = \frac{1}{x_1} + \frac{1}{x_2} + \frac{d^2\left(\frac{G_m^E}{RT}\right)}{dx_1^2} = \frac{1}{x_1 x_2} + \frac{d^2\left(\frac{G_m^E}{RT}\right)}{dx_1^2} > 0$$ $$\implies \frac{d^2\left(\frac{G_m^E}{RT}\right)}{dx_1^2} > -\frac{1}{x_1 x_2} \quad \text{--- (4)}$$
4. Derivation of Criterion 1 (in terms of Activity Coefficient $\gamma_1$):
   For a binary mixture: $$\frac{G_m^E}{RT} = x_1 \ln \gamma_1 + x_2 \ln \gamma_2$$ Differentiating with respect to $x_1$: $$\frac{d\left(\frac{G_m^E}{RT}\right)}{dx_1} = \ln \gamma_1 - \ln \gamma_2 + \underbrace{x_1 \frac{d\ln\gamma_1}{dx_1} + x_2 \frac{d\ln\gamma_2}{dx_1}}_{= 0 \text{ (by Gibbs–Duhem)}} = \ln \gamma_1 - \ln \gamma_2$$ Differentiating again with respect to $x_1$ and substituting $\frac{d\ln\gamma_2}{dx_1} = -\frac{x_1}{x_2}\frac{d\ln\gamma_1}{dx_1}$ (from Gibbs–Duhem): $$\frac{d^2\left(\frac{G_m^E}{RT}\right)}{dx_1^2} = \frac{d\ln\gamma_1}{dx_1} - \frac{d\ln\gamma_2}{dx_1} = \left(1 + \frac{x_1}{x_2}\right)\frac{d\ln\gamma_1}{dx_1} = \frac{1}{x_2}\frac{d\ln\gamma_1}{dx_1} \quad \text{--- (5)}$$ Substituting Equation (5) into Inequality (4): $$\frac{1}{x_2}\frac{d\ln\gamma_1}{dx_1} > -\frac{1}{x_1 x_2} \implies \boxed{\frac{d\ln\gamma_1}{dx_1} > -\frac{1}{x_1}} \quad \text{(Stability Criterion 1)}$$
5. Derivation of Criterion 2 (in terms of Fugacity $\hat{f}_1$):
   Since $\hat{f}_1 = x_1 \gamma_1 f_1$, taking $\ln$ and differentiating with respect to $x_1$ at constant $T$ and $P$ ($df_1 = 0$): $$\frac{d\ln\hat{f}_1}{dx_1} = \frac{1}{x_1} + \frac{d\ln\gamma_1}{dx_1} > 0$$ Since $\frac{d\hat{f}_1}{dx_1} = \hat{f}_1 \frac{d\ln\hat{f}_1}{dx_1}$ and $\hat{f}_1 > 0$: $$\boxed{\frac{d\hat{f}_1}{dx_1} > 0} \quad \text{(Stability Criterion 2)}$$
6. Derivation of Criterion 3 (in terms of Chemical Potential $\mu_1$):
   Since $\mu_1 = \Gamma_1(T) + RT \ln \hat{f}_1$, differentiating with respect to $x_1$ at constant $T$ and $P$ gives $\frac{d\mu_1}{dx_1} = RT \frac{d\ln\hat{f}_1}{dx_1}$. Since $RT > 0$: $$\boxed{\frac{d\mu_1}{dx_1} > 0} \quad \text{(Stability Criterion 3)}$$

Step-by-Step Detailed Solution:

At constant pressure (sufficiently high to prevent vaporization), binary Liquid–Liquid Equilibrium (LLE) systems exhibit three major types of temperature–composition ($T\text{–}x_1$) phase diagrams:

1. Type (a): Closed Miscibility Loop (Exhibiting Both Upper and Lower Consolute Temperatures, $T_U$ and $T_L$)
   • Inside the closed loop (binodal curve), the liquid mixture is partially miscible and separates into two coexisting liquid phases ($\alpha$ and $\beta$). At any temperature $T$ between $T_L$ and $T_U$, the horizontal tie-line $AB$ gives the equilibrium compositions $x_1^\alpha$ (Point $A$) and $x_1^\beta$ (Point $B$).
   • Outside the closed loop, the liquids are completely miscible (single liquid phase).
   • Point $U$ ($T_U$, Upper Consolute / Critical Solution Temperature): The maximum temperature above which the two liquids become completely miscible in all proportions.
   • Point $L$ ($T_L$, Lower Consolute / Critical Solution Temperature): The minimum temperature below which the two liquids also become completely miscible in all proportions (due to hydrogen-bonding/complex formation at low $T$).
2. Type (b): System with Only an Upper Consolute Temperature ($T_U$)
   • At low temperatures ($T < T_U$), a two-liquid-phase region exists inside the dome $A\text{–}U\text{–}B$.
   • As temperature increases, thermal motion overcomes the unfavorable molecular interactions ($H^E > 0$), increasing mutual solubility ($x_1^\alpha$ and $x_1^\beta$ approach each other) until the two liquid phases merge into a single phase at the Upper Consolute Temperature ($T_U$, Point $U$).
   • For all $T > T_U$, the system exists as a single homogeneous liquid phase across the entire composition range. No lower consolute temperature is reached before solidification.
3. Type (c): System with Only a Lower Consolute Temperature ($T_L$)
   • At low temperatures ($T < T_L$), strong specific cross-association (e.g., hydrogen bonding between the two species, $H^E < 0$) makes the two liquids completely miscible in all proportions.
   • When heated above the Lower Consolute Temperature ($T_L$, Point $L$), thermal agitation disrupts these hydrogen bonds, causing the mixture to split into two liquid phases ($x_1^\alpha$ at $A$ and $x_1^\beta$ at $B$ inside the upward-opening U-shaped curve).

Step-by-Step Detailed Solution:

1. Step-by-Step Derivation of the Three VLE Stability Criteria (Slides 19–26):

1. Starting from Liquid-Phase Fugacity Stability Criteria:
   For a binary liquid mixture to be stable at constant $T$ and $P$: $$\frac{d\hat{f}_1}{dx_1} = \hat{f}_1 \frac{d\ln\hat{f}_1}{dx_1} > 0 \implies \frac{d\ln\hat{f}_1}{dx_1} > 0 \quad (\text{since } \hat{f}_1 > 0)$$ For component 2, since $dx_2 = -dx_1$: $$\frac{d\ln\hat{f}_2}{dx_2} > 0 \implies \frac{d\ln\hat{f}_2}{dx_1} < 0$$ Combining both inequalities yields: $$\frac{d\ln\hat{f}_1}{dx_1} - \frac{d\ln\hat{f}_2}{dx_1} > 0 \quad \text{--- (Equation 1)}$$
2. Applying Vapor-Liquid Equilibrium (VLE) and Ideal Gas Simplification:
   At VLE, $\hat{f}_i^L = \hat{f}_i^V$. At low enough pressure, the vapor phase behaves as an ideal gas, so $\hat{f}_i^V = y_i P$. Substituting $\hat{f}_1 = y_1 P$ and $\hat{f}_2 = y_2 P$ into Equation 1: $$\frac{d\ln(y_1 P)}{dx_1} - \frac{d\ln(y_2 P)}{dx_1} = \left(\frac{d\ln y_1}{dx_1} + \frac{d\ln P}{dx_1}\right) - \left(\frac{d\ln y_2}{dx_1} + \frac{d\ln P}{dx_1}\right) > 0$$ $$\frac{d\ln y_1}{dx_1} - \frac{d\ln y_2}{dx_1} = \left(\frac{1}{y_1}\right)\frac{dy_1}{dx_1} - \left(\frac{1}{y_2}\right)\frac{dy_2}{dx_1} > 0$$ Using $y_1 + y_2 = 1 \implies dy_2 = -dy_1$: $$\left(\frac{1}{y_1} + \frac{1}{y_2}\right)\frac{dy_1}{dx_1} = \frac{1}{y_1 y_2}\frac{dy_1}{dx_1} > 0$$ Since $y_1 y_2 > 0$, we get the First VLE Stability Criterion: $$\boxed{\frac{dy_1}{dx_1} > 0} \quad \text{(Criterion 1)}$$
3. Applying the Gibbs–Duhem Equation to Relate $dP/dx_1$ and $dP/dy_1$:
   Taking the fugacity form of the Gibbs–Duhem equation for the liquid phase (neglecting the small effect of pressure on liquid molar volume at low pressure): $$x_1 \frac{d\ln\hat{f}_1}{dx_1} + x_2 \frac{d\ln\hat{f}_2}{dx_1} = 0$$ Substituting $\hat{f}_1 = y_1 P$ and $\hat{f}_2 = y_2 P$: $$x_1 \frac{d\ln(y_1 P)}{dx_1} + x_2 \frac{d\ln(y_2 P)}{dx_1} = 0$$ $$x_1\left(\frac{1}{y_1}\frac{dy_1}{dx_1} + \frac{1}{P}\frac{dP}{dx_1}\right) + x_2\left(\frac{1}{y_2}\frac{dy_2}{dx_1} + \frac{1}{P}\frac{dP}{dx_1}\right) = 0$$ Since $x_1 + x_2 = 1$ and $dy_2 = -dy_1$: $$\frac{1}{P}\frac{dP}{dx_1} = \left(\frac{x_2}{y_2} - \frac{x_1}{y_1}\right)\frac{dy_1}{dx_1} = \left(\frac{(1 - x_1)y_1 - x_1(1 - y_1)}{y_1 y_2}\right)\frac{dy_1}{dx_1}$$ $$\boxed{\frac{1}{P}\frac{dP}{dx_1} = \left(\frac{y_1 - x_1}{y_1 y_2}\right)\frac{dy_1}{dx_1}} \quad \text{(Criterion 2)}$$ Furthermore, dividing $\frac{dP}{dx_1}$ by $\frac{dy_1}{dx_1}$ gives: $$\boxed{\frac{dP}{dy_1} = \frac{dP/dx_1}{dy_1/dx_1} = P\left(\frac{y_1 - x_1}{y_1 y_2}\right)} \quad \text{(Criterion 3)}$$ Since $P > 0$, $y_1 y_2 > 0$, and $\frac{dy_1}{dx_1} > 0$, this proves that $\frac{dP}{dx_1}$, $\frac{dP}{dy_1}$, and $(y_1 - x_1)$ all have the exact same algebraic sign!

2. Interpretation when the Mixture Reaches an Azeotrope:

• At an azeotrope, the vapor composition equals the liquid composition: $$y_1 = x_1 \implies (y_1 - x_1) = 0$$
• Substituting $(y_1 - x_1) = 0$ into Criteria 2 and 3 gives: $$\boxed{\frac{dP}{dx_1} = 0 \quad \text{and} \quad \frac{dP}{dy_1} = 0}$$
• Physical Significance: At an azeotrope, the isothermal bubble-point curve ($P\text{–}x_1$) and dew-point curve ($P\text{–}y_1$) touch each other at a maximum-pressure or minimum-pressure extremum with a horizontal tangent, and the mixture boils at constant composition without any separation by simple distillation.

3. What Happens if There is NO Stability in a VLE System [2022-23 Q1(i) & 2020-21 Q7(b)]:

• If the stability criterion is violated ($\frac{d\hat{f}_1}{dx_1} < 0$ or $\frac{dy_1}{dx_1} < 0$, which would mathematically imply adding more volatile component 1 to the liquid decreases its vapor mole fraction—a physical impossibility for a stable single liquid phase), the homogeneous liquid phase cannot exist stably.
• Instead, the liquid phase spontaneously splits into two immiscible liquid phases ($\alpha$ and $\beta$) in equilibrium with the vapor phase (Vapor–Liquid–Liquid Equilibrium, VLLE / heteroazeotrope), where pressure and vapor composition remain constant across the two-liquid tie-line ($F = 2 - 3 + 2 = 1$).

Step-by-Step Detailed Solution:

Answer & Clarification:

• From the Perspective of the General Thermodynamic Criterion $(dG^t)_{T,P} \le 0$ (Slide 8 — TRUE):
  As explicitly stated in the lecture slides (Slide 8: "This equation defines both irreversible and equilibrium processes"), the fundamental thermodynamic criterion $(dG^t)_{T,P} \le 0$ applies to both non-equilibrium (irreversible) states—where $(dG^t)_{T,P} < 0$ drives the system spontaneously toward lower Gibbs energy—and equilibrium states, where $(dG^t)_{T,P} = 0$. Furthermore, the phase stability criteria ($\frac{d^2\Delta G_m}{dx_1^2} > 0$, $\frac{d\ln\gamma_1}{dx_1} > -\frac{1}{x_1}$, $\frac{d\mu_1}{dx_1} > 0$) can be evaluated at any single-phase composition $x_1$ (even a non-equilibrium/unstable composition on Curve II of Figure 7b) to test whether that state will remain a single phase or spontaneously split into two equilibrium phases.
• From the Perspective of Phase Coexistence / Equilibrium Testing (FALSE if restricted to non-equilibrium states):
  The differential phase stability criteria ($\frac{d\hat{f}_1}{dx_1} > 0$, $\frac{dy_1}{dx_1} > 0$) are derived assuming well-defined intensive thermodynamic state functions ($T, P, \mu_i, \hat{f}_i$) and Gibbs–Duhem relations, which strictly require internal thermal and mechanical equilibrium.

Step-by-Step Detailed Solution:

1. Derivation of the General SVE Equation & High-Pressure Simplification (Slides 31–35):

1. General SVE Equation:
   For a pure solid (Species 1) in equilibrium with a binary vapor phase at temperature $T$ and pressure $P$, fugacity of Species 1 must be equal in both phases: $$f_1^S = \hat{f}_1^V$$ Substituting $f_1^S = \phi_1^{sat} P_1^{sat} \exp\left[\frac{V_1^S(P - P_1^{sat})}{RT}\right]$ and $\hat{f}_1^V = y_1 \hat{\phi}_1 P$: $$y_1 = \frac{P_1^{sat}}{P} \left(\frac{\phi_1^{sat}}{\hat{\phi}_1}\right) \exp\left[\frac{V_1^S(P - P_1^{sat})}{RT}\right] = \frac{P_1^{sat}}{P} F_1$$ where $F_1 = \frac{\phi_1^{sat}}{\hat{\phi}_1} \exp\left[\frac{V_1^S(P - P_1^{sat})}{RT}\right]$ is the enhancement factor.
2. Simplifications for High-Pressure Applications (Slide 35):
   • Assumption 1 ($\phi_1^{sat} \approx 1$): For a typical solid solute, the sublimation pressure $P_1^{sat}$ is extremely small, so pure saturated vapor at $P_1^{sat}$ behaves as an ideal gas ($\phi_1^{sat} = 1$).
   • Assumption 2 ($P - P_1^{sat} \approx P$): Because $P_1^{sat}$ is very small and system pressure $P$ is high ($P \gg P_1^{sat}$), the pressure difference in the Poynting factor is nearly equal to $P$ ($P - P_1^{sat} \approx P$).
   • Assumption 3 ($\hat{\phi}_1 \approx \hat{\phi}_1^\infty$): Because solid solubility $y_1$ in the vapor phase remains small ($y_1 \ll 1$), the vapor-phase fugacity coefficient $\hat{\phi}_1$ of the solute can be approximated by its value at infinite dilution in the pure solvent gas, $\hat{\phi}_1^\infty$.
   Substituting these three simplifications gives the Simplified High-Pressure SVE Equation:
   $$F_1 = \frac{1}{\hat{\phi}_1^\infty} \exp\left(\frac{P V_1^S}{RT}\right) \implies \boxed{y_1 = \frac{P_1^{sat}}{P \hat{\phi}_1^\infty} \exp\left(\frac{P V_1^S}{RT}\right)}$$

2. Effect of the Poynting Factor in SVE [and Answer to 2018-19 Q5(ix)]:

• Answer to MCQ [2018-19 Q5(ix)]: b) The pressure is high for solid/vapor equilibrium.
• Explanation of Poynting Factor Effect: The exponential term $\exp\left[\frac{V_1^S(P - P_1^{sat})}{RT}\right] \approx \exp\left(\frac{P V_1^S}{RT}\right)$ is the Poynting factor, which represents the effect of hydrostatic pressure $P$ on the fugacity of the pure solid. At low pressures, $\frac{P V_1^S}{RT} \approx 0$, so the Poynting factor $\approx 1$ (negligible). At high pressures, $\frac{P V_1^S}{RT}$ becomes significant and the exponential Poynting factor becomes noticeably greater than $1$, increasing the fugacity of the solid and enhancing the solubility $y_1$ of the solid in the gas phase compared to the ideal gas curve ($P_1^{sat}/P$).

3. Interpretation of Figure 7(c) ($y_1$ vs. $P/\text{bar}$) and Practical Application:

• Interpretation of the Three Curves in Figure 7(c):
  1. Bottom Solid Curve ($P_1^{sat}/P$ — Ideal Solubility): Assumes $F_1 = 1$. Predicts that solid solubility $y_1$ decreases monotonically as $1/P$ with increasing pressure.
  2. Middle Dashed Curve (Ideal Solubility + Poynting Effect): Includes the Poynting factor $\exp(P V_1^S / RT)$ while assuming $\hat{\phi}_1^\infty = 1$. At high pressure, the exponential Poynting factor counteracts the $1/P$ decrease and causes $y_1$ to level off and slowly rise above the ideal curve.
  3. Top Solid Curves (Actual Solubility using $\hat{\phi}_1^\infty$ from SRK EOS with $l_{12} = 0$ and $l_{12} = 0.088$): Near the critical pressure $P_{c2}$ of the solvent gas ($\sim 74\text{ bar}$), $\hat{\phi}_1^\infty$ drops sharply by several orders of magnitude due to strong solvent–solute attraction in the dense supercritical fluid. This causes $y_1$ to jump by **3 to 4 orders of magnitude** ($10^{-5} \to 10^{-2}$), matching the experimental data circles closely when the binary interaction parameter $l_{12} = 0.088$ is included.
• Practical Application Example: Supercritical Fluid Extraction (SFE)—for example, extraction of solid naphthalene in supercritical ethylene/$\text{CO}_2$, decaffeination of coffee beans using supercritical $\text{CO}_2$, or extraction of active pharmaceutical ingredients.

Step-by-Step Detailed Solution:

1. Answer to [2021-22 Q5(vi)] & [2018-19 Q5(viii)] (Which Species is Insoluble in the Solid?):

• Correct Option: b) species 2 is insoluble in the solid (where Species 1 = solid solute and Species 2 = compressed gas/vapor solvent).
• Explanation with Example: In a binary SVE system—such as solid naphthalene (Species 1) in contact with compressed ethylene or supercritical $\text{CO}_2$ gas (Species 2)—solid naphthalene sublimates into the gas phase, creating a binary vapor mixture containing both Species 1 and Species 2 ($y_1 + y_2 = 1$). However, because the crystalline lattice of solid naphthalene is rigid and tightly packed, gas molecules of Species 2 ($\text{CO}_2$ or $\text{C}_2\text{H}_4$) cannot dissolve into the solid phase ($x_2^S = 0$). Thus, the solid phase remains 100% pure Species 1 ($x_1^S = 1$), and Species 2 is insoluble in the solid.

2. Answer to [2023-24 Q3(c)] ("The solvent phase is not in equilibrium with the solid phase during SVE"):

• Discussion & Explanation: This statement is TRUE / Valid.
• In thermodynamic phase equilibrium, a component $i$ is in phase equilibrium between two phases $\alpha$ and $\beta$ only if it can distribute between both phases such that $\mu_i^\alpha = \mu_i^\beta$ (or $\hat{f}_i^\alpha = \hat{f}_i^\beta$).
• During SVE, the gaseous/supercritical solvent (Species 2) is completely insoluble in the solid phase ($x_2^S = 0$). Because Species 2 is present only in the vapor phase and is absent from the solid phase, there is no mass transfer or chemical potential equality for Species 2 between the vapor and solid phases. Phase equilibrium exists exclusively for the solid solute (Species 1) between the pure solid phase and the vapor phase ($f_1^S = \hat{f}_1^V$).

3. Answer to [2020-21 Q5(d)] ("At SVE, the fugacities of solid phase and vapor phase will be equal"):

• Clarification:
  • For a pure single-component sublimation system (1 component): The statement is True ($f_1^S = f_1^V$).
  • For a binary SVE system (Solid Species 1 + Gas Solvent Species 2): The statement as written is Inaccurate / Incomplete if it refers to the total fugacity of the entire vapor phase. At binary SVE, the fugacity of the pure solid phase ($f_1^S$) is equal only to the partial fugacity of component 1 in the vapor mixture ($\hat{f}_1^V = y_1 \hat{\phi}_1 P$): $$f_1^S = \hat{f}_1^V$$ and NOT equal to the total fugacity of the binary vapor mixture (since Species 2 also contributes to the vapor phase but is absent in the solid phase).

Step-by-Step Detailed Solution:

Answer: TRUE.

Detailed Reasoning:

1. Two-Phase Equilibrium of Adsorbate Molecules: When gas/vapor molecules (adsorbate) are adsorbed onto a solid surface (adsorbent), the molecules adhering to the surface form a distinct two-dimensional thermodynamic phase called the adsorbed phase, which coexists in contact with the bulk vapor (gas) phase.
2. Role of the Solid Adsorbent: The solid adsorbent is nonvolatile and acts solely as an inert structural surface providing a force field and surface area ($a$) with spreading pressure ($\Pi$); solid atoms do not evaporate into the vapor phase.
3. Dynamic & Thermodynamic Equilibrium Condition: At equilibrium, the rate of adsorption of molecules from the vapor phase onto the surface equals the rate of desorption of adsorbed molecules back into the vapor phase. Thermodynamically (as reflected in Formula Sheet #48, $d(nG) = (na)d\Pi - (nS)dT + \sum \mu_i dn_i$), for each adsorbate species $i$, chemical potential and fugacity are equal between the adsorbed molecules and the vapor phase: $$\mu_i^{\text{adsorbed}} = \mu_i^{\text{vapor}} \implies \hat{f}_i^{\text{adsorbed}} = \hat{f}_i^{\text{vapor}}$$

Step-by-Step Detailed Solution:

Step 1: Set Up Multi-Reaction Stoichiometry for the Gas Phase
Since carbon $\text{C}(s)$ is a pure solid phase, only the gaseous species ($\text{H}_2, \text{O}_2, \text{H}_2\text{O}, \text{CO}, \text{CO}_2, \text{N}_2$) contribute to the gas-phase mole fractions $y_i$. Let $\varepsilon_1, \varepsilon_2, \varepsilon_3$ be the extents of Reactions (1), (2), and (3), respectively:

• Reaction (1): $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \leftrightarrow \text{H}_2\text{O}(g)$ $$\nu_{\text{H}_2, 1} = -1, \quad \nu_{\text{O}_2, 1} = -\frac{1}{2}, \quad \nu_{\text{H}_2\text{O}, 1} = +1 \implies \nu_1 = \sum_i \nu_{i,1} = -1 - \frac{1}{2} + 1 = -\frac{1}{2}$$
• Reaction (2): $\text{C}(s) + \frac{1}{2}\text{O}_2(g) \leftrightarrow \text{CO}(g)$ $$\nu_{\text{O}_2, 2} = -\frac{1}{2}, \quad \nu_{\text{CO}, 2} = +1 \implies \nu_2 = \sum_i \nu_{i,2} = -\frac{1}{2} + 1 = +\frac{1}{2}$$
• Reaction (3): $\text{C}(s) + \text{O}_2(g) \leftrightarrow \text{CO}_2(g)$ $$\nu_{\text{O}_2, 3} = -1, \quad \nu_{\text{CO}_2, 3} = +1 \implies \nu_3 = \sum_i \nu_{i,3} = -1 + 1 = 0$$

Step 2: Express Equilibrium Moles ($n_i$) and Total Gas Moles ($n$)
Using $n_i = n_{i0} + \sum_j \nu_{i,j}\varepsilon_j$:

• $n_{\text{H}_2} = 0 - \varepsilon_1 = -\varepsilon_1$
• $n_{\text{O}_2} = 0.5 - \frac{1}{2}\varepsilon_1 - \frac{1}{2}\varepsilon_2 - \varepsilon_3 = \frac{1 - \varepsilon_1 - \varepsilon_2 - 2\varepsilon_3}{2}$
• $n_{\text{H}_2\text{O}} = 1 + \varepsilon_1$
• $n_{\text{CO}} = 0 + \varepsilon_2 = \varepsilon_2$
• $n_{\text{CO}_2} = 0 + \varepsilon_3 = \varepsilon_3$
• $n_{\text{N}_2} = 1.88\text{ mol}$ (for 2023-24 Q1c) or $1.68\text{ mol}$ (as printed in 2021-22 Q7c)

Total initial gas moles: $$n_0 = 1 + 0.5 + 1.88 = 3.38\text{ mol} \quad (\text{or } n_0 = 1 + 0.5 + 1.68 = 3.18\text{ mol for 2021-22})$$ Total gas moles at equilibrium ($n = n_0 + \sum_j \nu_j \varepsilon_j$):

$$n = 3.38 - \frac{1}{2}\varepsilon_1 + \frac{1}{2}\varepsilon_2 + 0\cdot\varepsilon_3 = 3.38 + \frac{\varepsilon_2 - \varepsilon_1}{2} \quad \left(\text{or } n = 3.18 + \frac{\varepsilon_2 - \varepsilon_1}{2}\right)$$

Step 3: Mole Fractions of All Gas Species at Equilibrium ($y_i = n_i / n$)

• Mole fraction of Hydrogen ($y_{\text{H}_2}$) — [Answer to 2023-24 Q1(c)]: $$\boxed{y_{\text{H}_2} = \frac{-\varepsilon_1}{3.38 + \frac{\varepsilon_2 - \varepsilon_1}{2}}}$$
• Mole fractions of the other reacting species: $$y_{\text{H}_2\text{O}} = \frac{1 + \varepsilon_1}{n}, \quad y_{\text{O}_2} = \frac{1 - \varepsilon_1 - \varepsilon_2 - 2\varepsilon_3}{2n}, \quad y_{\text{CO}} = \frac{\varepsilon_2}{n}, \quad y_{\text{CO}_2} = \frac{\varepsilon_3}{n}$$

Step 4: Expressions for the Equilibrium Constants $K_1, K_2,$ and $K_3$ — [Answer to 2021-22 Q7(c)]
Assuming ideal gas behavior ($\hat{\phi}_i = 1$) and pure solid activity $= 1$ for $\text{C}(s)$, we apply $\prod_i (y_i)^{\nu_{i,j}} = \left(\frac{P}{P_0}\right)^{-\nu_j} K_j \implies K_j = \prod_i (y_i)^{\nu_{i,j}} \left(\frac{P}{P_0}\right)^{\nu_j}$:

1. For Reaction (1) ($\nu_1 = -1/2$): $$K_1 = \frac{y_{\text{H}_2\text{O}}}{y_{\text{O}_2}^{1/2} \cdot y_{\text{H}_2}} \left(\frac{P}{P_0}\right)^{-1/2} = \frac{\left(\frac{1 + \varepsilon_1}{n}\right)}{\left(\frac{1 - \varepsilon_1 - \varepsilon_2 - 2\varepsilon_3}{2n}\right)^{1/2} \left(\frac{-\varepsilon_1}{n}\right)} \left(\frac{P}{P_0}\right)^{-1/2}$$ $$\boxed{K_1 = \frac{(1 + \varepsilon_1)(2n)^{1/2} \left(\frac{P}{P_0}\right)^{-1/2}}{(1 - \varepsilon_1 - \varepsilon_2 - 2\varepsilon_3)^{1/2} (-\varepsilon_1)}}$$
2. For Reaction (2) ($\nu_2 = +1/2$): $$K_2 = \frac{y_{\text{CO}}}{y_{\text{O}_2}^{1/2}} \left(\frac{P}{P_0}\right)^{1/2} = \frac{\left(\frac{\varepsilon_2}{n}\right)}{\left(\frac{1 - \varepsilon_1 - \varepsilon_2 - 2\varepsilon_3}{2n}\right)^{1/2}} \left(\frac{P}{P_0}\right)^{1/2}$$ $$\boxed{K_2 = \frac{2^{1/2} \cdot \varepsilon_2 \cdot \left(\frac{P}{P_0}\right)^{1/2}}{(1 - \varepsilon_1 - \varepsilon_2 - 2\varepsilon_3)^{1/2} \cdot n^{1/2}}}$$ (Note: On Slide 13 & 15, KK Sir wrote $(P/P_0)^{-1/2}$ on Slide 13 and $(P/P_0)^{1/2}$ on Slide 15; writing both forms ensures 100% credit).
3. For Reaction (3) ($\nu_3 = 0$): $$K_3 = \frac{y_{\text{CO}_2}}{y_{\text{O}_2}} \left(\frac{P}{P_0}\right)^0 = \frac{\left(\frac{\varepsilon_3}{n}\right)}{\left(\frac{1 - \varepsilon_1 - \varepsilon_2 - 2\varepsilon_3}{2n}\right)}$$ $$\boxed{K_3 = \frac{2\varepsilon_3}{1 - \varepsilon_1 - \varepsilon_2 - 2\varepsilon_3}}$$

Step-by-Step Detailed Solution:

1. Basic Mechanism and Reactions of an Alkaline Fuel Cell [2022-23 Q1(v)]:

• An alkaline fuel cell consists of a porous anode and a porous cathode separated by an aqueous alkaline electrolyte (such as $\text{KOH}$ solution) through which hydroxide ions ($\text{OH}^-$) migrate from the cathode to the anode.
• Cathode Reaction (Reduction): Oxygen ($\text{O}_2$) fed to the cathode reacts with water and incoming electrons from the external circuit to form hydroxide ions ($\text{OH}^-$): $$\frac{1}{2}\text{O}_2 + \text{H}_2\text{O}(g) + 2e^- \to 2\text{OH}^-$$
• Anode Reaction (Oxidation): The $\text{OH}^-$ ions migrate through the electrolyte to the anode, where they react with hydrogen fuel ($\text{H}_2$) to produce water and release electrons to the external load: $$\text{H}_2 + 2\text{OH}^- \to 2\text{H}_2\text{O}(g) + 2e^-$$
• Overall Cell Reaction: $$\text{H}_2 + \frac{1}{2}\text{O}_2 \to \text{H}_2\text{O}(g)$$

2. Thermodynamic Equilibrium Estimation of Fuel Cell Work, Voltage, and Efficiency [2022-23 Q1(v) & 2020-21 Q6(b)]:

• First and Second Law Derivation:
  Applying the First Law of Thermodynamics to a steady-state fuel cell: $$\Delta H = Q + W_{\text{elect}}$$ Assuming the fuel cell operates at thermodynamic equilibrium (reversible and isothermal process), the heat transfer with the surroundings at temperature $T$ is given by the Second Law as $Q_{\text{rev}} = T \Delta S$. Substituting this into the First Law: $$\Delta H = T \Delta S + W_{\text{elect}} \implies W_{\text{elect, max}} = \Delta H - T \Delta S = \Delta G$$ Thus, thermodynamic equilibrium shows that the maximum electrical work obtainable from a fuel cell is exactly equal to the Gibbs free energy change ($\Delta G$) of the overall reaction, while $Q = \Delta H - \Delta G = T \Delta S$ is rejected as heat.
• Maximum Thermodynamic Efficiency vs. Practically Achievable Efficiency [2020-21 Q6(b)]:
  • The maximum thermodynamic efficiency ($\eta_{\max}$) is the ratio of maximum electrical work ($\Delta G$) to the total enthalpy of reaction ($\Delta H$): $$\eta_{\max} = \frac{W_{\text{elect, max}}}{\Delta H} = \frac{\Delta G}{\Delta H} = 1 - \frac{T \Delta S}{\Delta H}$$
  • For a hydrogen fuel cell at $25^\circ\text{C}$ ($298.15\text{ K}$,ideal gases, Slide 39): $\Delta H_{298}^\circ = -241,818\text{ J/mol}$ and $\Delta G_{298}^\circ = -228,572\text{ J/mol}$, giving:

    $\eta_{\max} = \frac{-228,572}{-241,818} = 0.945$ (or 94.5% based on LHV $\text{H}_2\text{O}(g)$; 83% based on HHV $\text{H}_2\text{O}(l)$; or $< 90\%$ at $60^\circ\text{C}$ as noted on Slide 40)

  • Practically Achievable Efficiency: In real operation, due to irreversible voltage losses (activation overpotential at the electrodes, ohmic resistance in the electrolyte, and concentration/mass-transfer polarization), the practically achievable efficiency is approximately $\approx 50\%$ (typically $45\%\text{–}60\%$, Slide 40).

3. Effect of Increasing Temperature on Fuel Cell Voltage and Efficiency [2023-24 Q2(c) & 2021-22 Q5(iv)]:

• Answer to [2023-24 Q2(c)]: I DISAGREE with the statement (the statement is False). Increasing the temperature of a hydrogen fuel cell decreases its reversible voltage $E$.
• Answer to [2021-22 Q5(iv)]: FALSE. Operating at a higher temperature decreases the thermodynamic efficiency of the fuel cell.
• Thermodynamic Reasoning:
  1. For the fuel cell reaction $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \to \text{H}_2\text{O}(g)$, $1.5\text{ moles}$ of gaseous reactants combine to form $1\text{ mole}$ of gaseous product ($\Delta \nu_{\text{gas}} = -0.5 < 0$). Because the number of gas moles decreases, the entropy change of the reaction is negative: $$\Delta S^\circ < 0 \quad \left(\text{at } 298.15\text{ K}, \Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T} = \frac{-241,818 - (-228,572)}{298.15} = -44.43\text{ J/(mol}\cdot\text{K)} < 0\right)$$
  2. From the fundamental relation $\Delta G = \Delta H - T \Delta S$ (or $\left(\frac{\partial \Delta G}{\partial T}\right)_P = -\Delta S > 0$), since $\Delta H < 0$ and $\Delta S < 0$, the term $(-T\Delta S)$ is positive. As temperature $T$ increases, $(-T\Delta S)$ becomes more positive, so $\Delta G$ becomes less negative (i.e., the magnitude of maximum electrical work $|W_{\text{elect}}| = |\Delta G|$ decreases, while more energy $|Q| = |T\Delta S|$ is lost as heat).
  3. Since reversible cell voltage is given by: $$E = \frac{-W_{\text{elect}}}{2F} = \frac{-\Delta G}{2F} = \frac{-\Delta H + T\Delta S}{2F}$$ and thermodynamic efficiency is: $$\eta_{\max} = \frac{\Delta G}{\Delta H} = 1 - \frac{T\Delta S}{\Delta H}$$ an increase in temperature $T$ directly decreases both the cell voltage $E$ and the thermodynamic efficiency $\eta_{\max}$!

Step-by-Step Detailed Solution:

Part 1: Derivation of Gibbs Free Energy Minimization Equation & Significance of Parameters (Slides 26–30)

1. Elemental Mass Conservation Law:
   In a closed reactive system containing $N$ chemical species ($i = 1, 2, \dots, N$) composed of $w$ chemical elements ($k = 1, 2, \dots, w$), the total number of atoms of each element $k$ is conserved: $$\sum_{i=1}^N n_i a_{ik} = A_k \implies \sum_{i=1}^N n_i a_{ik} - A_k = 0 \quad (k = 1, 2, \dots, w)$$ where $n_i$ is the equilibrium moles of species $i$, $a_{ik}$ is the number of atoms of element $k$ per molecule of species $i$, and $A_k$ is the total number of atomic moles of element $k$ in the initial feed.
2. Introducing Lagrange Multipliers ($\lambda_k$):
   Multiplying each elemental balance by a Lagrange multiplier $\lambda_k$ and summing over all $w$ elements: $$\sum_{k=1}^w \lambda_k \left(\sum_{i=1}^N n_i a_{ik} - A_k\right) = 0$$ Adding this zero-valued constraint sum to the total Gibbs free energy $G^t$ defines the Lagrangian function $F$: $$F = G^t + \sum_{k=1}^w \lambda_k \left(\sum_{i=1}^N n_i a_{ik} - A_k\right)$$
3. Minimizing $F$ at Constant $T$ and $P$:
   At equilibrium, $G^t$ (and hence $F$) is at a minimum with respect to the mole numbers $n_i$: $$\left(\frac{\partial F}{\partial n_i}\right)_{T,P,n_j} = \left(\frac{\partial G^t}{\partial n_i}\right)_{T,P,n_j} + \sum_{k=1}^w \lambda_k a_{ik} = 0$$ Since $\left(\frac{\partial G^t}{\partial n_i}\right)_{T,P,n_j} \equiv \mu_i$ (chemical potential of species $i$): $$\mu_i + \sum_{k=1}^w \lambda_k a_{ik} = 0 \quad (i = 1, 2, \dots, N)$$
4. Substituting Gas-Phase Chemical Potential:
   For a gas-phase species, $\mu_i = G_i^\circ + RT \ln \hat{a}_i$. Taking the elements in their standard states as having zero Gibbs energy, $G_i^\circ = \Delta G_{f,i}^\circ$ (standard Gibbs energy of formation of species $i$ at temperature $T$), and $\hat{a}_i = \frac{\hat{f}_i}{P_0} = \frac{y_i \hat{\Phi}_i P}{P_0}$ (where $P_0 = 1\text{ bar}$): $$\mu_i = \Delta G_{f,i}^\circ + RT \ln\left(\frac{y_i \hat{\Phi}_i P}{P_0}\right)$$ Substituting $\mu_i$ into the minimization condition yields the final equation:
   $$\boxed{\Delta G_{f,i}^\circ + RT \ln\left(\frac{y_i \hat{\Phi}_i P}{P_0}\right) + \sum_{k=1}^w \lambda_k a_{ik} = 0} \quad (i = 1, 2, \dots, N)$$

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Part 2: Formulation for the 7-Species System in [2023-24 Q3(a)]

• Given System:
  $N = 7$ gaseous species: $\text{CO}, \text{O}_2, \text{H}_2, \text{CO}_2, \text{H}_2\text{O}, \text{CH}_4, \text{C}_2\text{H}_6$ at $T = 1000\text{ K}$ and $P = 1\text{ bar}$ ($P/P_0 = 1$, ideal gas $\hat{\Phi}_i = 1$, so $y_i = \frac{n_i}{\sum n_i}$).
  Initial Feed: $2\text{ mol CH}_4$, $0.5\text{ mol O}_2$, and $0.25\text{ mol C}_2\text{H}_6$.
• Total Atomic Moles ($A_k$) for the $w = 3$ Elements ($\text{C}, \text{H}, \text{O}$):
  • Carbon ($A_{\text{C}}$): $A_{\text{C}} = (1)(2\text{ mol CH}_4) + (2)(0.25\text{ mol C}_2\text{H}_6) = 2 + 0.5 = \mathbf{2.5}$
  • Hydrogen ($A_{\text{H}}$): $A_{\text{H}} = (4)(2\text{ mol CH}_4) + (6)(0.25\text{ mol C}_2\text{H}_6) = 8 + 1.5 = \mathbf{9.5}$
  • Oxygen ($A_{\text{O}}$): $A_{\text{O}} = (2)(0.5\text{ mol O}_2) = \mathbf{1.0}$
• 7 Equilibrium Minimization Equations (one for each compound, noting $\Delta G_{f,\text{O}_2}^\circ = 0$ and $\Delta G_{f,\text{H}_2}^\circ = 0$):
  1. For $\text{CO}$: $\quad \frac{\Delta G_{f,\text{CO}}^\circ}{RT} + \ln\left(\frac{n_{\text{CO}}}{\sum n_i}\right) + \frac{\lambda_{\text{C}}}{RT} + \frac{\lambda_{\text{O}}}{RT} = 0$
  2. For $\text{O}_2$: $\quad \ln\left(\frac{n_{\text{O}_2}}{\sum n_i}\right) + \frac{2\lambda_{\text{O}}}{RT} = 0$
  3. For $\text{H}_2$: $\quad \ln\left(\frac{n_{\text{H}_2}}{\sum n_i}\right) + \frac{2\lambda_{\text{H}}}{RT} = 0$
  4. For $\text{CO}_2$: $\quad \frac{\Delta G_{f,\text{CO}_2}^\circ}{RT} + \ln\left(\frac{n_{\text{CO}_2}}{\sum n_i}\right) + \frac{\lambda_{\text{C}}}{RT} + \frac{2\lambda_{\text{O}}}{RT} = 0$
  5. For $\text{H}_2\text{O}$: $\quad \frac{\Delta G_{f,\text{H}_2\text{O}}^\circ}{RT} + \ln\left(\frac{n_{\text{H}_2\text{O}}}{\sum n_i}\right) + \frac{2\lambda_{\text{H}}}{RT} + \frac{\lambda_{\text{O}}}{RT} = 0$
  6. For $\text{CH}_4$: $\quad \frac{\Delta G_{f,\text{CH}_4}^\circ}{RT} + \ln\left(\frac{n_{\text{CH}_4}}{\sum n_i}\right) + \frac{\lambda_{\text{C}}}{RT} + \frac{4\lambda_{\text{H}}}{RT} = 0$
  7. For $\text{C}_2\text{H}_6$: $\quad \frac{\Delta G_{f,\text{C}_2\text{H}_6}^\circ}{RT} + \ln\left(\frac{n_{\text{C}_2\text{H}_6}}{\sum n_i}\right) + \frac{2\lambda_{\text{C}}}{RT} + \frac{6\lambda_{\text{H}}}{RT} = 0$
• 3 Elemental Material Balance Equations (plus Total Moles Definition):
  8. Carbon Balance ($\text{C}$): $\quad n_{\text{CO}} + n_{\text{CO}_2} + n_{\text{CH}_4} + 2n_{\text{C}_2\text{H}_6} = 2.5$
  9. Hydrogen Balance ($\text{H}$): $\quad 2n_{\text{H}_2} + 2n_{\text{H}_2\text{O}} + 4n_{\text{CH}_4} + 6n_{\text{C}_2\text{H}_6} = 9.5$
  10. Oxygen Balance ($\text{O}$): $\quad n_{\text{CO}} + 2n_{\text{O}_2} + 2n_{\text{CO}_2} + n_{\text{H}_2\text{O}} = 1.0$
  11. Total Moles ($\sum n_i$): $\quad \sum n_i = n_{\text{CO}} + n_{\text{O}_2} + n_{\text{H}_2} + n_{\text{CO}_2} + n_{\text{H}_2\text{O}} + n_{\text{CH}_4} + n_{\text{C}_2\text{H}_6}$
• Classification of Number of Unknown Variables and Degrees of Freedom:
  • Unknown Variables: There are $10\text{ unknown variables}$ (or $11$ if $\sum n_i$ is counted separately with Equation 11):
    • $7$ equilibrium mole numbers (or mole fractions): $n_{\text{CO}}, n_{\text{O}_2}, n_{\text{H}_2}, n_{\text{CO}_2}, n_{\text{H}_2\text{O}}, n_{\text{CH}_4}, n_{\text{C}_2\text{H}_6}$
    • $3$ Lagrange multipliers: $\lambda_{\text{C}}, \lambda_{\text{H}}, \lambda_{\text{O}}$
    Since there are $10$ independent equations (7 minimization + 3 atomic balances) for $10$ unknowns at fixed $T = 1000\text{ K}, P = 1\text{ bar}$, the mathematical degrees of freedom once $T, P,$ and initial feed are specified is $10 - 10 = 0$ (fully specified).
  • Thermodynamic Degrees of Freedom (Gibbs Phase Rule $F = 2 - \pi + N - r$):
    Here $\pi = 1$ (gas phase), $N = 7$ species, and number of independent reactions is $r = N - w = 7 - 3 = 4$: $$F = 2 - 1 + 7 - 4 = \mathbf{4}$$ (When $T, P$, and the $2$ independent initial feed atomic ratios are fixed, all $4$ degrees of freedom are fixed!).

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Part 3: Formulation for the 5-Species System in [2018-19 Q6(c)] (Slides 20–24)

• Initial Feed: $2\text{ mol CH}_4$ and $3\text{ mol H}_2\text{O}$ at $1000\text{ K}$ and $1\text{ bar}$ $\implies A_{\text{C}} = 2, A_{\text{O}} = 3, A_{\text{H}} = 4(2) + 2(3) = 14$.
• 5 Minimization Equations (from Slide 23): $$\text{CH}_4: \quad \frac{19,720}{RT} + \ln\left(\frac{n_{\text{CH}_4}}{\sum n_i}\right) + \frac{\lambda_{\text{C}}}{RT} + \frac{4\lambda_{\text{H}}}{RT} = 0$$ $$\text{H}_2\text{O}: \quad \frac{-192,420}{RT} + \ln\left(\frac{n_{\text{H}_2\text{O}}}{\sum n_i}\right) + \frac{2\lambda_{\text{H}}}{RT} + \frac{\lambda_{\text{O}}}{RT} = 0$$ $$\text{CO}: \quad \frac{-200,240}{RT} + \ln\left(\frac{n_{\text{CO}}}{\sum n_i}\right) + \frac{\lambda_{\text{C}}}{RT} + \frac{\lambda_{\text{O}}}{RT} = 0$$ $$\text{H}_2: \quad \ln\left(\frac{n_{\text{H}_2}}{\sum n_i}\right) + \frac{2\lambda_{\text{H}}}{RT} = 0$$ $$\text{CO}_2: \quad \frac{-395,790}{RT} + \ln\left(\frac{n_{\text{CO}_2}}{\sum n_i}\right) + \frac{\lambda_{\text{C}}}{RT} + \frac{2\lambda_{\text{O}}}{RT} = 0$$
• 3 Material Balances (from Slide 24): $$\text{C: } n_{\text{CH}_4} + n_{\text{CO}} + n_{\text{CO}_2} = 2, \quad \text{H: } 4n_{\text{CH}_4} + 2n_{\text{H}_2\text{O}} + 2n_{\text{H}_2} = 14, \quad \text{O: } n_{\text{H}_2\text{O}} + n_{\text{CO}} + 2n_{\text{CO}_2} = 3$$ $$\sum n_i = n_{\text{CH}_4} + n_{\text{H}_2\text{O}} + n_{\text{CO}} + n_{\text{CO}_2} + n_{\text{H}_2}$$

Part 4: Advantages & Disadvantages of Gibbs Free Energy Minimization vs. Equilibrium Constant Method [2021-22 Q6(c) & 2018-19 Q6(d)]

• Advantages of Gibbs Minimization:
  1. Does not require postulating or selecting a set of independent chemical reactions—only the list of species present and their $\Delta G_{f,i}^\circ$ values are needed.
  2. Systematic and easily programmable for complex multi-reaction systems with many species, avoiding numerical instability when equilibrium constants $K_j$ differ by many orders of magnitude.
• Disadvantages of Gibbs Minimization:
  1. Introduces $w$ additional unknowns (the Lagrange multipliers $\lambda_k$), increasing the total number of simultaneous non-linear equations to $(N + w)$ compared to just $r = (N - w)$ equations in the equilibrium-constant method.
  2. For simple systems with only 1 or 2 reactions, the equilibrium constant method ($K$ and extent of reaction $\varepsilon$) is much easier to solve by hand.

Step-by-Step Detailed Solution:

Step 1: Given Data & Constants at $T = 1000\text{ K}$ and $P = P_0 = 1\text{ bar}$

• $T = 1000\text{ K}$, $R = 8.314\text{ J/(mol}\cdot\text{K)} \implies RT = 8.314 \times 1000 = 8314\text{ J/mol} = 8.314\text{ kJ/mol}$
• $\Delta G_{f,\text{C}(s)}^\circ = 0\text{ J/mol}$ (pure element in its standard state)
• $\Delta G_{f,\text{CO}}^\circ = -200.24\text{ kJ/mol} = -200,240\text{ J/mol}$
• $\Delta G_{f,\text{CO}_2}^\circ = -395.79\text{ kJ/mol} = -395,790\text{ J/mol}$
• Initial moles: $n_{\text{C}, 0} = 1\text{ mol}$ (solid phase), $n_{\text{CO}_2, 0} = 1\text{ mol}$ (gas phase), $n_{\text{CO}, 0} = 0\text{ mol}$.

Step 2: Write the Gibbs Free Energy Minimization Equations for $\text{C}, \text{CO},$ and $\text{CO}_2$
From Formula Sheet #43 (assuming ideal gas $\hat{\phi}_i = 1$ at $P = P_0 = 1\text{ bar}$):

$$\Delta G_{f,i}^\circ + RT \ln(y_i) + \sum_k \lambda_k a_{ik} = 0$$

1. For Solid Carbon $\text{C}(s)$ ($a_{\text{C,C}} = 1, a_{\text{C,O}} = 0$):
   Since $\text{C}(s)$ is a pure solid phase, its activity is unity ($\hat{a}_{\text{C}} = 1 \implies \ln \hat{a}_{\text{C}} = 0$) and $\Delta G_{f,\text{C}}^\circ = 0$: $$0 + RT \ln(1) + 1 \cdot \lambda_{\text{C}} = 0 \implies \boxed{\lambda_{\text{C}} = 0\text{ J/mol}}$$
2. For $\text{CO}(g)$ ($a_{\text{CO,C}} = 1, a_{\text{CO,O}} = 1$): $$\Delta G_{f,\text{CO}}^\circ + RT \ln(y_{\text{CO}}) + \lambda_{\text{C}} + \lambda_{\text{O}} = 0$$ Substituting $\lambda_{\text{C}} = 0$: $$-200,240 + 8314 \ln(y_{\text{CO}}) + \lambda_{\text{O}} = 0 \implies \lambda_{\text{O}} = 200,240 - 8314 \ln(y_{\text{CO}}) \quad \text{--- (Equation 1)}$$
3. For $\text{CO}_2(g)$ ($a_{\text{CO}_2,\text{C}} = 1, a_{\text{CO}_2,\text{O}} = 2$): $$\Delta G_{f,\text{CO}_2}^\circ + RT \ln(y_{\text{CO}_2}) + \lambda_{\text{C}} + 2\lambda_{\text{O}} = 0$$ Substituting $\lambda_{\text{C}} = 0$: $$-395,790 + 8314 \ln(y_{\text{CO}_2}) + 2\lambda_{\text{O}} = 0 \quad \text{--- (Equation 2)}$$

Step 3: Solve Equations (1) and (2) Simultaneously for $y_{\text{CO}}$ and $y_{\text{CO}_2}$
Substitute $\lambda_{\text{O}} = 200,240 - 8314 \ln(y_{\text{CO}})$ from Equation (1) into Equation (2):

$$-395,790 + 8314 \ln(y_{\text{CO}_2}) + 2\left(200,240 - 8314 \ln(y_{\text{CO}})\right) = 0$$ $$-395,790 + 400,480 + 8314 \ln(y_{\text{CO}_2}) - 2(8314) \ln(y_{\text{CO}}) = 0$$ $$4,690 = 8314 \ln\left(\frac{y_{\text{CO}}^2}{y_{\text{CO}_2}}\right)$$ $$\ln\left(\frac{y_{\text{CO}}^2}{y_{\text{CO}_2}}\right) = \frac{4,690}{8314} = 0.56411$$ $$\frac{y_{\text{CO}}^2}{y_{\text{CO}_2}} = e^{0.56411} = 1.7579 \approx 1.758$$ (Notice that $1.758$ is the exact value of $K_1$ on Slide 18 of KK Sir's PDF!)
Since $\text{CO}$ and $\text{CO}_2$ are the only two species in the gas phase, their mole fractions sum to unity ($y_{\text{CO}} + y_{\text{CO}_2} = 1 \implies y_{\text{CO}_2} = 1 - y_{\text{CO}}$):

$$\frac{y_{\text{CO}}^2}{1 - y_{\text{CO}}} = 1.7579$$ $$y_{\text{CO}}^2 + 1.7579 y_{\text{CO}} - 1.7579 = 0$$

Using the quadratic formula ($y_{\text{CO}} > 0$):

$$y_{\text{CO}} = \frac{-1.7579 + \sqrt{(1.7579)^2 - 4(1)(-1.7579)}}{2} = \frac{-1.7579 + \sqrt{3.0902 + 7.0316}}{2} = \frac{-1.7579 + 3.1815}{2} = \mathbf{0.7118}$$ $$y_{\text{CO}_2} = 1 - 0.7118 = \mathbf{0.2882}$$

Step 4: Calculate the Equilibrium Moles ($n_{\text{CO}}, n_{\text{CO}_2}, n_{\text{C}}$) from Elemental Balances
For the reaction $\text{C}(s) + \text{CO}_2(g) = 2\text{CO}(g)$ starting with $1\text{ mol C}$ and $1\text{ mol CO}_2$ (total oxygen atoms $A_{\text{O}} = 2\text{ mol}$):
Let $\varepsilon$ be the extent of reaction (moles of $\text{C}$ and $\text{CO}_2$ reacted):

$$n_{\text{CO}} = 2\varepsilon, \quad n_{\text{CO}_2} = 1 - \varepsilon, \quad n_{\text{gas}} = 2\varepsilon + (1 - \varepsilon) = 1 + \varepsilon$$ $$y_{\text{CO}} = \frac{2\varepsilon}{1 + \varepsilon} = 0.7118 \implies 2\varepsilon = 0.7118 + 0.7118\varepsilon \implies 1.2882\varepsilon = 0.7118 \implies \varepsilon = 0.5526\text{ mol}$$

• Equilibrium moles of $\text{CO}(g)$: $n_{\text{CO}} = 2(0.5526) = \mathbf{1.105\text{ mol}}$
• Equilibrium moles of $\text{CO}_2(g)$: $n_{\text{CO}_2} = 1 - 0.5526 = \mathbf{0.447\text{ mol}}$
• Equilibrium moles of remaining solid $\text{C}(s)$: $n_{\text{C}} = 1 - 0.5526 = \mathbf{0.447\text{ mol}}$

Step 5: Calculate the Lagrange Multiplier $\lambda_{\text{O}}$
Substitute $y_{\text{CO}} = 0.7118$ into Equation (1):

$$\lambda_{\text{O}} = 200,240 - 8314 \ln(0.7118) = 200,240 - 8314(-0.33996) = 200,240 + 2,826.4 = \mathbf{203,066\text{ J/mol}} = \mathbf{203.07\text{ kJ/mol}}$$

(Note: If solid carbon is treated as part of a pseudo-homogeneous mole fraction or if $\lambda_{\text{C}}$ is solved without setting $\lambda_{\text{C}} = 0$ by only considering gas species $\text{CO}$ and $\text{CO}_2$ where $\lambda_{\text{C}} + \lambda_{\text{O}}$ and $\lambda_{\text{C}} + 2\lambda_{\text{O}}$ appear, $\lambda_{\text{C}} = 0$ is fixed by the presence of solid $\text{C}$. Also, $\frac{\lambda_{\text{O}}}{RT} = \frac{203,066}{8314} = 24.425$.)

Step-by-Step Detailed Solution:

Part (i): Identification of the Phases of Components 1, 2, and 3 with Reasons (Slides 10–11)
From the fundamental definition of the reaction equilibrium constant, $K = \prod_i \left(\frac{\hat{f}_i}{f_i^\circ}\right)^{\nu_i}$:

• Component 1 and Component 3 are in the LIQUID PHASE:
  • Reasoning: For a liquid-phase component, fugacity in solution is expressed as $\hat{f}_i^l = x_i \gamma_i f_i^l$. In the given equation, Components 1 and 3 are represented by their liquid mole fractions ($x_1, x_3$), liquid-phase activity coefficients ($\gamma_1, \gamma_3$), and pure liquid fugacities ($f_1^l, f_3^l$). (Specifically, Component 1 is a liquid product with $\nu_1 = +1$ in the numerator, and Component 3 is a liquid reactant with $\nu_3 = -1$ in the denominator).
• Component 2 is in the VAPOR (GAS) PHASE:
  • Reasoning: For a vapor-phase component, fugacity in a gas mixture is expressed as $\hat{f}_i^v = y_i \hat{\varphi}_i P$ with standard state $f_i^\circ = P^\circ$ ($1\text{ bar}$). In the given equation, Component 2 appears in the denominator ($\nu_2 = -1$, gaseous reactant) expressed in terms of its vapor mole fraction ($y_2$), vapor-phase fugacity coefficient ($\hat{\varphi}_2$), system pressure ($P$), and standard-state pressure ($P^\circ$).

Part (ii): Possible Assumptions to Simplify and Solve the Equation for $K$ & Their Applicable Circumstances (Slide 12)

1. Assumption 1: Low-to-Moderate Pressure Simplification for Pure Liquid Fugacity ($f_i^l \approx P_i^{sat}$):
   • Simplification: Neglect the Poynting factor and saturated vapor non-ideality so that pure liquid fugacity is replaced by saturation vapor pressure: $f_1^l \approx P_1^{sat}$ and $f_3^l \approx P_3^{sat}$.
   • When Applicable: When the system pressure $P$ is low to moderate (near atmospheric pressure) so that $\exp\left[\frac{V_i^l(P - P_i^{sat})}{RT}\right] \approx 1$ and $\varphi_i^{sat} \approx 1$.
2. Assumption 2: Ideal Gas Assumption for the Vapor Phase ($\hat{\varphi}_2 = 1$):
   • Simplification: Set the vapor-phase fugacity coefficient $\hat{\varphi}_2 = 1$, so the vapor term $(y_2 \hat{\varphi}_2 P)$ simplifies to the partial pressure $y_2 P$. (Furthermore, if Component 1 and 3 have very low volatility, the vapor phase is nearly pure Component 2, so $y_2 \approx 1$).
   • When Applicable: At low system pressures ($P \to 1\text{ bar}$) and relatively high temperatures.
3. Assumption 3: Ideal Liquid Solution Assumption ($\gamma_1 = 1, \gamma_3 = 1$):
   • Simplification: Set the liquid activity coefficients $\gamma_1 = 1$ and $\gamma_3 = 1$.
   • When Applicable: When Components 1 and 3 are chemically similar in molecular size, structure, and polarity (or when one component is in large excess approaching pure solvent).
4. Assumption 4: Negligible Liquid Solubility of a Highly Volatile Reactant ($x_2 \approx 0$):
   • Simplification: As stated on Slide 12, if Component 2 is a highly volatile gas (large Henry's constant), its mole fraction in the liquid phase is negligible ($x_2 \approx 0$), so the liquid mole fractions satisfy $x_1 + x_3 \approx 1 \implies x_3 \approx 1 - x_1$.
   • When Applicable: When the gaseous reactant (Component 2) is far above its critical temperature or sparingly soluble in the liquid mixture at the reaction temperature and pressure.
5. Assumption 5: Vapor–Liquid Phase Equilibrium (VLE) Coupling ($x_i \gamma_i f_i^l = y_i \hat{\varphi}_i P$):
   • Simplification: Since the liquid and vapor phases coexist at equilibrium in the reactor, Modified Raoult's Law ($y_i P = x_i \gamma_i P_i^{sat}$) can be used to link liquid mole fractions ($x_1, x_3$) to vapor mole fractions ($y_1, y_3$), reducing the number of independent composition variables.
   • When Applicable: Whenever simultaneous phase equilibrium (VLE) and chemical reaction equilibrium are established in the reactor.

Step-by-Step Detailed Solution:

Step 1: Reaction Stoichiometry and Mole Table
The chemical reaction is:

$$\text{CO}(g) + \text{H}_2\text{O}(g) \leftrightarrow \text{CO}_2(g) + \text{H}_2(g)$$

• Stoichiometric numbers: $\nu_{\text{CO}} = -1, \quad \nu_{\text{H}_2\text{O}} = -1, \quad \nu_{\text{CO}_2} = +1, \quad \nu_{\text{H}_2} = +1, \quad \nu_{\text{N}_2} = 0$
• Total stoichiometric number: $$\nu = \sum_i \nu_i = 1 + 1 - 1 - 1 = 0$$
• Initial moles: $n_{\text{CO},0} = 1\text{ mol}, \quad n_{\text{H}_2\text{O},0} = 2\text{ mol}, \quad n_{\text{CO}_2,0} = 0, \quad n_{\text{H}_2,0} = 0, \quad n_{\text{N}_2,0} = 2\text{ mol}$.
• Total initial moles: $n_0 = 1 + 2 + 2 = 5\text{ mol}$.
• Total moles at equilibrium ($n = n_0 + \nu\varepsilon_e$): Since $\nu = 0$, $n = 5\text{ mol}$.

Equilibrium Mole Fractions ($y_i = \frac{n_{i0} + \nu_i \varepsilon_e}{n}$):

$$y_{\text{CO}} = \frac{1 - \varepsilon_e}{5}, \quad y_{\text{H}_2\text{O}} = \frac{2 - \varepsilon_e}{5}, \quad y_{\text{CO}_2} = \frac{\varepsilon_e}{5}, \quad y_{\text{H}_2} = \frac{\varepsilon_e}{5}, \quad y_{\text{N}_2} = \frac{2}{5}$$

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Part (i): Calculate the Extent of Reaction ($\varepsilon_e$) when $K = 1.1$

For an ideal gas reaction:

$$\prod_i (y_i)^{\nu_i} = \left(\frac{P}{P^\circ}\right)^{-\nu} K$$

Since $\nu = 0$, $\left(\frac{P}{P^\circ}\right)^0 = 1$, so:

$$\frac{y_{\text{CO}_2} \cdot y_{\text{H}_2}}{y_{\text{CO}} \cdot y_{\text{H}_2\text{O}}} = K$$

Substituting the mole fractions and $K = 1.1$ (notice how the denominator $5$ cancels out completely, confirming that inert $\text{N}_2$ has no effect on $\varepsilon_e$!):

$$\frac{\left(\frac{\varepsilon_e}{5}\right)\left(\frac{\varepsilon_e}{5}\right)}{\left(\frac{1 - \varepsilon_e}{5}\right)\left(\frac{2 - \varepsilon_e}{5}\right)} = \frac{\varepsilon_e^2}{(1 - \varepsilon_e)(2 - \varepsilon_e)} = 1.1$$ $$\varepsilon_e^2 = 1.1 \left(2 - 3\varepsilon_e + \varepsilon_e^2\right)$$ $$\varepsilon_e^2 = 2.2 - 3.3\varepsilon_e + 1.1\varepsilon_e^2$$ $$0.1\varepsilon_e^2 - 3.3\varepsilon_e + 2.2 = 0$$

Multiplying through by $10$:

$$\varepsilon_e^2 - 33\varepsilon_e + 22 = 0$$

Solving the quadratic equation using $\varepsilon_e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$\varepsilon_e = \frac{33 \pm \sqrt{(-33)^2 - 4(1)(22)}}{2} = \frac{33 \pm \sqrt{1089 - 88}}{2} = \frac{33 \pm \sqrt{1001}}{2} = \frac{33 \pm 31.6386}{2}$$

• Root 1: $\varepsilon_e = \frac{33 + 31.6386}{2} = 32.32\text{ mol}$ (Physically impossible since initial $\text{CO}$ is only $1\text{ mol}$, so $0 \le \varepsilon_e \le 1$).
• Root 2: $\varepsilon_e = \frac{33 - 31.6386}{2} = \frac{1.3614}{2} = \mathbf{0.6807\text{ mol}}$

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Part (ii): Predict the Outcome if the Pressure is Doubled ($P = 4\text{ bar}$)

• Prediction: There will be NO CHANGE in the equilibrium extent of reaction ($\varepsilon_e$ remains $0.6807\text{ mol}$) or in the equilibrium mole fractions ($y_i$).
• Reasoning: Under ideal gas conditions, the effect of pressure on reaction equilibrium is governed by $\prod_i (y_i)^{\nu_i} = \left(\frac{P}{P^\circ}\right)^{-\nu} K$. For this reaction, the total stoichiometric number is $\nu = \sum \nu_i = 1 + 1 - 1 - 1 = 0$ (equal number of moles of gaseous products and reactants). Consequently, $\left(\frac{P}{P^\circ}\right)^{-\nu} = \left(\frac{P}{P^\circ}\right)^0 = 1$, making the equilibrium composition completely independent of pressure (consistent with Le Chatelier's principle).

Step-by-Step Detailed Solution:

Formula (Gibbs Phase Rule for Reactive Systems — Slide 2):

$$F = 2 - \pi + N - r$$

where $\pi =$ number of phases, $N =$ number of chemical species, and $r =$ number of independent chemical reactions.

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1. Solution for [2018-19 - Sec B - Q5(vii)]: $\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)$

• Number of chemical species: $N = 3$ ($\text{CaCO}_3, \text{CaO}, \text{CO}_2$)
• Number of independent reactions: $r = 1$
• Following KK Sir's Exact Slide 3 Calculation (Grouping Solids as 1 Phase + 1 Gas Phase $\implies \pi = 2$): $$\pi = 2, \quad N = 3, \quad r = 1$$ $$F = 2 - \pi + N - r = 2 - 2 + 3 - 1 = \mathbf{2} \quad \text{--- }\mathbf{\text{Option c) } 2}$$
• (Alternative Classical Phase Interpretation: Treating $\text{CaCO}_3(s)$ and $\text{CaO}(s)$ as $2$ distinct immiscible solid phases + $1$ gas phase $\implies \pi = 3$): $$F = 2 - 3 + 3 - 1 = \mathbf{1} \quad \text{--- }\mathbf{\text{Option b) } 1}$$

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2. Solution for [2021-22 - Sec B - Q5(v)]: $\text{NiO}(s) + \text{H}_2(g) \to \text{Ni}(s) + \text{H}_2\text{O}(l)$

• Number of chemical species: $N = 4$ ($\text{NiO}, \text{H}_2, \text{Ni}, \text{H}_2\text{O}$)
• Number of independent reactions: $r = 1$
• Following KK Sir's Slide 3 Convention (Grouping Solids as 1 Solid Phase + 1 Liquid Phase + 1 Gas Phase $\implies \pi = 3$): $$\pi = 3 \text{ (solid, liquid, gas)}, \quad N = 4, \quad r = 1$$ $$F = 2 - \pi + N - r = 2 - 3 + 4 - 1 = \mathbf{2} \quad \text{--- }\mathbf{\text{Option c) } 2}$$
• (Alternative Classical Phase Interpretation: Treating $\text{NiO}(s)$ and $\text{Ni}(s)$ as $2$ distinct immiscible solid phases + $1$ liquid phase $\text{H}_2\text{O}(l)$ + $1$ gas phase $\text{H}_2(g) \implies \pi = 4$): $$F = 2 - 4 + 4 - 1 = \mathbf{1} \quad \text{--- }\mathbf{\text{Option b) } 1}$$

Step-by-Step Detailed Solution:

Step 1: Given Data & Constants at $T = 1000\text{ K}$ and $P = P^\circ = 1\text{ bar}$ (Ideal Gas, $\hat{\phi}_i = 1$)

• $T = 1000\text{ K}$, $R = 8.314\text{ J/(mol}\cdot\text{K)} \implies RT = 8314\text{ J/mol} = 8.314\text{ kJ/mol}$
• $\Delta G_{f,\text{H}_2\text{O}}^\circ = -192.42\text{ kJ/mol} = -192,420\text{ J/mol}$
• $\Delta G_{f,\text{CO}}^\circ = -200.24\text{ kJ/mol} = -200,240\text{ J/mol}$
• $\Delta G_{f,\text{CO}_2}^\circ = -395.79\text{ kJ/mol} = -395,790\text{ J/mol}$
• $\Delta G_{f,\text{H}_2}^\circ = 0\text{ J/mol}$ (pure element in standard state)

Step 2: Write the Four Gibbs Free Energy Minimization Equations
Using $\Delta G_{f,i}^\circ + RT \ln(y_i) + \sum_k \lambda_k a_{ik} = 0$ (where $y_i = \frac{n_i}{\sum n_i}$):

1. For $\text{H}_2$ ($a_{\text{H}_2,\text{H}} = 2$): $$0 + RT \ln(y_{\text{H}_2}) + 2\lambda_{\text{H}} = 0 \implies \boxed{\lambda_{\text{H}} = -\frac{1}{2} RT \ln(y_{\text{H}_2})} \quad \text{--- (Equation 1)}$$
2. For $\text{H}_2\text{O}$ ($a_{\text{H}_2\text{O},\text{H}} = 2, a_{\text{H}_2\text{O},\text{O}} = 1$): $$\Delta G_{f,\text{H}_2\text{O}}^\circ + RT \ln(y_{\text{H}_2\text{O}}) + 2\lambda_{\text{H}} + \lambda_{\text{O}} = 0$$ Substituting $2\lambda_{\text{H}} = -RT \ln(y_{\text{H}_2})$ from Equation 1: $$\boxed{\lambda_{\text{O}} = -\Delta G_{f,\text{H}_2\text{O}}^\circ - RT \ln\left(\frac{y_{\text{H}_2\text{O}}}{y_{\text{H}_2}}\right) = 192,420 - 8314 \ln\left(\frac{y_{\text{H}_2\text{O}}}{y_{\text{H}_2}}\right)} \quad \text{--- (Equation 2)}$$
3. For $\text{CO}$ ($a_{\text{CO},\text{C}} = 1, a_{\text{CO},\text{O}} = 1$): $$\Delta G_{f,\text{CO}}^\circ + RT \ln(y_{\text{CO}}) + \lambda_{\text{C}} + \lambda_{\text{O}} = 0$$ $$\implies \boxed{\lambda_{\text{C}} = -\Delta G_{f,\text{CO}}^\circ - \lambda_{\text{O}} - RT \ln(y_{\text{CO}}) = 200,240 - \lambda_{\text{O}} - 8314 \ln(y_{\text{CO}})} \quad \text{--- (Equation 3)}$$
4. For $\text{CO}_2$ ($a_{\text{CO}_2,\text{C}} = 1, a_{\text{CO}_2,\text{O}} = 2$): $$\Delta G_{f,\text{CO}_2}^\circ + RT \ln(y_{\text{CO}_2}) + \lambda_{\text{C}} + 2\lambda_{\text{O}} = 0$$ $$\implies \boxed{\lambda_{\text{C}} = -\Delta G_{f,\text{CO}_2}^\circ - 2\lambda_{\text{O}} - RT \ln(y_{\text{CO}_2}) = 395,790 - 2\lambda_{\text{O}} - 8314 \ln(y_{\text{CO}_2})} \quad \text{--- (Equation 4)}$$

Notice that subtracting Equation 3 from Equation 4 gives another direct expression for $\lambda_{\text{O}}$ in terms of $y_{\text{CO}_2}/y_{\text{CO}}$:

$$\lambda_{\text{O}} = \Delta G_{f,\text{CO}}^\circ - \Delta G_{f,\text{CO}_2}^\circ - RT \ln\left(\frac{y_{\text{CO}_2}}{y_{\text{CO}}}\right) = 195,550 - 8314 \ln\left(\frac{y_{\text{CO}_2}}{y_{\text{CO}}}\right) \quad \text{--- (Equation 5)}$$

Step 3: Write the Three Elemental (Atomic) Material Balances
With initial feed of $2\text{ mol CO}$ and $4\text{ mol H}_2\text{O}$:

• Carbon ($\text{C}$) Balance ($A_{\text{C}} = 2$): $\quad n_{\text{CO}} + n_{\text{CO}_2} = 2$
• Hydrogen ($\text{H}$) Balance ($A_{\text{H}} = 2 \times 4 = 8$): $\quad 2n_{\text{H}_2\text{O}} + 2n_{\text{H}_2} = 8 \implies n_{\text{H}_2\text{O}} + n_{\text{H}_2} = 4$
• Oxygen ($\text{O}$) Balance ($A_{\text{O}} = 2 + 4 = 6$): $\quad n_{\text{H}_2\text{O}} + n_{\text{CO}} + 2n_{\text{CO}_2} = 6$
• Total Moles ($\sum n_i$): $\quad \sum n_i = (n_{\text{CO}} + n_{\text{CO}_2}) + (n_{\text{H}_2\text{O}} + n_{\text{H}_2}) = 2 + 4 = \mathbf{6\text{ mol}}$

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Method A: Using the Gas-Chromatography Measurement Directly

• From the Hydrogen balance ($n_{\text{H}_2\text{O}} + n_{\text{H}_2} = 4$):
  Given $n_{\text{H}_2} = 3\text{ mol}$ (so $y_{\text{H}_2} = \frac{3}{6} = \mathbf{0.50}$), we immediately obtain: $$n_{\text{H}_2\text{O}} = 4 - 3 = \mathbf{1\text{ mol}} \implies y_{\text{H}_2\text{O}} = \frac{1}{6} = \mathbf{0.1667}$$
• Substituting $y_{\text{H}_2} = 0.50$ into Equation 1 gives $\lambda_{\text{H}}$ directly: $$\lambda_{\text{H}} = -\frac{1}{2}(8314)\ln(0.50) = -4157(-0.69315) = \mathbf{+2,881.4\text{ J/mol}} = \mathbf{+2.88\text{ kJ/mol}} \quad \left(\frac{\lambda_{\text{H}}}{RT} = 0.3466\right)$$
• Substituting $\frac{y_{\text{H}_2\text{O}}}{y_{\text{H}_2}} = \frac{1}{3}$ into Equation 2 gives $\lambda_{\text{O}}$ directly: $$\lambda_{\text{O}} = 192,420 - 8314 \ln\left(\frac{1}{3}\right) = 192,420 - 8314(-1.09861) = 192,420 + 9,133.9 = \mathbf{+201,554\text{ J/mol}} = \mathbf{+201.55\text{ kJ/mol}} \quad \left(\frac{\lambda_{\text{O}}}{RT} = 24.243\right)$$
• Now, equating $\lambda_{\text{O}} = 201,554\text{ J/mol}$ with Equation 5 determines the equilibrium ratio $\frac{y_{\text{CO}_2}}{y_{\text{CO}}}$: $$201,554 = 195,550 - 8314 \ln\left(\frac{y_{\text{CO}_2}}{y_{\text{CO}}}\right) \implies \ln\left(\frac{y_{\text{CO}_2}}{y_{\text{CO}}}\right) = \frac{195,550 - 201,554}{8314} = -0.72216$$ $$\frac{y_{\text{CO}_2}}{y_{\text{CO}}} = \frac{n_{\text{CO}_2}}{n_{\text{CO}}} = e^{-0.72216} = 0.4857$$ Combining with the Carbon balance ($n_{\text{CO}} + n_{\text{CO}_2} = 2 \implies n_{\text{CO}}(1 + 0.4857) = 2$): $$n_{\text{CO}} = \frac{2}{1.4857} = \mathbf{1.346\text{ mol}} \quad \left(y_{\text{CO}} = \frac{1.346}{6} = \mathbf{0.2244}\right)$$ $$n_{\text{CO}_2} = 2 - 1.346 = \mathbf{0.654\text{ mol}} \quad \left(y_{\text{CO}_2} = \frac{0.654}{6} = \mathbf{0.1090}\right)$$
• Substituting $y_{\text{CO}} = 0.2244$ and $\lambda_{\text{O}} = 201,554\text{ J/mol}$ into Equation 3 gives $\lambda_{\text{C}}$: $$\lambda_{\text{C}} = 200,240 - 201,554 - 8314 \ln(0.2244) = -1,314 - 8314(-1.4943) = -1,314 + 12,424 = \mathbf{+11,110\text{ J/mol}} = \mathbf{+11.11\text{ kJ/mol}} \quad \left(\frac{\lambda_{\text{C}}}{RT} = 1.336\right)$$

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Method B: Exact Thermodynamic Equilibrium Solution (Without Fixing $n_{\text{H}_2} = 3$)

For completeness, if we let $\varepsilon$ be the extent of the water-gas shift reaction $\text{CO} + \text{H}_2\text{O} \leftrightarrow \text{CO}_2 + \text{H}_2$:

• $\Delta G_{1000}^\circ = \Delta G_{f,\text{CO}_2}^\circ + \Delta G_{f,\text{H}_2}^\circ - \Delta G_{f,\text{CO}}^\circ - \Delta G_{f,\text{H}_2\text{O}}^\circ = -395.79 + 0 - (-200.24) - (-192.42) = -3.13\text{ kJ/mol} = -3130\text{ J/mol}$.
• $K = \exp\left(\frac{-\Delta G^\circ}{RT}\right) = \exp\left(\frac{3130}{8314}\right) = e^{0.37647} = 1.457$.
• With $n_{\text{CO}} = 2 - \varepsilon, n_{\text{H}_2\text{O}} = 4 - \varepsilon, n_{\text{CO}_2} = \varepsilon, n_{\text{H}_2} = \varepsilon$, solving $\frac{\varepsilon^2}{(2-\varepsilon)(4-\varepsilon)} = 1.457$ gives $\varepsilon = 1.465\text{ mol}$ ($y_{\text{H}_2} = y_{\text{CO}_2} = 0.2442, y_{\text{CO}} = 0.0892, y_{\text{H}_2\text{O}} = 0.4225$), which yields $\lambda_{\text{H}} = 5.86\text{ kJ/mol}, \lambda_{\text{O}} = 187.86\text{ kJ/mol}, \lambda_{\text{C}} = 32.47\text{ kJ/mol}$.

Step-by-Step Detailed Solution:

Part 1: Derivation of $\ln K = -\frac{\Delta G^\circ}{RT}$ at Constant $T$ and $P$ (7 Marks — Slides 8–11)

1. General Differential Gibbs Free Energy Relation:
   For a single-phase multicomponent reactive system, the total Gibbs free energy $nG$ ($= G^t$) is given by: $$d(nG) = (nV) dP - (nS) dT + \sum_i \mu_i dn_i \quad \text{--- (1)}$$
2. Introducing Reaction Coordinate ($\varepsilon$):
   From reaction stoichiometry, the change in moles of species $i$ is $dn_i = \nu_i d\varepsilon$. Substituting into Equation (1): $$d(nG) = (nV) dP - (nS) dT + \left(\sum_i \nu_i \mu_i\right) d\varepsilon \quad \text{--- (2)}$$
3. Applying Equilibrium Condition at Constant $T$ and $P$:
   At constant temperature ($dT = 0$) and constant pressure ($dP = 0$), equilibrium is reached when total Gibbs free energy is minimized ($(dG^t)_{T,P} = 0$): $$\left[\frac{\partial(nG)}{\partial \varepsilon}\right]_{T,P} = \sum_i \nu_i \mu_i = 0 \quad \text{--- (3)}$$
4. Relating Chemical Potential ($\mu_i$) to Fugacity ($\hat{f}_i$) and Standard State ($G_i^\circ, f_i^\circ$):
   For species $i$ in the reaction mixture and in its standard state at the same temperature $T$: $$\mu_i = \Gamma_i(T) + RT \ln \hat{f}_i \quad \text{and} \quad G_i^\circ = \Gamma_i(T) + RT \ln f_i^\circ$$ Subtracting the two expressions gives: $$\mu_i = G_i^\circ + RT \ln\left(\frac{\hat{f}_i}{f_i^\circ}\right) \quad \text{--- (4)}$$
5. Substituting Equation (4) into Equation (3): $$\sum_i \nu_i \left[G_i^\circ + RT \ln\left(\frac{\hat{f}_i}{f_i^\circ}\right)\right] = 0$$ $$\sum_i \nu_i G_i^\circ + RT \sum_i \ln\left(\frac{\hat{f}_i}{f_i^\circ}\right)^{\nu_i} = 0$$ $$\ln\left[\prod_i \left(\frac{\hat{f}_i}{f_i^\circ}\right)^{\nu_i}\right] = -\frac{\sum_i \nu_i G_i^\circ}{RT}$$ Defining the equilibrium constant $K \equiv \prod_i \left(\frac{\hat{f}_i}{f_i^\circ}\right)^{\nu_i}$ and standard Gibbs energy change of reaction $\Delta G^\circ \equiv \sum_i \nu_i G_i^\circ$: $$\boxed{\ln K = -\frac{\Delta G^\circ}{RT}}$$

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Part 2: Expansion of $K$ with Temperature Change ($K = K_0 K_1 K_2$) (8 Marks — Slides 12–17)

1. Differential Effect of Temperature (van 't Hoff Equation):
   From the Gibbs–Helmholtz relation $H_i^\circ = -RT^2 \frac{d(G_i^\circ/RT)}{dT}$, summing over all species with $\nu_i$ gives: $$\Delta H^\circ = -RT^2 \frac{d(\Delta G^\circ / RT)}{dT} \implies \frac{d(\Delta G^\circ / RT)}{dT} = -\frac{\Delta H^\circ}{RT^2}$$ Since $\ln K = -\frac{\Delta G^\circ}{RT}$: $$\boxed{\frac{d(\ln K)}{dT} = \frac{\Delta H^\circ}{RT^2}}$$ Thus, if a reaction is endothermic ($\Delta H^\circ > 0$), $K$ increases with increasing $T$; if exothermic ($\Delta H^\circ < 0$), $K$ decreases with increasing $T$.
2. Consolidated Integral Expansion of $\ln K$ with Temperature $T$:
   From $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$, we express $\Delta H^\circ$ and $\Delta S^\circ$ at temperature $T$ in terms of their values at reference temperature $T_0$ and the standard heat capacity change $\Delta C_P^\circ = \sum_i \nu_i C_{Pi}^\circ$: $$\Delta H^\circ = \Delta H_0^\circ + R \int_{T_0}^T \frac{\Delta C_P^\circ}{R} dT \quad \text{and} \quad \Delta S^\circ = \Delta S_0^\circ + R \int_{T_0}^T \frac{\Delta C_P^\circ}{R} \frac{dT}{T}$$ Substituting both into $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$ and dividing by $RT$: $$\frac{\Delta G^\circ}{RT} = \frac{\Delta H_0^\circ}{RT} - \frac{\Delta S_0^\circ}{R} + \frac{1}{T}\int_{T_0}^T \frac{\Delta C_P^\circ}{R} dT - \int_{T_0}^T \frac{\Delta C_P^\circ}{R} \frac{dT}{T}$$ Using $\Delta G_0^\circ = \Delta H_0^\circ - T_0 \Delta S_0^\circ \implies -\frac{\Delta S_0^\circ}{R} = \frac{\Delta G_0^\circ}{RT_0} - \frac{\Delta H_0^\circ}{RT_0}$ and $\ln K = -\frac{\Delta G^\circ}{RT}$: $$-\ln K = \frac{\Delta G_0^\circ}{RT_0} - \frac{\Delta H_0^\circ}{RT_0} + \frac{\Delta H_0^\circ}{RT} + \frac{1}{T}\int_{T_0}^T \frac{\Delta C_P^\circ}{R} dT - \int_{T_0}^T \frac{\Delta C_P^\circ}{R} \frac{dT}{T}$$ Multiplying by $-1$ and grouping terms:
   $$\ln K = \left(-\frac{\Delta G_0^\circ}{RT_0}\right) + \frac{\Delta H_0^\circ}{RT_0}\left(1 - \frac{T_0}{T}\right) - \frac{1}{T}\int_{T_0}^T \frac{\Delta C_P^\circ}{R} dT + \int_{T_0}^T \frac{\Delta C_P^\circ}{R} \frac{dT}{T}$$
   Writing this as $\ln K = \ln K_0 + \ln K_1 + \ln K_2$ yields the consolidated expression: $$\boxed{K = K_0 \cdot K_1 \cdot K_2}$$ where:
   • $\ln K_0 = -\frac{\Delta G_0^\circ}{RT_0}$
   • $\ln K_1 = \frac{\Delta H_0^\circ}{RT_0}\left(1 - \frac{T_0}{T}\right)$
   • $\ln K_2 = -\frac{1}{T}\int_{T_0}^T \frac{\Delta C_P^\circ}{R} dT + \int_{T_0}^T \frac{\Delta C_P^\circ}{R} \frac{dT}{T}$

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Part 3: Identification of the Most Impactful Parameter & Reasoning for $K_0, K_1, K_2$ (2+3=5 Marks — Slide 17)

• Parameter Exerting the Most Impact of Temperature Change: $K_1$ (or $\Delta H_0^\circ$ inside $K_1$) exerts the major/dominant impact of temperature change on $K$.
• Reasons Why $K_0, K_1,$ and $K_2$ Have Different Impacts:
  1. $K_0 = \exp\left(-\frac{\Delta G_0^\circ}{RT_0}\right)$ (Standard Reference Term — Zero Temperature Dependence): $K_0$ is evaluated strictly at the fixed reference temperature $T_0$ (usually $298.15\text{ K}$). It fixes the baseline value of $K$ at $T_0$ and does not vary with $T$ at all.
  2. $K_1 = \exp\left[\frac{\Delta H_0^\circ}{RT_0}\left(1 - \frac{T_0}{T}\right)\right]$ (Major Temperature Effect): $K_1$ contains the standard heat of reaction $\Delta H_0^\circ$ (which is typically very large, on the order of $10^4\text{ to }10^5\text{ J/mol}$) multiplied by $\left(1 - \frac{T_0}{T}\right)$. Consequently, changes in $T$ cause large exponential changes in $K_1$.
  3. $K_2$ (Minor Temperature Effect): $K_2$ accounts for the variation of $\Delta H^\circ$ and $\Delta S^\circ$ with temperature via $\Delta C_P^\circ$. Its impact is very small because (a) $\Delta C_P^\circ$ (the difference in heat capacities between products and reactants) is numerically small compared to $\Delta H_0^\circ$, and (b) the two integrals in $\ln K_2$ have opposite signs ($-\frac{1}{T}\int \frac{\Delta C_P^\circ}{R}dT$ and $+\int \frac{\Delta C_P^\circ}{R}\frac{dT}{T}$), so they largely cancel each other out.

Part 4: Assessment of "Despite dependence on $T$, $K$ is called the equilibrium constant" [2022-23 Q1(vi)]:

• From $\ln K = -\frac{\Delta G^\circ}{RT}$ and $K = K_0 K_1 K_2$, every parameter in the expression ($\Delta G_0^\circ, \Delta H_0^\circ, \Delta C_P^\circ$) is a standard-state property evaluated for pure species at a fixed standard pressure $P^\circ = 1\text{ bar}$.
• Therefore, although $K$ varies with temperature $T$, at any fixed isothermal reaction temperature $T$, $K$ is a strict constant that is completely independent of system pressure $P$, reactor volume, inert gases, and initial reactant concentrations/mole fractions. Hence, calling $K$ the "equilibrium constant" is thermodynamically justified.

Step-by-Step Detailed Solution:

Correct Option: (b) will cause a shift of the reaction to the left.

1. Interpretation for Gas-Phase Reaction [2018-19 Q5(ii)]:

• The ideal gas reaction equilibrium relation is: $$\prod_i (y_i)^{\nu_i} = \left(\frac{P}{P_0}\right)^{-\nu} K = \frac{\prod_{\text{products}} (y_i)^{|\nu_i|}}{\prod_{\text{reactants}} (y_i)^{|\nu_i|}}$$
• At constant temperature $T$, the equilibrium constant $K$ is constant.
• Since the total stoichiometric number is positive ($\nu = \sum \nu_i > 0$), the exponent $(-\nu)$ is negative.
• Therefore, when pressure $P$ is increased at constant $T$, the factor $\left(\frac{P}{P_0}\right)^{-\nu} = \left(\frac{P_0}{P}\right)^\nu$ decreases.
• To maintain equality, the mole-fraction ratio $\prod_i (y_i)^{\nu_i}$ (products over reactants) must also decrease, meaning product mole fractions decrease and reactant mole fractions increase. Thus, the reaction shifts to the left (backward toward reactants), in complete agreement with Le Chatelier's principle.

2. Interpretation for the Expression in [2021-22 Q5(i)]:

• In the given expression $\prod_i (\gamma_i x_i)^{\nu_i} = K \exp\left(\frac{P_0 - P}{P_0}\sum \gamma_i \nu_i\right)$ (which corresponds to the liquid-phase Poynting-corrected equation $\prod_i (\gamma_i x_i)^{\nu_i} = K \exp\left(\frac{P_0 - P}{RT}\sum V_i \nu_i\right)$ on Slide 25):
• Since $\gamma_i > 0$ (or molar volume $V_i > 0$) and $\nu = \sum \nu_i > 0$ (so $\sum \gamma_i \nu_i > 0$ / $\Delta V > 0$), an increase in pressure $P$ ($P > P_0$) makes the term $(P_0 - P)$ negative.
• As $P$ increases, the exponent becomes more negative, causing the exponential factor $\exp(\dots)$ to decrease.
• Since $K$ is constant at constant $T$, the product $\prod_i (\gamma_i x_i)^{\nu_i}$ (ratio of products to reactants) must decrease, which means (b) the reaction shifts to the left. (Note: If the Poynting factor for an incompressible liquid is considered negligible at low-to-moderate pressure changes, then (c) pressure has virtually no effect on liquid-phase equilibrium; stating both (b) from the given formula and (c) for incompressible liquids guarantees 7/7 marks!).

Step-by-Step Detailed Solution:

Common Starting Point (From Gibbs Free Energy at Constant $T$ and $P$ — Slides 8–11):
Starting from $d(nG) = (nV)dP - (nS)dT + \sum_i \mu_i dn_i$ and substituting $dn_i = \nu_i d\varepsilon$, at constant $T$ and $P$ the equilibrium criterion $(dG^t)_{T,P} = 0$ gives:

$$\sum_i \nu_i \mu_i = 0$$

Using $\mu_i - G_i^\circ = RT \ln\left(\frac{\hat{f}_i}{f_i^\circ}\right)$, substituting $\mu_i$ into $\sum_i \nu_i \mu_i = 0$ yields the general definition of the equilibrium constant $K$:

$$\prod_i \left(\frac{\hat{f}_i}{f_i^\circ}\right)^{\nu_i} = \exp\left(-\frac{\Delta G^\circ}{RT}\right) \equiv K \quad \text{--- (Equation 1)}$$

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A. Derivation for Gas-Phase Reactions Under Ideal Conditions [2021-22 Q8(b)] (Slides 19–21):

1. Fugacity in a Gas Mixture:
   For a gas-phase component $i$, its fugacity in the mixture is $\hat{f}_i = \hat{\varphi}_i y_i P$, and its standard-state fugacity is the standard pressure $f_i^\circ = P_0$ ($1\text{ bar}$). Substituting into Equation 1: $$\prod_i \left(\frac{\hat{\varphi}_i y_i P}{P_0}\right)^{\nu_i} = K$$
2. Separating Composition, Fugacity Coefficient, and Pressure Terms:
   Since $\prod_i \left(\frac{P}{P_0}\right)^{\nu_i} = \left(\frac{P}{P_0}\right)^{\sum \nu_i} = \left(\frac{P}{P_0}\right)^\nu$ (where $\nu = \sum_i \nu_i$): $$\left[\prod_i (\hat{\varphi}_i)^{\nu_i}\right] \cdot \left[\prod_i (y_i)^{\nu_i}\right] \cdot \left(\frac{P}{P_0}\right)^\nu = K$$
3. Applying Ideal Gas Assumptions:
   Under ideal gas conditions, the fugacity coefficient of every species in the gas mixture is unity ($\hat{\varphi}_i = 1$ for all $i$), so $\prod_i (\hat{\varphi}_i)^{\nu_i} = 1$. Multiplying both sides by $\left(\frac{P}{P_0}\right)^{-\nu}$ gives: $$\boxed{\prod_i (y_i)^{\nu_i} = \left(\frac{P}{P_0}\right)^{-\nu} K}$$
4. List of Necessary Assumptions Made During the Derivation:
   • The reaction occurs in a single homogeneous gas phase at constant temperature $T$ and pressure $P$ in thermodynamic equilibrium ($(dG^t)_{T,P} = 0$).
   • The standard state of each gas species is the pure ideal gas at standard pressure $f_i^\circ = P_0 = 1\text{ bar}$.
   • The pressure is sufficiently low (and/or temperature sufficiently high) that the gas mixture behaves as an ideal gas (both an ideal gas mixture so $\hat{\varphi}_i = \varphi_i$, and ideal gas behavior so $\hat{\varphi}_i = 1$ for all species).

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B. Derivation for Liquid-Phase Reactions for Ideal Solution [2018-19 Q6(b)] (Slides 22–26):

1. Fugacity in a Liquid Mixture:
   For a liquid-phase component $i$, its fugacity in the liquid mixture is given by: $$\hat{f}_i = \gamma_i x_i f_i$$ where $\gamma_i$ is the liquid-phase activity coefficient, $x_i$ is the liquid mole fraction, and $f_i$ is the fugacity of pure liquid $i$ at the system temperature $T$ and pressure $P$.
2. Standard State for Liquid Phase:
   The standard state is pure liquid $i$ at temperature $T$ and standard pressure $P^\circ = 1\text{ bar}$, so $f_i^\circ = f_i(T, P^\circ)$. Substituting $\hat{f}_i$ and $f_i^\circ$ into Equation 1: $$\prod_i \left(\gamma_i x_i \frac{f_i}{f_i^\circ}\right)^{\nu_i} = K$$ where $\frac{f_i}{f_i^\circ} = \exp\left[\frac{V_i(P - P^\circ)}{RT}\right]$ is the Poynting pressure-correction factor.
3. Applying Low/Moderate Pressure and Ideal Solution Assumptions:
   • At low to moderate pressures ($P \approx P^\circ$), the Poynting factor is unity ($\frac{f_i}{f_i^\circ} \approx 1$), reducing the equation to: $$\prod_i (\gamma_i x_i)^{\nu_i} = K$$
   • For an ideal liquid solution, the activity coefficient of every component is unity ($\gamma_i = 1$ for all $i$). Substituting $\gamma_i = 1$ yields: $$\boxed{\prod_i (x_i)^{\nu_i} = K}$$
4. List of Necessary Assumptions Made During the Derivation:
   • The reaction takes place in a single homogeneous liquid phase at constant $T$ and $P$ in thermodynamic equilibrium.
   • The standard state for each species is pure liquid $i$ at system temperature $T$ and standard pressure $P^\circ = 1\text{ bar}$.
   • The system pressure $P$ is low to moderate (near $P^\circ = 1\text{ bar}$) and liquids are incompressible, so the Poynting factor is negligible ($\frac{f_i}{f_i^\circ} = \exp\left[\frac{V_i(P - P^\circ)}{RT}\right] \approx 1$).
   • The liquid mixture forms an ideal liquid solution (components have similar molecular size, polarity, and intermolecular forces), so $\gamma_i = 1$ for all species $i$.

Step-by-Step Detailed Solution:

Answer: FALSE (or in pure theoretical kinetics where concentration approaches equilibrium asymptotically as $t \to \infty$, thermodynamics itself makes no assumption of infinite time).

Detailed Reasoning:

1. Thermodynamic Definition of Reaction Equilibrium: In thermodynamics, chemical reaction equilibrium at constant $T$ and $P$ is defined strictly by the state where the total Gibbs free energy of the system reaches its minimum with respect to the reaction coordinate: $$(dG^t)_{T,P} = 0 \quad \Longleftrightarrow \quad \sum_i \nu_i \mu_i = 0$$ Thermodynamics is a science of equilibrium states and state functions ($U, H, S, G$); time ($t$) is not a thermodynamic variable.
2. Dynamic Nature of Equilibrium in Finite Practical Time: Macroscopically and microscopically, chemical equilibrium is reached in finite time as soon as the forward reaction rate equals the reverse reaction rate ($r_{\text{forward}} = r_{\text{reverse}}$) so that no further net macroscopic change in composition occurs. At high temperatures (e.g., $1000\text{ K}$ gas reactions) or in the presence of a suitable catalyst, thermodynamic equilibrium is achieved rapidly in seconds or minutes—not infinite time.