Step 1: Thermodynamic Foundation

The spontaneity of a chemical reaction is governed by the Gibbs Free Energy equation, which relates enthalpy ($\Delta H$), entropy ($\Delta S$), and absolute temperature ($T$):

$\Delta G = \Delta H - T\Delta S$ [Per the Second Law of Thermodynamics and the Gibbs-Helmholtz Equation]

Step 2: Analyzing the Given Parameters

The provided reaction is: $A + B \rightarrow C + D + q$.

• Enthalpy ($\Delta H$): Since heat ($q$) is released as a product, the reaction is exothermic. By convention, for an exothermic process, the change in enthalpy is negative ($\Delta H < 0$).
• Entropy ($\Delta S$): The problem explicitly states that the reaction has a positive entropy change ($\Delta S > 0$).

Step 3: Calculating Spontaneity ($\Delta G$)

Substitute the signs of the variables into the Gibbs Free Energy equation:

$\Delta G = (\text{negative}) - T(\text{positive})$

Since absolute temperature ($T$) is always positive ($T > 0$ in Kelvin), the term $T\Delta S$ is positive. Subtracting a positive value from a negative value results in a negative outcome:

$\Delta G = (\text{negative}) - (\text{positive}) = \text{negative}$

Step 4: Logical Deduction of Spontaneity

According to the criteria for thermodynamic spontaneity:

Because $\Delta H < 0$ and $\Delta S > 0$, the value of $\Delta G$ will be negative at all temperatures. Therefore, the reaction is spontaneous at all temperatures.

Final Solution: The reaction will be spontaneous at all temperatures because both the enthalpy change ($\Delta H < 0$) and the entropy change ($\Delta S > 0$) favor spontaneity, resulting in a negative Gibbs free energy change ($\Delta G < 0$).

Step 1: Defining the Thermodynamic System

To analyze the condition for an adiabatic process, we refer to the First Law of Thermodynamics, which is expressed by the mathematical equation:

$\Delta U = q + w$

Where:

• $\Delta U$ represents the change in internal energy of the system.
• $q$ represents the heat exchanged between the system and the surroundings.
• $w$ represents the work done on or by the system.

Step 2: Defining Adiabatic Constraints

An adiabatic process is defined as a thermodynamic process in which there is no transfer of heat ($q$) into or out of the system. The system is considered to be thermally isolated from its surroundings.

Mathematically, for any adiabatic process:

$q = 0$

Step 3: Synthesizing the Result

Substituting the adiabatic condition ($q = 0$) into the First Law of Thermodynamics equation:

$\Delta U = 0 + w$

$\Delta U = w$

This implies that in an adiabatic process, any change in internal energy is exclusively due to the work done on or by the system.

Step 4: Comparative Analysis

Final Solution: The correct condition for a process to occur under adiabatic conditions is $q = 0$.

Given Variables & Initial Setup

The chemical reaction for the combustion of cyanamide is given by:

$NH_2CN(s) + \frac{3}{2}O_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(l)$

• Internal energy change ($\Delta U$): $-742.7 \text{ kJ mol}^{-1}$
• Temperature ($T$): $298 \text{ K}$
• Universal gas constant ($R$): $8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}$
• State of reactants/products: $NH_2CN(s)$, $O_2(g)$, $N_2(g)$, $CO_2(g)$, $H_2O(l)$

Step 1: Determining the change in gaseous moles ($\Delta n_g$)

The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) is governed by the equation [Per the First Law of Thermodynamics, $\Delta H = \Delta U + \Delta n_g RT$].

First, identify the stoichiometric coefficients of gaseous species:

Calculating $\Delta n_g = \sum n_{products(g)} - \sum n_{reactants(g)}$:

$\Delta n_g = (1 + 1) - (1.5) = 2 - 1.5 = 0.5 \text{ mol}$

Step 2: Calculation of Enthalpy Change ($\Delta H$)

Substitute the known values into the enthalpy relation:

$\Delta H = \Delta U + \Delta n_g RT$

Substituting values:

$\Delta H = -742.7 \text{ kJ mol}^{-1} + (0.5 \text{ mol}) \times (8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K})$

Step 3: Arithmetic Resolution

Perform the product calculation for the work component:

Work component $= 0.5 \times 8.314 \times 10^{-3} \times 298$

Work component $= 0.5 \times 2.477572 = 1.238786 \text{ kJ mol}^{-1}$

Final summation:

$\Delta H = -742.7 + 1.2388$

$\Delta H = -741.4612 \text{ kJ mol}^{-1}$

Final Solution: The enthalpy change for the reaction at 298 K is $-741.46 \text{ kJ mol}^{-1}$.

Step 1: Given Variables & Conversion to Standard SI Units

To ensure consistency in energy units, all values must be converted to Joules (J) and the temperature must be defined in Kelvin (K). The standard temperature for thermodynamic data is generally $T = 298 \text{ K}$ unless otherwise specified.

• $\Delta U^\circ = -10.5 \text{ kJ} = -10500 \text{ J}$ [Conversion factor: $1 \text{ kJ} = 10^3 \text{ J}$]
• $\Delta S^\circ = -44.1 \text{ JK}^{-1}$
• $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - (2 + 1) = -1 \text{ mol}$
• $T = 298 \text{ K}$
• $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$ [Universal Gas Constant]

Step 2: Calculating Enthalpy Change ($\Delta H^\circ$)

The relationship between internal energy change ($\Delta U^\circ$) and enthalpy change ($\Delta H^\circ$) is governed by the equation [Per the First Law of Thermodynamics and the definition of Enthalpy]:

$\Delta H^\circ = \Delta U^\circ + \Delta n_g RT$

Substituting the values:

$\Delta H^\circ = -10500 \text{ J} + (-1 \text{ mol}) \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K})$

$\Delta H^\circ = -10500 \text{ J} - 2477.57 \text{ J}$

$\Delta H^\circ = -12977.57 \text{ J}$

Step 3: Calculating Gibbs Free Energy ($\Delta G^\circ$)

The standard Gibbs free energy is calculated using the Gibbs-Helmholtz equation [Gibbs-Helmholtz relation]:

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

Substituting the values:

$\Delta G^\circ = -12977.57 \text{ J} - (298 \text{ K}) \times (-44.1 \text{ J K}^{-1})$

$\Delta G^\circ = -12977.57 \text{ J} + 13141.8 \text{ J}$

$\Delta G^\circ = +164.23 \text{ J}$

Step 4: Predict Spontaneity

The spontaneity of a reaction at constant temperature and pressure is determined by the sign of $\Delta G^\circ$ [Criterion for spontaneity]:

• If $\Delta G^\circ < 0$, the reaction is spontaneous.
• If $\Delta G^\circ > 0$, the reaction is non-spontaneous.

Since $\Delta G^\circ = +164.23 \text{ J} > 0$, the reaction is non-spontaneous under standard conditions at 298 K.

Final Solution: The standard Gibbs free energy change $\Delta G^\circ$ is $+164.23 \text{ J}$, and the reaction is non-spontaneous.

Step 1: Given Variables & Initial Parameters

• Mass of Aluminium ($m$): $60.0 \text{ g}$
• Initial Temperature ($T_i$): $35^\circ\text{C}$
• Final Temperature ($T_f$): $55^\circ\text{C}$
• Molar Heat Capacity ($C_m$): $24 \text{ J mol}^{-1} \text{ K}^{-1}$
• Molar Mass of Aluminium ($M$): $26.98 \text{ g mol}^{-1}$ [Based on Periodic Table standard atomic weight]

Step 2: Conversion to Molar Quantity

To use the molar heat capacity, the mass must be converted into moles ($n$):

$n = \frac{m}{M} = \frac{60.0 \text{ g}}{26.98 \text{ g mol}^{-1}}$

$n \approx 2.2238 \text{ mol}$

Step 3: Temperature Change Calculation

The change in temperature ($\Delta T$) is identical in Kelvin and Celsius scales [Per the definition of the Kelvin scale offset]:

$\Delta T = T_f - T_i = 55^\circ\text{C} - 35^\circ\text{C} = 20 \text{ K}$

Step 4: Application of Heat Capacity Formula

The total heat ($q$) required is given by the product of moles, molar heat capacity, and change in temperature [Per the definition of $q = n \cdot C_m \cdot \Delta T$]:

$q = n \cdot C_m \cdot \Delta T$

$q = 2.2238 \text{ mol} \times 24 \text{ J mol}^{-1} \text{ K}^{-1} \times 20 \text{ K}$

$q = 2.2238 \times 480 \text{ J}$

$q = 1067.424 \text{ J}$

Step 5: Unit Conversion to Kilojoules

To obtain the result in kJ, convert Joules to Kilojoules using the factor $1 \text{ kJ} = 1000 \text{ J}$:

$q (\text{kJ}) = \frac{1067.424 \text{ J}}{1000}$

$q \approx 1.067 \text{ kJ}$

Visual Representation of Thermodynamic Energy Absorption

Final Solution: The total heat necessary is 1.067 kJ.

Step 1: Identifying Variables and Sign Conventions

To determine the change in internal energy ($\Delta U$), we apply the First Law of Thermodynamics. We must first establish the sign conventions based on the IUPAC system:

• Heat ($q$) absorbed by the system is positive: $q = +701\text{ J}$ [Per the IUPAC sign convention for thermodynamic heat transfer].
• Work ($w$) done by the system is negative: $w = -394\text{ J}$ [Per the IUPAC sign convention, energy leaving the system as work is negative].

Step 2: Mathematical Formulation

The First Law of Thermodynamics defines the change in internal energy as the sum of heat exchange and work done on/by the system:

$\Delta U = q + w$ [By the First Law of Thermodynamics, which states the conservation of energy in a closed system].

Step 3: Execution of Calculation

Substituting the identified values into the governing equation:

$\Delta U = (+701\text{ J}) + (-394\text{ J})$

$\Delta U = 701\text{ J} - 394\text{ J}$

$\Delta U = 307\text{ J}$

Visual Representation of Energy Flux

Analytical Summary

The positive value of $\Delta U$ indicates that the internal energy of the system has increased during the process. This confirms that the energy absorbed as heat ($701\text{ J}$) was greater than the energy expended by the system to perform work ($394\text{ J}$), resulting in a net gain of $307\text{ J}$.

Final Solution: The change in internal energy ($\Delta U$) for the process is 307 J.

Step 1: Establishing the Thermochemical Equations

To determine the enthalpy change for the reaction $\text{CCl}_4(g) \rightarrow \text{C}(g) + 4\text{Cl}(g)$, we must utilize Hess's Law of Constant Heat Summation [Hess's Law]. We are given the following standard enthalpy values ($kJ \cdot mol^{-1}$):

• $\Delta_f H^{\circ}(\text{CCl}_4, l) = -135.5$
• $\Delta_{vap} H^{\circ}(\text{CCl}_4) = 30.5$
• $\Delta_a H^{\circ}(\text{C}, s) = 715.0$
• $\Delta_a H^{\circ}(\text{Cl}_2, g) = 242.0$

First, we determine the enthalpy of formation for gaseous carbon tetrachloride $\text{CCl}_4(g)$ using the enthalpy of vaporization: $\Delta_f H^{\circ}(\text{CCl}_4, g) = \Delta_f H^{\circ}(\text{CCl}_4, l) + \Delta_{vap} H^{\circ}(\text{CCl}_4) = -135.5 + 30.5 = -105.0\text{ kJ mol}^{-1}$.

Step 2: Constructing the Enthalpy Cycle

The total enthalpy change ($\Delta_r H^{\circ}$) for the process $\text{CCl}_4(g) \rightarrow \text{C}(g) + 4\text{Cl}(g)$ corresponds to the sum of atomization energies minus the enthalpy of formation of the reactant gas:

$\Delta_r H^{\circ} = [\Delta_a H^{\circ}(\text{C}) + 4 \times \Delta_a H^{\circ}(\text{Cl})] - \Delta_f H^{\circ}(\text{CCl}_4, g)$

Note: The enthalpy of atomization of $\text{Cl}$ is half the bond dissociation energy of $\text{Cl}_2$: $\Delta_a H^{\circ}(\text{Cl}) = \frac{1}{2} \Delta_a H^{\circ}(\text{Cl}_2) = \frac{242}{2} = 121.0\text{ kJ mol}^{-1}$.

Step 3: Numerical Calculation

Substituting the values into the equation derived in Step 2:

$\Delta_r H^{\circ} = [715.0 + 4(121.0)] - (-105.0)$

$\Delta_r H^{\circ} = [715.0 + 484.0] + 105.0$

$\Delta_r H^{\circ} = 1199.0 + 105.0 = 1304.0\text{ kJ mol}^{-1}$

Step 4: Calculation of Average Bond Enthalpy

In $\text{CCl}_4(g)$, there are four identical $\text{C-Cl}$ bonds. The average bond enthalpy is calculated by dividing the total enthalpy change for the atomization of the gaseous molecule by the number of bonds [Definition of Mean Bond Enthalpy]:

$\text{Bond Enthalpy (C-Cl)} = \frac{\Delta_r H^{\circ}}{4} = \frac{1304.0}{4} = 326.0\text{ kJ mol}^{-1}$

Visual Representation of the Bond Breaking Process

Final Solution: The enthalpy change for the process is $1304.0\text{ kJ mol}^{-1}$ and the average bond enthalpy of (C–Cl) is $326.0\text{ kJ mol}^{-1}$.

Step 1: Analyzing the Enthalpy Change ($\Delta H$)

The reaction involves the formation of a chemical bond between two gaseous chlorine atoms to form a diatomic chlorine molecule: $2\text{Cl}(g) \rightarrow \text{Cl}_2(g)$.

• Principle: Bond formation is an exothermic process. Energy is released when electrons move into a more stable, lower-energy orbital configuration within a covalent bond.
• Theoretical Justification: Per the definition of bond dissociation enthalpy, breaking a bond requires energy ($\Delta H > 0$); conversely, forming a bond releases energy [By the Law of Conservation of Energy].
• Conclusion: Since the system releases heat to the surroundings, $\Delta H$ is negative ($\Delta H < 0$).

Step 2: Analyzing the Entropy Change ($\Delta S$)

Entropy ($S$) is a measure of the degree of randomness or disorder in a system. We compare the state of the system before and after the reaction:

• Reactants: $2$ moles of independent, monatomic chlorine gas particles.
• Products: $1$ mole of a diatomic chlorine gas molecule.
• Analytical Assessment: The total number of independent particles decreases from $2$ to $1$. A reduction in the number of particles directly correlates to a decrease in the number of available microstates and spatial configurations [Per the Third Law of Thermodynamics and statistical mechanics].
• Conclusion: Because the system progresses toward a more ordered state (lower number of particles), the change in entropy is negative ($\Delta S < 0$).

Step 3: Visual Representation of Atomic Transition

Step 4: Summary Table

Final Solution: For the reaction $2\text{Cl}(g) \rightarrow \text{Cl}_2(g)$, the enthalpy change ($\Delta H$) is negative and the entropy change ($\Delta S$) is negative.

Theoretical Foundation: The Standard State Convention

In chemical thermodynamics, the absolute enthalpy ($H$) of a substance cannot be measured directly because it includes various forms of internal energy (electronic, vibrational, rotational, and translational). Consequently, thermochemistry relies on enthalpy changes ($\Delta H$) measured relative to a defined reference point.

Step 1: Defining the Reference Point

By international scientific convention established by the IUPAC (International Union of Pure and Applied Chemistry), the enthalpy of formation ($\Delta_fH^\circ$) for any element in its most stable, natural physical state at standard conditions ($T = 298.15 \, \text{K}$, $P = 1 \, \text{bar}$) is defined as zero.

• Standard state: The most stable form of an element at $1 \, \text{bar}$ pressure and a specified temperature (usually $298.15 \, \text{K}$).
• Mathematical assignment: $\Delta_fH^\circ (\text{element, standard state}) = 0 \, \text{kJ/mol}$.

Step 2: Justification and Logical Basis

[Per the First Law of Thermodynamics and the Principle of State Functions], since enthalpy is a state function dependent only on the current configuration, we require a "zero-point" scale to calculate the enthalpy of compounds. By assigning zero to the elements from which these compounds are formed, we simplify the calculation of enthalpy changes for reactions: $\Delta_rH^\circ = \sum \Delta_fH^\circ (\text{products}) - \sum \Delta_fH^\circ (\text{reactants})$.

Step 3: Comparative Analysis of Elemental States

Visual Representation of the Convention

Final Solution: The enthalpies of all elements in their standard states are defined as zero.

Step 1: Establishing the Thermodynamic Relationship

The relationship between the standard Gibbs free energy change ($\Delta G^\circ$) and the equilibrium constant ($K_{eq}$) for a chemical reaction is defined by the Gibbs isotherm equation at equilibrium [Per the Second Law of Thermodynamics and the definition of chemical equilibrium]:

$\Delta G^\circ = -RT \ln K_{eq}$

Step 2: Identification of Given Variables

• Gas Constant ($R$) = $8.314 \, \text{J K}^{-1} \text{ mol}^{-1}$
• Temperature ($T$) = $300 \, \text{K}$
• Equilibrium Constant ($K_{eq}$) = $10$

Step 3: Conversion of Natural Logarithm

To compute the value, we convert the natural logarithm ($\ln$) to base-10 logarithm ($\log_{10}$) using the conversion factor $\ln(x) = 2.303 \log_{10}(x)$:

$\Delta G^\circ = -2.303 \cdot R \cdot T \cdot \log_{10}(K_{eq})$

Step 4: Substitution and Calculation

Substitute the given values into the derived equation:

$\Delta G^\circ = -2.303 \times 8.314 \, \text{J K}^{-1} \text{ mol}^{-1} \times 300 \, \text{K} \times \log_{10}(10)$

Since $\log_{10}(10) = 1$ [By logarithmic identity $\log_b(b) = 1$]:

$\Delta G^\circ = -2.303 \times 8.314 \times 300 \times 1$

$\Delta G^\circ = -2.303 \times 2494.2$

$\Delta G^\circ = -5744.1426 \, \text{J mol}^{-1}$

Step 5: Unit Conversion to Kilojoules

Expressing the result in $\text{kJ mol}^{-1}$ for standard scientific reporting:

$\Delta G^\circ = \frac{-5744.1426}{1000} \, \text{kJ mol}^{-1}$

$\Delta G^\circ \approx -5.744 \, \text{kJ mol}^{-1}$

Visual Representation of Gibbs Energy Directionality

Final Solution: $\Delta G^\circ = -5.744 \text{ kJ mol}^{-1}$

Step 1: Establishing the Target Equation

The standard enthalpy of formation ($\Delta_{f}H^{\circ}$) for methanol ($\text{CH}_{3}\text{OH(l)}$) is defined as the enthalpy change when $1\text{ mole}$ of the compound is formed from its constituent elements in their most stable standard states. The target thermochemical equation is:

$\text{C(graphite)} + 2\text{H}_{2}\text{g)} + \frac{1}{2}\text{O}_{2}\text{g)} \rightarrow \text{CH}_{3}\text{OH(l)}$

Step 2: Defining the Given Thermochemical Equations

We are provided with the following three equations:

• (i) $\text{CH}_{3}\text{OH(l)} + \frac{3}{2}\text{O}_{2}\text{g)} \rightarrow \text{CO}_{2}\text{g)} + 2\text{H}_{2}\text{O(l)}$; $\Delta_{r}H^{\circ}_{1} = -726\text{ kJ mol}^{-1}$
• (ii) $\text{C(graphite)} + \text{O}_{2}\text{g)} \rightarrow \text{CO}_{2}\text{g)}$; $\Delta_{r}H^{\circ}_{2} = -393\text{ kJ mol}^{-1}$
• (iii) $\text{H}_{2}\text{g)} + \frac{1}{2}\text{O}_{2}\text{g)} \rightarrow \text{H}_{2}\text{O(l)}$; $\Delta_{r}H^{\circ}_{3} = -286\text{ kJ mol}^{-1}$

Step 3: Manipulating Equations via Hess's Law

To obtain the target equation, we apply algebraic operations to the given equations [Per Hess's Law of Constant Heat Summation]:

1. Retain Equation (ii) to place $\text{C(graphite)}$ on the reactant side:
   $\text{C(graphite)} + \text{O}_{2}\text{g)} \rightarrow \text{CO}_{2}\text{g)}$; $\Delta H = -393\text{ kJ mol}^{-1}$
2. Multiply Equation (iii) by 2 to obtain $2\text{H}_{2}\text{g)}$:
   $2\text{H}_{2}\text{g)} + \text{O}_{2}\text{g)} \rightarrow 2\text{H}_{2}\text{O(l)}$; $\Delta H = 2 \times (-286) = -572\text{ kJ mol}^{-1}$
3. Reverse Equation (i) to place $\text{CH}_{3}\text{OH(l)}$ on the product side:
   $\text{CO}_{2}\text{g)} + 2\text{H}_{2}\text{O(l)} \rightarrow \text{CH}_{3}\text{OH(l)} + \frac{3}{2}\text{O}_{2}\text{g)}$; $\Delta H = +726\text{ kJ mol}^{-1}$

Step 4: Algebraic Summation

Adding the modified equations and their respective enthalpies:

Canceling common species on both sides:
$(\text{O}_{2} + \text{O}_{2} = 2\text{O}_{2})$ on the left and $\frac{3}{2}\text{O}_{2}$ on the right leaves $\frac{1}{2}\text{O}_{2}$ on the left.
$\text{CO}_{2}$ and $2\text{H}_{2}\text{O}$ cancel entirely.

Step 5: Final Calculation

$\Delta_{f}H^{\circ} = (-393) + (-572) + (726)$

$\Delta_{f}H^{\circ} = -965 + 726$

$\Delta_{f}H^{\circ} = -239\text{ kJ mol}^{-1}$

Final Solution: The standard enthalpy of formation of $\text{CH}_{3}\text{OH(l)}$ is $-239\text{ kJ mol}^{-1}$.

1. Theoretical Foundation of Adiabatic Systems

An adiabatic process is a thermodynamic process in which there is no heat transfer between the system and its surroundings. By definition, the boundary of an adiabatic system is thermally insulated [Per the First Law of Thermodynamics].

2. Applying the First Law of Thermodynamics

The First Law of Thermodynamics is mathematically expressed as:

$\Delta U = q + w$

Where:

• $\Delta U$: Change in internal energy of the system.
• $q$: Heat exchanged by the system with the surroundings.
• $w$: Work done on or by the system.

3. Deriving the Adiabatic Condition

Since the process is adiabatic, the system is perfectly insulated, ensuring that no thermal energy crosses the boundary. Therefore, the heat exchange is constrained to:

$q = 0$

Substituting this constraint into the First Law of Thermodynamics:

$\Delta U = 0 + w$

$\Delta U = w$

4. Visual Representation of the Adiabatic Boundary

5. Analytical Summary Table

Final Solution: The correct condition for a process occurring under adiabatic conditions is $q = 0$.

Step 1: Theoretical Definition of Enthalpy of Formation

The standard enthalpy of formation ($\Delta_fH^o$) of a substance is defined as the change in enthalpy when one mole of a compound is formed from its constituent elements in their most stable reference states under standard conditions ($1 \text{ bar}$ pressure and usually $298.15 \text{ K}$).

Step 2: Analysis of the Given Thermochemical Equation

The provided equation is:

$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$; $\Delta_rH^o = -92.4 \text{ kJ mol}^{-1}$

[Per the definition of reaction enthalpy, the value $-92.4 \text{ kJ mol}^{-1}$ represents the heat evolved when $2$ moles of $NH_3$ are formed from $1$ mole of $N_2$ and $3$ moles of $H_2$.]

Step 3: Calculating Enthalpy per Mole

To find the standard enthalpy of formation of $NH_3$, we must normalize the enthalpy change for the production of exactly $1 \text{ mole}$ of the product. Since the formation of $2 \text{ moles}$ of $NH_3$ releases $92.4 \text{ kJ}$, we divide the total enthalpy change by the stoichiometric coefficient of $NH_3$.

$$\Delta_fH^o(NH_3) = \frac{\Delta_rH^o}{\nu_{NH_3}}$$

[By the Principle of Stoichiometric Proportionality]

Substituting the values:

$$\Delta_fH^o(NH_3) = \frac{-92.4 \text{ kJ mol}^{-1}}{2}$$

$$\Delta_fH^o(NH_3) = -46.2 \text{ kJ mol}^{-1}$$

Step 4: Verification of Standard States

The reactants $N_2(g)$ and $H_2(g)$ are in their most stable elemental forms at standard conditions. The enthalpy of formation for elements in their standard state is defined as zero. Thus, the change in enthalpy for the reaction represents the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants:

$$\Delta_rH^o = [2 \times \Delta_fH^o(NH_3)] - [\Delta_fH^o(N_2) + 3 \times \Delta_fH^o(H_2)]$$

Since $\Delta_fH^o(N_2) = 0$ and $\Delta_fH^o(H_2) = 0$ [By the Standard Enthalpy Convention],

$$-92.4 \text{ kJ mol}^{-1} = 2 \times \Delta_fH^o(NH_3)$$

$$\Delta_fH^o(NH_3) = -46.2 \text{ kJ mol}^{-1}$$

Final Solution: The standard enthalpy of formation of $NH_3$ gas is $-46.2 \text{ kJ mol}^{-1}$.

Step 1: Identifying Given Parameters and Thermodynamic Relations

Given the chemical reaction: $2\text{A(g)} + \text{B(g)} \rightarrow 2\text{D(g)}$

• $\Delta U^{\circ} = -10.5 \text{ kJ mol}^{-1} = -10500 \text{ J mol}^{-1}$
• $\Delta S^{\circ} = -44.1 \text{ J K}^{-1} \text{ mol}^{-1}$
• Temperature ($T$) is assumed to be $298 \text{ K}$ (Standard Ambient Temperature and Pressure, unless otherwise specified).
• Change in moles of gas ($\Delta n_g$) = $(\sum n_g)_{\text{products}} - (\sum n_g)_{\text{reactants}} = 2 - (2 + 1) = -1 \text{ mol}$.

Step 2: Calculation of Enthalpy Change ($\Delta H^{\circ}$)

The relationship between enthalpy change and internal energy change is given by the formula [Per the First Law of Thermodynamics and definition of Enthalpy]:

$\Delta H^{\circ} = \Delta U^{\circ} + \Delta n_g RT$

Substituting the values with $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$:

$\Delta H^{\circ} = -10500 \text{ J} + (-1 \text{ mol}) \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K})$

$\Delta H^{\circ} = -10500 - 2477.57 \text{ J}$

$\Delta H^{\circ} = -12977.57 \text{ J mol}^{-1} = -12.978 \text{ kJ mol}^{-1}$

Step 3: Calculation of Gibbs Free Energy Change ($\Delta G^{\circ}$)

The Gibbs-Helmholtz equation defines the relationship between enthalpy, entropy, and Gibbs free energy [Per the Second Law of Thermodynamics]:

$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$

$\Delta G^{\circ} = -12977.57 \text{ J} - (298 \text{ K} \times -44.1 \text{ J K}^{-1})$

$\Delta G^{\circ} = -12977.57 + 13141.8 \text{ J}$

$\Delta G^{\circ} = +164.23 \text{ J mol}^{-1} = +0.164 \text{ kJ mol}^{-1}$

Step 4: Analysis of Spontaneity

The spontaneity of a reaction at constant temperature and pressure is determined by the sign of $\Delta G$ [Criterion for Spontaneity]:

Since $\Delta G^{\circ} = +0.164 \text{ kJ mol}^{-1} > 0$, the reaction is non-spontaneous under standard conditions at $298 \text{ K}$.

Final Solution: $\Delta G^{\circ} = +0.164 \text{ kJ mol}^{-1}$. The reaction is non-spontaneous at 298 K.

Theoretical Foundation: The Second Law of Thermodynamics

In an isolated system, the First Law of Thermodynamics states that the internal energy ($U$) remains constant, leading to $\Delta U = 0$. To determine the change in entropy ($\Delta S$) for such a system, we must invoke the Second Law of Thermodynamics, which dictates the directionality of spontaneous processes.

Step 1: Defining the Constraints of an Isolated System

An isolated system is defined by its inability to exchange both matter and energy (heat or work) with its surroundings. Therefore, for an isolated system:

• $q = 0$ [Per the definition of an isolated boundary]
• $w = 0$ [As no work is done on or by the surroundings]
• $\Delta U = q + w = 0$ [Per the First Law of Thermodynamics]

Step 2: Applying the Clausius Inequality

The entropy change for any process is governed by the relation $\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings}$. For an isolated system, the surroundings are unaffected by the processes occurring within the system, meaning $\Delta S_{surroundings} = 0$. Thus, $\Delta S_{total} = \Delta S_{system}$.

The Second Law states that for any spontaneous process occurring in an isolated system, the total entropy must increase:

$\Delta S_{total} > 0$ [Criterion for spontaneous processes]

Step 3: Equilibrium Conditions

If the isolated system is at equilibrium, no further spontaneous changes occur. In this state, the entropy reaches its maximum possible value. Consequently, the change in entropy for a reversible process (at equilibrium) is zero:

$\Delta S = 0$ [Criterion for equilibrium]

Analytical Visualization of Entropy Evolution

Summary of Deductive Logic

Final Solution: For an isolated system, the change in entropy ($\Delta S$) will be positive ($\Delta S > 0$) if the process is spontaneous, and zero ($\Delta S = 0$) if the system is at thermodynamic equilibrium.

Theoretical Foundation: Standard Enthalpy of Formation

In chemical thermodynamics, the enthalpy of an element is a relative property. Absolute enthalpies cannot be determined directly; therefore, a reference point is established [Per the conventions of the International Union of Pure and Applied Chemistry (IUPAC)].

Step 1: Defining the Standard State

The standard state of an element is defined as its most stable physical form at a pressure of $1 \, \text{bar}$ ($10^5 \, \text{Pa}$) and a specified temperature, commonly $298.15 \, \text{K}$. For example:

• Oxygen exists as $O_{2(g)}$
• Carbon exists as graphite $C_{(graphite)}$
• Sodium exists as $Na_{(s)}$

Step 2: The Convention of Zero Enthalpy

By international convention, the enthalpy of formation ($\Delta_fH^\ominus$) of an element in its most stable standard state is defined as zero. This serves as the baseline for calculating the enthalpy changes of chemical reactions ($\Delta_rH^\ominus$) [Based on the Enthalpy of Formation Convention].

Step 3: Analytical Verification

Consider the formation of Carbon Dioxide:

$C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}$

The standard enthalpy of reaction is calculated as:

$\Delta_rH^\ominus = \sum \Delta_fH^\ominus (\text{products}) - \sum \Delta_fH^\ominus (\text{reactants})$

$\Delta_rH^\ominus = [\Delta_fH^\ominus(CO_{2(g)})] - [\Delta_fH^\ominus(C_{(graphite)}) + \Delta_fH^\ominus(O_{2(g)})]$

Since $\Delta_fH^\ominus(C_{(graphite)}) = 0$ and $\Delta_fH^\ominus(O_{2(g)}) = 0$, then $\Delta_rH^\ominus = \Delta_fH^\ominus(CO_{2(g)})$ [By the Hess's Law of Constant Heat Summation].

Final Solution: The enthalpies of all elements in their standard states are defined as zero ($0$).

Step 1: Theoretical Foundation of Standard Enthalpy of Formation

The standard enthalpy of formation ($\Delta_{f}H^{\circ}$) of a compound is defined as the change in enthalpy when one mole of the substance is formed from its constituent elements in their most stable reference states at a pressure of $1 \text{ bar}$ and a specified temperature (usually $298.15 \text{ K}$). [Per IUPAC definition of standard enthalpy of formation].

Step 2: Analysis of the Given Thermochemical Equation

The provided equation is:

$\text{N}_{2}(g) + 3\text{H}_{2}(g) \rightarrow 2\text{NH}_{3}(g); \Delta_{r}H^{\circ} = -92.4 \text{ kJ mol}^{-1}$

In this balanced equation, the enthalpy change ($\Delta_{r}H^{\circ}$) corresponds to the formation of two moles of $\text{NH}_{3}(g)$ from its elements, nitrogen gas ($\text{N}_{2}$) and hydrogen gas ($\text{H}_{2}$), which are in their standard states.

Step 3: Algebraic Normalization for One Mole

To determine the enthalpy of formation for one mole of $\text{NH}_{3}$, the thermochemical equation must be divided by the stoichiometric coefficient of the product ($\text{NH}_{3}$), which is $2$.

$\frac{1}{2}\text{N}_{2}(g) + \frac{3}{2}\text{H}_{2}(g) \rightarrow 1\text{NH}_{3}(g)$

The enthalpy change for this normalized reaction is given by the relation:

$\Delta_{f}H^{\circ}(\text{NH}_{3}) = \frac{\Delta_{r}H^{\circ}}{n}$

Where $n$ is the number of moles of the product formed.

Step 4: Calculation

Substituting the given values into the equation:

$\Delta_{f}H^{\circ}(\text{NH}_{3}) = \frac{-92.4 \text{ kJ mol}^{-1}}{2}$

$\Delta_{f}H^{\circ}(\text{NH}_{3}) = -46.2 \text{ kJ mol}^{-1}$

Visual Representation of Reaction Coordinate

Final Solution: The standard enthalpy of formation of $\text{NH}_{3}(g)$ is $-46.2 \text{ kJ mol}^{-1}$.

Step 1: Define the Target Reaction

The standard enthalpy of formation ($\Delta_fH^\circ$) for $CH_3OH(l)$ refers to the enthalpy change when one mole of $CH_3OH(l)$ is formed from its constituent elements in their standard states. The balanced thermochemical equation is:

$C(graphite) + 2H_2(g) + \frac{1}{2}O_2(g) \rightarrow CH_3OH(l)$

Step 2: List Given Thermochemical Equations

We are provided with the following data:

• (i) $CH_3OH(l) + \frac{3}{2}O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$ ; $\Delta_rH^\circ_1 = -726 \text{ kJ mol}^{-1}$
• (ii) $C(graphite) + O_2(g) \rightarrow CO_2(g)$ ; $\Delta_rH^\circ_2 = -393 \text{ kJ mol}^{-1}$
• (iii) $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$ ; $\Delta_rH^\circ_3 = -286 \text{ kJ mol}^{-1}$

Step 3: Application of Hess's Law of Constant Heat Summation

[Per Hess's Law, the total enthalpy change for a reaction is independent of the pathway taken. We manipulate the given equations to yield the target reaction.]

To obtain the target reaction, we perform the following operations:

1. Take Equation (ii) as it is (provides $C$):
   $C(graphite) + O_2(g) \rightarrow CO_2(g)$ ; $\Delta H = -393 \text{ kJ mol}^{-1}$
2. Multiply Equation (iii) by 2 (provides $2H_2$):
   $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$ ; $\Delta H = 2 \times (-286) = -572 \text{ kJ mol}^{-1}$
3. Reverse Equation (i) (moves $CH_3OH$ to product side):
   $CO_2(g) + 2H_2O(l) \rightarrow CH_3OH(l) + \frac{3}{2}O_2(g)$ ; $\Delta H = +726 \text{ kJ mol}^{-1}$

Step 4: Summation of Enthalpy Changes

Adding the modified equations:

By cancelling common species on both sides ($CO_2$, $2H_2O$, and simplifying $O_2$):

$(1 + 1 - 1.5)O_2(g) = 0.5 O_2(g)$

The resulting equation is: $C(graphite) + 2H_2(g) + \frac{1}{2}O_2(g) \rightarrow CH_3OH(l)$

The total enthalpy is: $\Delta_fH^\circ = (-393 - 572 + 726) \text{ kJ mol}^{-1}$

$\Delta_fH^\circ = -965 + 726 = -239 \text{ kJ mol}^{-1}$

Visual Representation of Enthalpy Cycle

Final Solution: The standard enthalpy of formation of $CH_3OH(l)$ is $-239 \text{ kJ mol}^{-1}$.

Step 1: Given Data and Standard Enthalpies of Formation

The standard enthalpy of reaction, $\Delta_{r}H^{\circ}$, is determined using the enthalpies of formation ($\Delta_{f}H^{\circ}$) of the reactants and products. The provided values are as follows:

• $\Delta_{f}H^{\circ} [\text{CO(g)}] = -110 \, \text{kJ mol}^{-1}$
• $\Delta_{f}H^{\circ} [\text{CO}_{2}(\text{g})] = -393 \, \text{kJ mol}^{-1}$
• $\Delta_{f}H^{\circ} [\text{N}_{2}\text{O(g)}] = 81 \, \text{kJ mol}^{-1}$
• $\Delta_{f}H^{\circ} [\text{N}_{2}\text{O}_{4}(\text{g})] = 9.7 \, \text{kJ mol}^{-1}$

Step 2: Defining the Reaction Equation

The chemical reaction is given by:

$\text{N}_{2}\text{O}_{4}(\text{g}) + 3\text{CO(g)} \rightarrow \text{N}_{2}\text{O(g)} + 3\text{CO}_{2}(\text{g})$

Step 3: Application of Hess’s Law

[Per Hess's Law of Constant Heat Summation], the standard enthalpy of reaction is calculated by the difference between the sum of enthalpies of formation of products and the sum of enthalpies of formation of reactants, adjusted by their stoichiometric coefficients ($\nu$):

$\Delta_{r}H^{\circ} = \sum (\nu_p \Delta_{f}H^{\circ}_{\text{products}}) - \sum (\nu_r \Delta_{f}H^{\circ}_{\text{reactants}})$

Step 4: Calculation of Summation

Substitute the stoichiometric coefficients into the formula:

$\Delta_{r}H^{\circ} = [\Delta_{f}H^{\circ}(\text{N}_{2}\text{O}) + 3 \times \Delta_{f}H^{\circ}(\text{CO}_{2})] - [\Delta_{f}H^{\circ}(\text{N}_{2}\text{O}_{4}) + 3 \times \Delta_{f}H^{\circ}(\text{CO})]$

Insert the numerical values:

$\Delta_{r}H^{\circ} = [81 + 3(-393)] - [9.7 + 3(-110)]$

Perform the multiplications:

$\Delta_{r}H^{\circ} = [81 - 1179] - [9.7 - 330]$

Solve the inner expressions:

$\Delta_{r}H^{\circ} = [-1098] - [-320.3]$

Final algebraic summation:

$\Delta_{r}H^{\circ} = -1098 + 320.3$

$\Delta_{r}H^{\circ} = -777.7 \, \text{kJ mol}^{-1}$

Final Solution: The standard enthalpy of reaction $\Delta_{r}H^{\circ}$ is $-777.7 \, \text{kJ mol}^{-1}$.

Definition and Theoretical Foundation

A thermodynamic state function is a macroscopic property of a system whose value depends exclusively on the current state of the system, independent of the path or history by which that state was reached. Mathematically, for a state function $X$, the change in its value $\Delta X$ between two states $1$ and $2$ is defined as:

$\Delta X = X_{final} - X_{initial}$ [Per the definition of exact differentials in multivariable calculus]

Step 1: Evaluation of Thermodynamic Criteria

• Path Independence: The integral of the differential of a state function around a closed cycle is zero: $\oint dX = 0$.
• Path Dependence: Quantities such as heat ($q$) and work ($w$) are path functions, meaning their values depend on the specific mechanism of the process [Per the First Law of Thermodynamics: $\Delta U = q + w$].

Step 2: Graphical Representation of State vs. Path Functions

Step 3: Comparative Analysis of Thermodynamic Quantities

Based on the analysis, a thermodynamic state function is defined as a quantity whose value depends only on the state of the system and not on the path by which the state is attained.

Final Solution: The correct description is that a thermodynamic state function is a quantity whose value depends only on the state of the system and is independent of the path by which the state is reached.