Class 11 Chemistry (Chemistry Part-I) Chapter 5 Thermodynamics

Key Concept: Adiabatic Process

b) No heat transfer

[Solution Description] An adiabatic process is characterized by no heat transfer between the system and its surroundings.

Key Concept: System Types

b) Adiabatic Process

[Solution Description] An adiabatic process involves changes in internal energy solely through work done, with no heat exchange occurring with the surroundings.

Key Concept: Basic Concept

b) Negative

[Solution Description] During an expansion, the system does work on the surroundings. According to the sign convention, work done by the system is negative. Therefore, the correct answer is negative.

Key Concept: Gibbs Energy and Spontaneity

c) $\Delta G < 0$ [Solution Description] A reaction is considered spontaneous when the Gibbs free energy change ($\Delta G$) is negative: $\Delta G < 0$ Therefore, the correct condition for spontaneity is $\Delta G < 0$.

Key Concept: Example Identification

c) A thermos flask

[Solution Description] A thermos flask is a well-known example of an isolated system as it minimizes heat transfer with its surroundings, maintaining the temperature inside.

Key Concept: Thermochemical Equation

b) It is exothermic

[Solution Description] The negative sign of $\Delta_rH^0$ indicates that heat is released during the reaction, which characterizes an exothermic process. Therefore, the reaction is exothermic.

Key Concept: State Function Understanding

c) Work

[Solution Description] State functions depend only on the state of the system and not on how it got there. Work depends on the path taken to reach a particular state, hence it is not a state function. Pressure, volume, and enthalpy are state functions.

Key Concept: Application of Mean Bond Enthalpy

c) $3328 \text{ kJ/mol}$

[Solution Description] Propane $(C_3H_8)$ consists of a total of 8 C-H bonds. Using the mean bond enthalpy of 416 kJ/mol for each C-H bond, the total energy required is calculated as follows:

$\text{Total Energy} = 8 \times 416 \text{ kJ/mol} = 3328 \text{ kJ/mol}$

Key Concept: Complex Reaction Enthalpy, Real-World Application

a) -888 kJ/mol

[Solution Description] To calculate the standard enthalpy change $(\Delta_r H^0)$ for the reaction, we need to consider the bonds broken and formed.

Bonds broken:

– For two moles of $H_2$, four H-H bonds will be broken.

– For one mole of $O_2$, two O=O bonds will be broken.

The energy required to break these bonds is:

$$\text{Total energy for breaking} = 4 \times 436 , \text{kJ/mol} + 2 \times 498 , \text{kJ/mol}$$

$$= 1744 , \text{kJ/mol} + 996 , \text{kJ/mol} = 2740 \, \text{kJ/mol}$$

Bonds formed:

– For two moles of $H_2O$, four O-H bonds are formed.

The energy released in forming these bonds is:

$\text{Total energy for forming} = 4 \times 463 , \text{kJ/mol} = 1852 , \text{kJ/mol}$

The standard enthalpy change for the reaction is calculated as:

$\Delta_r H^0 = (\text{Energy for breaking}) – (\text{Energy for forming})$

$\Delta_r H^0 = 2740 , \text{kJ/mol} – 1852 , \text{kJ/mol} = 888 , \text{kJ/mol}$

Since it’s an exothermic reaction, the value is negative:

$\Delta_r H^0 = -888, \text{kJ/mol}$

Key Concept: Adiabatic Process

b) No heat is transferred between the system and surroundings.

[Solution Description] An adiabatic process occurs when no heat is exchanged between the system and its surroundings. Thus, the statement “No heat is transferred” describes an adiabatic process accurately.

Key Concept: Entropy Definition

c) Measure of disorder

[Solution Description] Entropy is a thermodynamic quantity that represents the degree of disorder or randomness in a system. It is a key concept in understanding how energy conversions occur in natural processes.

Key Concept: State Variables

c) Pressure

[Solution Description] A state variable is a property of a system that depends only on the current state of the system, not on how it reached that state. Common examples include pressure, volume, temperature, and internal energy.

Key Concept: Standard Enthalpy of Formation

c) -857.49 kJ

[Solution Description] The enthalpy change for forming 3 moles of water is calculated as:

$\Delta_{f}H = 3 \times (-285.83 , \text{kJ/mol})$

This gives:

$\Delta_{f}H = -857.49 , \text{kJ}$

Hence, the total enthalpy change is $-857.49 , \text{kJ}$.

Key Concept: First Law Mathematical Expression, Enthalpy Change Calculation

c) 94.7 J

[Solution Description]

To calculate the change in internal energy ($\Delta U$), we use the first law of thermodynamics:

$\Delta U = q + w$

where $q$ is the heat absorbed and $w$ is the work done by the system.

The work done by the gas can be calculated using:

$$w = -P_{\text{ext}} \times \Delta V$$

Here, $$P_{\text{ext}} = 2 \text{ atm} = 2 \times 101.325 \text{ J/L}$$ (since 1 atm = 101.325 J/L) and $$\Delta V = 3 \text{ L} – 1 \text{ L} = 2 \text{ L}$$.

$$w = -(2 \times 101.325) \times 2 = -405.3 \text{ J}$$

Now substitute the values into the first law equation:

$$\Delta U = 500 \text{ J} – 405.3 \text{ J} = 94.7 \text{ J}$$

Therefore, the change in internal energy is 94.7 J.

Key Concept: Basic Concept

c) 70 J

[Solution Description] According to the first law of thermodynamics, the change in internal energy ($\Delta U$) can be calculated using the formula: $\Delta U = q + w$, where $q$ is the heat absorbed by the system and $w$ is the work done on it.

Substituting the given values:

$\Delta U = 50\ \text{J} + 20\ \text{J} = 70\ \text{J}$

Thus, the change in internal energy is 70 J.

Key Concept: State Function

c) Because it does not depend on the path taken

[Solution Description] Entropy is a state function because it depends only on the initial and final states of a system, not on the path taken to reach these states. Therefore, changes in entropy are independent of the specific processes used.

Key Concept: Comparative Analysis

c) Isothermal reversible expansion does work; free expansion does not.

[Solution Description]

Free expansion occurs without any external pressure so no work is done and no heat is exchanged, meaning $w = 0$, $q = 0$, and $\Delta U = 0$. In contrast, during an isothermal reversible expansion, work is done on or by the system which can be calculated using $-nRT \ln \frac{V_f}{V_i}$.

Therefore, in a free expansion, no work is done while in an isothermal reversible expansion, work is non-zero.

Key Concept: Advanced Calculation, Concept Integration

b) 476 K

[Solution Description]

Start by using the relation between temperatures and volumes in an adiabatic process for an ideal gas:

$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma – 1}$

where $T_1 = 300 \, \text{K}$, $V_2 = \frac{1}{2} V_1$, and $\gamma = \frac{5}{3}$. Plugging these values into the equation gives:

$T_2 = 300 \left(2\right)^{\frac{5}{3} – 1} = 300 \times 2^{\frac{2}{3}}$

Calculate $2^{\frac{2}{3}}$ which is approximately $1.5874$:

$T_2 = 300 \times 1.5874 \approx 476.22 \, \text{K}$

Hence, the final temperature $T_2$ is approximately $476 \, \text{K}$.

Key Concept: Basic Definition

a) The enthalpy change when one mole of an ionic compound dissociates into its ions in the gaseous state.

[Solution Description] Lattice enthalpy is defined as the enthalpy change when one mole of an ionic compound dissociates into its ions in the gaseous state.

Key Concept: Reaction Analysis

c) $C(graphite)+O_2(g)\rightarrow CO_2(g)$

[Solution Description] The standard enthalpy of formation is defined for the formation of 1 mole of compound from its elements in their most stable forms at 25°C and 1 bar. Among the given reactions, $C(graphite)+O_2(g)\rightarrow CO_2(g)$ is the correct representation as it forms 1 mole of $CO_2$ from its elemental states.

Key Concept: Application of Hess’s Law

a) 1.9 kJ/mol

[Solution Description] Using Hess’s Law, we have:

$$C_{graphite}(s) + O_2(g) \rightarrow CO_2(g) \quad (\Delta H^0 = -393.5\, \text{kJ/mol})$$

$$C_{diamond}(s) + O_2(g) \rightarrow CO_2(g) \quad (\Delta H^0 = -395.4\, \text{kJ/mol})$$

Rearranging gives:

$$C_{diamond}(s) \rightarrow C_{graphite}(s) = -393.5 – (-395.4)$$

Which simplifies to:

$$\Delta H^0 = -393.5 + 395.4 = 1.9 \, \text{kJ/mol}$$

Key Concept: Enthalpy

c) Zero

[Solution Description]

An adiabatic process is characterized by no heat transfer between the system and its surroundings. Thus, the value of heat transfer ($q$) in an adiabatic process is zero.

Key Concept: Enthalpy and Phase Change

b) Enthalpy change is always positive.

[Solution Description]

Enthalpy changes during phase transformations are always positive because they require energy input to overcome intermolecular forces during endothermic processes such as melting or vaporization. This means these processes absorb heat.

Key Concept: Basic Definition

d) Degree of randomness

[Solution Description] Entropy is a thermodynamic quantity that measures the degree of disorder or randomness in a system. It reflects how energy is dispersed in a system and provides insight into its ability to do work.

Key Concept: Basic Concept

c) Entropy increases

[Solution Description]

The second law of thermodynamics states that the entropy of an isolated system always increases during a spontaneous process.

Key Concept: Intermediate Calculation

a) -2303.77 J

[Solution Description] The work done for a reversible isothermal expansion is given by:

$w = -nRT \ln \frac{V_f}{V_i}$

where $n = 1 \, \text{mol}$, $R = 8.314 \, \text{J/mol K}$, $T = 400 \, \text{K}$, $V_f = 20 \, \text{L}$, and $V_i = 10 \, \text{L}$.

Substituting these values,

$w = -(1)(8.314)(400) \ln \left(\frac{20}{10}\right)$

$w = -3325.6 \times \ln(2)$

$w = -3325.6 \times 0.693$

$w = -2303.77 \, \text{J}$

Thus, the work done is approximately $-2303.77 \, \text{J}$.

Key Concept: Heat Capacity Proportionality

b) It doubles.

[Solution Description] Heat capacity is an extensive property, meaning it depends on the amount of substance. Therefore, if you double the amount of substance, its heat capacity will also double because it is directly proportional to the quantity.

Key Concept: Enthalpy Calculation

c) 147.5 kJ

[Solution Description] To calculate the enthalpy change, we use the formula: $\Delta H = \Delta U + \Delta ngRT$

First, convert $R$ to kJ by dividing by 1000: $R = 0.008314 \, \text{kJ/mol K}$.

Next, substitute the given values into the equation:

$$\Delta H = 150 + (-1)(0.008314)(298)$$

$$\Delta H = 150 – 2.478$$

$$\Delta H = 147.522 \, \text{kJ}$$

Thus, the enthalpy change is approximately $147.5 \, \text{kJ}$.

Key Concept: Basic Definition

b) $G = H – TS$

[Solution Description] The definition of Gibbs energy is given by the formula $G = H – TS$. Here, $G$ is the Gibbs energy, $H$ is enthalpy, $T$ is temperature, and $S$ is entropy.

Key Concept: Conceptual Application, Real-World Problem Solving

c) $\gamma = 1.39$, significant effect due to rapid compression

[Solution Description]

The ratio of specific heats $\gamma$ is defined as $\gamma = C_p/C_V$. For the given values:

$\gamma = \frac{C_p}{C_V} = \frac{29.1}{20.8} \approx 1.399$

For an adiabatic process, the temperature change is characterized by $TV^{\gamma-1} = \text{constant}$. Therefore, when the volume decreases, the temperature increases inversely based on that ratio. With initial temperature $T_1 = 310 \text{ K}$, any observed temperature rise would depend on the exact decrease in volume ratio. This ratio affects how much work is performed and results in heating or cooling internally without external heat transfer.

Key Concept: Joule’s Experiment

d) Internal energy is independent of the path taken.

[Solution Description] In Joule’s experiment, it was demonstrated that the change in internal energy is independent of the specific pathway taken between two states. It depends only on the initial and final states of the system.

Key Concept: Boundary Concept

c) They allow exchange of energy but not matter

[Solution Description] In thermodynamics, a closed system is defined by its boundary which allows the exchange of energy (such as heat or work) but not matter with its surroundings. The boundary acts as a barrier for matter while still allowing energy transfer.

Key Concept: Bond Enthalpy Calculation

c) 408 kJ $mol^{-1}$

[Solution Description] Ethane $(C_2H_6)$ contains six C–H bonds and one C–C bond. If the mean bond enthalpy of the C–C bond is 350 kJ mol^{-1}$$, we first calculate the combined enthalpy for all C–H bonds by subtracting the C–C bond enthalpy from the total atomization enthalpy:

$\text{Total C–H Bond Enthalpy} = 2800 – 350 = 2450 \text{ kJ mol}^{-1}$

Next, to find the mean bond enthalpy of a single C–H bond, divide this result by 6:

$\Delta_{\text{C–H}} H^{\circ} = \frac{2450}{6} \approx 408.33 \text{ kJ mol}^{-1}$

Thus, the mean bond enthalpy of a C–H bond in $C_2H_6$ is approximately 408 kJ $mol^{-1}$.

Key Concept: Enthalpy Changes

c) The combustion of methane

[Solution Description]

A negative enthalpy change indicates an exothermic process, where energy is released to the surroundings. Combustion processes typically release energy in the form of heat.

Key Concept: Entropy and Enthalpy

a) Spontaneous

[Solution Description] To find if the process is spontaneous, we need to calculate the Gibbs free energy change $(\Delta G)$ using the formula $\Delta G = \Delta H – T \Delta S$. Make sure the units are consistent; convert $\Delta S$ from J/K·mol to kJ/K·mol by dividing by $1000$:

$\Delta S = \frac{50 \text{ J/K·mol}}{1000} = 0.05 \text{ kJ/K·mol}$

Substitute the values into the formula:

$\Delta G = -150 \text{ kJ/mol} – (298 \text{ K}) (0.05 \text{ kJ/K·mol})$

Calculate:

$\Delta G = -150 – 14.9 = -164.9 \text{ kJ/mol}$

Since $\Delta G < 0$, the process is spontaneous.

Key Concept: Real-World Application, Temperature Effect on Spontaneity

c) 450 K

[Solution Description]

The reaction becomes spontaneous when $\Delta_r G = 0$, which gives us:

$\Delta_r H = T \Delta_r S$

Solving for $T$:

$T = \frac{\Delta_r H}{\Delta_r S} = \frac{180,000 \text{ J/mol}}{400 \text{ J/K mol}}$

$T = 450 \text{ K}$

Therefore, the reaction becomes spontaneous at temperatures above 450 K.

Key Concept: Comparative Solubility

c) Increased kinetic energy

[Solution Description] At higher temperatures, the increased kinetic energy helps to overcome high lattice enthalpies, making the dissolution process more favorable. Therefore, solubility generally increases as temperature increases due to the overcoming of lattice forces.

Key Concept: Ideal Gas Law Application

d) $900R$

[Solution Description]

The expression for $p\Delta V$ using the ideal gas law for a change in moles of gases ($\Delta n_g$) is:

$p\Delta V = \Delta n_gRT$

Substituting the given values:

$p\Delta V = 3 \times R \times 300 = 900R$

Therefore, $p\Delta V = 900R$.

Key Concept: True Concept

a) True

[Solution Description]

By definition, for diatomic molecules, where there is only one bond to break, the enthalpy of atomization is equal to the bond dissociation enthalpy. Therefore, this statement is true.

Key Concept: Heat Capacity

c) 250 J

[Solution Description] The formula for heat absorbed is $q = C \Delta T$, where $C$ is the heat capacity and $\Delta T$ is the change in temperature. Here, $C = 50 \text{ J/°C}$ and $\Delta T = 25 – 20 = 5 \text{ °C}$. Thus, $q = 50 \times 5 = 250 \text{ J}$.

Key Concept: Heat and Work Relationship

d) In a reversible isothermal expansion and compression

[Solution Description]

For an isothermal process involving an ideal gas, the internal energy change $\Delta U = 0$. Thus, using the first law of thermodynamics $\Delta U = q + w_{ad}$, we have:

$0 = q + w_{ad}$

So $q = -w_{ad}$. However, since both are numerically equal but opposite in sign, none of these options fit without specifying the complete balance.

Correct scenario: No real scenario exists where $q=w_{ad}$ directly unless explicitly stated with negative work balancing positive heat precisely.

Key Concept: Diatomic Example

b) 242 kJ/mol

[Solution Description] To find the bond dissociation enthalpy for a diatomic molecule like $Cl_2$, we use known values from data tables. The bond dissociation enthalpy for the Cl-Cl bond is typically around 242 kJ/mol.

Key Concept: Comparative Analysis

b) Internal Energy and Pressure

[Solution Description] Extensive properties depend on the size or amount of matter such as volume and internal energy, while intensive properties like density do not change with quantity. Thus, internal energy (extensive) and pressure (intensive) form a correct pair as they represent different types of properties according to their dependencies.

Key Concept: Gibbs Energy and Equilibrium, Entropy and Second Law

b) Spontaneous at 500 K; might turn non-spontaneous as temperature increases.

[Solution Description] To determine spontaneity, we calculate $\Delta G$ using $\Delta G = \Delta H – T\Delta S$.

Let’s convert $\Delta S$ to kJ by dividing by 1000:

$\Delta S = -0.3$ kJ/mol K.

Calculating $\Delta G$:

$\Delta G = (-200) – (500)(-0.3)$

$= -200 + 150 = -50 \text{ kJ/mol}$

Since $\Delta G < 0$, the reaction is spontaneous at 500 K. To assess the effect of increasing temperature, note that $\Delta S$ is negative. As $T$ increases, $T\Delta S$ becomes more negative, meaning $\Delta G$ could become positive or less negative, potentially leading to non-spontaneity. Thus, at higher temperatures, the reaction may become non-spontaneous.

Key Concept: Definition Recall

b) Energy of an isolated system is constant.

[Solution Description] The first law of thermodynamics states that the energy of an isolated system remains constant since it cannot exchange heat or work with its surroundings.

Key Concept: Constant Pressure

b) Enthalpy change

[Solution Description] In calorimetry, $q_p$ represents the heat absorbed or evolved at constant pressure. This is significant as it relates directly to enthalpy change ($\Delta H$), which is defined under conditions of constant pressure.

Key Concept: State Functions

c) Temperature

[Solution Description]

Internal energy is a state function because it depends only on the initial and final states of a system, not on how the system got from one state to another. It is independent of the path taken.

Key Concept: Enthalpy vs. Internal Energy

c) -97.4 kJ/mol

[Solution Description]

To find the enthalpy change ($\Delta H$), we use the relation:

$\Delta H = \Delta U + \Delta n_gRT$

where $\Delta n_g$ is the change in moles of gas. Here, $\Delta n_g = 2 – (1+3) = -2$. Assuming the reaction occurs at standard temperature $T = 298\ K$, and using $R = 8.314\ J/mol\cdot K$,

$$\Delta H = -92.4 \times 1000\ J/mol + (-2)(8.314\ J/mol\cdot K)(298\ K)$$

Simplifying,$\Delta H = -92400\ J/mol – 4955.3\ J/mol = -97355.3\ J/mol$

Converting back to kJ,

$\Delta H = -97.36\ kJ/mol$

Hence, the correct answer is approximately $-97.4\ kJ/mol$.

Key Concept: System-Surroundings Relation

c) The system + The surroundings

[Solution Description] In thermodynamics, the universe is defined as the combination of the system and its surroundings. This means the total quantity or space consisting of both the part we are interested in (system) and the rest (surroundings).

Key Concept: Adiabatic Process

b) No heat is transferred to or from the system

[Solution Description] An adiabatic process is characterized by no heat transfer between the system and its surroundings.

Key Concept: Advanced Calculation, Concept Integration

b) 34.3°C

[Solution Description]

The heat lost by the iron will be equal to the heat gained by the water:

$$m_{\text{iron}} \cdot c_{\text{iron}} \cdot (\Delta T_{\text{iron}}) = m_{\text{water}} \cdot c_{\text{water}} \cdot (\Delta T_{\text{water}})$$

Given: $m_{\text{iron}} = 500 \, \text{g}$$, $$T_{\text{initial, iron}} = 300^\circ\text{C}$,

$m_{\text{water}} = 1000 \, \text{g}$$, $$T_{\text{initial, water}} = 20^\circ\text{C}$.

Let $T_f$ be the final temperature. For the iron:

$500 \times 0.45 \times (300 – T_f) = 1000 \times 4.18 \times (T_f – 20)$

Solving for $T_f$: $225(300 – T_f) = 4180(T_f – 20)$

Expanding both sides: $67500 – 225T_f = 4180T_f – 83600$

Bringing terms involving $T_f$ together: $67500 + 83600 = 4180T_f + 225T_f$

Simplifying: $151100 = 4405T_f$

Finally: $T_f = \frac{151100}{4405} \approx 34.3$

Hence, the final equilibrium temperature $T_f$ is approximately $34.3^\circ \text{C}$.

Key Concept: Calculation of Reaction Enthalpy

c) -482 kJ/mol

[Solution Description] To find the reaction enthalpy, we use the formula:

$\Delta_r H^0 = \sum(\text{Bond enthalpies of reactants}) – \sum(\text{Bond enthalpies of products})$

Breaking bonds in reactants:

$$2 \times \text{H-H} = 2 \times 436 = 872 \, \text{kJ/mol}$$

$$1 \times \text{O=O} = 1 \times 498 = 498 \, \text{kJ/mol}$$

Total energy to break bonds = 872 + 498 = 1370 kJ/mol

Forming bonds in products:

$$4 \times \text{O-H} = 4 \times 463 = 1852 \, \text{kJ/mol}$$

Energy released forming bonds = 1852 kJ/mol

Thus, $\Delta_r H^0 = 1370 – 1852 = -482$ kJ/mol.

Key Concept: Theoretical Implications, Advanced Conceptual

c) Because volume remains constant, hence $W = 0$

[Solution Description]

In a bomb calorimeter, reactions occur at constant volume. According to thermodynamic principles, work is related to volume change by the equation:

$W = -P\Delta V,$

where $W$ is work, $P$ is pressure, and $\Delta V$ is the change in volume. Since the volume remains constant ($\Delta V = 0$), there is no work done ($W = 0$). Energy transfer occurs purely through heat exchange, and this heat absorbed or released at constant volume reflects the change in internal energy ($\Delta U$), not work.

Key Concept: Conceptual Understanding, Complex Scenarios

a) Isothermal Reversible Process

[Solution Description] For an ideal gas, the internal energy change $\Delta U$ is primarily dependent on temperature change. In an isothermal process, the temperature remains constant. Thus, the internal energy change $\Delta U$ will be zero. An example of such a process is isothermal reversible or irreversible expansion or compression.

Key Concept: State Variables

c) Pressure

[Solution Description] State variables describe the state of a system and include properties like pressure, volume, and temperature. Work and heat are forms of energy transfer, not state variables.

Key Concept: Enthalpy of Atomization

d) They are exactly equal

[Solution Description] The enthalpy of atomization refers to the energy required to convert one mole of molecules into atoms. For diatomic hydrogen $(H_2)$, this is equivalent to the bond dissociation enthalpy because it involves breaking one H-H bond to form two hydrogen atoms. Therefore, the enthalpy of atomization of diatomic hydrogen is equal to its bond dissociation enthalpy.

Key Concept: Isolated System Characteristics

d) Exchanges neither matter nor energy

[Solution Description] An isolated system does not exchange any matter or energy with its surroundings. This means it is completely insulated and self-contained.

Key Concept: Physical State Importance

a) It affects the $\Delta_rH$ value

[Solution Description] Specifying the physical state (solid, liquid, gas) in thermochemical equations is crucial because the enthalpy change depends on the different states due to varying energy levels associated with each phase.

Key Concept: Equation Balancing

b) 6

[Solution Description]

To balance the equation for the combustion of glucose, we start with:

$C_6H_{12}O_6(s) + xO_2(g) \rightarrow yCO_2(g) + zH_2O(l)$

First, balance the carbon atoms: 6 in glucose on both sides.

$C_6H_{12}O_6(s) + xO_2(g) \rightarrow 6CO_2(g) + zH_2O(l)$

Next, balance the hydrogen atoms: 12 in glucose on both sides.

$C_6H_{12}O_6(s) + xO_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)$

Finally, balance the oxygen atoms: 18 needed for products (12 from $CO_2$ and 6 from $H_2O$). Glucose contains 6. Therefore,

$x = \frac{18 – 6}{2} = 6$

The balanced equation is:

$C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)$

So, the coefficient of $O_2$ is 6.

Key Concept: Real World Application, Enthalpy Changes during Phase Transformations

c) At high temperatures, equilibrium shifts left, reducing yield.

[Solution Description] The synthesis of ammonia in the Haber process is exothermic, which means it releases heat. According to Le Chatelier’s Principle, increasing the temperature will favor the endothermic direction to absorb the excess heat. Therefore, at higher temperatures, the equilibrium shifts towards the reactants ($N_2$ and $H_2$), reducing the yield of $NH_3$. Thus the correct answer is related to how temperature affects an exothermic reaction by shifting equilibrium.

Key Concept: Real World Application, Stoichiometry and Enthalpy

b) -1425 kJ

[Solution Description]

The change in enthalpy for the reaction can be calculated by multiplying the number of moles by the molar enthalpy change. Given that the enthalpy change per mole is -2850 kJ/mol and the amount of hydrocarbon used is 0.5 moles, we compute the total change in enthalpy as follows:

$\Delta H = 0.5 \, \text{mol} \times (-2850 \, \text{kJ/mol}) = -1425 \, \text{kJ}$

Thus, the total change in enthalpy for the combustion process is -1425 kJ.

Key Concept: State Comparison

c) Gas

[Solution Description] Gases have the highest entropy compared to solids and liquids due to their high degree of freedom and random motion, resulting in higher disorder.

Key Concept: State Functions

c) It is independent of the path taken between two states

[Solution Description]

A state function depends only on the initial and final states of a system and not on the path taken to reach those states. Internal energy, being a state function, is therefore independent of the path.

Key Concept: Adiabatic Process

b) No heat is exchanged

[Solution Description] An adiabatic process is one in which no heat is transferred to or from the system. Hence, no heat exchange takes place between the system and its surroundings.

Key Concept: Phase Transformation

c) It is the energy required to convert one mole of solid to liquid at constant temperature and pressure.

[Solution Description] The enthalpy of fusion $(\Delta_{fus}H^0)$ refers to the amount of energy required to change one mole of a solid substance to a liquid at constant temperature and pressure.

Key Concept: Exothermic Reactions

d) All of the above

[Solution Description] An exothermic reaction releases heat, leading to a negative change in enthalpy $(\Delta_r H^0 < 0)$. The reaction given as $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l); \Delta_r H^0 = -285.8, \text{kJ mol}^{-1}$ has a negative enthalpy indicating it is exothermic.

Key Concept: Complex Reaction Enthalpy, Multi-Step Reaction Enthalpy

a) 132 kJ/mol

[Solution Description] To determine the enthalpy change of the target reaction, we use Hess’s Law by combining the given reactions to form the target equation:

$\text{Target: } C_2H_4(g) + H_2O(g) \rightarrow C_2H_5OH(g)$

Rearrange the provided reactions appropriately:

Original Reaction 1 reversed:

$2CO_2(g) + 2H_2O(l) \rightarrow C_2H_4(g) + 3O_2(g), \Delta_rH^\circ = +1411\, \text{kJ/mol}$

Use Reaction 2 as is: $C_2H_5OH(g) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l), \Delta_rH^\circ = -1367\, \text{kJ/mol}$

Water phase change: $H_2O(l) \rightarrow H_2O(g), \Delta_rH^\circ = +44\, \text{kJ/mol}$

Adding these equations gives the target reaction: $C_2H_4(g) + H_2O(g) \rightarrow C_2H_5OH(g)$

Calculating the net enthalpy change: $\Delta_rH^\circ = (+1411) + (-1367) + (2 \times 44) = 132\, \text{kJ/mol}$

Key Concept: Enthalpy of Atomization

b) $\mathrm{H_2(g)} \rightarrow \mathrm{2H(g)}$

[Solution Description] The enthalpy of atomization refers to the transformation of an element from its standard state into gaseous atoms. Among the provided options, the reaction $\mathrm{H_2(g)} \rightarrow \mathrm{2H(g)}$ corresponds to breaking the $\mathrm{H-H}$ bond.

Key Concept: Reaction Combination

b) 1469.7 kJ/mol

[Solution Description] Reverse the third reaction and multiply by 2 to match the product side with two $NH_3$ molecules:

$2 \times (\text{N} (\text{g}) + 3 \text{H} (\text{g}) \rightarrow \text{NH}_3 (\text{g})); \Delta_rH^\circ = 2 \times (-391.5, \text{kJ/mol}) = -783.0, \text{kJ/mol}$

Add all reactions accordingly:

$\Delta_rH^\circ = (+945.0, \text{kJ/mol}) + 3 \times (+435.9, \text{kJ/mol}) + (-783.0, \text{kJ/mol})$

$\Delta_rH^\circ = 945.0 + 1307.7 – 783.0 = 1469.7, \text{kJ/mol}$

Therefore, the enthalpy change is $\Delta_rH^\circ = 1469.7, \text{kJ/mol}$.