Q: Q1. Choose the correct statement.\n(i) The presence of reacting species in a covered beaker is an example of an open system.\n(ii) There is an exchange of energy as well as matter betw

The correct statement is (iii).\n\n**(iii) The presence of reactants in a closed insulated vessel is an example of an isolated system.**\n\n**Explanation:**\n* **Open System:** Exchanges both matter and energy with the surroundings (e.g., an open beaker). Statement (i) is incorrect because a covered beaker restricts matter exchange. \n* **Closed System:** Exchanges energy but NOT matter with the surroundings (e.g., a covered beaker, a sealed container). Statement (ii) is incorrect because a closed system does not exchange matter. Statement (iv) is also a correct statement describing a closed system, as a covered beaker prevents significant matter exchange (like evaporation) but allows heat exchange. \n* **Isolated System:** Exchanges neither matter nor energy with the surroundings (e.g., a perfectly sealed and insulated thermos flask). Statement (iii) accurately describes an isolated system.

Q: Q3. The enthalpy of all elements in their standard states is:\n(i) unity\n(ii) zero\n(iii) < 0\n(iv) different for each element

The correct answer is (ii).\n\n**(ii) zero**\n\n**Explanation:**\nBy convention, the standard enthalpy of formation (ΔfH°) of an element in its most stable physical state at standard conditions (298 K and 1 atm pressure) is defined as zero. This serves as a reference point for calculating enthalpies of formation of compounds. For example, the standard enthalpy of formation of O₂(g), H₂(g), C(graphite), Na(s), Fe(s) are all zero.

Q: Q4. What are the limitations of the first law of thermodynamics?

The first law of thermodynamics, which states that energy can neither be created nor destroyed, but can be converted from one form to another, has certain limitations:\n\n1. **Direction of Spontaneity:** The first law does not predict the direction of a spontaneous process. For example, heat always flows spontaneously from a hotter body to a colder body, but the first law does not explain why the reverse process (heat flowing from cold to hot without external work) does not occur spontaneously, even though energy conservation would still hold.\n2. **Feasibility of a Process:** It does not provide any information about the feasibility or spontaneity of a process. It only states that if a process occurs, energy will be conserved. It cannot tell us whether a given process will occur under a specific set of conditions.\n3. **Extent of a Process:** The first law does not indicate the extent to which a reaction or process will proceed before reaching equilibrium. It doesn't explain why some reactions go to completion while others reach an equilibrium state.\n4. **Quality of Energy:** It does not distinguish between different forms of energy or the 'quality' of energy. For instance, electrical energy can be completely converted into heat, but heat energy cannot be completely converted into electrical energy without some loss or external work. The first law treats all forms of energy equally, without considering their capacity to do useful work.\n5. **Irreversibility:** Most natural processes are irreversible. The first law does not explain this irreversibility or the degradation of energy (increase in entropy) that accompanies such processes.

Q: Q10. Calculate the maximum work obtained when 0.75 mol of an ideal gas expands isothermally and reversibly at 27°C from a volume of 15 L to 25 L.

To calculate the maximum work obtained during an isothermal reversible expansion of an ideal gas, we use the following formula:\n\n$W_{rev} = -2.303 nRT \log \frac{V_2}{V_1}$\n\nGiven values:\nNumber of moles, $n = 0.75$ mol\nTemperature, $T = 27°C = 27 + 273.15 = 300.15$ K (approximately 300 K for calculations)\nInitial volume, $V_1 = 15$ L\nFinal volume, $V_2 = 25$ L\nGas constant, $R = 8.314$ J K⁻¹ mol⁻¹\n\n**Step 1: Convert temperature to Kelvin.**\n$T = 27°C + 273.15 = 300.15$ K\n\n**Step 2: Substitute the values into the formula.**\n$W_{rev} = -2.303 \times (0.75 \text{ mol}) \times (8.314 \text{ J K⁻¹ mol⁻¹}) \times (300.15 \text{ K}) \times \log \frac{25 \text{ L}}{15 \text{ L}}$\n\n**Step 3: Calculate the ratio of volumes and its logarithm.**\n$\frac{V_2}{V_1} = \frac{25}{15} = \frac{5}{3} \approx 1.6667$\n$\log(1.6667) \approx 0.2218$\n\n**Step 4: Perform the multiplication.**\n$W_{rev} = -2.303 \times 0.75 \times 8.314 \times 300.15 \times 0.2218$\n$W_{rev} = -1.72725 \times 8.314 \times 300.15 \times 0.2218$\n$W_{rev} = -14.354 \times 300.15 \times 0.2218$\n$W_{rev} = -4308.3 \times 0.2218$\n$W_{rev} = -955.48$ J\n\nRounding to appropriate significant figures, the work obtained is approximately -955.5 J.\n\nThe negative sign indicates that work is done *by* the system (gas expands). The maximum work obtained is the magnitude of this value.\n\n**Final Answer:** The maximum work obtained is **955.5 J** (or -955.5 J if considering the sign convention for work done by the system).

Q: Q11. Calculate the standard enthalpy of formation of CH3OH(l) from the following data:\nCH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l); ΔrH° = –726 kJ mol–1\nC(graphite) + O2(g) → CO2(g); ΔcH°

The standard enthalpy of formation of CH3OH(l) is -239 kJ mol⁻¹.

Q: Q12. Calculate the enthalpy change for the process CCl4(g) → C(g) + 4Cl(g) and calculate bond enthalpy of C–Cl in CCl4(g).\nΔvapH°(CCl4) = 30.5 kJ mol–1.\nΔfH°(CCl4(l)) = –135.5 kJ mol–

The enthalpy change for the process CCl4(g) → C(g) + 4Cl(g) is 1304 kJ mol⁻¹. The average bond enthalpy of C–Cl in CCl4(g) is 326 kJ mol⁻¹.

Q: Q13. For an isolated system, ΔU = 0, what will be ΔS?

For an isolated system, ΔU = 0. According to the Second Law of Thermodynamics, for a spontaneous process occurring in an isolated system, the entropy (ΔS) of the system must increase, i.e., ΔS > 0. If the process is at equilibrium, then ΔS = 0. Therefore, for an isolated system, ΔS ≥ 0.

Question 1

Q1. Choose the correct statement.
(i) The presence of reacting species in a covered beaker is an example of an open system.
(ii) There is an exchange of energy as well as matter betw

Accepted Answer

The correct statement is (iii).

**(iii) The presence of reactants in a closed insulated vessel is an example of an isolated system.**

**Explanation:**
*   **Open System:** Exchanges both matter and energy with the surroundings (e.g., an open beaker). Statement (i) is incorrect because a covered beaker restricts matter exchange. 
*   **Closed System:** Exchanges energy but NOT matter with the surroundings (e.g., a covered beaker, a sealed container). Statement (ii) is incorrect because a closed system does not exchange matter. Statement (iv) is also a correct statement describing a closed system, as a covered beaker prevents significant matter exchange (like evaporation) but allows heat exchange. 
*   **Isolated System:** Exchanges neither matter nor energy with the surroundings (e.g., a perfectly sealed and insulated thermos flask). Statement (iii) accurately describes an isolated system.

Question 2

Q2. For the process to occur under adiabatic conditions, the correct condition is:
(i) ΔT = 0
(ii) Δp = 0
(iii) q = 0
(iv) w = 0

Accepted Answer

The correct condition for a process to occur under adiabatic conditions is (iii).

**(iii) q = 0**

**Explanation:**
An adiabatic process is defined as a thermodynamic process in which there is no exchange of heat (q) between the system and its surroundings. Therefore, the heat exchanged, q, is zero.

Question 3

Q3. The enthalpy of all elements in their standard states is:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element

Accepted Answer

The correct answer is (ii).

**(ii) zero**

**Explanation:**
By convention, the standard enthalpy of formation (ΔfH°) of an element in its most stable physical state at standard conditions (298 K and 1 atm pressure) is defined as zero. This serves as a reference point for calculating enthalpies of formation of compounds. For example, the standard enthalpy of formation of O₂(g), H₂(g), C(graphite), Na(s), Fe(s) are all zero.

Question 4

Q4. What are the limitations of the first law of thermodynamics?

Accepted Answer

The first law of thermodynamics, which states that energy can neither be created nor destroyed, but can be converted from one form to another, has certain limitations:

1.  **Direction of Spontaneity:** The first law does not predict the direction of a spontaneous process. For example, heat always flows spontaneously from a hotter body to a colder body, but the first law does not explain why the reverse process (heat flowing from cold to hot without external work) does not occur spontaneously, even though energy conservation would still hold.
2.  **Feasibility of a Process:** It does not provide any information about the feasibility or spontaneity of a process. It only states that if a process occurs, energy will be conserved. It cannot tell us whether a given process will occur under a specific set of conditions.
3.  **Extent of a Process:** The first law does not indicate the extent to which a reaction or process will proceed before reaching equilibrium. It doesn't explain why some reactions go to completion while others reach an equilibrium state.
4.  **Quality of Energy:** It does not distinguish between different forms of energy or the 'quality' of energy. For instance, electrical energy can be completely converted into heat, but heat energy cannot be completely converted into electrical energy without some loss or external work. The first law treats all forms of energy equally, without considering their capacity to do useful work.
5.  **Irreversibility:** Most natural processes are irreversible. The first law does not explain this irreversibility or the degradation of energy (increase in entropy) that accompanies such processes.

Question 5

Q5. Define the term entropy. What is its physical significance?

Accepted Answer

### Definition of Entropy (S)
Entropy is a thermodynamic state function that is a measure of the degree of randomness or disorder of a system. It quantifies the number of possible microscopic arrangements (microstates) of the particles in a system that correspond to a given macroscopic state. The greater the disorder or randomness in a system, the higher is its entropy.

Mathematically, for a reversible process, the change in entropy (ΔS) is defined as the heat absorbed isothermally and reversibly (q_rev) divided by the absolute temperature (T) at which the absorption occurs:

$\Delta S = \frac{q_{rev}}{T}$

Its SI unit is Joules per Kelvin (J K⁻¹).

### Physical Significance of Entropy

1.  **Measure of Disorder/Randomness:** Entropy is fundamentally a measure of the disorder or randomness of a system. For example, gases have higher entropy than liquids, and liquids have higher entropy than solids, because particles in gases have more freedom of movement and thus more possible arrangements.
2.  **Criterion for Spontaneity (Second Law of Thermodynamics):** The second law of thermodynamics states that for a spontaneous process in an isolated system, the total entropy of the universe (system + surroundings) always increases (ΔS_universe > 0). If ΔS_universe = 0, the system is at equilibrium. If ΔS_universe < 0, the process is non-spontaneous in that direction.
3.  **Direction of Natural Processes:** Entropy helps explain why natural processes tend to proceed in a particular direction. Systems naturally evolve towards states of higher probability, which correspond to higher entropy (greater disorder).
4.  **Energy Dispersal:** Entropy can also be understood as a measure of the dispersal of energy. In any spontaneous process, energy tends to spread out from more concentrated forms to more dispersed forms, leading to an increase in the overall entropy of the universe.
5.  **Predicting Phase Transitions:** Entropy changes are crucial in understanding phase transitions. For instance, melting (solid to liquid) and vaporization (liquid to gas) involve an increase in entropy because the particles gain more freedom of movement and disorder.

Question 6

Q6. Define Gibbs energy. What is its physical significance?

Accepted Answer

### Definition of Gibbs Energy (G)
Gibbs energy (also known as Gibbs free energy or Gibbs function) is a thermodynamic potential that measures the 'useful' or process-initiating work obtainable from an isothermal, isobaric thermodynamic system. It is a state function defined as:

$G = H - TS$

Where:
*   $G$ is the Gibbs energy
*   $H$ is the enthalpy of the system
*   $T$ is the absolute temperature
*   $S$ is the entropy of the system

The change in Gibbs energy (ΔG) for a process occurring at constant temperature and pressure is given by the Gibbs-Helmholtz equation:

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

Its SI unit is Joules (J) or kiloJoules (kJ).

### Physical Significance of Gibbs Energy

1.  **Criterion for Spontaneity (at Constant T and P):** Gibbs energy is the most important criterion for predicting the spontaneity of a process occurring at constant temperature and pressure, which are common conditions in chemical reactions and biological systems.
    *   If $\Delta G < 0$: The process is spontaneous (feasible) in the forward direction.
    *   If $\Delta G > 0$: The process is non-spontaneous in the forward direction, but the reverse process is spontaneous.
    *   If $\Delta G = 0$: The system is at equilibrium, and no net change occurs.
2.  **Maximum Useful Work:** The decrease in Gibbs energy ($\Delta G$) for a process represents the maximum amount of non-expansion work (e.g., electrical work, chemical work) that can be extracted from the system at constant temperature and pressure.
3.  **Relationship with Equilibrium Constant:** Gibbs energy change is directly related to the equilibrium constant (K) of a reaction:
    $\Delta G^\circ = -RT \ln K$
    Where $\Delta G^\circ$ is the standard Gibbs energy change, R is the gas constant, and T is the absolute temperature. This relationship allows us to calculate the equilibrium constant from thermodynamic data and vice-versa.
4.  **Predicting Reaction Direction:** By analyzing the sign of $\Delta G$, chemists can predict whether a reaction will proceed spontaneously, require energy input, or is already at equilibrium under given conditions.
5.  **Temperature Dependence of Spontaneity:** The Gibbs-Helmholtz equation ($\Delta G = \Delta H - T\Delta S$) clearly shows how temperature influences the spontaneity of a reaction by affecting the relative contributions of enthalpy and entropy changes.

Question 7

Q7. Spontaneity in terms of Gibbs energy change. Explain.

Accepted Answer

### Spontaneity in Terms of Gibbs Energy Change (ΔG)

The Gibbs energy change (ΔG) is the most reliable criterion for predicting the spontaneity of a process or a chemical reaction occurring at constant temperature (T) and pressure (P). This is because it combines the effects of both enthalpy change (ΔH) and entropy change (ΔS) of the system, which are the two driving forces for spontaneity.

The relationship between Gibbs energy change, enthalpy change, and entropy change is given by the Gibbs-Helmholtz equation:

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

Where:
*   $\Delta G$ = Change in Gibbs energy of the system
*   $\Delta H$ = Change in enthalpy of the system
*   $T$ = Absolute temperature (in Kelvin)
*   $\Delta S$ = Change in entropy of the system

Based on the sign of $\Delta G$, the spontaneity of a process at constant temperature and pressure can be predicted as follows:

1.  **If $\Delta G < 0$ (Negative):**
    *   The process is **spontaneous** (or feasible) in the forward direction. This means the reaction will proceed on its own without any external intervention under the given conditions of temperature and pressure.
    *   A negative $\Delta G$ indicates that the system can do useful work.

2.  **If $\Delta G > 0$ (Positive):**
    *   The process is **non-spontaneous** in the forward direction. This means the reaction will not occur on its own under the given conditions.
    *   However, the reverse process would be spontaneous. To make a non-spontaneous process occur, external energy input (work) is required.

3.  **If $\Delta G = 0$ (Zero):**
    *   The system is at **equilibrium**. This means there is no net change occurring in the system; the rates of the forward and reverse processes are equal.
    *   At equilibrium, the system has no tendency to change spontaneously in either direction.

### Factors Influencing Spontaneity (from $\Delta G = \Delta H - T\Delta S$)

The spontaneity of a reaction depends on the signs and magnitudes of $\Delta H$ and $\Delta S$, and the absolute temperature $T$. Let's analyze the four possible cases:

Question 8

Q8. The reaction $X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)$ is an exothermic reaction. Predict the direction of the reaction if (a) temperature is increased (b) pressure is increased

Accepted Answer

According to Le Chatelier's Principle, if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

The given reaction is: $X_2(g) + Y_2(g) ightleftharpoons 2XY(g)$; $\Delta H < 0$ (exothermic)

(a) **Temperature is increased:**
Since the reaction is exothermic ($\Delta H < 0$), an increase in temperature will favor the endothermic direction to absorb the added heat. Therefore, the equilibrium will shift in the **backward direction** (towards reactants).

(b) **Pressure is increased:**
We need to compare the number of moles of gaseous reactants and products.
Number of moles of gaseous reactants = 1 (for $X_2$) + 1 (for $Y_2$) = 2 moles
Number of moles of gaseous products = 2 (for $2XY$) = 2 moles
Since the number of moles of gaseous reactants is equal to the number of moles of gaseous products ($\Delta n_g = 0$), a change in pressure will have **no effect** on the equilibrium position.

(c) **Catalyst is added:**
A catalyst increases the rate of both forward and backward reactions equally. It helps the system to reach equilibrium faster but does **not change the equilibrium position** or the value of the equilibrium constant. Therefore, there will be **no effect** on the direction of the reaction at equilibrium.

(d) **Inert gas is added at constant volume:**
When an inert gas is added at constant volume, the total pressure of the system increases, but the partial pressures of the reactants and products remain unchanged. Since the equilibrium position depends on the partial pressures of the reacting gases, adding an inert gas at constant volume will have **no effect** on the equilibrium position.

Question 9

Q9. Derive the relationship between $\Delta H$ and $\Delta U$.

Accepted Answer

## Derivation of the Relationship between Enthalpy Change ($\Delta H$) and Internal Energy Change ($\Delta U$)

**1. Definition of Enthalpy (H):**
Enthalpy ($H$) is defined as the sum of the internal energy ($U$) and the product of pressure ($P$) and volume ($V$) of a system.
$H = U + PV$  (Equation 1)

**2. Change in Enthalpy ($\Delta H$):**
For a process occurring at constant pressure, the change in enthalpy ($\Delta H$) can be expressed as:
$\Delta H = H_{final} - H_{initial}$
Using Equation 1, we can write:
$\Delta H = (U_2 + P_2V_2) - (U_1 + P_1V_1)$
Where subscripts 1 and 2 refer to initial and final states, respectively.
If the process occurs at constant pressure, $P_1 = P_2 = P$. So,
$\Delta H = (U_2 + PV_2) - (U_1 + PV_1)$
$\Delta H = (U_2 - U_1) + P(V_2 - V_1)$
$\Delta H = \Delta U + P\Delta V$  (Equation 2)

**3. First Law of Thermodynamics:**
The First Law of Thermodynamics states that the change in internal energy ($\Delta U$) of a system is equal to the heat ($q$) absorbed by the system plus the work ($w$) done on the system.
$\Delta U = q + w$  (Equation 3)

**4. Work Done (w):**
For a process involving expansion or compression, the work done by the system against a constant external pressure ($P_{ext}$) is given by:
$w = -P_{ext}\Delta V$
If the process is carried out reversibly or against a constant external pressure equal to the system's pressure, then $P_{ext} = P$.
So, $w = -P\Delta V$  (Equation 4)

**5. Substituting Work into First Law:**
Substitute Equation 4 into Equation 3:
$\Delta U = q - P\Delta V$
Rearranging for $q$:
$q = \Delta U + P\Delta V$  (Equation 5)

**6. Relating $q_p$ to $\Delta H$:**
When a process occurs at constant pressure, the heat absorbed or released is denoted as $q_p$. From Equation 5, we can see that:
$q_p = \Delta U + P\Delta V$
Comparing this with Equation 2:
$\Delta H = \Delta U + P\Delta V$
Therefore, $q_p = \Delta H$. This means that the enthalpy change ($\Delta H$) represents the heat exchanged by the system at constant pressure.

**7. Relationship for Gaseous Reactions (Ideal Gas):**
For reactions involving gases, especially when the number of moles of gaseous reactants and products changes, the $P\Delta V$ term can be related to the change in the number of moles of gas.
Assuming ideal gas behavior, $PV = nRT$.
If the reaction occurs at constant temperature ($T$) and constant pressure ($P$):
$P V_1 = n_1 RT$ (initial state)
$P V_2 = n_2 RT$ (final state)
Subtracting the initial state from the final state:
$P(V_2 - V_1) = (n_2 - n_1)RT$
$P\Delta V = \Delta n_g RT$
Where $\Delta n_g$ is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.
$\Delta n_g = \sum n_{g,products} - \sum n_{g,reactants}$

**8. Final Relationship:**
Substitute $P\Delta V = \Delta n_g RT$ into Equation 2:
$\Delta H = \Delta U + \Delta n_g RT$

This is the final relationship between $\Delta H$ and $\Delta U$, particularly useful for reactions involving gases.

Question 10

Q10. Calculate the maximum work obtained when 0.75 mol of an ideal gas expands isothermally and reversibly at 27°C from a volume of 15 L to 25 L.

Accepted Answer

To calculate the maximum work obtained during an isothermal reversible expansion of an ideal gas, we use the following formula:

$W_{rev} = -2.303 nRT \log \frac{V_2}{V_1}$

Given values:
Number of moles, $n = 0.75$ mol
Temperature, $T = 27°C = 27 + 273.15 = 300.15$ K (approximately 300 K for calculations)
Initial volume, $V_1 = 15$ L
Final volume, $V_2 = 25$ L
Gas constant, $R = 8.314$ J K⁻¹ mol⁻¹

**Step 1: Convert temperature to Kelvin.**
$T = 27°C + 273.15 = 300.15$ K

**Step 2: Substitute the values into the formula.**
$W_{rev} = -2.303 	imes (0.75 	ext{ mol}) 	imes (8.314 	ext{ J K⁻¹ mol⁻¹}) 	imes (300.15 	ext{ K}) 	imes \log \frac{25 	ext{ L}}{15 	ext{ L}}$

**Step 3: Calculate the ratio of volumes and its logarithm.**
$\frac{V_2}{V_1} = \frac{25}{15} = \frac{5}{3} \approx 1.6667$
$\log(1.6667) \approx 0.2218$

**Step 4: Perform the multiplication.**
$W_{rev} = -2.303 	imes 0.75 	imes 8.314 	imes 300.15 	imes 0.2218$
$W_{rev} = -1.72725 	imes 8.314 	imes 300.15 	imes 0.2218$
$W_{rev} = -14.354 	imes 300.15 	imes 0.2218$
$W_{rev} = -4308.3 	imes 0.2218$
$W_{rev} = -955.48$ J

Rounding to appropriate significant figures, the work obtained is approximately -955.5 J.

The negative sign indicates that work is done *by* the system (gas expands). The maximum work obtained is the magnitude of this value.

**Final Answer:** The maximum work obtained is **955.5 J** (or -955.5 J if considering the sign convention for work done by the system).

Question 11

Q11. Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l); ΔrH° = –726 kJ mol–1
C(graphite) + O2(g) → CO2(g); ΔcH°

Accepted Answer

The standard enthalpy of formation of CH3OH(l) is -239 kJ mol⁻¹.

Question 12

Q12. Calculate the enthalpy change for the process CCl4(g) → C(g) + 4Cl(g) and calculate bond enthalpy of C–Cl in CCl4(g).
ΔvapH°(CCl4) = 30.5 kJ mol–1.
ΔfH°(CCl4(l)) = –135.5 kJ mol–

Accepted Answer

The enthalpy change for the process CCl4(g) → C(g) + 4Cl(g) is 1304 kJ mol⁻¹. The average bond enthalpy of C–Cl in CCl4(g) is 326 kJ mol⁻¹.

Question 13

Q13. For an isolated system, ΔU = 0, what will be ΔS?

Accepted Answer

For an isolated system, ΔU = 0. According to the Second Law of Thermodynamics, for a spontaneous process occurring in an isolated system, the entropy (ΔS) of the system must increase, i.e., ΔS > 0. If the process is at equilibrium, then ΔS = 0. Therefore, for an isolated system, ΔS ≥ 0.

Question 14

Q14. Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:
CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l); ΔᵣH° = –726 kJ mol⁻¹
C(graphite) + O₂(g) → CO₂(g); ΔᵣH°

Accepted Answer

The standard enthalpy of formation of CH₃OH(l) is -239 kJ mol⁻¹.

Question 15

Q15. Calculate the enthalpy change for the process
CCl₄(g) → C(g) + 4 Cl(g)
and calculate bond enthalpy of C – Cl in CCl₄(g).
ΔᵥₐₚH⊖(CCl₄) = 30.5 kJ mol⁻¹
ΔբH⊖(CCl₄) = –135.5 kJ mol

Accepted Answer

The enthalpy change for the process CCl₄(g) → C(g) + 4 Cl(g) is the enthalpy of atomisation of CCl₄(g). We can calculate this using the given data and Hess's Law.

**Step 1: Calculate the standard enthalpy of formation of CCl₄(g).**
We are given the enthalpy of formation of liquid CCl₄ and its enthalpy of vaporisation.
CCl₄(l) → CCl₄(g); ΔᵥₐₚH⊖ = 30.5 kJ mol⁻¹
C(s, graphite) + 2Cl₂(g) → CCl₄(l); ΔբH⊖ = –135.5 kJ mol⁻¹
Adding these two equations gives the formation of gaseous CCl₄:
C(s, graphite) + 2Cl₂(g) → CCl₄(g)
ΔբH⊖(CCl₄, g) = ΔբH⊖(CCl₄, l) + ΔᵥₐₚH⊖(CCl₄)
ΔբH⊖(CCl₄, g) = (–135.5) + (30.5) = –105.0 kJ mol⁻¹

**Step 2: Calculate the enthalpy of atomisation for CCl₄(g).**
The target reaction is: CCl₄(g) → C(g) + 4 Cl(g). This is the enthalpy of atomisation, ΔₐH⊖(CCl₄, g).
We can use the following relationship:
ΔᵣH⊖ = Σ (Enthalpy of atomisation of reactants) - Σ (Enthalpy of atomisation of products)
For the formation reaction of CCl₄(g): C(s) + 2Cl₂(g) → CCl₄(g)
ΔբH⊖(CCl₄, g) = [ΔₐH⊖(C, s) + 2 × ΔₐH⊖(Cl₂, g)] - [ΔₐH⊖(CCl₄, g)]
Note that the enthalpy of atomisation of Cl₂ is for the reaction Cl₂(g) → 2Cl(g), so ΔₐH⊖(Cl₂, g) is 242 kJ mol⁻¹.
-105.0 kJ mol⁻¹ = [715.0 kJ mol⁻¹ + 2 × (242 kJ mol⁻¹)] - ΔₐH⊖(CCl₄, g)
-105.0 = [715.0 + 484] - ΔₐH⊖(CCl₄, g)
-105.0 = 1199 - ΔₐH⊖(CCl₄, g)
ΔₐH⊖(CCl₄, g) = 1199 + 105.0 = 1304 kJ mol⁻¹
So, the enthalpy change for the process CCl₄(g) → C(g) + 4 Cl(g) is **1304 kJ mol⁻¹**.

**Step 3: Calculate the bond enthalpy of the C-Cl bond.**
The CCl₄ molecule has one central carbon atom bonded to four chlorine atoms, meaning there are four C-Cl single bonds. The enthalpy of atomisation is the energy required to break all these bonds.
ΔₐH⊖(CCl₄, g) = 4 × (Average bond enthalpy of C-Cl)
Average C-Cl bond enthalpy = ΔₐH⊖(CCl₄, g) / 4
Average C-Cl bond enthalpy = 1304 kJ mol⁻¹ / 4 = **326 kJ mol⁻¹**.

Question 16

Q16. For an isolated system, ΔU = 0, what will be ΔS ?

Accepted Answer

For an isolated system, there is no exchange of energy or matter with the surroundings. The condition ΔU = 0 is a consequence of the first law of thermodynamics for an isolated system (since q=0 and w=0, ΔU = q + w = 0).

The change in entropy (ΔS) is governed by the second law of thermodynamics, which states that the entropy of the universe (or an isolated system) always increases for a spontaneous process and remains constant for a process at equilibrium.

Therefore, for any process occurring in an isolated system:
1.  **For a spontaneous (irreversible) process:** The entropy of the system must increase. So, ΔS > 0.
2.  **For a reversible process (at equilibrium):** The entropy of the system remains constant. So, ΔS = 0.

A process for which ΔS < 0 is non-spontaneous and will not occur in an isolated system.

Combining these possibilities, the change in entropy for any feasible process in an isolated system must be greater than or equal to zero.

**ΔS ≥ 0**

Thus, if a process is occurring, ΔS will be positive. If the system is already at equilibrium, ΔS will be zero.

Question 17

Q17. For the reaction at 298 K,
2A + B → C
ΔH = 400 kJ mol⁻¹ and ΔS = 0.2 kJ K⁻¹ mol⁻¹
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant o

Accepted Answer

A reaction is spontaneous when the change in Gibbs free energy (ΔG) is negative (ΔG < 0). The relationship between ΔG, ΔH, and ΔS is given by the Gibbs-Helmholtz equation: ΔG = ΔH - TΔS **Step 1: Set up the condition for spontaneity.** For the reaction to be spontaneous, we must have: ΔG < 0 ΔH - TΔS < 0 **Step 2: Find the equilibrium temperature (T_eq).** At equilibrium, ΔG = 0. The temperature at which this occurs is the equilibrium temperature. 0 = ΔH - T_eqΔS T_eq = ΔH / ΔS **Step 3: Substitute the given values to find T_eq.** Given: ΔH = 400 kJ mol⁻¹ ΔS = 0.2 kJ K⁻¹ mol⁻¹ T_eq = (400 kJ mol⁻¹) / (0.2 kJ K⁻¹ mol⁻¹) = 2000 K **Step 4: Determine the temperature range for spontaneity.** The reaction has a positive ΔH (endothermic) and a positive ΔS (increase in entropy). Such reactions become spontaneous at high temperatures. From the spontaneity condition in Step 1: ΔH - TΔS < 0 TΔS > ΔH Since ΔS is positive, we can divide by it without changing the inequality direction: T > ΔH / ΔS T > 2000 K Therefore, the reaction will become spontaneous at temperatures **above 2000 K**.

Question 18

Q18. The equilibrium constant for a reaction is 10. What will be the value of ΔG°? R = 8.314 JK⁻¹mol⁻¹, T = 300 K.

Accepted Answer

The relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (K) is given by the equation:
ΔG° = -2.303 RT log K

Given values:
Equilibrium constant, K = 10
Universal gas constant, R = 8.314 J K⁻¹ mol⁻¹
Temperature, T = 300 K

Substituting these values into the equation:
ΔG° = -2.303 × (8.314 J K⁻¹ mol⁻¹) × (300 K) × log(10)

Since log(10) = 1:
ΔG° = -2.303 × 8.314 × 300 × 1 J mol⁻¹
ΔG° = -5744.14 J mol⁻¹

To express the answer in kJ mol⁻¹, we divide by 1000:
ΔG° = -5744.14 / 1000 kJ mol⁻¹
ΔG° = -5.744 kJ mol⁻¹

Thus, the value of ΔG° for the reaction is -5.744 kJ mol⁻¹.

Question 19

Q19. Comment on the thermodynamic stability of NO(g), given
½ N₂(g) + ½ O₂(g) → NO(g) ; ΔᵣH° = 90 kJ mol⁻¹
NO(g) + ½ O₂(g) → NO₂(g) ; ΔᵣH° = –74 kJ mol⁻¹

Accepted Answer

To comment on the thermodynamic stability of NO(g), we need to analyze its tendency to decompose into its constituent elements and its tendency to react to form other compounds, based on the given enthalpy changes.

**1. Stability of NO(g) with respect to its elements (N₂ and O₂):**
The first reaction represents the formation of nitric oxide (NO) from its elements, nitrogen (N₂) and oxygen (O₂).
½ N₂(g) + ½ O₂(g) → NO(g) ; ΔᵣH° = +90 kJ mol⁻¹

The standard enthalpy of formation (ΔfH°) of NO(g) is positive (+90 kJ mol⁻¹). This means the reaction is endothermic. An endothermic formation indicates that the product (NO) is at a higher energy level than the reactants (N₂ and O₂). Compounds with positive enthalpy of formation are thermodynamically unstable relative to their constituent elements. Therefore, NO(g) is thermodynamically unstable and has a tendency to decompose back into N₂(g) and O₂(g). The reverse reaction (decomposition) would be exothermic (ΔH° = -90 kJ mol⁻¹).

**2. Stability of NO(g) with respect to NO₂(g):**
The second reaction shows the oxidation of NO(g) to nitrogen dioxide (NO₂(g)).
NO(g) + ½ O₂(g) → NO₂(g) ; ΔᵣH° = –74 kJ mol⁻¹

The enthalpy change for this reaction is negative (–74 kJ mol⁻¹), which means the reaction is highly exothermic. An exothermic reaction indicates that the products are more stable (at a lower energy level) than the reactants. This large negative enthalpy change implies that NO(g) has a strong tendency to react with oxygen to form the more stable NO₂(g).

**Conclusion:**
Based on both reactions, we can conclude that nitric oxide, NO(g), is a thermodynamically unstable species. 
- It is unstable with respect to its decomposition into N₂ and O₂ because its formation is endothermic.
- It is also unstable with respect to its oxidation to NO₂ because this process is highly exothermic.

Despite its thermodynamic instability, NO can exist because its decomposition is kinetically slow at normal temperatures.

Question 20

Q20. Calculate the enthalpy change for the process
CCl₄(g) → C(g) + 4Cl(g)
and calculate bond enthalpy of C – Cl in CCl₄(g).
ΔᵥₐₚH⊖(CCl₄) = 30.5 kJ mol⁻¹
ΔբH⊖(CCl₄) = –135.5 kJ mol⁻

Accepted Answer

The enthalpy change for the process CCl₄(g) → C(g) + 4Cl(g) is 1304.5 kJ mol⁻¹. This represents the total enthalpy required to break all four C-Cl bonds in one mole of gaseous CCl₄. The average bond enthalpy of a C-Cl bond in CCl₄ is calculated by dividing this total enthalpy change by the number of bonds, which is 4. Therefore, the bond enthalpy of the C-Cl bond is 326.125 kJ mol⁻¹.

Question 21

Q21. For an isolated system, ΔU = 0, what will be ΔS ?

Accepted Answer

For an isolated system, the change in internal energy (ΔU) is zero because there is no exchange of energy (heat or work) with the surroundings. According to the second law of thermodynamics, the entropy of an isolated system can never decrease.

For any spontaneous process occurring within the isolated system, the entropy will increase, so ΔS > 0.

If the system is at equilibrium, there is no change in entropy, so ΔS = 0.

Therefore, the overall condition for the change in entropy (ΔS) for an isolated system is ΔS ≥ 0. It will be greater than zero for a spontaneous process and zero for a reversible process (at equilibrium).

Question 22

Q22. Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:
CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l); ΔᵣH° = –726 kJ mol⁻¹
C(graphite) + O₂(g) → CO₂(g); ΔᵣH°

Accepted Answer

The standard enthalpy of formation of CH₃OH(l) is -239 kJ mol⁻¹.

Chapter 6: Thermodynamics