Chapter 4: Chemical Thermodynamics - 50 Subjective Questions for HSC Revision | Chemca.in

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Chapter 4: Chemical Thermodynamics

Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions

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1. Define system, surroundings, and universe in thermodynamics.

System: A specific part of the universe under thermodynamic investigation is called the system.

Surroundings: The rest of the universe outside the system, which can interact with it, is called the surroundings.

Universe: The system and surroundings together constitute the universe. (Universe = System + Surroundings).

2. Classify systems based on the exchange of matter and energy. Give one example of each.

Open System: A system that can exchange both matter and energy with its surroundings. Example: Hot coffee in an open cup.
Closed System: A system that can exchange energy but not matter with its surroundings. Example: Hot coffee in a sealed, conducting flask.
Isolated System: A system that can exchange neither matter nor energy with its surroundings. Example: Hot coffee in a perfectly insulated thermos flask.

3. Define extensive and intensive properties. Give two examples of each.

Extensive Property: A property whose value depends on the quantity or size of matter present in the system. Examples: Mass, Volume, Internal Energy, Heat Capacity.

Intensive Property: A property whose value is independent of the quantity or size of matter present in the system. Examples: Temperature, Pressure, Density, Surface Tension, Refractive Index.

4. What are State Functions and Path Functions? Give examples.

State Function: A thermodynamic property whose value depends only on the current state of the system (initial and final states) and is independent of the path followed to reach that state. Examples: Internal energy ($U$), Enthalpy ($H$), Entropy ($S$), Temperature ($T$), Pressure ($P$).

Path Function: A thermodynamic property whose value depends on the specific path or mechanism taken during a change of state. Examples: Work ($W$) and Heat ($q$).

5. Define Isothermal and Adiabatic processes.

Isothermal Process: A process in which the temperature of the system remains constant throughout the change ($\Delta T = 0$). The internal energy remains constant ($\Delta U = 0$). Heat is exchanged with the surroundings to maintain temperature.
Adiabatic Process: A process in which no heat is exchanged between the system and its surroundings ($q = 0$). The system is completely insulated. Temperature changes during this process.

6. Define Isobaric and Isochoric processes.

Isobaric Process: A process in which the pressure of the system remains constant throughout the change ($\Delta P = 0$). Generally occurs in open vessels.
Isochoric Process: A process in which the volume of the system remains constant throughout the change ($\Delta V = 0$). The work done in an isochoric process is zero ($W = -P_{ext}\Delta V = 0$).

7. Distinguish between a reversible process and an irreversible process.

Reversible Process	Irreversible Process
Driving force is infinitesimally greater than opposing force.	Driving force is significantly greater than opposing force.
It takes infinite time for completion.	It takes finite time for completion.
It is an ideal/hypothetical process.	All natural/spontaneous processes are irreversible.
Maximum work is obtained.	Maximum work cannot be obtained.

8. Derive the expression for pressure-volume ($PV$) work.

Consider an ideal gas enclosed in a cylinder fitted with a weightless, frictionless piston of area $A$. Let $P_{ext}$ be the external opposing pressure.

If the gas expands, the piston moves outward by a distance $d$.

Work done by the system ($W$) = Force $\times$ displacement = $f \times (-d)$ (negative sign because expansion is work done *by* the system against opposing force).

Since Pressure ($P_{ext}$) = Force / Area $\Rightarrow f = P_{ext} \times A$

Substituting $f$: $W = - P_{ext} \times A \times d$

Since Area $\times$ displacement = Change in volume ($\Delta V = V_2 - V_1$)

$$ W = -P_{ext} \cdot \Delta V $$

9. What is free expansion? What is the work done during free expansion of an ideal gas?

Free Expansion: The expansion of a gas into a vacuum is called free expansion.

Work Done: In a vacuum, there is no opposing external pressure ($P_{ext} = 0$).

According to the equation of work: $W = -P_{ext} \Delta V$.

Since $P_{ext} = 0$, $W = -0 \times \Delta V = 0$.

Therefore, the work done during the free expansion of an ideal gas is zero.

10. Write the formula for maximum work ($W_{max}$) done during isothermal and reversible expansion of an ideal gas.

The maximum work ($W_{max}$) obtained when $n$ moles of an ideal gas expand isothermally and reversibly from volume $V_1$ to $V_2$ at temperature $T$ is given by:

$$ W_{max} = -2.303 \, nRT \log_{10} \left(\frac{V_2}{V_1}\right) $$

Since $P_1V_1 = P_2V_2$ at constant $T$, it can also be written as:

$$ W_{max} = -2.303 \, nRT \log_{10} \left(\frac{P_1}{P_2}\right) $$

11. State the sign conventions for heat ($q$) and work ($W$) in thermodynamics.

$+q$: Heat is absorbed by the system from the surroundings (endothermic).
$-q$: Heat is released by the system to the surroundings (exothermic).
$+W$: Work is done on the system by the surroundings (compression).
$-W$: Work is done by the system on the surroundings (expansion).

12. Define Internal Energy ($U$). Is it a state function?

Internal Energy ($U$): Every substance possesses a definite amount of energy associated with its chemical nature and thermodynamic state (due to kinetic and potential energies of its atoms/molecules). This is called internal energy.

Yes, Internal Energy is a state function and an extensive property. Its absolute value cannot be determined, but the change ($\Delta U = U_{final} - U_{initial}$) can be measured.

13. State the First Law of Thermodynamics.

The First Law of Thermodynamics (Law of Conservation of Energy) can be stated in several ways:

Energy can neither be created nor destroyed, but can be converted from one form to another.
The total energy of the universe remains constant.
The total internal energy of an isolated system is constant.

14. Write the mathematical expression for the First Law of Thermodynamics. Explain the terms.

The mathematical formulation is:

$$ \Delta U = q + W $$

Where:

$\Delta U$ = Change in internal energy of the system
$q$ = Heat supplied to or removed from the system
$W$ = Work done on or by the system

15. Deduce the First Law of Thermodynamics for an isothermal process.

In an isothermal process, the temperature is constant ($\Delta T = 0$). Since internal energy depends on temperature, $\Delta U = 0$.

According to the first law: $\Delta U = q + W$

Substituting $\Delta U = 0$:

$$ 0 = q + W \implies q = -W \text{ or } W = -q $$

This means the heat absorbed by the system is entirely used to do work of expansion on the surroundings.

16. Deduce the First Law of Thermodynamics for an adiabatic process.

In an adiabatic process, there is no exchange of heat between the system and surroundings, so $q = 0$.

According to the first law: $\Delta U = q + W$

Substituting $q = 0$:

$$ \Delta U = W $$

This means the work done on the system ($+W$) increases its internal energy, causing a temperature rise. If the system does work (expansion, $-W$), its internal energy decreases, causing cooling.

17. Deduce the First Law of Thermodynamics for an isochoric process.

In an isochoric process, volume remains constant ($\Delta V = 0$).

Therefore, PV work $W = -P_{ext}\Delta V = 0$.

According to the first law: $\Delta U = q + W$

Substituting $W = 0$:

$$ \Delta U = q_v $$

This means that heat absorbed at constant volume ($q_v$) is exactly equal to the increase in internal energy.

18. Define Enthalpy ($H$). Is it a state function?

Enthalpy ($H$): Enthalpy is defined as the sum of the internal energy ($U$) of a system and the energy equivalent to pressure-volume work ($PV$). It represents the total heat content of a system.

$$ H = U + PV $$

Yes, Enthalpy is a state function and an extensive property because $U, P,$ and $V$ are all state functions.

19. Show that change in enthalpy ($\Delta H$) is equal to the heat transferred at constant pressure ($q_p$).

By definition, $H = U + PV$.

Change in enthalpy at constant pressure is: $\Delta H = \Delta U + P\Delta V$.

From the First Law, $\Delta U = q + W$. At constant pressure, $W = -P\Delta V$ and $q = q_p$.

So, $\Delta U = q_p - P\Delta V$.

Substituting $\Delta U$ into the enthalpy equation:

$\Delta H = (q_p - P\Delta V) + P\Delta V$

$$ \Delta H = q_p $$

Thus, heat absorbed at constant pressure equals the change in enthalpy.

20. Derive the relationship between $\Delta H$ and $\Delta U$ for chemical reactions involving gases.

We know that $\Delta H = \Delta U + P\Delta V$ at constant pressure.

For ideal gases, $PV = nRT$.

For gaseous reactants and products at constant $T$ and $P$:

$PV_1 = n_1RT$ and $PV_2 = n_2RT$

$P(V_2 - V_1) = (n_2 - n_1)RT \implies P\Delta V = \Delta n_g RT$

Where $\Delta n_g$ = (moles of gaseous products) - (moles of gaseous reactants).

Substituting $P\Delta V$ in the enthalpy equation:

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

21. When is $\Delta H$ equal to $\Delta U$?

Since $\Delta H = \Delta U + \Delta n_g RT$, $\Delta H = \Delta U$ under the following conditions:

When the reaction is carried out in a closed vessel (isochoric, $\Delta V = 0$).
When the reaction involves only solids and liquids (since their volume change is negligible, $\Delta V \approx 0$).
When a gaseous reaction occurs such that the number of moles of gaseous products equals gaseous reactants ($\Delta n_g = 0$). Example: $H_{2(g)} + I_{2(g)} \rightarrow 2HI_{(g)}$.

22. Define Enthalpy of Fusion ($\Delta_{fus}H$).

Enthalpy of Fusion: The enthalpy change that accompanies the fusion (melting) of one mole of a solid directly into liquid at constant temperature and pressure.

Example: $H_2O_{(s)} \rightarrow H_2O_{(l)} \quad \Delta_{fus}H = +6.01 \text{ kJ mol}^{-1}$

23. Define Enthalpy of Vaporization ($\Delta_{vap}H$).

Enthalpy of Vaporization: The enthalpy change that accompanies the vaporization of one mole of a liquid into gas at constant temperature and pressure.

Example: $H_2O_{(l)} \rightarrow H_2O_{(g)} \quad \Delta_{vap}H = +40.7 \text{ kJ mol}^{-1}$ (at 373 K)

24. Define Enthalpy of Sublimation ($\Delta_{sub}H$) and relate it to fusion and vaporization.

Enthalpy of Sublimation: The enthalpy change that accompanies the conversion of one mole of a solid directly into a gas at constant temperature and pressure.

Since enthalpy is a state function, sublimation can be viewed in two steps: melting then boiling. Therefore:

$$ \Delta_{sub}H = \Delta_{fus}H + \Delta_{vap}H $$

25. What is the standard state of a substance?

The standard state of a substance is its most stable pure form under standard conditions, which are defined as 1 bar pressure and a specified temperature, usually 298 K (25°C). For solutions, the standard concentration is 1 M.

26. Define Standard Enthalpy of Formation ($\Delta_f H^\circ$). What is its value for elements in standard states?

Standard Enthalpy of Formation: The enthalpy change that accompanies the formation of one mole of a compound from its constituent elements, with all substances in their standard states.

By convention, the standard enthalpy of formation of any element in its most stable standard state is zero. (e.g., $\Delta_f H^\circ(O_{2(g)}) = 0$, $\Delta_f H^\circ(C_{(graphite)}) = 0$).

27. Define Standard Enthalpy of Combustion ($\Delta_c H^\circ$).

Standard Enthalpy of Combustion: The enthalpy change accompanying the complete oxidation (burning) of one mole of a substance in presence of excess oxygen, with all substances in their standard states.

It is always negative (exothermic process). Example: burning of methane.

28. How do you calculate the standard enthalpy of a reaction ($\Delta_r H^\circ$) from enthalpies of formation?

The standard enthalpy of a reaction is the difference between the sum of standard enthalpies of formation of products and reactants.

$$ \Delta_r H^\circ = \sum \Delta_f H^\circ (\text{products}) - \sum \Delta_f H^\circ (\text{reactants}) $$

Where the sums are multiplied by the respective stoichiometric coefficients from the balanced equation.

29. Define Enthalpy of Neutralization. Why is its value constant (-57.3 kJ/mol) for any strong acid and strong base?

Enthalpy of Neutralization: The enthalpy change when one gram equivalent of an acid is completely neutralized by one gram equivalent of a base in dilute solutions.

Reason for constant value: Strong acids and strong bases are completely ionized in water. The neutralization of any strong acid by a strong base is simply the reaction between $H^+_{(aq)}$ and $OH^-_{(aq)}$ to form 1 mole of $H_2O_{(l)}$. Since the actual chemical reaction is always the same ($H^+ + OH^- \rightarrow H_2O$), the enthalpy change is constant ($-57.3 \text{ kJ mol}^{-1}$).

30. Define Bond Enthalpy.

Bond Enthalpy: The average enthalpy change required to break one mole of a specific type of covalent bond in a gaseous molecule to separate it into gaseous atoms or radicals.

It is always positive because energy is required to break bonds (endothermic).

31. How is the enthalpy of reaction calculated using bond enthalpies?

The standard enthalpy of a gaseous reaction can be estimated by taking the difference between the bond enthalpies of the bonds broken in the reactants and the bonds formed in the products.

$$ \Delta_r H^\circ = \sum \Delta H^\circ (\text{reactant bonds broken}) - \sum \Delta H^\circ (\text{product bonds formed}) $$

(Note: Reactants minus Products, unlike the formation formula).

32. State Hess's Law of Constant Heat Summation.

Hess's Law: It states that the overall enthalpy change for a chemical reaction is the same regardless of whether the reaction takes place in a single step or in a series of steps, provided the initial and final states are the same.

This law is a direct consequence of enthalpy being a state function.

33. Write two applications of Hess's Law.

To calculate the enthalpy of formation of compounds (like $CO, C_6H_6$) which cannot be synthesized directly from their elements easily.
To calculate enthalpies of extremely slow reactions, or enthalpies of transition (e.g., C(diamond) to C(graphite)).

34. What is a spontaneous process? Give two examples.

Spontaneous Process: A process that occurs on its own, without the intervention of an outside continuous forcing agency, under a given set of conditions. It has a natural tendency to occur in a specific direction.

Examples:

Heat flowing from a hot body to a cold body.
Dissolution of sugar or salt in water.
Expansion of a gas into a vacuum.

35. Is a decrease in enthalpy ($\Delta H < 0$) the sole criterion for spontaneity? Justify.

No. While many spontaneous processes are exothermic ($\Delta H < 0$, energy decreases), there are endothermic processes ($\Delta H > 0$) that are also spontaneous.

For example, the melting of ice above $0^\circ \text{C}$ or dissolution of $NH_4Cl$ in water are endothermic but spontaneous. Therefore, lowering of energy is a driving force, but not the sole criterion. Another factor (entropy) must be considered.

36. Define Entropy ($S$).

Entropy ($S$): Entropy is a thermodynamic state function that measures the degree of randomness, disorder, or chaos in a system. The greater the disorder, the higher the entropy.

Solid < Liquid < Gas (Entropy increases in this order).

37. State the mathematical expression for entropy change ($\Delta S$). Mention its unit.

Entropy change ($\Delta S$) is equal to the heat transferred reversibly ($q_{rev}$) divided by the absolute temperature ($T$) at which the transfer takes place.

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

Unit: Joules per Kelvin per mole ($\text{J K}^{-1} \text{ mol}^{-1}$).

38. Predict the sign of $\Delta S$ for: (a) Freezing of water (b) Sublimation of Iodine.

(a) Freezing of water: Liquid changes to solid. Randomness decreases. Therefore, $\Delta S$ is negative.
(b) Sublimation of Iodine: Solid changes to gas. Randomness increases greatly. Therefore, $\Delta S$ is positive.

39. State the Second Law of Thermodynamics in terms of entropy.

Second Law of Thermodynamics: The total entropy of the universe (system + surroundings) increases continuously during any spontaneous process.

$$ \Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} > 0 \quad \text{(for spontaneous process)} $$

40. Define Gibbs Free Energy ($G$).

Gibbs Energy ($G$): It is a thermodynamic state function defined as $G = H - TS$, where $H$ is enthalpy, $T$ is absolute temperature, and $S$ is entropy.

Physically, the change in Gibbs free energy ($\Delta G$) represents the maximum amount of useful, non-expansion work that can be extracted from a closed system at constant temperature and pressure.

41. Write the Gibbs-Helmholtz equation.

For a process occurring at constant temperature, the change in Gibbs free energy ($\Delta G$) is given by the Gibbs-Helmholtz equation:

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

42. Relate the sign of $\Delta G$ with the spontaneity of a reaction.

At constant temperature and pressure:

If $\Delta G < 0$ (negative), the process is spontaneous.
If $\Delta G > 0$ (positive), the process is non-spontaneous (spontaneous in the reverse direction).
If $\Delta G = 0$, the process is at equilibrium.

43. Under what conditions is an endothermic reaction spontaneous?

For an endothermic reaction, $\Delta H$ is positive ($+$). For it to be spontaneous, $\Delta G$ must be negative.

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

Since $\Delta H$ is positive, the only way $\Delta G$ can be negative is if $\Delta S$ is positive (increase in randomness) AND the temperature $T$ is high enough such that the magnitude of $T\Delta S$ is greater than $\Delta H$ ($|T\Delta S| > |\Delta H|$).

Thus, it is spontaneous at high temperatures.

44. Write the relationship between standard Gibbs energy change ($\Delta G^\circ$) and the equilibrium constant ($K$).

The relationship is given by the equation:

$$ \Delta G^\circ = -RT \ln K $$

Converting natural log to base 10:

$$ \Delta G^\circ = -2.303 \, RT \log_{10} K $$

Where $R$ is the universal gas constant, $T$ is absolute temperature, and $K$ is the equilibrium constant.

45. State the Third Law of Thermodynamics.

Third Law of Thermodynamics: The entropy of a perfectly crystalline substance is zero at absolute zero temperature ($0 \text{ K}$).

This law allows the calculation of absolute entropies of substances at any temperature $T$ by measuring heat capacities.

46. 2 moles of an ideal gas expand isothermally and reversibly from 10 L to 20 L at 300 K. Write the formula to calculate the work done.

Formula for max work: $W_{max} = -2.303 \, nRT \log_{10}(V_2/V_1)$

Here, $n = 2$, $T = 300 \text{ K}$, $V_1 = 10 \text{ L}$, $V_2 = 20 \text{ L}$, $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

Calculation setup: $W = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10}(20/10)$

47. If the internal energy of a system increases by 200 J when 400 J of heat is added to it, how much work was done and by whom?

From First Law: $\Delta U = q + W$

Given: $\Delta U = +200 \text{ J}$, $q = +400 \text{ J}$ (heat added).

$200 = 400 + W \implies W = 200 - 400 = -200 \text{ J}$.

Since $W$ is negative, 200 J of work was done by the system on the surroundings (expansion).

48. What is the value of $\Delta H$ and $\Delta U$ for an isothermal reversible expansion of an ideal gas?

For an ideal gas, Internal Energy ($U$) and Enthalpy ($H$) depend only on Temperature.

In an isothermal process, $\Delta T = 0$.

Therefore, $\Delta U = 0$ and $\Delta H = 0$.

49. Calculate $\Delta n_g$ for the reaction: $C_3H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_2O_{(l)}$

$\Delta n_g$ = (moles of gaseous products) - (moles of gaseous reactants)

Products: 3 moles $CO_{2(g)}$. (Water is liquid, so ignore it).

Reactants: 1 mole $C_3H_{8(g)}$ + 5 moles $O_{2(g)}$ = 6 moles gas.

$\Delta n_g = 3 - 6 = -3$ moles.

50. Why does entropy increase when a solid dissolves in a liquid?

In a solid state, particles are arranged in a highly ordered, rigid crystal lattice (low entropy). When it dissolves in a liquid (solvent), the crystal lattice breaks down, and the solute particles become free to move randomly throughout the liquid phase. This massive increase in randomness and disorder leads to an increase in entropy ($\Delta S > 0$).

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