Short & Long Question Answers: Thermodynamics | Chemistry Class 11 - NEET PDF Download

Short Answer Type Questions

Q1. Under what conditions the heat evolved or absorbed is equal to the internal energy change?

Answer. The heat evolved or absorbed is equal to the internal energy change at constant volume.

Q2. Why enthalpy of neutralization of HF is greater than 57.1 kJ mol^–1?

Answer. This is due to the high hydration energy of fluoride ions.

Q3. Define a cyclic process.

Answer. A cyclic process is one in which the initial and final states are the same. It is a series of processes that end with the system in the same state in which it began.

The initial and final internal energies of a system are equal when it goes through a cyclic process. As a result, in any cyclic process, the internal energy change is zero.

Q4. Name two intensive and extensive properties of a system.

Answer. The intensive and extensive properties are as follows:

Intensive properties: Viscosity, refractive index.

Extensive properties: Mass, volume, heat capacity, etc.

Q5. Which of the following statements is correct?

1. The presence of reacting species in a covered beaker is an example of an open system.
2. There is an exchange of energy as well as a matter between the system and the surroundings in a closed system.
3. The presence of reactants in a closed vessel made up of copper is an example of a closed system.
4. The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.

Correct Answer: (c) The presence of reactants in a closed vessel made up of copper is an example of a closed system.

Explanation: There is no exchange of matter in a closed system (for example, the presence of reactants in a closed vessel made of conducting material, such as copper), but there is an exchange of energy between the system and its surroundings.

Q6. What are the applications of Hess’s Law of constant heat summation?

Answer. Hess’s law is used-

• To calculate the heat of formation, combustion, neutralisation, ionisation, and other processes.
• Determine the enthalpies of the reactants and products.
• Calculate the bond enthalpies.
• Calculate the lattice energies of crystalline solids.

Q7. What are heat capacities at constant volume and constant pressure? What is the relationship between them?

Answer. Heat capacity at constant volume (Cv): The amount of heat supplied to a system to raise its temperature by one degree Celsius while keeping the volume of the system constant is referred to as its heat capacity at constant volume (Cv).

Heat capacity at constant pressure (Cp): The amount of heat supplied to a system to raise its temperature by one degree Celsius while keeping the external pressure constant is referred to as its heat capacity at constant pressure (Cp).

Cp – Cv = R is the relationship between Cp and Cv.

Q8. (i) For a reaction both ΔH and ΔS are negative. Under what conditions does the reaction occur spontaneously?

(ii) For a reaction both ΔH and ΔS are positive. Under what conditions does the reaction occur spontaneously?

Answer. ΔG = ΔH−TΔS

For a reaction to be spontaneous ΔG should be negative.

(i) Both ΔH and ΔS are negative, ΔG can be negative only if TΔS < ΔH in magnitude.

ΔG = ΔH − TΔS

ΔG = (−) − T(−)

This is possible only if either ΔH has a large negative value or T is so low that TΔS < ΔH.

(ii) Both ΔH and ΔS are positive so ΔG will be negative only if TΔS > ΔH in magnitude.

ΔG = ΔH−TΔS

ΔG = (+) − T(+)

Thus either ΔS should be very large so that even if T is low, TΔS is greater than ΔH, or if ΔS is small, T should be high so that TΔS > ΔH.

Q9. How will you get to the relationship q_p = q_v+ Δn_gRT?

Answer: The relationship can be derived as follows-

Enthalpy change ΔH = q_p, where q_p is the heat change at constant pressure,

Internal energy change ΔU = q_v, where q_v is the heat change at constant volume.

Now ΔH = ΔU + PΔV

For ideal gases PV = nRT

∴ ΔH = ΔU + (PV₂ – PV₁)

= ΔU + P(V₂ – V₁) = ΔU + (n₂RT – n₁RT)

= ΔU + RT(n₂ – n₁) = ΔU + Δn_gRT

or

q_p = q_v + Δn_gRT

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

Answer.

For an isothermal reversible expansion of an ideal gas

w = – nRT log V₂/V₁ = – 2.303 nRT log V₂/V₁

Putting n = 0.75 mol; V1 = 15 L; V2 = 25 L, T = 27 + 273 = 300 K R = 8.314 JK-1 mol-1.

w = – 2.303 × 0.75 × 8.314 × 300 log 25/15

w = -955.5J

Long Answer Type Questions

Q1. Define-

(i) Standard enthalpy of formation.

(ii) Standard enthalpy of combustion

(iii) Enthalpy of atomization

(iv) Enthalpy of solution

(v) Lattice enthalpy

Answer. (i) Standard enthalpy of formation.

The change in enthalpy when one mole of a compound is formed from its elements in their standard states under standard conditions, i.e. at 298K and 101.3kPa pressure, is referred to as the standard enthalpy of formation.

(ii) Standard enthalpy of combustion

The enthalpy change when one mole of a compound is completely burned in oxygen with all reactants and products in their standard state under standard conditions is defined as standard enthalpy of combustion (298K and 1 bar pressure).

(iii) Enthalpy of atomization

This is the enthalpy change that occurs when one mole of a substance is completely broken down into its atoms in the gas phase.

(iv) Enthalpy of solution

The heat change that occurs when one mole of a substance dissolves in a specified amount of a solvent is defined as the enthalpy of solution. The enthalpy of solution at infinite dilution is the enthalpy change observed when dissolving two moles of a substance in an infinite amount of solvent.

(v) Lattice enthalpy

The enthalpy change that occurs when one mole of an ionic compound dissociates into its ions in a gaseous state is referred to as the lattice enthalpy of an ionic compound.

Q2. 1 g or graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure according to the reaction

C(graphite) + O₂(g) → CO₂(g).

During the reaction, the temperature rises from 298 K to 299 K. If the heat, the capacity of the bomb calorimeter is 20.7 kJ K^-1 what is the enthalpy change for the above reaction at 298 K and 1 atm?

Answer. q = Heat change = Cv × ΔT, where q is the heat absorbed by the calorimeter.

The quantity of heat from the reaction will have the same magnitude, but the opposite sign, because heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.

∴ q = – Cv × ΔT = – 20.7 kJ K-1 × (299 – 298) K

= –20.7kJ.

Here, the negative sign indicates the exothermic nature of the reaction.

Thus, ΔU for the combustion of 1 g of graphite = – 20.7 kJ K^–1

For combustion of 1 mol of graphite

= – [12.0×(−20.7)]/1 = – 2.48 × 102 kJ mol^–1

Since Δng_g = 0

∴ ΔH = ΔU = – 2.48 × 102 kJ mol^–1.

Q3. For the reaction Ag₂O(s) → 2Ag(s) + ½ O₂(g) : ΔH = 30.56 kJ mol^–1 and ΔS = 6.66JK^–1 mol^–1 (at 1 atm). Calculate the temperature at which ΔG is equal to zero. Also predict the direction of the reaction (i) at this temperature and (ii) below this temperature.

Answer. Given:

ΔH = 30.56 kJmol^–1 = 30560Jmol^–1

ΔS = 6.66 ×10−3kJ K^–1mol^–1

T = ? at which ΔG = 0

ΔG = ΔH − TΔS

0 = ΔH − TΔS

T = ΔH / ΔS

T = (30.56 kJmol^–1) / (6.66×10−3 kJ K^–1mol^–1)

T = 4589K

(i) At 4589K ; ΔG = 0 the reaction is in equilibrium.

(ii) At temperature below 4589k, ΔH > TΔSΔG = ΔH −TΔS > 0, the reaction in the forward direction, is non spontaneous. In other words, the reaction occurs in the backward direction.