Important Questions for Class 11 Chemistry Chapter 5 Thermodynamics

Important Questions for Class 11 Chemistry Chapter 5: Chapter 5 of Class 11 Chemistry, Thermodynamics, focuses on energy changes in chemical processes and their feasibility. Key topics include system, surroundings, types of systems (open, closed, isolated), state functions, and the first law of thermodynamics.

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Important Questions for Class 11 Chemistry Chapter 5 Thermodynamics

1. Define a system

Ans: A system in thermodynamics is a part of the universe within a specified boundary.

2. Define surroundings

Ans: The remaining of the universe which might be in a position to exchange energy and matter with the system is termed its surroundings.

3. State the first law of thermodynamics

Ans: The first law of thermodynamics states that energy can neither be created nor be destroyed but can be transformed from one form to another.

4. What kind of system is the coffee held in a cup?

Ans: Coffee held in a cup is an open system because it can exchange matter (water vapour and energy (heat) with the surroundings.

5. Give an example of an isolated system

Ans: Coffee held in a thermos flask is an example of an isolated system because it can neither exchange energy nor matter with the surroundings.

6. Name the different types of the system.

Ans: There are three types of system

1. Open system
2. Closed system
3. Isolated system

7. What will happen to internal energy if work is done by the system?

Ans: There is a decrease in internal energy of the system if work is done by the system.

8. From a thermodynamic point of view, to which system the animals and plants belong?

Ans: Open system

9. How may the state of the thermodynamic system be defined?

Ans: The state of the thermodynamic system can be defined by specifying values of state variables like temperature, pressure and volume.

10. Define enthalpy

Ans: It is defined as the total heat content of the system.

11. Give the mathematical expression of enthalpy.

Ans: Mathematically H=U+PVH=U+PV where UU is internal energy

12. When is enthalpy change ΔHΔH -

(i) positive (ii) negative.

Ans:

1. In an endothermic reaction it absorbs heat from the surroundings so enthalpy change in positive
2. In an exothermic reaction heat is evolved so enthalpy change is negative.

13. Give the expression for

(i) isothermal irreversible change,

and isothermal reversible change

Ans: For isothermal reversible change

Q=   −wQ=   −w

Q=pext(Vf−Vi)   Q=pext(Vf−Vi)

For isothermal reversible change

Q=   −wQ=   −w

Q=2.303nRTlog(VfVi)   Q=2.303nRTlog⁡(VfVi)

14. Define Heat capacity

Ans: Specific heat /specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or one Kelvin).

15. Define specific heat.

Ans: Specific heat /specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or one Kelvin).

16. Give the mathematical expression of heat capacity.

Ans: The mathematical expression of heat capacity is as follows

q=mcΔTq=mcΔT

If,

m=1m=1

q=CΔT

17. With the help of first law of thermodynamics and H = U + pV H = U + pV prove Δ H = q p ΔH=qp

Ans: Enthalpy is defined as follows

H = U + p V H=U+pV

Δ H = Δ ( U + p V ) ΔH=Δ(U+pV)

Δ H = Δ U + Δ ( p V ) ΔH=ΔU+Δ(pV)

Δ H = Δ U + p Δ V + V Δ p ΔH=ΔU+pΔV+VΔp ……(i)

From first law of thermodynamics,

Δ U = q + w ΔU=q+w

= q − p Δ V =q−pΔV

From equation (i) and (ii), we get

Δ H = q − p Δ V + p Δ V + V Δ p ΔH=q−pΔV+pΔV+VΔp

Δ H = q + V Δ p ΔH=q+VΔp

At constant pressure,

Δ p = 0 Δp=0

V Δ p = 0 VΔp=0

Δ H = q p ΔH=qp at constant pressure

Therefore, Δ H = q p ΔH=qp

18. Why is the difference between Δ H ΔH and Δ U ΔU not significant for solids or liquids?

Ans: The difference between Δ H ΔH and Δ U ΔU is not significant for solids or liquids because systems made up entirely of solids and/or liquids do not experience significant volume changes when heated, the difference between and is usually insignificant.

19. What is an extensive and intensive property?

Ans: Extensive property is defined as the property which depends on the quantity or size of the matter present in the system.

Intensive property is defined as the property which depends on the quantity or size of matter present in the system.

20. Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to 3 2 R 32R .

Ans: For an ideal gas, the average kinetic energy per mole of the gas at any temperature is given by E k = 3 2 R T Ek=32RT

Hence increase in average kinetic energy of gas for 1 o C 1oC rise in temperature is

Δ E k −→ = 3 2 R ( T + 1 ) − 3 2 R T ΔEk→=32R(T+1)−32RT

By definition, E K −→ EK→ is the molar heat capacity of gas at constant volume C V CV

∴ C V = 3 2 R ∴CV=32R

21. A 1.25 1.25 g sample of octane ( C 18 H 18 C18H18 ) is burnt in excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rises from 294.05 294.05 to 300.78 300.78 K K . If heat capacity of the calorimeter is 8.93 KJ K - 1 8.93KJK - 1 . find the heat transferred to calorimeter.

Ans: Mass of octane,

M = 1.250 g M=1.250g

M = 0.00125 M=0.00125

Heat capacity, c = 8.93 KJ K - 1 c=8.93KJK - 1

Increase in temperature, Δ T = 300.78 − 294.05 ΔT=300.78−294.05

Δ T = 6.73 K ΔT=6.73K

So, the heat transferred to calorimeter is

m c Δ T = 0.00125 × 8.93 × 6.73 mcΔT=0.00125×8.93×6.73

= 0.075 KJ =0.075KJ

22. What is the relation between the enthalpy of reaction and bond enthalpy?

Ans: In a chemical reaction the breaking of bonds and formation of new bonds in products takes place. The heat of reaction is dependent on the values needed to break the bond formation. Thus

Heat of reaction = Heat required for breaking of bonds in reactants − − Heat required for breaking of bonds in products

1. Δ H o = ΔHo= Bond energy is required to break the bonds - Bond energy required to form the bonds = Bond energy of reactants – Bond energy of products.

Benefits of Solving Important Questions for Class 11 Chemistry Chapter 5

Enhanced Conceptual Understanding

Application of Theoretical Knowledge

Familiarity with Question Patterns

Improved Problem-Solving Skills