MPBSE Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Mpbse Class 11 Chemistry Chapter 6 Solutions

MPBSE Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics Question And Answers

Question 1. Give two examples of path-dependent quantities. Are they properties ofa system?
Answer:

Two path-dependent quantities are heat (q) and work ( w). These are not the properties ofa system.

Question 2. Under what conditions will a system be in thermodynamic equilibrium?
Answer:

A system will be in thermodynamic equilibrium if it simultaneously maintains mechanical equilibrium, thermal equilibrium and chemical equilibrium.

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 3. Why is a process occurring in an open container considered to be an isobaric? What is the origin of the internal energy of a system? Why cannot the absolute value of internal energy be determined?
Answer:

A process in an open container takes place under constant atmospheric pressure. Thus, it is an isobaric process.

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Question 4. Is the internal energy of a system at 25°C greater or less than its internal energy at 50°C?
Answer:

The internal energy of a system increases with the temperature rise. So, the internal energy ofa system will be greater at 50 °C than that at 25 °C.

Question 5. Under which condition will the pressure-volume work be, \(w=-\int_{V_1}^{V_2} P d V?\) pressure of the gas.
Answer:

⇒ \(w=-\int_{V_1}^{V_2} P_{e x} d V\)

So, in case ofa reversible process, \(w=-\int_{V_1}^{V_2} P d V\)

Question 6. According to the first law of thermodynamics, AU = q + w. Write down the form of this equation for the following processes: Cyclic process Adiabatic process Isothermal expansion of an ideal gas Process occurring in an isolated system.
Answer:

An isolated system does not exchange energy or matter with its surroundings. So, for a process occurring in an isolated system, q = 0 and w = 0. Therefore, ΔU = q + w or, ΔU= 0 + 0 or, ΔU = 0

Equilibrium Class 11 Chemistry Solutions

Question 7. The definition of enthalpy shows that for n mol of an ideal gas H = U + nRT.
Answer:

If the enthalpy, internal energy, pressure and volume of ‘n’ mol of an ideal gas at a temperature of TK are, U, P and V respectively, then H = U+PV. For the ‘n’ mol of an ideal gas, PV = nRT. So, H = U+nRT.

Question 8. Prove that for an Ideal gas undergoing an isothermal change, AH = 0.
Answer:

The change in enthalpy of an ideal gas undergoing a process, ΔH = ΔU+nRAT. In an isothermal process, ΔT = O. So, ΔH = A{Again, in an isothermal process of an ideal gas, A U = 0 and hence AH = 0.

Question 9. Under what conditions are

1. ΔU = qv
2. ΔH = qp

Answer:

ΔU = qv; Conditions: Closed system, constant volume, only P-V work is considered

ΔH = qp; Conditions: Closed system, constant pressure, only P-V work is considered.

Question 10. Give an example ofa combustion reaction whose standard enthalpy change is equal to the standard enthalpy of formation ofthe compound formed in the reaction.
Answer:

At 25 C the standard heat of combustion of solid naphthalene [C₁₀H₈(s)] is 5147 kj. mol-1. this means that at 25 c and 1 atm pressure when 1 mol of solid naphthalene is completely burnt in the presence of oxygen the enthalpy change that occurs is 51747kj.

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 11. The standard enthalpy of combustion of C_xH_y) at 25°C is Q kj. mol^-1 . Write down the thermochemical equation for the combustion reaction of this compound.
Answer:

The combustion reaction for C(s, graphite) is:

This reaction also C(s, graphite) + O₂(g)→CO₂(g). represents the formation reaction of CO₂(g). Therefore, at 25°C, the standard heat of combustion of C(s, graphite) = the standard heat formation of CO₂(g)

Question 12. Consider the given enthalpy diagram, and > calculate the unknown AH by applying Hess’s law.
Answer:

Following the given diagram, we have

1. A+B→ C + 2D ; ΔH = ?
2. A + B→ E + 2D; ΔH = +27kJ
3. E + 2D →C+ 2D;ΔH = -13kJ

Adding equation 2 and equation 3, we get A + B→C+2D; ΔH = (27- 13)kJ

= 14 kJ

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 13.  For the reaction, A + B →D, AH is -30 kj. Suppose, D is prepared from A and B and then it is again converted into A and B by following the stages D → E → A + B. Calculate the total enthalpy change in these two stages.
Answer:

A + B → D; ΔH = -30 kj ……………………….(1)

The process D→E →A + B comprises the following two steps:

D→E ⋅⋅⋅⋅⋅(2) and E→A + B

Overall reaction: D→A + B The reaction (4) is the opposite ofthe reaction (1).

So, the enthalpy change in reaction (4) is +30 kj.

Hence, the total enthalpy change in steps (2) and (3) is +30 kJ

Question 14. Water remains in equilibrium with its vapour at 100°C and atm. Will the transformation of water into its vapour be spontaneous at this pressure and temperature?
Answer:

No. Since water and its vapour are in equilibrium, neither the forward process (water→vapour) nor the reverse process (vapour →water) is favourable

Question 15. If the process A→B occurs reversibly, then the change in entropy of the system is ΔS₁. When the same process occurs irreversibly, the change in entropy of the system is ΔS₂. Will the value of ΔS₁ be greater than, less than or equal to the value of ΔS₂?
Answer:

Since entropy is a state function, the change in entropy ofa system in a process does not depend upon whether the process is carried out reversibly or irreversibly.

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 17. Write the relation between ΔS_sys & AS_surrwhen a process reaches equilibrium. What will be the value of ΔS_univ?
Answer:

For a process at equilibrium

⇒ \(\Delta S_{\text {system }}+\Delta S_{\text {surr }}=0\text { But } \Delta S_{s y g}+\Delta S_{\text {surr }}=\Delta S_{\text {univ }} \text {. So, } \Delta S_{\text {univ }}=0\)

Question 18. Give two examples of spontaneous processes in which the disorderliness of the system decreases.
Answer:

Transformation of water into ice at 1 atm pressure and below (T’C temperature. Condensation of water vapor at1 atm pressure and below 100 C temperature.

Question 19. What do you mean by a perpetual motion machine of the second kind?
Answer:

A machine working in a cyclic process absorbs heat from a single thermal reservoir and completely converts the heat into the equivalent amount of work, is called a perpetual motion machine of a second kind. This type of machine contradicts the second law of thermodynamics, & it is impossible to construct.

Question 20. A certain amount of gas is enclosed in a container with permeable and diathermal walls. Which type of system does the gas belong to?
Answer:

The walls of the container are diathermal. So the system can exchange heat with its surroundings. Again, the walls ofthe container are permeable. So, the system can also exchange matter with its surroundings. Hence, the gas belongs to an open system.

Chemical Equilibrium Class 11 Ncert Solutions

Question 21. Does the volume of a closed system remain fixed?
Answer:

If the walls of a closed system are non-rigid or movable, then the volume of the system does not remain fixed. For example, a gas enclosed in a cylinder fitted with a movable piston is considered a closed system. Here, the volume of the gas (system) can be increased or decreased by altering the pressure of the gas (system)

Question 22. Give an example of a thermodynamic quantity which is not a state function. Is it a property of a system?
Answer:

Heat is not a state function because heat absorbed or by a system in a process depends upon the path of the realised It is not a property of the system.

Question 23. Give an example of a process which Is simultaneously isothermal and adiabatic.
Answer:

Adiabatic free expansion of an ideal gas (or isothermal free expansion of an ideal gas). The reason is that no exchange of heat occurs between the system and surroundings in this process and the temperature of the tire system remains constant throughout the process.

Question 24. At 25°C, the standard reaction enthalpy for the reaction AB₃(g)→1/2A₂(g)+3/2(g) is. find the standard reaction enthalpy for the reaction.
Answer:

Writing this equation. in reverse manner and multiplying both sides by 2, we get,

A₂(g) + 3B₂(g)→2AB₃(g); -2AH°. So, at 25°C, the standard

Question 25. Mention the standard state of sulphur and iodine at 25
Answer: 

At 25°C, the standard state of sulphur is solid rhombic sulphur [S(rhombic, s)] and that of iodine is solid iodine [1₂(s)].

Question 26. What do you mean by ‘the enthalpy of solidification of water at 0°C and 1 atm pressure = -6.02 kj-mol^-1 .’?
Answer:

This means that 6.02 kj of heat is released when one mole of water completely freezes to ice at 0°C and 1 atm pressure.

Question 27. Why are spontaneous natural processes irreversible?
Answer: ‘

The spontaneous or natural processes are irreversible because the thermodynamic equilibrium of the system is not maintained in such types of processes.

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 28. Which of the following will have a greater entropy?

1. 1 mol of H₂ gas (T = 300 K, V = 5ml, )
2. 1 mol of H₂ gas (T = 300 K, V = 10mL).

Answer:

As the entropy of a gas increases with the increase in its volume,

1. The entropy of mol of H₂ (T = 300 K, V = 10 mL)
2. Will be greater than that of l mol of H₂ (T = 300 K, V = 5 mL).

Question 29. For a process, ΔS_sys = -15 J.K^-1 .. For what value of ASsurr will the process be non-spontaneous?
Answer:

The condition of non-spontaneity of a process is \(\Delta S_{s y s}+\Delta S_{s u r r}<0.\) \(\Delta S_{s y s}=-15 \mathrm{~J} \cdot \mathrm{K}^{-1}\), then will be non-spontaneous

Question 30. A gas is allowed to expand against zero external pressure. Explain with reason whether the process is reversible or irreversible.
Answer:

The expansion ofa gas against zero external pressure is an irreversible process. As the opposing pressure is zero, the gas expands rapidly, and it cannot maintain thermodynamic equilibrium during its expansion.

Question 31. In a process, 701 J of heat is absorbed by a system and 394J of work is done by the system. What is the change in internal energy for the process?
Answer:

Given: q = +701J and w = -394 J {-ve sign as the work is done by the system)

Now, ΔU = q + w or, A U = (701- 394)J = +307 J.

So, the change in internal energy of the system = +307 J.

Question 32. The latent heat of the vaporization of water at a normal boiling point is 40.75 kJ. mol^-1 . . Calculate the change in entropy of vaporization.
Answer:

Given:

⇒ \(\Delta H_{\text {vap }}\) = 4075 kJ. mol^-1,

Tb =100C = 375k

⇒ \(\Delta S=\frac{\Delta H_{v a p}}{T_b}=\frac{40.75 \times 10^3 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{373}=109.25 \mathrm{~J} \cdot \mathrm{mol}^{-1}\)

Question 33. Due mole of ideal gas is expanded isothermally. In this process, which of the quantity (or quantities) among w. q, ΔH , ΔH is(are) zero or >0 or <0?
Answer:

During isothermal expansion, heat is absorbed and work is done by the gas. So q > 0 and w < 0. Again internal energy and enthalpy remain the same during isothermal expansion of an ideal gas. Thus, ΔU = 0 and ΔH = 0 . For the isothermal expansion of mol of ideal gas q>0, w< 0, ΔU = 0, ΔH = 0.

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 34. Give examples of two processes by which the internal energy of a gas can be increased.
Answer:

The internal energy of a gas can be increased by increasing the die temperature ofthe gas. If a gas is compressed adiabatically (considering only P-V work) its internal energy increases.

Question 35. Give examples of three processes in which the change in internal energy of the system is zero.
Answer:

The change in internal energy of any cyclic process is zero. The change in internal energy of an ideal gas during isothermal expansion or compression is zero. In adiabatic free expansion of an ideal gas, the change in internal energy is equal to zero.

Question 36. What do you mean by the standard enthalpy of atomisation of chlorine at 25°C = + 121 kj.mol^-1 .?
Answer:

This means that at 25°C and 1 atm pressure, 121 kj of heat is required to produce 1 mol of gaseous Cl-atom from Cl₂(g). Thus, the change in enthalpy for the process,

⇒ \(\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{Cl}(\mathrm{g}); \Delta H_{\text {atom }}^0=+121 \mathrm{~kJ}\)

Solutions For Class 11 Chemistry Chapter 6 Chemical Thermodynamics

Question 37. When does the entropy of the system attain maximum value for a spontaneous or irreversible process occurring in an isolated system? Under this condition, what will be the change in entropy of the system?
Answer:

When a spontaneous or irreversible process occurs in an isolated system, the entropy of the system increases with the progress of the process towards equilibrium. The value of entropy becomes maximum when the process attains equilibrium, and there occurs no further change in the entropy ofthe system. Thus, the value of entropy is maximum at the equilibrium state ofthe process and under this condition, the change in entropy ofthe system is zero.

Question 38. Is the entropy change of a system influenced by the change in temperature? Explain.
Answer:

The entropy of a system is highly dependent on temperature. With the increase or decrease in temperature, the randomness of the constituent particles (atoms, molecules or ions) of a system increases or decreases. Now, the entropy of a system is a measure of the randomness of its constituent particles. Thus, the entropy change ofa system is influenced by the change in temperature.