NCERT Solutions For Class 11 Chemistry chapter 6- Thermodynamics

NCERT Solutions For Class 11 Chemistry chapter 6- Thermodynamics

NCERT Solutions for class-11 Chemistry Chapter 6 Thermodynamics is prepared by our senior and renown teachers of Physics Wallah primary focus while solving these questions of class-11 in NCERT textbook, also do read theory of this Chapter 6 Thermodynamics while going before solving the NCERT questions. Our Physics Wallah team Prepared Other Subjects NCERT Solutions for class 11.

Chapter 6 Thermodynamics

Answer the following Questions.

1. Choose the correct answer. A thermodynamic state function is a quantity

(i) used to determine heat changes

(ii) whose value is independent of path

(iii) used to determine pressure volume work

(iv) whose value depends on temperature only

Solution :

A thermodynamic state function is a quantity Whose value is independent of a path. Functions like p, V, T etc. depend only on the state of a system and not on the path.

Hence, alternative (ii) is correct.

2. For the process to occur under adiabatic conditions, the correct condition is:

(i) ∆T = 0

(ii) ∆p = 0

(iii) q = 0

(iv) w = 0

Solution : A system is said to be under adiabatic conditions if there is no exchange Of heat between the system and its surroundings. Hence, under adiabatic conditions, q = 0.

Therefore, alternative (iii) is correct,

3. The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element

Solution : The enthalpy of all elements in their standard state is zero. Therefore, alternative (ii) is correct

4. ΔU ^⊖ of combustion of methane is −X kJ mol ⁻¹ . The value of ΔH⊖ is

(i)=ΔU ^⊖

(ii) >ΔU ^⊖

(iii) <=ΔU ^⊖

(iv) = 0

Solution :

SinceΔH ^θ =ΔU ^θ +Δn _g RT and ΔU ^θ =−Xkmol ⁻¹

ΔH ^θ =(−X)+Δn _g RT

⇒△H ^θ <ΔU ^θ

Therefore, alternative (iii) is correct.

5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3kJmol ⁻¹ −393.5kJmol ⁻¹ , and −285.8kJmol ⁻¹

respectively. Enthalpy of formation of CH ₄ will be

(i)−74.8kJmol ⁻¹

(ii)−52.27kJmol ⁻¹

(iii)+74.8kJmol ⁻¹

(iv)+52.26kJmol ⁻¹

Solution :

According to the question,

(i)CH ₄ (g)+2O ₂ (g)⟶CO ₂ (z)+2H ₂ O(g)ΔH=−890.3kJmol ⁻¹

(ii)C(x)+O ₂ (y)⟶CO ₂ (g)ΔH=−393.5kJmol ⁻¹

(iii)2H ₂ (g)+O ₂ (z)⟶2H ₂ O(g)

ΔH=−285.8kJmol ⁻¹

Thus, the desired equation is the one that represents the formation of CH ₄ (g) i.e..,

= [-395.5 + 2(-285.8) - (-890.3)] kJ Mol ^-1

= -74.8 kJ Mol ^-1

∴ Enthalpy of formation of CH ₄ (g)=−74.8kJmol ⁻¹ Hence, alternative (i) is correct.

6. A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature

Solution :
For a reaction to be spontaneous, ΔG should be negative.
ΔG = ΔH – TΔS
According to the question, for the given reaction,
ΔS = positive
ΔH = negative (since heat is evolved)
⇒ ΔG = negative
Therefore, the reaction is spontaneous at any temperature.
Hence, alternative (iv) is correct.

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

Solution :

According to the first law of thermodynamics,
ΔU = q + W    (i)
Where,
ΔU = change in internal energy for a process
q = heat
W = work
Given,
q = + 701 J (Since heat is absorbed)
W = –394 J (Since work is done by the system)
Substituting the values in expression (i), we get
ΔU = 701 J + (–394 J)
ΔU = 307 J
Hence, the change in internal energy for the given process is 307 J.

8. The reaction of cyanamide,NH ₂ CN(s) with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7kJmol ⁻¹ at 298 K. Calculate enthalpy change for the reaction at 298 K.

NH ₂ CN(g) + 3/2O ₂ (g)→N ₂ (g)+CO ₂ (g)+H ₂ O(l)

Solution :

Enthalpy change for a reaction (ΔH) is given by the expression,
ΔH = ΔU + Δn _g RT
Where,
ΔU = change in internal energy
Δn _g = change in number of moles
For the given reaction,
Δng = ∑n _g (products) – ∑n _g (reactants)
= (2 – 1.5) moles
Δn _g = 0.5 moles
And,
ΔU = –742.7 kJ mol ^–1
T = 298 K
R = 8.314 × 10 ^–3 kJ mol ^–1 K ^–1
Substituting the values in the expression of ΔH:
ΔH = (–742.7 kJ mol ^–1 ) + (0.5 mol) (298 K) (8.314 × 10 ^–3 kJ mol ^–1 K ^–1 )
= –742.7 + 1.2
ΔH = –741.5 kJ mol ^–1

9. Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 Jmol ⁻¹ K ⁻¹ .

Solution :

From the expression of heat (q) q=m⋅ c. ΔT

Where,

c= molar heat capacity

m= mass of substance

ΔT= change in temperature

Substituting the values in the expression of q:

q=(60/27mol)(24Jmol ⁻¹ K ⁻¹ )(20K)

q=1066.7J

q=1.07k

10. Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at

–10.0°C.ΔfusH=6.03kJmol−1 at 0 ^∘ C

C _p [H ₂ O(I)]=75.3Jmol ₋₁ K−1

C _ρ [H ₂ O(s)]=36.8Jmol ⁻¹ K ⁻¹

Solution :

Total enthalpy change involved in the transformation is the of the following changes:

(a) Energy change involved in the transformation of 1 mol of water at 10 ^∘ C to 1 mol of water at 0 C.

(b) Energy change involved in the transformation of 1 mol of water at 0∘  to 1 mol of ice at 0 ^∘ C

(c) Energy change involved in the transformation of 1 mol of ice at0∘C to 1 mol of ice at−10 ^∘ C.

ΔH= C _p [ H ₂ OCl]ΔT+ ΔH _fivering + C _ρ [H ₂ O _(s) ]ΔT

=(75.3] mol ⁻¹ K ⁻¹ )(0−10)K + (−6.03 × 10 ³ Jmol ⁻¹ )+(36.8] mol ⁻¹ K ⁻¹ )(−10 −0)K

=−7533 mol ⁻¹ − 6030Jmol ⁻¹ − 368Jmol ⁻¹

=−7151J mol ⁻¹

=−7.151kJmol ⁻¹

Hence, the enthalpy change involved in the transformation is−7.151kJmol ⁻¹ .

11. Enthalpy of combustion of carbon to CO ₂ is –393.5−7.151kJmol ⁻¹ . Calculate the heat released upon formation of 35.2 g ofCO ₂ from carbon and dioxygen gas.

Solution :

Formation of CO ₂ from carbon and dioxygen gas can be represented as:

C(s) + O ₂ (g)⟶CO ₂ (g)

Δ _f H=−393.5kJmol ⁻¹

(1 mole =44g) Heat released on formation of 44gCO ₂ =−393.5kJmol−1

∴ Heat released on formation of 35.2gCO ₂

=−314.8kJmol ⁻¹

12. Enthalpies of formation of CO(g), CO ₂ (g), N ₂ O(g) and N ₂ O ₄ (g) are −110,−393,81 and 9.7kJmol ⁻¹ respectively.

Find the value of ∆H for the reaction:

N ₂ O ₄ (g)+3CO(g)→N ₂ O(g+3CO ₂ (g]

Solution :

Δ _r H for a reaction is defined as the difference between ΔH value of products and ΔH value of reactants.

Δ,H=∑Δ,H( products )−∑Δ _f H( reactants )

For the given reaction,

N ₂ O _4(g) + 3CO _(g) ⟶ N ₂ O _(g) + 3CO ₂ (g)

Δ _r H=[ {ΔfH(N ₂ O)+3ΔJH(CO ₂ )}−{ΔfH (N ₂ O ₄ ) + 3ΔjH(CO)} ]

Substituting the values ofΔH for N ₂ O,CO ₂ ,N ₂ O ₄ , and CO

From the question, we get:

Δ _r H=[ { 81kJmol ⁻¹ + 3(−393)kJmol ⁻¹ } − {9.7kJmol ⁻¹ +3(−110) kJmol ⁻¹ }]

Δ _r H=−7777kJmol ⁻¹

Hence, the value ofΔ _r H

for the reaction is −777.7 kJmol ⁻¹ .

13. Given

N ₂ (g) + 3H ₂ (g) ⟶ 2NH ₃ (y);

Δ _r Hθ=−92.4kJmol ⁻¹

What is the standard enthalpy of formation ofNH ₃ gas?

Solution : Standard of formation of a compound is the charge in enthalpy that takes place during the formation of 1 mole Of a substance in its standard form from its constituent elements in their standard state.

Re-writing the given equation for 1 mole of NH _3(g).

1/2N ₂ (g)+3/2H ₂ (g)⟶NH _3(g)

∴ Standard enthalpy of formation of NH _3(g)

=1/2 Δ _r Hθ = 1/2(−92.4 kJmol ⁻¹ )= −46.2kJmol ⁻¹

14. Calculate the standard enthalpy of formation ofCH ₃ OH(l)  from the following data:

CH ₃ OH(l)+3/2O ₂ (g]→CO ₂ (g)+2H ₂ O(l): Δ,H∘=−726kJmol ⁻¹

C(graphite) +O ₂ (g)→CO2(g]: ΔeH=−393kJmol ⁻¹

H ₂ (g)+1/2O ₂ (g)→H ₂ O(1); Δ,H=−286kJmol ⁻¹

Solution :

The reaction that takes place during the formation ofCH ₃ OH(l) can be written as:

C(s) + 2H ₂ O(g) + 1/2O ₂ (G), ⟶ CH ₃ OH( _η) (1)

The reaction (I) can be obtained from the given reactions by following the algebraic calculations as:

Equation (ii) + 2 × equation (iii) – equation (i)
Δ _f Hθ [CH ₃ OH(l)] = ΔcH ^θ + 2Δ _f H ^θ [H ₂ O(l)] – Δ _r H ^θ
= (–393 kJ mol ^–1 ) + 2(–286 kJ mol ^–1 ) – (–726 kJ mol ^–1 )
= (–393 – 572 + 726) kJ mol ^–1
Δ _f H ^θ [CH ₃ OH(l)] = –239 kJ mol ^–1

15. Calculate the enthalpy change for the process CCl ₄ (g)→C(g)+4Cl(g) and calculated bond enthalpy of C−Cl in CCl ₄ (g)

Δ _va pH ^θ (CC|4) = 30.5kJmol ⁻¹ Δ _f H ^θ (CCl4) =−135.5kJmol−1

Δ _a H ^θ (C) = 715.0kJmol ⁻¹ , where Δ _a H ^θ is enthalpy of atomisation

Δ ₂ H ^θ (Cl ₂ ) = 242kJmol ⁻¹

Solution :

The chemical equations implying to the given values of enthalpies” are:
(1) CCl _4(l) à CCl _4(g) ; ΔvapH ^Θ = 30.5 kJmol ⁻¹
(2) C _(s) à C _(g) ΔaH ^Θ = 715 kJmol ⁻¹
(3) Cl2 _(g) à 2Cl _(g) ; Δ _a H ^Θ = 242 kJmol ⁻¹
(4) C _(g) + 4Cl _(g) à CCl _4(g) ; ΔfH ^Θ

= -135.5 kJmol ⁻¹ ΔH for the process CCl _4(g) à C _(g) + 4Cl _(g) can be measured as:
ΔH=Δ _a H ^Θ (C) + 2ΔaH ^Θ (Cl ₂ ) – Δ _vap H ^Θ –ΔfH
= (715kJmol ⁻¹ ) + 2(kJmol ⁻¹ ) – (30.5kJmol ⁻¹ ) – (-135.5kJmol ⁻¹ )
Therefore, H= 1304kJmol ⁻¹
The value of bond enthalpy for C-Cl in CCl ₄ (g)
= 1304/4kJmol ⁻¹
= 326 kJmol ⁻¹

16. For an isolated system, ∆U = 0, what will be ∆S ?

Solution :
ΔS will be positive i.e., greater than zero
Since ΔU = 0, ΔS will be positive and the reaction will be spontaneous.

17. For the reaction at 298 K,

2A + B → C

∆H = 400kJmol ⁻¹

and ∆S = 0.2kJmol ⁻¹

At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range.

Solution :

From the expression

ΔG= ΔH−TΔS

Assuming the reaction at equilibrium,δ

T for the reaction would be:

T=(ΔH−ΔG)1/ΔS=ΔH/ΔS(ΔG=0 at equilibrium)

=400kJmol ⁻¹ 0.2 kJK ⁻¹ mol ⁻¹ T=2000K

For the reaction to be spontaneous,ΔG must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.

18. For the reaction,2Cl(g)→Cl ₂ (g) , what are the signs of ∆H and ∆S ?

Solution :

∆H and ∆S are negative

The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy Is being released. Hence ∆H is negative.

Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ∆S is negative for the given reaction.

19. For the reaction

2A(g)+B(g)→2D(g)ΔUe=−10.5kJ and ΔS∘=−44.1JK ⁻¹

Calculate ΔG ^⊖ for the reaction, and predict whether the reaction may occur spontaneously

Solution :

For the given reaction,

2A(g)+B(g)→2D(g)Δηg=2−(3)=−1 mole

Substituting the value of ΔU ^θ

in the expression of ΔH:

ΔH ^θ =ΔU ^θ +Δn _g RT

=(−10.5kJ)−(−1)(8.314×10 ⁻³ kJK−1 mol ⁻¹ ) (298K) = −10.5kJ ⁻² .48kJΔH ^⊖ =−12.98kJ

Substituting the values ofΔH ^⊖ and ΔS ^⊖ in the expression of ΔG ^⊖

ΔG ^θ = △H ^θ −TΔS ^θ

=−12.98kJ − ( 298K) (−44.1JK ⁻¹ ) = −12.98kJ + 13.14 kJ ΔG ^⊖ =+0.16kJ

SinceΔG ^θ for the reaction is positive, the reaction will not occur spontaneously.

20. The equilibrium constant for a reaction is 10. What will be the value of ∆G ^⊖ ? R = 8.314JK ⁻¹ mol ⁻¹

T = 300 K.

Solution :

From the expression, ΔG ^θ = −2.303 RT logk _eq

ΔG ^θ for the reaction,

=(2.303) (8.314JK ⁻¹ mol ⁻¹ ) (300K) log10=−5744.14Jmol ⁻¹

=−5.744kkmol ⁻¹

21. Comment on the thermodynamic stability of NO(g) , given

12N _2(g) +12O _2(g) →NO _(g) ;

Δ _r H ^⊖ = 90kJmol ⁻¹ NO _(g) + 12O _2(g) →NO ₂ (g):

Δ _r H ^e = −74kJmol ⁻¹

Solution :

The positive value of Δ _r H indicates that heat is absorbed during the formation of NO(g), j. This means that NO(g) has higher than the reactants(N2 and O2) .

Hence, NO _(g) is unstable.  The negative value o _f Δ _r H

H indicates that heat is evolved during the formation ofNO _2(g) from NO _(g) and O _2(g)

. The product,NO _2(g) is stabilized with minimum energy.

Hence, unstableNO _(g) changes to unstableNO _2(g).

22. Calculate the entropy change in surroundings when 1.00 mol ofH ₂ O(l) is formed under standard conditions.ΔH ^θ =−286kJ mol ⁻¹

Solution :

It is given that 286 kJmol ⁻¹ of heat is evolved the formation of 1 mol ofH ₂ O(l).

Thus, an equal amount of heat will be absorbed by the surroundings.

q _surr = +286 kJmol ⁻¹

Entropy change(ΔS _surr )  for the surroundings = q _surr / 7

=286kJmol ⁻¹ / 298k

∴ΔS _surt =959.73 Jmol ⁻¹ K ⁻¹

NCERT Solutions For Class 11 Chemistry Chapter Wise.

1. Chapter 1 Some Basic Concepts of Chemistry

2. Chapter 2 Structure of Atom

3. Chapter 3 Classification of Elements and Periodicity in Properties

4. Chapter 4 Chemical Bonding and Molecular Structure

5. Chapter 5 States of Matter

6. Chapter 6 Thermodynamics

7. Chapter 7 Equilibrium

8. Chapter 8 Reodx Reactions

9. Chapter 9 Hydrogen

10. Chapter 10 S-Block Elements

11. Chapter 11 P-Block Elements

12. Chapter 12 Some Basics Principles and Techniques

13. Chapter 13 Hydrocarbons

14. Chapter 14 Environmental Chemistry