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## NCERT SOLUTIONS FOR CLASS 11 CHEMISTRY CHAPTER 6- THERMODYNAMICS – Exercises

### Chapter-6 Thermodynamics

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 :

SinceΔHθ=ΔUθ+ΔngRT and ΔUθ=−Xkmol−1

ΔHθ=(−X)+ΔngRT

⇒△Hθ<ΔUθ

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 will be

(i)−74.8kJmol−1

(ii)−52.27kJmol−1

(iii)+74.8kJmol−1

(iv)+52.26kJmol−1

Solution :

According to the question,

(i)CH4(g)+2O2(g)⟶CO2(z)+2H2O(g)ΔH=−890.3kJmol−1

(ii)C(x)+O2(y)⟶CO2(g)ΔH=−393.5kJmol−1

(iii)2H2(g)+O2(z)⟶2H2O(g)

ΔH=−285.8kJmol−1

Thus, the desired equation is the one that represents the formation of CH4(g) i.e..,

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

= -74.8 kJ Mol-1

∴ Enthalpy of formation of CH4(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,NH2CN(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.

NH2CN(g) + 3/2O2(g)→N2(g)+CO2(g)+H2O(l)

Solution :

Enthalpy change for a reaction (ΔH) is given by the expression,
ΔH = ΔU + ΔngRT
Where,
ΔU = change in internal energy
Δng = change in number of moles
For the given reaction,
Δng = ∑ng (products) – ∑ng (reactants)
= (2 – 1.5) moles
Δn= 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−1K−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−1K−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 0C

Cp[H2O(I)]=75.3Jmol−1K−1

Cρ[H2O(s)]=36.8Jmol−1K−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 0C

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

ΔH= Cp[ H2OCl]ΔT+ ΔH fivering + Cρ[H2O(s)]ΔT

=(75.3] mol−1K−1)(0−10)K + (−6.03 × 103Jmol−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 COis –393.5−7.151kJmol−1 . Calculate the heat released upon formation of 35.2 g ofCO2  from carbon and dioxygen gas.

Solution :

Formation of CO2  from carbon and dioxygen gas can be represented as:

C(s) + O2(g)⟶CO2(g)

ΔfH=−393.5kJmol−1

(1 mole =44g) Heat released on formation of 44gCO2=−393.5kJmol−1

∴ Heat released on formation of 35.2gCO2

=−314.8kJmol−1

12. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are −110,−393,81 and 9.7kJmol−1 respectively.

Find the value of ∆H for the reaction:

N2O4(g)+3CO(g)→N2O(g+3CO2(g]

Solution :

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

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

For the given reaction,

N2O4(g) + 3CO(g)⟶ N2O(g) + 3CO2(g)

ΔrH=[ {ΔfH(N2O)+3ΔJH(CO2)}−{ΔfH (N2O4) + 3ΔjH(CO)} ]

Substituting the values ofΔH for N2O,CO2,N2O4 , and CO

From the question, we get:

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

ΔrH=−7777kJmol−1

Hence, the value ofΔrH

for the reaction is −777.7 kJmol−1.

13. Given

N2(g) + 3H2(g) ⟶ 2NH3(y);

ΔrHθ=−92.4kJmol−1

What is the standard enthalpy of formation ofNHgas?

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 NH3(g).

1/2N2(g)+3/2H2(g)⟶NH3(g)

∴ Standard enthalpy of formation of NH3(g)

=1/2 ΔrHθ = 1/2(−92.4 kJmol−1 )= −46.2kJmol−1

14. Calculate the standard enthalpy of formation ofCH3OH(l)  from the following data:

CH3OH(l)+3/2O2(g]→CO2(g)+2H2O(l): Δ,H∘=−726kJmol−1

C(graphite) +O2(g)→CO2(g]: ΔeH=−393kJmol−1

H2(g)+1/2O2(g)→H2O(1); Δ,H=−286kJmol−1

Solution :

The reaction that takes place during the formation ofCH3OH(l) can be written as:

C(s) + 2H2O(g) + 1/2O2(G), ⟶ CH3OH(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)
ΔfHθ [CH3OH(l)] = ΔcHθ + 2ΔfHθ [H2O(l)] – ΔrHθ
= (–393 kJ mol–1) + 2(–286 kJ mol–1) – (–726 kJ mol–1)
= (–393 – 572 + 726) kJ mol–1
ΔfHθ [CH3 OH(l)] = –239 kJ mol–1

15. Calculate the enthalpy change for the process CCl4(g)→C(g)+4Cl(g) and calculated bond enthalpy of C−Cl in CCl4(g)

ΔvapHθ (CC|4) = 30.5kJmol−1 ΔfHθ(CCl4) =−135.5kJmol−1

ΔaHθ(C) = 715.0kJmol−1, where ΔaHθ  is enthalpy of atomisation

Δ2Hθ(Cl2) = 242kJmol−1

Solution :

The chemical equations implying to the given values of enthalpies” are:
(1) CCl4(l) à CCl4(g) ; ΔvapHΘ = 30.5 kJmol−1
(2) C(s) à C(g) ΔaHΘ = 715 kJmol−1
(3) Cl2(g) à 2Cl(g) ; ΔaHΘ = 242 kJmol−1
(4) C(g) + 4Cl(g) à CCl4(g); ΔfHΘ

= -135.5 kJmol−1 ΔH for the process CCl4(g) à C(g) + 4Cl(g) can be measured as:
ΔH=ΔaHΘ(C) + 2ΔaHΘ(Cl2) – ΔvapHΘ–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 CCl4(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)→Cl2(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 ΔUθ

in the expression of ΔH:

ΔHθ=ΔUθ+ΔngRT

=(−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

ΔGθ= △Hθ−TΔSθ

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

SinceΔGθ  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, ΔGθ= −2.303 RT logk eq

ΔGθ for the reaction,

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

=−5.744kkmol−1

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

12N2(g)+12O2(g)→NO(g);

ΔrH= 90kJmol−1NO(g) + 12O2(g)→NO2(g):

ΔrHe= −74kJmol−1

Solution :

The positive value of ΔrH 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 ofΔrH

H indicates that heat is evolved during the formation ofNO2(g) from NO(g) and O2(g)

. The product,NO2(g)  is stabilized with minimum energy.

Hence, unstableNO(g) changes to unstableNO2(g).

22. Calculate the entropy change in surroundings when 1.00 mol ofH2O(l) is formed under standard conditions.ΔHθ=−286kJ mol−1

Solution :

It is given that 286 kJmol−1 of heat is evolved the formation of 1 mol ofH2O(l).

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

qsurr= +286 kJmol−1

Entropy change(ΔSsurr)  for the surroundings = qsurr / 7

=286kJmol−1/ 298k

∴ΔSsurt =959.73 Jmol−1K−1

## Conclusions for NCERT SOLUTIONS FOR CLASS 11 CHEMISTRY CHAPTER 6- THERMODYNAMICS

An academic team of knowledgeable members of SWC has produced and published the NCERT Solutions for class 11’s Chemistry chapter for your use as a reference. You can get answers to all of the chapters of the NCERT Chemistry class 11 here at SWC. Please make use of the following NCERT answers that were created by SWC as a reference for this chapter. In addition to that, study the chapter’s theory before attempting to solve the NCERT problems.