**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.

**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}