# NCERT Solution Class-11 Chapter-6 Thermodynamics

**Exercise : Solutions of Questions on Page Number : 182**

** Q1 :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.**

** Answer :**

A thermodynamic state function is a quantity whose value is independent of a path.

Functions like p, V, Tetc. depend only on the state of a system and not on the path.

Hence, alternative (ii) is correct.

**Q2 :For the process to occur under adiabatic conditions, the correct condition is:**

** (i) ΔT = 0**

** (ii) Δp = 0**

** (iii) q = 0**

** (iv) w= 0**

** Answer :**

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.

**Q3 :The enthalpies of all elements in their standard states are:**

** (i) unity**

** (ii) zero**

** (iii) < 0**

** (iv) different for each element**

** Answer :**

The enthalpy of all elements in their standard state is zero.

Therefore, alternative (ii) is correct.

**Q4 :ΔU ^{ø}¸of combustion of methane is – XkJ mol^{-1}. The value of ΔHÃŽÂ¸is**

**(i) = ΔU**

^{ø}¸**(ii) > ΔU**

^{ø}¸**(iii) < ΔU**

^{ø}¸**(iv) = 0**

**Answer :**

Since ΔH

^{ø}¸= ΔU

^{ø}+ Δn

_{g}RT and ΔU

^{ø}¸= -XkJ mol-1,

ΔH

^{ø}¸= (-X) + Δn

_{g}RT.

⇒ ΔH

^{ø}< ΔU

^{ø}

Therefore, alternative (iii) is correct.

**Q5 :The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol ^{-1} -393.5 kJ mol-1, and -285.8 kJ mol^{-1} respectively. Enthalpy of formation of CH4(g) will be**

**(i) -74.8 kJ mol**

^{-1}(ii) -52.27 kJ mol^{-1}**(iii) +74.8 kJ mol**

^{-1}(iv) +52.26 kJ mol^{-1}.**Answer :**

According to the question,

Thus, the desired equation is the one that represents the formation of CH

_{4}(g)i.e.,

Enthalpy of formation of CH

_{4}(g) = -74.8 kJ mol

^{-1}

Hence, alternative (i) is correct.

**Q6 :A reaction, A + B → C + D + qis 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**

** Answer :**

For a reaction to be spontaneous, ΔGshould 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.

**Q7 :In a process, 701 J of heat is absorbed by a system and 394 J ofwork is done by the system. What is the change in internal energy for the process?**

** Answer :**

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.

**Q8 :The reaction of cyanamide, NH _{2}CN(s),with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be -742.7 kJ mol^{-1}at 298 K. Calculate enthalpy change for the reaction at 298 K.**

**Answer :**

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,

Δn

_{g}= ∑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 x 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 x 10-3 kJ mol

^{-1}K

^{-1})

= -742.7 + 1.2

ΔH = -741.5 kJ mol

^{-1}

**Q9 :Calculate the number of kJ of heat necessary to raise the temperatureof 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol ^{-1} K^{-1}.**

**Answer :**

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 = 1066.7 J

q = 1.07 kJ

**Q10 :Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. Δ _{fus}H = 6.03 kJ mol^{-1} at 0°C.**

**C**

_{p}[H_{2}O(l)] = 75.3 J mol^{-1}K^{-1}**C**

_{p}[H_{2}O(s)] = 36.8 J mol^{-1}K^{-1}**Answer :**

Total enthalpy change involved in the transformation is the sum 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 at 0°C to 1 mol of ice at -10°C.

= (75.3 J mol

^{-1}K

^{-1}) (0 – 10)K + (-6.03 × 103 J mol

^{-1}) + (36.8 J mol

^{-1}K

^{-1}) (-10-0)K

= -753 J mol

^{-1}– 6030 J mol

^{-1}-368 J mol

^{-1}

= -7151 J mol

^{-1}

= -7.151 kJ mol

^{-1}

Hence, the enthalpy change involved in the transformation is -7.151 kJ mol

^{-1}.

**Q11 :Enthalpy of combustion of carbon to CO _{2 }is -393.5 kJ mol^{-1}. Calculate the heat released upon formation of 35.2 g of CO_{2 }from carbon and dioxygen gas.**

**Answer :**

Formation of CO

_{2}from carbon and dioxygen gas can be represented as:

(1 mole = 44 g)

Heat released on formation of 44 g CO

_{2}= -393.5 kJ mol

^{-1}

Heat released on formation of 35.2 g CO

_{2 }

= -314.8 kJ mol

^{-1}

**Q12 :Enthalpies of formation of CO(g), CO _{2}(g), N_{2}O(g) and N_{2}O_{4}(g) are -110 kJ mol^{-1}, -393 kJ mol^{-1}, 81 kJ mol^{-1}and 9.7 kJ mol^{-1}respectively. Find the value of Δ_{r}H for the reaction:**

**N**

_{2}O_{4}(g)+ 3CO(g) N_{2}O(g)+ 3CO_{2}(g)**Answer :**

Δ

_{r}Hfor a reaction is defined as the difference between ΔfHvalue of products and Δ

_{f}H value of reactants.

For the given reaction,

N

_{2}O

_{4}(g)+ 3CO(g) N

_{2}O(g)+ 3CO

_{2}(g)

Substituting the values of Δ

_{f}H for N

_{2}O, CO2, N

_{2}O

_{4},and CO from the question, we get:

Hence, the value of Δ

_{r}H for the reaction is.

**Q13 :Given**

** ; Δ _{r}Hθ= -92.4 kJ mol-1**

**What is the standard enthalpy of formation of NH**

_{3 }gas?**Answer :**

Standard enthalpy of formation of a compound is the change 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),

Standard enthalpy of formation of NH

_{3}(g)

= ½ ΔrHθ

= ½ (-92.4 kJ mol-1)

= -46.2 kJ mol-1

**Q14 :Calculate the standard enthalpy of formation of CH _{3}OH(l) from the following data:**

**CH**

_{3}OH(l) + O_{2}(g) CO_{2}(g) + 2H_{2}O(l) ; Δ_{r}Hθ = -726 kJ mol^{-1}**C(g) + O**

_{2}(g) CO_{2}(g) ; Δ_{c}Hθ =-393 kJ mol-1**H**

_{2}(g) +**Answer :**

The reaction that takes place during the formation of CH

_{3}OH(l)can be written as:

C(s)+ 2H

_{2}O(g)+ O

_{2}(g) CH

_{3}OH(l) (1)

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:

Equation (ii) + 2 × equation (iii) – equation (i)

Δ

_{f}H

^{θ }[CH

_{3}OH(l)] = Δ

_{c}H

^{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

**Q15 :Calculate the enthalpy change for the process**

** CCl _{4}(g) → C(g) + 4Cl(g)**

**and calculate bond enthalpy of C-Cl in CCl**

_{4}(g).**Δ**

_{vap}H^{θ}¸ (CCl_{4}) = 30.5 kJ mol^{-1}.**Δ**

_{f}H^{θ}¸ (CCl_{4}) = -135.5 kJ mol^{-1.}**Δ**

_{a}H^{θ}¸ (C) = 715.0 kJ mol^{-1}, where Δ_{a}H^{θ}¸ is enthalpy of atomisation**Δ**

_{a}H^{θ}¸ (Cl_{2}) = 242 kJ mol^{-1}**Answer :**

The chemical equations implying to the given values of enthalpies are:

Δ

_{vap}H

^{θ}= 30.5 kJ mol

^{-1}

Δ

_{a}H

^{θ}= 715.0 kJ mol

^{-1}

Δ

_{a}H

^{θ}= 242 kJ mol

^{-1}

Δ

_{f}H

^{θ}= -135.5 kJ mol

^{-1}

Enthalpy change for the given process can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) – Equation (i)- Equation (iv)

ΔH = Δ

_{a}H

^{θ}(C) + 2Δ

_{a}H

^{θ}(Cl

_{2}) – Δ

_{vap}H

^{θ}– Δ

_{f}H

= (715.0 kJ mol

^{-1}) + 2(242 kJ mol

^{-1}) – (30.5 kJ mol

^{-1}) -(-135.5 kJ mol

^{-1})

ΔH = 1304 kJ mol

^{-1}

Bond enthalpy of C-Cl bond in CCl

_{4}(g)

= 326 kJ mol

^{-1}

**Q16 :For an isolated system, ΔU = 0, what will be ΔS?**

** Answer :**

ΔS will be positive i.e., greater than zero

Since ΔU= 0, ΔS will be positive and the reaction will be spontaneous.

**Q17 :For the reaction at 298 K,**

** 2A + B → C**

** ΔH = 400 kJ mol-1and ΔS = 0.2 kJ K-1mol-1**

** At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?**

** Answer :**

From the expression,

ΔG= ΔH- TΔS

Assuming the reaction at equilibrium, ΔT for the reaction would be:

(ΔG = 0 at equilibrium)

T= 2000 K

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

**Q18 :For the reaction,**

** 2Cl(g) → Cl _{2}(g),what are the signs of ΔH and ΔS ?**

**Answer :**

ΔHand ΔS 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, ΔH is negative.

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

**Q19 :For the reaction**

** 2A(g) + B(g) → 2D(g)**

** ΔU ^{θ}¸ = -10.5 kJ and ΔS^{θ}¸= -44.1 JK^{-1}.**

**Calculate ΔGÃŽÂ¸ for the reaction, and predict whether the reaction may occur spontaneously.**

**Answer :**

For the given reaction,

2 A(g) + B(g) → 2D(g)

Δn

_{g}= 2 – (3)

= -1 mole

Substituting the value of ΔU

^{θ}¸ in the expression of ΔH:

ΔH

^{θ}¸ = ΔU

^{θ}¸ + Δn

_{g}RT

= (-10.5 kJ) – (-1) (8.314 x 10-3 kJ K

^{-1}mol

^{-1}) (298 K)

= -10.5 kJ – 2.48 kJ

ΔH

^{θ}= -12.98 kJ

Substituting the values of ΔH

^{θ}and ΔS

^{θ}in the expression of ΔG

^{θ}:

ΔG

^{θ}= ΔH

^{θ}– TΔS

^{θ}

= -12.98 kJ – (298 K) (-44.1 J K-1)

= -12.98 kJ + 13.14 kJ

ΔG

^{θ}= + 0.16 kJ

Since ΔG

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

**Q20 :The equilibrium constant for a reaction is 10. What will be the value of ΔG ^{θ}? R = 8.314 JK^{-1}mol^{-1}, T= 300 K.**

**Answer :**

From the expression,

ΔG

^{θ}= -2.303 RT log K eq

ΔG

^{θ}for the reaction,

= (2.303) (8.314 JK-1mol-1) (300 K) log10

= -5744.14 Jmol

^{-1}

= -5.744 kJ mol

^{-1}

**Q21 :Comment on the thermodynamic stability of NO(g),given**

** N _{2}(g) + O_{2}(g)→ NO(g); Δ_{r}H^{θ}= 90 kJ mol^{-1}**

**NO(g)+O**

_{2}(g)→NO_{2}(g) : Δ_{r}H^{θ}= -74 kJ mol^{-1}**Answer :**

The positive value of ΔrH indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants (N

_{2}and O

_{2}). Hence, NO(g) is unstable.

The negative value of ΔrH indicates that heat is evolved during the formation of NO

_{2}(g) from NO(g) and O

_{2}(g). The product, NO

_{2}(g) is stabilized with minimum energy.

Hence, unstable NO(g) changes to stable NO

_{2}(g).

**Q22 :Calculate the entropy change in surroundings when 1.00 mol of H _{2}O(l)is formed under standard conditions. Δ_{f}H^{θ}= -286 kJ mol^{-1}.**

**Answer :**

It is given that 286 kJ mol

^{-1}of heat is evolved on the formation of 1 mol of H

_{2}O(l). Thus, an equal amount of heat will be absorbed by the surroundings.

q

_{surr}= +286 kJ mol

^{-1}

Entropy change (ΔSsurr) for the surroundings =

ΔS

_{surr}= 959.73 J mol

^{-1}K

^{-1}