Page No 183: - Chapter 6 - Thermodynamics - Ncert Solutions class 11 - Chemistry

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Question 6.9:

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 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

Question 6.10:

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. ΔfusH = 6.03 kJ mol^–1 at 0°C.

Cp[H₂O(l)] = 75.3 J mol^–1 K^–1

Cp[H₂O(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.

Question 6.11:

Enthalpy of combustion of carbon to CO₂ is –393.5 kJ mol^–1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.

Answer:

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

(1 mole = 44 g)

Heat released on formation of 44 g CO₂ = –393.5 kJ mol^–1

Heat released on formation of 35.2 g CO₂

= –314.8 kJ mol^–1

Question 6.12:

Enthalpies of formation of CO₍g_), CO₂₍g_), N₂O₍g₎ and N₂O₄₍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 ΔrH for the reaction:

N₂O₄₍g₎ + 3CO₍g₎

N₂O₍g₎ + 3CO₂₍g₎

Answer:

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

For the given reaction,

N₂O₄₍g₎ + 3CO₍g₎

 N₂O₍g₎ + 3CO₂₍g₎

Substituting the values of ΔfH for N₂O, CO₂, N₂O_4, and CO from the Question, we get:

Hence, the value of ΔrH for the reaction is .

Question 6.13:

Given

; ΔrH^θ = –92.4 kJ mol^–1

What is the standard enthalpy of formation of NH₃ 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 NH₃₍g_),

Standard enthalpy of formation of NH₃₍g₎

= ½ ΔrH^θ

= ½ (–92.4 kJ mol^–1)

= –46.2 kJ mol^–1

Question 6.14:

Calculate the standard enthalpy of formation of CH₃OH₍l₎ from the following data:

CH₃OH₍l₎ +  O₂₍g₎   CO₂₍g₎ + 2H₂O₍l₎ ; ΔrH^θ = –726 kJ mol^–1

C₍g₎ + O₂₍g₎   CO₂₍g₎ ; ΔcH^θ = –393 kJ mol^–1

H₂₍g₎ + O₂₍g₎   H₂O₍l₎ ; ΔfH^θ = –286 kJ mol^–1.

Answer:

The reaction that takes place during the formation of CH₃OH₍l₎ can be written as:

C₍s₎ + 2H₂O₍g₎ +  O₂₍g₎

 CH₃OH₍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)

ΔfH^θ [CH₃OH₍l₎] = Δ_cH^θ + 2ΔfH^θ [H₂O₍l₎] – ΔrH^θ

= (–393 kJ mol^–1) + 2(–286 kJ mol^–1) – (–726 kJ mol^–1)

= (–393 – 572 + 726) kJ mol^–1

ΔfH^θ [CH₃OH₍l₎] = –239 kJ mol^–1

Question 6.15:

Calculate the enthalpy change for the process

CCl₄₍g₎ → C₍g₎ + 4Cl₍g₎

and calculate bond enthalpy of C–Cl in CCl₄₍g_).

ΔvapH^θ (CCl₄) = 30.5 kJ mol^–1.

ΔfH^θ (CCl₄) = –135.5 kJ mol^–1.

ΔaH^θ (C) = 715.0 kJ mol^–1, where ΔaH^θ is enthalpy of atomisation

ΔaH^θ (Cl₂) = 242 kJ mol^–1

Answer:

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

 ΔvapH^θ = 30.5 kJ mol^–1

 ΔaH^θ = 715.0 kJ mol^–1

 ΔaH^θ = 242 kJ mol^–1

 ΔfH = –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 = ΔaH^θ(C) + 2ΔaH^θ (Cl₂) – ΔvapH^θ – ΔfH

= (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

Question 6.16:

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.

Question 6.17:

For the reaction at 298 K,

2A + B → C

ΔH = 400 kJ mol^–1 and ΔS = 0.2 kJ K^–1 mol^–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, T should be greater than 2000 K.

Question 6.18:

For the reaction,

2Cl₍g) → Cl₂₍g_), what are the signs of ΔH and ΔS ?

Answer:

ΔH and 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.

Question 6.19:

For the reaction

2A₍g₎ + B₍g₎ → 2D₍g₎

ΔU^θ = –10.5 kJ and ΔS^θ= –44.1 JK^–1.

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

Answer:

For the given reaction,

2 A₍g₎ + B₍g₎ → 2D₍g₎

Δng = 2 – (3)

= –1 mole

Substituting the value of ΔU^θ in the expression of ΔH:

ΔH^θ = ΔU^θ + ΔngRT

= (–10.5 kJ) – (–1) (8.314 × 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.