NCERT Solutions for Class 11 Chemistry Chapter 6 – Chemical Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 6 – Chemical Thermodynamics

Page No 182:

Question 6.1:

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.

Answer:

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.

Question 6.2:

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.

Question 6.3:

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.

Question 6.4:

ΔU^θof combustion of methane is – X kJ mol^–1. The value of ΔH^θ is

(i) = ΔU^θ

(ii) > ΔU^θ

(iii) < ΔU^θ

(iv) = 0

Answer:

Since ΔH^θ = ΔU^θ + Δn_gRT and ΔU^θ = –X kJ mol^–1,

ΔH^θ = (–X) + Δn_gRT.

⇒ ΔH^θ < ΔU^θ

Therefore, alternative (iii) is correct.

Question 6.5:

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 CH_4(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.

Question 6.6:

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

Answer:

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.

Question 6.7:

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?

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.

Question 6.8:

The reaction of cyanamide, NH₂CN_(s),with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol^–1at 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 + Δn_gRT

Where,

ΔU = change in internal energy

Δn_g = 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 × 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

Page No 183:

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. Δ_fusH = 6.03 kJ mol^–1 at 0°C.

C_p[H₂O(l)] = 75.3 J mol^–1 K^–1

C_p[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), CO_2(g), N₂O_(g) and N₂O_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_rH for the reaction:

N₂O_4(g) + 3CO_(g)

N₂O_(g) + 3CO_2(g)

Answer:

Δ_rH for a reaction is defined as the difference between Δ_fH value of products and Δ_fH value of reactants.

For the given reaction,

N₂O_4(g) + 3CO_(g)

 N₂O_(g) + 3CO_2(g)

Substituting the values of Δ_fH for N₂O, CO₂, N₂O_4, and CO from the question, we get:

Hence, the value of Δ_rH for the reaction is.

Question 6.13:

Given

; Δ_rH^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 NH_3(g),

Standard enthalpy of formation of NH_3(g)

= ½ Δ_rH^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) + O_2(g)  CO_2(g) + 2H₂O_(l) ; Δ_rH^θ = –726 kJ mol^–1

C_(g) + O_2(g)  CO_2(g) ; Δ_cH^θ = –393 kJ mol^–1

H_2(g) +O_2(g)  H₂O_(l) ; Δ_fH^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) + O_2(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)

Δ_fH^θ [CH₃OH_(l)] = Δ_cH^θ + 2Δ_fH^θ [H₂O_(l)] – Δ_rH^θ

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

= (–393 – 572 + 726) kJ mol^–1

Δ_fH^θ [CH₃OH_(l)] = –239 kJ mol^–1

Question 6.15:

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_vapH^θ (CCl₄) = 30.5 kJ mol^–1.

Δ_fH^θ (CCl₄) = –135.5 kJ mol^–1.

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

Δ_aH^θ (Cl₂) = 242 kJ mol^–1

Answer:

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

 Δ_vapH^θ = 30.5 kJ mol^–1

 Δ_aH^θ = 715.0 kJ mol^–1

 Δ_aH^θ = 242 kJ mol^–1

 Δ_fH = –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_aH^θ(C) + 2Δ_aH^θ (Cl₂) – Δ_vapH^θ – Δ_fH

= (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, Tshould be greater than 2000 K.

Question 6.18:

For the reaction,

2Cl_(g) → Cl_2(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)

Δn_g = 2 – (3)

= –1 mole

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

ΔH^θ = ΔU^θ + Δn_gRT

= (–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.

Page No 184:

Question 6.20:

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 logK_eq

ΔG^θ for the reaction,

= (2.303) (8.314 JK^–1 mol^–1) (300 K) log10

= –5744.14 Jmol^–1

= –5.744 kJ mol^–1

Question 6.21:

Comment on the thermodynamic stability of NO_(g), given

N_2(g) + O_2(g) → NO_(g) ; Δ_rH^θ = 90 kJ mol^–1

NO_(g) +O_2(g) → NO_2(g) : Δ_rH^θ= –74 kJ mol^–1

Answer:

The positive value of Δ_rH indicates that heat is absorbed during the formation of NO_(g). This means that NO_(g) has higher energy than the reactants (N₂ and O₂). Hence, NO_(g) is unstable.

The negative value of Δ_rH 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).

Question 6.22:

Calculate the entropy change in surroundings when 1.00 mol of H₂O_(l) is formed under standard conditions. Δ_fH^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 H₂O_(l). Thus, an equal amount of heat will be absorbed by the surroundings.

q_surr = +286 kJ mol^–1

Entropy change (ΔS_surr) for the surroundings = 

ΔS_surr = 959.73 J mol^–1 K^–1