NCERT Solutions for Class 11 Science Chemistry Chapter 6 Thermodynamics are provided here with simple step-by-step explanations. These solutions for Thermodynamics are extremely popular among class 11 Science students for Chemistry Thermodynamics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 11 Science Chemistry Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 11 Science Chemistry are prepared by experts and are 100% accurate.

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.

Page No 182:

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.

Page No 182:

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.

Page No 182:

Question 6.4:

ΔUθof combustion of methane is – X kJ mol–1. The value of ΔHθ is

(i) = ΔUθ

(ii) > ΔUθ

(iii) < ΔUθ

(iv) = 0

Answer:

Since ΔHθ = ΔUθ + ΔngRT and ΔUθ = –X kJ mol–1,

ΔHθ = (–X) + ΔngRT.

⇒ ΔHθ < ΔUθ

Therefore, alternative (iii) is correct.

Page No 182:

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 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 CH4 (g) i.e.,

Enthalpy of formation of CH4(g) = –74.8 kJ mol–1

Hence, alternative (i) is correct.

Page No 182:

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 = ΔHTΔS

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.

Page No 182:

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.

Page No 182:

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 + 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

Δng = 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

Page No 183:

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.

Page No 183:

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

Page No 183:

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, N2O4, and CO from the question, we get:

Hence, the value of ΔrH for the reaction is.

Page No 183:

Question 6.13:

Given

; ΔrHθ = –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 ΔrHθ

= ½ (–92.4 kJ mol–1)

= –46.2 kJ mol–1

Page No 183:

Question 6.14:

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

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

C(g) + O2(g) CO2(g) ; ΔcHθ = –393 kJ mol–1

H2(g) +O2(g) H2O(l) ; ΔfHθ = –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)

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 [CH3OH(l)] = –239 kJ mol–1

Page No 183:

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).

ΔvapHθ (CCl4) = 30.5 kJ mol–1.

ΔfHθ (CCl4) = –135.5 kJ mol–1.

ΔaHθ (C) = 715.0 kJ mol–1, where ΔaHθ is enthalpy of atomisation

ΔaHθ (Cl2) = 242 kJ mol–1

Answer:

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

ΔvapHθ = 30.5 kJ mol–1

ΔaHθ = 715.0 kJ mol–1

ΔaHθ = 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 = ΔaHθ(C) + 2ΔaHθ (Cl2) – ΔvapHθ – 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 CCl4 (g)

= 326 kJ mol–1

Page No 183:

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.

Page No 183:

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 = ΔHTΔS

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

G = 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.

Page No 183:

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.

Page No 183:

Question 6.19:

For the reaction

2A(g) + B(g) → 2D(g)

ΔUθ = –10.5 kJ and ΔSθ= –44.1 JK–1.

Calculate ΔGθ 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 ΔUθ in the expression of ΔH:

ΔHθ = ΔUθ + ΔngRT

= (–10.5 kJ) – (–1) (8.314 × 10–3 kJ K–1 mol–1) (298 K)

= –10.5 kJ – 2.48 kJ

ΔHθ = –12.98 kJ

Substituting the values of ΔHθ and ΔSθ in the expression of ΔGθ:

ΔGθ = ΔHθTΔSθ

= –12.98 kJ – (298 K) (–44.1 J K–1)

= –12.98 kJ + 13.14 kJ

ΔGθ = + 0.16 kJ

Since ΔGθ 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 ΔGθ? R = 8.314 JK–1 mol–1, T = 300 K.

Answer:

From the expression,

ΔGθ = –2.303 RT logKeq

ΔGθ for the reaction,

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

= –5744.14 Jmol–1

= –5.744 kJ mol–1

Page No 184:

Question 6.21:

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

N2(g) + O2(g) → NO(g) ; ΔrHθ = 90 kJ mol–1

NO(g) +O2(g) → NO2(g) : ΔrHθ= –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 (N2 and O2). Hence, NO(g) is unstable.

The negative value of ΔrH indicates that heat is evolved during the formation of NO2(g) from NO(g) and O2(g). The product, NO2(g) is stabilized with minimum energy.

Hence, unstable NO(g) changes to stable NO2(g).

Page No 184:

Question 6.22:

Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ΔfHθ = –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.

qsurr = +286 kJ mol–1

Entropy change (ΔSsurr) for the surroundings =

ΔSsurr = 959.73 J mol–1 K–1



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