Thermodynamics

Class 11 > Chemistry > NCERT-I > Thermodynamics

Thermodynamics

Book Solutions

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.

Exercise

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.
Exercise

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.

 

Exercise

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.

Exercise

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) +52.27 kJ mol–1             (ii) –52.27 kJ mol–1

(iii) +74.8 kJ mol–1           (iv) –74.8 kJ mol–1.

Answer

Exercise

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.

Exercise

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.

Exercise

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.

NH2CN(g) +3/2O2(g) → N2(g) + CO2(g) + H2O(l)

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

Exercise

9

Calculate the heat (in kJ) required for 60.0 g aluminium to raise the temperature from 35∘C to 65∘C. For aluminium molar heat capacity is 24 Jmol−1K−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=(6027 Exercise

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.

ΔHtotal=Cp[H2OCl]ΔT+ΔHfreezingCp[H2Os]ΔT

= (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.

Exercise

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

Exercise

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

Exercise

13

N2(g) + 3H2(g) → 2NH3(g); 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 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.”

Dividing the chemical equation given in the question by 2, we get

(0.5)N2(g) + (1.5)H2(g) → 2NH3(g)

Therefore, Standard Enthalpy for formation of ammonia gas

= (0.5) ΔrHΘ

= (0.5)(-92.4 kJmol−1)

= -46.2 kJmol−1

Exercise

14

Determine Standard Enthalpy of formation for CH3OH(l) from the data given below:

Answer

Exercise

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


Exercise

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.


Exercise

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

Now,

ΔG=ΔH–TΔS

Let, the given reaction is at equilibrium, then ΔT will be:

T = (ΔH–ΔG)1ΔS ΔHΔS; (ΔG = 0 at equilibrium)

= 400kJmol−1/0.2kJmol−1K−1

Therefore, T = 2000K

Thus, for the spontaneous, ΔG must be –ve and T > 2000K.


Exercise

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.


Exercise

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 = ΔHθ – TΔSθ

= –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.

Exercise

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


Exercise

21

What can be said about the thermodynamic stability of NO(g), given

(1/2)N2(g) + (1/2)O2(g) → NO(g); ΔrHθ = 90 kJ mol–1

NO(g) + (1/2)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).

Exercise

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

ΔrHΘ=−286kJmol−1 is given so that amount of heat is evolved during the formation of 1 mole of H2O(l).

Thus, the same heat will be absorbed by surrounding. Qsurr = +286kJmol−1.

Now, ΔSsurr = Qsurr/7

= 286kJmol−1298K

Therefore, ΔSsurr=959.73Jmol−1K−1

Exercise

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