(i) Energy factor opposes .So enthalpy factor must favour , i.e. ,ΔS must be positive. 

(ii) For A----> B , ΔG is –ve, therefore, for B----->A, ΔG will be +ve. 

(iii) For B-----> A, ΔH = - ve and ΔS is also –ve , i.e. ,Δs opposes the process but at low temp., TΔS may be so low that ΔH is greater in magnitude then TΔS and the process will be spontaneous.

Option : (d) free energy

Given :

4CO_(g) + 2NO_2(g)  → 4CO_2(g)  + N_2(g)

Molar mass of CO = 28 g mol^-1

Δ_rH⁰ = -1200 kJ;

Molar mass of CO = 28 g mol^-1

m_co = 12 g, 

ΔH = ?

From the reaction,

∵ For 4 × 28 g CO, 

ΔH⁰ = – 1200 kJ

∵ For 12g CO 

ΔH⁰ = \(\frac{(-1200)\times 12}{4\times 28}\)

= -128.6 kJ

∴ Heat evolved = 128.6 kJ

(a) N₂O_4(g) → 2NO_2(g)

Since 1 mole N₂O₄ on dissociation gives two moles of NO₂, the number of molecules increase, disorder increases hence entropy increases, ΔS > 0.

(b) Fe₂O_3(s) + 3H_2(g) → 2Fe(s) + 3H₂O_(g)

In the reaction number of moles of gaseous reactants and products are same, hence ΔS = 0.

(c) N₂(g) + 3H_2(g) → 2NH_3(g)

In the reaction, 4 moles of gaseous reactants form 2 moles of gaseous products (Δn < 0). Therefore disorder decreases and hence entropy decreases, ΔS < 0.

(d) MgCO_3(s) → MgO_(s) + CO_2(g)

In this 1 mole of orderly solid MgCO₃ gives 1 mole of solid MgO and 1 mole of gaseous CO₂(Δn > 0) with more disorder. Hence entropy increases, ΔS > 0.

(e) CO_2(g) → CO_2(s)

In this system from higher disorder in gaseous state changes to less disorder in the solid state, hence entropy decreases, ΔS < 0.

(f) Cl_2(g) → 2Cl_(g)

Since the dissociation of Cl₂ gas gives double Cl atoms, the number of atoms increases (Δn >0) increasing the disorder of the system. Hence ΔS > 0.

Phase transformation (or transition) :

• The transition of one phase (physical state) of a matter to another phase at constant temperature and pressure without change in chemical composition is called phase transformation.
• During phase transformation, both the phases exist at equilibrium. Solids ⇌ Liquid; Liquid ⇌ Vapour.

There are following types of processes : 

(1) Isothermal process : It is defined as a process in which the temperature of the system remains constant throughout the change of a state of the system. 

In this, 

ΔT = 0.

Features :

• In this process, the temperature at initial state, final state and throughout the process remains constant.
• In this process, system exchanges heat energy with its surroundings to maintain constant temperature. E.g., in case of exothermic process liberated heat is given to the surroundings and in case of endothermic process heat is absorbed from the surroundings so that temperature of the system remains constant and ΔT = 0.
• Isothermal process is carried out with a closed system.
• Internal energy (U) of the system remains constant, hence, Δ U = 0.
• In this process, pressure and volume of a gaseous system change.

(2) Isobaric process : It is defined as a process which is carried out at constant pressure. Hence, Δ P = 0.

Features :

• In this process, the volume (of gaseous system) changes against a constant pressure.
• Since the external atmospheric pressure remains always constant, all the processes carried out in open vessels, or in the laboratory are isobaric processes.
• In this volume and temperature change.
• Internal energy of a system changes, hence, ΔU ≠ 0.

(3) Isochoric process : It is defined as a process which is carried out at constant volume of the system.

Features :

• In this process, temperature and pressure of the system change but volume remains constant.
• Since ΔV = 0, no mechanical work is performed.
• In this internal energy (U) of the system changes. The example of this process in cooking takes place in a pressure cooker.

(4) Adiabatic process : It is defined as a process in which there is no exchange of heat energy between the system and its surroundings. Hence, Q = 0.

Features :

• An adiabatic process is carried out in an isolated system.
• In this process, temperature and internal energy of a system change, ΔT ≠ 0, Δ U ≠ 0.
• During expansion, temperature and energy decrease and during compression,temperature and energy increase.
• If the process is exothermic, the temperature rises and if the process is endothermic the temperature decreases in the adiabatic process.

(5) Reversible process : A process carried out in such a manner that at every stage, the driving force is only infinitesimally greater than the opposing force and it can be reversed by an infinitesimal increase in force and the system exists in equilibrium with its surroundings throughout, is called a reversible process.

Features :

• This is a hypothetical process.
•  Driving force is infinitesimally greater than the opposing force throughout the change.
• The process can be reversed at any point by making infinitesimal changes in the conditions.
• The process takes place infinitesimally slowly involving infinite number of steps.
• At the end of every step of the process, the system attains mechanical equilibrium, hence, throughout the process, the system exists in temperature-pressure equilibrium with its surroundings.
• In this process, maximum work is obtained.
• Temperature remains constant throughout the isothermal reversible process.

(6) Irreversible process : it is defined as the unidirectional process which proceeds in a definite direction and cannot be reversed at any stage and in which driving force and opposing force differ in a large magnitude. It is also called a spontaneous process.

Features :

• It takes place without the aid of external agency. 
• All irreversible processes are spontaneous. 
• All natural processes are irreversible processes. 
• Equilibrium is attained at the end of process. 
• They are real processes and are not hypothetical.

 Examples :

• Flow of heat from a matter at higher temperature to a matter at lower temperature. 
• Flow of a gas from higher to lower pressure. 
• Flow of water from higher level to lower level. 
• Flow of a solvent into a solution through a semipermeable membrane due to osmosis. 
• Flow of electricity from higher potential terminal to lower potential terminal. 

The following are the types of phase changes : 

(1) Fusion : This involves the change of a matter from solid state to liquid state. In this heat is absorbed, hence it is endothermic (ΔH > 0).

H₂O_(S)  → H₂O_(I)

(2) Vaporisation or evaporation : This involves the change of a matter from liquid state to gaseous state. In this heat is absorbed, hence it is endothermic (ΔH > 0).

H₂O_(I)  → H₂O_(g)

(3) Sublimation : This involves the change of matter from solid state directly into gaseous state. In this heat is absorbed, hence it is endothermic (ΔH > 0).

Camphor_(g) → Camphor_(g)

Following are the types of systems : 

1. Open system 

2. Closed system 

3. Isolated system 

4. Homogeneous system 

5. Heterogeneous system.

Given that

18gH₂O(l) \(\overset{vaporisation}{\rightarrow}\) 18gH₂O (g)

No. of moles in 18 g H₂O(l) =\(\frac{18g}{18g\,mol^{-1}}\)

∆_vapU^Θ = ∆_vapH^Θ − p∆V

= ∆_vapH^Θ − ∆n_gRT

(Assuming steam behaves as an ideal gas)

= 40.66 kJ mol⁻¹ − 1 × 8.314 JK⁻¹ mol⁻¹ × 373K × 103 kJ J⁻¹

= 40.66 kJ mol⁻¹ − 3.10 kJ mol⁻¹

= 37.56 kJ mol⁻¹

Given : 

Number of moles of a gas = n = 3 mol

Initial volume = V₁ = 300 cm³

= 0.3 dm³

Final volume = V₂ = 2.5 dm³

External pressure = P_ex = 1.9 bar

Temperature = T = 300 K

∵ W = -P_ex (V₂ – V₁)

= -1.9 (2.5 – 0.3)

= - 4.18 dm³bar

Now, 

1 dm³bar = 100 J

∴ W = -4.18 × 100

= -4180 J

∴ Work of expansion = W = -4180 J

(a) If the given thermochemical equation is multiplied by 3 then,

Δ_rH⁰ = 3ΔH⁰ = 3 × (-323) = -969 kJ

(b) If the direction of equation is reversed, then the reaction will be,

O_2(g) + 2H_F(g) → OF_2(g) + H₂O_(g)

∴ Δ_rH⁰ = – ΔH⁰ 

= – (- 323) 

= + 323 kJ

If the given thermochemical equation is multiplied by 3 then,
ΔrH0 = 3ΔH0 = 3 × (-323) = -969 kJ
(b) If the direction of equation is reversed, then the reaction will be,
O2(g) + 2HF(g) → OF2(g) + H2O(g)
∴ ΔrH0 = – ΔH0 
= – (- 323) 
= + 323 kJ

(a) ∆H = ∆U + P∆V 

when ∆V = 0, ∆H = ∆U 

(b) ∆H = ∆U + ∆nRT 

when ∆n = 0, ∆H = ∆U

∆_fusH^Θ = 2.6 kJ mol⁻¹

∆_vapH^Θ = 98.2 kJ mol⁻¹

^{\(\therefore\)} Δ_subH^Θ = Δ_fusH^Θ + Δ_vapH^Θ

= 2.6 + 98.2

= 100.8 kJ mol⁻¹

Given : 

Q = 1500 J W = -850 J 

(For expansion work is negative) 

ΔU = ? 

By first law of thermodynamics, 

ΔU = Q + W 

= 1500 + (- 850)

= 650 J

∴ Change in internal energy = Δ U = 650 J

Given : 

Mass of ethanol (C₂H₅OH) = m = 6.24 g 

Heat energy supplied = ΔH = 5.89 kJ

Heat of vaporisation of ethanol = Δ_vapH = ?

Molar mass of ethanol,

C₂H₅OH = 46 g mol^-1

∵ For 6.24 g C₂H₅OH  ΔH = 5.89kJ

∴ For 1 mole C₂H₅OH = 46 g C₂H₅OH

ΔH = \(\frac{5.89\times 46}{6.24}\)

= 43.42 kJ mol^-1

∴ Enthalpy of vaporisation of C₂H₅OH_(I)

= 43.42 kJ

∴ Enthalpy of vaporisation of C₂H₅OH_(I)

= 43.42 kJ

Correct answer is

(a) 43.4 kJ mol^-1

Given : 

Number of moles of ethanol, 

(C₂H₅OH) = n = 1 mol 

Δ_fH⁰(CH₃OH) = -238.9 kJ mol^-1

= -238.9 × 10³J mol^-1

Temperature = T = 298 K

ΔS = ?

ΔS_surr = ?

Since Δ_fH⁰ is negative, 

The reaction for the formation of one mole of C₂H₅OH is exothermic. 

As heat is released to the surroundings,

ΔH⁰_surr = + 238.9 kJ mol^-1

∴ ΔS_surr = \(\frac{ΔH^0_{surr}}{T}\) = \(\frac{+238.9\times 10^3}{298}\)

= +801.7JK^-1

Thus,

Entropy of the surroundings increases.

∴ ΔS_surr = +801.7JK^-1

Correct answer is

(b) 2 and 3

Correct answer is

(b) 801.7 JK^-1

Enthalpy of formation of carbon in the form of graphite is taken as zero because it is a more commonly found in stable form of carbon.

(a)If both (A) and (R)are true and (R) is the true explanation of (A).

Given : 

Mass of C₂H₅OH_(I) = m = 10 g

Heat liberated = ΔH₁ = -51 kJ

Molar mass of C₂H₅OH = 46 gmol^-1

Standard enthalpy of formation of C₂H₅OH_(I)

= Δ_fH = ?

Standard enthalpy of formation is the enthalpy change for the formation of 1 mole C₂H₅OH_(I) 

i.e., 46 g C₂H₅OH_(I).

Now,

∵ For the formation of 10 g C₂H₅OH_(I)

ΔH₁ = -51 kJ

∴ For the formation of 46 g C₂H₅OH,

Δ_fH⁰ = \(\frac{-51\times 46}{10}\) = – 234.6 kJ mol^-1

∴ Standard enthalpy of formation of C₂H₅OH

= Δ_fH⁰ = – 234.6 kJ mol^-1

The standard enthalpy of formation of diamond is not zero because it is slightly less stable form of carbon. Its ∆_fH^Θ is 1.89 kJ mol⁻¹ .

Option : (b) 170 J

Option : (b) zero

It is defined as the standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation. Its symbol is ∆_fH^Θ.

Option : (c) ΔH < ΔU

(1) System : The portion of the universe under thermo-dynamic consideration to study thermodynamic properties is called a system.

Explanation :

• As such any portion of the universe under thermodynamic consideration is a system. The thermodynamic consideration involves the study of thermodynamic parameters like pressure, volume, temperature, energy, etc.
• The system may be very small or very large
• The system is confined by a real or an imaginary boundary.

(2) Surroundings : The remaining portion of the universe other than under thermodynamics study i.e„ the system is called the surroundings.

Explanation :

• Surroundings represent a large stock of mass and energy and can be exchanged with the system when allowed.
• For a liquid in an open vessel, the surrounding atmosphere around it represents the surroundings.

(3) Boundary : The wall or interface separating the system from its surrounding is called a boundary.

Explanation :

• This boundary may be either real or imaginary.
• Through this boundary, exchange of heat and matter between the system and surroundings can take place, e.g. when a liquid is placed in a beaker the walls of beaker represent real boundaries while open portion of the beaker is imaginary boundary.
• Everything outside the boundary represents surroundings. 

Option : (b) ΔH < ΔU

Option : (b) ΔU = 0

Option : (c) 3.94 kJ

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

In this equation, volume decreases. So, the sign of work done will be positive i.e. work will be done on the system.

Option : (c) < ΔU°

Option : (b) zero

Option : (a) Hg_(I) → Hg_(g)

∆H_f(H₂O) = -572/2 = - 286 kJ/mol. 

Option : (d) \(\frac{1}{2}\)H_2(g) + \(\frac{1}{2}\)Cl_2(g) → HCl_(g)

By the first law of thermodynamics,

ΔU = Q + W 

Where ΔU is the change in internal energy 

Q is heat supplied to the system 

W is the work obtained.

∴ Q = ΔU – W

If Q_P is the heat absorbed at constant pressure by the system, so that the volume changes by ΔV against constant pressure P then,

W = -PΔV 

∴ Q = ΔU – (-PΔV) 

∴ Q = ΔU + PΔV ……… (1)

If ΔH is the enthalpy change for the system, then 

ΔH = ΔU + PΔV ……….. (2)

By comparing above equations, (1) and (2), we can write, 

Q_P = ΔH

Hence heat absorbed at constant pressure is equal to the enthalpy change for the system. Since by increase in enthalpy heat content of the system increases, enthalpy is also called as the heat content of the system.

As the expansion takes place at a constant pressure, the change in enthalpy is given by

ΔH = ΔU + P(V₂ – V₁)

ΔH = Change in enthalpy = ?

ΔU = Change in internal energy = 418 J

P = Constant pressure = 2.026 × 10⁵ Pa

V₂ = 15 dm³ = 15 × 10^-3 m³

V₁ = 5 dm³ = 5 × 10^-3 m³

∴ ΔH = 418 + 2.026 × 10⁵ × (15 × 10^-3 – 5 × 10^-3)

= 418 + 2026

∴ ΔH = 2.444 × 10³ J 

= 2.444 kJ

∴ Change in enthalpy = 2.444 × 10³ J

= 2.444 kJ

Option : (d) -200J

Given :

1/2N_2(g) + 3/2H_2(g) → NH_3(g)

ΔH= -46.0 kJ mol^-1 

ΔH = Heat of formation of NH₃ at constant pressure 

= -46.0 kJ mol^-1 = -4600 J mol^-1

ΔU = Change in internal energy = ? 

Δn = (Number of moles of ammonia) – (Number of moles of hydrogen + Number of moles of nitrogen)

= [1 – ( \(\frac{1}{2}\)+ \(\frac{3}{2}\))]= -1 mol

R = 8.314 JK^-1 mol^-1

T = Temperature in kelvin = 298 K

ΔH = Δ U + ΔnRT

∴ -46000 = ΔU + (-1 × 8.314 × 298)

∴ -46000 = ΔU – 2477.0

∴ ΔU = -46000 + 2477.0

= -43523 J 

= -43.523 kJ

∴ Change in internal energy = -43.523 kJ

Given : 

Q = -2.5 kJ 

(since heat is released) 

W= + 5.5 kJ 

(since the work will be of compression) 

ΔU = ? 

ΔU = Q + W 

= -2.5 + 5.5 

= +3 kJ 

Internal energy of the system will increase by 3 kJ.

∴ ΔU = 3 kJ

Given : 

Q = + 4000 kJ 

(since heat is absorbed)

(a) Since volume remains constant, 

ΔV = 0.

W = -P_ex(V₂ – V₁) 

= -P_exΔV = -P_ex(0) = 0 

∴ ΔU = Q + W 

= 4000 + 0 

= 4000 kJ

(b) Q = + 4000 kJ 

W = + 2000 kJ 

(Work done on the system) 

ΔU = Q + W 

= 4000 + 2000 

= 6000 kJ

(c) W = -600 kJ 

(Work of expansion)

ΔU = Q + W 

ΔU = 4000 + (-600) 

= 3400 kJ

1. For any thermodynamic process or a chemical reaction at constant volume, 

ΔV = 0.

2. Since ΔH = ΔU + PΔV, at constant volume ΔH = ΔU.

3. In the reactions, involving only solids and liquids, Δ V is negligibly small, hence ΔH = ΔH.

4. In a chemical reaction, in which number of moles of gaseous reactants and gaseous products are equal, then change in number of moles, 

Δn = n – n = 0. 

Since ΔH = ΔU + ΔnRT, as Δn = 0, ΔH = ΔU.

Given : 

Latent heat of evaporation = ΔH 

= 80 kJ mol^-1 of water 

Temperature = T = 273 + 100 = 373 K 

Pressure = P = 1 atm 

Mass of water = m = 100 g 

Molar mass of water = 18 g mol^-1

W = ?, 

ΔH = 1, 

U = ?, 

Q = ?

Number of moles of water = \(\frac{m}{M}\) = \(\frac{100}{8}\) = 5.556 mol

H₂O_(I) → H₂O_(g)

5.556 mol

Change in number of moles = Δn 

= 5.556 – 0 

= 5.556 mol

For evaporation of 1 mol H₂O, 

ΔH = 80 kJ 

For 5.556 mol H₂O, 

ΔH= 80 × 5.556 

= 444.5 kJ

In this reaction, the work will be of expansion. 

W= -ΔnRT 

= -5.556 × 8.314 × 373 

= – 17230 J 

= -17.23 kJ

Now,

ΔH = ΔU + ΔnRT 

ΔU = ΔH – ΔnRT

= 444.5 – 5.556 × 8.314 × 373 × 10 

= 444.5 – 17.23 

= 427.27 kJ

In this, Q = Q_P  = ΔH = 444.5 kJ

∴ W= -17.23 kJ; 

ΔH = 444.5 kJ 

ΔU= 427.21 kJ, 

Q = 444.5 kJ

Given :

Q = + 8 kJ 

(since heat is absorbed by the system) 

W = -2.7 kJ 

(It will be a work of expansion) 

ΔH = ?, 

ΔU = ? 

ΔU = Q + W 

= 8 + (-2.7) 

= 5.3 kJ 

Internal energy of the system will increase by 5.3 kJ.

Due to expansion, 

ΔV > 0,

ΔH = ΔU + PΔV 

= 5.3 + 2.7 

= 8 kJ 

Enthalpy of the system will increase by 8 kJ

∴ ΔU = 5.3 kJ, ΔH = 8 kJ

(1) Homogeneous system : A system consisting of only one uniform phase is called a homogeneous system.

Explanation :

(1) The properties of homogeneous system are uniform throughout the phase or system.

(2) The homogeneous systems are :

• Solutions of miscible liquids (water and alcohol) or soluble solids in liquids, (NaCl in water), etc.
•  Mixture of gases. H₂ and N₂, NH₃ and H₂, etc.

(2) Heterogeneous system : A system consisting of two or more phases separated by interfacial boundaries is called a heterogeneous system. 

Explanation : 

These systems are :

• Mixture of two or more immiscible liquids. E.g. Water and benzene. 
• Solid in equilibrium with liquid. E.g. Ice ⇌ water. 
• Liquid in equilibrium with vapour. E.g. Water ⇌ vapour. 

Closed system :

1. A closed system can exchange only energy, but not matter with the surroundings.

2. In this, the total amount of energy does not remain constant.

3. Example : Hot water kept in a sealed glass flask.

Isolated system :

1. An isolated system can exchange neither matter nor energy with the surroundings. 

2. In this, the total amount of energy remains constant. 

3. Example : Hot water kept in a thermos flask.

Universe represents an isolated system due to the following reasons :

• The total mass and energy of the universe remain constant. 
• The universe has no boundary. 
• The universe has no surroundings.

The free energy change of a reaction is given by ΔG = ΔH – TΔS 

For a reaction to be spontaneous, ΔG should be –ve. 

If both ΔH and ΔS are positive ΔG can be –ve if TΔS>ΔH in magnitude. Thus, entropy factor should dominate over enthalpy factor. Such reactions are therefore, called entropy driven. This can happen in either of the following two ways: 

(i) ΔS should be so large that even if T is low, TΔS should be greater than ΔH 

(ii) If ΔS is small, T should be so large that TΔS>ΔH.