Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 4 Chemical Thermodynamics Important Questions and Answers.

## Maharashtra State Board 12th Chemistry Important Questions Chapter 4 Chemical Thermodynamics

Question 1.

Define the term energy.

Answer:

The energy of a system is defined as its capacity to perform the work. A system with higher energy can perform more work.

Question 2.

What are different forms of energy?

Answer:

The energy of a system has many different forms as follows :

- Kinetic energy arises due to motion, like rotational, vibrational and translational.
- Potential energy arises due to position and state of a matter. If depends upon the temperature of the system.
- Heat energy (or thermal energy) which is transferred from the hotter body to the colder body.
- Radiant energy is associated with electro-magnetic or light radiation.
- Electrical energy is produced in the galvanic cells.
- Chemical energy is stored in chemical substances.

All these various forms of energy can be converted from one form to another without any loss.

Question 3.

Explain the concept of interconversion of different forms of energy.

Answer:

There are various forms of energy like kinetic energy, potential energy, heat or thermal energy, radiant energy, electrical energy and chemical energy.

All these forms of energy are interconvertible. For example, a body at very high level possesses higher potential energy. When it falls down, potential energy is converted into kinetic energy. Falling of water from high level is used to drive turbines converting potential energy into kinetic energy which is further converted into electrical energy.

In galvanic cells, chemical energy is converted into electrical energy.

In electrolytic cells, chemical energy is converted into electrical energy. But during interconversion, the energy can neither be created nor destroyed, and there is a conservation of energy.

Question 4.

What is thermodynamics ? What are its drawbacks?

Answer:

Thermodynamics : It is concerned with the energy changes in physical and chemical changes.

Drawbacks :

- It does not give information on the rates of physical or chemical changes.
- It does not explain mechanisms involved in physical and chemical processes.

Question 5.

Define and explain :

(1) System (2) Surroundings (3) Boundary.

Answer:

(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.

Question 6.

What are the types of systems ?

Answer:

Following are the types of systems :

- Open system
- Closed system
- Isolated system
- Homogeneous system
- Heterogeneous system.

Question 7.

Define and explain the following :

(1) Open system

(2) Closed system

(3) Isolated system.

Answer:

(1) Open system :it is defined as a system which can exchange both matter and energy with its surroundings, e.g. a beaker containing water. The water continuously absorbs energy from the surroundings and forms vapour which diffuse in the surroundings. So that this system exchanges energy and matter (or mass), with the surroundings.

Fig. 4.1 : Open system

(2) Closed system : it is defined as a system which can exchange only energy but not the matter with its surroundings, e.g. A closed vessel containing hot water so that only heat is lost to the surroundings and not the matter.

Fig. 4.2 : Closed system

(3) Isolated system : it is defined as a system which can neither exchange energy nor matter with its surroundings, e.g. hot water filled in a thermally insulated closed vessel like thermos flask.

In actual practice, perfectly isolated system is not possible.

Universe represents an isolated system.

Question 8.

‘Universe is an isolated system’. Explain.

Answer:

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.

Question 9.

Define and explain :

(1) Homogeneous system

(2) Heterogeneous system.

Answer:

(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. H
_{2}and N_{2}, NH_{3}and H_{2}, 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.

Question 10.

Explain : (A) Extensive property (B) Intensive property of a system.

OR

What is the difference between extensive and intensive properties?

Answer:

The properties of a system are classified as (A) Extensive property and (B) Intesive property.

(A) Extensive property : It is defined as a property of a system whose magnitude depends on the amount of matter present in the system.

Explanation :

- More the quantity (or amount) of the matter of the system, more is the magnitude of extensive property, e.g., mass, volume, heat, energy, enthalpy, etc.
- The extensive properties are additive.

(B) Intensive property : It is defined as a property of a system whose magnitude is independent of the amount of matter present in the system.

Explanation :

- Intensive property is characteristic of the system, e.g., refractive index, density, viscosity, temperature, pressure, boiling point, melting point, freezing point of a pure liquid, surface tension, etc.
- The intensive properties are not additive.

Question 11.

Select extensive and intensive properties in the following :

Moles, molar heat capacity, entropy, heat capacity.

Answer:

Extensive property : Moles, entropy, heat capacity.

Intensive property : Molar heat capacity.

Question 12.

What is a state function ? Give examples.

Answer:

State function : The property which depends on the state of the system and independent of the path followed by the system to attain the final state is called a state function.

For example, pressure, volume, temperature, etc.

Question 13.

Classify the following properties as intensive or extensive :

(i) Temperature (ii) Density (iii) Enthalpy (iv) Mass (v) Energy (vi) Refractive index (vii) Pressure (viii) Viscosity (ix) Volume (x) Weight.

Answer:

(1) Intensive properties : Temperature, Density, Refractive index. Pressure, Viscosity.

(2) Extensive properties : Enthalpy, Mass, Energy, Volume, Weight.

Question 14.

What are path functions?

Answer:

Path functions : The properties which depend on the path of the process are called path functions. For example, work (W) and heat (Q).

Question 15.

Define thermodynamic equilibrium. Mention different types of thermodynamic equilibria.

Answer:

Thermodynamic equilibrium : A system is said to have attained a state of thermodynamic equilibrium if there is no change in any thermodynamic functions or state functions like energy, pressure, volume, etc. with time.

For a system to be in thermodynamic equilibrium, it has to attain following three types of equilibrium :

- Thermal equilibrium
- Chemical equilibrium
- Mechanical equilibrium

Question 16.

Distinguish between :

(1) Open system and Closed system :

Open system:

- An open system can exchange both matter and energy with the surroundings.
- In this, the total amount of matter does not remain constant.
- Example : Hot water kept in an open beaker.

Closed system:

- A closed system can exchange only energy, but not matter with the surroundings.
- In this, the total amount of matter remains constant.
- Example : Hot water kept in a closed glass flask.

(2) Closed system and Isolated system :

Closed system:

- A closed system can exchange only energy, but not matter with the surroundings.
- In this, the total amount of energy does not remain constant.
- Example : Hot water kept in a sealed glass flask.

Isolated system:

- An isolated system can exchange neither matter nor energy with the surroundings.
- In this, the total amount of energy remains constant.
- Example : Hot water kept in a thermos flask.

(3) Open system and Isolated system :

Open system:

- An open system can exchange matter with the surroundings.
- It can exchange energy with the surroundings.
- In this, the total amount of energy does not remain constant.
- Example : Hot water kept in an open beaker.

Isolated system

- An isolated system cannot exchange matter with the surroundings.
- It cannot exchange energy with the surroundings.
- In this, the total amount of energy remains constant.
- Example : Hot water kept in a thermos flask.

Question 17.

What is a thermodynamic process ? What are different types of processes ?

Answer:

(i) Thermodynamic process : It is defined as a transition by which a state of a system changes from initial equilibrium state to final equilibrium state.

The process is carried out by changing the state functions or thermodynamic variables like pressure, volume and temperature. During the process one or more properties of the system change.

(ii) Types of processes :

- Isothermal process
- Isobaric process
- Isochoric process
- Adiabatic process
- Reversible process
- Irreversible (spontaneous) process.

Question 18.

Define and explain different types of processes.

Answer:

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.

Question 19.

Distinguish between :

(1) Isothermal process and Adiabatic process.

(2) Reversible and irreversible processes.

Answer:

Isothermal process:

- In an isothermal process, the temperature of the system remains constant. ΔT = 0
- In this process, the system exchanges heat with the surroundings. Q ≠ 0 (Closed system)
- The total internal energy of the system remains constant.
- In this process, the system is not thermally isolated.
- In this process, Q = -W as ΔU = 0.
- ΔH = 0.

Adiabatic process:

- In an adiabatic process, the temperature of the system changes. ΔT ≠ 0
- In this process, the system does not exchange heat with the surroundings. Q = 0 (Isolated system)
- The total internal energy of the system changes. ΔU ≠ 0
- In this process, the system is thermally isolated.
- In this process, W = ΔU.
- ΔH ≠ 0.

(2) Reversible and irreversible processes.

Reversible process:

- The process whose direction can be reversed at any stage by an infinitesimal increase in the opposing force is called a reversible process.
- Such a process is not spontaneous and takes place infinitesimally slowly and takes infinite time for completion.
- In this process, the thermodynamic equilibrium is always maintained between the system and the surroundings at every step.
- The opposing force is infinitesimally less than the driving force.
- It is an ideal or hypothetical process.
- Maximum work can be derived from such a process.

Irreversible process:

- The process whose direction cannot be reversed by an infinitesimal increase in the opposing force is called an irreversible process.
- Such a process is spontaneous and takes finite time for completion.
- The thermodynamic equilibrium is attained only at the end of the process.
- The opposing force is significantly less than the driving force.
- It is a practical or real and spontaneous process.
- Work derived from such a process is always less than the maximum work.

Question 20.

Show that pressure times volume (PV) is equal to work.

Answer:

The work is defined as the energy by which a body is displaced through a distance d by applying a force f.

∴ W = f × d

If area is A = d^{2} and volume V = d^{3} then,

PV = \(\frac{f}{A}\) × d^{3} = \(\frac{f}{d^{2}}\) × d^{3} = f × d = W

∴ The term PV represents the pressure-volume work.

Question 21.

Explain the process of (A) expansion and (B) compression with suitable examples.

Answer:

(A ) Expansion : Consider an ideal cylinder fitted with a piston and filled with H_{2}O_{2(l)}.

2H_{2}O_{2(l)} → 2H_{2}O_{(l)} + O_{2(l)}

Fig. 4.5 : Decomposition of H_{2}O_{2}

The oxygen gas produced pushes the piston upwards lifting the mass. Thus, the system performs the work on the surroundings and loses energy by expansion. In this work is done by the system.

(B) Compression : Consider an ideal cylinder fitted with a piston and containing gaseous NH_{3(g)} and HCl_{(g)}.

Fig. 4.6 : Reaction between NH_{3}_{(g)} and HCl_{(g)}

NH_{3}_{(g)} + HCl_{(g)} → NH_{4}Cl_{(s)}

As the reaction proceeds, due to consumption of gases, the volume decreases and there is work due to compression. In this work is done on the system by surroundings and the system gains energy.

Question 22.

What are the sign conventions for Q and W in (A) expansion, (B) compression?

Answer:

(A) For expansion, work is done by the system hence,

Q = -ve and W = -ve

(B) For compression, work is done on the system hence,

Q = -ve and W = +ve

Fig. 4.7 : Sign conventions

Question 23.

Explain sign convention of work during expansion and compression.

OR

Explain +W and -W.

Answer:

(A) Expansion of a gas :

(1) When a gas expands against a constant pressure, P_{ex} changing the volume from initial volume V_{1} to final volume V_{2},

Change in volume, Δ V = V_{2} – V_{1}

The mechanical work = W = -P_{ex} × Δ V

= -P_{ex} (V_{2} – V_{1})

(2) During expansion V_{2} > V_{1}. The work is said to be performed by the system on the surroundings. This results in the decrease in the (work) energy of the system. Hence the work is negative, i.e. W is -ve.

(B) Compression of a gas : During compression, V_{2} < V_{1}. The work is said to be performed on the system by the surroundings. This results in the increase in the (work) energy of the system. Hence the work is positive, i.e. W is + ve.

Question 24.

What are different units of energy and work ?

Answer:

Question 25.

What are the characteristics of maximum work?

Answer:

(1) The process is carried out at constant temperature.

(2) During the complete process, driving force is infinitesimally greater than opposing force.

(3) Throughout the process, the system exists in equilibrium with its surroundings.

(4) The work obtained is maximum. This is given by,

W_{max} = -2.303 nRT log_{10} \(\frac{V_{2}}{V_{1}}\)

OR

W_{max} = -2.303 nRT log_{10} \(\frac{P_{1}}{P_{2}}\)

where n, P, V and T represent number of moles, pressure, volume and temperature respectively.

(5) ΔU = 0, ΔH = 0.

(6) The heat absorbed in reversible manner

Q_{rev}, is completely converted into work.

Q_{rev} = -W_{max}.

Hence work obtained is maximum.

Solved Examples 4.3

Question 26.

Solve the following :

(1) 2.5 moles of an ideal gas are expanded isothermally from 12 dm^{3} to 25 dm^{3} against a pressure of 3.0 bar. Calculate the work obtained.

Solution :

Given : n = 2.5 mol; V_{1} = 12 dm^{3};

V_{2} = 25 dm^{3}

P_{ext} = 3.0 bar; W = ?

W = -P_{ext} × (V_{2} – V_{1})

= – 3 × (25 – 12)

= -39 dm^{3} bar

∵ V_{1} dm^{3} = 100 J

∴ W = -39 × 100 = -3900 J = -3.9kJ

Ans. W= -3.9 kJ

(2) When 2.2 moles of an ideal gas are expanded from 3.5 dm^{3} to 12 dm^{3} against a constant pressure, the work obtained is 3910 J. Estimate the external pressure.

Solution :

Given : n = 2.2 mol; V_{1} = 3.5 dm^{3}; V_{2} = 12 dm^{3}

W= -3910 J = \(\frac{-3910}{100}\)dm^{3} bar

W = -P_{ex} (V_{2} – V_{1})

= 4.6 bar

Ans. P_{ex} = 4.6 bar

(3) Three moles of an ideal gas are expanded isothermally from a volume of 300 cm^{3} to 2.5 dm^{3} at 300 K against a pressure of 1.9 bar. Calculate the work done in L atm and joules.

Solution :

Given : Number of moles of a gas = n = 3 mol

Initial volume = V_{1} = 300 cm^{3}

= 0.3 dm^{3}

Final volume = V_{2} = 2.5 dm^{3}

External pressure = P_{ex} =1.9 bar

Temperature = T = 300 K

∵ W = -P_{ex} (V_{2} – V_{1})

= -1.9 (2.5 – 0.3)

= -4.18 dm^{3} bar

Now, 1 dm^{3} bar = 100 J

∴ W = -4.18 × 100

= -4180 J

Ans. Work of expansion = W = -4180 J

(4) Calculate the constant external pressure needed to compress an ideal gas from 25 dm^{3} to 15 dm^{3}. The amount of work done in the compression process is 3500 joules.

Solution :

As the compression of the gas takes place against a constant pressure, the work done is given by

W = -P_{ex}(V_{2} – V_{1})

W = Work done by the gas against the external pressure = 3500 J

∴ W = \(\frac{3500}{100}\) = 35 dm^{3} bar

P = Constant external pressure = ?

V_{2} = Final volume = 15 dm^{3}

V_{1} = Initial volume = 25 dm^{3}

∴ 35= -P × (15 – 25)

∴ 35 = 10 × P

∴ P = \(\frac{35}{10}\) = 3.5 bar

Ans. External pressure = 3.5 bar

(5) Three moles of an ideal gas are compressed isothermally and reversibly to a volume of 2 dm^{3}. The work done is 2.983 kJ at 22°C. Calculate the initial volume of the gas.

Solution :

Given : Number of moles of a gas = n = 3 mol

Final volume = V_{2} = 2 dm^{3}

Initial volume = V_{1} = ?

For compression,

Wmax = +2.983 kJ = 2983 J

Temperature = T = (273 + 22) K = 295 K

Wmax= -2.303 nRT log_{10}\(\frac{V_{2}}{V_{1}}\)

∴ \(\frac{2983}{2.303 \times 3 \times 8.314 \times 295}\)

= -[log_{10}2 – log_{10}V_{1}]

0.1760 = -log_{10}2 + log_{10} V_{1}

= -0.3010 + log_{10}V_{1}

∴ log_{10} V_{1} = 0.1760 + 0.3010 = 0.4770

∴ V_{1} = Antilog 0.4770

= 3.0 dm^{3}

Ans. Initial volume of the gas = 3.0 dm^{3}

(6) A chemical reaction takes place in a container of cross sectional area 100 cm^{2}. As a result of the reaction, a piston is pushed out through 10 cm against an external pressure of 1.0 bar. Calculate the work done by the system.

Solution :

Given : Cross sectional area = A = 100 cm^{2}

Displacement of a piston = 1 = 10 cm

External pressure = P = 1.0 bar

Work = W = ?

Volume change = A × l

∴ ΔV = 100 × 10

= 1000 cm^{3}

= 1 dm^{3}

Work = W = -P × ΔV

= -1 × 1

= -1 dm^{3} bar

= – 1 × 100 J

= -100 J

Ans. Work = W = -100 J

(7) 5 moles of helium expand isothermally and reversibly from a pressure 4 atm to 0.4 at 300 K. Calculate the work done, change in internal energy and heat absorbed during the expansion. (R = 8.314 J k^{-1} mol^{-1})

Solution :

Given : n = 5 mol

P_{1} = 4 atm

P_{2} = 0.4 atm

T = 300 K

W_{max} = ?, ΔU = ?, Q = ?

W_{max} = -2.303 nRT log_{10} (\(\frac{P_{1}}{P_{2}}\))

= -2.303 × 5 × 8.314 × 300 log_{10} \(\frac{4}{0.4}\)

= -2.303 × 5 × 8.314 × 300 × 1

= -28720 J

= -28.72 kJ

For an isothermal process, ΔU = 0

By first law, ΔU = Q + Wmax

∴ Q = -W_{max}

= – (-28.72) = 28.72 kJ

Ans. W_{max} = – 28.72 kJ; ΔU = 0;

Q = 28.72 kJ

(8) 2.8 × 10^{-2} kg of nitrogen is expanded isothermally and reversibly at 300 K from 15.15 × 10^{5} Pa when the work done is found to be -17.33 kJ. Find the final pressure.

Solution :

Given : Mass of nitrogen = m = 2.8 × 10^{-2} kg

Temperature = T = 300 K

Work obtained in expansion = Wmax = -17.33 kJ

= – 17330 J

Initial pressure = P_{1} = 15.15 × 10^{5} Pa

= 1.515 × 10^{6} Pa

Molar mass of nitrogen (N_{2}) = M_{N2}

= 28 × 10^{-3} kg mol^{-1}

Final pressure = P_{2} = ?

Number of moles of N_{2} = n = \(\frac{m}{M_{\mathrm{N}_{2}}}\)

= \(\frac{2.8 \times 10^{-2}}{28 \times 10^{-3}}=1 \mathrm{~mol}\)

W_{max} = -2.303 × nRT log_{10} \(\frac{P_{1}}{P_{2}}\)

17330 = 2.303 × 1 × 8.314 × 300 × \(\log _{10} \frac{1.515 \times 10^{6}}{P_{2}}\)

∴ \(\frac{17330}{2.303 \times 1 \times 8.314 \times 300}\)

= [log_{10} 1.515 × 10^{6} – log_{10}P_{2}]

3.017 = 6.1804 – log_{10}P_{2}

∴ log_{10}P_{2} = 6.1804 – 3.017 = 3.1634

∴ P_{2} = Antilog 3.1634

= 1456.8 Pa

Ans. Final pressure = 1456.8 Pa

(9) Carbon monoxide expands isothermally and reversibly at 300 K doing 4.754 kJ of work. If the initial volume changes from 10 dm^{3} to 20 dm^{3}, calculate the number of moles of carbon monoxide. (R = 8.314 JK^{-1} mol^{-1})

Solution :

W_{max} = -2.303 nRT log_{10} \(\frac{V_{2}}{V_{1}}\)

W_{max} = Maximum work done = -4.754 kJ

= -4754 J, n = Number of moles = ?,

= 8.314 JK^{-1} mol^{-1}

T = 300 K, V_{1} = Initial volume of carbon monoxide = 10 dm^{3}

V_{2} = Final volume of carbon monoxide = 20 dm^{3}

∴ -4754 = 22.303 × n × 8.314 × 300 log_{10} \(\frac {20}{10}\)

∴ -4754 = – 2.303 × n × 8.314 × 300 × log_{10}2

∴ – 4754 = – 2.303 × n × 8.314 × 300 × 0.3010

∴ n = \(\frac{-4754}{-2.303 \times 8.314 \times 300 \times 0.3010}\)

= 2.75 mol

Ans. Number of mol = 2.75 mol

(10) Given that the work done in isothermal and reversible expansion is 6.4 kJ when 2 moles of an ideal gas expanded to double its volume. Calculate the temperature at which expansion takes place.

Solution :

Given : Work = W_{max} = -6.4 kJ (For expansion)

= – 6400 J

Number of moles = 2

If _{1} = x L

V_{2} = 2 L

Temperature = T = ?

For isothermal reversible expansion,

(11) 300 mmol of perfect gas occupies 13 dm^{3} at 320 K. Calculate the work done in joules when the gas expands :

(a) Isothermally against a constant external pressure of 0.20 bar,

(b) Isothermal and reversible process,

(c) Into vacuum until the volume of gas is increased by 3 dm^{3}. (R = 8.314 J mol^{-1}K^{-1})

Solution :

Given : Number of moles of a gas = n

= 300 mmol = 0.3 mol

Initial volume = V_{1} = 13 dm^{3}

Increase in volume = ΔV = 3 dm^{3}

Pressure = P_{ex} = 0.2 atm

Temperature = 320 K

(a) Expansion against constant pressure is an irreversible process.

∴ W = -P_{ex} × ΔV

= -0.2 × 3

= -0.6 dm^{3} bar

= -0.6 × 100 J

= -60 J

(b) For isothermal reversible process,

W_{max} = 2.303 nRT log_{10} \(\frac{V_{2}}{V_{1}}\)

Now, V_{2} = V_{1} + ΔV = 13 + 3 = 16 dm^{3}

W_{max} = – 2.303 × 0.3 × 8.314 × 320 log_{10} \(\frac {16}{13}\)

= -165.4 J

(c) In vacuum, P_{ex} = 0

∴ W = -P_{ex} × ΔV

= -0 × 3

= 0

Ans. (a) W= -60.78 J

(b) W_{max} = -165.4 J

(c) W = 0.

Question 27.

Define and explain the term internal energy.

Answer:

Internal energy : It is defined as the total energy constituting potential energy and kinetic energy of the molecules present in the system.

Explanation :

- The internal energy of a system is a state function and thermodynamic function. It is denoted by U.
- Its value depends on the state of a system.
- The change in internal energy (Δ U) depends only on the initial state and the final state of the system.

Δ U = U_{2}– U_{1} - It is an extensive property of the system.
- It has same unit as heat and work.
- Total internal energy U of the system is,

Total energy = Potential energy + Kinetic energy

Question 28.

Explain the formulation of first law of thermodynamics.

OR

Deduce mathematical equation for the first law of thermodynamics. Justify its expression.

Answer:

(1) The first law of thermodynamics is based on the principle of conservation of energy.

(2) If Q is the heat absorbed by the system and if W is the work done by surroundings on the system then the internal energy of the system will increase by Δ U.

(3) From the conservation of energy we can write,

Increase in internal energy of the system = Quantity of heat absorbed by the system + Work done on the system

∴ ΔU = Q + W

(4) For an infinitesimal change,

dU = dQ + dW

Question 29.

Deduce the mathematical expression of first law of thermodynamics for the following processes :

(1) Isothermal process

(2) Isobaric process

(3) Isochoric process

(4) Adiabatic process.

Answer:

(1) Isothermal process :This is a process which is carried out at constant temperature. Since internal energy, U of the system depends on temperature there is no change in the internal energy U of the system. Hence ΔU = 0.

By first law of thermodynamics,

ΔU = Q +W

∴ 0 = Q + W

∴ Q = -W or W = -Q.

- Hence in expansion, the heat absorbed by the system is entirely converted into work on the surroundings.
- In compression, the work done on the system is converted into heat which is transferred to the surroundings by the system, keeping temperature constant.

(2) Isobaric process : In this, throughout the process pressure remains constant. Hence the system performs the work of expansion due to volume change ΔV.

W= -P_{ext} × ΔV

Let Q_{P} be the heat absorbed by the system at constant pressure.

By first law of thermodynamics,

ΔU = Q_{P} + W.

∴ ΔU = Q_{P} – P_{ex}ΔV

or Q_{P} = ΔU + P_{ex}ΔV

In this process, the heat absorbed Q_{P} is used to increase the internal energy (ΔU) of the system.

(3) Isochoric process : In this process the volume of the system remains constant. Hence ΔV = 0. Therefore, the system does not perform mechanical work.

∴ W = -PΔV = -P × (0) = 0

Let Q_{V} be the heat absorbed at constant volume.

By first law of thermodynamics,

ΔU = Q + W

∴ ΔU = Q_{V}.

(4) Adiabatic process : In this process, the system does not exchange heat, Q with its surroundings.

∴ Q = 0.

Since by first law of thermodynamics,

ΔU = Q + W

∴ ΔU = W_{ad}.

Hence,

(i) the increase in internal energy ΔU is due to the work done on the system by surroundings. This results in increase in energy and temperature of the system.

(ii) if the work is done by the system on surroundings, like expansion, then there is a decrease in internal energy (-ΔU) and temperature of the system decreases.

Question 30.

What are the IUPAC sign conventions of Q, U and W?

Answer:

In thermodynamics, the sign conventions are adopted according to IUPAC convention, based on acquisition of energy.

(i) Heat absorbed = +Q

Heat evolved = -Q

(ii) Internal energy change :

Increase in energy = + Δ U

Decrease in energy = – Δ U

(iii) Work done by the system = – W

Work done on the system = + W

Question 31.

Define and explain the term enthalpy.

OR

What is meant by enthalpy of a system ?

Answer:

Enthalpy (H) : It is defined as the total energy of a system consisting of internal energy (U) and pressure – volume (P × V) type of energy, i.e. enthalpy represents the sum of internal energy U and product PV energy. It is denoted by H and is represented as

Explanation :

- Enthalpy represents total heat content of the system.
- Enthalpy is a thermodynamic state function.
- Enthalpy is an extensive property.
- The absorption of heat by a system increases its enthalpy. Hence enthalpy is called heat content of the system.

Question 32.

Derive the expression, ΔH = ΔU + PΔV.

Answer:

Enthalpy (H) of a system is defined as

H = U + PV

where U is internal energy

P is pressure and V is volume.

Consider a process in which a state of a system changes from an initial state A to a final state B. Let H_{1}, U_{1}, P_{1}, V_{1} and H_{2}, U_{2}, P_{2}, V_{2} be the state functions of the system in initial and final states.

Then,

H_{1} = U_{1} + P_{2}V_{2} and H_{2} = U_{2} + P_{2}V_{2}

The enthalpy change ΔH is given by,

ΔH = H_{2} – H_{1}

= (U_{2} + P_{2}V_{2}) – (U_{1} + P_{1}V_{1})

= (U_{2} – U1) + (P_{2}V_{2} – P_{1}V_{1})

= ΔU + ΔPV

where ΔU = U_{2} – U_{1}

At constant pressure, P_{1} = P_{2} = P

∴ P_{2}V_{2} – P_{1}V_{1} = PV_{2} – PV_{1}

= P(V_{2} – V_{1})

= P × ΔV

Hence, ΔH = ΔU + PΔV

This is a relation for enthalpy change.

Question 33.

Show that the heat absorbed at constant pressure is equal to the change in enthalpy of the system.

OR

Why is enthalpy called heat content of the system?

Answer:

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 Δ V against constant pressure P then,

W = -PΔV

∴ Q_{P} = ΔU – (-PΔV)

∴ Q_{P} = 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.

Question 34.

What are the conditions under which ΔH = ΔU?

Answer:

- For any thermodynamic process or a chemical reaction at constant volume, Δ V = 0.
- Since ΔH = ΔU + PΔV, at constant volume ΔH = ΔU.
- In the reactions, involving only solids and liquids, Δ V is negligibly small, hence ΔH = ΔH.
- In a chemical reaction, in which number of moles of gaseous reactants and gaseous products are equal, then change in number of moles, Δ n = n
_{2}– n_{1}= 0. Since ΔH = ΔU + ΔnRT, as Δn = 0, ΔH = ΔU.

Solved Examples 4.6 – 4.8

Question 35.

Solve the following :

(1) Calculate the change in internal energy when a gas is expanded by supplying 1500 J of heat energy. Work done in expansion is 850 J.

Solution :

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

Ans. Change in internal energy = Δ U = 650 J

(2) A system absorbs 520 J of heat and performs work of 210 J. Calculate the change in internal energy.

Solution :

Given : Since the heat is absorbed by the system, the work is of expansion.

Q = 520 J

W= -210 J

ΔV = ?

ΔU = Q + W

= 520 + (- 210)

= 310 J

Ans. Internal energy change = Δ U = 310 J

(3) A gas expands from 6 litres to 20 dm^{3} at constant pressure 2.5 atmosphere. If the system is supplied with 5000 J of heat, calculate W and ΔU.

Solution :

Given : V_{1} = 6 dm^{3}

V_{2} = 20 dm^{3}

P = 2.5 atm

Q = 5000 J

W = ?; ΔU = ?

For expansion,

W = -P_{ex}(V_{2} – V_{1})

= -2.5 (20 – 6)

= – 35

= – 35 × 100 J

= – 3500 J

ΔU = Q + W

= 5000 + (-3500)

= -1500 J

Ans. W= -3500 J; ΔU = – 1500 J

(4) An ideal gas expands against a constant pressure of 2.026 × 10^{5} Pa from 5 dm^{3} to 15 dm^{3}. If the change in the internal energy is 418 J, calculate the change in enthalpy.

Solution :

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

ΔH = ΔU + P(V_{2} – V_{1})

ΔH = Change in enthalpy = ?

ΔU = Change in internal energy = 418 J

P = Constant pressure = 2.026 × 10^{5} Pa

V_{2} = 15 dm^{3} = 15 × 10^{-3} m^{3}

V_{1} = 5 dm^{3} = 5 × 10^{-3} m^{3}

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

= 418 + 2026

∴ ΔH = 2.444 × 10^{3} J = 2.444 kJ

Ans. Change in enthalpy = 2.444 × 10^{3} J

= 2.444 kJ

(5) In a reaction, 2.5 kJ of heat is released from the system and 5.5 kJ of work is done on the system. Find ΔU.

Solution :

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.

Ans. Δ U = 3 kJ

(6) A chemical reaction is carried out by supplying 8 kJ of heat. The system performs the work of 2.7 kJ. Calculate ΔH and ΔU.

Solution :

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, Δ V > 0,

∴ PΔV = +2.7 kJ

ΔH = ΔU + PΔV = 5.3 + 2.7 = 8 kJ

Enthalpy of the system will increase by 8 kJ

Ans. ΔU = 5.3 kJ, ΔH = 8 kJ

(7) A sample of gas absorbs 4000 kJ of heat, (a) if volume remains constant. What is ΔU? (b) Suppose that in addition to absorption of heat by the sample, the surroundings does 2000 kJ of work on the sample. What is Δ U ? (c) Suppose that as the original sample absorbs heat, it expands against atmospheric pressure and does 600 kJ of work on its surroundings. What is ΔU ?

Solution :

Given : Q = + 4000 kJ (since heat is absorbed)

(a) Since volume remains constant, Δ V = 0.

W = -P_{ex} (V_{2} – V_{1})

= -P_{ex}ΔV = -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

Ans. (a) Δ U = 4000 kJ

(b) Δ U = 6000 kJ

(c) Δ U = 3400 kJ

(8) Calculate the internal energy change at 298 K for the following reaction :

\(\frac{1}{2} \mathbf{N}_{2(\mathrm{~g})}+\frac{3}{2} \mathbf{H}_{2(\mathrm{~g})} \rightarrow \mathbf{N H}_{3(\mathrm{~g})}\)

The enthalpy change at constant pressure is -46.0 kJ mol^{-1}. (R = 8.314 JK^{-1} mol^{-1})

Solution :

Given : \(\frac{1}{2} \mathbf{N}_{2(\mathrm{~g})}+\frac{3}{2} \mathbf{H}_{2(\mathrm{~g})} \rightarrow \mathbf{N H}_{3(\mathrm{~g})}\)

ΔH= -46.0 kJ mol^{-1}

ΔH = Heat of formation of NH_{3} at constant pressure

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

Δ U = 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

Ans. Change in internal energy = -43.523 kJ

(9) 5 moles of helium expand isothermally and reversibly from a pressure 40 × 10^{-5} Nm^{-2} to 4 × 10^{-5} Nm^{-2} at 300 K. Calculate the work done, change in internal energy and heat absorbed during the expansion. (R = 8.314 JK^{-1} mol^{-1})

Solution :

As the expansion takes place isothermally and reversibly, the work done is given by

Wmax = -2.203 nRT log\(\frac{P_{1}}{P_{2}}\)

Wmax = Maximum work done

n = Number of moles of helium = 5 moles

R = Gas constant = 8.314 JK^{-1} mol^{-1}

T = 300 K

P_{1} = Initial pressure = 40 × 10^{-5} Nm^{-2}

P_{2} = Final pressure = 4 × 10^{-5} Nm^{-2}

∴ W_{max} = -2.303 × 5 × 8.314 × 300 × log \(\frac {40}{4}\)

= – 2.303 × 5 × 8.314 × 300 × log 10

= -2.303 × 5 × 8.314 × 300 × 1

= – 28720J

As the expansion takes place isothermally, i.e., at the same temperature, there is no change in the internal energy of the system.

Q = ΔU + W

∴ Q = – W= + 28720 J as ΔU = 0

Ans. Work done = -28720 J, Heat absorbed = 28720 J, Change in internal energy = 0

(10) Calculate the work done in each of the following reactions. State whether work is done on or by the system.

(a) The oxidation of one mole of SO_{2} at 50°C.

2SO_{2(g)} + O_{2(g)} → 2SO_{3(g)}

(b) Decomposition of 2 moles of NH_{4}NO_{3} at 100°C

NH_{4}NO_{3(s)} → N_{2}O_{(g)} + 2H_{2}O_{(g)}

Solution :

(a) Given reaction :

2SO_{2(g)} + O_{2(g)} → 2SO_{3(g)}

For 1 mole of SO_{2},

SO_{2(g)} + \(\frac {1}{2}\)O_{2}(g) → SO_{2(g)}

∴ Δn = (n_{2})_{gaseous products} – (n_{1})_{gaseous reactants}

= 1 – (1 + \(\frac {1}{2}\))

= -0.5 mol

Since there is decrease in number of moles of gases, there will be compression, hence, the work will be done on the system by the surroundings.

Work is given by,

∴ W = – ΔnRT

= – (- 0.5) × 8.314 × (273 + 50)

= + 1342.7 J

(b) Given reaction :

NH_{4}NO_{3(s)} → N_{2}O_{(g)} + 2H_{2}O_{(g)}

For 2 moles of NH_{4}NO_{3},

2NH_{4}NO_{3(s)} → 2N_{2}O_{(g)} + 4H_{2}O_{(g)}

∴ Change in number of moles,

Δn = (n_{2})_{gaseous products} – (n_{1})_{gaseous reactants}

= (2 + 4) – 0

= 6 mol

Since there is an increase in number of moles of gases, work of expansion is done by the system on the surroundings.

∴ W = -ΔnRT

= – 6 × 8.314 × (273+ 100)

= – 18606 J

= – 18.606 kJ

Ans. (a) W = 1342.7 J (b) W= -18.606 kJ

(11) The amount of heat evolved for the combustion of ethane is -900kJ mol^{-1} at 300K and 1 atm.

C_{2}H_{6(g)} + \(\frac {7}{2}\)O_{2(g)} → 2CO_{2(g)} + 3H_{2}O_{(l)}

Calculate W, ΔH and ΔU for the combustion of 12 × 10^{-3} kg ethane.

Solution :

Given : ΔH = -900 kJ mol^{-1}

Temperature = T = 300 K

Pressure = P = 1 atm

Mass of ethane = m = 12 × 10^{-3} kg

Molar mass of ethane (C_{2}H_{6}) = 30 × 10^{-3} kg mol^{-1}

ΔH = ? Δ U = ? for given ethane.

Number of moles of C_{2}H_{6} = n = \(\frac{m \mathrm{~kg}}{M \mathrm{~kg} \mathrm{~mol}^{-1}}\)

= \(\frac{12 \times 10^{-3}}{30 \times 10^{-3}}\)

= 0.4 mol

For the given reaction,

C_{2}H_{6(g)} + \(\frac {7}{2}\)O_{2(g)} → 2CO_{2(g)} + 3H_{2}O_{(l)}

Δn = (n_{2})_{gaseous products} – (n_{1})_{gaseous reactants}

= 2 – (1 + \(\frac {7}{2}\)) = -2.5 mol

For 1 mol of C_{2}H_{6}, Δn = -2.5 mol

∴ For 0.4 mol of C_{2}H_{6}, Δn = -2.5 × 0.4

= -1 mol

Since there is a decrease in number of moles, the work is of compression on the system.

W = -ΔnRT

= – (-1) × 8.314 × 300

= + 2494 J

= + 2.494 kJ

For 1 mol of C_{2}H_{6} ΔH = -900 kJ

∴ For 0.4 mol of C_{2}H_{6}, ΔH= – 900 × 0.4

= – 360 kJ

ΔH = ΔU + ΔnRT

ΔU = ΔH – ΔnRT

= -360 – (-1) × 8.314 × 300× 10^{-3}

= – 360 + 2.494

= – 357.506 kJ

Ans. W = + 2.494 kJ, ΔH = -360 kJ;

ΔU= – 357.506 kJ

(12) The latent heat of evaporation of water is 80 kJ mol^{-1}. If 100 g water are evaporated at 100 °C and 1 atm, calculate W, ΔH, ΔU and Q.

Solution :

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 = \(\frac{m}{M}=\frac{100}{18}\) = 5.556 mol

H_{2}O_{(l)} → H_{2}O_{(g)}

5.556 mol 5.556 mol

Change in number of moles = Δn = 5.556 – 0

= 5.556 mol

For evaporation of 1 mol H_{2}O, ΔH = 80 kJ

For 5.556 mol H_{2}O, 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^{-3}

= 444.5 – 17.23

= 427.27 kJ

In this, Q = Q_{P} = ΔH = 444.5 kJ

Ans. W= -17.23 kJ; ΔH = 444.5 kJ

ΔU= 427.21 kJ, Q = 444.5 kJ

(13) Oxidation of propane is represented as

C_{3}H_{8(g)} + 5O_{2(g)} → 3CO_{2(g)} + 4H_{2}O_{(g)}, ΔH^{0} = -2043 kJ

How much pressure volume work is done and what is the value of AU at constant pressure of 1 bar when the volume change is + 22.4 dm^{3}.

Solution :

Given :

ΔH^{0} = – 2043 kJ

Change in volume = ΔV = +22.4 L

ΔU = ?

C_{3}H_{8(g)} + 5O_{2(g)} → 3CO_{2(g)} + 4H_{2}O_{(g)}

Δn = (n_{2})_{gaseous products} – (n_{1})_{gaseous reactants}

= (3 + 4) – (1 + 5)

= 1 mol

Since there is an increase in number of moles, the work will be of expansion.

W = -P × ΔV dm^{3} bar

= – 1 × 22.4

= – 1 × 22.4 × 100 J

= – 2240 J

= -2.240

ΔH = ΔU + PΔV

ΔU = ΔH – PΔV

= – 2043 – (2.24)

= – 2040.7 kJ

Ans. W = -2.27 kJ, ΔU = -2040.7 kJ

(14) How much heat is evolved when 12 g of CO reacts with NO_{2} ? The reaction is :

4CO_{(g)} + 2NO_{2(g)} → 4CO_{2(g)} + N_{2(g)},

Δ_{r}H^{0} = -1200 kJ

Solution :

Given : 4CO_{(g)} + 2NO_{2(g)} → 4CO_{2(g)} + N_{2(g)}

Molar mass of CO = 28 g mol^{-1}

Δ_{r}H^{0} = – 1200 kJ;

Molar mass of CO = 28 g mol^{-1}

m_{co} = 12 g, ΔH = ?

From the reaction,

∵ For 4 × 28 g CO, ΔH^{0} = – 1200 kJ

∵ For 12g CO ΔH^{0} = \(\frac{(-1200) \times 12}{4 \times 28}\)

= -128.6 kJ

Ans. Heat evolved = 128.6 kJ

Question 36.

What is phase transformation?

Answer:

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.

Question 37.

Mention different types of phase transitions.

Answer:

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).

H_{2}O_{(s)} → H_{2}O_{(l)}

(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).

H_{2}O_{(l)} → H_{2}O_{(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_{(s)} → Camphor_{(g)}

Question 38.

Define and explain enthalpy of freezing.

Answer:

Enthalpy of freezing (Δ_{freez}H) : The enthalpy change that accompanies the solidification of one mole of a liquid into solid at constant temperature and pressure is called enthalpy of freezing.

For example,

\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \stackrel{1 \mathrm{~atm}, 273 \mathrm{~K}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})}\)

Δ_{freez}H = -6.01 kJ mol^{-1}

This equation describes that when one mole of water freezes (solidifies) at 0 °C (273 K) and 1 atmosphere, 6.01 kJ of heat will be released to the surroundings.

Question 39.

Define and explain the following :

(A) Enthalpy of vaporisation.

(B) Enthalpy of sublimation.

Answer:

(A) Enthalpy of vaporisation (Δ_{vap}H) : The enthalpy change that accompanies the vaporisation of one mole of a liquid at constant temperature and pressure is called heat of vaporisation or evaporation. For example,

\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \stackrel{1 \mathrm{~atm}, 373 \mathrm{~K}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\)

Δ_{vap}H = +40.7 kJ mol^{-1}

This equation describes that when one mole of water is evaporated at 100 °C (373 K) and 1 atmosphere, 40.7 kJ of heat will be absorbed.

(B) Enthalpy of sublimation (Δ_{sub}H) : The enthalpy change or the amount of heat absorbed that accompanies the sublimation of one mole of a solid directly into its vapour at constant temperature and pressure is called enthalpy of sublimation.

For example,

\(\mathrm{CO}_{2(\mathrm{~s})} \stackrel{1 \mathrm{~atm}, 195 \mathrm{~K}}{\longrightarrow} \mathrm{CO}_{2(\mathrm{~g})}\)

Δ_{sub}H = 25.2 kJ mol^{-1}

This equation describes that when 1 mole of dry solid carbon dioxide, CO_{2(s)} sublimes forming gaseous CO_{2(g)}, 25.2 kJ of heat will be absorbed.

Question 40.

Explain process of sublimation and enthalpy of sublimation ?

OR

How is enthalpy of sublimation related to enthalpy of fusion and enthalpy of vaporisation ?

Answer:

(1) The sublimation involves the conversion of a solid into vapour at constant temperature and pressure. For example,

H_{2}O_{(s)} → H_{2}O_{(g)}, Δ_{sub}H = 51.08 kJ mol^{-1} at 0°C.

(2) This conversion takes place in two steps, first melting of solid into liquid and second vaporisation of the liquid.

Hence we can write,

Δ_{sub}H = Δ_{fus}H + Δ_{vap}H

Question 41.

Arrange the following in order of increasing enthalpy :

H_{2}O_{(s)}, H_{2}O_{(g)}, H_{2}O_{(l)}

Answer:

The increasing order of enthalpy of the given substance will be,

H_{H2O(g)}, < H_{H2O(l)}, < H_{H2O(s)}

This is because the conversion of H_{2}O_{(s)} to H_{2}O_{(l)} and further to H_{2}O_{(g)} involves absorption of heat.

Question 42.

Define and explain :

(A) Enthalpy of atomisation

(B) Enthalpy of ionisation.

Answer:

(A) Enthalpy of atomisation (Δ_{ato}H) : The enthalpy change or amount of heat absorbed accompanying the dissociation of the molecules in one mole of a gaseous substance into free gaseous atoms at constant temperature and pressure is called enthalpy of atomisation.

For example,

Cl_{2(g)} → 2Cl_{(g)}, Δ_{ato}H = 242 kJ mol^{-1}

CH_{4(g)} → C_{(g)} + 4H_{(g)}, Δ_{ato}H = 1660 kJ mol^{-1}.

(B) Enthalpy of ionisation (Δ_{ion}H) : The enthalpy change or amount of heat absorbed accompanying the removal of one electron from each atom or ion in one mole of gaseous atoms or ions is called enthalpy of ionisation.

For example,

Na_{(g)} → Na^{+}_{(g)} + e^{–} Δ_{ion}H = 494 kJ mol^{-1}

This equation describes that when one mole of gaseous sodium atoms, Na_{(g)} are ionised forming gaseous ions, Na^{+}_{(g)}, the energy required is 494 kJ.

Question 43.

Define and explain electron gain enthalpy.

Answer:

Electron gain enthalpy (Δ_{eg}H) : It is defined as the enthalpy change, when mole of gaseous atoms of an element accept electrons to form gaseous ion.

E.g. Cl_{(g)} + e^{–} → Cl^{– }_{(g)} Δ_{eg}H = – 349 kJ mol^{-1}.

It is the reverse of ionisation process.

Question 44.

Define enthalpy of solution.

Answer:

Enthalpy of solution : It is defined as the enthalpy change in a process when one mole of a substance is dissolved in specified amount of a solvent.

NaCl_{(s)} + aq ⇌ NaCl_{(aq)} Δ_{soln}H = 4 kJ mol^{-1}

Question 45.

Define enthalpy of solution at infinite dilution.

Answer:

Enthalpy of solution (Δ_{soln}H) : It is defined as the enthalpy change when one mole of a substance is dissolved in a large excess of a solvent, so that further dilution will not change the enthalpy at constant temperature and pressure.

For example,

HCl_{(g)} + aq → HCl_{(aq)} Δ_{soln}H

= -75.14 kJ mol^{-1}

Question 46.

Explain the enthalpy of solution of an ionic compound.

Answer:

An ionic compound dissolves in a polar solvent like water in two steps as follows :

Step-I : The ions are separated from the molecule involving crystal lattice enthalpy Δ_{L}H.

\(\mathrm{MX}_{(\mathrm{s})} \rightarrow \mathrm{M}_{(\mathrm{g})}^{+}+\mathrm{X}_{(\mathrm{g})}^{-} \quad \Delta_{\mathrm{L}} H\)

Δ_{L}H is always positive.

Step-II : The gaseous ions are hydrated with water molecules involving hydration energy, Δ_{hyd}H.

\(\mathbf{M}_{(\mathrm{g})}^{+}\) + xH_{2}O_{(l)} → [M(H_{2}O)_{x}]^{+}

\(\mathrm{X}_{(\mathrm{g})}^{-}\) + yH_{2}O_{(l)} → [X(H_{2}O)_{y}]^{–}

Δ_{hyd}H is always negative.

The enthalpy change ΔsolnH of solution is given by,

Δ_{soln}H = Δ_{L}H + Δ_{hyd}H

For example, consider enthalpy of solution of NaCl(s).

Δ_{L}H_{NaCl} = 790 kJ mol^{-1}

Δ_{hyd}H_{NaCl} = -786 kJ mol^{-1}

Hence enthalpy change for solution of NaCl(s) is,

Δ_{soln}H = Δ_{L}H_{NaCl} + Δ_{hyd}H_{NaCl}

= 790 + (-786)

= + 4 kJ mol^{-1}

Therefore dissolution of NaCl in water is an endothermic process.

Solved Examples 4.9

Question 47.

Solve the following :

(1) Heat of fusion of ice at 0 °C and 1 atmosphere is 6.1 kJ mol^{-1} and heat of evaporation of water at 100 °C is 40.7 kJ mol^{-1}. Calculate the enthalpy change for the conversion of 1 mole of ice at 0 °C into vapour at 100 °C. Heat capacity of water is 4.184 JK^{-1} mol^{-1}.

Solution :

Given : Heat of fusion of ice = Δ_{fus}H = 6.01 kJ mol^{-1}

Heat of evaporation of water = Δ_{vap}H = 40.7 kJ mol^{-1}

Temperature of ice = 273 K

Temperature of vapour = (273 + 100) K = 373 K

Heat capacity of water = 4.184 JK^{-1} g^{-1}

Heat capacity of 1 mole of water

= C_{H2O} = 4.184 × 18

= 75.312 JK^{-1} mol^{-1}

The conversion of ice at 0 °C to water at 100°C takes place in three steps as follows:

ΔH_{1} = n_{ice} × Δ_{fus}H = 1_{mol} × 6.01_{kJ mol-1} = 6.01 kJ

ΔH_{2} is the enthalpy change for raising the temperature from 273 K to 373 K.

ΔH_{2} = n_{water} × C_{H2O} × (T_{2} – T_{1})

= 1_{mol} × 75.312_{JK-1 mol-1} × (373 – 273)K

= 7531 J

= 7.531 kJ

ΔH_{3} = n_{water} × Δ_{vap}H

= 1 × 40.7

= 40.7 kJ

Hence total enthalpy change will be ΔH = ΔH_{1} + ΔH_{2} + ΔH_{3}

= 6.01 + 7.531 + 40.7

= 54.241 kJ

Ans. ΔH = 54.241 kJ

(2) Heat of sublimation of ice at 0°C and 1 atmosphere is 6.01 kJ mol^{-1} and heat of evaporation of water at 0 °C and 1 atmosphere is 45.07 kJ mol^{-1}. Calculate the heat of sublimation of one mole of ice at 0 °C and 1 atmosphere. Write the equation for the same.

Solution :

The sublimation of ice can be represented by following equation,

\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})} \stackrel{1 \mathrm{~atm}, 0^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\)

This is a process of two steps.

ΔH_{1} = n_{H2O} × Δ_{fus}H

= 1 × 6.01 =6.01 kJ

ΔH_{2} = n_{H2O} × Δ_{vap}H

= 1 × 45.07 = 45.07

Heat of sublimation = ΔH = ΔH_{1} + ΔH_{2}

= 6.01 + 45.07

= 51.08 kJ

Ans. Heat of sublimation of ice = 51.08 kJ

(3) Heat of evaporation of ethyl alcohol at 78.5 °C and 1 atmosphere is 38.6 kJ mol^{-1}. If 100 g ethyl alcohol vapour is condensed, what will be ΔH ?

Solution :

Given : Δ_{vap}H_{C2H5OH} = 38.6 kJ mol-1

Mass of C_{2}H_{5}OH = m = 100 g

Molar mass of C_{2}H_{5}OH = M = 46 g mol-1

Δ_{cond}H_{C2H5OH} = ?

Number of moles of C_{2}H_{5}OH = \(\frac{m}{M}=\frac{100}{46}\)

= 2.174 mol

Heat of condensation = Δ_{cond}H = -38.6 kJ mol^{-1}

∴ Δ_{cond}H = n × Δ_{cond}H

= 2.174_{mol} × (-38.6)_{kJ mol-1} kJ = – 83.9 kJ

Ans. Heat of condensation = Δ_{cond}H = -83.9 kJ

(4) The hydration enthalpies of Li^{+}_{(g)}, and Br^{–}_{(g)} are -500 kJ mol^{-1} and -350 kJ mol^{-1} respectively and the lattice energy of LiBr_{(s)} is 807 kJmol^{-1}. Write the thermochemical equations for enthalpy of solution of LiBr(s) and calculate its value.

Solution :

Given : Enthalpy of hydration of Li^{+}_{(g)}

= Δ_{hyd}H1

= -500 kJmol^{-1}

Enthalpy of hydration of Br^{–}_{(g)} = Δ_{hyd}H_{2}

= -350 kJ mol^{-1}

Lattice energy of LiBr_{(s)} = Δ_{L}H_{3} = 807 kJ mol^{-1}

Enthalpy of solution of LiBr_{(s)} = Δ_{soln}ΔH = ?

The thermochemical equation for the dissolution of LiBr_{(s)} forming a solution is,

LiBr_{(s)} + aq → Li^{+}_{(aq)} + Br^{–}_{(aq)} (I) Δ_{sol}H = ?

This takes place in two steps as follows :

(i) LiBr_{(s)} → Li^{+}_{(g)} + Br^{–}_{(g)} Δ_{L}H_{3}

(ii) (a) Li^{+}_{(g)} + aq → Li^{+}_{(aq)} Δ_{hyd}H_{1}

(b) Br^{–}_{(g)} + aq → Br^{–}_{(aq)} Δ_{hyd}H_{2}

Hence by adding equations (i) and (ii) (a) and (b) we get equation I.

∴ Δ_{sol}H = Δ_{L}H_{3} + Δ_{hyd}H_{1} + Δ_{hyd}H_{2}

= 807 + (-500) + (-350)

= -43 kJ mol^{-1}

Ans. Heat of solution of LiBr_{(s)} = Δ_{sol}H = -43 kJ mol^{-1}

(5) Heat of solution of NaCl is 3.9 kJ mol^{-1}. If the lattice energy of NaCl is 787 kJ mol^{-1}, calculate the hydration energies of ions of the salt.

Solution :

Given : Heat of solution of NaCl

= Δ_{soln}H^{0}

= ΔH_{1} = 3.9 kJmol^{-1}

Lattice energy of NaCl = ΔLH

= ΔH_{2} = 787 kJ mol^{-1}

Hydration energy of Na^{+}_{(g)} and Cl^{–}_{(g)}

= Δ_{hyd}H(Na^{+} + Cl^{–})

= ΔH_{3} = ?

Thermochemical equation for dissolution of NaCl(s) is;

NaCl_{(s)} + aq → Na^{+}_{(aq)} + Cl^{–}_{(aq)}…ΔH_{1}

NaCl_{(s)} → Na^{+}_{(g)} + Cl^{–}_{(g)}… ΔH_{2}

Na^{+}_{(g)} + Cl^{–}_{(g) }+ aq → Na^{+}_{(aq)} + Cl^{–}_{(aq)} ΔH_{3}

∴ ΔH_{1 }= ΔH_{2} + AH_{3}

3.9 = 787 + AH3

∴ ΔH_{3} = -787 + 3.9= -783.1 kJmol^{-1}

Ans. Hydration energy of Na^{+}_{(g)} and Cl^{–}_{(g)}

= -783.1 kJmol^{-1}

(6) Enthalpies of solution are given as follows :

CuSO_{4(s)} + 10H_{2}O → CUSO_{4}(10H_{2}O)

ΔH_{1} = -54.5 kJ mol^{-1}

CuSO_{4}(s) + 100 H_{2}O → CUSO_{4}(100H_{2}O)

ΔH_{2 }= -68.4 kJ mol^{-1}

A solution contains 1 mol of CuSO_{4} in 180 g water at 25 °C. If it is diluted by adding 1620 g water, calculate the enthalpy of dilution.

Solution :

Given : Enthalpy of solution of CuSO_{4} in 10 mol H_{2}O

= Δ_{soln}H = ΔH_{1} = -54.5 kJ mol^{-1}

Enthalpy of solution of CuSO_{4} in 100 mol

H_{2}O = ΔH_{2}

= -68.4 kJmol^{-1}

Mass of water = 1620 g

For dilution, Δ_{dil}H = ?

Now 180 g H_{2}O = \(\frac {180}{18}\) = 10 mol H_{2}O

And, 1620 g H_{2}O = \(\frac {1620}{18}\) = 90 mol H_{2}O

Hence for heat of dilution,

CUSO_{4}(10H_{2}O) + 90H_{2}O_{(l)} → CUSO_{4}(100H_{2}O) Δ_{dil}H = ?

∴ Δ_{dil}H = ΔH_{2} -ΔH_{1}

= -68.4 – (54.5)

= -13.9 kJmol^{-1}

Ans. Heat of dilution = Δ_{dil}H = -13.9 kJ mol^{-1}

(7) Heat of solution and heat of hydration of AgF are -20.5kJmol^{-1} and -930kJmol^{-1} respectively. Calculate lattice energy of AgF.

Solution :

Given : Heat of solution of AgF = Δ_{soln}H

= -ΔH_{1} = -20.5 kJmol^{-1}

Heat of hydration of AgF = Δ_{hyd}H = ΔH_{2}

= -930 kJ mol^{-1}

Lattice energy of AgF = Δ_{L}H = ΔH_{3} = ?

For heat of solution, AgF_{(s)} + aq → AgF_{(aq)} ΔH_{1}

For heat of hydration,

Ag^{+}_{(g)} + F^{–}_{(g)} + aq → Ag^{+}_{(aq)} + F^{–}_{(aq)} ΔH_{2}

For Lattice energy, Ag^{+}_{(g)} + F^{–}_{(g)} → AgF_{(s)}

Δ_{L}H = ?

From above equations,

∴ ΔH_{3} = ΔH_{2} – ΔH_{1}

= -930 – (-20.5)

= -909.5 kJmol^{-1}

Ans. Lattice energy of AgF_{(s)} = -909.5 kJ mol^{-1}

(8) Bond enthalpy of H_{2} is 436kJmol^{-1} while hydration energy of hydrogen ion is -1075 kJ mol^{-1}. Calculate the enthalpy of formation of H^{+}_{(aq)}. (Ionisation energy of hydrogen is 1312 kJ mol^{-1}

Solution :

Given : Bond enthalpy of H_{2(g)} = ΔH^{0}H_{2(g)}

= 436 kJ mol^{-1}

Hydration energy of H^{+}_{(g)} = ΔH_{2} = -1075 kJmol^{-1}

Ionisation energy of H_{(g)} = ΔH_{3} = 1312 kJ mol^{-1}

Enthalpy of formation of H^{+}_{(aq)} = Δ_{f}H = ?

Thermochemical equation for the formation of H^{+}_{(aq)}

\(\frac {1}{2}\)H_{2(g)} + aq → H^{+}_{(aq)}Δ_{f}H

This takes place in three steps as follows :

Hence heat of formation of H^{+}_{(aq)} is

Δ_{f}H = ΔH_{1} + ΔH_{2} + ΔH_{3}

= \(\frac {1}{2}\) × 436 + (-1075) + 1312

= 218 – 1075 + 1312.

= 455 kJ mol^{-1}

Ans. Enthalpy of formation of H^{+}_{(aq)} = 455 kJmol^{-1}

(9) Calculate lattice energy of crystalline sodium chloride from the following data :

Solution :

Given : Bond enthalpy of Cl_{2} = ΔH^{0} = 244 kJmol^{-1}

Thermochemical equation for the formation of 1 mole of NaCl_{(s)},

\(\mathrm{Na}_{(\mathrm{s})}+\frac{1}{2} \mathrm{Cl}_{2} \longrightarrow \mathrm{NaCl}_{(s)} \quad \Delta_{\mathrm{f}} H_{\mathrm{NaCl}}^{0}=-411 \mathrm{~kJ}\)

Lattice energy, Δ_{L}H = ?

Since enthalpy is a state function, this reaction can be written in various steps as follows :

By Hess’s law,

Δ_{L}H^{0} = Δ_{sub}H^{0}+ Δ_{ion}H^{0} + \(\frac {1}{2}\)ΔH^{0}cl_{2} + ΔegH^{0} + Δ_{L}H^{0}

-411 = 109 + 496 + \(\frac {1}{2}\) × 244 + (-348) + Δ_{L}H

= 109 + 496 + 122 – 348 + Δ_{L}H

∴ Δ_{L}H= -790 kJ mol^{-1}.

Ans. Lattice energy of NaCl_{(s)} = -790 kJ mol^{–}

(10) Calculate the internal energy at 298 K for the formation of one mole of ammonia, if the enthalpy change at constant pressure is -42.0 kJ mol^{-1}.

(Given : R = 8.314 J K^{-1} mol^{-1})

Solution :

Given : ΔH = -42.0 kJ mol^{-1}, T = 298 K, ΔU = ?

\(\frac{1}{2} \mathrm{~N}_{2(\mathrm{~g})}+\frac{3}{2} \mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{NH}_{3(\mathrm{~g})}\)

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

ΔH = ΔU + ΔnRT

∴ ΔU = ΔH – ΔnRT

= -42 – (- 1) × 8.314 × 298 × 10^{-3}

= -42 + 2.477

= -39.523 kJ

Ans. ΔU = -39.523 kJ

Question 48.

What is thermochemistry ? Explain.

Answer:

Thermochemistry : Thermodynamic study of heat changes or enthalpy changes during the chemical reactions is called thermochemistry.

Consider a reaction, Reactants → Products

The heat changes ΔH for the reaction may be represented as,

ΔH_{reaction} = Σ H_{products} – Σ H_{reactants}

where H represents enthalpy.

The energy released or absorbed during a chemical change appears in the form of heat energy.

Question 49.

Define and explain the term, enthalpy or heat of reaction.

Answer:

Enthalpy or heat of reaction : The enthalpy of a chemical reaction is the difference between the sum of the enthalpies of products and that of the reactants with every substance in a definite physical state and in the amounts (moles) represented by the coefficients in the balanced equation.

Explanation : Consider the following general reaction,

aA + bB → cC + dD

The heat of the reaction ΔH is the difference in sum of enthalpies of products and sum of enthalpies of reactants.

∴ ΔH = Σ H_{products} – Σ H_{reactants}

= [cH_{C} + dH_{D}] – [aH_{A} + bH_{B}]

= Σ_{P}H – Σ_{R}Z

where H represents enthalpy of the substance.

For endothermic reaction, ΔH is positive, (ΔH > 0).

For exothermic reaction, ΔH is negative, (ΔH < 0).

Question 50.

Explain the sign convention used for ΔH.

Answer:

The change in enthalpy or heat of reaction ΔH is given by,

(i) If the sum of enthalpies of products, ΣH and reactants, Σ_{R}H are equal then ΔH for the reaction is zero, (ΔH = 0).

i. e. Σ_{P}H = Σ_{R}H

∴ ΔH = Σ_{P}H – Σ_{R}H = 0

(ii) If the sum of enthalpies of products Σ_{P}H is greater than the sum of enthalpies of reactants Σ_{R}H, then ΔH is positive, (ΔH > 0). Since such reactions take place with the absorption of heat from surroundings, they are called endothermic reactions.

∴ ΣH_{products} > ΣH_{reactants}

∴ ΔH > 0

(iii) If the sum of enthalpies of products Σ_{P}H is less than the sum of enthalpies of reactants, Σ_{R}H then ΔH is negative, (ΔH < 0). Since in such reactions heat is given out to the surroundings, they are called exothermic reactions.

∴ Σ_{P}H < Σ_{R}H

∴ ΔH < 0

Question 51.

Define : (i) Exothermic process (ii) Endothermic process.

Answer:

(i) Exothermic process : A process taking place with the evolution of heat is called exothermic process.

For this process, Q is -ve, ΔH is -ve.

(ii) Endothermic process : A process taking place with the absorption of heat (from the surroundings) is called endothermic process.

For this process, Q is +ve, ΔH is +ve.

Question 52.

Distinguish between Endothermic reaction and Exothermic reaction.

Answer:

Endothermic reaction:

- In endothermic reaction heat is absorbed from suroundings.
- Sum of enthalpies of products is greater than sum of enthalpies of reactants i.e. Σ
_{P}H > Σ_{R}H - Heat of reaction, ΔH is positive.
- Products are less stable than reactants.
- C
_{(s)}+ O_{2(g)}→ CO_{2(g)}

ΔH = -394 kJ - This reaction requires supply of thermal energy.

Exothermic reaction:

- In exothermic reaction heat is given out to surroundings.
- Sum of enthalpies of products is less than sum of enthalpies of reactants.

i.e. Σ_{P}H < Σ_{R}H - Heat of reaction, ΔH is negative.
- Products are more stable than reactants.
- N
_{2(g)}+ O_{2(g)}→ 2NO

ΔH = + 180 kJ - This reaction does not require supply of thermal energy.

Question 53.

Explain the standard state of an element.

Answer:

Standard state of an element : It is defined as the most stable state of an element at 298 K and 1 atmosphere (or 1 bar).

In this state, the enthalpy of the element is assumed to be zero.

∴ H^{0}_{element} or in general H_{element} = 0

∴ H^{0}_{graphite} = H_{H2(g)} = 0; H_{Na(s)} = 0; H_{Hg(l)} = 0

Question 54.

What is a thermochemical equation? Explain with an example.

Answer:

Thermochemical equation : It is defined as a balanced chemical equation along with the corresponding heat of reaction (ΔH) and physical states and number of moles of all reactants and all products appropriately mentioned.

E.g. C_{6}H_{12}O_{6(s)} + 6O_{2(g)} = 6CO_{2(g)} + 6H_{2}O_{(l)}

ΔH = -2808 kJ mol^{-1}

Question 55.

What are the guidelines followed for writing thermochemical equations?

Answer:

According to IUPAC conventions, while writing thermochemical equations, following rules must be followed :

(1) Reaction is represented by balanced chemical equation for the number of moles of the reactants and the products. E.g.

CH_{4(g)} + 2O_{2(g)} = CO_{2(g)} + 2H_{2}O_{(l)}

ΔrH°= -890 kJ mol^{-1}

(2) The physical states of all the substances in the reaction must be mentioned. E.g. (s) for solid, (l) for liquid and (g) for gas.

(3) Heat or enthalpy changes are measured at 298 K and 1 atmosphere (or 1 bar).

(4) ΔH^{0} is written at right hand side of thermochemical equation.

(5) Proper sign must be indicated for ΔH^{0}. For endothermic reaction ΔH^{0} is positive, (+ΔH^{0}) and for exothermic reaction ΔH is negative, (-ΔH^{0}).

(6) The enthalpy of the elements in their standard states is taken as zero. (H^{0}_{Element} = 0; H^{0}_{C(s)} = 0, H^{0}_{H2(g)} = 0)

(7) When all the substances taking part in the reaction are in their standard states, the enthalpy change is written as ΔH^{0}.

(8) The enthalpy of any compound is equal to its heat of formation.

(9) In case of elements, the allotropic form must be mentioned. E.g. C_{(graphite)}, S_{(rhombic)}, Sn_{(white)}

(10) For the reverse reaction, ΔH^{0} value has equal magnitude but opposite sign.

Question 56.

Define the following terms giving examples :

(1) Standard enthalpy of reaction.

(2) Standard enthalpy of formation or standard heat of formation

(3) Standard enthalpy of combustion or standard heat of combustion.

Answer:

(1) Standard enthalpy of reaction : it is defined as the difference between the sum of enthalpies of products and that of the reactants with every substance in its standard state at constant temperature (298 K) and pressure (1 atm).

Reactants → Products

ΔH^{0}_{reaction} = ΣH^{0}_{products} – ΣH^{0}_{reactants}

(2) Standard enthalpy of formation or standard heat of formation (Δ_{f}H^{0}) : It is defined as the enthalpy change ΔH^{0} when one mole of a pure compound is formed in its standard state from its constituent elements in their standard states at constant temperature (298 K) and pressure (1 atmosphere or 1 bar). It is denoted by Δ_{f}H^{0}. E.g.

C_{(s)} + O_{2}_{(g)} = CO_{2}_{(g)} Δ_{f}H^{0}= -394 kJ mol^{-1}

(Δ_{f}H^{0} may be positive or negative.)

(3) Standard enthalpy of combustion or standard heat of combustion : it is defined as the enthalpy change when one mole of a substance in the standard state undergoes complete combustion in a sufficient amount of oxygen at constant temperature (298 K) and pressure (1 atmosphere or 1 bar). It is denoted by Δ_{c}H^{0}.

E.g. CH_{3}OH_{(l)} + \(\frac {3}{2}\)O_{2(g)} = CO_{2(g)} + 2H_{2}O

ΔCH^{0 }= -726 kJ mol^{-1}

(Δ_{c}H^{0} is always negative.)

[Note : Calorific value : It is the enthalpy change or amount of heat liberated when one gram of a substance undergoes combustion.

Question 57.

Show that the standard heat of formation of a compound is equal to its enthalpy.

Answer:

Consider the formation of one mole of gaseous CO_{2} in the standard state at 298 K and 1 atmosphere. The thermochemical equation for formation can be represented as,

C_{(s)} + O_{2(g)} = CO_{2(g)} Δ_{f}H^{0} = -394 kJ mol^{-1}

Now heat of this reaction, ΔH^{0} is,

ΔH^{0}_{reaction} = Σ_{P}H^{0} – Σ_{R}H^{0}

∴ Δ_{f}H^{0}_{co2(g)} = H^{0}_{co2(g)} – [H^{0}c_{(s)} + H^{0}O_{2(g)}]

Since the enthalpies of elements in their standard states are zero,

i.e.

H^{0}c_{(s)} = o, H^{0}O_{2(g)} = 0

∴ Δ_{f}H^{0}co_{2(g)} = H^{0}co_{2(g)} – [0 + 0]

∴ Δ_{f}H^{0}co2 = Hco_{2(g)}

Therefore standard heat of formation of a compound is equal to its enthalpy.

Question 58.

Standard enthalpy of formation of various following compounds are given. Write thermochemical equation for each :

Compound | Δ_{f}H^{0 }KJ mol^{-1} |

Cao_{(s)} |
-635.1 |

Al_{2}Cl_{6(s)} |
-1300 |

C_{2}H_{6(g)} |
-84.7 |

CH_{3}COOH_{(l)} |
-484.7 |

C_{2}H_{5}OH_{(l)} |
-277.7 |

NaNO_{3(s)} |
-950.8 |

Answer:

(Hint: Write one mole of the compound on right hand side and corresponding constituent elements along with their standard physical states on left hand side.)

Question 59.

Standard enthalpy of combustion of different substances are given. Write thermochemical equation for each.

Substance | Δ_{C}H^{0} KJ mol^{-1} |

C_{(graphite)} |
-393.5 |

C_{6}H_{6(l)} |
-3268 |

C_{2}H_{5}OH_{(l)} |
-1409 |

CH_{3}CHO |
-1166 |

Answer:

In the combustion reaction, C forms CO_{2(g)},

H forms H_{2}O_{(l)}, etc.

(1) C_{(graphite)} + O_{2(g)} → CO_{2(g)}

Δ_{C}H^{0 }= -393.5 kJ mol^{-1}

Question 60.

Write the thermochemical equations for enthalpy of solution of :

(1) Glucose (C_{6}H_{12}O_{6})

(2) NaCl_{(s)}

(3) CaBr_{2(s)}

Answer:

Question 61.

How is standard enthalpy of formation useful to calculate standard enthalpy of reaction ?

Answer:

(1) The standard enthalpies of formation, Δ_{f}H^{0} of the compounds can be used to determine the standard enthalpy of reaction (Δ_{r}H^{0}).

(2) Δ_{r}H^{0} of a reaction can be obtained by subtracting the sum of Δ_{f}H^{0} values of all the reactants from the sum of Δ_{f}H^{0} values of all the products with each Δ_{f}H^{0} value multiplied by the appropriate coefficient of that substance in the balanced thermochemical equation.

(3) Consider following reaction :

aA + bB → cC + dD

The standard enthalpy of the reaction is given by,

where a, b, c and d are the coefficients (moles) of the substances A, B, C and D respectively.

Question 62.

Write the balanced chemical equation that have ΔH^{0} value equal to Δ_{f}H^{0} for each of the following substances :

(1) C_{2}H_{2(g)}

(2) KCIO_{3(s)}

(3) C_{12}H_{22}O_{11(s)} (4) CH_{3}-CH_{2}-OH_{(1)}

Answer:

Δ_{f}H^{0} represents the standard enthalpy of formation of each given substance. Hence it is necessary to write thermochemical equation for the formation of each substance. ΔH^{0} of this formation reaction is equal to standard heat of formation, Δ_{f}H^{0}.

Question 63.

Consider the chemical reaction,

OF_{2(g)} + H_{2}O_{(g)} → O_{2(g)} + 2HF_{(g)} ΔH^{0} = -323 kJ

What is ΔH^{0} of the reaction if (a) the equation is multiplied by 3, (b) direction of reaction is reversed?

Answer:

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

Δ_{r}H^{0} = 3ΔH^{0} = 3 × (-323) = -969 kJ

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

O_{2(g)} + 2HF_{(g)} → OF_{2(g)} + H_{2}O_{(g)}

∴ Δ_{r}H^{0} = – ΔH^{0} = – (- 323) = + 323 kJ

Question 64.

Define bond enthalpy (or bond energy).

Answer:

Bond enthalpy (or Bond energy) : The enthalpy change or amount of heat energy required to break one mole of particular covalent bonds of gaseous molecules forming free gaseous atoms or radicals at constant temperature (298 K) and pressure (1 atmosphere) is called bond enthalpy or bond energy. For example, bond enthalpy of H_{2} is 436.4 kJ mol^{-1}.

Question 65.

Explain bond enthalpy of diatomic molecules.

Answer:

In case of diatomic molecules, since there is only one bond, the bond enthalpy is equal to heat of atomisation. For example, heat of atomisation of

HCl_{(g)} is 431.9 kJ mol^{-1}.

(Bond enthalpy is generally denoted by D).

Question 66.

Explain bond enthalpy in polyatomic molecules.

Answer:

Consider bond enthalpy in H_{2}O. The thermochemical equation for dissociation of H_{2}O_{(g)} is,

H_{2}O_{(g)} → 2H_{(g)} + O_{(g)}, Δ_{r}H^{0} = 927 kJ mol^{-1}

In this, two O – H bonds are broken. It can be represented in stepwise as follows :

In above, even if two identical O – H bonds are borken, the energies required to break each bond are different.

The average bond enthalpy of O – H bond is,

Δ_{r}H^{0} = \(\frac {927}{2}\) = 463.5 kJ mol^{-1}

Solved Examples 4.10

Question 67.

Solve the following :

(1) Standard enthalpy of formation of ethane, C_{2}H_{6(g)} is -84.7 kJ mol^{-1}. Calculate the enthalpy change for the formation of 0.1 kg ethane.

Solution :

Given : Enthalpy of formation of C_{2}H_{6(g)}

= Δ_{f}H^{0} = ΔH_{1} = -84.7 kJ mol^{-1}

Mass of C_{2}H_{6(g)} = 0.1 kg = 100 g

Molar mass ofC_{2}H_{6} = 30 g mol^{-1}

ΔH^{0} for the formation of 0.1 kg C_{2}H_{6} = 100 g

C_{2}H_{6} = ΔH_{2} = ?

Ans. Heat of formation = -282.3 kJ

(2) When 10 g C_{2}H_{5}OH_{(l)} are formed, 51 kJ heat is liberated. Calculate standard enthalpy of formation of C_{2}H_{5}OH_{(l)}.

Solution :

Given : Mass of C_{2}H_{5}OH_{(l)} = m = 10 g

Heat liberated = ΔH1 = -51 kJ

Molar mass of C_{2}H_{5}OH = 46 gmol^{-1}

Standard enthalpy of formation of C_{2}H_{5}OH_{(l)}

= Δ_{f}H = ?

Standard enthalpy of formation is the enthalpy change for the formation of 1 mole C_{2}H_{5}OH_{(l)} i.e., 46 g C_{2}H_{5}OH_{(l)}.

Now,

∵ For the formation of 10 g C_{2}H_{5}OH_{(l)}

ΔH_{1} = -51 kJ

∴ For the formation of 46 g C_{2}H_{5}OH,

Δ_{f}H^{0} = \(\frac{-51 \times 46}{10}\) = – 234.6 kJ mol^{-1}

Ans. Standard enthalpy of formation of C_{2}H_{5}OH = Δ_{f}H^{0} = – 234.6 kJ mol^{-1}

(3) Standard enthalpy of combustion of CH_{3}OH is -726kJ mol^{-1}. Calculate enthalpy change for the combustion of 0.5 kg CH_{3}OH.

Solution :

Given : Standard enthalpy of combustion of

CH_{3}OH = Δ_{C}H^{0} = ΔH_{1} = -726 kJ mol^{-1}

Mass of CH_{3}OH = m = 0.5 kg = 500 g

Molar mass of CH_{3}OH = 32 g mol^{-1}

Enthalpy of combustion = Δ_{C}H = ΔH_{2} = ?

Now,

Enthalpy of combustion is ΔH for the combustion of 1 mole CH_{3}OH = 32 g CH_{3}OH.

∵ For 1 mole CH_{3}OH = 32g CH_{3}OH

ΔH_{1} = – 726 kJ

∴ For 500 g CH_{3}OH, ΔH_{2} = \(\frac{-726 \times 500}{32}\)

= -11344 kJ

Ans. Enthalpy change for combustion of 0.5 kg CH_{3}OH = – 11344 kJ

(4) The heat evolved in a reaction of 7.5 g of Fe_{2}O_{3} with enough CO is 1.164 kJ.

Calculate ΔH^{0} for the reaction,

Fe_{2}O_{3(s)} + 3CO_{(g)} → 2Fe_{(s)} + 3CO_{2}_{(g)}

Solution :

Fe_{2}O_{3(s)} + 3CO_{(g)} → 2Fe_{(s)} + 3CO_{2}_{(g)}

ΔH = -1.164 kJ

Atomic mass of Fe = 56 g mol^{-1}

Atomic mass of O = 16 g mol^{-1}

Mass of Fe_{2}O_{3} = 7.5 g

ΔH = -1.164 kJ

ΔH^{0} for reaction = ?

Molar mass of Fe_{2}O_{3} = 2 × 56 + 3 × 16

= 160 g mol^{-1}

∵ For 7.5 g Fe_{2}O_{3} ΔH= – 1.164 kJ

∴ For 160 g Fe_{2}O_{3}

ΔH^{0} = \(\frac{-1.164 \times 160}{7.5}\) = -24.86 kJ mol^{-1}

Ans. ΔH^{0} for the reaction = -24.83 kJ mol^{-1}

(5) Calculate the standard enthalpy of the reaction,

2C (graphite) + 3H_{2(g)} → C_{2}H_{6(g)}, ΔH^{0} = ? from the following ΔH^{0 }values :

Solution :

Ans. Standard enthalpy of formation of C_{2}H_{6 }= -84.4 kJ mol^{-1}

(6) The enthalpy of combustion of ethane is -1300 kJ. How much heat will be evolved by combustion of 1.3 × 10^{-3} kg of ethane?

Solution :

Given : Δ_{C}H_{C2H6(g)} = -1300 kJ mol^{-1}

ΔH = ?

Amount of C_{2}H_{6(g)} = 1.3 × 10^{-3} kg

Molar mass of C_{2}H_{6} = 30 × 10^{-3} kg mol^{-1}

Number of moles of C_{2}H_{6}

= nC_{2}H_{6} = \(\frac{1.3 \times 10^{-3}}{30 \times 10^{-3}}\) = 4,333 × 10^{-2} mol

For, combustion of 1 mol C_{2}H_{6} ΔH = -1300 kJ

∴ For combustion of 4.333 × 10^{-2} mol C_{2}H_{6},

ΔH = 4.333 × 10^{-2} × ( -1300) = – 56.33 kJ

Ans. Heat evolved is -56.33 kJ

(7) Calculate heat of formation of pentane from the following data :

(i) C_{(s)} + O_{2(g)} = CO_{2(g)} ΔH^{0} = -393.51 kJ

(ii) H_{2(g)} + \(\frac {1}{2}\)O_{2(g)} = H_{2}O_{(l)} ΔH^{0} = -285.80 kJ

(iii) C_{5}H_{12} + 😯_{2(g)} = 5CO_{2(g)} + 6H_{2}O_{1} ΔH^{0} = -3537 kJ

Solution :

Given :

(i) CO_{(s)} + O_{2(g)} = CO_{2(g)} ….. (1)

\(\Delta H_{1}^{0}\) = -393.51 kJ mol^{-1}

(ii) H_{2(g)} + \(\frac {1}{2}\)O_{2(g)} = H_{2}O_{(l)} … (2)

\(\Delta H_{2}^{0}\) = – 285.80 kJ mol^{-1}

(iii) C_{5}H_{12(g)} + 😯_{2(g)} = 5CO_{2(g)} + 6H_{2}O_{(l)} ….. (3)

\(\Delta H_{3}^{0}\) = -3537 kJ mol^{-1}

Required thermochemical equation :

5C_{(s) }+ 6H_{2(g)} → C_{5}H_{12(g)} – ΔH = ?

Add 5 × equation (1) and 6 × equation (2) and subtract equation (3), then we get the required equation.

∴ ΔH^{0} = 5 \(\Delta H_{1}^{0}\) + 6 \(\Delta H_{2}^{0}\) – \(\Delta H_{3}^{0}\)

= 5( -393.52) + 6( -285.8) – (-3537)

= -1967.6 – 1714.8 + 3537

= -145.4 kJ mol^{-1}

Ans. Δ_{f}H^{0}C_{5}H_{12} = -145.4 kJ mol^{-1}

(8) How much heat is evolved when 12 g of CO react with NO_{2} according to the following reaction,

4CO_{(g)} + 2NO_{2(g)} → 4CO_{2(g)} + N_{2(g)}, ΔH^{0} = -1198 kJ ?

Solution :

Given : Mass of CO(g) = m = 12 g

Molar mass of CO = 28 g mol^{-1}

4CO_{(g)} + 2NO_{2(g)} → 4CO_{2(g)} + N_{2(g)}

Δ_{r}H^{0 }= -1198 kJ

Mass of 4 moles of CO = 4 × 28 g CO = 112 g CO

Ans. Heat evolved during combustion of 12 g CO = 128.4 kJ, or Δ_{C}H = -128.4 kJ

(9) The heats of formation of C_{12}H_{2}_{2}O_{11(S)}, CO_{2(g)} and H_{2}O_{(l)} are -2271.82, – 393.50 and 285.76 kJ respectively. Calculate the amount of cane sugar (C_{12}H_{2}_{2}O_{11(S)}) which will supply 11296.8 kJ of energy.

Solution :

Given : Δ_{f}H_{C12H22O11(S)} = -2271.82 J mol^{-1}

Δ_{f}H_{CO2(g)} = – 393.5 kJ mol^{-1}

Δ_{f}H_{H2O(l)} = – 285.76 kJ mol^{-1}

Energy required = 11296.8 kJ

Thermochemical equation for combustion of C_{12}H_{2}_{2}O_{11(S)} is represented as,

= [ 12(-393.5) + 11 (-285.76] – [-2271.82 + 12(0)]

= [ -4722 – 3143.36] + 2271.82

= -5593.54 kJ mol^{-1}

Molar mass of C_{12}H_{2}_{2}O_{11(S)} = 342

To obtain 5593.5 kJ energy, C_{12}H_{2}_{2}O_{11(S)} required is 342 gram.

Hence for 11296.8 energy, the amount of C_{12}H_{2}_{2}O_{11(S)} required as = \(\frac{11296.8 \times 342}{5593.54}\)

= 690.7 g

Ans. Amount of sugar required = 690.7 g

(10) 6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat energy. What is enthalpy of vaporization of ethanol?

Solution :

Given : Mass of ethanol (C_{2}H_{5}OH) = m = 6.24 g

Heat energy supplied = ΔH = 5.89 kJ

Heat of vaporisation of ethanol = Δ_{vap}H = ?

Molar mass of ethanol, C_{2}H_{5}OH = 46 g mol^{-1}

∵ For 6.24 g C_{2}H_{5}OH ΔH = 5.89kJ

∴ For 1 mole C_{2}H_{5}OH = 46 g C_{2}H_{5}OH

ΔH = \(\frac{5.89 \times 46}{6.24}\)

= 43.42 kJ mol^{-1}

∴ Enthalpy of vaporisation of C_{2}H_{5}OH_{(l)}

= 43.42 kJ

Ans. Enthalpy of vaporisation of C_{2}H_{5}OH_{(l)}

= 43.42 kJ

(11) Given the following equations calculate the standard enthalpy of the reaction :

Solution :

By subtracting eq. (ii) from eq. (iii), we get eq. (i)

∴ eq. (i) = eq. (iii) – eq. (ii)

\(\Delta H_{1}^{0}=\Delta H_{3}^{0}-\Delta H_{2}^{0}\)

= -1670 – (-847.6)

= – 822.4 kJ

∴ Δ_{r}H^{0} = ΔH^{0}_{1} = -822.4 kJ

Ans. Standrad enthalpy of the reaction = Δ_{r}H^{0} = -822.4 kJ

(12) Calculate the standard enthalpy of combustion of CH_{2}COOH_{(l)} from the following data : Δ_{f}H^{0}(CO_{2}) = -393.3 kJ mol^{-1}

Δ_{f}H^{0}(H_{2}O) = -285.8 kJ mol^{-1}

Δ_{f}H^{0}(CH_{3}COOH) = -483.2 kJ mol^{-1}

Solution :

∴ ΔH_{1} = 2ΔH_{2} + 2ΔH_{3} – ΔH_{4}

= 2(-393.3) + 2(-285.8) – (-483.2)

= -786.6 – 571.6 + 483.2

= -875 kJ mol^{-1}

Ans. Standard enthalpy of combustion of CH_{3}COOH = -875 kJ mol^{-1}.

(13) The bond enthalpies of H_{2(g)}, Br_{2(g)} and HBr_{(g)} are 436 kJ mol^{-1}, 193 kJ mol^{-1} and 366 kJ mol^{-1} respectively. Calculate the enthalpy change for the following reaction,

H_{2(g)} + Br_{2(g)} → 2HBr_{(g)}.

Solution :

Given : Bond enthalpy of H_{2(g)} = ΔH^{0}H_{2(g)}

= 436 kJ mol-1

Bond enthalpy of Br_{2(g)} = ΔH^{0}Br_{2(g)} = 193 kJ mol^{-1}

Bond enthalpy of HBr_{(g)} = ΔH^{0}HBr_{(g)} = 366 kJ mol^{-1}

Given reaction,

H_{2(g)} + Br_{2(g)} → 2HBr_{(g)}

OR

H-H_{(g)} + Br-Br_{(g)} → 2H-Br_{(g)}

The enthalpy change of the reaction is,

= [436 + 193] – 2[366]

= 629 – 732

= -103 kJ

Ans. Enthalpy change for the reaction = Δ_{r}H^{0}

= -103 kJ

(14) Calculate Δ_{r}H^{0} of the reaction

CH_{4(g)} + O_{2(g)} → CH_{2}O_{(g)} + H_{2}O_{(g)}

From the following data:

Solution:

Given:

Standard enthalpy change for the reaction = Δ_{r}H^{0} = ?

= [ 2ΔH^{0}_{C-H} + ΔH^{0}_{o=o} ] – [ΔH^{0}_{C=o} + 2ΔH^{0}_{o-H}]

= [2 × 414 + 499] – [745 + 2 × 464]

= [828 + 499] – [745 + 928]

= -346 kJ

Ans. Standard enthalpy change for the reaction = ΔrH^{0 }= -346 kJ

(15) Calculate C-Cl bond enthalpy from the following data :

CH_{3}Cl_{(g)} + Cl_{2(g)} → CH_{2}Cl_{2(g)} + HCl_{(g)} ΔH^{0} = – 104 kJ

Bond | C–H | Cl–Cl | H–Cl |

ΔH^{0}/KJ mol^{-1} |
414 | 243 | 431 |

(330 kJ mol^{-1})

Solution :

Given :

Bond | C–H | Cl–Cl | H–Cl |

ΔH^{0}/KJ mol^{-1} |
414 | 243 | 431 |

For the given reaction, Δ_{r}H^{0} = -104 kJ

Bond enthalpy of C-Cl = ΔH^{0}C–Cl] = ?

In this reaction, 1 C–H, 1 Cl–Cl bonds of the reactants are broken while 1C–Cl and 1H–Cl bonds of the products are formed.

Sum of bond enthalpies of bonds formed of the products

Ans. Bond enthalpy of C–Cl = ΔH^{0}C–Cl

= 330 kJ mol^{-1}

(16) The enthalpy change for the atomisation of 10^{10} molecules of ammonia is 1.94 × 10^{-11} kJ. Calculate the bond enthalpy of N-H bond.

Solution :

Given : Enthalpy change for atomisation of 10^{10} molecules = 1.94 × 10^{-11} kJ

Number of NH_{3} molecules dissociate = 10^{10}

Bond enthalpy of N-H = ΔH = ?

1 mole of NH_{3} contains 6.022 × 10^{23} NH_{3} molecules.

∵ For atomisation of 1010 molecules of NH_{3}

ΔH = 1.94 × 10^{-11} kJ

∴ For atomisation of 6.022 × 10^{23} molecules of NH_{3},

ΔH = \(\frac{1.94 \times 10^{-11} \times 6.022 \times 10^{23}}{10^{10}}\)

= 1168 kJ mol^{-1}

In NH_{3} three N-H bonds are broken on atomisation.

NH_{3(g)} → N_{(g)} + 3H_{(g)} ΔH = 1168 kJ mol^{-1}

∴ Average bond enthalpy of N-H bond is,

ΔH = \(\frac{1168}{3}\) = 389.3 kJ mol^{-1}

Ans. Bond enthalpy of N-H bond

= 389.3 kJ mol^{-1}

(17) Calculate the enthalpy of atomisation (or dissociation) of CH_{2}Br_{2(g)} at 25°C from the following data :

Bond enthalpies | C-H | C-Br |

ΔH^{0} kJ mol^{-1} |
414 | 352 |

Solution :

Given : Bond enthalpies : ΔH^{0}_{C-H}

= 414 kJ mol-1;

ΔH^{0}_{C-Br} = 352 kJ mol^{-1}

Enthalpy of atomisation of CH_{2}Br_{2(g)} = ?

Thermochemical equation for atomisation (or dissociation) of CH_{2}Br_{2} is,

ΔatomH° = sum of bond enthalpies of all bonds broken

= 2ΔH^{0}_{C-H} + 2ΔH^{0}_{C-Br}

= 2 × 414 + 2 × 352

= 828 + 704

= 1532 kJ mol^{-1}

Ans. Enthalpy of atomisation of CH_{2}Br_{2(g)}

= 1532 kJ mol^{-1}

(18) Enthalpy of sublimation of graphite is 716 kJ mol^{-1}.

Bond enthalpy | H-H | C-H |

ΔH^{0} kJ mol^{-1} |
436.4 | 414 |

Calculate standard enthalpy of formation of CH_{4}.

Solution :

Given : Δ_{sub}H^{0}_{graphite} = 716 kJ mol^{-1}

Bond enthalpy | H-H | C-H |

ΔH^{0} kJ mol^{-1} |
436.4 | 414 |

Thermochemical equation for the formation of CH_{4},

= [716 + 2 × 436.4] – [4 × 414]

= [716+ 872.8] – [1656]

= 1588.8 – 1656

= -67.2 kJ mol^{-1}

Ans. Standard enthalpy of formation of CH4 = Δ_{f}H^{0}_{CH4(g)} = -67.2 kJ mol^{-1}

(19) Calculate enthalpy of formation of propane from the following data :

Heat of sublimation of graphite is 716 kJ mo^{-1}.

Bond enthalpy | H-H | C-H | C-C |

ΔH^{0} kJ mol^{-1} |
436.4 | 414 | 350 |

Solution :

Given: Enthalpy of sublimation of graphite = Δ_{sub}H^{0}_{C}

= 716 kJmol^{-1}

Bond enthalpy | H-H | C-H | C-C |

ΔH^{0} kJ mol^{-1} |
436.4 | 414 | 350 |

Enthalpy of formation of propane = Δ_{f}H^{0} = ?

Thermochemical equation of the formation of propane, CH_{3}-CH_{2}-CH_{3},

= [3 × 716 + 4 × 436.4] – [2 × 350 + 8 × 414]

= [2148 + 1745.6] – [700 + 3312]

= -118.4 kJmol^{-1}

Ans. Enthalpy of formation of propane (C_{3}H_{8})

= -118.4 kJmol^{-1}

(20) The standard enthalpy of formation of propene, CH_{3}-CH = CH_{2} is -13.2 kJ mol^{-1}. Enthalpy of sublimation (atomisation) of graphite is 716 kJmol^{-1}.

Bond enthalpy | H-H | C-H | C-C |

ΔH^{0} kJ mol^{-1} |
436.4 | 414 | 350 |

Calculate bond enthalpy of C = C

Solution :

Bond enthalpy of C = C = ΔH^{0}_{C=C} = ?

For the formation of propene, (CH_{3} – CH = CH_{2}),

13.2 = [3 × 716 + 3 × 436.4] – [6 × 414 + 350 + \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) ]

= [2148 + 1309.2] – [2484 + 350 + \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) ]

= [3457.2] – [2834 + \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) ]

= 3457.2 – 2834 – \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\)

\(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) = 3457.2 – 2834 – 13.2

= 610 kJmol^{-1}

Ans. Bond enthalpy of C = C = \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\)

= 610 kJ mol^{-1}

(21) Calculate the enthalpy of the reaction,

CH_{3}COOH_{(g)} + CH_{3}CH_{2}OH_{(g)} → CH_{3}COOCH_{2}CH_{3(g)} + H_{2}O_{(g)}

Bond enthalpies of O-H, C-O, in kJmol^{-1} are 464, 351 respectively.

Solution :

In this reaction 1O-H and 1C-0 bond of the reactants are broken while 1C-0 and 1O-H bonds of the products are formed. Enthalpy of reaction,

Ans. Hence Enthalpy change for the reaction = Δ_{r}H^{0} = 0.

Question 68.

(1) What is a spontaneous process?

(2) What are its characteristics?

Answer:

(1) Spontaneous process : It is defined as a process that takes place on its own or without the intervention of the external agency or influence. For example, expansion of a gas or flow of a gas from higher pressure to low pressure or a flow of heat from higher temperature to lower temperature.

(2) Characteristics :

- It occurs on its own and doesn’t require external agency.
- It proceeds in one direction and can’t be completely reversed by external stimulant.
- These processes may be fast or slow.
- These processes proceed until an equilibrium is reached.

Question 69.

Give the examples of spontaneous processes.

Answer:

The examples of the spontaneous processes are as follows :

- All natural processes are spontaneous.
- A flow of gas from higher pressure to lower pressure.
- Flow of water on its own from higher level to lower level.
- Flow of heat from hotter body to colder body.
- Acid-base neutralisation is a spontaneous reaction.

Question 70.

What is relation between spontaneity and energy of a system ?

Answer:

(1) The spontaneous process takes place in a direction in which energy of the system decreases. For example, neutralisation reaction between NaOH and HCl solution is exothermic with release of energy.

(2) The spontaneous process also takes place with the increase in energy by absorbing heat. For example,

(a) Melting of ice at 0 °C by absorption of heat

(b) Dissolution of NaCl,

NaCl_{(s)} + aq → NaCl_{(aq)} → Na^{+}_{(aq)} + Cl^{–}_{(aq)}

ΔH^{0} = + 3.9 kJ mol^{-1}

Question 71.

Which of the following are spontaneous ?

(a) Dissolving sugar in hot coffee.

(b) Separation of Ar and Kr from their mixture.

(c) Spreading of fragrance when a bottle of perfume is opened.

(d) Flow of heat from cold object to hot object.

(e) Heat transfer from ice to room temperature at 25 °C.

Answer:

The spontaneous processes are :

(a) Dissolving sugar in hot coffee.

(b) Separation of Ar and Kr from their mixture.

(c) Spreading of fragrance when a bottle of perfume is opened.

The non-spontaneous processes are :

(d) Flow of heat from cold object to hot object.

(e) Heat transfer from ice to room temperature at 25 °C.

Question 72.

Explain : (a) Order in a system.

(b) Disorder in a system.

Answer:

(a) (i) When the atoms, molecules or ions constituting the system are arranged in a perfect order then the system is said to be in order. For example, in the solid state, the constituent atoms, molecules or ions are tightly placed at lattice points in the crystal lattice.

(ii) When solid melts forming a liquid or when a liquid vaporises, the constituents are separated and are in random motion imparting maximum disorder.

(iii) As energy of the system decreases order increases.

(b) Increase in entropy is a measure of disorder in the system. Consider following process :

Fig. 4.11 : Order decreases and disorder increases, Entropy increases

Question 73.

What is the change in order and entropy in the following :

(i) Dissolution of solid I_{2} in water.

(ii) Dissociation of H_{2(g)} into atoms ?

Answer:

(i) For dissolution of solid I_{2},

In the solid I2, there is ordered arrangement which collapses in solution increasing disorder and entropy, hence ΔS is positive.

(ii) In the dissociation of H_{2(g)}

H_{2(g)} → 2H_{(g)} (ΔS > 0)

In the molecular state, two H atoms in every molecule are together but in atomic state the disorder is increased with the increase in entropy and hence ΔS > 0.

Question 74.

How does addition of heat to a system at different temperatures changes disorder or ΔS ?

Answer:

- The amount of heat added to a system at higher temperature causes less disorder than when the heat is added at lower temperature.
- Since disorder depends on the temperature at which heat is added, ΔS relates reciprocally to temperature.
- This can also be explained from equation,

ΔS = \(\frac{Q_{\text {rev }}}{T}\)

Question 75.

Explain the change in entropy for the following processes :

(i) 2H_{2}O_{2(l)} → 2H_{2}O_{(l)} + O_{2(g)}

(ii) 2H_{2(g)} + O_{2(g)} → 2H_{2}O_{(l)}

(iii) When ice melts at 0 °C and water vaporises at 100 °C.

Answer:

(i) In the following reaction,

2H_{2}O_{2(l)} → 2H_{2}O_{(l)} + O_{2(g)} ΔS = + 126 JK

Due to formation of O_{2} gas from liquid, entropy increases.

(ii) In the reaction, entropy decreases due to formation of liquid H_{2}O from gaseous H_{2} and O_{2}.

(iii) \(\text { Ice } \stackrel{0^{\circ} \mathrm{C}}{\longrightarrow} \text { water } \stackrel{100^{\circ} \mathrm{C}}{\longrightarrow} \text { vapour }\)

In these two steps, entropy increases due to increase in disorder from solid ice to liquid water and further to gaseous state.

Question 76.

How does entropy change in the following processes ? Explain.

(a) freezing of a liquid

(b) sublimation of a solid

(c) dissolving sugar in water

(d) condensation of vapour.

Answer:

(a) Freezing of a liquid results in decrease in randomness and disorder, hence entropy decreases, ΔS < 0.

(b) Sublimation of a solid converts it into vapour where the molecules or atoms are free to move randomly. Hence disorder increases accompanying increase in entropy, ΔS > 0.

(c) Dissolving sugar in water separates the molecules of sugar in the solution increasing disorder and entropy, ΔS > 0.

(d) Condensation of vapour decreases the disorder and randomness, hence entropy, ΔS < 0.

Question 77.

Predict the sign of ΔS in the following processes. Give reasons for your answer :

OR

Explain with reason sign conventions of ΔS in the following reactions.

Answer:

(a) N_{2}O_{4(g)} → 2NO_{2(g)}

Since 1 mole N_{2}O_{4} on dissociation gives two moles of NO_{2}, the number of molecules increase, disorder increases hence entropy increases, ΔS > 0.

(b) Fe_{2}O_{3(s)} + 3H_{2(g)} → 2Fe_{(s)} + 3H_{2}O_{(g)}

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

(c) N_{2(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 MgCO_{3} gives 1 mole of solid MgO and 1 mole of gaseous CO_{2} (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 Cl_{2} gas gives double Cl atoms, the number of atoms increases (Δn >0) increasing the disorder of the system. Hence ΔS > 0.

Question 78.

Identify which of the following pairs has larger entropy ? Why ?

(a) He_{(g)} in a volume of 1 L or He_{(g)} in a volume of 5 L both at 25 °C.

(b) O_{2}_{(g)} at 1 atm or O_{2(g)} at 10 atm both at the same temperature.

(c) C_{2}H_{5}OH_{(l)} or C_{2}H_{5}OH_{(g)}

(d) 5 mol of Ne or 2 mol of Ne.

Answer:

(a) Atoms of He in 5 L at 25 °C occupy more volume than in 1 L. Hence, the randomness and disorder is more in 5 L. Expansion of a gas always increases its entropy. Therefore He_{(g)} in 5L will have larger entropy.

(b) O_{2(g)} at 1 atm will occupy more volume than O_{2(g)} at 10 atm at the same temperature. Hence at 1 atm O_{2(g)} will have higher disorder and hence higher entropy.

(c) The molecules of gaseous C_{2}H_{5}OH_{(g)} will have more disorder and randomness due to free motion of molecules than C_{2}H_{5}OH_{(l)}. Hence entropy of C_{2}H_{5}OH_{(g)} will be larger.

(d) 5 mol Ne will contain more Ne atoms than 2 mol Ne. Hence disorder in 5 mol will be more. Therefore 5 mol Ne will have larger entropy.

Question 79.

Mention entropy change (ΔS) for :

(i) spontaneous process

(ii) nonspontaneous process

(iii) at equilibrium.

Answer:

(a) ΔS_{total} > 0, the process is spontaneous

(b) ΔS_{total} < 0, the process is non-spontaneous

(c) ΔS_{total} = 0, the process is at equilibrium.

Question 80.

Define Gibbs free energy and change in free energy. What are the units of Gibbs free energy ?

OR

Derive the relation between ΔG and ΔS Total.

Answer:

(i) Gibbs free energy, G is defined as,

G = H – TS

where H is the enthalpy, S is the entropy of the system at absolute temperature T.

Since H, T and S are state functions, G is a state function and a thermodynamic function.

(ii) At constant temperature and pressure, change in free energy ΔG for the system is represented as, ΔG = ΔH – TΔS

This is called Gibbs free energy equation for ΔG. In this ΔS is total entropy change, i.e., ΔS_{Total}.

(iii) The SI units of ΔG are J or kJ (or Jmol^{-1} or kJmol^{-1}).

The c.g.s. units of ΔG are cal or kcal (or cal mol^{-1} or kcal mol^{-1}.)

Question 81.

Explain Gibbs free energy and spontaneity of the process.

Answer:

The total entropy change for a system and its surroundings accompanying a process is given by,

ΔS_{Total} = ΔS_{system} + ΔS_{surr}

By second law, for a spontaneous process,

ΔS_{Total} > 0. If + ΔH is the enthalpy change (or enthalpy increase) for the process, or a reaction at constant temperature (T) and pressure, then enthalpy change (or enthalpy decrease) for the surroundings will be -ΔH.

By Gibbs equation,

ΔG = ΔH – TΔS

By comparing above two equations,

∴ ΔG = -TΔS_{Total}

As ΔS_{Total} increases, ΔG decreases.

For a spontaneous process, ΔS_{Total} > 0

which is according to second law of thermodynamics.

∴ ΔG < 0.

Hence in a spontaneous process, Gibbs free energy decreases (ΔG < 0) while entropy increases (ΔS > 0).

Therefore for a non-spontaneous process Gibbs free energy increases (Δ G > 0).

It can be concluded that for a process at equilibrium, ΔG=0.

Hence,

- For the spontaneous process, Δ G < 0
- For the non-spontaneous process, Δ G > 0
- For the process at equilibrium, Δ G = 0.

Question 82.

How does second law of thermodynamics explain the conditions of spontaneity ?

Answer:

The second law explains the conditions of spontaneity as below :

(i) ΔS_{total} > 0 and ΔG < 0, the process is spontaneous.

(ii) ΔS_{total} < 0 and ΔG > 0, the process is nonspontaneous.

(iii) ΔS_{total} = 0 and ΔG = 0, the process is at equilibrium.

Question 83.

Discuss the factors, ΔH, ΔS and ΔG for spontaneous and non-spontaneous processes.

OR

What can be said about the spontaneity of reactions when (1) ΔH and ΔS are both positive (2) ΔH and ΔS are both negative (3) ΔH is positive and ΔS is negative (4) ΔH is negative and ΔS is positive.

Answer:

For a spontaneous or a non-spontaneous process, ΔH and ΔS may be positive or negative (ΔH < 0 or ΔH > 0; ΔS < 0 or ΔS > 0). But ΔG must decrease, i.e., ΔG < 0. If ΔG > 0, the process or a reaction will definitely be non-spontaneous. This can be explained by Gibbs equation, ΔG = ΔH – TΔS.

(1) If ΔH and ΔS are both negative, then ΔG will be negative only when TΔS < ΔH or when temperature T is low. Such reactions must be carried out at low temperatures.

(2) If ΔH and ΔS are both positive then ΔG will be negative if, TΔS > ΔH; such reactions must be carried out at high temperature.

(3) If ΔH is negative (ΔH < 0) and ΔS is positive (ΔS > 0) then for all temperatures ΔG will be negative and the reaction will be spontaneous. But as temperature increase, ΔG will be more negative, hence the reaction will be more spontaneous at higher temperature.

(4) If ΔH is positive, (ΔH > 0) and ΔS is negative (ΔS < 0), ΔG will be always positive (ΔG > 0) and hence the reaction will be non-spontaneous at all temperatures.

This can be summarised in the following table :

Question 84.

Obtain a temperature condition for equilibrium.

Answer:

For a system at equilibrium, free energy change ΔG is,

ΔG = ΔH – TΔS

where ΔH is enthalpy change, ΔS is entropy change at temperature, T. Since ΔG = 0 at equilibrium,

O = ΔH – TΔS

∴ TΔS = ΔH

OR T = \(\frac{\Delta H}{\Delta S}\)

Hence at temperature T, changeover between forward spontaneous step and backward non-spontaneous step occurs and the system attains an equilibrium.

Here ΔH and ΔS are assumed to be independent of temperature.

Question 85.

Predict the signs of ΔH, ΔS and ΔG of the system when a solid melts at 1 atmosphere and at (a) -55 °C (b) -95 °C (c) -77 °C, if the normal melting point of the solid is -77 °C.

Answer:

Since the normal melting point of the solid is -77°C, to melt the solid at any temperature other than at -77 °C, the pressure is required to be changed. During the phase change, the system will be at equilibrium, hence Δ G = 0.

(a) In case a solid at -55 °C, the temperature should be decreased (ΔH < 0, ΔS < 0) to -77 °C and then it will melt, so ΔH > 0, ΔS > 0, ΔG = 0.

(b) In case of a solid at -95 °C, it represents supercooled system and the temperature should be raised to -77 °C (ΔH > 0, ΔS > 0) and then it will melt so ΔH > 0, ΔS > 0, ΔG = 0.

(c) At -77 °C, solid will melt, solid and liquid will be at equilibrium. Melting involves absorption of heat,

ΔH >0, ΔS > 0, ΔG = 0.

Solved Examples 4.11

Question 86.

Solve the following :

(1) In an isothermal reversible process, 6 kJ heat is absorbed at 27 °C. Calculate the entropy change.

Solution :

Given : Temperature = T = 273 + 21 = 300 K

Heat absorbed = Q_{rev} = 6 kJ = 6000 J

Entropy change = ΔS = ?

ΔS = \(\frac{Q_{\text {rev }}}{T}=\frac{6000}{300}\) = 20 JK^{-1}

Ans. Entropy change = ΔS = 20 JK^{-1}

(2) The latent heat of evaporation of water is 2.26 kJ g^{-1} at 1 atm and 100 °C. Calculate the entropy change for evaporation of 1 mole of water at 100 °C.

Solution :

Given : Latent heat of evaporation = Δ_{vap}H^{0}

= 2.26 kJ g^{-1}

Temperature = T = 273 + 100 = 373 K

Molar mass of water = 18 g mol^{-1}

ΔS = ?

For 1g H_{2}O_{(l)} Δ_{vap}H^{0} = 2.26 KJ

∴ For 1 mol H_{2}O_{(l)} = 18 g H_{2}O_{(l)}

Δ_{vap}H^{0} = 2.26 × 18

= 40.68 kJ

= 40680 J

Entropy change, ΔS is given by,

ΔS = \(\frac{\Delta_{\mathrm{vap}} H^{0}}{T}=\frac{40680}{373}\) = 109.06 JK^{-1} mol^{-1}

Ans. Entropy change = ΔS = 109.06 JK^{-1} mol^{-1}

(3) Calculate the standard (absolute) entropy change for the formation of CO_{2(g)}.

Substance | C_{(graphite)} |
O_{2(g)} |
CO_{2(g)} |

Standard molar enthalpy JK^{-1} mol^{-1} |
5.74 | 205 | 213.7 |

Solution:

Given:

Substance | C_{(graphite)} |
O_{2(g)} |
CO_{2(g)} |

Standard molar enthalpy S^{0} JK^{-1} mol^{-1} |
5.74 | 205 | 213.7 |

For the formation of CO_{2(g)},

(4) The standard entropies of H_{2(g)}, O_{2(g)} and H_{2}O_{(g)} in JK^{-1} mol^{-1} are 130, 205 and 189 respectively. The heat of formation of H_{2}O_{(g)} is -242 kJ mol^{-1}. Calculate ΔS for formation of H_{2}O_{(g)}, for the surroundings and the universe at 298 K. Mention whether the reaction is spontaneous or non-spontaneous.

Solution :

Given :

Substance | H_{2(g)} |
O_{2(g)} |
H_{2}O_{(g)} |

Standard entropy S^{0} JK^{-1 }mol^{-1} |
130 | 205 | 189 |

Δ_{f}H^{0 }= -242 kJ mol^{-1}

ΔS_{universe} = ?, ΔS_{surr} = ?

Thermochemical equation for the formation of H_{2}O_{(g)}

H_{2(g)} + \(\frac {1}{2}\)O_{2(g)} → H_{2}O_{(g)}

ΔS^{0} = [S^{0}_{H2O}] – [H^{0}_{H2} + \(\frac {1}{2}\) H^{0}_{O2}]

= 189 – [130 + \(\frac {1}{2}\)(205)]

= 189 – [232.5]

= -43.5 JK^{-1} mol^{-1}

Hence, ΔSsystem = -43.5 JK-1 mol^{-1}

Since for the formation of H_{2}O_{(g)}

Δ_{f}H^{0} = -242 kJmol^{-1} = -242 × 10^{3} Jmol^{-1}, the reaction is exothermic. Hence the surroundings gains heat energy 242 × 10^{3}J. Therefore entropy of the system decreases while entropy of surroundings increases.

Hence, ΔS_{sys} < 0 but ΔS_{universe} > 0, hence the reaction is spontaneous.

Ans. ΔS_{H2O(g)} = -43.5 JK^{-1} mol^{-1}

ΔS_{surr} = 813 JK^{-1} mol^{-1}

ΔS_{universe} = 769.5 JK^{-1}.

(5) Calculate ΔS_{Total} and hence show whether the following reaction is spontaneous at 25 °C.

Hg_{(s)} + O_{2(g)} → Hg_{(l)} + SO_{2(g)} ΔH^{0} = – 238.6 kJ

ΔS^{0} = +36.7 JK^{-1}

Solution :

Given : Hg_{(s)} + O_{2(g)} → Hg_{(l)} + SO_{2(g)}

Δ_{r}H^{0} = – 238.6 kJ

ΔS^{0} = +36.7 JK^{-1}

T = 273 + 25 = 298 K

ΔS_{Total} = ?

ΔS_{Total} = ΔS_{sys} + Δ S_{surr}

Now, ΔS_{sys} = 36.7 JK^{-1}

Since the reaction is exothermic, system loses heat to surroundings. Hence the entropy of the surroundings increases.

ΔH_{surr} = + 238.6 kJ = 238600 J

∵ ΔS_{Total} > 0, the reaction is spontaneous.

Ans. ΔS_{Total} = 837.4 JK^{-1}

The reaction is spontaneous.

(6) What is the value of AS_{surr} for the following reaction at 298 K ?

6CO_{2(g)} + 6H_{2}O_{(l)} → C_{6}H_{12}O_{6(s)} + 6O_{2(g)},

ΔG^{0} = 2879 kJ mol^{-1}, ΔS^{0} = -210 JK^{-1} mol^{-1}.

Solution :

Given :

6CO_{2(g)} + 6H_{2}O_{(l)} → C_{6}H_{12}O_{6(s)} + 6O_{2(g)},

ΔG^{0} = 2879 kJmol^{-1};

ΔS^{0} = -210 JK^{-1}mol^{-1} = -0.210 kJ K^{-1} mol^{-1}

T = 298 K ΔH^{0} = ?

ΔG^{0} = ΔH^{0} – TΔS^{0}

∴ ΔH^{0} = ΔG^{0} + TΔS^{0}

= 2879 + 298(-0.210)

= 2879 – 62.58

= 2816.42 kJ mol^{-1}

Since ΔH^{0} > 0, the reaction is endothermic, and system absorbs heat from surroundings. Hence surroundings loses heat,

Ans. ΔS^{0}surr = -9.45 kJ K^{-1}

(7) Calculate ΔS_{surr} when on mole of methanol (CH_{3}OH) is formed from its elements under standard conditions if Δ_{f}H^{0}(CH_{3}OH) = -238.9 J mol^{-1}.

Solution :

Given : Number of moles of ethanol,

(C_{2}H_{5}OH) = n = 1 mol

Δ_{f}H^{0}(CH_{3}OH) = -238.9 kJ mol^{-1}

= -238.9 × 10^{3}J mol^{-1}

Temperature = T = 298 K

ΔS = ?

ΔS_{surr} = ?

Since Δ_{f}H^{0} is negative, the reaction for the formation of one mole of C_{2}H_{5}OH is exothermic.

As heat is released to the surroundings,

ΔH^{0}_{surr} = + 238.9 kJ mol^{-1}

∴ ΔS_{surr} = \(\frac{\Delta H_{\text {surr }}^{0}}{T}=\frac{+238.9 \times 10^{3}}{298}\)

= +801.7 JK^{-1}

Thus entropy of the surroundings increases.

Ans. ΔS_{surr} = +801.7 JK^{-1}

(8) What is the value of ΔSsurr for the following reaction at 298 K –

6CO_{2(g)} + 6H_{2}O_{(l)} → C_{6}H_{12}O_{6(s)} + 6O_{2(g)}

Given that: ΔG° = 2879 KJ mol^{-1}

ΔS^{0} = -210 J K^{-1} mol^{-1}

Solution :

Given : ΔG^{0} = 2879 KJ mol^{-1} = 2879 × 10^{3} J mol^{-1}

ΔS^{0} = -210 JK^{-1} mol^{-1}

ΔS_{surr} = ?

ΔG^{0} = ΔH^{0} – TΔS^{0}

∴ ΔH^{0} = ΔG^{0} + TΔS^{0}

= 2879 × 10^{3} + 298 × (- 210)

= 2879 × 10^{3} – 62580

= 2816420 J

Since, for a system, ΔH^{0} is +2816420 J, the surrounding loses heat to system,

∴ ΔH^{0}_{surr} = – 2816420 J

∴ ΔS^{0}_{surr} = \(\frac{\Delta H_{\text {surr }}^{0}}{T}\)

= \(\frac{-2816420}{298}\)

= -9451 JK^{-1}

= -9.451 kJ K^{-1}

Ans. ΔS_{surr} = -9.451 kJ K^{-1}

(9) Determine whether the reactions with the following ΔH and ΔS values are spontaneous or non-spontaneous. State whether they are exothermic or endothermic.

(a) ΔH = -110 kJ and ΔS = +40 JK^{-1} at 400 K

(b) ΔH = +50 kJ and ΔS = -130 JK^{-1} at 250 K.

Solution :

(a) Given : ΔH= -110 kJ ΔS = 40 JK^{-1} = 0.04 kJK^{-1}

Temperature = T = 400 K ΔG = ?

Since ΔH is negative, the reaction is exothermic

ΔG = ΔH – TΔS

= -110 – 400 × 0.04

= -110 – 16

= -126 kJ

Since ΔG is negative, the reaction is spontaneous.

(b) Given : ΔH=50 kJ,

ΔS= -130 JK^{-1} = -0.13 kJ K^{-1}

Temperature = T = 250 K

ΔG = ?

Since ΔH is positive, the reaction is endothermic.

ΔG = ΔH – TΔS

= 50 – 250 × (-0.13)

= 50 + 32.5

= 82.5 kJ

Since ΔG > 0, the reaction is non-spontaneous.

Ans. (a) ΔG = -126 kJ; The reaction is exothermic and spontaneous.

(b) ΔG = 82.5 kJ; The reaction is endothermic and non-spontaneous.

(10) For a certain reaction, ΔH^{0} = -224 kJ and ΔS^{0} = -153 JK^{-1}. At what temperature will it change from spontaneous to non-spontaneous ?

Solution :

Given : ΔH^{0} = – 224 kJ = – 224000 J

ΔS^{0} = – 153 JK^{-1}

Temperature (T) at which, reaction changes from spontaneous to non-spontaneous = ?

Find the temperature at equilibrium, where ΔG^{0} = 0

ΔG^{0} = ΔH^{0} – TΔS^{0}

0 = ΔH^{0} – TΔS^{0}

∴ TΔS^{0} = ΔH^{0}

∴ T = \(\frac{\Delta H^{0}}{\Delta S^{0}}\)

= \(\frac{224000}{153}\)

= 1464 K.

Hence reaction will be spontaneous below 1464 K. It will be at equilibrium at 1464 K and non-spontaneous above 1464 K.

Ans. Change over temperature from spontaneous to non-spontaneous = 1464 K.

(11) Determine whether the reactions with the following ΔH and ΔS values are spontaneous or non-spontaneous. State whether the reactions are exothermic or endothermic.

(a) ΔH = -110 kJ, ΔS = +40 JK^{-1} at 400 K

(b) ΔH = + 40 kJ, ΔS = – 120 JK^{-1} at 250 K

Solution :

(a) Given : ΔH = -110 kJ, ΔS = +40 JK^{-1} at T = 400K

ΔG = ΔH – TΔS

= -110 – 16

= -126 kJ

Since ΔG is negative, the reaction is spontaneous.

Since ΔH is negative, the reaction is exothermic.

(b) Given : ΔH = + 40 kJ, ΔS = -120 JK^{-1} at T = 250 K

ΔG = ΔH – TΔS

= 40 + 30

= 70 kJ

Since ΔG is negative, the reaction is spontaneous.

Since ΔH is negative, the reaction is exothermic.

(12) Determine whether the following reaction will be spontaneous or non-spontaneous under standard conditions.

Zn_{(s)} + Cu^{2+} → Zn^{2+} +Cu_{(s)} ΔH^{0 }= -219 kJ, ΔS^{0 }= -21 JK^{-1}

Solution :

Given : ΔH^{0} = -219 kJ;

ΔS^{0} = -21 JK^{-1}= 0.021 kJ K^{-1}

ΔG^{0} = ?

For standard conditions : Pressure = 1 atm

Temperature = T = 298 K

ΔG^{0} = ΔH^{0} – TΔS^{0}

= -219 – 298 × (-0.021)

= -219 + 6.258

= -212.742 kJ

Since ΔG < 0, the reaction is spontaneous.

Ans. The reaction is spontaneous.

(13) The equilibrium constant for a gaseous reversible reaction at 200 °C is 1.64 × 10^{3} atm^{2}. Calculate ΔG° for the reaction.

Solution :

Given : Equilibrium constant = KP = 1.64 × 10^{3} atm^{2}

Temperature = T = 273 + 200 = 473 K

ΔG^{0} = ?

ΔG^{0} = -2.303 RTlog_{10} K_{p}

= – 2.303 × 8.314 × 473 × log_{10} 1.64 × 10^{3}

= – 2.303 × 8.314 × 473 × (3.2148)

= -29115 J

= -29.115 kJ

Ans. ΔG^{0} = -29.115 kJ

(14) Calculate ΔG for the reaction at 25°C

CO_{(g)} + 2H_{2}_{(g)} ⇌ CH_{3}OH_{(g)}

ΔG^{0} = -24.8 kJ mol^{-1}.

if P_{co} = 4 atm, P_{H2} = 2 atm, P_{CH3OH} = 2 atm.

Solution :

Given : Partial pressures : pco = 4 atm,

P_{H2} = 2 atm,

P_{CH3OH} = 2 atm

Temperature = T = 273 + 25 = 298 K

ΔG^{0 }= -24.8 kJ mol^{-1}

CO_{(g)} + 2H_{2}_{(g)} ⇌ CH_{3}OH_{(g)}

The reaction quotient, Q is,

= – 24.8 + 2.303 × 8.314 × 298 × log_{10} 0.125

= – 24.8 + 2.303 × 8.314 × 298 × (\(\overline{1} \cdot 09691\))

= – 24.8 + 2.303 × 8.314 × 298 × (- 0.90709)

= – 24.8 – 2.303 × 8.314 × 298 × 0.90709 × 10^{-3}

= -24.8 – 5.176

= -29.976 kJ mol^{-1}.

Ans. ΔG = – 29.976 kJ mol^{-1}

(15) Calculate K_{P} for the reaction,

C_{2}H_{4(g)} + H_{2(g)} ⇌ C_{2}H_{6(g)},

ΔG^{0} = -100 kJ mol^{-1}, at 25°C.

Solution :

Given : ΔG^{0} = – 100 kJ mol^{-1} = – 100 × 10^{3} J mol^{-1}

= -1 × 10^{5} Jmol^{-1}

Temperature = T = 273 + 25 = 298 K

Equilibrium constant = K_{P} = ?

(16) K_{P} for the reaction,

MgCO_{3(s)} → MgO_{(s)} + CO_{2(g)} is 9 × 10^{-10}.

Calculate ΔG^{0} for the reaction at 25 °C.

Solution :

Given : K_{P} = 9 Δ 10^{-10}

Temperature = T = 273 + 25 = 298 K

ΔG^{0} = ?

ΔG^{0} = -2.303 RTlog_{10}K_{P}

= -2.303 × 8.314 × 298 × log109 × 10^{-10}

= -2.303 × 8.314 × 298 × \([\overline{10} \cdot 9542]\)

= – 2.303 × 8.314 × 298 × [ – 9.0458]

= 51683 Jmol^{-1}

= 51.683 kJmol^{-1}

Ans. ΔG^{0} = 51.653 kJ mol^{-1}

(17) Calculate ΔH, ΔS and ΔG for melting of 10 g ice at 0 °C and 1 atm. (Δ_{fus}H^{0} = 6.02 kJ mol^{-1} for ice)

Solution :

Given : Δ_{fus}H^{0} = 6.02 kJ mol^{-1} = 6.02 × 10^{3} Jmol^{-1}

Temperature = T = 273 + 0 = 273 K

Mass of ice = 10 g

Molar mass of H_{2}O = 18 g mol^{-1}

ΔH= ?, ΔS = ?, ΔG = ?

For melting of ice,

H_{2}O_{(s)} ⇌ H_{2}O_{(l)}

For 1 mol ice = 18 g ice Δ_{fusion}H = 6.05 kJ

∴ For 10 g ice

ΔH = \(\frac{6.02 \times 10}{18}\)

= 3.344 kJ

ΔH = 3.344 kJ = 3.344 × 10^{3} J

∴ ΔS = \(\frac{\Delta H}{T}=\frac{3.344 \times 10^{3}}{273}\) = 12.25 JK^{-1}

ΔG = ΔH – TΔS

= 3.344 – 273 × 12.25 ×10^{-3} kJ

= 3.344 – 3.344

= 0

Since ΔG = 0, the system is at equilibrium.

Ans. ΔH = 3.344 kJ; ΔS = 12.25 JK^{-1}; ΔG = 0

(18) Calculate Kp, ΔG^{0} for the reaction,

C_{(s)} + H_{2}O_{(g)} ⇌ CO_{(g)} + H_{2}_{(g)}

at 990 K if the equilibrium concentrations are as follows :

[H_{2}O] = 1.10 mol dm-^{-3},

[CO] = [H_{2}] = 0.2 mol dm^{-3},

R = 0.08206 L atm K^{-1} mol^{-1}.

Solution :

Given : [H_{2}O] = 1.1 mol dm^{-3},

[CO] = 0.2 mol dm^{-3},

[H_{2}] = 0.2 mol dm^{-3}, T = 990 K,

R = 0.08206 L atm K^{-1} mol^{-1}

K_{P} = ? ΔG^{0} = ?

C_{(s)} + H_{2}O_{(g)} ⇌ CO_{(g)} + H_{2(g)}

K_{P} = K_{C} × (RT)^{Δn}

= 0.03636 × (0.08206 × 990)

= 2.954 atm

ΔG^{0 }= -2.303 RTlog_{10}K_{P}

= -2.303 × 8.314 × 990 × log_{10} 2.954

= -2.303 × 8.314 × 990 × 0.4704

= -8917 J

= -8.917 kJ

Ans. K_{P} = 2.954 atm; ΔG^{0} = -8.917 kJ

Multiple Choice Questions

Question 87.

Select and write the most appropriate answer from the given alternatives for each subquestion :

1. For an isochoric process, the change in

(a) pressure is zero

(b) volume is negative

(c) volume is zero

(d) temperature is zero

Answer:

(c) volume is zero

2. Which of the following is an extensive property ?

(a) Surface tension

(b) Refractive index

(c) Energy

(d) Temperature

Answer:

(c) Energy

3. Which of the following is an intensive property ?

(a) Enthalpy

(b) Weight

(c) Refractive index

(d) Volume

Answer:

(c) Refractive index

4. Which of the following pairs is an intensive property ?

(a) Density, viscosity

(b) Surface tension, mass

(c) Viscosity, internal energy

(d) Heat capacity, volume

Answer:

(a) Density, viscosity

5. The property which is not intensive is

(a) freezing point

(b) viscosity

(c) temperature

(d) free energy

Answer:

(d) free energy

6. Which of the following is not an extensive property ?

(a) molarity

(b) molar heat capacity

(c) mass

(d) volume

Answer:

(b) molar heat capacity

7. Which of the following is NOT a state function ?

(a) Work

(b) Enthalpy

(c) Temperature

(d) Pressure

Answer:

(a) Work

8. In an adiabatic process

(a) ΔT ≠ 0

(b) ΔU ≠ 0

(c) Q = 0

(d) All of these

Answer:

(d) All of these

9. For an isothermal and reversible process

(a) P_{1}V_{1} = P_{2}V_{2}

(b) P_{1}V_{1} ≠ P_{2}V_{2}

(c) ΔV ≠ 0

(d) ΔH ≠ 0

Answer:

(a) P_{1}V_{1} = P_{2}V_{2}

10. For the process to occur under adiabatic conditions, the correct condition is :

(a) ΔT = 0

(b) Δp = 0

(c) Q = 0

(d) W = 0

Answer:

(c) Q = 0

11. What is true for an adiabatic process ?

(a) ΔT = 0

(b) ΔU

(c) ΔH = ΔU

(d) Q = 0

Answer:

(d) Q = 0

12. ΔU = 0 is true for

(a) Adiabatic process

(b) Isothermal process

(c) Isobaric process

(d) Isochoric process

Answer:

(b) Isothermal process

13. When a gas expands in vacuum, the work done by the gas is

(a) maximum

(b) zero

(c) less than zero

(d) greater than zero

Answer:

(b) zero

14. When a sample of an ideal gas is allowed to expand at constant temperature against an atomospheric pressure,

(a) surroundings does work on the system

(b) ΔU = 0

(c) no heat exchange takes place between the system and surroundings

(d) internal energy of the system increases

Answer:

(b) ΔU = 0

15. In what reaction of the following work is done by the system on the surroundings ?

(a) Hg_{(l)} → Hg_{(g)}

(b) 3O_{2(g)} → 2O_{3(g)}

(c) H_{2(g)} + Cl_{2(g)} → 2HCl_{(g)}

(d) N_{2(g)} + 3H_{2(g)} → 2NH_{3(g)}

Answer:

(a) Hg_{(l)} → Hg_{(g)}

16. A gas does 0.320 kJ of work on its surroundings and absorbs 120 J of heat from the surroundings. Hence, ΔU is

(a) 440 kJ

(b) 200 J

(c) 120.32 J

(d) -200J

Answer:

(d) -200J

17. For an isothermal and reversible expansion of 0.5 mol of an ideal gas, Wmax is -3.918 kJ. The value of ΔU is

(a) 3.918 kJ

(b) zero

(c) 1.959 kJ

(d) 3918 J

Answer:

(b) zero

18. The mathematical expression of the first law of thermodynamics for an adiabatic process is

(a) W = Q

(b) W = -ΔU

(c) W = +ΔU

(d ) W = -Q

Answer:

(c) W = +ΔU

19. A gaseous system absorbs 600 kJ of heat and performs the work of expansion equal to 130 kJ. The internal energy change is

(a) 730 kJ

(b) -470 kJ

(c) -730 kJ

(d) 470 kJ

Answer:

(d) 470 kJ

20. When a gas is compressed, the work obtained is 360 J while the heat transferred is 190 J. Hence the change in internal energy is

(a) -170 J

(b) 170 J

(c) 550 J

(d) -550 J

Answer:

(b) 170 J

21. For the reaction N_{2(g)} + 3H_{2(g)} = 2NH_{3(g)}; Which of the following is valid ?

(a) ΔH = ΔU

(b) ΔH < ΔU

(c) ΔH > ΔU

(d) ΔH = 2ΔH

Answer:

(b) ΔH < ΔU

22. For which reaction ΔH = ΔU ?

Answer:

(b) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6(\mathrm{~s})}+6 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 6 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)

23. For the following reaction at 298 K

H_{2(g)} + \(\frac {1}{8}\)O_{2(g)} = H_{2}O_{(l)}

Which of the following alternative is correct ?

(a) ΔH = ΔU

(b) ΔH > ΔU

(c) ΔH < ΔU

(d) ΔH = 1.5 ΔU

Answer:

(c) ΔH < ΔU

24. The heat of combustion of carbon is 394 kJ mol^{-1}. The heat evolved in combustion of 6.023 × 10^{21} atoms of carbon is

(a) 3940 kJ

(b) 3940.0 kJ

(c) 3.94 kJ

(d) 0.394 kJ

Answer:

(c) 3.94 kJ

25. Which of the reactions defines the heat of formation ?

Answer:

(d) \(\frac{1}{2} \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{HCl}_{(\mathrm{g})}\)

26. ΔU^{o} of combustion of methane is -X kJ mol^{-1}. The value of ΔH^{0} is

(a) = ΔU^{o}

(b) > ΔU^{o}

(c) < ΔU^{o}

(d) =0

Answer:

(c) < ΔU^{o}

27. The enthalpy of combustion of 5(rhomic)is -297.4 kJ mol^{-1}. The amount of sulphur required to produce 29.74 kJ of heat is

(a) 32 × 10^{-2} kg

(b) 3.2 × 10^{-3} kg

(c) 3.2 × 10^{-2} kg

(d) 6.4 × 10^{-3} kg

Answer:

(b) 3.2 × 10^{-3} kg

28. The heat of formation of SO_{2(g)} and SO_{3(g)} are -269 kJ mol^{-1} and -395 kJ mol^{-1} respectively the value of ΔH for the reaction

SO_{2(g)} + \(\frac {1}{2}\)O_{2(g)} → SO_{3(g)} is

(a) -664 kJ mol^{-1}

(b) -126 kJ mol^{-1}

(c) 63 kJ mol^{-1}

(d) 126 kJ mol^{-1}

Answer:

(b) -126 kJ mol^{-1}

29. 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

(a) – 74.8 kJ mol^{-1}

(b) – 52.27 kJ mol^{-1}

(c) + 74.8 kJ mol^{-1}

(d) + 52.26 kJ mol^{-1}

Answer:

(a) – 74.8 kJ mol^{-1}

30. The enthalpies of formation of N_{2}O_{(g)} and NO_{(g)} are 82 kJ mol^{-1} and 90 kJ mol^{-1} respectively. Then enthalpy of a reaction 2N_{2}O_{(g)} + O_{2(g)} → 4NO_{(g)} is …………

(a) 8 kJ

(b) -16 kJ

(c) 88 kJ

(d) 196 kJ

Answer:

(d) 196 kJ

31. The heat of combustion of naphthalene (C_{10}H_{8}) to CO_{2} gas and water vapour at 298 K and at constant pressure is -5.1567 × 10^{6} J. The heat of combustion at constant volume at 298 K is (R = 8.314 JK^{-1} mol^{-1})

(a) -5.1567 × 10^{6} J

(b) -5.6161 × 10^{6} J

(c) -5.1616 × 10^{6} J

(d) -5.7161 × 10^{6} J

Answer:

(c) -5.1616 × 10^{6} J

32. Given the reaction,

2NH_{3(g)} → N_{2(g)} + 3H_{2(g)} ΔH = 92.6 kJ

The enthalpy of formation of NH_{3} is

(a) -92.6 kJ

(b) 92.6 kJ mol^{-1}

(c) -46.3 kJmol^{-1}

(d) -185.2 kJmol^{-1}

Answer:

(c) -46.3 kJmol^{-1}

33. Calculate the heat of reaction at 298 K for the reaction C_{2}H_{4(g)} + H_{2(g)} = C_{2}H_{6(g)}

Given that the heats of combustion of ethylene, hydrogen and ethane are 337.0, 68.4 and 373.0 kcal respectively.

(a) 23.4 kcal

(b) 62.2 kcal

(c) 32.4 kcal

(d) 34.2 kcal

Answer:

(c) 32.4 kcal

34. Entropy change for a process is given by,

(a) Q_{rev} × T

(b) Q_{rev}/T

(c) \(\frac{T}{Q_{\text {rev }}}\)

(d) ΔH_{rev} × T

Answer:

(b) Q_{rev}/T

35. For a spontaneous process, total entropy change for a system and its surroundings is

(a) ΔS_{total} < 0

(b) ΔS_{total} = 0

(c) ΔS_{total} > 0

(d) ΔS_{total} ≤ 0

Answer:

(c) ΔS_{total} > 0

36. For a system at equilibrium,

(a) ΔS_{total} = 0

(b) ΔS_{total} > 0

(c) ΔS_{total} < 0

(d) ΔS_{total} ≥ 0

Answer:

(a) ΔS_{total} = 0

37. If the enthalpy of vaporisation of water at 100°C is 186.5 J·mol^{-1}, the entropy of vaporization will be-

(a) 4.0 J·K^{-1} mol^{-1}

(b) 3.0 J·K^{-1} mol^{-1}

(c) 1.5 J·K^{-1} mol^{-1}

(d) 0.5 J·K^{-1} mol^{-1}

Answer:

(d) 0.5 J·K^{-1} mol^{-1}

38. Heat of fusion of ice is 6.02kJmol-1 at 0 °C. If 100 g water is frozen at 0 °C, entropy change will be

(a) -0.1225 JK^{-1}

(b) 310.6 JK^{-1}

(c) -122.6 JK^{-1}

(d) 92.8 JK^{-1}

Answer:

(c) -122.6 JK^{-1}

39. If for a reaction ΔH is negative and ΔS is positive then the reaction is

(a) spontaneous at all temperatures

(b) non-spontaneous at all temperatures

(c) spontaneous only at high temperatures

(d) spontaneous only at low temperature

Answer:

(a) spontaneous at all temperatures

40. The relationship between ΔG^{o} of a reaction and its equilibrium constant is

Answer:

(c) \(\frac{R T \ln K}{\Delta G^{0}}=-1\)

41. Which of the following has highest entropy?

(a) Al_{(s)}

(b) CaCO_{3(s)}

(c) H_{2}O_{(l)}

(d) CO_{2(g)}

Answer:

(d) CO_{2(g)}

42. The entropy change for the formation of 3.5 mol NO_{(g)} from the following data will be,

Answer:

(b) 42.875 JK^{-1}

43. Gibbs free energy change at equilibrium is

(a) ΔG = 0

(b) ΔG > 0

(c) ΔG < 0

(d) ΔG ≤ 0

Answer:

(a) ΔG = 0

44. For spontaneous process,

(a) ΔG = 0

(b) ΔG > 0

(c) ΔG < 0

(d) ΔG ≤ 0

Answer:

(c) ΔG < 0

45. A substance which shows highest entropy is

(a) SrCO_{3(S)}

(b) Cu_{(S)}

(c) NaC_{(aq)}

(d) Cl_{2(g)}

Answer:

(d) Cl_{2(g)}

46. For which of the following reactions ΔS is negative ?

Answer:

(a) \(\mathrm{Mg}_{(s)}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{MgCl}_{2(s)}\)

47. For a reaction, at 300K enthalpy is 138 kJ and entropy change is 115 JK^{-1}. Hence the free energy change of the reaction is

(a) 130.5 kJ

(b) 103.5 kJ

(c) 82.8 kJ

(d) – 60.5 kJ

Answer:

(b) 103.5 kJ

48. Bond enthalpies of H_{2-1}, I_{2}_{(g)} and HI are 436, 151 and 298 kJ mol^{-1} respectively. Hence enthalpy of formation of HI_{(g)} is

(a) -9 kJmol^{-1}

(b) -4.5kJmol^{-1}

(c) 4.5 kJ mol^{-1}

(d) 9 kJ mol^{-1}

Answer:

(b) -4.5kJmol^{-1}

49. The average bond energy of C-H bond is 410 kJmol^{-1}. The enthalpy change of atomisation of 3.2 g CH_{4}_{(g)} is

(a) 1312 kJ

(b) 29.8 kJ

(c) 328 kJ

(d) 120 kJ

Answer:

(c) 328 kJ

50. For a chemical reaction ΔS = -0.035 kJ/K and ΔH = -20 kJ. At what temperature does the reaction turn non-spontaneous ?

(a) 5.14 K

(b) 57.14 K

(c) 571.4 K

(d) 5714.0 K

Answer:

(c) 571.4 K

51. For a certain reaction, ΔH = -50 kJ and ΔS = -80 JK^{-1}, at what temperature does the reaction turn from spontaneous to non-spontaneous.

(a) 6.25 K

(b) 62.5 K

(c) 625 K

(d) 6250 K

Answer:

(c) 625 K