AglaSem Class Notes

AglaSem NCERT Solutions

Chemical energy stored by molecules can be released as heat during chemical reactions when a fuel like methane, cooking gas or coal burns in air. The chemical energy may also be used to do mechanical work when a fuel burns in an engine or to provide electrical energy through a galvanic cell like dry cell. Thus, various forms of energy are interrelated and under certain conditions, these may be transformed from one form into another. The study of these energy transformations forms the subject matter of thermodynamics. The laws of thermodynamics deal with energy changes of macroscopic systems involving a large number of molecules rather than microscopic systems containing a few molecules. Thermodynamics is not concerned about how and at what rate these energy transformations are carried out, but is based on initial and final states of a system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state. Macroscopic properties like pressure and temperature do not change with time for a system in equilibrium state. In this unit, we would like to answer some of the important questions through thermodynamics, like:

How do we determine the energy changes involved in a chemical reaction/process? Will it occur or not?

What drives a chemical reaction/process?

To what extent do the chemical reactions proceed?


We are interested in chemical reactions and the energy changes accompanying them. For this we need to know certain thermodynamic terms. These are discussed below.

6.1.1 The System and the Surroundings

A system in thermodynamics refers to that part of universe in which observations are made and remaining universe constitutes the surroundings. The surroundings include everything other than the system. System and the surroundings together constitute the universe .
The universe = The system + The surroundings

However, the entire universe other than the system is not affected by the changes taking place in the system. Therefore, for all practical purposes, the surroundings are that portion of the remaining universe which can interact with the system. Usually, the region of space in the neighbourhood of the system constitutes its surroundings.

For example, if we are studying the reaction between two substances A and B kept in a beaker, the beaker containing the reaction mixture is the system and the room where the beaker is kept is the surroundings (Fig. 6.1).




Note that the system may be defined by physical boundaries, like beaker or test tube, or the system may simply be defined by a set of Cartesian coordinates specifying a particular volume in space. It is necessary to think of the system as separated from the surroundings by some sort of wall which may be real or imaginary. The wall that separates the system from the surroundings is called boundary. This is designed to allow us to control and keep track of all movements of matter and energy in or out of the system.

6.1.2 Types of the System

We, further classify the systems according to the movements of matter and energy in or out of the system.

1. Open System

In an open system, there is exchange of energy and matter between system and surroundings [Fig. 6.2 (a)]. The presence of reactants in an open beaker is an example of an open system*. Here the boundary is an imaginary surface enclosing the beaker and reactants.


2. Closed System

In a closed system, there is no exchange of matter, but exchange of energy is possible between system and the surroundings [Fig. 6.2 (b)]. The presence of reactants in a closed vessel made of conducting material e.g., copper or steel is an example of a closed system.



3. Isolated System

In an isolated system, there is no exchange of energy or matter between the system and the surroundings [Fig. 6.2 (c)]. The presence of reactants in a thermos flask or any other closed insulated vessel is an example of an isolated system.


* We could have chosen only the reactants as system then walls of the beakers will act as boundary.

6.1.3 The State of the System

The system must be described in order to make any useful calculations by specifying quantitatively each of the properties such as its pressure (p), volume (V), and temperature (T ) as well as the composition of the system. We need to describe the system by specifying it before and after the change. You would recall from your Physics course that the state of a system in mechanics is completely specified at a given instant of time, by the position and velocity of each mass point of the system. In thermodynamics, a different and much simpler concept of the state of a system is introduced. It does not need detailed knowledge of motion of each particle because, we deal with average measurable properties of the system. We specify the state of the system by state functions or state variables.

The state of a thermodynamic system is described by its measurable or macroscopic (bulk) properties. We can describe the state of a gas by quoting its pressure (p), volume (V), temperature (T ), amount (n) etc. Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached. In order to completely define the state of a system it is not necessary to define all the properties of the system; as only a certain number of properties can be varied independently. This number depends on the nature of the system. Once these minimum number of macroscopic properties are fixed, others automatically have definite values. The state of the surroundings can never be completely specified; fortunately it is not necessary to do so.

6.1.4 The Internal Energy as a State Function

When we talk about our chemical system losing or gaining energy, we need to introduce a quantity which represents the total energy of the system. It may be chemical, electrical, mechanical or any other type of energy you may think of, the sum of all these is the energy of the system. In thermodynamics, we call it the internal energy, U of the system, which may change, when

• heat passes into or out of the system,
• work is done on or by the system,
• matter enters or leaves the system.

These systems are classified accordingly as you have already studied in section 6.1.2.

(a) Work

Let us first examine a change in internal energy by doing work. We take a system containing some quantity of water in a thermos flask or in an insulated beaker. This would not allow exchange of heat between the system and surroundings through its boundary and we call this type of system as adiabatic. The manner in which the state of such a system may be changed will be called adiabatic process. Adiabatic process is a process in which there is no transfer of heat between the system and surroundings. Here, the wall separating the system and the surroundings is called the adiabatic wall (Fig 6.3).

Let us bring the change in the internal energy of the system by doing some work on it. Let us call the initial state of the system as state A and its temperature as TA. Let the internal energy of the system in state A be called UA. We can change the state of the system in two different ways.

One way: We do some mechanical work, say 1 kJ, by rotating a set of small paddles and thereby churning water. Let the new state be called B state and its temperature, as TB. It is found that TB > TA and the change in temperature, ΔT = TB-TA. Let the internal energy of the system in state B be UB and the change in internal energy, ΔU =UB– UA.

Second way: We now do an equal amount (i.e., 1kJ) electrical work with the help of an immersion rod and note down the temperature change. We find that the change in temperature is same as in the earlier case, say, TB – TA.

In fact, the experiments in the above manner were done by J. P. Joule between 1840–50 and he was able to show that a given amount of work done on the system, no matter how it was done (irrespective of path) produced the same change of state, as measured by the change in the temperature of the system.

So, it seems appropriate to define a quantity, the internal energy U, whose value is characteristic of the state of a system, whereby the adiabatic work, wad required to bring about a change of state is equal to the difference between the value of U in one state and that in another state, ΔU i.e.,

ΔU=U2−U1 =wad

Therefore, internal energy, U, of the system is a state function.

The positive sign expresses that wad is positive when work is done on the system. Similarly, if the work is done by the system,wad will be negative.

Can you name some other familiar state functions? Some of other familiar state functions are V, p, and T. For example, if we bring a change in temperature of the system from 25°C to 35°C, the change in temperature is 35°C-25°C = +10°C, whether we go straight up to 35°C or we cool the system for a few degrees, then take the system to the final temperature. Thus, T is a state function and the change in temperature is independent of the route taken. Volume of water in a pond, for example, is a state function, because change in volume of its water is independent of the route by which water is filled in the pond, either by rain or by tubewell or by both,

(b) Heat

We can also change the internal energy of a system by transfer of heat from the surroundings to the system or vice-versa without expenditure of work. This exchange of energy, which is a result of temperature difference is called heat, q. Let us consider bringing about the same change in temperature (the same initial and final states as before in section 6.1.4 (a) by transfer of heat through
thermally conducting walls instead of adiabatic walls (Fig. 6.4).



We take water at temperature, TA in a container having thermally conducting walls, say made up of copper and enclose it in a huge heat reservoir at temperature, TB. The heat absorbed by the system (water), q can be measured in terms of temperature difference , TB – TA. In this case change in internal energy, ΔU= q, when no work is done at constant volume.

The q is positive, when heat is transferred from the surroundings to the system and q is negative when heat is transferred from system to the surroundings.

(c) The general case

Let us consider the general case in which a change of state is brought about both by doing work and by transfer of heat. We write change in internal energy for this case as:

ΔU = q + w ————————————————(6.1)

For a given change in state, q and w can vary depending on how the change is carried out. However, q +w = ΔU will depend only on initial and final state. It will be independent of the way the change is carried out. If there is no transfer of energy as heat or as work (isolated system) i.e., if w = 0 and q = 0, then Δ U = 0.

The equation 6.1 i.e., ΔU = q + w is mathematical statement of the first law of thermodynamics, which states that

The energy of an isolated system is constant.

It is commonly stated as the law of conservation of energy i.e., energy can neither be created nor be destroyed.

Note: There is considerable difference between the character of the thermodynamic property energy and that of a mechanical property such as volume. We can specify an unambiguous (absolute) value for volume of a system in a particular state, but not the absolute value of the internal energy. However, we can measure only the changes in the internal energy, ΔU of the system.

Problem 6.1

Express the change in internal energy of a system when

(i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. What type of wall does the system have ?

(ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have?

(iii) w amount of work is done by the system and q amount of heat is supplied to the system. What type of system would it be?


(i) Δ U = wad, wall is adiabatic
(ii) Δ U = – q, thermally conducting walls
(iii) Δ U = q – w, closed system.


Many chemical reactions involve the generation of gases capable of doing mechanical work or the generation of heat. It is important for us to quantify these changes and relate them to the changes in the internal energy. Let us see how!

6.2.1 Work

First of all, let us concentrate on the nature of work a system can do. We will consider only mechanical work i.e., pressure-volume work.

For understanding pressure-volume work, let us consider a cylinder which contains one mole of an ideal gas fitted with a frictionless piston. Total volume of the gas is Vi and pressure of the gas inside is p. If external pressure is pex which is greater than p, piston is moved inward till the pressure inside becomes equal to pex. Let this change be achieved in a single step and the final volume be Vf . During this compression, suppose piston moves a distance, l and is cross-sectional area of the piston is A [Fig. 6.5(a)].


then, volume change = l x A = ΔV = (Vf – Vi )

We also know, pressure = force/area

Therefore, force on the piston = pex . A

If w is the work done on the system by movement of the piston then

w = force × distance = pex . A .l = pex . (-ΔV) = – pex ΔV = – pex (Vf – Vi ) ————————————-(6.2)

The negative sign of this expression is required to obtain conventional sign for w, which will be positive. It indicates that in case of compression work is done on the system. Here (Vf – Vi ) will be negative and negative multiplied by negative will be positive. Hence the sign obtained for the work will be positive.

If the pressure is not constant at every stage of compression, but changes in number of finite steps, work done on the gas will be summed over all the steps and will be equal to −ΣpΔV [Fig. 6.5 (b)]


If the pressure is not constant but changes during the process such that it is always infinitesimally greater than the pressure of the gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, dV. In such a case we can calculate the work done on the gas by the relation

——————————————-( 6.3)

Here, pex at each stage is equal to (pin + dp) in case of compression [Fig. 6.5(c)]. In an expansion process under similar conditions, the external pressure is always less than the pressure of the system i.e., pex = (pin-dp). In general case we can write, pex = (pin + dp). Such processes are called reversible processes.



A process or change is said to be reversible, if a change is brought out in such a way that the process could, at any moment, be reversed by an infinitesimal change. A reversible process proceeds infinitely slowly by a series of equilibrium states such that system and the surroundings are always in near equilibrium with each other. Processes other than reversible processes are known as irreversible processes.

In chemistry, we face problems that can be solved if we relate the work term to the internal pressure of the system. We can relate work to internal pressure of the system under reversible conditions by writing equation 6.3 as follows:

Since dp dV is very small we can write


Now, the pressure of the gas (pin which we can write as p now) can be expressed in terms of its volume through gas equation. For n mol of an ideal gas i.e., pV =nRT

⇒ p = nRT/V
Therefore, at constant temperature (isothermal process),

= – 2.303 nRT log Vf/Vi —————————————————(6.5)

Free expansion: Expansion of a gas in vacuum (pex = 0) is called free expansion. No work is done during free expansion of an ideal gas whether the process is reversible or irreversible (equation 6.2 and 6.3).

Now, we can write equation 6.1 in number of ways depending on the type of processes.

Let us substitute w = – pexΔV (eq. 6.2) in equation 6.1, and we get

ΔU = q − pexΔV

If a process is carried out at constant volume (ΔV = 0), then

ΔU = qV

the subscript V in qV denotes that heat is supplied at constant volume.

Isothermal and free expansion of an ideal gas

For isothermal (T = constant) expansion of an ideal gas into vacuum ; w = 0 since pex = 0. Also, Joule determined experimentally that q = 0; therefore, ΔU = 0

Equation 6.1, ΔU =q+w can be expressed for isothermal irreversible and reversible changes as follows:

1. For isothermal irreversible change q = -w = pex (V– Vi )

2. For isothermal reversible change q = – w = nRT ln V/Vi = 2.303 nRT log V/Vi

3. For adiabatic change, q = 0, ΔU = wad

Problem 6.2

Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?


We have q = – w = pex (10 – 2) = 0(8) = 0

No work is done; no heat is absorbed.

Problem 6.3

Consider the same expansion, but this time against a constant external pressure of 1 atm.


We have q = – w = pex (8) = 8 litre-atm

Problem 6.4

Consider the same expansion, to a final volume of 10 litres conducted reversibly.


We have q = – w = 2.303 x 10 log(10/2) = 16.1 litre-atm

6.2.2 Enthalpy, H

(a) A useful new state function

We know that the heat absorbed at constant volume is equal to change in the internal energy i.e., ΔU = qV. But most of chemical reactions are carried out not at constant volume, but in flasks or test tubes under constant atmospheric pressure. We need to define another state function which may be suitable under these conditions.

We may write equation (6.1) as ΔU = qp − pΔV at constant pressure, where qp is heat absorbed by the system and -pΔV represent expansion work done by the system.

Let us represent the initial state by subscript 1 and final state by 2

We can rewrite the above equation as
U2-U1 = qp – p (V2 – V1)

On rearranging, we get

qp = (U2 + pV2) – (U1 + pV1) ———————————————(6.6)

Now we can define another thermodynamic function, the enthalpy H [Greek word enthalpien, to warm or heat content] as :

H = U + pV —————————————–(6.7)

so, equation (6.6) becomes

qp= H2 – H1 = ΔH

Although q is a path dependent function, H is a state function because it depends on U, p and V, all of which are state functions. Therefore, ΔH is independent of path. Hence, qp is also independent of path.

For finite changes at constant pressure, we can write equation 6.7 as
ΔH = ΔU + ΔpV

Since p is constant, we can write
ΔH = ΔU + pΔV ——————————————-(6.8)

It is important to note that when heat is absorbed by the system at constant pressure, we are actually measuring changes in the enthalpy.

Remember ΔH = qp, heat absorbed by the system at constant pressure.

ΔH is negative for exothermic reactions which evolve heat during the reaction and ΔH is positive for endothermic reactions which absorb heat from the surroundings.

At constant volume (ΔV = 0), ΔU = qV, therefore equation 6.8 becomes

ΔH = ΔU = qV

The difference between ΔH and ΔU is not usually significant for systems consisting of only solids and / or liquids. Solids and liquids do not suffer any significant volume changes upon heating. The difference, however, becomes significant when gases are involved. Let us consider a reaction involving gases. If VA is the total volume of the gaseous reactants, VB is the total volume of the gaseous products, nA is the number of moles of gaseous reactants and nB is the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we write,

pVA = nART

and pVB = nBRT

Thus, pVB – pVA = nBRT – nART = (nB-nA)RT

or p (VB – VA) = (nB – nA) RT

or p ΔV = ΔngRT ——————————(6.9)

Here, Δng refers to the number of moles of gaseous products minus the number of moles of gaseous reactants.

Substituting the value of pΔV from equation 6.9 in equation 6.8, we get

ΔH = ΔU + ΔngRT —————————-(6.10)

The equation 6.10 is useful for calculating ΔH from ΔU and vice versa.

Problem 6.5

If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and 100°C is 41kJ mol-1. Calculate the internal energy change, when

(i) 1 mol of water is vaporised at 1 bar pressure and 100°C.

(ii) 1 mol of water is converted into ice.


(i) The change H2O(l) → H2O(g)

ΔH = =ΔU + ΔngRT

or ΔU = ΔH – ΔngRT, substituting the values, we get

ΔU = 41.00 kJmol-1 x 8.3 Jmol-1 K-1 x 373 K = 41.00 kJ mol-1−3.096 kJ mol-1 = 37.904 kJ mol-1

(ii) The change H2O(l) → H2O(s)

There is negligible change in volume, So, we can put pΔV = Δng RT ≈ 0 in this case,


so, ΔU =41.00kJ mol-1

(b) Extensive and Intensive Properties

In thermodynamics, a distinction is made between extensive properties and intensive properties. An extensive property is a property whose value depends on the quantity or size of matter present in the system. For example, mass, volume, internal energy, enthalpy, heat capacity, etc. are extensive properties.

Those properties which do not depend on the quantity or size of matter present are known as intensive properties. For example temperature, density, pressure etc. are intensive properties. A molar property, χm, is the value of an extensive property χ of the system for 1 mol of the substance. If n is the amount of matter, χm = χ/n is independent of the amount of matter. Other examples are molar volume, Vm and molar heat capacity, Cm. Let us understand the distinction between extensive and intensive properties by considering a gas enclosed in a container of volume V and at temperature T [Fig. 6.6(a)]. Let us make a partition such that volume is halved, each part [Fig. 6.6 (b)] now has one half of the original volume, V/2, but the temperature will still remain the same i.e., T. It is clear that volume is an extensive property and temperature is an intensive property.




(c) Heat Capacity

In this sub-section, let us see how to measure heat transferred to a system. This heat appears as a rise in temperature of the system in case of heat absorbed by the system.

The increase of temperature is proportional to the heat transferred
The magnitude of the coefficient depends on the size, composition and nature of the system. We can also write it as q = C ΔT

The coefficient, C is called the heat capacity.

Thus, we can measure the heat supplied by monitoring the temperature rise, provided we know the heat capacity.

When C is large, a given amount of heat results in only a small temperature rise. Water has a large heat capacity i.e., a lot of energy is needed to raise its temperature.

C is directly proportional to amount of substance. The molar heat capacity of a substance, Cm = C/n , is the heat capacity for one mole of the substance and is the quantity of heat needed to raise the temperature of one mole by one degree celsius (or one kelvin). Specific heat, also called specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree celsius (or one kelvin). For finding out the heat, q, required to raise the temperatures of a sample, we multiply the specific heat of the substance, c, by the mass m, and temperatures change, ΔT as

q = C× m ×ΔT =CΔT —————————————————(6.11)

(d) The relationship between Cp and CV for an ideal gas

At constant volume, the heat capacity, C is denoted by CV and at constant pressure, this is denoted by Cp . Let us find the relationship between the two.

We can write equation for heat, q

at constant volume as qV = CVΔT =ΔU

at constant pressure as qp = CpΔT =ΔH

The difference between Cp and CV can be derived for an ideal gas as:

For a mole of an ideal gas, ΔH = ΔU + Δ(pV ) = ΔU + Δ(RT ) = ΔU + RΔT

∴ ΔH= ΔU+RΔT ——————————————-(6.12)

On putting the values of ΔH and ΔU, we have

CpΔT = CVΔT + R ΔT —————————————————–(6.13)


We can measure energy changes associated with chemical or physical processes by an experimental technique called calorimetry. In calorimetry, the process is carried out in a vessel called calorimeter, which is immersed in a known volume of a liquid. Knowing the heat capacity of the liquid in which calorimeter is immersed and the heat capacity of calorimeter, it is possible to determine the heat evolved in the process by measuring temperature changes. Measurements are made under two different conditions:

i) at constant volume, qV
ii) at constant pressure, qp

(a) ΔU measurements

For chemical reactions, heat absorbed at constant volume, is measured in a bomb calorimeter (Fig. 6.7). Here, a steel vessel (the bomb) is immersed in a water bath. The whole device is called calorimeter. The steel vessel is immersed in water bath to ensure that no heat is lost to the surroundings. A combustible substance is burnt in pure dioxygen supplied in the steel bomb. Heat evolved during the reaction is transferred to the water around the bomb and its temperature is monitored. Since the bomb calorimeter is sealed, its volume does not change i.e., the energy changes associated with reactions are measured at constant volume. Under these conditions, no work is done as the reaction is carried out at constant volume in the bomb calorimeter. Even for reactions involving gases, there is no work done as ΔV = 0. Temperature change of the calorimeter produced by the completed reaction is then converted to qV, by using the known heat capacity of the calorimeter with the help of equation 6.11.


(b) ΔH measurements

Measurement of heat change at constant pressure (generally under atmospheric pressure) can be done in a calorimeter shown in Fig. 6.8. We know that ΔH = qp (at constant p) and, therefore, heat absorbed or evolved, qp at constant pressure is also called the heat of reaction or enthalpy of reaction, ΔrH.


In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, qp will be negative and ΔrH will also be negative. Similarly in an endothermic reaction, heat is absorbed, qp is positive and ΔrH will be positive.

Problem 6.6

1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation

C (graphite) + O2 (g) → CO2 (g)

During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?


Suppose q is the quantity of heat from the reaction mixture and CV is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter.

q = CV x ΔT

Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.

q = – CV x ΔT = 20.7 kJ/K x (299 – 298)K = – 20.7 kJ

(Here, negative sign indicates the exothermic nature of the reaction) Thus, ΔU for the combustion of the 1g of graphite = – 20.7 kJK-1

For combustion of 1 mol of graphite, = (12.0 gmol-1 x (20.7kJ))/1g
= – 2.48 102 kJ mol-1 , Since Δ ng = 0, Δ H = Δ U = – 2.48 102 kJ mol-1


In a chemical reaction, reactants are converted into products and is represented by,

Reactants → Products

The enthalpy change accompanying a reaction is called the reaction enthalpy. The enthalpy change of a chemical reaction, is given by the symbol ΔrH

ΔrH = (sum of enthalpies of products) – (sum of enthalpies of reactants)

=            —————————————————–(6.14)

(Here symbol Σ (sigma) is used for summation and ai and bi are the stoichiometric coefficients of the products and reactants respectively in the balanced chemical equation. For example, for the reaction
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)

= [Hm (CO2 ,g) + 2Hm (H2O, l)]- [Hm (CH4 , g) + 2Hm (O2, g)]

where Hm is the molar enthalpy.

Enthalpy change is a very useful quantity. Knowledge of this quantity is required when one needs to plan the heating or cooling required to maintain an industrial chemical reaction at constant temperature. It is also required to calculate temperature dependence of equilibrium constant.

(a) Standard enthalpy of reactions

Enthalpy of a reaction depends on the conditions under which a reaction is carried out. It is, therefore, necessary that we must
specify some standard conditions. The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states.

The standard state of a substance at a specified temperature is its pure form at 1 bar. For example, the standard state of liquid ethanol at 298 K is pure liquid ethanol at 1 bar; standard state of solid iron at 500 K is pure iron at 1 bar. Usually data are taken at 298 K.

Standard conditions are denoted by adding the superscriptVto the symbol ΔH, e.g., ΔHΘ

(b) Enthalpy changes during phase transformations

Phase transformations also involve energy changes. Ice, for example, requires heat for melting. Normally this melting takes place at constant pressure (atmospheric pressure) and during phase change, temperature remains constant (at 273 K).

H2O(s) → H2 O( l); ΔfusHΘ = 6.00 kJ mol−1

Here ΔfusHΘ is enthalpy of fusion in standard state. If water freezes, then process is reversed and equal amount of heat is given off to the surroundings.

The enthalpy change that accompanies melting of one mole of a solid substance in standard state is called standard enthalpy of fusion or molar enthalpy of fusion, ΔfusHΘ .

Melting of a solid is endothermic, so all enthalpies of fusion are positive. Water requires heat for evaporation. At constant temperature of its boiling point Tb and at constant pressure:

H2O(l) → H2O(g); ΔvapHΘ = + 40.79kJ mol-1

ΔvapHΘ is the standard enthalpy of vaporization.

Amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1bar) is called its standard enthalpy of vaporization or molar enthalpy of vaporization, ΔvapHΘ .

Sublimation is direct conversion of a solid into its vapour. Solid CO2 or ‘dry ice’ sublimes at 195K with ΔsubHΘ=25.2 kJ mol-1 ; naphthalene sublimes slowly and for this ΔsubHΘ = 73.0 kJ mol-1 .

Standard enthalpy of sublimation, ΔsubHΘ is the change in enthalpy when one mole of a solid substance sublimes at a constant temperature and under standard pressure (1bar).

The magnitude of the enthalpy change depends on the strength of the intermolecular interactions in the substance undergoing the phase transfomations. For example, the strong hydrogen bonds between water molecules hold them tightly in liquid phase. For an organic liquid, such as acetone, the intermolecular dipole-dipole interactions are significantly weaker. Thus, it requires less heat to vaporise 1 mol of acetone than it does to vaporize 1 mol of water. Table 6.1 gives values of standard enthalpy changes of fusion and vaporisation for some substances.


Table 6.1 Standard Enthalpy Changes of Fusion and Vaporisation
N2 63.15 0.72 77.35 5.59
NH3 195.40 5.65 239.73 23.35
HCl 159.0 1.992 188.0 16.15
CO 68.0 6.836 82.0 6.04
CH2COCH3 177.8 5.72 329.4 29.1
CCl4 250.16 2.5 349.69 30.0
H2O 273.15 6.01 373.15 40.79
NaCl 1081.0 28.8 1665.0 170.0
C6H6 278.65 9.83 353.25 30.8

Problem 6.7

A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaporisation at 100°C.

ΔvapHΘ for water at 373K = 40.66 kJ mol-1


We can represent the process of evaporation as evaporisation

No. of moles in 18 g H2O(l) is = 18g/18 gmol-1 = 1mol

ΔvapU =ΔvapHΘ − pΔV = ΔvapHΘ −ΔngRT

(assuming steam behaving as an ideal gas).

ΔvapHΘ – ΔngRT = 40.66 kJ mol-1 -(1)(8.314 JK-1 mol-1)(373k)(10-3KJ J-1)

ΔvapUΘ = 40.66 kJ mol-1 – 3.10 kJ mol-1 = 37.56kJ mol-1

(c) Standard enthalpy of formation

The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called Standard Molar Enthalpy of Formation. Its symbol is ΔfHΘ, where the subscript ‘ f ’ indicates that one mole of the compound in question has been formed in its standard state from its elements in their most stable states of aggregation. The reference state of an element is its most stable state of aggregation at 25°C and 1 bar pressure. For example, the reference state of dihydrogen
is H2 gas and those of dioxygen, carbon and sulphur are O2 gas, Cgraphite and Srhombic respectively. Some reactions with standard molar enthalpies of formation are given below.

H2 (g) + ½O2 (g) → H2 O(1);
ΔfHΘ = -285.8 kJ mol−1

C (graphite, s) + 2H2 (g)→CH4 (g);

ΔfHΘ = − 74.81kJmol−1

2C graphite,s + 3H2(g) + ½O2(g) → C2H5OH(1);
ΔfHΘ = − 277.7kJmol−1

It is important to understand that a standard molar enthalpy of formation, ΔfHΘ, is just a special case of ΔrHΘ, where one mole of a compound is formed from its constituent elements, as in the above three equations, where 1 mol of each, water, methane and ethanol is formed. In contrast, the enthalpy change for an exothermic reaction:

CaO(s)+CO2 (g)→CaCO3 (s);

ΔrHΘ = -178.3kJ mol−1

is not an enthalpy of formation of calcium carbonate, since calcium carbonate has been formed from other compounds, and not from
its constituent elements. Also, for the reaction given below, enthalpy change is not standard enthalpy of formation,
ΔfHΘ for HBr(g).

H2(g) + Br2( l) → 2HBr(g);
ΔrHΘ = -72.8 kJmol−1

Here two moles, instead of one mole of the product is formed from the elements, i.e.,

ΔrHΘ = 2ΔfHΘ.

Therefore, by dividing all coefficients in the balanced equation by 2, expression for enthalpy of formation of HBr (g) is written as

½H2(g) + ½Br2(1) → HBr(g );
ΔfHΘ = -36.4 kJ mol−1

Standard enthalpies of formation of some common substances are given in Table 6.2.

Table 6.2 Standard Molar Enthalpies of Formation (Δf HΘ) at 298K of a Few Selected Substances
Al2O3(s) -1675.7 Hl(g) +26.48
BaCO3(s) -1216.3 KCl(s) -436.75
Br2(l) 0 KBr(s) -393.8
Br2(g) +30.91 MgO(s) -601.70
CaCO3(S) -1206.92 Mg(OH)2(s) -924.54
C (diamond) +1.89 NaF(s) -573.65
C (graphite) 0 NaCl(s) -411.15
CaO(s) -635.09 NaBr(s) -361.06
CH4(g) -74.81 NaI(s) -287.78
C2H4(g) 52.26 NH3(g) -46.11
CH3OH(l) -238.86 NO(g) +90.25
C2H5OH(l) -277.69 NO2(g) +33.18
C6H6(l) +49.0 PCl3(l) -319.70
CO(g) -110.53 PCl5(s) -443.5
CO2(g) -393.51 SiO2(s) (quartz) -910.94
C2H6(g) -84.63 SnCl2(s) -325.1
Cl2(g) 0 SnCl4(l) -511.3
C3H8(g) -103.85 SO2(g) -296.83
n-C4H10(g) -126.15 SO3(g) -395.72
HgS(s) red -58.2 SiH4(g) +34
H2(g) 0 SiCl4(g) -657.0
H2O(g) -241.82 C(g) +716.68
H2O(l) -285.83 H(g) +217.97
HF(g) -271.1 Cl(g) +121.68
HCl(g) -92.31 Fe2O3(s) -824.2
HBr(g) -36.40

By convention, standard enthalpy for formation, ΔfHΘ, of an element in reference state, i.e., its most stable state of aggregation is taken as zero.

Suppose, you are a chemical engineer and want to know how much heat is required to decompose calcium carbonate to lime and carbon dioxide, with all the substances in their standard state.

CaCO3(s) → CaO(s) + CO2 (g); ΔrHΘ = ?

Here, we can make use of standard enthalpy of formation and calculate the enthalpy change for the reaction. The following general equation can be used for the enthalpy change calculation.

ΔrHΘ = aiΔfHΘ(products) + biΔfHΘ(reactants) —————————————————(6.15)

where a and b represent the coefficients of the products and reactants in the balanced equation. Let us apply the above equation for decomposition of calcium carbonate. Here, coefficients ‘a’ and ‘b’ are 1 each.


ΔrHΘ = ΔfHΘ[CaO(s)] + ΔfHΘ[CO2(g)] – ΔfHΘ[CaCO3(s)] = 1( -635.1 kJ mol-1) – 1( -393.5 kJ mol-1 ) -1( -1206.9 kJ mol-1 ) = 178.3 kJ mol-1

Thus, the decomposition of CaCO3 (s) is an endothermic process and you have to heat it for getting the desired products.

(d) Thermochemical equations

A balanced chemical equation together with the value of its ΔrH is called a thermochemical equation. We specify the physical state (alongwith allotropic state) of the substance in an equation. For example:

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O( l):
ΔrHΘ = -1367 kJ mol-1

The abov e equation describes the combustion of liquid ethanol at constant temperature and pressure. The negative sign of enthalpy change indicates that this is an exothermic reaction.

It would be necessary to remember the following conventions regarding thermochemical equations.

1. The coefficients in a balanced thermochemical equation refer to the number of moles (never molecules) of reactants and products involved in the reaction.

2. The numerical value of ΔrHΘ refers to the number of moles of substances specified by an equation. Standard enthalpy change ΔrHΘ will have units as kJ mol-1.

To illustrate the concept, let us consider the calculation of heat of reaction for the following reaction :

Fe2O3 s + 3H2 g→2Fe(s) + 3H2O(l) ,

From the Table (6.2) of standard enthalpy of formation (ΔfHΘ), we find :

ΔfHΘ(H2O, l) = -285.83 kJ mol-1;
ΔfHΘ(Fe2O3, s)= – 824.2 kJ mol-1;

Also ΔfHΘ (Fe2O3, s) = 0 and ΔfHΘ(H2 ,g ) = 0 as per convention


ΔrH1Θ= 3(-285.83 kJ mol-1) – 1(- 824.2 kJ mol-1) = (-857.5 + 824.2) kJ mol-1 = -33.3 kJ mol-1

Note that the coefficients used in these calculations are pure numbers, which are equal to the respective stoichiometric coefficients. The unit for ΔrHΘ is kJ mol-1, which means per mole of reaction. Once we balance the chemical equation in a particular way, as above, this defines the mole of reaction. If we had balanced the equation differently, for example,

(1/2)Fe2O3(s) + (3/2)H2(g)→Fe(s) + (3/2)H2O(1)

then this amount of reaction would be one mole of reaction and ΔrHΘ would be

ΔrH2Θ = (3/2)( -285.83 kJmol-1) – (1/2)(-824.2 kJmol-1) = (- 428.7 + 412.1) kJ mol-1 = -16.6 kJ mol-1 =  1/2ΔrH1Θ

It shows that enthalpy is an extensive quantity.

3. When a chemical equation is reversed, the value of ΔrHΘ is reversed in sign. For example

N2 (g) + 3H2 (g) → 2NH3 (g);
ΔrHΘ = -91.8kJ mol-1

2NH3 (g) → N2(g) + 3H2 (g);
ΔrHΘ = +91.8kJ mol-1

(e) Hess’s Law of Constant Heat Summation

We know that enthalpy is a state function, therefore the change in enthalpy is independent of the path between initial state (reactants) and final state (products). In other words, enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps. This may be stated as follows in the form of Hess’s Law.

If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.

Let us understand the importance of this law with the help of an example.

Consider the enthalpy change for the reaction

C (graphite,s) + 1/2 O2 (g) → CO (g);ΔrHΘ =?

Although CO(g) is the major product, some CO2 gas is always produced in this reaction. Therefore, we cannot measure enthalpy change for the above reaction directly. However, if we
can find some other reactions involving related species, it is possible to calculate the enthalpy change for the above reaction.

Let us consider the following reactions:

C (graphite,s) + O2 (g) → CO2 (g); ΔrHΘ = -393.5 kJmol−1————————————————(i)

CO (g) + 1/2 O2 (g) → CO2 (g);
ΔrHΘ = -283.0 kJmol−1 ———————————–(ii)

We can combine the above two reactions in such a way so as to obtain the desired reaction. To get one mole of CO(g) on the right,
we reverse equation (ii). In this, heat is absorbed instead of being released, so we change sign of ΔrHΘ value

CO2 (g) → CO (g) + 1/2 O2 (g);
ΔrHΘ =+283.0 kJmol−1 ———————–(iii)

Adding equation (i) and (iii), we get the desired equation,

C (graphite,s) + + 1/2 O2 (g) → CO (g);
for which ΔrHΘ = ( -393.5 + 283.0) = – 110.5 kJ mol−1

In general, if enthalpy of an overall reaction A→B along one route is ΔrH and ΔrH1, ΔrH2rH3….. representing enthalpies of reactions leading to same product, B along another route,then we have

ΔrH = ΔrH1 + ΔrH2 + ΔrH3 … —————————————————(6.16)

It can be represented as:



It is convenient to give name to enthalpies specifying the types of reactions.

(a) Standard enthalpy of combustion (symbol : ΔcHΘ )

Combustion reactions are exothermic in nature. These are important in industry, rocketry, and other walks of life. Standard enthalpy of combustion is defined as the enthalpy change per mole (or per unit amount) of a substance, when it undergoes combustion and all the reactants and products being in their standard states at the specified temperature.

Cooking gas in cylinders contains mostly butane (C4H10). During complete combustion of one mole of butane, 2658 kJ of heat is released. We can write the thermochemical reactions for this as:

C4H10 (g) + 13/2 O2 (g) → 4CO2(g) + 5H2O(1);
ΔcHΘ = -2658.0 kJ mol−1

Similarly, combustion of glucose gives out 2802.0 kJ/mol of heat, for which the overall equation is :

C6H12O6(g) + 6O2(g) → 6CO2 (g) + 6H2 O(1);
ΔcHΘ = -2802.0 kJ mol−1

Our body also generates energy from food by the same overall process as combustion, although the final products are produced after a series of complex bio-chemical reactions involving enzymes.

Problem 6.8

The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O (1) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, ΔfHΘ of benzene. Standard enthalpies of formation of CO2(g) and H2O(l) are -393.5 kJ mol−1 and – 285.83
kJ mol−1 respectively.


The formation reaction of benezene is given by :
6C (graphite) + 3H2 g → C6H6 (l) ;

ΔfHΘ = ? … —————————————–(i)

The enthalpy of combustion of 1 mol of benzene is :

C6H6 (l) + 15/2 O2 → 6CO2 (g) + 3H2O (l);
ΔcHΘ = -3267kJ mol−1 … ——————————–(ii)

The enthalpy of formation of 1 mol of CO2(g) :

C (graphite) + O2(g) → CO2 g ;
ΔfHΘ = -393.5kJ mol−1 … ———————————————(iii)

The enthalpy of formation of 1 mol of H2O(l) is :

H2 (g) + 1/2 O2 (g) → H2O (l);

ΔfHΘ = -285.83kJ mol−1 … ——————————————–(iv)

multiplying eqn. (iii) by 6 and eqn. (iv) by 3 we get:

6C (graphite) + 6O2 g → 6CO2 g ;
ΔfHΘ = -2361kJ mol−1

3H2 (g) + 3/2 O2 (g) → 3H2O (l);
ΔfHΘ = -857.49kJ mol−1

Summing up the above two equations :

6C (graphite) + 3H2 (g) + 15/2 O2 g → 6CO2 (g) + 3H2O(l);

ΔfHΘ = −3218.49 kJ mol-1 … ————————–(v )

Reversing equation (ii);

6CO2 (g) + 3H2O(l) → C6H6(l) + 15/2 O2;

ΔfHΘ = −3267.0 kJ mol-1 … ————————–(vi )

Adding equations (v) and (vi), we get

6C (graphite) + 3H2 (g) → C6H6(l);

ΔfHΘ = 48.51 kJ mol-1

(b) Enthalpy of atomization (symbol: ΔaHΘ )

Consider the following example of atomization of dihydrogen

H2(g) → 2H(g); ΔaHΘ = 435.0 kJ mol-1

You can see that H atoms are formed by breaking H—H bonds in dihydrogen. The enthalpy change in this process is known as
enthalpy of atomization, ΔaHΘ. It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.

In case of diatomic molecules, like dihydrogen (given above), the enthalpy of atomization is also the bond dissociation enthalpy. The other examples of enthalpy of atomization can be

CH4(g) → C(g) + 4H(g); ΔaHΘ = 1665 kJ mol-1

Note that the products are only atoms of C and H in gaseous phase. Now see the following reaction:

Na(s) → Na(g) ; ΔaHΘ = 108.4 kJ mol-1

In this case, the enthalpy of atomization is same as the enthalpy of sublimation.

(c) Bond Enthalpy (symbol: ΔbondHΘ)

Chemical reactions involve the breaking and making of chemical bonds. Energy is required to break a bond and energy is released when a bond is formed. It is possible to relate heat of reaction to changes in energy associated with breaking and making of chemical bonds. With reference to the enthalpy changes associated with chemical bonds, two different terms are used in thermodynamics.

(i) Bond dissociation enthalpy

(ii) Mean bond enthalpy

Let us discuss these terms with reference to diatomic and polyatomic molecules.
Diatomic Molecules: Consider the following process in which the bonds in one mole of dihydrogen gas (H2) are broken:

H2(g) → 2H(g) ; ΔH-HHΘ = 435.0 kJ mol-1

The enthalpy change involved in this process is the bond dissociation enthalpy of H-H bond. The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase.

Note that it is the same as the enthalpy of atomization of dihydrogen. This is true for all diatomic molecules. For example:

Cl2(g) → 2Cl(g) ; ΔCl-ClHΘ = 242 kJ mol-1

OΘ(g) → 2O(g) ; ΔO=OHΘ = 428 kJ mol-1

In the case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the same molecule. Polyatomic Molecules: Let us now consider a polyatomic molecule like methane, CH4. The overall thermochemical equation for its atomization reaction is given below:

CH4 (g) → C(g) + 4H(g);
ΔaHΘ = 1665 kJ mol-1

In methane, all the four C — H bonds are identical in bond length and energy. However, the energies required to break the individual C — H bonds in each successive step differ :

CH4 (g)→CH3 (g)+H(g); ΔbondHΘ = +427 kJ mol−1

CH3 (g) →CH2 (g)+ H(g); ΔbondHΘ = +439kJ mol−1

CH2 (g)→ CH(g)+ H(g); ΔbondHΘ = +452kJ mol−1

CH(g)→C(g)+ H(g); ΔbondHΘ= +347kJ mol−1


CH4 (g) → C(g) + 4H(g); ΔaHΘ = 1665kJ mol−1

In such cases we use mean bond enthalpy of C — H bond.

For example in CH4, ΔC-HHΘ is calculated as:

ΔC-HHΘ = ¼(ΔaHΘ) = ¼ (1665kJ mol−1 )

= 416 kJ mol−1

We find that mean C—H bond enthalpy in methane is 416 kJ/mol. It has been found that mean C—H bond enthalpies differ slightly from compound to compound, as in CH3CH2Cl,CH3NO2 , etc, but it does not differ in a great deal*. Using Hess’s law, bond enthalpies can be calculated. Bond enthalpy values of some single and multiple bonds are given in Table 6.3. The reaction enthalpies are very important quantities as these arise from the changes that accompany the breaking of old bonds and formation of the new bonds. We can predict enthalpy of a reaction in gas phase, if we know different bond enthalpies. The standard enthalpy of reaction, ΔrHΘ is related to bond enthalpies of the reactants and products in gas phase reactions as:

ΔrHΘ = Σbond enthalpiesreactants – Σbond enthalpiesproducts—————————————————————————–(6.17)**

This relationship is particularly more useful when the required values of ΔfHΘ are not available. The net enthalpy change of a reaction is the amount of energy required to break all the bonds in the reactant molecules minus the amount of energy required to break all the bonds in the product molecules. Remember that this relationship is approximate and is valid when all substances (reactants and products) in the reaction are in gaseous state.

Table 6.3(a) Some Mean Single Bond Enthalpies in kJ mol-1 at 298 K
H C N 0 F Si P S Cl Br I
435.8 414 389 404 569 293 318 339 431 308 297 H
347 293 351 439 289 264 259 330 276 238 C
159 201 272
201 243
138 184 368 351
201 O
155 540 490 327 255 197
176 213 226 360 289 213 Si
213 230 331 272 213 P
213 251 213
243 218 209 Cl
192 180 Br
151 I
Table 6.3(b) Some Mean Multiple Bond Enthalpies in kJ mol-1 at 298 K
N=N 418 C=C 611 O=O 498
N=N 946 C=C 837
C=N 615 C=O 741
C=N 891 C=O 1070
* Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same. 

**If we use enthalpy of bond formation, (ΔfHbondΘ ), which is the enthalpy change when one mole of a particular type of bond is formed from gaseous atom, then ΔrHΘ = ΣΔfHbonds of productsΘ – ΣΔfHbonds of reactantsΘ

(d) Enthalpy of Solution (symbol : ΔsolHΘ )

Enthalpy of solution of a substance is the enthalpy change when one mole of it dissolves in a specified amount of solvent. The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions (or solute
molecules) are negligible.

When an ionic compound dissolves in a solvent, the ions leave their ordered positions on the crystal lattice. These are now more free in solution. But solvation of these ions (hydration in case solvent is water) also occurs at the same time. This is shown diagrammatically, for an ionic compound, AB (s)


The enthalpy of solution of AB(s), ΔsolHΘ, in water is, therefore, determined by the selective values of the lattice enthalpy, ΔlatticeHΘ and enthalpy of hydration of ions, ΔhydHΘ as

ΔsolHΘ = ΔlatticeHΘ + ΔhydHΘ

For most of the ionic compounds, ΔsolHΘ is positive and the dissociation process is endothermic. Therefore the solubility of most salts in water increases with rise of temperature. If the lattice enthalpy is very high, the dissolution of the compound may not take place at all. Why do many fluorides tend to be less soluble than the corresponding chlorides? Estimates of the magnitudes of enthalpy changes may be made by using tables of bond energies (enthalpies) and lattice energies (enthalpies).

Lattice Enthalpy

The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.

Na+Cl(s) → Na+(g) + Cl(g) ;

ΔlatticeHΘ = +788 kJmol-1

Since it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber Cycle (Fig. 6.9).



Let us now calculate the lattice enthalpy of Na+Cl(s) by following steps given below :

1. Na(s)→ Na(g) , sublimation of sodium metal, ΔsubHΘ = 108.4 kJ mol−1

2. Na(g)→Na+(g) + e−1(g) , the ionization of sodium atoms, ionization enthalpy

ΔiHΘ = 496 kJ mol−1

3. 1/2 Cl2 (g) → Cl(g), the dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy.

1/2 ΔbondHΘ = 121kJ mol−1
4. Cl(g) + e−1(g) → Cl(g) electron gained by chlorine atoms. The electron gain enthalpy,

ΔegHΘ = -348.6 kJ mol−1 .

You have learnt about ionization enthalpy and electron gain enthalpy in Unit 3. In fact, these terms have been taken from thermodynamics. Earlier terms, ionization energy and electron affinity were in practice in place of the above terms (see the box for justification).

Ionization Energy and Electron Affinity

Ionization energy and electron affinity are defined at absolute zero. At any other temperature, heat capacities for the reactants and the products have to be taken into account. Enthalpies of reactions for

M(g) → M+(g) + e (for ionization)
M(g) + e → M(g) (for electron gain)

at temperature, T is

The value of Cp for each species in the above reaction is 5/2 R (CV = 3/2R)
So,ΔrCpΘ = + 5/2 R (for ionization)
ΔrCpΘ = – 5/2 R (for electron gain)


ΔrHΘ (ionization enthalpy) = E0 (ionization energy) + 5/2 RT

ΔrHΘ (electron gain enthalpy) = – A( electron affinity) – 5/2 RT

5. Na+ (g) + Cl (g) → Na+Cl (s)

The sequence of steps is shown in Fig. 6.9, and is known as a Born-Haber cycle. The importance of the cycle is that, the sum of the enthalpy changes round a cycle is zero.

Applying Hess’s law, we get,

ΔlatticeHΘ = 411.2 +108.4 +121+ 496 − 348.6 ΔlatticeHΘ = +788 kJ

for NaCl(s)→ Na+ (g) + Cl(g)

Internal energy is smaller by 2RT ( because Δng = 2) and is equal to + 783 kJ mol-1.

Now we use the value of lattice enthalpy to calculate enthalpy of solution from the expression:

ΔsolHΘ = ΔlatticeHΘ + ΔhydHΘ

For one mole of NaCl(s),

lattice enthalpy = + 788 kJ mol-1

and ΔhydHΘ = – 784 kJ mol-1( from the literature)
ΔsolHΘ = + 788 kJ mol-1 – 784 kJ mol-1 = + 4 kJ mol-1

The dissolution of NaCl(s) is accompanied by very little heat change.


The first law of thermodynamics tells us about the relationship between the heat absorbed and the work performed on or by a system. It puts no restrictions on the direction of heat flow. However, the flow of heat is unidirectional from higher temperature to lower temperature. In fact, all naturally occurring processes whether chemical or physical will tend to proceed spontaneously in one direction only. For example, a gas expanding to fill the available volume, burning carbon in dioxygen giving carbon dioxide.

But heat will not flow from colder body to warmer body on its own, the gas in a container will not spontaneously contract into one corner or carbon dioxide will not form carbon and dioxygen spontaneously. These and many other spontaneously occurring changes show unidirectional change. We may ask ‘what is the driving force of spontaneously occurring changes ? What determines the direction of a spontaneous change ? In this section, we shall establish some criterion for these processes whether these will take place or not.

Let us first understand what do we mean by spontaneous reaction or change ? You may think by your common observation that spontaneous reaction is one which occurs immediately when contact is made between the reactants. Take the case of combination of hydrogen and oxygen. These gases may be mixed at room temperature and left for many years without observing any perceptible change. Although the reaction is taking place between them, it is at an extremely slow rate. It is still called spontaneous reaction. So spontaneity means ‘having the potential to proceed without the assistance of external agency’. However, it does not tell about the rate of the reaction or process. Another aspect of spontaneous reaction or process, as we see is that these cannot reverse their direction on their own. We may summarise it as follows:

A spontaneous process is an irreversible process and may only be reversed by some external agency.

(a) Is decrease in enthalpy a criterion for spontaneity ?

If we examine the phenomenon like flow of water down hill or fall of a stone on to the ground, we find that there is a net decrease in potential energy in the direction of change. By analogy, we may be tempted to state that a chemical reaction is spontaneous in a given direction, because decrease in energy has taken place, as in the case of exothermic reactions. For example:

1/2 N2(g) + 3/2 H2(g) = NH3(g) ; ΔrHΘ = – 46.1 kJ mol-1

1/2 H2(g) + 1/2 Cl2(g) = HCl (g) ; ΔrHΘ = – 92.32 kJ mol-1
H2(g) + 1/2 O2(g) → H2O(l) ; ΔrHΘ = -285.8 kJ mol-1

The decrease in enthalpy in passing from reactants to products may be shown for any exothermic reaction on an enthalpy diagram as shown in Fig. 6.10(a).

Thus, the postulate that driving force for a chemical reaction may be due to decrease in energy sounds ‘reasonable’ as the basis of evidence so far !

Now let us examine the following reactions:

1/2 N2(g) + O2(g) → NO2(g);

ΔrHΘ= +33.2 kJ mol-1

C(graphite, s) + 2 S(l) → CS2(l);

ΔrHΘ = +128.5 kJ mol-1

These reactions though endothermic, are spontaneous. The increase in enthalpy may be represented on an enthalpy diagram as shown in Fig. 6.10(b).


it becomes obvious that while decrease in enthalpy may be a contributory factor for spontaneity, but it is not true for all cases.

(b) Entropy and spontaneity

Then, what drives the spontaneous process in a given direction ? Let us examine such a case in which ΔH = 0 i.e., there is no change in enthalpy, but still the process is spontaneous.

Let us consider diffusion of two gases into each other in a closed container which is The two gases, say, gas A and gas B are represented by black dots and white dots respectively and separated by a movable partition [Fig. 6.11 (a)]. When the partition is withdrawn [Fig.6.11( b)], the gases begin to diffuse into each other and after a period of time, diffusion will be complete.

Let us examine the process. Before partition, if we were to pick up the gas molecules from left container, we would be sure that these will be molecules of gas A and similarly if we were to pick up the gas molecules from right container, we would be sure that these will be molecules of gas B. But, if we were to pick up molecules from container when partition is removed, we are not sure whether the molecules picked are of gas A or gas B. We say that the system has become less predictable or more chaotic.

We may now formulate another postulate: in an isolated system, there is always a tendency for the systems’ energy to become more disordered or chaotic and this could be a criterion for spontaneous change !

this point, we introduce another thermodynamic function, entropy denoted as S. The above mentioned disorder is the manifestation of entropy. To form a mental picture, one can think of entropy as a measure of the degree of randomness or disorder in the system. The greater the disorder in an isolated system, the higher is the entropy. As far as a chemical reaction is concerned, this entropy change can be attributed to rearrangement of atoms or ions from one pattern in the eactants to another (in the products). If the structure of the products is very much disordered than that of the reactants, there will be a resultant increase in entropy. The change in entropy accompanying a chemical reaction may be estimated qualitatively by a consideration of the structures of the species taking part in the reaction. Decrease of regularity in structure would mean increase in entropy. For a given substance, the crystalline solid state is the state of lowest entropy (most ordered), The gaseous state is state of highest entropy.

Now let us try to quantify entropy. One way to calculate the degree of disorder or chaotic distribution of energy among molecules would be through statistical method which is beyond the scope of this treatment. Other way would be to relate this process to the heat involved in a process which would make entropy a
thermodynamic concept. Entropy, like any other thermodynamic property such as internal energy U and enthalpy H is a state function and ΔS is independent of path.

Whenever heat is added to the system, it increases molecular motions causing increased randomness in the system. Thus heat (q) has randomising influence on the system. Can we then equate ΔS with q ? Wait ! Experience suggests us that the distribution of heat also depends on the temperature at which heat is added to the system. A system at higher temperature has greater randomness in it than one at lower temperature. Thus, temperature is the measure of average chaotic motion of particles in the system. Heat added to a system at lower temperature causes greater randomness than when the same quantity of heat is added to it at higher temperature. This suggests that the entropy change is inversely proportional to the temperature. ΔS is related with q and T for a reversible reaction as :

ΔS = qrev/T ————————————————————-(6.18)

The total entropy change ( ΔStotal) for the system and surroundings of a spontaneous process is given by surrtotal > 0 ——————————————————————–(6.19)

When a system is in equilibrium, the entropy is maximum, and the change in entropy, ΔS = 0.

We can say that entropy for a spontaneous process increases till it reaches maximum and at equilibrium the change in entropy is zero. Since entropy is a state property, we can calculate the change in entropy of a reversible process by ΔSsys = qsys,rev /T

We find that both for reversible and irreversible expansion for an ideal gas, under isothermal conditions, ΔU = 0, but ΔStotal i.e., ΔSsys + ΔSsurr is not zero for irreversible process. Thus, ΔU does not discriminate between reversible and irreversible process, whereas ΔS does.

Problem 6.9

Predict in which of the following, entropy increases/decreases :

(i) A liquid crystallizes into a solid.
(ii) Temperature of a crystalline solid is raised from 0 K to 115 K.

(iii) 2NaHCO3 (s) → Na2CO3 (s) + CO2 (g) + H2O (g)

(iv) H2 g → 2H g


(i) After freezing, the molecules attain an ordered state and therefore, entropy decreases.

(ii) At 0 K, the contituent particles are static and entropy is minimum. If temperature is raised to 115 K, these begin to move and oscillate about their equilibrium positions in the lattice and system becomes more disordered, therefore entropy increases.

(iii) Reactant, NaHCO3 is a solid and it has low entropy. Among products there are one solid and two gases. Therefore, the products represent a condition of higher entropy.

(iv) Here one molecule gives two atoms i.e., number of particles increases leading to more disordered state. Two moles of H atoms have higher entropy than one mole of dihydrogen molecule.

Problem 6.10

For oxidation of iron,

4Fe (s) +3O2 (g) → 2Fe2O3 (s)

entropy change is – 549.4 JK-1mol-1at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous?

rHΘ for this reaction is -1648 103 J mol-1)


One decides the spontaneity of a reaction by considering ΔStotal (ΔSsys + ΔSsurr . For calculating ΔSsurr, we have to consider the heat absorbed by the surroundings which is equal to -ΔrHΘ. At temperature T, entropy change of the surroundings is

ΔSsurr = -(ΔrHΘ/T)(at constant pressure) = (-1648 x 103 Jmol-1 )/298 K = 5530 JK-1mol-1

Thus, total entropy change for this reaction

ΔrStotal = 5530JK-1mol-1 + (-549.4JK-1mol-1)=4980.6JK-1mol-1

This shows that the above reaction is spontaneous.

(c) Gibbs energy and spontaneity

We have seen that for a system, it is the total entropy change, ΔStotal which decides the spontaneity of the process. But most of the chemical reactions fall into the category of either closed systems or open systems. Therefore, for most of the chemical reactions there are changes in both enthalpy and entropy. It is clear from the discussion in previous sections that neither decrease in enthalpy nor increase in entropy alone can determine the direction of spontaneous change for these systems.

For this purpose, we define a new thermodynamic function the Gibbs energy or Gibbs function, G, as G = H – TS ——————————————————————-(6.20)

Gibbs function, G is an extensive property and a state function.

The change in Gibbs energy for the system, ΔGsys can be written as

ΔGsys = ΔHsys − T ΔGsys −SsysΔT

At constant temperature, ΔT = 0

∴ΔGsys = ΔHsys − TΔSsys

Usually the subscript ‘system’ is dropped and we simply write this equation as

ΔG = ΔH – TΔS ——————————————————————-(6.21)

Thus, Gibbs energy change = enthalpy change – temperature x entropy change, and is referred to as the Gibbs equation, one of the most important equations in chemistry. Here, we have considered both terms together for spontaneity: energy (in terms of ΔH) and entropy (ΔS, a measure of disorder) as indicated earlier. Dimensionally if we analyse, we find that ΔG has units of energy because, both ΔH and the TΔS are energy terms, since TΔS = (K) (J/K) = J.

Now let us consider how G is related to reaction spontaneity.

We know,

ΔStotal = ΔSsys + ΔSsurr

If the system is in thermal equilibrium with the surrounding, then the temperature of the surrounding is same as that of the system. Also, increase in enthalpy of the surrounding is equal to decrease in the enthalpy of the system.

Therefore, entropy change of surroundings,

ΔSsurr = ΔHsurr/T = -ΔHsurr/T

ΔStotal =− ΔSsys + (-ΔHsys/T)

Rearranging the above equation:

TΔStotal = TΔSsys – ΔHsys

For spontanious process, ΔStotal > 0 , so

TΔSsys – ΔHsys > 0
⇒ −(ΔHsys − TΔSsys ) > 0

Using equation 6.21, the above equation can be written as

−ΔG > 0
ΔG = ΔH − T ΔS < 0 ——————————————-(6.22)

ΔHsys is the enthalpy change of a reaction, TΔSsys is the energy which is not available to do useful work. So ΔG is the net energy available to do useful work and is thus a measure of the ‘free energy’. For this reason, it is also known as the free energy of the reaction.

ΔG gives a criteria for spontaneity at constant pressure and temperature.

(i) If ΔG is negative (< 0), the process is spontaneous. (ii) If ΔG is positive (> 0), the process is non spontaneous.

Note : If a reaction has a positive enthalpy change and positive entropy change, it can be spontaneous when TΔS is large enough to
outweigh ΔH. This can happen in two ways;

(a) The positive entropy change of the system can be ‘small’ in which case T must be large.

(b) The positive entropy change of the system can be ‘large’, in which case T may be small. The former is one of the reasons why reactions are often carried out at high temperature. Table 6.4 summarises the effect of temperature on spontaneity of reactions.

Table 6.4 Effect of Temperature on Spontaneity of Reactions
+ Reaction spontaneous at all temperature
-(at low T) Reaction spontaneous at low temperature
+(at high T) Reaction nonspontaneous at high temperature
+ + +(at low T) Reaction nonspontaneous at low temperature
+ + -(T high T) Reaction spontaneous at high temperature
+ +(at all T) Reaction nonspontaneous at all temperature


We have seen how a knowledge of the sign and magnitude of the free energy change of a chemical reaction allows:

(i) Prediction of the spontaneity of the chemical reaction.

(ii) Prediction of the useful work that could be extracted from it.

So far we have considered free energy changes in irreversible reactions. Let us now examine the free energy changes in reversible reactions.

‘Reversible’ under strict thermodynamic sense is a special way of carrying out a process such that system is at all times in perfect equilibrium with its surroundings. When applied to a chemical reaction, the term ‘reversible’ indicates that a given reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up. This means that the reactions in both the directions should proceed with a decrease in free energy,
which seems impossible. It is possible only if at equilibrium the free energy of the system is minimum. If it is not, the system would
spontaneously change to configuration of lower free energy.

So, the criterion for equilibrium

A + B  C + D ; is

ΔrG = 0

Gibbs energy for a reaction in which all reactants and products are in standard state, ΔrG0 is related to the equilibrium constant of the reaction as follows:

0 = ΔrGΘ + RT ln K

or ΔrGΘ = – RT ln K

or ΔrGΘ = – 2.303 RT log K ——————————————————–(6.23)

We also know that
ΔrGΘ = ΔrHΘ – TΔrSΘ = -RT ln K —————————————————–(6.24)

For strongly endothermic reactions, the value of ΔrHΘ may be large and positive. In such a case, value of K will be much smaller than 1 and the reaction is unlikely to form much product. In case of exothermic reactions, ΔrHΘ is large and negative, and ΔrGΘ is likely to be large and negative too. In such cases, K will be much larger than 1. We may expect strongly exothermic reactions to have a large K, and hence can go to near completion. ΔrGΘ also depends upon ΔrSΘ, if the changes in the entropy of reaction is also taken into account, the value of K or extent of chemical reaction will also be affected, depending upon whether ΔrSΘ is positive or negative.

* The term low temperature and high temperature are relative. For a particular reaction, high temperature could even mean room temperature.

Using equation (6.24),

(i) It is possible to obtain an estimate of ΔGTheta; from the measurement of ΔHTheta; and ΔSTheta;, and then calculate K at any temperature for economic yields of the products.

(ii) If K is measured directly in the laboratory, value of ΔGTheta; at any other temperature can be calculated.

Problem 6.11

Calculate ΔrGTheta; for conversion of oxygen to ozone, 3/2 O2(g) → O3(g) at 298 K. if Kp for this conversion is 2.47 x 10–29 .


We know ΔrGTheta; = – 2.303 RT log Kp and R = 8.314 JK-1mol-1
Therefore, ΔrGTheta; = – 2.303 (8.314 JK-1mol-1) x (298 K) (log 2.47 10–-29) = 163000 J mol-1= 163 kJ mol-1.

Problem 6.12

Find out the value of equilibrium constant for the following reaction at 298 K.

Standard Gibbs energy change, ΔrGTheta; at the given temperature is -13.6 kJ mol-1.


We know, log K = –ΔrGTheta;/2.303 RT
= (–13.6 x 103 J mol-1)/2.303 (8.314 JK-1 mol-1) (298K)

= 2.38

Hence K = antilog 2.38 = 2.4 x 102.

Problem 6.13

At 60°C, dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.


If N2O4 is 50% dissociated, the mole
fraction of both the substances is given by

xN2O4 = (1 – 0.5)/(1- 0.5);
xNO2 = (2 x 0.5)/(1 + 0.5)

pN2O4 = (0.5/1.5) x 1atm,
pNO2 = (1/1.5) x 1 atm.

The equilibrium constant Kp is given by Kp = (pNO2)2/pN2O4 = 1.5/(1.5)2(0.5)

= 1.33 atm.

ΔrGTheta; = -RT ln Kp
ΔrGTheta; = (- 8.314 JK-1 mol-1) x (333 K) x (2.303) x (0.1239) = – 763.8 kJ mol-1


Thermodynamics deals with energy changes in chemical or physical processes and enables us to study these changes quantitatively and to make useful predictions. For these purposes, we divide the universe into the system and the surroundings. Chemical or physical processes lead to evolution or absorption of heat (q), part of which may be converted into work (w). These quantities are related through the first law of thermodynamics via ΔU = q + w. ΔU, change in internal energy, depends on initial and final states only and is a state function, whereas q and w depend on the path and are not the state functions. We follow sign conventions of q and w by giving the positive sign to these quantities when these are added to the system. We can measure the transfer of heat from one system to another which causes the change in temperature. The magnitude of rise in temperature depends on the heat capacity (C) of a substance. Therefore, heat absorbed or evolved is q = CΔT. Work can be measured by w = -pexΔV, in case of expansion of gases. Under reversible process, we can put pex = p for infinitesimal changes in the volume making wrev = – p dV. In this condition, we can use gas equation, pV = nRT.

At constant volume, w = 0, then ΔU = qV , heat transfer at constant volume. But in study of chemical reactions, we usually have constant pressure. We define another state function enthalpy. Enthalpy change, ΔH = ΔU + ΔngRT, can be found directly from
the heat changes at constant pressure, ΔH = qp.

There are varieties of enthalpy changes. Changes of phase such as melting, vaporization and sublimation usually occur at constant temperature and can be characterized by enthalpy changes which are always positive. Enthalpy of formation, combustion and other enthalpy changes can be calculated using Hess’s law. Enthalpy
change for chemical reactions can be determined by

and in gaseous state by

ΔrHΘ = Σ bond enthalpies of the reactants – Σ bond enthalpies of the products

First law of thermodynamics does not guide us about the direction of chemical reactions i.e., what is the driving force of a chemica reaction. For isolated systems, ΔU = 0. We define another state function, S, entropy for this purpose. Entropy is a measure of disorder or randomness. For a spontaneous change, total entropy change is positive. Therefore, for an isolated system, ΔU = 0, ΔS > 0, so entropy change distinguishes a spontaneous change, while energy change does not. Entropy changes can be measured by the equation ΔS = qrev/T for a reversible process. qrev/T is independent of path. Chemical reactions are generally carried at constant pressure, so we define another state function Gibbs energy, G, which is related to entropy and enthalpy changes of the system by the equation:

ΔrG = ΔrH – T ΔrS

For a spontaneous change, ΔGsys < 0 and at equilibrium, ΔGsys = 0.

Standard Gibbs energy change is related to equilibrium constant by
ΔrGΘ = – RT ln K.

K can be calculated from this equation, if we know ΔrGTheta; which can be found from   ΔrGΘ = ΔrHΘ − TΔrSΘ . Temperature is an important factor in the equation. Many reactions which are non-spontaneous at low temperature, are made spontaneous at high temperature for systems having positive entropy of reaction.


6.1 Choose the correct answer. A thermodynamic state function is a
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.

6.2 For the process to occur under adiabatic conditions, the correct
condition is:
(i) ΔT = 0
(ii) Δp = 0
(iii) q = 0
(iv) w = 0

6.3 The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element

6.4 ΔU0of combustion of methane is – X kJ mol-1. The value of ΔHΘ is
(i) = ΔUΘ
(ii) > ΔUΘ
(iii) < ΔUΘ
(iv) = 0

6.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol-1 -393.5 kJ mol-1, and -285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will be

(i) -74.8 kJ mol-1 (ii) -52.27 kJ mol-1
(iii) +74.8 kJ mol-1 (iv) +52.26 kJ mol-1.

6.6 A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be

(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(v) possible at any temperature

6.7 In a process, 701 J of heat is absorbed by a system and 394 J of
work is done by the system. What is the change in internal energy
for the process?

6.8 The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be -742.7 kJ mol-1 at 298 K. Calculate enthalpy change for the reaction at 298 K.

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

6.9 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol-1 K-1.

6.10 Calculate the enthalpy change on freezing of 1.0 mol of water
at10.0°C to ice at -10.0°C. ΔfusH = 6.03 kJ mol-1 at 0°C.
Cp [H2O(l)] = 75.3 J mol-1 K-1

Cp [H2O(s)] = 36.8 J mol-1 K-1

6.11 Enthalpy of combustion of carbon to CO2 is -393.5 kJ mol-1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.

6.12 Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are -110, – 393, 81 and 9.7 kJ mol-1 respectively. Find the value of ΔrH for the reaction:

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

6.13 Given N2(g) + 3H2(g) → 2NH3(g) ; ΔrHTheta; = -92.4 kJ mol-1

What is the standard enthalpy of formation of NH3 gas?

6.14 Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
CH3OH (l) + 3/2 O2(g) → CO2(g) + 2H2O(l) ; ΔrHTheta; = -726 kJ mol-1
C(graphite) + O2(g) → CO2(g) ; ΔcHTheta; = -393 kJ mol-1
H2(g) + 1/2 O2(g) → H2O(l) ; ΔfHTheta; = -286 kJ mol-1.

6.15 Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g).

ΔvapHΘ(CCl4) = 30.5 kJ mol-1.
ΔfHΘ (CCl4) = -135.5 kJ mol-1.
ΔaHΘ (C) = 715.0 kJ mol-1 , where ΔaH0 is enthalpy of atomisation ΔaHΘ (Cl2) = 242 kJ mol-1

6.16 For an isolated system, ΔU = 0, what will be ΔS ?

6.17 For the reaction at 298 K,
2A + B → C
ΔH = 400 kJ mol-1 and ΔS = 0.2 kJ K-1 mol-1

At what temperature will the reaction become spontaneous
considering ΔH and ΔS to be constant over the temperature range.

6.18 For the reaction,
2 Cl(g) → Cl2(g), what are the signs of ΔH and ΔS ?

6.19 For the reaction
2 A(g) + B(g) → 2D(g)
ΔUΘ = -10.5 kJ and ΔSΘ = -44.1 JK-1.
Calculate ΔG0 for the reaction, and predict whether the reaction
may occur spontaneously.

6.20 The equilibrium constant for a reaction is 10. What will be the value of ΔGΘ ? R = 8.314 JK-1 mol-1, T = 300 K.

6.21 Comment on the thermodynamic stability of NO(g), given
1/2 N2(g) + 1/2 O2(g) → NO(g) ; ΔrHΘ = 90 kJ mol-1
NO(g) + 1/2 O2(g) → NO2(g) : ΔrHΘ = -74 kJ mol-1

6.22 Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ΔfHTheta; = -286 kJ mol-1.

Answer to Some Selected Problems

6.1 (ii)

6.2 (iii)

6.3 (ii)

6.4 (iii)

6.5 (i)

6.6 (iv)

6.7 q = + 701 J

w = – 394 J, since work is done by the system

ΔU = 307 J

6.8 –741.5 kJ

6.9 1.09 kJ

6.10 ΔH = –6.415 kJ mol–1

6.11 –315 kJ

6.12 ΔrH = –778 kJ

6.13 – 46.2 kJ mol–1

6.14 –239 kJ mol–1

6.15 327 kJ mol–1

6.16 ΔS > 0

6.17 2000 K

6.18 ΔH is negative (bond energy is released) and ΔS is negative (There is less randomness among the molecules than among the atoms)

6.19 0.164 kJ, the reaction is not spontaneous.

6.20 –5.744 kJ mol–1

6.21 NO(g) is unstable, but NO2(g) is formed.

6.22 qsurr = + 286 kJ mol–1 ΔSsurr = 959.73 J K–1


I. Multiple Choice Questions (Type-I)

1. Thermodynamics is not concerned about______.

(i) energy changes involved in a chemical reaction.
(ii) the extent to which a chemical reaction proceeds.
(iii) the rate at which a reaction proceeds.
(iv) the feasibility of a chemical reaction.

2. Which of the following statements is correct?

(i) The presence of reacting species in a covered beaker is an example of open system.
(ii) There is an exchange of energy as well as matter between the system and the surroundings in a closed system.
(iii) The presence of reactants in a closed vessel made up of copper is an example of a closed system.
(iv) The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.

3. The state of a gas can be described by quoting the relationship between___.

(i) pressure, volume, temperature
(ii) temperature, amount, pressure
(iii) amount, volume, temperature
(iv) pressure, volume, temperature, amount

4. The volume of gas is reduced to half from its original volume. The specific heat will be  ______.

(i) reduce to half
(ii) be doubled
(iii) remain constant
(iv) increase four times

5. During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is

6. ΔfUΘ of formation of CH4(g) at certain temperature is –393 kJ mol–1. The value of ΔfHΘis

(i) zero
(ii) < ΔfUΘ
(iii) > ΔfUΘ
(iv) equal to ΔfUΘ

7. In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from  the following.

(i) q = 0, ΔT ≠ 0, w = 0
(ii) q ≠ 0, ΔT = 0, w = 0
(iii) q = 0, ΔT = 0, w = 0
(iv) q = 0, ΔT < 0, w ≠ 0

8. The pressure-volume work for an ideal gas can be calculated by using the expression  . The work can also be calculated from the pV– plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from volume Vi to Vf . choose the correct option.

(i) w (reversible) = w (irreversible)
(ii) w (reversible) < w (irreversible) (iii) w (reversible) > w (irreversible)
(iv) w (reversible) = w (irreversible) + pex.ΔV

9. The entropy change can be calculated by using the expression ΔS = qrev/T. When water freezes in a glass beaker, choose the correct statement amongst the following :

(i) ΔS (system) decreases but ΔS (surroundings) remains the same.
(ii) ΔS (system) increases but ΔS (surroundings) decreases.
(iii) ΔS (system) decreases but ΔS (surroundings) increases.
(iv) ΔS (system) decreases and ΔS (surroundings) also decreases.

10. On the basis of thermochemical equations (a), (b) and (c), find out which of the algebric relationships given in options (i) to (iv) is correct.

(a) C(graphite) + O2 (g) → CO2 (g) ; ΔrH = x kJ mol–1
(b) C(graphite) + 12 O2 (g) → CO (g) ; ΔrH = y kJ mol–1
(c) CO(g) + 12 O2 (g) → CO2 (g) ; ΔrH = z kJ mol–1

(i) z = x + y
(ii) x = y – z
(iii) x = y + z
(iv) y = 2z – x

11. Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options (i) to (iv) is correct?

(a) C(g) + 4H(g) → CH4 (g); ΔrH = x kJ mol–1
(b) C(graphite,s) + 2H2 (g) → CH4 (g); ΔrH = y kJ mol–1

(i) x = y
(ii) x = 2y
(iii) x > y
(iv) x < y

12. The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound

(i) is always negative
(ii) is always positive
(iii) may be positive or negative
(iv) is never negative

13. Enthalpy of sublimation of a substance is equal to

(i) enthalpy of fusion + enthalpy of vapourisation
(ii) enthalpy of fusion
(iii) enthalpy of vapourisation
(iv) twice the enthalpy of vapourisation

14. Which of the following is not correct?

(i) ΔG is zero for a reversible reaction
(ii) ΔG is positive for a spontaneous reaction
(iii)ΔG is negative for a spontaneous reaction
(iv) ΔG is positive for a non-spontaneous reaction

II. Multiple Choice Questions (Type-II)

In the following questions two or more options may be correct.

15. Thermodynamics mainly deals with

(i) interrelation of various forms of energy and their transformation from one form to another.
(ii) energy changes in the processes which depend only on initial and final states of the microscopic systems containing a few molecules.
(iii) how and at what rate these energy transformations are carried out.
(iv) the system in equilibrium state or moving from one equilibrium state to another  quilibrium state.

16. In an exothermic reaction, heat is evolved, and system loses heat to the surrounding. For such system

(i) qp will be negative
(ii) ΔrH will be negative
(iii) qp will be positive
(iv) ΔrH will be positive

17. The spontaneity means, having the potential to proceed without the assistance of external agency. The processes which occur spontaneously are

(i) flow of heat from colder to warmer body.
(ii) gas in a container contracting into one corner.
(iii) gas expanding to fill the available volume.
(iv) burning carbon in oxygen to give carbon dioxide.

18. For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression w = – nRT ln Vf/Vi

A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.

(i) Work done at 600 K is 20 times the work done at 300 K.
(ii) Work done at 300 K is twice the work done at 600 K.
(iii) Work done at 600 K is twice the work done at 300 K.
(iv) ΔU = 0 in both cases.

19. Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below :

2Zn (s) + O2 (g) &rarr; 2ZnO (s); ΔH = – 693.8 kJ mol–1

(i) The enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
(ii) The enthalpy of two moles of ZnO is more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
(iii) 693.8 kJ mol–1 energy is evolved in the reaction.
(iv) 693.8 kJ mol–1 energy is absorbed in the reaction.

III. Short Answer Type

20. 18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol–1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?

21. One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?

22. Standard molar enthalpy of formation, ΔfHΘ is just a special case of enthalpy of reaction, ΔrHΘ. Is the ΔrHΘ for the following reaction same as ΔfHΘ? Give reason for your answer.

CaO(s) + CO2(g) → CaCO3(s); ΔfHΘ = –178.3 kJ mol–1

23. The value of ΔfHΘ for NH3 is –91.8 kJ mol–1. Calculate enthalpy change for the following reaction :

2NH3(g) → N2(g) + 3H2(g)

24. Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is ΔrH and ΔrH1, ΔrH2, ΔrH3 ….. represent enthalpies of intermediate reactions leading to product B. What will be the relation between ΔrH for overall reaction and ΔrH1 , ΔrH2 ….. etc. for intermediate reactions.

25. The enthalpy of atomisation for the reaction CH4(g)→ C(g) + 4H(g) is 1665 kJ mol–1. What is the bond energy of C–H bond?

26. Use the following data to calculate ΔlatticeHΘ for NaBr. ΔsubHΘ for sodium metal = 108.4 kJ mol–1 Ionization enthalpy of sodium = 496 kJ mol–1 Electron gain enthalpy of bromine = – 325 kJ mol–1 Bond dissociation enthalpy of bromine = 192 kJ mol–1 ΔfHΘ for NaBr (s) = – 360.1 kJ mol–1

27. Given that ΔH = 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?

28. Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.

29. Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium?

30. At 298 K. Kp for the reaction  is 0.98. Predict whether the reaction is spontaneous or not.

31. A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in Fig. 6.1. What will be the value of ΔH for the cycle as a whole?

32. The standard molar entropy of H2O (l) is 70 J K–1 mol–1. Will the standard molar entropy of H2O(s) be more, or less than 70 J K–1 mol–1?

33. Identify the state functions and path functions out of the following : enthalpy, entropy, heat, temperature, work, free energy.

34. The molar enthalpy of vapourisation of acetone is less than that of water. Why?

35. Which quantity out of ΔrG and ΔrGΘ will be zero at equilibrium?

36. Predict the change in internal energy for an isolated system at constant volume.

37. Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.

38. Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre?

39. Heat capacity (Cp ) is an extensive property but specific heat (c) is an intensive property. What will be the relation between Cp and c for 1 mol of water?

40. The difference between Cp and CV can be derived using the empirical relation H = U + pV. Calculate the difference between Cp and CV for 10 moles of an ideal gas.

41. If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.

42. The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction.

H2(g) + Br2(g) → 2HBr(g)

Given that Bond energy of H2, Br2 and HBr is 435 kJ mol–1, 192 kJ mol–1 and 368 kJ mol–1respectively.

43. The enthalpy of vapourisation of CCl4 is 30.5 kJ mol–1. Calculate the heat required for the vapourisation of 284 g of CCl4 at constant pressure. (Molar mass of CCl4 = 154 g mol–1).

44. The enthalpy of reaction for the reaction :

2H2(g) + O2(g) → 2H2O(l) is ΔrHΘ = – 572 kJ mol–1.

What will be standard enthalpy of formation of H2O (l) ?

45. What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, pext in a single step as shown in Fig. 6.2. Explain graphically.

46. How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps?

47. Represent the potential energy/enthalpy change in the following processes graphically.

(a) Throwing a stone from the ground to roof.

In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?

48. Enthalpy diagram for a particular reaction is given in Fig. 6.3. Is it possible to decide spontaneity of a reaction from given diagram. Explain.

49. 1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in Fig. 6.4. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.

50. An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L inone step. Calculate the amount of work done by the gas. If the same expansion were carried outreversibly, will the work done be higher or lower than the earlier case? (Given that 1 L bar = 100 J)

IV. Matching Type

In the following questions more than one correlation is possible between options of both columns.

51. Match the following :

(i) Adiabatic process (a) Heat
(ii) Isolated system (b) At constant volume
(iii) Isothermal change (c) First law of thermodynamics
(iv) Path function (d) No exchange of energy and matter
(v) State function (e) No transfer of heat
(vi) ΔU = q (f) Constant temperature
(vii) Law of conservation of energy (g) Internal energy
(viii) Reversible process (h) pext = 0
(ix) Free expansion (i) At constant pressure
(x) ΔH = q (j) Infinitely slow process which proceeds through a series of equilibrium states.
(xi) Intensive property (k) Entropy
(xii) Extensive property (l) Pressure
(m) Specific heat

52. Match the following processes with entropy change:

Reaction Entropy change
(i) A liquid vapourises (a) ΔS = 0
(ii) Reaction is non-spontaneous at all temperatures and ΔH is positive (b) ΔS = positive
(iii) Reversible expansion of an ideal gas (c) ΔS = negative

53. Match the following parameters with description for spontaneity :



Δ (Parameters) Description
(i) + + (a) Non-spontaneous at high 


(ii) + at high T (b) Spontaneous at all temperatures
(iii) + (c) Non-spontaneous at all temperatures

54. Match the following :

(i) Entropy of vapourisation (a) decreases
(ii) K for spontaneous process (b) is always positive
(iii) Crystalline solid state (c) lowest entropy
(iv) ΔU in adiabatic expansion of ideal gas (d) ΔHvap/Tb

V. Assertion and Reason Type

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

55. Assertion (A): Combustion of all organic compounds is an exothermic reaction.

Reason (R) : The enthalpies of all elements in their standard state are zero.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.

56. Assertion (A) : Spontaneous process is an irreversible process and may be reversed by some external agency.
Reason (R) : Decrease in enthalpy is a contributory factor for spontaneity.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.

57. Assertion (A) : A liquid crystallises into a solid and is accompanied by decrease in entropy.
Reason (R) : In crystals, molecules organise in an ordered manner.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.

VI. Long Answer Type

58. Derive the relationship between ΔH and ΔU for an ideal gas. Explain each term involved in the equation.

59. Extensive properties depend on the quantity of matter but intensive properties do not. Explain whether the following properties are extensive or intensive.

Mass, internal energy, pressure, heat capacity, molar heat capacity, density, mole fraction, specific heat, temperature and molarity.

60. The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure lattice enthalpy of NaCl(s).

61. ΔG is net energy available to do useful work and is thus a measure of “free energy”. Show mathematically that ΔG is a measure of free energy. Find the unit of ΔG. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous?

62. Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (pi , Vi ) to (pf , Vf ). With the help of a pV plot compare the work done in the above case with that carried out against a constant external pressure pf .


I. Multiple Choice Questions (Type-I)

1. (iii)      2. (iii)      3. (iv)      4. (iii)       5. (iii)      6. (ii)
7. (iii) Justification : free expansion w = 0
adiabatic process q = 0
ΔU = q + w = 0, this means that internal energy remains constant. Therefore, ΔT = 0.

In ideal gas there is no intermolecular attraction. Hence when such a gas expands under adiabatic conditions into a vaccum no heat is absorbed or evolved since no external work is done to separate the molecules.

8. (ii) w (reversible) < w (irreversible)
Justification : Area under the curve is always more in irreversible compression as can be seen from Fig. 6.5 (a) and (b).

9. (iii)
Justification : Freezing is exothermic process. The heat released increases the entropy of surrounding.
10. (iii)
11. (iii)
Justification : Same bonds are formed in reaction (a) and (b) but bonds between reactant molecules are broken only in reaction (b).

12. (iii)      13. (i)      14. (ii)

II. Multiple Choice Questions (Type-II)

15. (i), (iv)      16. (i), (ii)      17. (iii), (iv)
18. (iii), (iv)

For isothermal expansion of ideal gases, ΔU = 0

since temperature is constant this means there is no change in internal energy. Therefore, ΔU = 0

19. (i), (iii)

III. Short Answer Type

20. + 81.58 kJ, ΔvapHΘ = + 40.79 kJ mol–1
21. Water
22. No, since CaCO3 has been formed from other compounds and not from its constituent elements.
23. ΔrHΘ = +91.8 kJ mol–1
24. ΔrH = ΔrH1+ ΔrH2 + ΔrH3 …..
25. 1665/4 kJ mol–1= 416.2 kJ mol–1
26. +735.5 kJ mol–1
27. It is spontaneous process. Although enthalpy change is zero but randomness or disorder (i.e., ΔS) increases. Therefore, in equation ΔG = ΔH – TΔS, the term TΔS will be negative. Hence, ΔG will be negative.

28. ΔS = qrev/T
29. Yes
30. The reaction is spontaneous ΔrGΘ= – RT ln Kp
31. ΔH (cycle) = 0
32. Less, because ice is more ordered than H2O (l).
33. State Functions : Enthalpy, Entropy, Temperature, Free energy Path Functions : Heat, Work

34. Because of strong hydrogen bonding in water, its enthalpy of vapourisation is more.
35. ΔrG will always be zero.
ΔrGΘ is zero for K = 1 because ΔGΘ = – RT lnK, ΔGΘ will be non zero for other values of K.

36. For isolated system, there is no transfer of energy as heat or as work i.e., w=0 and q=0. According to the first law of thermodynamics.
ΔU = q + w
= 0 + 0 = 0
∴ ΔU = 0

37. At constant volume By first law of thermodynamics:
q = ΔU + (–w)
(–w) = pΔV
∴ q = ΔU + pΔV
ΔV = 0, since volume is constant.
∴ qV = ΔU + 0
⇒ qV = ΔU = change in internal energy

At constant pressure

qp = ΔU + pΔV
But, ΔU + pΔV = ΔH
∴ qp = ΔH = change in enthalpy.

So, at a constant volume and at constant pressure heat change is a state function because it is equal to change in internal energy and change in enthalpy respectively which are state functions.

38. (–w) = pext (V2–V1) = 0 × (5 – 1) = 0
For isothermal expansion q = 0
By first law of thermodynamics
q = ΔU + (–w)
⇒ 0 = ΔU + 0 so ΔU = 0

39. For water, heat capacity = 18 × specific heat or Cp = 18 × c
Specific heat = c = 4.18 Jg–1K–1
Heat capacity = Cp = 18 × 4.18 JK–1 = 75.3 J K–1
40. CP – CV = nR = 10 × 4.184 J

41. Molar enthalpy change of graphite = enthalpy change for 1 g carbon × molar mass of carbon
= – 20.7 kJ g–1 × 12g mol–1
∴ ΔH = – 2.48 × 102 kJ mol–1

Negative value of ΔH ⇒ exothermic reaction.

42. ΔrHΘ= Bond energy of H2 + Bond energy of Br2 – 2 × Bond energy of HBr
= 435 + 192 – (2 × 368) kJ mol–1
⇒ ΔrHΘ
= –109 kJ mol–1

43. qp = ΔH = 30.5 kJ mol–1
∴ Heat required for vapourisation of 284 g of CCl4 = 284 g x 30.5 kJ mol–1/154 g mol–1 = 56.2 kJ

44. According to the definition of standard enthalpy of formation, the enthalpy change for the following reaction will be standard enthalpy of formation of H2O (l)

H2(g) + 1/2 O2(g) → H2O(l ).

or the standard enthalpy of formation of H2O(l) will be half of the enthalpy of the given equation i.e., ΔrHΘ is also halved.

ΔfHΘH2O(l ) = 1/2 × ΔrHΘ = − – 572 kJ mol–1/2 = – 286 kJ/mol.

45. Work done on an ideal gas can be calculated from p-V graph shown in Fig. 6.6. Work done is equal to the shaded area ABVIVII .

46. The work done can be calculated with the help of p–V plot. A p–V plot of the work of compression which is carried out by change in pressure in infinite steps, is given in Fig. 6.7. Shaded area represents the work done on the gas.


48. No.
Enthalpy is one of the contributory factors in deciding spontaneity but it is not the only factor. One must look for contribution of another factor i.e., entropy also, for getting the correct result.

49. It is clear from the figure that the process has been carried out in infinite steps, hence it is isothermal reversible expansion.
w = – 2.303nRT log V2/V1

But, p1V1 = p2V2 ⇒ V2/V1 = p1/p2 = 2/1 = 2
∴ w = – 2.303 nRT log p1/p2
= – 2.303 × 1 mol × 8.314 J mol–1 K–1 × 298 K–1 × log 2
= – 2.303 × 8.314 × 298 × 0.3010 J = –1717.46 J

50. w = – pex (Vf –Vi ) = –2 × 40 = – 80 L bar = – 8 kJ

The negative sign shows that work is done by the system on the surrounding. Work done will be more in the reversible expansion because internal pressure and exernal pressure are almost same at every step.

IV. Matching Type

51. (i) → (e) (ii) → (d) (iii) → (f) (iv) → (a) (v) → (g), (k), (l) (vi) → (b) (vii) → (c) (viii) → (j) (ix) → (h) (x) → (i) (xi) → (a), (l), (m) (xii) → (g), (k)
52. (i) →(b) (ii) → (c) (iii) → (a)
53. (i) → (c) (ii) → (a) (iii) → (b)
54. (i) → (b), (d) (ii) → (b) (iii) → (c) (iv) → (a)

V. Assertion and Reason Type

55. (ii) 56. (ii) 57. (i)

VI. Long Answer Type

59. Hint : Ratio of two extensive properties is always intensive
Extensive/Extensive = Intensive.

e.g., Mole fraction = Moles/Total number of moles = (Extensive)/ (Extensive)

60. • Na (s) + 1/2 Cl2 (g) → Na+(g) + Cl(g) ;              ΔlatticeHΘ

• Bonn – Haber Cycle
• Steps to measure lattice enthalpy from Bonn – Haber cycle
• Sublimation of sodium metal
(1) Na(s) → Na (g) ;              ΔsubHΘ
(2) Ionisation of sodium atoms
Na(g) → Na+(g) + e(g) ;              ΔtHΘ i.e., ionisation enthalpy
(3) Dissociation of chlorine molecule
1/2 Cl2(g) → Cl(g) ;              1/2 Δbond HΘ i.e., One-half of bond dissociation enthalpy.

(4) Cl(g) + e(g) → Cl(g) ;              ΔegHΘ i.e., electron gain enthalpy.

61. ΔSTotal = ΔSsys + ΔSsurr
ΔSTotal = ΔSsys + (-ΔHsys/T)
T ΔSTotal = T ΔSsys – ΔHsys
For spontaneous change, ΔStotal > 0
∴ T ΔSsys – ΔHsys > 0
⇒ – (ΔHsys – T Δ Ssys ) > 0

But, ΔHsys – T ΔSsys = ΔGsys
∴ – ΔGsys > 0
⇒ ΔGsys = ΔHsys – T ΔSsys < 0
ΔHsys= Enthalpy change of a reaction.
T ΔSsys = Energy which is not available to do useful work.
ΔGsys = Energy available for doing useful work.
• Unit of ΔG is Joule
• The reaction will be spontaneous at high temperature.


(i) Reversible Work is represented by the combined areas  and 
(ii) Work against constant pressure, pf is represented by the area 
Work (i) > Work (ii)

Some Useful Links



MOST of the reactions are carried out at atmospheric pressure, hence heat changes noted for these reactions are enthalpy changes. Enthalpy changes are directly related to the temperature changes by the relation:

ΔH = qp
= mCp ΔT
= VdCp ΔT … (1)
where V = Volume of the solution.
d = Density of the solution
Cp = Heat capacity
ΔT = Change in temperature

Measurement of heat changes are carried out in vessels called calorimeters. Reactions may also be carried out in beakers placed in thermos flask or in thermally insulated box or in styrofoam cup. Metallic calorimeters are not used
for measuring thermochemical changes because metals may react with substances. Stainless steel or gold plated copper calorimeters may be used. During measurement of heat changes, calorimeter, thermometer and stirrer also absorb some heat; this amount of heat should also be known. It is called calorimeter constant. In the case of a glass vessel, (e.g. beaker) calorimeter constant for that part is found, which is actually in contact with the reaction mixture. This is so because when thermal conductivity of the material of calorimeter is low, only the area of the calorimeter in contact with the liquid absorbs maximum heat. Method of mixtures is used to determine the calorimeter constant. To determine calorimeter constant, known volume of hot water at a specified temperature is added to known volume of water contained in the calorimeter at room temperature. Since energy is conserved, the heat taken by calorimeter and cold water should be equal to heat given by hot water. Thus, we can write the following equation :

ΔH1             +          ΔH2       =             –ΔH3 … (2)
Enthalpy change      Enthalpy              Enthalpy
of calorimeter,        change of              change of
stirrer and                cold water                hot water

Let tc, th and tm be temperatures of cold water, hot water and mixture respectively. Then, in view of the definition of enthalpy change given in equation (1) we can rewrite equation (2) as

m1 Cp1 (tm–tc) + m2Cp(tm–tc) + m3Cp (tm–th) = 0 … (3)

where m1, m2 and m3 are masses of calorimeter, cold water and hot water respectively and Cp1 and Cp are heat capacities of calorimeter and water respectively. Since, thermal conductivity of glass is low, only that part of the beaker gains maximum heat which comes in contact with water therefore, we can calculate only effective m1 Cp1 (i.e. calorimeter constant, W). On rewriting equation (3) we get

W (tm– tc) + m2Cp (tm– tc) + m3Cp (tm– th) = 0

W =(m2Cp (tm– tc) + m3Cp (tm– th))/(tm– tc) … (4)

but mCp = VdCp, where V, d and Cp are volume, density and heat capacity of water respectively. By definition, heat capacity of a substance is the amount of energy required to raise the temperature of 1 g of substance by 1 K (or 1°C). The amount of energy required to raise the temperature of 1 g of water by 1 K (or 1°C) is 4.184 Joules. This means that for 1 g water for rise of 1 Kelven temperature VdCp = 4.184 JK–1. Therefore, product of density and heat capacity can be taken as 4.184 J.mL–1.K–1. Thus, equation (4) can be written as :

W = (4.184) [Vc(tm – tc) + Vh(tm – th) J K–1 / (tm – tc) … (5)

where Vc = volume of cold water
Vh = volume of hot water

Technique for measuring the enthalpy changes are given in the following experiments.


To determine the enthalpy of dissolution of copper sulphate/potassium nitrate.

In thermochemical measurements generally aqueous solutions are mixed therefore, water in the reaction medium and the temperature changes result due to the chemical reactions taking place in solution.

According to law of conservation of energy, the sum of enthalpy changes taking place in the calorimeter (loss and gain of energy) must be zero. Thus, we can write the following equation-

Heat gained by
and stirrer
+ (ΔH2) Enthalpy
change of
in calorimeter
+ (ΔH3) Enthalpy change
of added solution/
water in
+ (ΔH4) Enthalpy
change of
= 0


In these reactions we take the product of density and heat capacity of solutions, dCp, to be 4.184 J.mL–1.K–1, nearly the same as that of pure water.*

Solution formation often accompanies heat changes. Enthalpy of solution is the amount of heat liberated or absorbed when one mole of a solute (solid/liquid) is dissolved in such a large quantity of solvent (usually water) that further dilution does not make any heat changes.

Material Required


A. Determination of Calorimeter constant of calorimeter (Beaker)

(i) Take 100 mL of water in a 250 mL beaker marked ‘A’.
(ii) Place this beaker on a wooden block kept in a larger beaker of capacity 500 mL (Fig. 3.1).
(iii) Pack the empty space between the large and the small beaker with cotton wool. Cover the beaker with a cardboard.
Insert thermometer and stirrer in the beaker through it.

* Density of the solutions is 4 to 6% higher than that of pure water and heat capacity is about 4 to 8% less than
pure water so the product of density and heat capacity (dCp ) is nearly the same as the product of pure water.

(iv) Record the temperature of water. Let this temperature be tc°C.
(v) In another beaker of 250 mL capacity marked ‘B’ take 100 mL of hot water (50-60°C).
(vi) Note the exact temperature of hot water. Let this temperature be th°C.
(vii) Lift the card board and pour the hot water contained in beaker B into beaker A. Stir the mixed water and note the
temperature. Let this temperature be tm°C.
(viii) Calculate the calorimeter constant of the beaker by using the expression (5) given above.
(Remember the three temperatures are in the order th > tm >tc).

B. Determination of Enthalpy of Dissolution
(i) Take 100 mL of distilled water in the beaker of which calorimeter constant has been determined and place it on
a wooden block kept in a larger beaker of capacity 500 mL (Fig. 3.1).

(ii) Pack the empty space between the larger and the smaller beaker with cotton wool and cover with a cardboard.
(iii) Record the temperature of water already taken in the small beaker. Let this be t’1°C.
(iv) Add weighed amount, say W1 g of well powdered copper sulphate in water and stir the solution with a stirrer till
the entire amount of copper sulphate dissolves.
(v) Note down the temperature attained by the solution after the addition of copper sulphate. Let this be t’2°C. Calculate the enthalpy of dissolution of copper sulphate as follows:

Total mass of the solution = Mass of Solvent + Mass of Solute
= (100 + W1) g
(Assuming density of water to be equal to 1 gL–1 at the
experimental temperature)
Change in temperature = (t’2 – t’1) °C
Enthalpy change of the calorimeter (beaker) = W (t’2 – t’1)
where, W = Calorimeter constant
Enthalpy change of solution = [(100 + W1) (t’2 – t’1)]x4.184 J
for (t’2 – t’1) °C rise in temperature
Total enthalpy change
of the Calorimeter = [W (t’2 – t’1) + (100 + W1) (t’2 – t’1)]x4.184 J
(beaker) and solution

Heat liberated
on dissolution = [W (t’2 – t’1) + (100 + W1) (t’2 – t’1)]x4.184 J/W1
of 1 g copper

Since 1 mol of copper sulphate weighs 249.5 g. Therefore,

ΔsolH of CuSO4.5H2O = 249.5 × ([W (t’2 – t’1) + (100 + W1) (t’2 – t’1)]x4.184 J mol–1)/W1


Enthalpy change in the dissolution of copper sulphate/potassium nitrate is _______ Jmol–1.


(a) To record the temperature of water, use a thermometer with 0.1°C graduation.
(b) In the determination of calorimeter constant record the temperature of hot water just before mixing.
(c) Avoid using very large amounts of copper sulphate/potassium nitrate.
(d) Stir the solution well to dissolve the solid and record the temperature. Avoid too much stirring, it may produce heat due to friction.
(e) Weigh copper sulphate carefully as it is hygroscopic in nature.
(f) Use cotton wool to create insulation between the two beakers.

Discussion Questions

(i) What is meant by the term, calorimeter constant?
(ii) Why is ΔSolH for some substances negative while for others it is positive?
(iii) How does ΔSolH vary with temperature?
(iv) Will the enthalpy change for dissolution of same amount of anhydrous copper sulphate and hydrated copper sulphate in the same amount of water be the same or different? Explain.
(v) How will the solubility of copper sulphate and potassium nitrate be affected on increasing the temperature? Explain.


To determine the enthalpy of neutralisation of a strong acid (HCl) with a strong base (NaOH).

A neutralisation reaction involves the combination of H+(aq) ions furnished by an acid and OH(aq) ions furnished by a base, evidently leading to the formation of H2O (l). Since the reaction envisages bond formation, therefore, this reaction is always exothermic. Enthalpy of neutralisation is defined as the amount of heat liberated when 1mol of H+ ions furnished by acid combine with 1 mole of OH ions furnished by base to form water. Thus:

H+(aq) + OH(aq) → H2O (l), Δneut H is negative
(Acid)   (Base)

where Δneut H is known as enthalpy of neutralisation.

If both the acid and the base are strong then for the formation of 1 mol H2O (l), always a fixed amount of heat, viz, 57 kJ mol–1 is liberated. If any one of the acid or the base is weak or if both of these are weak, then some of the heat liberated is used for the ionisation of the acid or base or both of them (as the case may be) and the amount of heat liberated is less than 57 kJ mol–1.

Material Required


A. Determination of calorimeter constant
This may be determined by following the procedure, as detailed in experiment 3.1.

B. Determination of Enthalpy of Neutralisation

(i) Take 100 mL of 1.0 M HCl solution in the calorimeter (beaker) and cover it with cardboard. In another beaker of 250 mL capacity take 100 mL of 1.0 M NaOH solution.
(ii) Note down the temperature of both the solutions, which is likely to be the same. Let it be t1°C.
(iii) Pour 100 mL 1 M NaOH solution into the calorimeter containing 100 mL of 1.0 M HCl solution.
(v) Mix the solutions by stirring and note the final temperature of the mixture. Let it be t2°C.

Calculate the enthalpy of neutralisation as follows:
(i) Note the rise in temperature of the mixture, which in this case is (t2-t1) °C.
(ii) Calculate the total amount of heat produced during the neutralisation process, using the following expression
Heat evolved = (100 + 100 + W) (t2 – t1) 4.18 J (where W, is the calorimeter constant)
(iii) Finally calculate the heat evolved when 1000 mL of 1M HCl is allowed to neutralise 1000 mL of 1M NaOH. This quantity would be ten times the quantity obtained in step (ii).
(iv) Express the quantity of heat evolved in kJ mol–1.

Enthalpy change in the neutralisation of hydrochloric acid solution with sodium hydroxide solution _______ kJmol–1.

(a) Record the temperature carefully with the help of a thermometer graduated up to 0.1°C.
(b) Measure the volume of hydrochloric acid and sodium hydroxide solution to be taken for the experiment carefully.
(c) Proper insulation should be made between the two beakers.
(d) Avoid unnecessary and excessive stirring to prevent heating due to friction.

Discussion Questions

(i) Why do we calculate the heat evolved for the neutralisation of 1000 mL of a (1 M) acid by 1000 mL of a (1 M) monoacidic base?
(ii) In comparison to heat evolved in neutralisation reaction between a strong acid and a strong base. Why is lesser quantity of heat evolved when any one of the acid or the base is weak and still less when both are weak?
(iii) Why does the reaction:  proceed in the forward direction with rise in temperature of the system?


To determine the enthalpy change for the interaction between acetone and chloroform (hydrogen bond formation).

On mixing, liquid pairs show departure from ideal behaviour. Acetone and chloroform form non-ideal liquid pair system, which shows a negative deviation from Raoult’s law. This negative deviation from Raoult’s law implies that the two components are strongly held together in liquid state on mixing due to hydrogen bonding. On the other hand in the pure state, only weak Van der waal’s forces hold molecules of chloroform as well as acetone. The hydrogen bonding between the molecules of acetone and chloroform is depicted as follows:

In this process enthalpy change takes place due to hydrogen bond formation. The enthalpy change is an extensive
thermodynamic property, therefore, the heat evolved from the system depends upon the amount of the liquid components
mixed. It is for this reason that the heat change is reported for specified amount. Therefore, enthalpy change during mixing of 1 mol chloroform with 1 mol acetone is reported.


Heat gained by
and stirrer

+ (ΔH2) Enthalpy
change of
+ (ΔH3) Enthalpy change
for acetone
+ (ΔH4) Enthalpy
change of
= 0

ΔH4 = – (ΔH1+ΔH2+ΔH3)

Material Required


A. Determination of calorimeter constant

This may be determined in a manner detailed in previous experiments; except that here instead of a beaker, boiling tube
may be taken and 8 mL of cold and 7.5 mL of hot water can be used instead of 100 mL.

B. Determination of Enthalpy Change on Mixing Chloroform and Acetone*

(i) Transfer the volume of chloroform equivalent to 0.1 mol (≈ 8.14 mL) after measuring from a measuring cylinder into the insulated boiling tube as shown in Fig. 3.2. Let the mass of chloroform taken be m1 grams.

(ii) Record the temperature of chloroform. Let it be t1°C.
(iii) Transfer the volume of acetone equivalent to 0.1 mol of acetone (≈ 7.34 mL) in a clean measuring cylinder. Let its mass be m2 grams.
(iv) Record the temperature of acetone. Let it be t2°C.
(v) Pour acetone from the measuring cylinder into the chloroform contained in the insulated boiling tube.
(vi) Stir gently the mixture of chloroform and acetone carefully with the help of a stirrer.
(vii) Record the temperature of the mixture of chloroform and acetone. Let it be t3°C.

*Volume of one mole of CHCl3 = Molar mass of CHCl3/Density of CHCl3
Volume of 0.1 mole = 1/10 th of the above volume
(Similarly you can calculate the volume of 0.1 mole of acetone).
Density of chloroform= 1.47 g /mL
Molar mass of chloroform = 119.5 g
1.47 g = 1 mL volume
119.5 g = 119.5/1.47 mL
1 mole = 81.4 mL
0.1 mole = 8.14 mL

Density of acetone = 0.79 g /mL
Molar mass of acetone = 58.0
0.79 g = 1 mL
58 g = 58/0.79 mL
1 mole = 73.4 mL
0.1 mole = 7.34 mL

Total volume of acetone and chloroform = 8.14 +7.34 = 15.48 mL

Calculate the enthalpy of interaction as follows :

(i) Let the room temperature be t°C, then heat gained by calorimeter (boiling tube) is W (t3– t), where W is the calorimeter constant, i.e. boiling tube in this experiment.
(ii) Note the value of specific heat of chloroform from literature. Let it be q1.
Then heat gained by chloroform = m1 x q1 (t3 – t1).
(iii) Note the value of the specific heat for acetone from literature. Let it be q2. Thus heat gained by acetone = m2 x q2 (t3 – t2).
(iv) Total heat gained by all the three components, i.e. boiling tube, chloroform and acetone = – {W(t3 – t1) + m1q1(t3 – t1) + m2q2 (t3 – t2)}. This in fact is the enthalpy change of interaction, on mixing 0.1 mol chloroform with 0.1 mol acetone.

The negative sign simply implies that the mixing of chloroform and acetone is an exothermic process.

Note : Here, care should be taken that the total volume of acetone and chloroform is equal to the volume of water for which water equivalent of the calorimeter has been calculated.


(a) Measure chloroform and acetone carefully.
(b) Record the temperature very carefully with a thermometer graduated up to 0.1°C.

Discussion Questions

(i) Chloroform and acetone do not form an ideal liquid pair, whereas acetone and benzene do form. Why?
(ii) Why does liquid pair of ethanol and water show positive deviation from Raoult’s law?
(iii) Give two examples of each of the liquid pairs for which ΔMixing H is negative and positive respectively.
(vi) How is the vapour pressure of the liquids related to interaction pattern between the molecules of the components of a liquid mixture?
(v) How can you correlate the heat evolved from the system with the strength of the hydrogen bond?

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